Post by: jeffreyH on 19/02/2015 00:16:56
Post by: chiralSPO on 19/02/2015 02:14:42
Post by: yor_on on 19/02/2015 10:59:59
Post by: Ethos_ on 19/02/2015 13:31:57
When approaching infinity would there be an exponential loss of photon energy? This is not a trivial question to answer in my view and should be thought provoking.Thought provoking indeed.
From our frame of reference, the photon will red shift because of the expansion but take an infinity to be infinitely red shifted. However, because of time dilation, the photon will experience no passage of time between it's creation and that proposed infinity. From the photon's perspective, it will simply not have enough time to loose any energy.
I recognize that this view will probably not be met with much enthusiastic support but is nevertheless my opinion.
Post by: Bill S on 19/02/2015 14:30:55
Quote from: Jeff
When approaching infinity would there be an exponential loss of photon energy? This is not a trivial question to answer in my view and should be thought provoking.
I keep thinking I must stop posting about infinity; then someone posts a gem like this! [;D]
I’m assuming the infinity you refer to is infinite wavelength. You talk of “approaching infinity”, so you are starting with a finite wave. Not only can something finite never become infinite, it cannot approach infinity. However much it progresses, it is still infinitely far away. So the answer must be “no, it can never lose all its energy”.
Things are never that simple though, are they? If it is losing energy, and energy is quantized, does it reach a point where the last quantum of energy is lost? If so, what happens to the wavelength? Would there still be a photon?
If something is infinite, it must always have been infinite. (There’s a good example of how difficult it is to talk about infinity without using the terminology of the finite like “…always have been”.) Could an eternal photon exist that had infinite wavelength? As Chiral pointed out, with infinite wavelength it would have zero energy. Is it a photon?
Quote from: Ethos
the photon will experience no passage of time between it's creation and that proposed infinity.
10/10 for bravery in stirring up that hornets’ nest.
Post by: PmbPhy on 19/02/2015 16:45:39
When approaching infinity would there be an exponential loss of photon energy? This is not a trivial question to answer in my view and should be thought provoking.I agree with chiralSPO in that photons will continually decrease in energy the further they travel. But I don't see it being exponential. Why "exponential"?
Post by: jeffreyH on 19/02/2015 17:02:31
When approaching infinity would there be an exponential loss of photon energy? This is not a trivial question to answer in my view and should be thought provoking.I agree with chiralSPO in that photons will continually decrease in energy the further they travel. But I don't see it being exponential. Why "exponential"?
I was looking for opinions on this and included the idea of an exponential decline to see what the reaction would be. I don't have that view myself. It was a bit provocative.
Post by: jeffreyH on 19/02/2015 19:23:02
Post by: evan_au on 19/02/2015 19:26:28
Quote from: Bill S
If [a photon] is losing energy, and energy is quantized, does it reach a point where the last quantum of energy is lost? ...Would there still be a photon?I think this might be confusing a light source emitting many photons with the behaviour of a single photon?
- Let's say a light source is emitting (say) a billion photons per second for an hour into space. A detector at Alpha Centauri would be very unlikely to detect one of these photons. From the viewpoint of a detector at the Andromeda galaxy, it (almost certainly) wouldn't see any photons from such a light source; from it's viewpoint, there wasn't a photon! This applies quantisation of light to a source emitting many particles.
- However, I understand the original post to be asking about the behaviour of a single photon. If you could follow one of the billions of photons, it would continue to propagate indefinitely through space until it impacts matter or a detector of some sort. If a local observer in a distant galaxy detected this photon, the energy of the photon would be lower than when it was emitted, due to cosmic redshift. I know of no lower limit to this photon energy, but it is still quantised as 1 photon, which still exists. (I guess in the extreme, if the photon wavelength was comparable to the size of the universe, then that is some sort of limit; but it would become undetectable by our equipment long before that!)
Post by: jeffreyH on 19/02/2015 19:44:05
When approaching infinity would there be an exponential loss of photon energy? This is not a trivial question to answer in my view and should be thought provoking.Thought provoking indeed.
From our frame of reference, the photon will red shift because of the expansion but take an infinity to be infinitely red shifted. However, because of time dilation, the photon will experience no passage of time between it's creation and that proposed infinity. From the photon's perspective, it will simply not have enough time to loose any energy.
I recognize that this view will probably not be met with much enthusiastic support but is nevertheless my opinion.
The wavelength changes over time and so can we say the wave 'experiences' time? I doubt it.
Post by: evan_au on 19/02/2015 19:48:51
Quote from: jeffreyH
If the field removes energy from the photon at source and extends an infinite distance then there is an infinite opportunity for the field to carry on removing photon energy.The scenario imagined here seems to be something like: a single Sun in the universe, and looking at the photon energy at large distances from the Sun.
Quote from: To paraphrase
If you follow one photon from the Sun, and measure the photon's energy in the Sun's frame of reference (no cosmic redshift), will the photon energy approach zero, due to Einstein redshift?As I understand it, the answer is "No": although the Sun's gravitational field extends to "infinity", it decays as the square of distance, so the total effect on photon energy is "finite". The photon will lose only a tiny fraction of its energy as it travels to infinity*.
There is a critical mass & density where the Sun's gravitational field is so strong that the photon energy would become zero, and it does this in a finite distance - the Swarzchild radius of a black hole.
*This is similar to the concept of an "escape velocity" from the Solar System; although the Sun's gravitational field extends to infinity, it will subtract at most a finite amount of energy from a body moving away from the Sun. Using Newton's physics, Laplace (http://en.wikipedia.org/wiki/Pierre-Simon_Laplace) deduced that black holes existed, because the escape velocity of a massive Sun would exceed the speed of light! Pretty impressive for someone who didn't know about Doppler Shift, Einstein Shift or relativistic time dilation!
Post by: jeffreyH on 19/02/2015 20:03:13
Oops! Crossed posts...Quote from: jeffreyHIf the field removes energy from the photon at source and extends an infinite distance then there is an infinite opportunity for the field to carry on removing photon energy.The scenario imagined here seems to be something like: a single Sun in the universe, and looking at the photon energy at large distances from the Sun.Quote from: To paraphraseIf you follow one photon from the Sun, and measure the photon's energy in the Sun's frame of reference (no cosmic redshift), will the photon energy approach zero, due to Einstein redshift?As I understand it, although the Sun's gravitational field extends to "infinity", it decays as the square of distance, so the total effect on photon energy is "finite"*.
There is a critical mass & density where the Sun's gravitational field is so strong that the photon energy would go to zero, and it does this in a finite distance - the Swarzchild radius of a black hole.
*This is similar to the concept of an "escape velocity" from the Solar System; although the Sun's gravitational field extends to infinity, it will subtract at most a finite amount of energy from a body moving away from the Sun. Using classical (Newton's) physics, Laplace (http://en.wikipedia.org/wiki/Pierre-Simon_Laplace) showed that black holes existed, because the escape velocity of a massive Sun would exceed the speed of light! Pretty impressive for someone who didn't know about Doppler Shift, Einstein Shift or relativistic time dilation!
I do agree that the gravitational field should only remove a finite amount of energy even over an infinite distance. I was simply stating an alternative view. Black holes are a different matter.
Post by: Ethos_ on 19/02/2015 20:09:52
We see the wavelength change in our frame but does that mean it also changes in the photon's frame? I'm guessing it doesn't. The photon recognizes no passage of time.When approaching infinity would there be an exponential loss of photon energy? This is not a trivial question to answer in my view and should be thought provoking.Thought provoking indeed.
From our frame of reference, the photon will red shift because of the expansion but take an infinity to be infinitely red shifted. However, because of time dilation, the photon will experience no passage of time between it's creation and that proposed infinity. From the photon's perspective, it will simply not have enough time to loose any energy.
I recognize that this view will probably not be met with much enthusiastic support but is nevertheless my opinion.
The wavelength changes over time and so can we say the wave 'experiences' time? I doubt it.
Post by: chiralSPO on 19/02/2015 20:41:15
What if, instead of a photon, we have an electron moving at 0.99 c? It has a wavelength too. If an electron and a photon are both emitted simultaneously from the same point source (bear with me here) and both arrive at a destination that was 100 ly away when they were emitted, the photon arrives first, and the electron about a year later. Was is the magnitude of the difference in their red-shifts?
Post by: jeffreyH on 19/02/2015 20:42:33
We see the wavelength change in our frame but does that mean it also changes in the photon's frame? I'm guessing it doesn't. The photon recognizes no passage of time.When approaching infinity would there be an exponential loss of photon energy? This is not a trivial question to answer in my view and should be thought provoking.Thought provoking indeed.
From our frame of reference, the photon will red shift because of the expansion but take an infinity to be infinitely red shifted. However, because of time dilation, the photon will experience no passage of time between it's creation and that proposed infinity. From the photon's perspective, it will simply not have enough time to loose any energy.
I recognize that this view will probably not be met with much enthusiastic support but is nevertheless my opinion.
The wavelength changes over time and so can we say the wave 'experiences' time? I doubt it.
That is a very interesting point. That suggests that the photon experiences no change in frequency. It would be the observer external to the photon's frame of reference that would see a change in frequency.
Post by: chiralSPO on 19/02/2015 20:44:24
Post by: PmbPhy on 19/02/2015 20:47:08
I don't think photons can "observe" space either...Correct. A lot of confusion arises when people start talking about the experiences of photons. :)
Post by: jeffreyH on 19/02/2015 21:23:04
Post by: JohnDuffield on 19/02/2015 22:14:21
The reason for posting this was the thought that a gravitational field extends to infinity. If the field removes energy from the photon...It doesn't. The ascending photon doesn't lose any energy. In similar vein the descending photon doesn't gain any energy. If you send a 511keV photon into a black hole, the black hole mass increases by 511kev/c². Conservation of energy applies. The descending photon appears to be blueshifted because you and your clocks go slower when you're lower.
Post by: Ethos_ on 19/02/2015 22:14:27
Experience is a very bad choice of word. Not very scientific really.Very true if we are only considering the photon as observer. However, if it were possible for one of us to travel with the photon, which we all know it is not, we could experience or observe the photon's frame and frequency. Because that is not possible, we can only surmise. And to be clear, my opinion about this subject is only that, an opinion.
Post by: yor_on on 19/02/2015 23:07:32
==
Da*n, Alternatively, it shouldn't exist any photons, or geodesics inside a black hole, except from incoming photons, that then from the observation of someone standing at the 'center' should disappear before reaching its 'center'. Or if you define everything inside a event horizon as a 'center', then no geodesics can exist inside that event horizon, as all photons must be 'gone', incoming or not. The redshift as a result of a gravitational potential is strongest at whatever we define as the gravity's 'center', so my first idea turned it about, ah well, probably comes from me reading too much sci fi, you know 'the guy looking out at the stars' sort of (well ok. The 'event horizon' if you now want to be picky:)
Any which way, still think Chiral's idea is interesting, what happens with the energy in a expansion? The thing is that 'photons' do redshift due to an expansion, it's not as some say that they get more 'spaced out' by it. You can measure those photons, and find each one of them of less energy. So even though a wave picture describes it best, 'photons' too loses energy there, independent of any observer dependencies as I gather.
Post by: jeffreyH on 20/02/2015 00:30:47
The reason for posting this was the thought that a gravitational field extends to infinity. If the field removes energy from the photon...It doesn't. The ascending photon doesn't lose any energy. In similar vein the descending photon doesn't gain any energy. If you send a 511keV photon into a black hole, the black hole mass increases by 511kev/c². Conservation of energy applies. The descending photon appears to be blueshifted because you and your clocks go slower when you're lower.
So if a constant stream of photons of identical wavelength are generated directly away from a black hole with each photon at a regular interval what will be seen? If we then station observation points outward at regular intervals along the photon path to measure the wavelength at each point what do you expect the results to be. All observation points will expect a speed of c which they should record in their local frame. It is the gradual change in wavelength that produce the important data points. The velocity can't change in a vacuum. So why does the wavelength gradually change for different observers at different radial distances?
If you consider the de Broglie equations we can relate frequency to energy, frequency to wavelength etc. You say no energy is lost. Why then the relationship between frequency and energy?
Post by: jeffreyH on 20/02/2015 00:37:07
Post by: PmbPhy on 20/02/2015 03:27:42
Quote from: jeffreyH
He then substituted v for c in E = mc^2 so that mv^2 = hv.Where did you get that from?
Post by: jeffreyH on 20/02/2015 10:30:55
Quote from: jeffreyHHe then substituted v for c in E = mc^2 so that mv^2 = hv.Where did you get that from?
It wasn't E= mc^2. It was just mc^2 that was changed otherwise the energy equation would be wrong. mv^2 is of course is related to kinetic energy through (1/2)mv^2.
The page I read through was http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/02._Fundamental_Concepts_of_Quantum_Mechanics/De_Broglie_Wavelength. The point I was making to John was that energy does change for the photon and what the reasons are.
Post by: yor_on on 20/02/2015 11:05:35
Post by: JohnDuffield on 20/02/2015 11:38:40
So if a constant stream of photons of identical wavelength are generated directly away from a black hole with each photon at a regular interval what will be seen?Nothing unusual. What you'd expect.
If we then station observation points outward at regular intervals along the photon path to measure the wavelength at each point what do you expect the results to be.The observers will give different measurements. Those further out will say the wavelength is longer than those closer in. From that you might conclude that the photons are redshifted. However you could contrive a similar gedankenexperiment with moving observers. The observers moving away from the photon source would report a redshift. But you know that this isn't because the photons are actually getting redshifted, and instead is because the observers are moving.
All observation points will expect a speed of c which they should record in their local frame. It is the gradual change in wavelength that produce the important data points. The velocity can't change in a vacuum. So why does the wavelength gradually change for different observers at different radial distances?Because the speed of light changes. There's this myth kicking around that it's absolutely constant, but it isn't. Check out the coordinate speed of light, and what Einstein said:
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.thenakedscientists.com%2Fforum%2Findex.php%3Faction%3Ddlattach%3Btopic%3D54160.0%3Battach%3D19455%3Bimage&hash=996897d8b077cbd11f692c13ad526541)
If you consider the de Broglie equations we can relate frequency to energy, frequency to wavelength etc. You say no energy is lost. Why then the relationship between frequency and energy?The energy doesn't change, and the frequency doesn't change. You and your clocks go slower when you're lower, so you measure the frequency to have changed. But it hasn't changed. You and your clocks changed, not the photon. See post #6 in this thread (http://www.thenakedscientists.com/forum/index.php?topic=54160.msg449568#msg449568) where PmbPhy said the coordinate speed of the photon will change and Gravitational redshift is only observed when local observers at different positions compare their measurements. The wavelength as measured by Schwarzschild observers remains unchanged.
Post by: jeffreyH on 20/02/2015 12:40:26
If you consider the de Broglie equations we can relate frequency to energy, frequency to wavelength etc. You say no energy is lost. Why then the relationship between frequency and energy?The energy doesn't change, and the frequency doesn't change. You and your clocks go slower when you're lower, so you measure the frequency to have changed. But it hasn't changed. You and your clocks changed, not the photon. See post #6 in this thread (http://www.thenakedscientists.com/forum/index.php?topic=54160.msg449568#msg449568) where PmbPhy said the coordinate speed of the photon will change and Gravitational redshift is only observed when local observers at different positions compare their measurements. The wavelength as measured by Schwarzschild observers remains unchanged.
This is by holding the frequency as constant while the coordinate speed changes, yes. But we live in a universe where local observations should always concur about measurements. 1 metre will still be 1 metre. The speed of light will still be c. 1 second will still be one second. The energy of the photon for local observers is then different. You can't get round it by looking at it from a 'Schwarzschild observer' perspective.
Post by: PmbPhy on 20/02/2015 13:25:04
I believe that paper is in error. Let me check on it and get back to you.Quote from: jeffreyHHe then substituted v for c in E = mc^2 so that mv^2 = hv.Where did you get that from?
It wasn't E= mc^2. It was just mc^2 that was changed otherwise the energy equation would be wrong. mv^2 is of course is related to kinetic energy through (1/2)mv^2.
The page I read through was http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/02._Fundamental_Concepts_of_Quantum_Mechanics/De_Broglie_Wavelength. The point I was making to John was that energy does change for the photon and what the reasons are.
Post by: JohnDuffield on 20/02/2015 14:49:21
This is by holding the frequency as constant while the coordinate speed changes, yes.The frequency is constant. So is the energy. When you send a 511keV photon into a black hole, the black hole mass increases by 511keV/c². Not a gazillion tonnes. Conservation of energy applies.
But we live in a universe where local observations should always concur about measurements. 1 metre will still be 1 metre. The speed of light will still be c. 1 second will still be one second. The energy of the photon for local observers is then different.The photon energy doesn't change. The observers change. It takes work to lift a brick. You have to add energy to it. It's the same for an observer. And once you've lifted the observer up, then because you have added energy to that observer, to him, the photon energy appears to have reduced. Even though it hasn't. You could do the same sort of thing by accelerating observers away from a photon source in gravity-free space.
You can't get round it by looking at it from a 'Schwarzschild observer' perspective.It isn't a matter of getting around it. It's a matter of getting it right.
Post by: jeffreyH on 20/02/2015 15:39:10
This is by holding the frequency as constant while the coordinate speed changes, yes.The frequency is constant. So is the energy. When you send a 511keV photon into a black hole, the black hole mass increases by 511keV/c². Not a gazillion tonnes. Conservation of energy applies.But we live in a universe where local observations should always concur about measurements. 1 metre will still be 1 metre. The speed of light will still be c. 1 second will still be one second. The energy of the photon for local observers is then different.The photon energy doesn't change. The observers change. It takes work to lift a brick. You have to add energy to it. It's the same for an observer. And once you've lifted the observer up, then because you have added energy to that observer, to him, the photon energy appears to have reduced. Even though it hasn't. You could do the same sort of thing by accelerating observers away from a photon source in gravity-free space.You can't get round it by looking at it from a 'Schwarzschild observer' perspective.It isn't a matter of getting around it. It's a matter of getting it right.
I think this says it all John.
http://en.wikipedia.org/wiki/Gravitational_redshift
In astrophysics, gravitational redshift or Einstein shift is the process by which electromagnetic radiation originating from a source that is in a gravitational field is reduced in frequency, or redshifted, when observed in a region of a weaker gravitational field. This is a direct result of gravitational time dilation - as one moves away from a source of gravitational field, the rate at which time passes is increased relative to the case when one is near the source. As frequency is inverse of time (specifically, time required for completing one wave oscillation), frequency of the electromagnetic radiation is reduced in an area of a higher gravitational potential (i.e., equivalently, of lower gravitational field) . There is a corresponding reduction in energy when electromagnetic radiation is red-shifted, as given by Planck's relation, due to the electromagnetic radiation propagating in opposition to the gravitational gradient. There also exists a corresponding blueshift when electromagnetic radiation propagates from an area of a weaker gravitational field to an area of a stronger gravitational field.
Post by: JohnDuffield on 20/02/2015 16:33:40
I think this says it all John.The article is wrong. Gravity is not a force in the Newtonian sense. It doesn't add energy to a falling body, it just converts potential energy into kinetic energy. It doesn't add energy to a descending photon either, or remove energy from an ascending photon. Not only that, the article confuses field and potential. The opening sentence should say this:
In astrophysics, gravitational redshift or Einstein shift is the process by which electromagnetic radiation originating from a source that is in a gravitational field appears to be reduced in frequency, or redshifted, when observed in a region of a higher gravitational potential.
Maybe I should talk to the guys who've been editing that page about this. The next bit isn't bad, but "is" should be "appears":
This is a direct result of gravitational time dilation - as one moves away from a source of gravitational field, the rate at which time passes is increased relative to the case when one is near the source. As frequency is inverse of time (specifically, time required for completing one wave oscillation), frequency of the electromagnetic radiation is reduced in an area of a higher gravitational potential".
The next bit is just wrong:
(i.e., equivalently, of lower gravitational field) . There is a corresponding reduction in energy when electromagnetic radiation is red-shifted, as given by Planck's relation, due to the electromagnetic radiation propagating in opposition to the gravitational gradient.
If it wasn't wrong, dropping a 511keV photon into a black hole would increase its mass by more than 511keV/c². The next bit confuses field and potential again:
There also exists a corresponding blueshift when electromagnetic radiation propagates from an area of a weaker gravitational field to an area of a stronger gravitational field.
This has been written by an amateur, Jeffrey. I recommend you ask around elsewhere about what I've said. Meanwhile ask yourself who you're going to believe, me and Einstein and Pete, or some guy who doesn't know the difference between gravitational potential and gravitational field.
