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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.

the photon will experience no passage of time between it's creation and that proposed infinity.

Quote from: jeffreyH on 19/02/2015 00:16:56When 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"?

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?

Quote from: jeffreyH on 19/02/2015 00:16:56When 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.

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.

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?

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, ...sorry, you cannot view external links. To see them, please REGISTER or LOGIN 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!

Quote from: Ethos_ on 19/02/2015 13:31:57Quote from: jeffreyH on 19/02/2015 00:16:56When 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.

Quote from: jeffreyH on 19/02/2015 19:44:05Quote from: Ethos_ on 19/02/2015 13:31:57Quote from: jeffreyH on 19/02/2015 00:16:56When 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.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.

I don't think photons can "observe" space either...

The reason for posting this was the thought that a gravitational field extends to infinity. If the field removes energy from the photon...

Experience is a very bad choice of word. Not very scientific really.

Quote from: jeffreyH on 19/02/2015 19:23:02The 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.

He then substituted v for c in E = mc^2 so that mv^2 = hv.

Quote from: jeffreyHHe then substituted v for c in E = mc^2 so that mv^2 = hv.Where did you get that from?

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?

Quote from: jeffreyH on 20/02/2015 00:30:47If 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 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.

Quote from: PmbPhy on 20/02/2015 03:27:42Quote 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 ...sorry, you cannot view external links. To see them, please REGISTER or LOGIN. The point I was making to John was that energy does change for the photon and what the reasons are.

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.

Quote from: jeffreyH on 20/02/2015 12:40:26This 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. Quote from: jeffreyH on 20/02/2015 12:40:26But 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. Quote from: jeffreyH on 20/02/2015 12:40:26You 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.

Gravity is not a force in the Newtonian sense.

It doesn't add energy to a descending photon either, or remove energy from an ascending photon.

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: ...sorry, you cannot view external links. To see them, please REGISTER or LOGIN

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

arrive at a destination that was 100 ly away when they were emitted

Wrong.

Wrong yet again! It's very easy to prove too

Thanks for that Pete I will read it later. I am done with John.

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.

It isn't wrong, and your article ends up saying this: The total energy of a photon moving through a gravitational field is constant.

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?

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.

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.

Quote from: PhysBang on 21/02/2015 16:57:42But 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.

Quote from: PhysBang on 21/02/2015 16:57:42Of 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.

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.

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.

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 can cherry-pick the conclusion if you would like, but that isn't good reasoning.

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.

if we replace "infinite distance" for "arbitrarily long distance" and "infinite speed" with "arbitrarily close to c"….

...My understanding is that the expansion of space is the major factor, if not the sole cause of the redshift...

Quote from: PhysBang on 21/02/2015 18:14:12Work 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.

Quote from: PhysBang on 21/02/2015 16:57:42I 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.

Quote from: PhysBang on 21/02/2015 16:57:42You 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.

Quote from: PhysBang on 21/02/2015 16:57:42I 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.