[JohnDuffield]

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.

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]

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.

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]

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]

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]

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.

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.

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.

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]

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.

[JohnDuffield]

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]

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.

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]

...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]

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.

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.

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.

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.

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.

[PhysBang]

...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.

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.

[JohnDuffield]

...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. 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?

[yor_on]

John, you can measure a expansionary redshift, and their photons too, and also find them having lost 'energy' due to it. And PhysBang haven't spoken out of his hat yet as I think 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. Both are in a way about 'motion'. It just that we are so used to thinking of it form of ordinary objects, as a ball thrown, that one seems a lot simpler than the other.

=
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'.
=

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.

[PhysBang]

...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.

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.

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. 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?

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.

[Bill S]

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?

[evan_au]

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. 

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.

[Bill S]

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?

[Toffo]

Before accelaration: An almost stationary star emitted the photon hundreds of years ago, almost no recoil energy went into the star.

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.

[jeffreyH (OP)]

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.

[JohnDuffield]

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.