Post by: PhysBang on 20/02/2015 18:02:26
If we just think in terms of time dilation, then we consider the output of energy in the rest frame of the emitter, then consider the time dilated version of that, conservation of energy requires that the energy output over time of the time dilation of the emitter must be lower, since it must be putting out the same energy over a longer period. Whether or not an emitter is time dilated or not is determined entirely by frame of reference, so that can't change the overall output of the emitter between any two given events.
Similarly, a blueshift leads to an increase in energy. But this is not really amazing or unexpected, since when one speaks of an object falling to the ground, one speaks of the potential energy that the object converts into other forms of energy. One cannot simply speak of a photon falling in to a black hole without giving relevant details and expect a coherent physical picture.
If one asks of a photon from a distant galaxy if there is a difference from its frequency when it was emitted versus when it was received, then the answer, "yes," makes a great deal of sense.
Post by: PmbPhy on 20/02/2015 18:16:38
Quote from: JohnDuffield
Gravity is not a force in the Newtonian sense.Wrong.
Quote from: JohnDuffield
It doesn't add energy to a descending photon either, or remove energy from an ascending photon.Wrong yet again! It's very easy to prove too:
Jeff: Please see: http://home.comcast.net/~peter.m.brown/gr/grav_red_shift.htm
Post by: yor_on on 20/02/2015 19:54:05
=
the way around it would then be to assume a length 'elongation' balancing it up :) Meaning that the reason it stays a constant locally measuring (presuming some 'slower time/clock') is that it locally measured now finds it a longer 'way' to propagate, balancing the slower clock, giving us a constant speed of light in a two mirror experiment, 'inertially' and uniformly measuring. You can't apply a Lorentz contraction locally on that as that would give a opposite effect.
But it depends, you could argue that if 'c' and the arrow (local clock) is equivalent, it won't be noticeable, as the speed of light then 'slows down' equivalently, locally measured. What that gives us is a constant that theoretically may differ but locally and measurably is invariant. In the end it comes down to what is most simple. Constants, or no constants? And naturally, how you want to define physics? As being the same everywhere? And 'repeatable experiments' too.
Post by: jeffreyH on 20/02/2015 21:54:34
Quote from: JohnDuffieldGravity is not a force in the Newtonian sense.Wrong.Quote from: JohnDuffieldIt doesn't add energy to a descending photon either, or remove energy from an ascending photon.Wrong yet again! It's very easy to prove too:
Jeff: Please see: http://home.comcast.net/~peter.m.brown/gr/grav_red_shift.htm
Thanks for that Pete I will read it later. I am done with John.
Post by: jeffreyH on 20/02/2015 21:56:10
That was a very nice link Jeffrey. Succinct and enlightening to how he thought. If you find more links able to compress ideas feel free to share them :)
Pete is looking into it. It may be wrong so beware.
Post by: jeffreyH on 20/02/2015 22:06:12
Quote from: JohnDuffieldGravity is not a force in the Newtonian sense.Wrong.Quote from: JohnDuffieldIt doesn't add energy to a descending photon either, or remove energy from an ascending photon.Wrong yet again! It's very easy to prove too:
Jeff: Please see: http://home.comcast.net/~peter.m.brown/gr/grav_red_shift.htm
I have seen that page before and it was very informative. Now that I am getting into Lagrangians the Schwarzschild metric equation makes much more sense.
Post by: evan_au on 21/02/2015 01:34:50
Quote from: chiralSPO
arrive at a destination that was 100 ly away when they were emittedA position 100 light-years away is in our galaxy, in fact in the same spiral arm of the Milky Way. This is too close to experience cosmic redshift.
Even the Andromeda Galaxy, at 2.5 million LY distance is part of our local galaxy cluster, and does not show cosmic redshift (in fact, it is moving towards a collision with our galaxy, and so exhibits Doppler blue-shift).
To see cosmic redshift, you need to go outside our local cluster of galaxies.
However, gravitational redshift has actually been demonstrated here on Earth, with photons "climbing" out of Earth's gravitational well, but the effect is incredibly small. It is also very slight and hard to measure on the Sun, since the high temperature of the Sun's surface means that the thermal motion of highly ionised atoms in the Sun's atmosphere is large compared to the gravitational redshift. Neutron stars would exhibit higher gravitational redshift, but their incredibly high surface temperature would completely ionise the atoms, and would impart a huge thermal spread, so it would be hard to collect a spectrum.
[Pedantic, I know; I also didn't answer the question about redshift of photons & electrons that was asked by chiralSPO [V] ]
Post by: chiralSPO on 21/02/2015 02:09:10
Is it expected just to be an issue of Doppler shift based on the velocity of the source when the wave was emitted and the velocity of the detector when the wave is measured? Or does the red-shift manifest as a result of the expansion of space through which the wave is propagating?
If the former is true, then a photon that has traveled infinitely far could have finite non-zero energy (or even infinite energy) as measured by an observer traveling at infinite speed towards the source of the light (don't tell me this is impossible, we are already talking about a photon that has traveled an infinite distance--if we replace "infinite distance" for "arbitrarily long distance" and "infinite speed" with "arbitrarily close to c" the results will be the same until the distance is great enough that the light will never be able to reach the detector because spacial expansion will have overtaken it, in which case... what photon?)
Post by: JohnDuffield on 21/02/2015 16:29:37
Wrong.No it isn't wrong. The force of gravity doesn't do any work. A falling brick doesn't acquire any energy. You add energy to the brick when you lift it up. When it falls potential energy is converted into kinetic energy. That's all.
Wrong yet again! It's very easy to prove tooIt isn't wrong, and your article ends up saying this: The total energy of a photon moving through a gravitational field is constant.
Quote from: Jeffreyh
Thanks for that Pete I will read it later. I am done with John.I said gravity doesn't add energy to a descending photon or remove energy from an ascending photon. Pete said that was wrong, and referred you to an article that said The total energy of a photon moving through a gravitational field is constant.. He's saying I'm wrong when I'm not.
Post by: PhysBang on 21/02/2015 16:57:42
No it isn't wrong. The force of gravity doesn't do any work. A falling brick doesn't acquire any energy. You add energy to the brick when you lift it up. When it falls potential energy is converted into kinetic energy. That's all.But there is a force, gravity, that causes a displacement, the falling of the brick, so, by definition, that's work.
Of course, thanks to general relativity, one can identify a system of coordinates where the brick is not displaced, i.e., it is assigned the same position (until the ground gets in the way).
This isn't really a problem, as people realized before 1900 that work is dependent on the system of coordinates used.
Quote
It isn't wrong, and your article ends up saying this: The total energy of a photon moving through a gravitational field is constant.Sure, the article also says, incorrectly and after a very bad argument, that, "Therefore the frequency of the light, as measured by any single observer, does not change as the light moves through the gravitational field!" So, yes, that article is incorrect, but that does not make your claims any more correct, either.
Post by: evan_au on 21/02/2015 17:33:12
Quote from: chiralSPO
Is it expected just to be an issue of Doppler shift based on the velocity of the source when the wave was emitted and the velocity of the detector when the wave is measured? Or does the red-shift manifest as a result of the expansion of space through which the wave is propagating?
I am going to go with a tentative "Yes" to both alternatives. [I would appreciate it if those more experienced than I could check this conclusion...]
For speeds well below c, the wavelength of light is inversely proportional to the velocity of an observer, but motion towards the observer reduces the wavelength (blueshift), while motion away increases the wavelength (redshift). For observers moving away from the source at relative velocities near c, the wavelength of light (http://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Motion_along_the_line_of_sight) approaches infinity, and the photon energy approaches zero.
The wavelength of an electron is inversely proportional to the momentum; for speeds well below the speed of light, it is inversely proportional to the velocity of the electron relative to an observer (towards or away from an observer). For relative speeds approaching c, the wavelength of an electron (http://en.wikipedia.org/wiki/Electron_diffraction#Wavelength_of_electrons) approaches 0. For observers moving away from the source at speeds approaching the velocity at which the electron was emitted, the wavelength approaches infinity (but the energy does not approach zero, because of the "rest-mass" of the electron).
To compare the two wavelengths, electron microscopes have resolution similar to an optical microscope operating at X-Ray frequencies (at least in theory).
So I deduce that the wavelength of an electron and a photon does not change by exactly the same proportion, since the photon always travels at c, and the electron never travels at c.
My guess is that for observers moving relative to the source at speeds much less than c, but electrons traveling at close to c, the percentage redshift of the photon would be similar to the percentage red-shift of the electron. It would not matter whether the observers were moving apart due to being in a fast rocket ship or due to the expansion of the universe.
Footnote: This is ignoring effects like the galaxy's magnetic field, which would bend the path of a charged electron differently from the path of the uncharged photon, so they are unlikely to end up at the same observer, even 1000 years apart....
Post by: JohnDuffield on 21/02/2015 17:39:23
But there is a force, gravity, that causes a displacement, the falling of the brick, so, by definition, that's work.The point is that gravity isn't adding any energy. You add energy to the brick when you lift it. Work is the transfer of energy, you did work on it. When it falls down, gravity isn't adding any more energy, it's just converting the energy you added into kinetic energy.
Of course, thanks to general relativity, one can identify a system of coordinates where the brick is not displaced, i.e., it is assigned the same position (until the ground gets in the way). This isn't really a problem, as people realized before 1900 that work is dependent on the system of coordinates used.The system of coordinates is just an abstract thing. The falling brick is falling. Its potential energy is being converted into kinetic energy, and its mass is reducing such that once the kinetic energy is dissipated, we're left with a mass deficit.
Sure, the article also says, incorrectly and after a very bad argument, that, "Therefore the frequency of the light, as measured by any single observer, does not change as the light moves through the gravitational field!" So, yes, that article is incorrect, but that does not make your claims any more correct, either.The article is correct in that the descending photon doesn't gain any energy. Conservation of energy applies to photons as well as bricks. You know this, because you know that when you send a 511keV photon into a black hole, the black hole mass increases by 511keV/c². In similar vein the ascending photon doesn't lose any energy.
Post by: PhysBang on 21/02/2015 18:14:12
Work has a very specific definition in physics. You should abandon your vague and condusing term and use the proper term. If you mean that there is no transfer of energy, then say that. Nobody claims that in a system of two bodies, one body falling towards the other increases the total energy of the system. However, one might speak of an increase in the energy of one body.But there is a force, gravity, that causes a displacement, the falling of the brick, so, by definition, that's work.The point is that gravity isn't adding any energy. You add energy to the brick when you lift it. Work is the transfer of energy, you did work on it. When it falls down, gravity isn't adding any more energy, it's just converting the energy you added into kinetic energy.
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I have no idea what you are claiming. Regardless, one must use a system of coordinates to properly describe motion and the choice of system bears on the amount of work done on an object.Of course, thanks to general relativity, one can identify a system of coordinates where the brick is not displaced, i.e., it is assigned the same position (until the ground gets in the way). This isn't really a problem, as people realized before 1900 that work is dependent on the system of coordinates used.The system of coordinates is just an abstract thing. The falling brick is falling. Its potential energy is being converted into kinetic energy, and its mass is reducing such that once the kinetic energy is dissipated, we're left with a mass deficit.
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The article is correct in that the descending photon doesn't gain any energy.You can cherry-pick the conclusion if you would like, but that isn't good reasoning.
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Conservation of energy applies to photons as well as bricks. You know this, because you know that when you send a 511keV photon into a black hole, the black hole mass increases by 511keV/c². In similar vein the ascending photon doesn't lose any energy.I agree that the energy content of a system does not increase through its internal physical development. However, you seem to be ignoring Newton's third law.
Post by: yor_on on 22/02/2015 09:34:46
I actually like my third definition in where we find light being a constant locally measured, but from a 'container reasoning' also find its speed to depend on mass. That will give you time dilations comparing between frames of reference, but constants locally. What one can notice from such a reasoning is that a 'propagation' become a very complex behavior, especially if we consider the experiments done by NIST. Also that there are several possible ways to define what a frame of reference would be, macroscopically versus microscopically. It still should place a 'real' (as in a proven 'twin experiment') time dilation to being a effect between frames of reference to make sense though. That as it presumes your local clock and 'c' to be one and the same, everywhere. And it also points out the difference between using a 'container model' versus a local interpretation. If you instead of this propagation postulate a field with excitations it should become different though, giving us a propagation as defined locally measurably, but theoretically becoming a expression of 'circumstances' and probabilities, defining the existence of a locally measured 'excitation', defined to some position in this field (time and space). The 'circumstances' both involving SpaceTime, as well as the experiments nature, what it is defined to measure.
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But I would expect us to need to change our ideas of what a 'propagation' is. What I mean by stating that you can define a frame of difference differently, is the difference between a ideal description, as me having my ideal 'local clock', versus one in where where we microscopically define frames of reference as consisting of quanta or 'bits', as when me using Planck scale to define a smallest 'unit' of time theoretically. Both are in some means ideal descriptions, but a 'quanta of time' states that going past it should change the nature of things defined. Furthermore, the 'quanta's' should then be the constants creating this SpaceTime macroscopically. I'm not sure if this change loops and strings, unless you define a local arrow to 'each one' of them? If you do, then time should become a smooth phenomena, with Planck scale as a first ordered pattern.
Actually I don't think you need to do that. If you define it the first way, our type of 'local arrow' ending at Planck scale, you come to a symmetry break. Or maybe you do, it's in a sense two ways of looking at time, them coexisting. One makes the arrow we measure, as described over a 'universal container'. The other is not a arrow in that sense, more like some 'keeping of a beat' to me. If you look at some 'SpaceTime unfolding' that way, it's more like static sheets of patterns, each one lighted up momentarily, 'flickering past', through that beat, the one that makes the anchor for our repeatable experiments. Depends on what you expect a arrow to be that one.
You can think of the one assuming a speed to automatically 'change' due to mass as being equivalent to the way your local arrow always tells you the same about your local life span, it never changes. If you do so a 'curved space' becomes a really tricky proposition as we find gravitational time dilations at centimeters. Then weight that against the fact that you only can measure light once, Using the description of a 'light path' is an assumption we make.
And yes it gives us two ways to define this universe of ours. One macroscopic, the other microscopic. One consisting of a light sphere, defining a time of the universe. The other describing it through scaling. From scaling the symmetry break constantly is here, it never went anywhere, it's just scales and patterns. From the other 'universal time exist' and you can prove it astronomically, anywhere you go.
and you definitely need 'patterns & sheets' to describe it, put them together and you gain your 'field'. The 'field' is an assumption made, resting on the way we observe dimensions existing, and assume a universe with a past, a present, and a future to exist. The 'sheet' being each 'instant of existence' that we can measure in. If it 'flickers' through that beat you won't notice it.
this type of universe sounds very deterministic, doesn't it :) But that's not what probabilities tells us. It's still probabilistic, in each measurement on a position in time and space. And from that you then may want to define something more, 'containing' all probabilities, or do as I and end it at Planck scale, defining what's behind that to have it all.
Finally, this should then be what makes the 'dimensions' we measure in. Myself, I've never been fully comfortable with the idea of 'premade dimensions' from where a SpaceTime unfolds, and if you look at Einstein his ideas tells us that the universe actually is observer dependent. It can 'shrink' in your direction of 'motion'. Looking at it from a 'field' it stops having to do with some 'energy' needed to contract it (a whole universe) for the observer. It's still energy needed as a coin of exchange naturally, but it changes our definition of what this 'infinite universe' is. It's not a container, it's more of 'limits'. Although to anyone consciously existing and comparing inside it it will have all qualities we expect a container to have, volumes and time, dimensions. The dimensions should be created through 'c' communicating, also making a local arrow in where each one of us unfold. But this last is a really difficult thing to define.
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What one really need to understand is the difference between a container model, and defining it strictly locally. Strictly locally 'c' is 'c'. You can split any acceleration infinitesimally, and if using Planck scale the way I do, presumably stop there. That becoming each 'static sheet' of pattern if we now instead use a global model. There the acceleration disappear, as in becoming unmeasurable, although we have to assume its 'property' to still exist. Doing so 'c' will hold everywhere. And locally defined you can ignore any ideas of a 'global container'.
So locally, 'c' is always 'c', using those definitions. In a normal container model you step away from that minuscule scale into a macroscopic definition of a arrow. That one defined by ideals, like 'the universal time defining a Big Bang', and 'frames of reference'. There you will find Lorentz contractions and time dilations. There you can define 'light paths', 'SpaceTime curvatures', and a 'infinite' but still 'contained' universe. And in that universe giving light a path, 'interacting' with mass, you also can define it as 'slowing down' as you might do in a acceleration. It's all in the equivalence principle.
but if you also find a equivalence between the local arrow and that speed it doesn't matter. 'c' is a constant both ways.
(hmm, that shouldn't be read as I consider mass to be a expression of 'slow time' though, you just need to consider observer dependencies and uniform motion to see where such an idea takes you.. It's just a statement stating that 'c' and your local clock are equivalent. So, no matter how a far observer defines my clock, locally it will behave as always, as will everything 'at rest' with me. And so will my local measurement of 'c' still be 'c'. :)
Although, find a way to redefine what happens in uniform motions, their time dilations and Lorentz contractions, to fit the idea of gravitational opposites, and I might become interested. But you need to do it from this equivalence of a local clock, to a local speed of speed of light in a vacuum. And using 'light clocks' may make sense geometrically but I think you need something different here.
To see it my way is really easy, as soon as you accept that your local clock equals 'c'. What might not be so easy to accept is the way I define 'c' to be a constant everywhere, 'ignoring' mass and accelerations. And use it as a stepping stone from where to define a universe.
Post by: JohnDuffield on 22/02/2015 12:44:19
Work has a very specific definition in physics. You should abandon your vague and condusing term and use the proper term. If you mean that there is no transfer of energy, then say that. Nobody claims that in a system of two bodies, one body falling towards the other increases the total energy of the system. However, one might speak of an increase in the energy of one body.The total energy of the falling brick does not increase. Again, gravity converts potential energy into kinetic energy, that's all. Then when the kinetic energy is dissipated, the brick has a mass deficit.
I have no idea what you are claiming. Regardless, one must use a system of coordinates to properly describe motion and the choice of system bears on the amount of work done on an object.You know full well what I'm saying. You start with a situation wherein the brick and the Earth are motionless relative to each other, and the brick is 10m above the ground. Then you drop the brick, and the brick ends up hitting the ground at 14m/s. You can't "choose some system" where the brick ends up hitting the ground at 1m/s.
You can cherry-pick the conclusion if you would like, but that isn't good reasoning.It's not cherry picking to point out something that's correct. You know it's correct, why are you trying to cast doubt upon it? You know that when you send a 511keV photon into a black hole, the black hole mass increases by 511keV/c².
I agree that the energy content of a system does not increase through its internal physical development. However, you seem to be ignoring Newton's third law.No I'm not. Momentum p=mv is shared equally between both objects, but kinetic energy KE=½mv² is not. The brick hits the ground at 14m/s, with 98 Joules of kinetic energy. The motion of the Earth towards the brick is not detectable, the Earth gains no detectable kinetic energy. It's similar for the black hole and the photon.
Post by: Bill S on 22/02/2015 14:17:49
Quote from: Chiral
Is it expected just to be an issue of Doppler shift based on the velocity of the source when the wave was emitted and the velocity of the detector when the wave is measured? Or does the red-shift manifest as a result of the expansion of space through which the wave is propagating?
I’m with evan here. It would seem logical that the Doppler effect would be influenced by the relative velocities of emitter and detector, but the fact that galaxies are not moving through space might negate this. My understanding is that the expansion of space is the major factor, if not the sole cause of the redshift.
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if we replace "infinite distance" for "arbitrarily long distance" and "infinite speed" with "arbitrarily close to c"….
An "infinite distance" and an "arbitrarily long distance” are not synonymous.
How do you define "infinite speed”? The nearest I have been able to come is the thought that “c” might be considered as infinite speed, but I would have to search my notes from a few years ago to remember how I got there. [:D]
Post by: JohnDuffield on 22/02/2015 14:40:45
...My understanding is that the expansion of space is the major factor, if not the sole cause of the redshift...I think there's a big issue here that you're missing, wherein the universe expanding over time can be likened to pulling away from a black hole through space. Note what Pete said: the total energy of a photon moving through a gravitational field is constant. And remember that if you accelerate away from the photon source, you measure the photons as redshifted, but they haven't lost any energy. If you climb away from the photon source, you measure the photons as redshifted, but they haven't lost any energy. So when the universe expands, the inference is this: you measure the CMB photons as redshifted, but they haven't lost any energy.
