# The Naked Scientists Forum

### Author Topic: Would the photon lose all its energy at infinity?  (Read 69642 times)

#### yor_on

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##### Re: Would the photon lose all its energy at infinity?
« Reply #25 on: 20/02/2015 11:05:35 »
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 :)

#### JohnDuffield

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##### Re: Would the photon lose all its energy at infinity?
« Reply #26 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:

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 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.
« Last Edit: 20/02/2015 11:45:41 by JohnDuffield »

#### jeffreyH

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##### Re: Would the photon lose all its energy at infinity?
« Reply #27 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 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.

#### PmbPhy

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##### Re: Would the photon lose all its energy at infinity?
« Reply #28 on: 20/02/2015 13:25:04 »
Quote from: jeffreyH
He 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.
I believe that paper is in error. Let me check on it and get back to you.

#### JohnDuffield

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##### Re: Would the photon lose all its energy at infinity?
« Reply #29 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.

#### jeffreyH

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##### Re: Would the photon lose all its energy at infinity?
« Reply #30 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.

#### JohnDuffield

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##### Re: Would the photon lose all its energy at infinity?
« Reply #31 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.
« Last Edit: 20/02/2015 16:44:47 by JohnDuffield »

#### PhysBang

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##### Re: Would the photon lose all its energy at infinity?
« Reply #32 on: 20/02/2015 18:02:26 »
It is amazingly uncontroversial that redshift leads to a loss in energy.

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.

#### PmbPhy

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##### Re: Would the photon lose all its energy at infinity?
« Reply #33 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:

« Last Edit: 21/02/2015 01:13:06 by evan_au »

#### yor_on

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##### Re: Would the photon lose all its energy at infinity?
« Reply #34 on: 20/02/2015 19:54:05 »
Maybe one can think of in form of a 'field' with photons as 'excitations' in it. Then the 'field' is what determines the energy you will find measuring somewhere in a photon 'propagation'. In other words it should be about the 'relations' defining it. If I was to argue that a black hole is time dilated intrinsically I also will be able to do that for any gravitational potential. That should then give us different speeds measured depending on the inertial platforms density (invariant mass). That would put into question any definition of a constant being true, not compensated for the gravitational potential that exist where you measure it. In effect, we should actually get different answers depending on mass. Do we?
=

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.
« Last Edit: 20/02/2015 20:20:34 by yor_on »

#### jeffreyH

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##### Re: Would the photon lose all its energy at infinity?
« Reply #35 on: 20/02/2015 21:54:34 »
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:

Thanks for that Pete I will read it later. I am done with John.
« Last Edit: 21/02/2015 01:15:25 by evan_au »

#### jeffreyH

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##### Re: Would the photon lose all its energy at infinity?
« Reply #36 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.

#### jeffreyH

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##### Re: Would the photon lose all its energy at infinity?
« Reply #37 on: 20/02/2015 22:06:12 »
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:

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.
« Last Edit: 21/02/2015 01:16:33 by evan_au »

#### evan_au

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##### Re: Would the photon lose all its energy at infinity?
« Reply #38 on: 21/02/2015 01:34:50 »
Quote from: chiralSPO
arrive at a destination that was 100 ly away when they were emitted
A 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] ]

#### chiralSPO

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##### Re: Would the photon lose all its energy at infinity?
« Reply #39 on: 21/02/2015 02:09:10 »
I think some pedantry is acceptable here, but allow me to ask the same question, adding a few orders of magnitude to the distance and speed of the electron: if the distance traveled were 10000000 ly, and the electron were moving at 0.9999 c (now arriving 1000 years later than the photon, if I counted the digits correctly), would there be a noticeable difference in the red-shift of each of the waves?

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?)

#### JohnDuffield

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##### Re: Would the photon lose all its energy at infinity?
« Reply #40 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 too
It 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.

#### PhysBang

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##### Re: Would the photon lose all its energy at infinity?
« Reply #41 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.

#### evan_au

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##### Re: Would the photon lose all its energy at infinity?
« Reply #42 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 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 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....

#### JohnDuffield

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##### Re: Would the photon lose all its energy at infinity?
« Reply #43 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.

#### PhysBang

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##### Re: Would the photon lose all its energy at infinity?
« Reply #44 on: 21/02/2015 18:14:12 »
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.
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.
Quote
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.
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.

Quote
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.
Quote
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.

#### yor_on

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##### Re: Would the photon lose all its energy at infinity?
« Reply #45 on: 22/02/2015 09:34:46 »
You know :)

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.
=

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.
=

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.

« Last Edit: 22/02/2015 16:03:57 by yor_on »

#### JohnDuffield

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##### Re: Would the photon lose all its energy at infinity?
« Reply #46 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.

#### Bill S

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##### Re: Would the photon lose all its energy at infinity?
« Reply #47 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.

Quote
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.

#### JohnDuffield

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##### Re: Would the photon lose all its energy at infinity?
« Reply #48 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

#### PhysBang

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##### Re: Would the photon lose all its energy at infinity?
« Reply #49 on: 22/02/2015 15:16:51 »
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.
OK, so you are committed to not using the correct terminology. Noted.
Quote
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.
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 am glad you chose this example, as it might help you make less serious mistakes in the future.
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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?
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.
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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.

Quote
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.
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.

#### The Naked Scientists Forum

##### Re: Would the photon lose all its energy at infinity?
« Reply #49 on: 22/02/2015 15:16:51 »