Post by: PhysBang on 22/02/2015 15:16:51
OK, so you are committed to not using the correct terminology. Noted.Work has a very specific definition in physics. You should abandon your vague and condusing term and use the proper term. If you mean that there is no transfer of energy, then say that. Nobody claims that in a system of two bodies, one body falling towards the other increases the total energy of the system. However, one might speak of an increase in the energy of one body.The total energy of the falling brick does not increase. Again, gravity converts potential energy into kinetic energy, that's all. Then when the kinetic energy is dissipated, the brick has a mass deficit.
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Yes. It's quite simple, you merely choose a system of coordinates with an overall motion of 13m/s, to the systems of coordinates you are naively using, in the opposite direction to the falling of the brick.I have no idea what you are claiming. Regardless, one must use a system of coordinates to properly describe motion and the choice of system bears on the amount of work done on an object.You know full well what I'm saying. You start with a situation wherein the brick and the Earth are motionless relative to each other, and the brick is 10m above the ground. Then you drop the brick, and the brick ends up hitting the ground at 14m/s. You can't "choose some system" where the brick ends up hitting the ground at 1m/s.
I am glad you chose this example, as it might help you make less serious mistakes in the future.
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Well, since the textbooks that I have available and the experiments say otherwise, I will have to stick with my belief that your claim is not correct.You can cherry-pick the conclusion if you would like, but that isn't good reasoning.It's not cherry picking to point out something that's correct. You know it's correct, why are you trying to cast doubt upon it?
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You know that when you send a 511keV photon into a black hole, the black hole mass increases by 511keV/c².This is irrelevant to the claim that you are making.
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In the scenario you describe, you want the gains of the black hole to be both detectable and not detectable. You cannot have it both ways.I agree that the energy content of a system does not increase through its internal physical development. However, you seem to be ignoring Newton's third law.No I'm not. Momentum p=mv is shared equally between both objects, but kinetic energy KE=½mv² is not. The brick hits the ground at 14m/s, with 98 Joules of kinetic energy. The motion of the Earth towards the brick is not detectable, the Earth gains no detectable kinetic energy. It's similar for the black hole and the photon.
Post by: PhysBang on 22/02/2015 15:18:40
This is all great stuff, but it contradicts every cosmology text that discusses the expansion of the universe and explicitly includes the loss of energy from redshift in calculating the energy density of photons....My understanding is that the expansion of space is the major factor, if not the sole cause of the redshift...I think there's a big issue here that you're missing, wherein the universe expanding over time can be likened to pulling away from a black hole through space. Note what Pete said: the total energy of a photon moving through a gravitational field is constant. And remember that if you accelerate away from the photon source, you measure the photons as redshifted, but they haven't lost any energy. If you climb away from the photon source, you measure the photons as redshifted, but they haven't lost any energy. So when the universe expands, the inference is this: you measure the CMB photons as redshifted, but they haven't lost any energy.
Post by: JohnDuffield on 22/02/2015 15:58:33
...Yes. It's quite simple, you merely choose a system of coordinates with an overall motion of 13m/s, to the systems of coordinates you are naively using, in the opposite direction to the falling of the brick.You're talking out of your hat. The brick and the ground still have a closing speed of 14m/s.
This is all great stuff, but it contradicts every cosmology text that discusses the expansion of the universe and explicitly includes the loss of energy from redshift in calculating the energy density of photons.But the fact remains that when you send a 511keV photon into a black hole, its mass increases by 511keV/c², not a zillion tonnes. Energy is conserved. We know of no situation where it isn't. If some of those cosmology texts say gravitational field energy is negative, they're at odds with Einstein, who said the energy of the gravitational field shall act gravitatively in the same way as any other kind of energy (http://einsteinpapers.press.princeton.edu/vol6-trans/197?highlightText=gravitatively). And the issue is this: how can a photon in space lose energy when it doesn't interact with anything, and where did that energy go?
Post by: yor_on on 22/02/2015 16:39:15
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You can argue that a photon due to you moving away loses energy as a result of its momentum becoming smaller in relation to your direction of motion, just as with the ball thrown to you. I don't see how I can argue the same with a expansion though? That supports what you see looking at a light source redshifting, moving away from you, relativistically speaking, and also supports your definition of a photon staying intrinsically the same. But a expansion is another matter to me, and not what I call 'observer dependent'.
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The point is that no matter ones direction, from, or towards the photon source, those photons will have a exact same speed. The only difference is one of energy lost, or gained. and that is one he* of a mystical thing to me. So we can use classical physics to describe it through the idea of momentum, but we can't use light slowing down or speeding up, relative ourselves. When it comes to a expansion I still don't know how a 'photon' is thought to lose energy in itself though. The best description is a wave there, and then just presuming the duality of light to be adhered too. Somewhat to how we define conservation laws and a 'photon recoil'. Those ideas make a great deal of sense to me.
Post by: PhysBang on 22/02/2015 17:25:15
OK, now you are talking about something different. It is important to be precise, especially when discussing quantities like work that can depend on choice of system of coordinates....Yes. It's quite simple, you merely choose a system of coordinates with an overall motion of 13m/s, to the systems of coordinates you are naively using, in the opposite direction to the falling of the brick.You're talking out of your hat. The brick and the ground still have a closing speed of 14m/s.
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You seem to think that you can stick to one aspect of one physical scenario and use it to defeat many unrelated aspects of physics. If you accept Newton's Third Law, then you accept that a photon falling into a black hole also attracts the black hole. So the energy increase in the photon is exactly matched by a loss in the black hole, just like for any gravitational interaction. One cannot simply measure the total energy of the system, that does not change, and then say that none of the energy for any part of the system never changes; that's simply the fallacy of division.This is all great stuff, but it contradicts every cosmology text that discusses the expansion of the universe and explicitly includes the loss of energy from redshift in calculating the energy density of photons.But the fact remains that when you send a 511keV photon into a black hole, its mass increases by 511keV/c², not a zillion tonnes. Energy is conserved. We know of no situation where it isn't. If some of those cosmology texts say gravitational field energy is negative, they're at odds with Einstein, who said the energy of the gravitational field shall act gravitatively in the same way as any other kind of energy (http://einsteinpapers.press.princeton.edu/vol6-trans/197?highlightText=gravitatively). And the issue is this: how can a photon in space lose energy when it doesn't interact with anything, and where did that energy go?
Post by: Bill S on 22/02/2015 18:42:22
I am stationary, relative to the Earth. A photon is approaching me at “c”, and at the start of the scenario is one light minute away.
I accelerate away from the photon at an appreciable % of “c”.
The photon is still closing the distance between us at “c”, but it is redshifted. That means that when it catches up with me I will measure a redshift.
Does the photon have less energy as a result of being redshifted?
How could my action take energy from a photon that was one light minute away?
Post by: evan_au on 22/02/2015 19:02:01
Quote from: Bill S
Does the photon have less energy as a result of being redshifted?Yes, the red-shifted photon will impart less energy to your detector.
- Viewed as a wave, the photon is emitted with a certain frequency at the source. But the wave crests will "catch up" with a moving spaceship at a slower rate. (Just like a ship will measure one frequency of ocean waves when it is stationary, but a different frequency when moving towards or away from the source...)
- Viewed as a stream of photons, the photon are emitted at a certain rate at the source. But they will "catch up" with a moving spaceship at a slower rate. This slower rate can be described via classical physics at low speeds, and by time dilation at relativistic speeds.
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How could my action take energy from a photon that was one light minute away?You only measure the energy of the photon when it strikes your detector. So the motion of the detector only affects the photon energy when it is 0 light-minutes away.
Post by: Bill S on 22/02/2015 19:16:08
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You only measure the energy of the photon when it strikes your detector. So the motion of the detector only affects the photon energy when it is 0 light-minutes away.
So, if the only factor causing the redshift is my velocity, and the photon misses my detector, it will go on its way unchanged from when it was emitted? There is no loss of energy, the photon was never redshifted, the only effect is in the measurement?
Post by: Toffo on 22/02/2015 19:31:31
After accelaration: A moving star emitted the photon thousands of years ago, some recoil energy went into the star.
Part of the "missing" energy can be found in the star.
Rest of the "missing" energy can be found ... somewhere else.
Post by: jeffreyH on 22/02/2015 19:50:35
Naïve questions.
I am stationary, relative to the Earth. A photon is approaching me at “c”, and at the start of the scenario is one light minute away.
I accelerate away from the photon at an appreciable % of “c”.
The photon is still closing the distance between us at “c”, but it is redshifted. That means that when it catches up with me I will measure a redshift.
Does the photon have less energy as a result of being redshifted?
How could my action take energy from a photon that was one light minute away?
At an appreciable % of c YOUR energy has increased relative to the photon. Plus your time has slowed down. Viewed this way then the photon has to appear to have less energy as your detection equipment is also in a more energetic state.
Post by: JohnDuffield on 23/02/2015 12:02:44
So, if the only factor causing the redshift is my velocity, and the photon misses my detector, it will go on its way unchanged from when it was emitted? There is no loss of energy, the photon was never redshifted, the only effect is in the measurement?Correct. And it's the same if you ascend. You measure the photo to be redshifted, but it didn't change, you and your measurement equipment changed.
Post by: JohnDuffield on 23/02/2015 12:28:57
Quote from: PhysBang
You seem to think that you can stick to one aspect of one physical scenario and use it to defeat many unrelated aspects of physics. If you accept Newton's Third Law, then you accept that a photon falling into a black hole also attracts the black hole. So the energy increase in the photon is exactly matched by a loss in the black holeNo it isn't. This is clearer if you use a falling brick. Momentum p=mv is shared equally, but kinetic energy KE=½mv² isn't.
Quote from: PhysBang
One cannot simply measure the total energy of the system, that does not change, and then say that none of the energy for any part of the system never changes; that's simply the fallacy of division.The brick's energy doesn't change. Gravity merely converts potential energy into kinetic energy. Read this (http://en.wikipedia.org/wiki/Binding_energy#Mass-energy_relation):
"As an illustration, consider two objects attracting each other in space through their gravitational field. The attraction force accelerates the objects and they gain some speed toward each other converting the potential (gravity) energy into kinetic (movement) energy..."
John, you can measure a expansionary redshift, and their photons too, and also find them having lost 'energy' due to it.True enough. But where did this energy go?
In fact any redshift of a photon means a loss of energy, and it's just as strange due to me moving away from its source than with a expansion.Like Bill was saying, your motion doesn't actually change the photon. It hasn't actually lost any energy.
I don't see how I can argue the same with a expansion though? That supports what you see looking at a light source redshifting, moving away from you, relativistically speaking, and also supports your definition of a photon staying intrinsically the same.Interesting, isn't it?
The point is that no matter ones direction, from, or towards the photon source, those photons will have a exact same speed.That's another can of worms. If you head towards a star that's two light years away at 0.99999c, and if that star goes nova just as you set off, you will see the flash when you're halfway there. Your local measurement of that light coming towards you is c, but you and the light covered the two light years in one year of my time. Your closing speed was 1.99999c.
The only difference is one of energy lost, or gained. and that is one he* of a mystical thing to me. So we can use classical physics to describe it through the idea of momentum, but we can't use light slowing down or speeding up, relative ourselves...What you measure is not always the way it is.
Post by: PhysBang on 23/02/2015 14:02:06
Sure, but the momentum of the brick changes if we examinie it alone. The momentum of the photon changes as well. And where is the momentum of a photon?Quote from: PhysBangYou seem to think that you can stick to one aspect of one physical scenario and use it to defeat many unrelated aspects of physics. If you accept Newton's Third Law, then you accept that a photon falling into a black hole also attracts the black hole. So the energy increase in the photon is exactly matched by a loss in the black holeNo it isn't. This is clearer if you use a falling brick. Momentum p=mv is shared equally, but kinetic energy KE=½mv² isn't.
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The brick's energy doesn't change. Gravity merely converts potential energy into kinetic energy. Read this (http://en.wikipedia.org/wiki/Binding_energy#Mass-energy_relation):OK, so you found a single sentence that supports what you are saying. But that sentence does not say that the momentum doesn't change. Nor is attempting textual analysis of individual quoatations a replacement for physics.
"As an illustration, consider two objects attracting each other in space through their gravitational field. The attraction force accelerates the objects and they gain some speed toward each other converting the potential (gravity) energy into kinetic (movement) energy..."
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Some people make a very basic mistake in relativity theory, thinking that the speed of light is always relative to some object. The speed of light is, in special relativity, constant relative to nice systems of coordinates where Newtonian mechanics works. In general relativity, the speed of light is constant in every infinitesimal region of every system of coordinates.The point is that no matter ones direction, from, or towards the photon source, those photons will have a exact same speed.That's another can of worms. If you head towards a star that's two light years away at 0.99999c, and if that star goes nova just as you set off, you will see the flash when you're halfway there. Your local measurement of that light coming towards you is c, but you and the light covered the two light years in one year of my time. Your closing speed was 1.99999c.
Post by: JohnDuffield on 23/02/2015 14:26:25
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.thenakedscientists.com%2Fforum%2Findex.php%3Faction%3Ddlattach%3Btopic%3D54160.0%3Battach%3D19455%3Bimage&hash=996897d8b077cbd11f692c13ad526541)
Post by: PhysBang on 23/02/2015 14:59:45
Post by: Bill S on 23/02/2015 15:09:02
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Again, cherry-picking quotations from Einstein does not help us understand physics.
That's true, but for non-experts like me some explanation of what Einstein meant by (for example) the last sentence of John's quote would be of great value.
Post by: PmbPhy on 23/02/2015 15:17:55
Again, cherry-picking quotations from Einstein does not help us understand physics.He was explaining that the coordinate speed of light only has the value c in an inertial frame, in a vacuum in the absence of a gravitational field. If there is a gravitational field present then the coordinate speed of light varies with the gravitational potential. This is a fact of general relativity and well tested.
I derived this here - http://home.comcast.net/~peter.m.brown/gr/c_in_gfield.htm
The derivation for a uniform gravitational field/uniformly accelerating frame of reference is the same as the one Einstein did in his 1911 paper on the subject.
And there's absolutely nothing wrong with quoting a physics source. It is doing physics in the sense that you're referring to the calculations of results obtained by others. Doing it correctly is the challenge. :)
Post by: PhysBang on 23/02/2015 15:47:25
Post by: PmbPhy on 23/02/2015 16:07:33
It is one thing to recognize that the coordinate speed of light changes over finite distances. It is another thing to deny that the speed of light is constant over infinitesimal regions.I see. Is that what he was claiming? In which post did he make such a remark?
By the way. If you're interested a derivation from a text there here's one from Gravitation and Spacetime - 3rd Ed. by Ohanian and Ruffini
http://home.comcast.net/~peter.m.brown/gr/Ohanian_sol.pdf
Post by: PmbPhy on 23/02/2015 16:31:15
Quote from: JohnDuffield
But the fact remains that when you send a 511keV photon into a black hole, its mass increases by 511keV/c², not a zillion tonnes.Do you know how to calculate the mass of a particle when it's in a gravitational field?
Quote from: JohnDuffield
We know of no situation where it isn't.There's no reason to assume that the energy of any particle, including photons, is conserved when its moving through a gravitational field. The law of conservation of energy applies to the total energy of a system, not just to single particles moving through a gravitational field. An example of when the energy of a particle changes when its moving through a gravitational field is when the gravitational potentials, i.e. the guv, are time dependent.
The derivation of this fact can be found in A First Course in General Relativity - 2nd Ed. by Bernard Schutz, page 176
I put it on my website at http://home.comcast.net/~peter.m.brown/gr/conserved_quantities.htm
Quote from: JohnDuffield
If some of those cosmology texts say gravitational field energy is negative, they're at odds with Einstein, ...They most certainly are not!
Quote from: JohnDuffield
...the energy of the gravitational field shall act gravitatively in the same way as any other kind of energyJust because the energy of a gravitational field acts the same way as other energy, it doesn't mean it has the have the same value of energy.
You claim that there's no such thing as negative energy in physics. I think that you don't know what energy is.
Post by: JohnDuffield on 23/02/2015 17:19:54
That's true, but for non-experts like me some explanation of what Einstein meant by (for example) the last sentence of John's quote would be of great value.He meant what he said. Light curves in the room you're in because the speed of light is spatially variable. The speed of light near the floor is less than the speed of light near the ceiling. If it wasn't, your pencil wouldn't fall down.
Post by: PmbPhy on 23/02/2015 17:30:39
Quote from: JohnDuffield
Light curves in the room you're in because the speed of light is spatially variable. The speed of light near the floor is less than the speed of light near the ceiling. If it wasn't, your pencil wouldn't fall down.So what? Everyone who knows GR knows this. So long as you know that this is not about infinitesimals. PhysBang was correct when he said
Quote
It is one thing to recognize that the coordinate speed of light changes over finite distances. It is another thing to deny that the speed of light is constant over infinitesimal regions.He's much better than you at physics. You'd better listen to him.
Post by: JohnDuffield on 23/02/2015 18:06:42
Do you know how to calculate the mass of a particle when it's in a gravitational field?You don't calculate it, you measure it. And I will reiterate: when you send a 511keV photon into a black hole, the black hole mass increases by 511keV/c². You know this.
See? This is what I mean by the errors you keep making. There's no reason to assume that the energy of any particle, including photons, is conserved when its moving through a gravitational field.I'm not making any errors, and you know it. Because in your article (http://home.comcast.net/~peter.m.brown/gr/grav_red_shift.htm) you said this:
"The total energy of a photon moving through a gravitational field is constant."
They most certainly are not!Yes they are. Einstein made it crystal clear that gravitational field energy is positive. It causes more gravity.
This is a good example of your poor understanding of what you read. Just because the energy of a gravitational field acts the same way. Just because it acts the same way as other energy it doesn't mean it has the have the same value of energy. That's just plain dumb.No, it's a good example of my understanding. When two bodies fall together some of their mass-energy is converted into kinetic energy which ends up being radiated away into space. We're then left with a mass deficit. People then refer to binding energy as negative energy, but there's no actual negative energy present, just less positive energy. The conservation of energy books balance.
You claim that there's no such thing as negative energy in physics. I think that you don't know what energy is.I know what energy is. See post #7 here (http://www.thenakedscientists.com/forum/index.php?topic=47166.0;nowap) where on 19/09/2013 you said this:
"My objection with French here is that he gives the impression that energy "exists" whereas physics uses it merely as a bookkeeping scheme which requires no physical system."
It does exist. Matter is made from it. Ah, I see you've changed your tune, because in your article (http://home.comcast.net/~peter.m.brown/mech/what_is_energy.htm) you say this:
"This means that there is a flow of energy going around the electron. What is it that's actually flowing? All we can do at this point is give it a name. And the name we give it is energy." Have you been reading my Energy Explained (http://www.ilovephilosophy.com/viewtopic.php?f=4&t=172025)?
Post by: Bill S on 23/02/2015 19:39:50
Light in a vacuum travels at “c”.
Light in a medium appears to travel more slowly, but this is because the photons interact with atoms in the medium (I know that’s an oversimplification). Between atoms, light still travels at “c”.
Light slows in a gravitational field. Unless gravitons are physically real, there is nothing in a gravitational field to take the place of atoms in other media.
What slows light in a gravitational field?
Post by: PhysBang on 23/02/2015 19:50:13
What slows light in a gravitational field?The causal structure of spacetime?
I think I need time to come up with a better answer.
Post by: JohnDuffield on 23/02/2015 19:55:50
Light slows in a gravitational field. Unless gravitons are physically real, there is nothing in a gravitational field to take the place of atoms in other media. What slows light in a gravitational field?The altered properties of space. In mechanics a shear wave travels at a speed v = √(G/ρ) where G is the shear modulus of elasticity and ρ is density. In electrodynamics the an electromagnetic wave travels at a speed c = √(1/ε0μ0) where ε0 is the permittivity of space and μ0 is the permeability.
Post by: yor_on on 23/02/2015 22:14:44
==
what I think one need to see there is that the description no longer is solely about 'speeds' and 'paths', but more about 'timings', as expressed through that possible 'field'. It's somewhat of a bother how to think of this field, as we have real observer dependencies existing, making my observation differ from yours. Also it's somewhat of a huddle trying to define it as 'deterministic', or not. Myself I would expect it to be microscopically probabilistic and macroscopically defined from observer dependencies though, just as the universe we think us see normally. So 'non linear' to me, and if you use time symmetries, also non linear both ways. If there is a symmetry that is.
The last one is weird, but it has its own logic. As long as you have a way to record every outcome 'everywhere' (and at all 'times') you can disregard that statement. but if you don't have that ability, then you will find the past bifurcate into possibilities, just as the future can be seen to do, until it becomes a 'now', as in you observing that specific outcome. If the universe is non linear though, then I don't see how we ever will be able to know all outcomes. And if that is true for a future, then that 'future' also is my 'now', and 'past'.
Ever heard of 'the fog of war'? How about 'the fog of history' :)
Post by: PmbPhy on 23/02/2015 22:21:27
The answer is that gravitational time dilation causes light to slow down. Think about it and you'll soon realize why.What slows light in a gravitational field?The causal structure of spacetime?
I think I need time to come up with a better answer.
Post by: PmbPhy on 23/02/2015 22:24:37
Quote from: JohnDuffield
The altered properties of space. In mechanics a shear wave travels at a speed v = √(G/ρ) where G is the shear modulus of elasticity and ρ is density. In electrodynamics the an electromagnetic wave travels at a speed c = √(1/ε0μ0) where ε0 is the permittivity of space and μ0 is the permeability.All wrong.
Post by: PmbPhy on 23/02/2015 22:31:34
Quote from: JohnDuffield
You don't calculate it, you measure it.I can't believe this!!! The mass is calculated using the theory of relativity. One measures the mass and compares it with what the theory predicts. But in all cases one can calculate what the mass should be. In fact sometimes you actually cannot measure it, you have to calculate it.
Post by: jeffreyH on 25/02/2015 01:24:20
Post by: jeffreyH on 25/02/2015 01:36:10
Light slows in a gravitational field. Unless gravitons are physically real, there is nothing in a gravitational field to take the place of atoms in other media. What slows light in a gravitational field?The altered properties of space. In mechanics a shear wave travels at a speed v = √(G/ρ) where G is the shear modulus of elasticity and ρ is density. In electrodynamics the an electromagnetic wave travels at a speed c = √(1/ε0μ0) where ε0 is the permittivity of space and μ0 is the permeability.
John! Really? Get your square roots right. You have just redefined the speed of light. By posting incorrect equations you are doing such a disservice to those struggling to learn physics and I just can't let this one go. It's just wrong. If you insist on posting equations then at least sanity check them.
Post by: PmbPhy on 25/02/2015 03:46:14
Being the OP I would like to pose a further question. Would the decrease in energy of the photon moving away from a gravitational field source be linear over distance?No. Take a look at the formula for it. First you have to keep in mind that you're comparing the results of two observers. The energy and thus frequency as measured by one observer remains constant as it moves through the field.
See Eq. 7 at http://home.comcast.net/~peter.m.brown/gr/grav_red_shift.htm
Post by: JohnDuffield on 25/02/2015 11:26:38
John! Really? Get your square roots right. You have just redefined the speed of light. By posting incorrect equations you are doing such a disservice to those struggling to learn physics and I just can't let this one go. It's just wrong. If you insist on posting equations then at least sanity check them.I haven't posted anything that's incorrect. See for example permeability (http://en.wikipedia.org/wiki/Vacuum_permeability#Significance_in_electromagnetism) on Wikipedia where you can see the expression written as
.Quote from: jeffreyh
Would the decrease in energy of the photon moving away from a gravitational field source be linear over distance?It doesn't lose any energy. And again: when you send a 511keV photon down into a black hole the black hole mass increases by 511keV/c². Conservation of energy applies. This is also true for an ascending photon. It appears to have lost energy because we added energy to you to lift you up. However it hasn't actually lose any energy.
Quote from: PmbPhy
I can't believe this!!! The mass is calculated using the theory of relativity. One measures the mass and compares it with what the theory predicts. But in all cases one can calculate what the mass should be. In fact sometimes you actually cannot measure it, you have to calculate it.You said the mass of a particle. There is no theory that predicts the mass of an electron, or any other particle. You know this. Do not be dishonest Pete. It does you no credit. And by the way, this is the wrong answer: The answer is that gravitational time dilation causes light to slow down. You measure time using say optical clocks. When your clock reading at one elevation doesn't match your clock reading at another elevation, it isn't because "time is going slower", it's because light is going slower. If you had read Einstein's original material (http://einsteinpapers.press.princeton.edu/vol7-trans/156?highlightText=%22speed%20of%20light%22) you would know this. Or are you saying Einstein was wrong?
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.thenakedscientists.com%2Fforum%2Findex.php%3Faction%3Ddlattach%3Btopic%3D54160.0%3Battach%3D19455%3Bimage&hash=996897d8b077cbd11f692c13ad526541)
Post by: PhysBang on 25/02/2015 14:03:36
This quotation comes from a section titled, "Some consequences of the equivalence hypothesis". It is a consequence of using systems of coordinates and their properties in Riemann geometry (that is, spacetime curvature as we tend to refer to it today) to represent gravity that we find that in some systems of coordinates, the speed of light is not constant over finite distances. Later in that same section, indeed, on the page that JohnDuffield has carefully cut from his screenshot, Einstein also writes, "Nevertheless, this limiting case <also> is of fundamental significance for the theory of general relativity; because the fact from which we started out, namely that no gravitational field exists in the vicinity of a free-falling observer, this very fact shows that in the vicinity of every world point the results of the special theory of relativity are valid (in the infinitesimal) for a suitably chosen local coordinate system."
[Please do not forget or ignore this very important point when someone tries to tell you that general relativity demands that time stops in some scenario. Especially if they tell you that Einstein said this.]
Post by: jeffreyH on 25/02/2015 17:51:39
Yes, Einstein writes, specifically, "the curvature of light rays occurs only in spaces where the speed of light is spatially variable." He says this because if one can introduce an accelerated reference frame relative to an inertial one, then the constant motion over time from the inertial frame will be accelerated (a spatially variable speed).
This quotation comes from a section titled, "Some consequences of the equivalence hypothesis". It is a consequence of using systems of coordinates and their properties in Riemann geometry (that is, spacetime curvature as we tend to refer to it today) to represent gravity that we find that in some systems of coordinates, the speed of light is not constant over finite distances. Later in that same section, indeed, on the page that JohnDuffield has carefully cut from his screenshot, Einstein also writes, "Nevertheless, this limiting case <also> is of fundamental significance for the theory of general relativity; because the fact from which we started out, namely that no gravitational field exists in the vicinity of a free-falling observer, this very fact shows that in the vicinity of every world point the results of the special theory of relativity are valid (in the infinitesimal) for a suitably chosen local coordinate system."
[Please do not forget or ignore this very important point when someone tries to tell you that general relativity demands that time stops in some scenario. Especially if they tell you that Einstein said this.]
Well the points I have been trying to discuss are lost in the mire so I am abandoning this thread and I'll just carry on without any reasonable answers. I want to learn. I don't want to be lectured and told why everything I read both in textbooks and online is wrong and then not be given any proof that it is wrong that I can reliably test. I won't be posting many more questions on this forum because it just isn't worth it.
Post by: JohnDuffield on 25/02/2015 18:16:49
Well the points I have been trying to discuss are lost in the mire so I am abandoning this thread and I'll just carry on without any reasonable answers.You've had some good answers.
I want to learnI don't think you do. I think you want confirmation of some idea you've come up with.
I don't want to be lectured and told why everything I read both in textbooks and online is wrong and then not be given any proof that it is wrong that I can reliably test. I won't be posting many more questions on this forum because it just isn't worth it.Both PmbPhy and I have told you that the ascending photon doesn't lose any energy. The proof is conservation of energy: you send a 511keV photon down into a black hole, and the black hole mass increases by 511keV/c². No energy is acquired by the descending photon. In similar vein no energy is lost by the ascending photon.
I won't be posting many more questions on this forum because it just isn't worth it.You ask a question, and you get an answer. Don't reject that answer just because it doesn't square with some popscience nonsense you've picked up. Pursue it, and/or ask the question elsewhere and compare answers.
Post by: jeffreyH on 25/02/2015 18:58:38
Well the points I have been trying to discuss are lost in the mire so I am abandoning this thread and I'll just carry on without any reasonable answers.You've had some good answers.I want to learnI don't think you do. I think you want confirmation of some idea you've come up with.I don't want to be lectured and told why everything I read both in textbooks and online is wrong and then not be given any proof that it is wrong that I can reliably test. I won't be posting many more questions on this forum because it just isn't worth it.Both PmbPhy and I have told you that the ascending photon doesn't lose any energy. The proof is conservation of energy: you send a 511keV photon down into a black hole, and the black hole mass increases by 511keV/c². No energy is acquired by the descending photon. In similar vein no energy is lost by the ascending photon.I won't be posting many more questions on this forum because it just isn't worth it.You ask a question, and you get an answer. Don't reject that answer just because it doesn't square with some popscience nonsense you've picked up. Pursue it, and/or ask the question elsewhere and compare answers.
Basically what you keep telling me is that all the things I am reading (lots of mathematics and physics textbooks) are basically pop science and that I have a pet theory to peddle. You quote Einstein willy nilly without even providing mathematical equations to demonstrate that what you say is correct. Usually you just copy and paste an easily recognizable equations derived by someone much cleverer than you. That is why I explicitly asked you how you derived a particular equation to which you replied it was a well known Schwarzschild metric equation. Anyone can do that. Mostly you post pretty pictures which I assume you do not compose yourself. What really irritates me is that I AM putting in the effort. Lots of it. So that I can actually get to a point where I have the tools I need to progress. I could be lazing on a beach somewhere drink in hand. However this is something I wish I had pursued when I was a lot younger. So don't start preaching to me please until you get your own house in order.
Post by: JohnDuffield on 25/02/2015 20:13:25
Basically what you keep telling me is that all the things I am reading (lots of mathematics and physics textbooks) are basically pop science and that I have a pet theory to peddle.I'm afraid I do think some of the things you say are popscience. I don't recall you referring to some textbook when asking a question.
You quote Einstein willy nilly without even providing mathematical equations to demonstrate that what you say is correct.A mathematical equation will not demonstrate that what I say is correct. Hard scientific evidence demonstrates that. And all the hard scientific evidence says there are no perpetual motion machines, and that energy is conserved.
Usually you just copy and paste an easily recognizable equations derived by someone much cleverer than you. That is why I explicitly asked you how you derived a particular equation to which you replied it was a well known Schwarzschild metric equation. Anyone can do that. Mostly you post pretty pictures which I assume you do not compose yourself. What really irritates me is that I AM putting in the effort. Lots of it. So that I can actually get to a point where I have the tools I need to progress. I could be lazing on a beach somewhere drink in hand. However this is something I wish I had pursued when I was a lot younger. So don't start preaching to me please until you get your own house in order.I will reiterate: when you ask a question, you get an answer. If you don't like that answer, ask your question elsewhere, and/or challenge the answer using your own references to Einstein and the evidence and the maths. Now, I apologise for causing offence, please can we get back to the physics.
Post by: PhysBang on 25/02/2015 20:51:27
I will reiterate: when you ask a question, you get an answer. If you don't like that answer, ask your question elsewhere, and/or challenge the answer using your own references to Einstein and the evidence and the maths. Now, I apologise for causing offence, please can we get back to the physics.References to Einstein are not how questions in physics are supposed to be answered.
Post by: jeffreyH on 25/02/2015 23:16:38
Light slows in a gravitational field. Unless gravitons are physically real, there is nothing in a gravitational field to take the place of atoms in other media. What slows light in a gravitational field?The altered properties of space. In mechanics a shear wave travels at a speed v = √(G/ρ) where G is the shear modulus of elasticity and ρ is density. In electrodynamics the an electromagnetic wave travels at a speed c = √(1/ε0μ0) where ε0 is the permittivity of space and μ0 is the permeability.
c = √(1/ε0μ0)
That is where you were wrong until you corrected it in a later post.
Post by: jeffreyH on 25/02/2015 23:47:12
http://bogpaper.com/science-sundays-with-john-duffield-bankrupting-physics/
So all you protectionist Phd wielding protectionists out there with your pop science theories had better watch out. He's coming for you. I think yor_on has the right idea. Quick, find a table to hide under.
Post by: jeffreyH on 26/02/2015 00:02:10
Smolin's "The Trouble With Physics" is on Amazon. I will not post a link. The outline and reviews should be read at least to see what Smolin's attitude really is. John holds him up as an example of someone "in his camp" so to speak. That is a disservice to Smolin.
Post by: jeffreyH on 26/02/2015 00:40:02
http://www.preposterousuniverse.com/blog/2013/07/29/talking-back-to-your-elders/
JohnDuffield, I think you are making misjudgments in your arguments.
And by the way. I too do not have the official credentials that so many people want to require of me. You need to put in the time to understand the issues in physics. It’s human nature to take a specific example (someone without the credentials) and then erroneously apply it to all people that do not have the credentials.
Post by: jeffreyH on 26/02/2015 01:18:36
Post by: PhysBang on 26/02/2015 12:48:06
But never mind John is here to fix physics.Seeing that there is too funny! An entire blog devoted to a school of economics that explicitly denies empirical research. A perfect place for that physics content.
http://bogpaper.com/science-sundays-with-john-duffield-bankrupting-physics/
So all you protectionist Phd wielding protectionists out there with your pop science theories had better watch out. He's coming for you. I think yor_on has the right idea. Quick, find a table to hide under.
Post by: JohnDuffield on 26/02/2015 14:42:13
c = √(1/ε0μ0)It isn't wrong. What's the square root of a sixteenth? A quarter. And what's one divided by the square root of sixteen? A quarter.
That is where you were wrong until you corrected it in a later post.
But never mind John is here to fix physics.No, I'm here to talk physics.
Post by: jeffreyH on 26/02/2015 15:44:58
c = √(1/ε0μ0)It isn't wrong. What's the square root of a sixteenth? A quarter. And what's one divided by the square root of sixteen? A quarter.
That is where you were wrong until you corrected it in a later post.But never mind John is here to fix physics.No, I'm here to talk physics.
Well talk about physics then. I posted an equation to help.
Post by: JohnDuffield on 26/02/2015 16:44:44
"Now use the Equivalence Principle to infer that in the room you are sitting in right now on Earth, where real gravity is present and you aren't really accelerating (we'll neglect Earth's rotation!), light and time must behave in the same way to a high approximation: light speeds up as it ascends from floor to ceiling (it doesn't slow down, as apparently quoted on your discussion site), and it slows down as it descends from ceiling to floor; it's not like a ball that slows on the way up and goes faster on the way down..."
Post by: PmbPhy on 26/02/2015 16:45:35
Quote from: JohnDuffield
It isn't wrong.Of course it's wrong.
In reply #74 you wrote claimed that
which is incorrect. The correct expression is
Post by: PhysBang on 26/02/2015 17:10:25
See for example hyperphysics (http://hyperphysics.phy-astr.gsu.edu/hbase/conser.html) where you can read this: "The conservation of energy principle is one of the foundation principles of all science disciplines. In varied areas of science there will be primary equations which can be seen to be just an appropriate reformulation of the principle of conservation of energy". You gave a Lagrangian for a massive particle in a gravitational field, wherein the first portion is the kinetic energy, and the mgz is the potential energy. This is not appropriate for a photon, because the photon is massless, and it's all kinetic energy. If you throw a massive particle upwards, kinetic energy is converted into potential energy. When all the kinetic energy is converted into potential energy, the particle has reached its highest point, and then it start falling back down, converting potential energy into kinetic energy. When you send a photon upwards, it doesn't slow down and stop. Instead, it speeds up. If you don't believe me, contact Don Koks, the editor of the Baez/PhysFAQ website (http://www.math.ucr.edu/home/baez/physics/), who said this:In General Relativity, we are free to use systems of coordinates in which the coordinate speed of light over finite distances can change. This is one way to represent the change in the energy of light from the effect of gravity on that light. In other systems of coordinates, we use the change in frequency of the light to represent the change in energy of the light due to gravity.
"Now use the Equivalence Principle to infer that in the room you are sitting in right now on Earth, where real gravity is present and you aren't really accelerating (we'll neglect Earth's rotation!), light and time must behave in the same way to a high approximation: light speeds up as it ascends from floor to ceiling (it doesn't slow down, as apparently quoted on your discussion site), and it slows down as it descends from ceiling to floor; it's not like a ball that slows on the way up and goes faster on the way down..."
Because the light can only be represented as kinetic energy, the only way to represent the change in energy is in the kinetic energy, either through speed or frequency. When it comes to an absorption event, in the system of coordinates in which the absorber is at rest, the light is absorbed at a higher or lower frequency depending on the way that gravity has changed the photon.
In a standard application of this, in a photon traveling away from or towards the Earth, for example, there is no concern about the conservation of energy as the energy of the entire system is conserved.
Post by: Bill S on 26/02/2015 18:46:51
Quote from: John
you send a 511keV photon down into a black hole, and the black hole mass increases by 511keV/c². No energy is acquired by the descending photon. In similar vein no energy is lost by the ascending photon.
There is something about this oft repeated assertion that puzzles me (so what’s new?)
If the 511keV photon [goes] down into a black hole, and the black hole mass increases by 511keV/c². That’s fine, and I wouldn’t argue with that; but does that mean that conservation of energy would prevent the photon from gaining, or losing, energy, as long as that energy came from, or passed to, the energy of the black hole? In either case the total energy would remain constant. Wouldn’t it?
NB, this thought came to me while dog walking, and by the time I came home PhysBang had posted #99, the two seemed to be linked.
Post by: yor_on on 26/02/2015 19:51:49
E. Noether's Discovery of the Deep Connection Between Symmetries and Conservation Laws: By Nina Byers (http://www.physics.ucla.edu/~cwp/articles/noether.asg/noether.html) It discusses "`Proper' and `Improper' Conservation Laws", and refers also to gravity.
"In special relativity these theories have a `proper energy theorem' in the sense of Hilbert and we will show how `proper energy theorems' give a principle of local energy conservation. In general relativity, on the other hand, the proper energy theorem becomes improper in that the energy-momentum tensor for which the theorem holds is gauge dependent. As will be shown below, there is transfer of energy to and from the gravitational field and it has not meaning to speak of a definite localization of the energy of the gravitational field in space. Consequently we do not have a principle of local energy conservation in spacetime regions in which there exist gravitational fields."
Maybe this one will help too?
As that is what it hinges on.
What is a gauge? (https://terrytao.wordpress.com/2008/09/27/what-is-a-gauge/)
Post by: David Cooper on 26/02/2015 22:12:54
c = √(1/ε0μ0)It isn't wrong. What's the square root of a sixteenth? A quarter. And what's one divided by the square root of sixteen? A quarter.
That is where you were wrong until you corrected it in a later post.
I too thought c = √(1/ε0μ0) was different from c = 1/√(ε0μ0), but it isn't - they are equivalent, so John made no error. When PMB was misled into saying John had got this wrong, his dyslexia made him read c = √(1/ε0μ0) as c = √(ε0μ0), so his "correction" was an honest mistake. As usual this is just one great big misunderstanding.
John's appears to have been right all the way through this thread (and certainly on the main issue) - there is no energy loss to the photon as it climbs out of a gravity well and no gain when it enters one.
Post by: PmbPhy on 26/02/2015 22:21:06
Quote from: David Cooper
I too thought c = √(1/ε0μ0) was different from c = 1/√(ε0μ0), but it isn't - they are equivalent, so John made no error. When PMB was misled into saying John had got this wrong, his dyslexia made him read c = √(1/ε0μ0) as c = √(ε0μ0), so his "correction" was an honest mistake. As usual this is just one great big misunderstanding.Thanks, David. Much appreciated. It wasn't so much my dyslexia as it was John expressing the speed of light in a manner which it's never expressed and not pointing this fact out when Jeff also objected to it.
Post by: yor_on on 27/02/2015 00:06:30
Post by: yor_on on 27/02/2015 00:56:14
Light slows in a gravitational field. Unless gravitons are physically real, there is nothing in a gravitational field to take the place of atoms in other media. What slows light in a gravitational field?The altered properties of space. In mechanics a shear wave travels at a speed v = √(G/ρ) where G is the shear modulus of elasticity and ρ is density. In electrodynamics the an electromagnetic wave travels at a speed c = √(1/ε0μ0) where ε0 is the permittivity of space and μ0 is the permeability.
John, How do you think there?
Are you arguing different permittivity to a perfect vacuum? Using a electromagnetic field to define it by I can accept, but the definition of permittivity is one where there is no friction and resistance, namely a vacuum. So either you mean that a electromagnetic field is a vacuum? Or that the vacuum we set as the standard to measure all other permittivity against varies? (or maybe both? Reading you again? although that would be highly contradictorily as it assumes Maxwell to be wrong, while still using his definitions)
"the parameter ε0 is a measurement-system constant. Its presence in the equations now used to define electromagnetic quantities is the result of the so-called "rationalization" process described below.
But the method of allocating a value to it is a consequence of the result that Maxwell's equations predict that, in free space, electromagnetic waves move with the speed of light. Understanding why ε0 has the value it does requires a brief understanding of the history."
And if you use Plank scale.
" Planck normalized to 1 the Coulomb force constant 1/(4πε0) (as does the cgs system of units). This sets the Planck impedance, ZP equal to Z0/4π, where Z0 is the characteristic impedance of free space.
Normalizing the permittivity of free space ε0 to 1: Sets the permeability of free space µ0 = 1, (because c = 1).
Sets the unit impedance or unit resistance to the characteristic impedance of free space, ZP = Z0 (or sets the characteristic impedance of free space Z0 to 1).
radius r). "
Finally "Both ε0 and μ0 appear in the Maxwell curl equations, so there are two free parameters, their product and their ratio. The two curl equations determine the propagation velocity of an electromagnetic wave in vacuum (c = 1/sqrt(μ0ε0)), and the ratio is related to the magnitude of E over H (sqrt(μ0/ε0) = 377 ohms). " By Bob S.
https://en.wikipedia.org/wiki/Vacuum_permittivity
https://en.wikipedia.org/wiki/Planck_units
https://www.physicsforums.com/threads/permittivity-and-permeability-of-free-space.122121/
==
the point is that a vacuum classically is 'empty'. A free charge can move through a vacuum without a EM field needed for it to propagate in. Quantum electrodynamics using the concept of a vacuum consisting of EM still have to follow that definition, although rationalizing it away as an effect of a 'vacuum ground state' equivalent to the classical counterpart without really explaining how it comes to be. And gravity is not electromagnetic. If you want to think of it as gravitons then this is interesting. http://www.physlink.com/Education/AskExperts/ae658.cfm
And QED is not really about waves, it's about quanta. so when it uses this concept it's conceptually different from defining it as waves, (although if you think as me the duality still should be there) But I'm having trouble following your reasoning here.
Post by: jeffreyH on 27/02/2015 02:19:19
Quote from: David CooperI too thought c = √(1/ε0μ0) was different from c = 1/√(ε0μ0), but it isn't - they are equivalent, so John made no error. When PMB was misled into saying John had got this wrong, his dyslexia made him read c = √(1/ε0μ0) as c = √(ε0μ0), so his "correction" was an honest mistake. As usual this is just one great big misunderstanding.Thanks, David. Much appreciated. It wasn't so much my dyslexia as it was John expressing the speed of light in a manner which it's never expressed and not pointing this fact out when Jeff also objected to it.
My objection is to the fact that this is OK with unity as a numerator but 15/SQRT(16) for instance is not the same as SQRT(15/16). So there is a distinction. It can matter. So to pose it in the correct form is not trivial. Especially if the numerator could be variable. In this case it is not so doesn't matter.
Post by: jeffreyH on 27/02/2015 02:31:04
See for example hyperphysics (http://hyperphysics.phy-astr.gsu.edu/hbase/conser.html) where you can read this: "The conservation of energy principle is one of the foundation principles of all science disciplines. In varied areas of science there will be primary equations which can be seen to be just an appropriate reformulation of the principle of conservation of energy". You gave a Lagrangian for a massive particle in a gravitational field, wherein the first portion is the kinetic energy, and the mgz is the potential energy. This is not appropriate for a photon, because the photon is massless, and it's all kinetic energy. If you throw a massive particle upwards, kinetic energy is converted into potential energy. When all the kinetic energy is converted into potential energy, the particle has reached its highest point, and then it start falling back down, converting potential energy into kinetic energy. When you send a photon upwards, it doesn't slow down and stop. Instead, it speeds up. If you don't believe me, contact Don Koks, the editor of the Baez/PhysFAQ website (http://www.math.ucr.edu/home/baez/physics/), who said this:In General Relativity, we are free to use systems of coordinates in which the coordinate speed of light over finite distances can change. This is one way to represent the change in the energy of light from the effect of gravity on that light. In other systems of coordinates, we use the change in frequency of the light to represent the change in energy of the light due to gravity.
"Now use the Equivalence Principle to infer that in the room you are sitting in right now on Earth, where real gravity is present and you aren't really accelerating (we'll neglect Earth's rotation!), light and time must behave in the same way to a high approximation: light speeds up as it ascends from floor to ceiling (it doesn't slow down, as apparently quoted on your discussion site), and it slows down as it descends from ceiling to floor; it's not like a ball that slows on the way up and goes faster on the way down..."
Because the light can only be represented as kinetic energy, the only way to represent the change in energy is in the kinetic energy, either through speed or frequency. When it comes to an absorption event, in the system of coordinates in which the absorber is at rest, the light is absorbed at a higher or lower frequency depending on the way that gravity has changed the photon.
In a standard application of this, in a photon traveling away from or towards the Earth, for example, there is no concern about the conservation of energy as the energy of the entire system is conserved.
Yes its the whole system's energy. As light is all kinetic energy you have a different system to that of other lower velocity particles. I am actually impressed by John for once I must admit. He actually did know what the Lagrangian was.
Post by: jeffreyH on 27/02/2015 02:53:56
Try this one Bill.
E. Noether's Discovery of the Deep Connection Between Symmetries and Conservation Laws: By Nina Byers (http://www.physics.ucla.edu/~cwp/articles/noether.asg/noether.html) It discusses "`Proper' and `Improper' Conservation Laws", and refers also to gravity.
"In special relativity these theories have a `proper energy theorem' in the sense of Hilbert and we will show how `proper energy theorems' give a principle of local energy conservation. In general relativity, on the other hand, the proper energy theorem becomes improper in that the energy-momentum tensor for which the theorem holds is gauge dependent. As will be shown below, there is transfer of energy to and from the gravitational field and it has not meaning to speak of a definite localization of the energy of the gravitational field in space. Consequently we do not have a principle of local energy conservation in spacetime regions in which there exist gravitational fields."
Maybe this one will help too?
As that is what it hinges on.
What is a gauge? (https://terrytao.wordpress.com/2008/09/27/what-is-a-gauge/)
Great post. I am just working through conservation laws and symmetries. The important point to note is "As will be shown below, there is transfer of energy to and from the gravitational field and it has not meaning to speak of a definite localization of the energy of the gravitational field in space. Consequently we do not have a principle of local energy conservation in spacetime regions in which there exist gravitational fields." When viewed in respect of the kinetic nature of the photon with zero rest mass the transfer is from the photon to the gravitational field. John declares that the energy is radiated away into space. How does this conserve energy in the system?
Post by: jeffreyH on 27/02/2015 02:59:36
Post by: yor_on on 27/02/2015 04:01:35
All photons leaving a 'gravity well' as a sun should redshift from the view of a thought up observer in that sun. But we're still 'visited' by photons traveling since the beginning of the Big Bang, Billions of light years, and they seem the same as any other 'photons', as far as I know? That is, if we adapt it for age-(distance) redshifts ('expansion' being one cause) taking place. If we want to take it to its extreme we'll have to assume some final eruption from a star, soon to become a 'black hole'. Would that mean that some photons then lose their energy and 'die out' propagating? As the gravity well (sun) balance on this edge of becoming a black hole? I think that is called a 'tired light theory' myself? And as gravity's reach is defined as 'infinite' you then can argue the same for its whole 'propagation'.
"A slightly different kind of supernova explosion occurs when even larger, hotter stars (blue giants and blue supergiants) reach the end of their short, dramatic lives. These stars are hot enough to burn not just hydrogen and helium as fuel, but also carbon, oxygen and silicon. Eventually, the fusion in these stars forms the element iron (which is the most stable of all nuclei, and will not easily fuse into heavier elements), which effectively ends the nuclear fusion process within the star. Lacking fuel for fusion, the temperature of the star decreases and the rate of collapse due to gravity increases, ..... until it collapse completely on itself, blowing out material in a massive supernova explosion..... "
Post by: jeffreyH on 27/02/2015 05:00:45
. This is not applicable to the photon. If we remove the term 
then we are working with a directional velocity only 
. In the absence of any gravitational field we can orient the path of the photon along the z-axis so that x and y terms vanish. The only way we can maintain this path is by somehow defining a flat spacetime that does not disturb the path. In this way we can then view the effect of gravity on a wave function without the added complexity. The potential and kinetic energy terms then need a relativistic mass equation for the photon. The wave function is not usually a one particle affair. However it is a very useful exercise.For an inertial frame the directional velocity is always c in a vacuum so this can be considered a constant. This only changes due to the gravitational field. With the inclusion of relativistic mass we can restore momentum to the photon.
Post by: jeffreyH on 27/02/2015 05:26:11
In basic form we have the relation E = mc^2. Since frequency is part of the relativistic mass of the photon we have an energy frequency relationship when viewed in this way. However this is used to illustrate the point and not to indicate that E = mc^2 can be directly applied to photons. Just in case anyone complains.
Post by: jeffreyH on 27/02/2015 05:38:37
Post by: jeffreyH on 27/02/2015 05:40:04
Post by: PmbPhy on 27/02/2015 07:40:29
Quote from: jeffreyH
Thanks to a page on Pete's site we have the relativistic mass defined as m = p/c.I'm glad I was able to be of service in that way. :)
Which page are you referring to if I may I ask?
Quote from: jeffreyH
However this is used to illustrate the point and not to indicate that E = mc^2 can be directly applied to photons. Just in case anyone complains.No. That is correct. You really can use it in that way. See:
http://home.comcast.net/~peter.m.brown/ref/relativistic_mass.htm
and notice that what you just described appears in these well known special relativity textbooks:
Relativity: Special, General and Cosmological by Rindler, Oxford Univ., Press, (2001), page 120
From Introducing Einstein's Relativity by Ray D'Inverno, Oxford Univ. Press, (1992), page 50
Special Relativity by A. P. French, MIT Press, page 20
Post by: jeffreyH on 27/02/2015 08:22:43
http://home.comcast.net/~peter.m.brown/ref/relativistic_mass/relativistic_mass.htm
Post by: jeffreyH on 27/02/2015 08:54:05
Quote from: jeffreyHThanks to a page on Pete's site we have the relativistic mass defined as m = p/c.I'm glad I was able to be of service in that way. :)
Which page are you referring to if I may I ask?Quote from: jeffreyHHowever this is used to illustrate the point and not to indicate that E = mc^2 can be directly applied to photons. Just in case anyone complains.No. That is correct. You really can use it in that way. See:
http://home.comcast.net/~peter.m.brown/ref/relativistic_mass.htm
and notice that what you just described appears in these well known special relativity textbooks:
Relativity: Special, General and Cosmological by Rindler, Oxford Univ., Press, (2001), page 120
From Introducing Einstein's Relativity by Ray D'Inverno, Oxford Univ. Press, (1992), page 50
Special Relativity by A. P. French, MIT Press, page 20
Thanks Pete. I tell you two things I don't like. 1) Being told I only talk Pop Science nonsense and 2) I am peddling some personal theory. I have worked hard and put in the time to get to the point I am at now. This stuff isn't easy so you have to be committed to learning it. This may mean you have to read some texts through multiple times or read various texts to pick things up. To be dismissed in such an offhand way is insulting.
Post by: JohnDuffield on 27/02/2015 09:49:44
If the 511keV photon [goes] down into a black hole, and the black hole mass increases by 511keV/c². That’s fine, and I wouldn’t argue with that;Good stuff. IMHO conservation of energy is just about the most important rule in physics.
but does that mean that conservation of energy would prevent the photon from gaining, or losing, energy, as long as that energy came from, or passed to, the energy of the black hole?No. If one thing gains energy, another thing loses it. But think about it. That photon is descending at the speed of light. If it's gaining energy, how is it getting it? If it starts off 93 million miles from the black hole, it takes 8 minutes for anything to get from the black hole to the photon. And things can't get out of a black hole. There is no magical mysterious mechanism by which energy can get from the black hole to the photon instantaneously.
In either case the total energy would remain constant. Wouldn’t it?Yes. This is the thing about gravity. There are websites and books out there that say gravity is negative energy, when it isn't. When two things fall together, energy is conserved. The books always balance. The total energy of the universe is not zero.
David: thanks re the c = √(1/ε0μ0) expression.
Quote from: yor_on
The interesting thing, to me then, is still how one should define it losing energy in a expansion...Me too. People say the CMBR photons have redshifted a thousandfold and lost most of their energy. But nobody seems to be able to say where it's gone. By the way, I think the Nina Byers paper is the sort of thing that causes confusion.
Quote from: yor_on
Are you arguing different permittivity to a perfect vacuum?No. It's a perfect vacuum, but like Einstein said, a concentration of energy in the guise of a massive star "conditions" the surrounding space, altering its metrical properties, such that "the speed of light is spatially variable". Then c = √(1/ε0μ0) at one location is not the same as c = √(1/ε0μ0) at another. The permittivity and/or permeability of space at one elevation is not the same as at another. But wherever you go you measure things like c and ε0 to be the same because it's an "immersive scale change". It's a bit like a flatlander trying to measure whether his world has been stretched using a ruler that's also been stretched.
Quote from: yor_on
the point is that a vacuum classically is 'empty'.That's not what Maxwell or Einstein thought. They thought of space as a gin-clear ghostly elastic thing, something that has properties, something that can be stressed, something that can wave. Check out this (http://www.rain.org/~karpeles/einsteindis.html) where you can read Einstein saying a field is "a state of space". Also see the arXiv (http://arxiv.org/find/grp_physics/1/ti:+aether/0/1/0/all/0/1) and this (http://en.wikipedia.org/wiki/Aether_theories#General_relativity) too.
Quote from: yor_on
And gravity is not electromagnetic. If you want to think of it as gravitons then this is interesting.Einstein struggled for years trying to combine gravity and electromagnetism. As for gravitons, they're "field quanta", they're virtual particles, not real particles. It's like you divide a field up into chunks and say each one is a virtual particle. And according to Einstein, a field is a state of space. Now, how many states of space are there where an electron is? One.
Quote from: Jeffrey
He actually did know what the Lagrangian was.My physics knowledge is extensive. That sometimes causes problems when it conflicts with some popscience book or science article, or even a textbook or paper.
Quote from: Jeffrey
Great post. I am just working through conservation laws and symmetries. The important point to note is "As will be shown below, there is transfer of energy to and from the gravitational field and it has not meaning to speak of a definite localization of the energy of the gravitational field in space. Consequently we do not have a principle of local energy conservation in spacetime regions in which there exist gravitational fields."I think this paper causes confusion, and I would recommend that you set it aside.
Quote from: Jeffrey
When viewed in respect of the kinetic nature of the photon with zero rest mass the transfer is from the photon to the gravitational field.There is no transfer of energy from the photon to the gravitational field.
Quote from: Jeffrey
John declares that the energy is radiated away into space. How does this conserve energy in the system?This applies to two massive bodies, not to the photon descending into the black hole.
Quote from: Jeffrey
One more thing. John talks of a mass deficit when a brick falls freely in a gravitational field. How does this square with the fact that a mass weighs more at lower altitudes.Because matter is affected half as much as light. If you replace the falling brick with a 511keV/c² electron and compare with a descending 511keV photon, the potential energy converted into kinetic energy is half the apparent energy gain of the photon. So imagine your brick has a mass x at some elevation. You weigh it using light in some guise. Then at the lower elevation, the brick has a mass which is less than x. But when you weigh it using light, you use light at what you think is the same frequency as before, when actually it's at a lower frequency. So the brick weighs heavier. Remember that at the lower elevation, you and your clocks are all going slower. So an descending photon appears to have a higher frequency, when it doesn't. It didn't gain energy, you lost it. You have a mass deficit too.
Post by: evan_au on 27/02/2015 10:25:24
Quote from: yor_on
the vacuum we set as the standard to measure all other permittivity againstAs I understand it, ε0 and μ0 are constants of the universe (like c), when you measure them locally.
In selecting the best material for a particular application, an electrical engineer is interested in the relative permittivity & permeability, εr and μr, which are both >1*.
The speed of light in this non-vacuum environment is now v= (1/√(ε0μ0))(1/√(εrμr)) = c/√(εrμr) =c/η
where η is the index of refraction (of great interest to opticians), and η=√(εrμr)
*except in certain metamaterials (http://en.wikipedia.org/wiki/Negative_index_metamaterials#Manipulating_permittivity_and_permeability), under quite restrictive conditions.
Post by: JohnDuffield on 27/02/2015 13:19:39
Inhomogeneous Vacuum: An Alternative Interpretation of Curved Spacetime (http://wenku.baidu.com/view/fbdd71fa941ea76e58fa0423.html)
"The strong similarities between the light propagation in a curved spacetime and that in a medium with graded refractive index are found. It is pointed out that a curved spacetime is equivalent to an inhomogeneous vacuum for light propagation. The corresponding graded refractive index of the vacuum in a static spherically symmetrical gravitational field is derived. This result provides a simple and convenient way to analyse the gravitational lensing in astrophysics."
Post by: PhysBang on 27/02/2015 13:32:42
No, there is just the operation of gravity. Do you believe in Newton's third law?but does that mean that conservation of energy would prevent the photon from gaining, or losing, energy, as long as that energy came from, or passed to, the energy of the black hole?No. If one thing gains energy, another thing loses it. But think about it. That photon is descending at the speed of light. If it's gaining energy, how is it getting it? If it starts off 93 million miles from the black hole, it takes 8 minutes for anything to get from the black hole to the photon. And things can't get out of a black hole. There is no magical mysterious mechanism by which energy can get from the black hole to the photon instantaneously.
Quote
Gravitational potential energy is a relative term that depends on the arbitrary choice of a zero point. Thus that can be negative. That's just a mathematical fact.In either case the total energy would remain constant. Wouldn’t it?Yes. This is the thing about gravity. There are websites and books out there that say gravity is negative energy, when it isn't. When two things fall together, energy is conserved. The books always balance. The total energy of the universe is not zero.
Quote
The real question is whether this can be used to do any physics. So far, I have not seen any evidence, including in the one paper that JohnDuffield cited, that one can use this equation to a description of motion that matches the observations we have of gravitational phenomena.Quote from: yor_onAre you arguing different permittivity to a perfect vacuum?No. It's a perfect vacuum, but like Einstein said, a concentration of energy in the guise of a massive star "conditions" the surrounding space, altering its metrical properties, such that "the speed of light is spatially variable". Then c = √(1/ε0μ0) at one location is not the same as c = √(1/ε0μ0) at another. The permittivity and/or permeability of space at one elevation is not the same as at another. But wherever you go you measure things like c and ε0 to be the same because it's an "immersive scale change". It's a bit like a flatlander trying to measure whether his world has been stretched using a ruler that's also been stretched.
Quote
I think it better to ignore Einstein's lecture, where he said that space was filled with stress-energy and read a textbook on GR instead. As Einstein knew when he describe the aether as the stress-energy tensor, this means that space, or rather spacetime (since the tensor is described only over spacetime), is filled not with stuff but with mathematical relationships.Quote from: yor_onthe point is that a vacuum classically is 'empty'.That's not what Maxwell or Einstein thought. They thought of space as a gin-clear ghostly elastic thing, something that has properties, something that can be stressed, something that can wave. Check out this (http://www.rain.org/~karpeles/einsteindis.html) where you can read Einstein saying a field is "a state of space". Also see the arXiv (http://arxiv.org/find/grp_physics/1/ti:+aether/0/1/0/all/0/1) and this (http://en.wikipedia.org/wiki/Aether_theories#General_relativity) too.
Quote
Then could you please show us an example of how to calculate, say, the fall of a pencil using your vacuum permeability equation? I have yet to see an example of how your ideas actually work. Given that you identify yourself as an authority here, I would love to see the examples.Quote from: JeffreyHe actually did know what the Lagrangian was.My physics knowledge is extensive. That sometimes causes problems when it conflicts with some popscience book or science article, or even a textbook or paper.
Quote from: Jeffrey
Great post. I am just working through conservation laws and symmetries. The important point to note is "As will be shown below, there is transfer of energy to and from the gravitational field and it has not meaning to speak of a definite localization of the energy of the gravitational field in space. Consequently we do not have a principle of local energy conservation in spacetime regions in which there exist gravitational fields."I think this paper causes confusion, and I would recommend that you set it aside. [/quote]
I agree that if one reads scientific articles, then one might be tempted to come to the conclusion that JohnDuffield's claims are false. If one wants merely to believe them, then please avaoid reading scientific articles or texts.
Post by: jeffreyH on 27/02/2015 14:00:56
The shame is you would probably make a very good science historian.
Post by: jeffreyH on 27/02/2015 14:09:46
Quote from: yor_onthe vacuum we set as the standard to measure all other permittivity againstAs I understand it, ε0 and μ0 are constants of the universe (like c), when you measure them locally.
In selecting the best material for a particular application, an electrical engineer is interested in the relative permittivity & permeability, εr and μr, which are both >1*.
The speed of light in this non-vacuum environment is now v= (1/√(ε0μ0))(1/√(εrμr)) = c/√(εrμr) =c/η
where η is the index of refraction (of great interest to opticians), and η=√(εrμr)
*except in certain metamaterials (http://en.wikipedia.org/wiki/Negative_index_metamaterials#Manipulating_permittivity_and_permeability), under quite restrictive conditions.
Above is an example of where the square root should be as in c/√(εrμr). It does matter and John should know this.
Post by: yor_on on 27/02/2015 14:12:55
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The point is that you don't like different 'paths' as I gather?
Post by: JohnDuffield on 27/02/2015 14:36:12
Well, either you assume a different permittivity John, or you are proposing some to me unknown, not observer dependent mechanism?The permittivity is different at a lower location. Only when you go there, you're different too, so the permittivity appears to be unchanged.
Because you are indeed telling me that light 'slows down'.Yes, that's what Einstein said, repeatedly. Check out the Einstein digital archives, and search on speed of light (http://einsteinpapers.press.princeton.edu/~searchResults?searchMode=advanced&context=-2&searchParam-SearchGroup=&lang=EN&searchText=%22speed+of+light%22) or velocity of light (http://einsteinpapers.press.princeton.edu/~searchResults?searchMode=advanced&context=-2&searchParam-SearchGroup=&lang=EN&searchText=%22velocity+of+light%22).
If you're using another definition that whole post I reacted on was unnecessary. I saw Peter using time dilations to define it but in your case I still don't know what you use? And you're starting to come on as arrogant, which I find slightly surprising.I meant to just give a straight answer that was factually correct and in line with Einstein and the evidence.
I don't mind people having pets, I have them too :) Writing "The permittivity and/or permeability of space at one elevation is not the same as at another." either seem to assume that there is one observer dependent redshift and another that solely belong to the 'photon' propagating, or that you still are thinking that all gravitational redshifts are outside observer dependencies?I'm sorry, I'm not clear what you mean here. Please restate.
The last one makes a joke of any idea defining a photon as intrinsically being the same in a gravitational redshift, the first one is new to me, two mechanisms for a redshift, and I think it has to be proved.We observe gravitational redshift, there's no issue in proving that. And IMHO conservation of energy says there shouldn't be any issue in proving that we observe it because we change. When you send a 511keV photon into a black hole, the black hole mass doesn't increase by a zillion tonnes. People say the descending photon is subject to an infinite blueshift, but its E=hf energy did not increase. But when you fall down, some of your mass-energy is converted into kinetic energy which is radiated away, leaving you with a mass deficit and less mass-energy. So the unchanged photon energy looks like it's increased.
The point is that you don't like different 'paths' as I gather?Again, I'm not clear what you mean. If you're referring to the photon taking many paths, I'm totally happy with that. I've used the analogy of a seismic wave to for that.
Post by: yor_on on 27/02/2015 14:46:45
Post by: yor_on on 27/02/2015 14:48:19
It's clear to me, what do you find unclear there?
Ignore 'still' btw, that was my fingers, not me :)
took it away as I saw it.
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Let's use the full text btw, because it's the conclusions that make a difference there.
"Writing "The permittivity and/or permeability of space at one elevation is not the same as at another." either seem to assume that there is one observer dependent redshift and another that solely belong to the 'photon' propagating, or that you are thinking that all gravitational redshifts are outside observer dependencies?.
The last one makes a joke of any idea defining a photon as intrinsically being the same in a gravitational redshift, the first one is new to me, two mechanisms for a gravitational redshift, and I think it has to be proved."
Post by: yor_on on 27/02/2015 15:15:40
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So to summarize, if I'm correct gravitational redshifts is observer dependent, or the equivalence principle has to be changed in some way. If it is correct, then there is no way, that I know of, giving this principle (gravitational redshifts) two different mechanisms, one observer dependent the other belonging to some 'intrinsic principle' embedded in the photon solely.
thats one
Two
No way this is 'outside' observer dependencies. If it was then we can throw away GR. As far as I understands it.
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and yes, it makes a 'photon' intrinsically the same, which is how I first read you John.
Post by: yor_on on 27/02/2015 16:31:55
A twin experiment is between two 'clocks', of a same origin, one accelerating to uniformly move away from that origin, to then return, finding one twins clock having been slower. Now, in my definitions this is a result of frames of reference interacting, but it is not a result of one clock ticking slower (or faster). Why I define it that way has to do with a lot of things, our definitions of repeatable experiments for one. observers in different uniform motions measuring each others clocks another, Also it has to do with you measuring, you won't ever find your clocks 'pace' to become 'slower' or 'faster', and your life span is the same, no matter what you do or how 'fast' you are. That one need a lot of reading up on to see how I think. But thinking that way, locality never change its 'pace', and that fit all experiments you can do, no matter NIST.
And it's not 'light paths', if I now remember you correct?
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the easiest way to see how I mean is to accept the definition of 'c', then define it as valid for all circumstances instead of arguing that 'c' has different values due to mass and accelerations. And make it equivalent your local 'clock'. If I would want it otherwise, still setting 'c' to my local clock (arrow), I would be forced to define 'c' to different values, not only in accelerations (and mass), but also in uniform motions. And 'c' and your local arrow is the same, ideally and experimentally, prove me wrong.
The only way you can go around the last would be arguing that a acceleration is the sole instigator of a time dilation, making any ideas of 'muon experiments' wrong, as we already have defined what a gravitational red (as well as blue) shift is. They are observer dependent, and so depending on your frame of reference (Earth, or 'free falling' uniformly moving Muon in this case) A gravitational acceleration is a 'free fall', no 'forces' acting upon you locally defined, and I define it locally. As well as the logic of defining it soley to accelerations won't work.
This one is really easy to understand. Just jump out a plane without a parachute, and see the Earth rush to meet you. Because that is what you will feel, ignoring wind and atmosphere tugging on you. It's frames of reference and in Einsteins world the Earth is constantly uniformly accelerating, at one gravity.
that doesn't mean that accelerations and mass doesn't count, everything count, mass energy, accelerations, uniform motion. To me it all seems to have to do with frames of reference interacting.
Post by: yor_on on 28/02/2015 10:37:52
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what that state to me is that we always will be looking through tinted glasses defining a position, scaling. We have our ideal clock and ruler, locally invariant, and we all agree on it being equivalent. If we didn't repeatable experiments cease to exist. The real point is that they are equivalent, that is what have made physics. Now consider that through scaling, would you expect this to become untrue, scaling something down? The terms of how it interact may change, but I would expect this to hold true all the way down under.
What I, very tentatively, see such a reasoning doing, is giving us a local time (clock) equivalent to 'c', created through interactions in a 'commonly agreeable on universe'. That's one of the things making scaling so incredibly fascinating. You can't free yourself from your local clock and ruler though, so even if I would define a 'bit' as some smallest common nominator of 'time', as I use Planck scale for, you still would find 'time' to tick for you in that experiment, if you now could measure at that scale, which you can't. You can't prove it, because you can't free yourself from the universe you exist in, and making a experiment not using time is a very weird proposition. I would call it another reality myself, coexistent with the one we observe.
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I would say that we have a lot of indirect proofs for it existing though. From the ideas of probabilities, to Feynman's sum over histories (http://www.einstein-online.info/spotlights/path_integrals), to entanglements, to the ideas behind a quantum computer. That has very little to do with the idea of 'c'. But a lot to do with what may happen as you scale it down. The point to take home from such a reasoning is that the clock and ruler is you, and that everything you measure use it. Scaling doesn't change that fact, which is why it will be very hard, probably impossible, to design a experiment proving it. And you live 'in time', if time is consisting of 'bits' then what's between them won't exist for you, because that's where you are not.
Looking at it this way it is a geometry, a weird one, shaped through scaling, up into a macroscopic world where we live and 'observe'. Although I have a distinct feeling that algebra will describe it better than a geometric formulation, and that was the way I understand Einstein to have thought of it too. I think he was right.
( actually this last is about whether you want to define it locally or globally. Globally ( a 'common universe') Algebra should be your best choice. Locally I expect geometry to do just as good, possibly even clearer?. That's my opinion at least, well, for now :)
Post by: yor_on on 28/02/2015 14:01:05
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We can turn it on its head by using one frame of reference, Earths. Then all gravitational accelerations belong to local accelerations as defined from some inertial point of view. now, is this right? Naaah, Einstein used locality, but also described it from the idea of a container. That's why we need different coordinate systems.
Post by: yor_on on 28/02/2015 14:05:59
Once you realize that one, the only thing you need to do, is to take it to its extremes and see where it leads you.
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Locally anchored, QM and relativity has a lot in common. Scaling is indeed about where a 'locality' ends, hoping to define it through quanta, or 'bits'. That is a good idea, as long as it in cooperates the other side of it, what comes before that quanta of 'motion', or 'time', or 'distance', or 'mass'. The other side don't use our arrow. Does that make you 'time less'? :) Nope, you're defined through a arrow and its outcomes, but what is 'you', is another question, that one belong to the exact same place as where we ask ourselves what 'thoughts' are? That's physics too (well to me it is:), but not one we can measure on. No way to measure that thought. The only thing we can measure on are those physical things that accompany it, but that's not the thought.
Yep, I see physics everywhere :)
And to my eyes, what stopped Einstein in his later years, and what made him argue against Entanglements and probability 'God does not play dice', was the way he thought of it, as a container. His work on a fifth dimension, unifying the universe, came from that exact position. He had the answer, but he wanted another. And that is what we all have John :) You and me, and all together. We have our pets.
Post by: jeffreyH on 28/02/2015 18:27:26
Post by: David Cooper on 28/02/2015 19:26:00
yor_on
It's streams of posts like this that make your writing useless. Almost nobody reads your posts because its like reading a book about nothing. It's a waste of our time so why do you do this? It makes everything else in the thread much harder to follow because you push other people's posts off the page into previous pages. Nobody wants to scroll through all this garbage to look for a small gem. Now please get with it.
Note: I know how selfish you are so I know that you're more interested in your own desires and less interested in everyone else's at our expense so I know that you don't care how we feel about it. Therefore when you try to justify your poor attitude you'll fail and just be wasting our time again.
+1 (with regard to the first paragraph)
Pete, you're picking fights all over. Believing that a expertise in one subject gives one reason to act rudely, or advertise ones own site as 'better', is a mistake on TNS. At least it used to be? Actually I don't know any site that finds the last to be acceptable? I'm getting pretty tired reading your attacks now.
I think Pete's been pushed beyond a point where all the irritation which he's held back for many years is suddenly being unleashed in a great blast as he tells everyone straight what he thinks. (I'm mindreading, but I hope in a good way.) I'm sure it'll all come good in the end, but we all need to think more carefully about how we treat each other and how we come across.
Post by: yor_on on 28/02/2015 19:31:29
Post by: Ethos_ on 28/02/2015 20:06:50
One faithful member stated not long ago: "We should all be better than this."
And I applaud those who ascribe to that notion. Let's all work on this goal.
Post by: PmbPhy on 28/02/2015 20:47:25
Quote from: jeffreyH
Yet when I take the time to read through them, which I usually don't, you make some very interesting points. It is a shame that these get missed.And that's one of the worst problems that he's causing. There's simply too much "filler" to want to wade through.
Post by: David Cooper on 28/02/2015 20:53:53
You read the rules and then you either accept them, or you leave. Is that complicated David?
Which rules are being broken? You are allowed to post long, rambling essays if you wish, and Pete's allowed to express annoyance with them, and to do so in a way that's rather stronger than might be socially acceptable. As for Pete's forum, it's trying to do something different which doesn't involve competing against this one, but to work as an addition to it so that disrupted conversations can be restarted in a more controlled space and better progress can be made with them. (I am involved in one such discussion there which has proved to be almost impossible to have on any other forum due to people who can't reason insisting on trying to derail it with a bombardment of illegal objections.) It is not designed to drag anyone away from here.
Post by: David Cooper on 28/02/2015 21:55:30
Post by: jeffreyH on 28/02/2015 22:50:30
Post by: PmbPhy on 28/02/2015 23:01:26
So is it a yes the photon does lose all its energy at infinity or a no?No. Of course not. Where on earth did you get that idea anyway? The energy of a photon remains constant as it moves through a gravitational field.
Post by: PhysBang on 28/02/2015 23:13:04
Look what I have to keep dealing with.I pointed out that I thought you were ignoring the very definition you were posting. It wasn't meant as an attack, just a point.
Post by: David Cooper on 28/02/2015 23:36:27
So is it a yes the photon does lose all its energy at infinity or a no?
Think about what happens with sound. There's someone plucking a string on a double bass at the bottom of a towerblock. You are able to watch and listen from the top of the towerblock, and you can hear that the note is flat. You go down in the lift and listen again, and it doesn't sound flat any more, so you go back up to the top and find that it's flat again. When you look down at the string from up there through a powerful telescope you can see it vibrating, and it appears to be vibrating exactly in sync with the sound you are hearing. You conclude from this that the sound has not reduced in frequency as it has climbed out of the gravity well, but that it is being generated at a lower frequency in the first place and it maintains that frequency all the way up as the sound climbs towards you. If it travels to infinity, the frequency will remain constant, but if you travel out to listen to it you will hear it as continuing to get more flat the further you get out of the gravity well. This reduction in the frequency you hear will become closer and closer to zero though (meaning the note doesn't appear to get much flatter at all) and the change will soon become impossible to measure - the strongest effect is deep down in the well, but once out in deep space you are in practical terms no longer in that well, even though you technically still are.
However, the frequency could still be reduced to zero over infinite distance because of dark energy and the expansion of space. It is this expansion that could decrease the energy of a photon infinitely, but not its journey out of a gravity well. [This then leads to other questions. Where then does that energy go to? Is the loss driven by the expansion of space, or do photons help to drive the expansion of space by throwing off some of their energy?]
Post by: Russell Crawford on 01/03/2015 12:33:04
That said if the wavelength could be infinite, the energy could, due to gravitational redshift, become increasingly smaller in an infinite sense.
I tend to believe that the particle property may be the best interpretation of the energy of a photon and that at some point it will have a "quanta" of energy that will be confirmed. But that is just an unfounded "belief".
With regard to "where the energy goes" it would seem to me (again a guess) that it would be lost over the distance from the source of the energy and the point at which it is measured. Perhaps trapped by gravity as planets are trapped and sorted? Perhaps it could be called energy over distance or redshift potential energy. Again, that is speculation. I hope this answer will help people think about this important subject.
Post by: yor_on on 01/03/2015 13:23:48
I'm sure it'll all come good in the end, but we all need to think more carefully about how we treat each other and how we come across.
You read the rules and then you either accept them, or you leave. Is that complicated?
Which rules are being broken? You are allowed to post long, rambling essays if you wish, and Pete's allowed to express annoyance with them, and to do so in a way that's rather stronger than might be socially acceptable. As for Pete's forum, it's trying to do something different which doesn't involve competing against this one, but to work as an addition to it so that disrupted conversations can be restarted in a more controlled space and better progress can be made with them. (I am involved in one such discussion there which has proved to be almost impossible to have on any other forum due to people who can't reason insisting on trying to derail it with a bombardment of illegal objections.) It is not designed to drag anyone away from here.
Well David. I'm sorry that it has come to this, and I actually Pm:ed Pete on the rules.
There are some rules concerning our behavior on this forum that we should try to follow. Keeping it friendly is one, another is about advertising ones own site. You can look it up or ask Pete. When it comes to writing long chunks :) Yep, got carried away there, it is as Jeffrey said, I think about it, then I see something I want to make better etc etc, ad infinitum. A dangerous thing, and one that shoot me in my ** here.
That doesn't mean that it excuses breaking other rules. It's in the end a Moderator decision whether to warn or not, so let us see what happens. If there is a lack of interest from the moderators point of view, I would suggest the rules to be rewritten though, so that they better fit the way this forum is going to act in the future.
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(Well. What'da'ya'now :) pm's Internet slang for 'personal message' so it's ok any which way. Weird stuff)
Post by: yor_on on 01/03/2015 15:45:10
So is it a yes the photon does lose all its energy at infinity or a no?
Depends, and it goes back to how you define it. How does a accelerating expansion act on light? As a wave it's possible to understand, as a 'point-like photon' it's not. The only way I see it working is accepting a existing duality, in all circumstances. And that takes us once again, to a place of 'no propagation', exchanging it to excitations in a field. That is also a question of whether this duality becomes a macroscopic phenomena, quanta microscopically, or not?
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that one is treating it 'practically' though. Theoretically I would guess no, it shouldn't lose its energy, unless we introduce this expansion acting on it. But all of this reasoning builds on you and me presuming a propagation. Exchanging that to excitations in a field it's no longer about 'unique photons propagating in a space' per se. To me it then seems to become more of a question of conservation laws. A propagation, or no propagation, that is the question :)
Post by: jccc on 01/03/2015 16:17:20
is it possible that when atoms exited, they give off gravitational waves? like a bell knocked gives off vibrating energy?
Post by: yor_on on 01/03/2015 16:41:38
Think it's called a 'Einstein-Rosen bridge' or nowadays 'wormhole'.
Post by: jeffreyH on 01/03/2015 16:49:54
So is it a yes the photon does lose all its energy at infinity or a no?
Think about what happens with sound. There's someone plucking a string on a double bass at the bottom of a towerblock. You are able to watch and listen from the top of the towerblock, and you can hear that the note is flat. You go down in the lift and listen again, and it doesn't sound flat any more, so you go back up to the top and find that it's flat again. When you look down at the string from up there through a powerful telescope you can see it vibrating, and it appears to be vibrating exactly in sync with the sound you are hearing. You conclude from this that the sound has not reduced in frequency as it has climbed out of the gravity well, but that it is being generated at a lower frequency in the first place and it maintains that frequency all the way up as the sound climbs towards you. If it travels to infinity, the frequency will remain constant, but if you travel out to listen to it you will hear it as continuing to get more flat the further you get out of the gravity well. This reduction in the frequency you hear will become closer and closer to zero though (meaning the note doesn't appear to get much flatter at all) and the change will soon become impossible to measure - the strongest effect is deep down in the well, but once out in deep space you are in practical terms no longer in that well, even though you technically still are.
However, the frequency could still be reduced to zero over infinite distance because of dark energy and the expansion of space. It is this expansion that could decrease the energy of a photon infinitely, but not its journey out of a gravity well. [This then leads to other questions. Where then does that energy go to? Is the loss driven by the expansion of space, or do photons help to drive the expansion of space by throwing off some of their energy?]
That's the sort of thing I was looking for. That sums up the situation exactly. It is unclear to many exactly what the true situation is. Thanks for this excellent post.
Post by: JohnDuffield on 01/03/2015 17:17:29
So is it a yes the photon does lose all its energy at infinity or a no?It's a no. The ascending photon doesn't lose any energy. It takes energy to lift you up, so the selfsame E=hf photon appears to have lost energy, when in fact, you've gained it. It's similar to what happens in gravity-free space when you accelerate away from a light source.
As to whether the CMBR photons have lost energy is another matter. I think the answer is no, but I can't be so confident about that. Things like the big bang and the early universe are tricky. In comparison, gravity is straightforward. Or ought to be.
Post by: yor_on on 01/03/2015 17:28:17
"This reduction in the frequency you hear will become closer and closer to zero though (meaning the note doesn't appear to get much flatter at all) and the change will soon become impossible to measure - the strongest effect is deep down in the well, but once out in deep space you are in practical terms no longer in that well, even though you technically still are.
However, the frequency could still be reduced to zero over infinite distance because of dark energy and the expansion of space."
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Let us assume that dark energy 'pushes' material bodies away from each other 'gravitationally'. Either you then have to define it to only existing outside/between galaxies, or explain how this new source of gravity interact inside solar systems, galaxies with the one we're accustomed to define. So now we have a unknown source of gravitation between galaxies, defining a redshift. But then it's not about a 'expanding space', it's just assembly's of 'bodies' separating from each other in a vacuum, in which case the redshift from a expansion becomes observer dependent again. But it leaves us to explain why this dark energy won't interact inside galaxies, solar systems etc, if so. Because that gravitational influence should be measurable. If we define the universe as infinite then it is a possible mechanism, although? Why only in between?
(also, such a behavior would separate a accelerating expansion from a inflations, that is unless we assume a vacuum to 'preexist', which is a interesting idea. You could then turn it around to a vacuum being the result of the distribution of matter, starting as some regime of 'temperature/energy'. Still needing it to be diffused though, linking itself into a 'common universe' becoming a 'inflationary period', as the cosmic 'light-sphere of age' we observe is presumed to be found no matter where you go)
The idea behind it, as I understand it, is that it is 'spread out' and 'diffused' everywhere, only becoming a 'strong negative force' over large distances counteracting gravity. Either as some wave building up from smaller 'pushing' galaxies apart, or as I think about it, as some 'upwelling' in each 'point' of a vacuum. If it was as a wave, you need to explain how it can 'push away' in all directions, using a 'upwelling' makes it a little easier to think about it, but not much :)
Post by: David Cooper on 01/03/2015 20:34:42
Get confused reading this David, are you associating a photon with a notes frequency? Then stating that dark energy interact with it? The note solely, or both note and photon? If you're thinking of gravitational red/blueshifts, assuming that dark energy interact through that, then that is observer dependent, and not about a photon 'energy' intrinsically, as far as I know?
I mentioned dark energy solely because I think that's were the confusion comes from - we see that the radiation from the big bang is of lower frequency than it must have set out with, and we predict that it will continue to be measured as having progressively lower energy over tens of billions of years to come, and infinite expansion would lead to infinite energy loss. This idea about photons losing energy can then potentially get tangled up in people's minds when they think about photons climbing out of gravity wells too, thereby leading to them thinking there is loss of enegy there as well, but there isn't - it isn't possible to change a frequency of anything if it travels between two points which remain a constant distance apart (unless they're both accelerating, or you lengthen the communication path in some other way or ramp up some other delay mechanism within it), and the energy of light is entirely wrapped up in its frequency, so it cannot be losing any energy if the frequency is unchanged. All that actually changes is the frequency that is produced when a device operates at different depths in a gravity well - the deeper it is, the lower the generated frequency will be, though from the point of view of the device it is producing exactly the same frequency on each occasion, and any device designed to measure the frequency locally would agree with it.
I used the musical note example because it allows you to see the movement of the string from far away and to see the frequency of the actual oscillation that produced the note. The situation is the same with sound and photon frequency, so it's a useful parallel. The note will get quieter over distance just as a beam of light will spread out and become dimmer, but the frequency is a constant in each case. The big difference is that sound travels more slowly, but we get the same apparent frequency reduction with it over the same change in altitude within the gravity well because the effect is entirely driven by time dilation. If we use an analogue radio communication to send the sound up from ground level to the top of the towerblock at the speed of light we will have the exact same reduction in the frequency of the note measured up there. (With a digital radio communication though, we would not hear the note flatten, but there would be occasional breaks in the stream instead.)
The other case worth thinking about is the light escaping from a black hole. If it comes from just outside the event horizon, there are different scenarios to consider which relate to how it is produced. If it comes from something that is stationary relative to the event horizon, time dilation will be so extreme that the photon will be generated at such a low frequency that I don't know if it would even be classed as a radio wave, but if you viewed it from where it was produced you would see it as light. By the time that photon has climbed up into deep space, it would be detected as having the extremely low frequency that it was actually produced with. Alternatively, if the photon came from some object that was racing into the black hole, and if it (the photon) was released just outside the event horizon, time dilation would not be so severe [take this bit with a pinch of salt - I'm taking that on trust from leading scientists in TV documentaries, but they often simplify the truth out of things and end up misleading people instead of informing them], so the light could be produced at a higher frequency, but the object's movement inwards during the production of the photon would spread the photon out and give it no higher effective frequency than the photon released from a stationary object hovering just outside the event horizon.
If we want to try releasing a photon from the event horizon itself, then an object suspended there would be completely frozen in time and could not produce a photon of greater frequency than zero, but an object falling in still could [repeat dose of a pinch of salt here]. In this case, the photon is going to have a frequency, but it will again be spread out, and either it will have the back end ripped into the black hole (thereby producing an effective frequency of zero) or the photon may just escape with next to no frequency, and it will be impossible to detect by the time it has reached deep space.
If a photon is released any distance inside the event horizon, it's impossible for it to be released from a stationary object, so there's only one case to consider, that being that it is being emitted from an object which is moving deeper into the black hole. In this case, the photon may be produced at a reasonable frequency [more salt required here], but it will be dragged backwards into the black hole and will therefore have no effective frequency at all from the point of view of an outside observer.
Post by: yor_on on 02/03/2015 15:16:59
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I split the idea of lights duality from this. Using a field(s) is a very simple idea actually, naively, ignoring observer dependencies, but it wrecks havoc with our ideas of a propagation, unless you presume a wave universe as where this is coming from. Doing so you now place the idea of a quanta as a secondary effect, which it is not, if you consider it in terms of scaling 'down under'. Myself I define this universe from 'quanta', waves and a duality being a complementary description. Waves was the first one we found, and it explained a lot to us, then Einstein had the temerity to come and destroy it (black body radiation:)
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It is possible to think of it this way, maybe? If there is discrete quanta, then they are not 'observer dependent', and what we define as observer dependent, when it comes to a clock and ruler, then becomes a result of the scales we use. That doesn't state that they are illusions. A illusion should be something you don't find to work, ah well, changing frame of reference to what you measured as being 'different' you then will be 'synchronized' with it :) So was it illusionary? In terms of your life span, not locally it was, but 'unifying the container' it becomes so. So strictly speaking, where was it you said you lived?
you only need waves when you want to connect it into what I call a 'commonly agreed on universe', that's also a container and a idea of something more needed to 'unify it'. Locally defined quanta works just fine I think. As do probability and statistics.
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A simple definition of the difference might be to think of a wave universe versus a 'quanta universe'. That's two different 'directions' in my mind. One is waves and light propagating in a container. The other is scaling.
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the question is how to get those two to agree with each other, scaling versus a container. You need something describing what we see. I think we will find it, but it will build on scaling.
Post by: Ethos_ on 03/03/2015 15:41:26
maybe photonists soon will be selling sneakoil?Yet more experimentally factual and effective than the "snakeoil" you're peddling.
Post by: Ethos_ on 03/03/2015 15:45:01
i am a college fall out, my theory is not. no?It might have been a wise move if you had stayed in college.
Post by: Robbo on 07/03/2015 18:57:12
this forum deleted some of my postings
maybe soon will ban my account
i recorded everything i posted
find true science at fuckedscience.com
if you see me here no more
I joined this forum last week and I have to say the most amazing thing I've encountered thus far aren't the dark mysteries of a black hole, or indeed the jaw-drop lunacy of Quantum Mechanics ... it is this thread.
I have a sports forum and so I'm well used to all types of contributor that a forum attracts - from the curious, to the arrogant, to the downright ignorant and to the disgustingly enlightened, all of whom just love to express their opinion in true Orwellian style.'
And with that thought sloshing around in my head, it reminded me of something my father once told me, 'You can deal with politicians, you can deal with bigots, you can deal with criminals and even terrorists but you cannot deal with an idiot'
This thread pays testament to the endless patience a lot of you guys have displayed while dealing with someone who's either being deliberately obtuse, or is just plain incapable of assimilating a whole slew of lucid clarifications that you guys have been kind enough to post in the somewhat vain hope of helping him.
At times, his posts have been disrespectful, impolite and on many occasions, inappropriately dismissive.
A lot of you guys deserve a virtual medal for such courtesy because I think most people would have reacted quite quickly to his attitude.
I apologise if I've offended anybody, it wasn't my intent - I was merely making an observation.
I'm extremely impressed with such a tolerant bunch of guys .. if only my forum could boast such a demographic :/
Post by: Ethos_ on 07/03/2015 21:35:30
break my legs if you want to, i am not leaving this place.It might have been more accurate to say: "I am not voluntarily leaving this place. Time will tell as to the alternate remedy.
Post by: jccc on 15/03/2015 08:16:41
should we start a poll?
Post by: jeffreyH on 16/03/2015 00:33:22
Post by: Ethos_ on 16/03/2015 01:35:40
Just look around you. Is it dark?There is none so blind as he who will not see.
Post by: jccc on 17/03/2015 19:18:15
how could electron emits photons? how many rounds of photon can an electron carry?
if photon is a real thing, when it passes water, it slows down, lose some energy, than comes out water, how could it gain energy and move faster?
energy exchange? how?
what's the difference between a green photon and a red photon? red light vibrates 4.2X10^14 times per second, it that mean red photon passes a single point 4.2x10^14 times? an electron emits 4.2x10^14 photons per second?
if light is gravitational wave produced by exited atoms, it is a force pause that acts on any matter on the way/field.
if a matter vibrates, its gravitational force vibrates, propagates at c.
thoughts?
Post by: chiralSPO on 17/03/2015 20:00:35
if energy is conserve, matter is conserve, charge is conserve.
how could electron emits photons? how many rounds of photon can an electron carry?
This is a good question. I don't know exactly how this works, and I am not entirely satisfied with any of the answers I have heard on this one yet...
if photon is a real thing, when it passes water, it slows down, lose some energy, than comes out water, how could it gain energy and move faster?
Because the photon has zero rest mass its energy and speed are unrelated. It does not lose energy as it slows down, or when it speeds back up. The energy is related to the frequency of the photon which is unchanged by this interaction.
what's the difference between a green photon and a red photon? red light vibrates 4.2X10^14 times per second, it that mean red photon passes a single point 4.2x10^14 times? an electron emits 4.2x10^14 photons per second?
No, electrons do not emit 4.2x10^14 photons per second. This is the frequency of the electromagnetic wave associated with the photon.
Post by: evan_au on 17/03/2015 20:48:03
Quote from: jccc
how many rounds of photon can an electron carry?An electron (or proton) does not "carry" any photons when it is traveling through free space.
- However, an electron (or proton) which is accelerated by some means does emit electromagnetic radiation (photons).
- So you do not want to be in the tunnel of the LHC when the beam is on, because the acceleration of the protons to bend them around in a circle emits intense X-Rays which would not be therapeutic.
- Similarly, a hot plasma in the Sun or in a hydrogen fusion reactor consists of protons and electrons flying around separately, at high velocity. However, if an electron approaches a proton, the electric field between them accelerates the electron, changing it's speed & direction and emitting photons (http://en.wikipedia.org/wiki/Bremsstrahlung). But because the speed is so high (due to high temperature), the proton can't hold on to the electron, and they continue their separate ways.
In empty space, an electron and proton which are far apart (physicists say "at infinity") will be attracted to each other and approach. The electron is likely to be captured by the proton, forming a hydrogen atom.
- The electron could fall straight to the lowest level (n=1), emitting a single ultraviolet photon with a wavelength of 91.2 nm. This is part of the Lyman series (http://en.wikipedia.org/wiki/Hydrogen_spectral_series#Lyman_series_.28n.E2.80.B2_.3D_1.29). Look at the line in the table labelled with "∞".
- Or, the electron could fall into the third-lowest level (n=3), emitting a single infra-red photon with a wavelength of 820nm (Paschen Series (http://en.wikipedia.org/wiki/Hydrogen_spectral_series#Paschen_series_.28Bohr_series.29_.28n.E2.80.B2_.3D_3.29)). This could be followed by a cascade down to the second level (n=2, Balmer Series (http://en.wikipedia.org/wiki/Hydrogen_spectral_series#Balmer_series_.28n.E2.80.B2_.3D_2.29)), emitting a single red photon of 656nm. This could then fall to the lowest level (n=1), emitting an ultraviolet photon of 121nm (from the Lyman series).
- Note that the total energy of the 3 photons is the same as the energy of the single photon emitted when the electron falls straight to the ground state.
- There are an infinite number of orbitals that the electron could fall into. Each one has its own characteristic energy level, and will emit a photon of a specific frequency and energy when transitioning to another level (although many of the energy levels are so close together that you can't easily distinguish them). This shows up in the spectrum of a Hydrogen atom (http://en.wikipedia.org/wiki/Hydrogen_spectral_series), which has an infinite number of lines in it.
Post by: jccc on 17/03/2015 21:53:47
now you only have 1 proton and 1 electron, how you build that atom as science said?
how a photon interacts with an atom? contact act? force act? which particle act with photon? what's the mechanism?
Post by: jccc on 17/03/2015 22:26:07
when a mass is vibrating, its gravitational force is vibrating, the mass emits gravitational wave.
when an atom vibrates at 10^14 or so per second, it emits visible light.
seems all correct?
Post by: PmbPhy on 19/03/2015 05:47:37
it was a long discussing, maybe we should make sure if photon is a real thing or not now.Poll on what? If its a poll about who wants you to be here then that's simple. Nobody wants you to be here.
should we start a poll?
Post by: PmbPhy on 19/03/2015 06:05:22
Quote from: jccc
if energy is conserve, matter is conserve, charge is conserve.So what? Those are independent laws of nature except for conservation of matter. There is no such law. Relativity proved that to be wrong. Conservation of charge cannot be derived from conservation of energy.
Why do you waste space and people's time with this nonsense?
Quote from: jccc
how could electron emits photons?Study quantum mechanics, particle physics and field theory and you'll learn how.
Quote from: jccc
how many rounds of photon can an electron carry?False notion. But to be expected from jccc.
Quote from: jccc
what's the difference between a green photon and a red photon? red light vibrates 4.2X10^14 times per second, it that mean red photon passes a single point 4.2x10^14 times? an electron emits 4.2x10^14 photons per second?Of course not. The energy and momentum is a function of frequency. That's part of its meaning. Another part is that it's related to wavelength and that's part of its meaning, i.e. wave properties. To understand the last part then one has to understand what a phasor is, and you don't know what that is because you keep ignoring everyone's suggestion to study math and physics. The phasor can be thought of as a rotating arrow and that rotating arrow can be used to visualize how wave functions add up. Feynman does this in his book QED (although he doesn't tell the reader that he's talking about phasors.
Quote from: jccc
if light is gravitational wave...Utter garbage!
Quote from: jccc
thoughts?Yes. Please leave us in peace.
Post by: jccc on 19/03/2015 12:31:25
if you make a 3 d model of an atom, a center piece, few energy levels, orbital shells, and an electron, you need a few legs to support the energy levels, or orbitals, you need something to support the electron, so everything has its position.
now you only have 1 proton and 1 electron, how you build that atom as science said?
how a photon interacts with an atom? contact act? force act? which particle act with photon? what's the mechanism?
debunk this?
Post by: jccc on 19/03/2015 12:32:35
when a charge is vibrating, its em force follows. the charge emits em wave.
when a mass is vibrating, its gravitational force is vibrating, the mass emits gravitational wave.
when an atom vibrates at 10^14 or so per second, it emits visible light.
seems all correct?
and this.
please don't delete my comment again, is this an open forum?
Post by: jccc on 19/03/2015 12:43:37
maybe photonists soon will be selling sneakoil?Yet more experimentally factual and effective than the "snakeoil" you're peddling.
this is mainstream science
this is my science
you be the judge.
Post by: jccc on 20/03/2015 02:25:11
Post by: jccc on 20/03/2015 03:51:19
Post by: JohnDuffield on 20/03/2015 17:45:10
This is a good question. I don't know exactly how this works, and I am not entirely satisfied with any of the answers I have heard on this one yet...Remember that you can make an electron (and a positron) out of a photon in pair production. Draw a circle on a piece of paper to emulate the electron, then without lifting your pen, draw another circle on top of the first, and another, and another. You are emulating electron spin. Now think about Compton scattering (http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/comptint.html) wherein the electron absorbs part of the photon, and as a result, moves. To emulate the moving electron with a smaller Compton wavelength (http://en.wikipedia.org/wiki/Compton_wavelength) draw an incomplete circle, then without lifting your pen, draw another incomplete circle, and another, and another.
Post by: jccc on 20/03/2015 20:22:36
all things must have precise mechanism, that's how physics law works.
can you build a car with 90% parts? write a program with 90% codes?
fundamental physics MUST be 100% accurate.
Post by: jccc on 20/03/2015 23:55:33
in atom world, electron and nucleus bound by em force, proton attracts electron, fluid ball repel electron, balanced at radius. when force applied, em wave produced.
no need any medium, no particle/photon emitted.
thoughts?
Post by: jccc on 21/03/2015 06:46:30
which means electron cannot be pushed into atom radius. so it is impossible for an electron to move toward nucleus, electron can only move on the surface of the atom ball, or away from atom radius.
seems to me all fit observation.
thoughts?
Post by: yor_on on 24/03/2015 23:12:46
Jccc, I'm sure you have your own unique way of looking at the universe. And you don't want people to destroy it, because you feel it has a elegance and a importance, even when found partially wrong. But you really need to preresent what it builds on, to make people give you the criticism you need to refine it.
Post by: jccc on 14/04/2015 05:27:58
when a charge is vibrating, its em force follows. the charge emits em wave.
when a mass is vibrating, its gravitational force is vibrating, the mass emits gravitational wave.
when an atom vibrates at 10^14 or so per second, it emits visible light.
seems all correct?
how do you think? anything wrong with logic?
Post by: Ethos_ on 14/04/2015 13:36:40
Depends upon how the "logic" was constructed. Logic built only from intuition invites error. Only when logic is built upon observable evidence can it be considered "Logical".when a charge is vibrating, its em force follows. the charge emits em wave.
when a mass is vibrating, its gravitational force is vibrating, the mass emits gravitational wave.
when an atom vibrates at 10^14 or so per second, it emits visible light.
seems all correct?
how do you think? anything wrong with logic?
Post by: jccc on 14/04/2015 16:00:48
Depends upon how the "logic" was constructed. Logic built only from intuition invites error. Only when logic is built upon observable evidence can it be considered "Logical".when a charge is vibrating, its em force follows. the charge emits em wave.
when a mass is vibrating, its gravitational force is vibrating, the mass emits gravitational wave.
when an atom vibrates at 10^14 or so per second, it emits visible light.
seems all correct?
how do you think? anything wrong with logic?
such as photon emitting? orbital changing? energy exchanging?
Post by: jccc on 16/04/2015 14:38:42
from charges, atoms able to from, from atoms, matters able to form. charges made chemical bounding, magnetic, gravity, light able to exist.
but how much we know about charges? is electron really fly around proton to form hydrogen atom?
if gravity becomes 1000 time stronger, can the moon still circling us? can the moon change orbit and release photons?
please help me to understand, please don't show me links, show me the logic/principle. 1 proton and 1 electron, how is hydrogen atom formed???
Post by: jccc on 03/06/2015 22:03:40
if energy is conserve, matter is conserve, charge is conserve.
how could electron emits photons? how many rounds of photon can an electron carry?
This is a good question. I don't know exactly how this works, and I am not entirely satisfied with any of the answers I have heard on this one yet...
how's going?
Post by: jeffreyH on 03/06/2015 22:11:59
Post by: chiralSPO on 04/06/2015 01:18:09
A good thing to remember though, is that this is only a model of how atoms can absorb and emit photons. It is great for predicting the energies involved, but it doesn't necessarily explain what is actually going on. Quantum electrodynamics is another model (group of models) that deals with that--but unfortunately, I don't know much about QED, and won't try to explain it.
Post by: PmbPhy on 04/06/2015 04:01:51
Quote from: jeffreyH
The larger issue is if an electron captures a photon then what velocity does the photon have?An electron cannot capture a photon. That'd make no sense in terms of quantum mechanics. It's the system of the nucleus and electron, i.e. the atom that captures it. The energy goes from the kinetic energy of the photon to the potential energy of the electron/nucleus system. In any case whenever a photon exists its moving with the speed c.
Quote from: jeffreyH
It is either orbiting the electron or it slows down.Not at all. There's no reason to think such a thing. You're trying to think of these things in classical terms when you have to think it quantum mechanical terms.
Quote from: jeffreyH
Another biggy is if the photon has no charge then how could an electron capture it in the first place?The electron doesn't capture the photon. The atom does.
Post by: jccc on 04/06/2015 04:32:11
so photon hits electron to produce current? point impact? isn't the theory won a nobel?
what if photon hits proton? will proton jumps out?
what is electron/nucleus system? you invented? what's the structure? the mechanism?
Post by: jeffreyH on 04/06/2015 14:33:29
Quote from: jeffreyHThe larger issue is if an electron captures a photon then what velocity does the photon have?An electron cannot capture a photon. That'd make no sense in terms of quantum mechanics. It's the system of the nucleus and electron, i.e. the atom that captures it. The energy goes from the kinetic energy of the photon to the potential energy of the electron/nucleus system. In any case whenever a photon exists its moving with the speed c.Quote from: jeffreyHIt is either orbiting the electron or it slows down.Not at all. There's no reason to think such a thing. You're trying to think of these things in classical terms when you have to think it quantum mechanical terms.Quote from: jeffreyHAnother biggy is if the photon has no charge then how could an electron capture it in the first place?The electron doesn't capture the photon. The atom does.
I know zip about quantum mechanics so I'll shut up now. lol
Post by: jeffreyH on 04/06/2015 15:26:39
if there is a fruit fight, i'm in.
I have taken on board what two fellow members have said. They are very knowledgeable. There is no fight. I can even understand why they said what they said. It is called learning. You might like to try it some time.
Post by: chiralSPO on 04/06/2015 15:59:30
isn't in solar cell electrons capture photons to produce current?
so photon hits electron to produce current? point impact? isn't the theory won a nobel?
what if photon hits proton? will proton jumps out?
what is electron/nucleus system? you invented? what's the structure? the mechanism?
Solar cells can operate by many different mechanisms, base on what they are made of and how they are constructed. The easiest way to think about it is:
A photon is absorbed by the solar panel. The energy of that photon increases the potential energy of an electron (1-2; promotes the electron to a higher energy level/band). This energy level is high enough that the electron can relax into the wire (3), go around the circuit, and end up back where it started (4). This is an overly simplistic model, but I think it gets the point across.
1 2 3 4
______ ___e___ ________ ________
-------- ------- ---e----> -----------
__e__ _______ ________ ___e____
Post by: jccc on 05/06/2015 06:29:40
what energy photon carries? zero mass no momentum?
how electron increase energy? move faster? orbit higher?
what is energy level/band?
how electron relax?
Thanks and good morning!
Post by: chiralSPO on 06/06/2015 18:04:55
how solar panel absorb photon? impact? em wave?Easiest to think of it as an electromagnetic interaction. The electric field of the photon can push the electron into a higher energy level if it (the photon) has the right amount of energy.
what energy photon carries? zero mass no momentum?a photon carries energy that is propotional to its frequency E = hv a photon has no rest mass, but it has momentum that is also proportional to its frequency p = hv/c
how electron increase energy? move faster? orbit higher?
Both(ish). There can be an increase in potential energy (analogous to orbiting higher) and/or an increase in momentum (like moving faster)
what is energy level/band?
In crystalline bulk materials like semiconductors and conductors (many of) the atomic energy levels of all the component atoms merge into energy bands that span the entire crystal (near complete delocalization of the electrons). The energy bands are defined by the potential energy of the electrons that occupy them.
how electron relax?
This means they fall down from a high energy level into a lower one.
Thanks and good morning!
Post by: jccc on 06/06/2015 18:42:00
photon has no charge, how could it produce em field?
if photon is a particle traveling at c in a straight line, how can it has frequency? is the photon vibrating in space?
is electron orbiting all the time? is a hydrogen atom has 2 d or 3 d orbital?
Post by: chiralSPO on 06/06/2015 19:52:02
for the last time: atoms are definitely 3D, not 2D! the probability densities of orbitals are 3D, though one could argue that the orbital itself (the wave function) is higher dimensional, because it is based on complex numbers.
Post by: jccc on 06/06/2015 20:10:26
a photon is neutral but it has an electric field that oscilates between negative and positive at a specific frequency (the average is zero) and this is the frequency that determines the energy of the photon.
for the last time: atoms are definitely 3D, not 2D! the probability densities of orbitals are 3D, though one could argue that the orbital itself (the wave function) is higher dimensional, because it is based on complex numbers.
neutral photon has an electric field?
what propability density? the only propability is proton attracting electron according to Coulomb's law.
what wave function? what is making waves? how?
we are discussing science right? not imagination, not assumption.
Thanks
Post by: chiralSPO on 06/06/2015 23:04:44
Post by: jccc on 06/06/2015 23:47:45
what's the propability for electron to stick with proton? 0?
what wave? electron moves along force, how it waves?
Thanks
Post by: PmbPhy on 07/06/2015 01:03:37
Quote from: chiralSPO
The electric field of the photon can push the electron into a higher energy level if it (the photon) has the right amount of energy.According to quantum electrodynamics photons don't have an electric or magnetic field.
Post by: jccc on 07/06/2015 04:58:15
Post by: lightarrow on 07/06/2015 12:31:16
if energy is conserve, matter is conserve, charge is conserve.Something is missing: a final "d". [:)]
Quote
how could electron emits photons? how many rounds of photon can an electron carry?Photons are created from the oscillation/acceleration of a charge or in an atom's transition from an energy state to another (or energy transitions of other electromagnetic systems). Energy is conserved because the electron/atom/system potential energy is transformed in energy of photons.
The number of photons is not a conserved quantity.
It's not magic, it's simple physics, you only have to study it.
--
lightarrow
Post by: Bill S on 07/06/2015 20:49:27
"Q:
Why are photons (all wavelengths) considered to be instruments of the so-called "electromagnetic force"? So far as I know, please correct me, photons have no electrical charge nor are they influenced by magnetic fields. The term "electromagnetic spectrum" seems to me to be very inappropriate and highly misleading. Perhaps I am missing something? Thank you!
- grahame (age 60)
PA
A:
Grahame- You�re right that electromagnetic waves, whether viewed classically or in terms of quantized photons, are not affected by static electrical or magnetic fields. They have no charge. Nevertheless, they do exert electrical and magnetic forces on charged particles and magnetic particles. Viewed classically, they consist of nothing but electrical and magnetic fields propagating through space, so it�s entirely appropriate to call them electromagnetic waves.
Mike W."
No wonder we !hitch-hikers" become confused!
Post by: jccc on 07/06/2015 22:11:55
1 atom vibrates in 1 direction, all other atoms within its gravitational field feel the vibration force according to distance and force direction. logically sounding. fact.
are gravitation waves carry energy? yes. is gravitation wave a particle? no. is gravitation wave needs medium to propagate? yes, mass and distance.
if em radiation is not gravitation radiation, what is it? photon emitted by atoms?
how hot gasses on the sun emit photons? electrons change orbitals? how many orbitals in a hot gas atom? why is sunlight spectrum continue?
the whole atomic structure is unclear, otherwise, why is all question on page 8 not answered fully?
Post by: jccc on 08/06/2015 02:25:09
both fields have force carrier. what is it? photon?
is this a joke?
Post by: jccc on 22/06/2015 19:35:11
Viewed classically, they consist of nothing but electrical and magnetic fields propagating through space, so it�s entirely appropriate to call them electromagnetic waves.
both fields have force carrier. what is it? photon?
is this a joke?
no 1 cares about photon anymore?
Post by: jccc on 23/06/2015 19:02:52
photon lost all its energy right here already?
Post by: jccc on 23/06/2015 20:37:14
when the moon was ringing from that impact, it produced gravitational waves, mostly absorbed by earth, then the near by matters. the nasa spaceship detected that gravity/shock waves, otherwise how do they know the moon is ringing?
is photon/em wave just mistake interpretation of gravitational wave?
rethink the results in the double slit experiment?
Post by: jccc on 24/06/2015 04:31:31
Post by: jccc on 24/06/2015 23:28:02
light is gravitational wave between the source atoms and the target/detector atoms.
Post by: jccc on 25/06/2015 23:54:06
einstein said: look, i told you God does not play dice.
newton said: i told you light is wave.
tesla said: vibration, vibration, vibration,
pete said: shut up and calculate!
Post by: jccc on 26/06/2015 00:56:30
alan said: alas
chiralSPO said : apparently
box said: i'm tired
bill said: me2
what do you say?
Post by: PmbPhy on 26/06/2015 10:44:46
Post by: jeffreyH on 26/06/2015 12:42:47
Post by: chiralSPO on 26/06/2015 14:02:04
Isn't it great. You start a thread to gather opinion and hopefully learn something. Then it is hijacked and all the opinion is lost in the dross. We might as well just all PM each other.
...yeah... it had a pretty good run before "somebody" knocked it way off topic. If I have time this weekend, I might try to clean this thread up a bit...
Post by: jccc on 26/06/2015 14:13:04
The larger issue is if an electron captures a photon then what velocity does the photon have? It is either orbiting the electron or it slows down. Another biggy is if the photon has no charge then how could an electron capture it in the first place? Isn't it then as likely to be captured by a proton or neutron?
clear now?
Post by: Colin2B on 26/06/2015 15:46:43
Isn't it great. You start a thread to gather opinion and hopefully learn something. Then it is hijacked and all the opinion is lost in the dross. We might as well just all PM each other.I'm convinced these random dross comments put newcomers off posting, it's almost as if the question isnt taken seriously but used as a new theory platform..
Post by: Bill S on 26/06/2015 16:46:00
Quote from: Colin
....but used as a new theory platform.
I'm glad you identified it. I would never have suspected it of being a theory!
I absolutely agree that it must be off-putting. I have been posting on three discussion forums for a few years, and TNS is still the best of those, but there is a limit to how much trolling any forum stand.
BTW Pete, your forum is not included in the three. I've not been with that anywhere near as long as the others, and your setup militates against trolling.
Post by: jccc on 26/06/2015 17:08:25
Isn't it great. You start a thread to gather opinion and hopefully learn something. Then it is hijacked and all the opinion is lost in the dross. We might as well just all PM each other.
...yeah... it had a pretty good run before "somebody" knocked it way off topic. If I have time this weekend, I might try to clean this thread up a bit...
off topic? photon?
Isn't it great. You start a thread to gather opinion and hopefully learn something. Then it is hijacked and all the opinion is lost in the dross. We might as well just all PM each other.I'm convinced these random dross comments put newcomers off posting, it's almost as if the question isnt taken seriously but used as a new theory platform..
my questions hang there for 2 weeks, any 1 answered any?
Quote from: Colin....but used as a new theory platform.
I'm glad you identified it. I would never have suspected it of being a theory!
what do you suspect it is? dross?
Post by: Colin2B on 26/06/2015 17:42:25
I use the word theory in it's very loosest sense. The same sense that most of the new theorists use it. (jefferyH, I don't include you in that, you know what a theory is)Quote from: Colin....but used as a new theory platform.
I'm glad you identified it. I would never have suspected it of being a theory!
Strange that the trollers don't understand why eventually folks stop responding to their deaf-eared questions.
Post by: Bill S on 26/06/2015 20:31:45
Quote from: jccc
what do you suspect it is? dross?
Why would you think I meant that?
Post by: jccc on 26/06/2015 20:39:05
Quote from: jcccwhat do you suspect it is? dross?
Why would you think I meant that?
my bad. i misinterpreted your comments.
Post by: Bill S on 27/06/2015 14:00:33
Quote from: jccc
i misinterpreted your comments.
Have you ever wondered about the wisdom of jumping to conclusions?
Post by: jccc on 27/06/2015 16:53:34
b. sunlight is vibrating hot atoms produced gravitational waves that propagate at c.
which 1 is your Pick? why?
Post by: jccc on 27/06/2015 18:01:27
does laser beam produce air flow?
if laser beam is beam of photon particles, it would bend flame and produce air flow.
from the videos i saw, never happen.
how do you think?
is solar sailing even possible?
Post by: Bill S on 27/06/2015 19:46:53
Quote from: jccc
is solar sailing even possible?
It would seem so; or is this part of a scientific conspiracy?
http://blogs.discovermagazine.com/d-brief/2015/06/08/bill-nyes-solar-sailing-spacecraft-unfurls-its-sails-finally/?utm_source=SilverpopMailing&utm_medium=email&utm_campaign=DSC_News_150611_Final&utm_content=#.VY7va0ZNTKt
Post by: jccc on 27/06/2015 20:01:21
based on observation fact and proven laws, i think light is gravity wave.
if you need mainstream community to make you believe my theory, you may need to wait a while.
Post by: Bill S on 27/06/2015 21:49:16
Quote
all the info out there, one has to make up his own mind about anything.
That is a “politician’s response”. It doesn’t even come close to answering the question.
Quote
based on observation fact and proven laws
Observation is only the interpretation of the observer.
You frequently make a statement and label it “Fact”, but is there such a thing as a “fact” in science? Even if there were, it would need to be interpreted.
“Proven laws” are working hypotheses that have passed all the tests we have been able to subject them to – so far.
I like the idea of challenging accepted wisdom, but if you are going to be taken seriously you have to do a number of things. These include:
1. Convince others that you have made accurate observations, or interpreted the observations of others correctly.
2. Ensure that what you claim to be facts have a sound basis, and that you have interpreted them appropriately.
3. Be sure your use and interpretation of the “laws of physics” make sense and can be understood by others.
Even just these three requirements are not easy for those of us who are non-scientists. This is one reason why I tend to ask questions, rather than make statements.
Of course, it is frustrating if you feel your questions are not being answered, but it must be equally frustrating for those who attempt to answer questions if their answers are ignored.
Post by: jccc on 27/06/2015 22:49:25
a. sunlight is electrons in hot atoms change orbital/energy level emitted photon particles that travel at c.
b. sunlight is vibrating hot atoms produced gravitational waves that propagate at c.
which 1 is your Pick? why?
your answer is? or still thinking? no wiki?
Post by: jccc on 03/07/2015 21:06:54
100 laser balloon popping
if laser beam is particle beam, the energy of the beam should be the same at different distance.
if laser beam is gravitational wave between the source atoms and the target atoms, the energy of the beam should decay by distance.
look how fast the 1st balloon pops and how slow the last 1 is?
Now are you convinced?
Post by: chiralSPO on 03/07/2015 21:29:18
This is why the furthest balloon takes longer to burst than the closest.
Why would you think that distance would effect gravity but not EM? Anything that spreads out into space will have decreased intensity with distance.
Post by: jccc on 03/07/2015 21:46:04
the m1 and m2 are the same, the r changed so the force changed, the energy a gravitational wave carries is proportional to f x frequency.
the laser beam didn't expend, even it does, very small, the light beam energy should about the same at 1st and last balloon.
More investigation is needed.
Post by: jccc on 03/07/2015 21:58:21
see the 2nd video, the flame is not bending by the laser beam.
Post by: jccc on 04/07/2015 06:55:27
The laser spreads out slightly as it goes (like a flashlight, but to a much smaller extent) so the laser "beam" is actually a laser "cone." The rate that the balloons pop has to do with the intensity of the laser light (unit power per unit area, for instance mW/cm2)
This is why the furthest balloon takes longer to burst than the closest.
Why would you think that distance would effect gravity but not EM? Anything that spreads out into space will have decreased intensity with distance.
if you measure the laser spot area, it is about the same. so the mw/cm^2 is about the same, the 1st and last balloon should take about same time to pop.
wave spread out in space, not particle beam. a bullet has about same momentum at 1 m and 20 m away. a laser beam is not, see the videos.
Post by: chiralSPO on 04/07/2015 14:48:07
The laser spreads out slightly as it goes (like a flashlight, but to a much smaller extent) so the laser "beam" is actually a laser "cone." The rate that the balloons pop has to do with the intensity of the laser light (unit power per unit area, for instance mW/cm2)
This is why the furthest balloon takes longer to burst than the closest.
Why would you think that distance would effect gravity but not EM? Anything that spreads out into space will have decreased intensity with distance.
if you measure the laser spot area, it is about the same. so the mw/cm^2 is about the same, the 1st and last balloon should take about same time to pop.
wave spread out in space, not particle beam. a bullet has about same momentum at 1 m and 20 m away. a laser beam is not, see the videos.
no, look closely at the video. You can see the size of the laser spot on the wall (1:29), and it looks to me like the brightest spot is about an inch in diameter, with a halo that is about two feet in diameter. A little bit later (1:50) they show the beam coming out of the laser, and it appears to be about 1/8 inch in diameter (maybe 1/4, but not 1 inch or two feet). Also it is obviously conical--you can see it spread out a lot. So all of the energy in small area near the laser is spread out over a significantly larger area far from the laser.
Post by: jccc on 04/07/2015 17:18:02
this video is only 1.42. see the spot on the wall after the last balloon popped. same size.
it is not believe me or not. it is the fact about the nature of light, all scientists should seek truth.
the double laser experiment is not tested yet, but the videos linked, support my theory. the laser beam's energy decay with distance is fact.
you can say air atoms on the path absorb some energy, then you can test it in vacuum. use same laser to heat up target at difference distance, the farther target will take longer to heat up.
Post by: chiralSPO on 04/07/2015 18:15:36
Post by: jccc on 04/07/2015 18:33:20
how you explain the beam is not bending the flame? is that enough to prove that the beam is not particle beam?
Post by: jccc on 04/07/2015 18:49:04
just 2 people care about light?
strange forum.
Post by: chiralSPO on 04/07/2015 19:48:00
Post by: jccc on 04/07/2015 20:11:20
do you care about true nature of light? still think light is photon emitted by electrons change orbitals?
Post by: jccc on 04/07/2015 20:28:32
1 atom vibrates in 1 direction, all other atoms within its gravitational field feel the vibration force according to distance and force direction. logically sounding. fact.
are gravitation waves carry energy? yes. is gravitation wave a particle? no.
if em radiation is not gravitation radiation, what is it? photon emitted by atoms?
how hot gasses/plasma on the sun emit photons? electrons change orbitals? how many orbitals in a hot gas atom/plasma ? why is sunlight spectrum continue?
Post by: chiralSPO on 04/07/2015 23:13:14
Post by: jccc on 04/07/2015 23:20:51
Post by: jccc on 04/07/2015 23:54:39
Post by: jccc on 06/07/2015 01:21:46
therefore, the first and the last balloon should take same time to pop. obviously, the videos showed opposite result.
same amount of photon, why carry different amount energy at different distance?
please help me to think, i'm lost.
Post by: chiralSPO on 06/07/2015 02:10:26
The laser spreads out slightly as it goes (like a flashlight, but to a much smaller extent) so the laser "beam" is actually a laser "cone." The rate that the balloons pop has to do with the intensity of the laser light (unit power per unit area, for instance mW/cm2)
This is why the furthest balloon takes longer to burst than the closest.
Why would you think that distance would effect gravity but not EM? Anything that spreads out into space will have decreased intensity with distance.
Post by: jccc on 06/07/2015 04:39:35
The laser spreads out slightly as it goes (like a flashlight, but to a much smaller extent) so the laser "beam" is actually a laser "cone." The rate that the balloons pop has to do with the intensity of the laser light (unit power per unit area, for instance mW/cm2)
This is why the furthest balloon takes longer to burst than the closest.
Why would you think that distance would effect gravity but not EM? Anything that spreads out into space will have decreased intensity with distance.
i think your laser cone argument is pretty weak. laser beam does not expend.
gravity is f=Gxm1m2/r^2, that's why distance effect gravity.
any matter/particle that spreads out into space will not decrease intensity with distance. newton's law says so. planet's momentum never decrease, comets momentum never decrease, why photons?
i understand, it is tough to accept, the whole world think light is photon particle, einstein got noble for it. qm based on quantum/photon. if it is not true, isn't the world turned up side down?
i really don't know how you feel, but i was shocked when i realized i was right.
well, we used to be the center of the stars.
Post by: jccc on 06/07/2015 09:20:48
strange thing is last 2 day many people here but only 2 nuts talked.
Post by: chiralSPO on 06/07/2015 22:01:21
The laser spreads out slightly as it goes (like a flashlight, but to a much smaller extent) so the laser "beam" is actually a laser "cone." The rate that the balloons pop has to do with the intensity of the laser light (unit power per unit area, for instance mW/cm2)
This is why the furthest balloon takes longer to burst than the closest.
Why would you think that distance would effect gravity but not EM? Anything that spreads out into space will have decreased intensity with distance.
i think your laser cone argument is pretty weak. laser beam does not expend.
gravity is f=Gxm1m2/r^2, that's why distance effect gravity.
any matter/particle that spreads out into space will not decrease intensity with distance. newton's law says so. planet's momentum never decrease, comets momentum never decrease, why photons?
i understand, it is tough to accept, the whole world think light is photon particle, einstein got noble for it. qm based on quantum/photon. if it is not true, isn't the world turned up side down?
i really don't know how you feel, but i was shocked when i realized i was right.
well, we used to be the center of the stars.
It's not that the momentum of each photon decreases with distance--that is certainly not true! What changes is the number of photons passing per unit area per unit time (photons per cm2 per second). Why would you say that the laser beam doesn't expand? For thing, you can clearly see the divergence in the video that YOU sent a link to (at the timepoints that I recommended). For another, you can look up:
http://www.quora.com/Is-the-light-from-lasers-reduced-by-the-inverse-square-law-as-distance-grows-similar-to-other-light-sources
http://vlab.amrita.edu/?sub=1&brch=189&sim=342&cnt=1
https://en.wikipedia.org/wiki/Beam_divergence
Post by: jccc on 06/07/2015 23:07:23
Laser light from gas or crystal lasers is highly collimated because it is formed in an optical cavity between two parallel mirrors, in addition to being coherent. In practice, gas lasers use slightly concave mirrors, otherwise the power output would be unstable due to mirror non-parallelism from thermal and mechanical stresses. The divergence of high-quality laser beams is commonly less than 1 milliradian, and can be much less for large-diameter beams. Laser diodes emit less-collimated light due to their short cavity, and therefore higher collimation requires a collimating lens.