What happens to the energy lost by photons in gravity?

[Hugh de Launay]

After I read Martin_K's post of 4:14 Oct. 30 on the frozen image of an object just before it fell through a black hole's event horizon, the next few minutes I was jumped by a handful of related ideas.

First, the frozen image scenario is illustrative because when photons are frozen in place, they will not travel to the observer, so the observer has to be omniscient to be aware of the frozen image.

Second, the black hole is moving or wobbling, so it will either release the photons or it will envelop them with its event horizon.

Third, the photons trapped at the event horizon will experience a constant loss of energy (Doppler effect) until all of their energy is lost to the curved space-time continuum (STC). Without energy, the photons cease to exist.
What happens to this energy?

Fourth, the photons are electromagnetic, so their electromagnetism is assimilated by the STC, or electromagnetism can be voided when energy is taken away from it. If the electromagnetism is also assimilated by the curved STC, then it united with the gravity field.

 

[Nugatory]

when photons are frozen in place

You need be to very cautious with this English-language description. It's easy to misinterpret "frozen in place" as meaning that that they aren't moving. Actually those words are being used to say something more like "Schwarzschild $r$ coordinate is not changing" so tell you more about the behavior of the Schwarzschild $r$ coordinate than the motion of the light. A good way to see what's really going on with the light is to use Kruskal instead of Schwarzschild coordinates to visualize the path of a flash of light at the event horizon; it will move along a lightlike path at speed $c$ relative to an observer in its path, just as it does everywhere else. (That observer will necessarily fall into the black hole, but that's their problem and doesn't have anything to do with the behavior of the light).

Second, the black hole is moving or wobbling, so it will either release the photons or it will envelop them with its event horizon.

If you're using coordinates in which the black hole is moving, then you aren't using Schwarzschild coordinates, so conclusions that you draw based on Schwarzschild coordinates don't carry over. If you actually work through the math, you will find that the light is just as trapped (that is, remains on the same lightlike hypersurface) whether you describe its path using coordinates in which the black hole is still or moving. An easy math-free way of seeing this is to consider that if the black hole is moving, it has to be moving relative to something else; we could just as well consider the black hole to be at rest and the something else to be moving. Either way, the path of the light through spacetime must be the same - it doesn't care whether observers elsewhere in the universe are moving or not.

Third, the photons trapped at the event horizon will experience a constant loss of energy (Doppler effect) until all of their energy is lost to the curved space-time continuum (STC). Without energy, the photons cease to exist.
What happens to this energy?

They do not lose energy. Anyone in their path will find that they have the same wavelength and energy, whether we measure it now or a million years from now (although you might want to try to precisely define what it means to say these measurements are made "a million years apart" - the exercise will expose some bogus hidden assumptions you're making).

Fourth, the photons are electromagnetic, so their electromagnetism is assimilated by the STC, or electromagnetism can be voided when energy is taken away from it. If the electromagnetism is also assimilated by the curved STC, then it united with the gravity field.

Except that they aren't losing any energy, at least not in any coordinate-independent way.

 

[Orodruin]

In addition to what Nugatory has already said, you need to define what you mean by "energy". There are many different definitions that could aspire to this name in GR. Even in SR, energy is not something invariant, meaning that it depends on the inertial frame you are using. This is also true in classical mechanics. Energy is not a form of substance that is gained or lost by an object.

 

[Hugh de Launay]

Hello Nugatory! I appreciate the time and effort you took to critique what I wrote. My reply is not a rejection of what you have just said, but is more a clarification of my perceptions. I sense your competence in physics and respect your views.

In the first item I took frozen in place to mean what it said. An image does not perpetuate itself if its object is under the event horizon. After the flash of light that showed it was frozen, there should be the black background of the black hole, not a frozen image. If there is a perpetual frozen image that means the photons are not moving forward. But I accept your criticism as valid. One should note, however, that just under the event horizon, a photon whose trajectory is perpendicular to the event horizon will be there as long as it exists.

I understand your relativity of motion point with the unknown motion of the black hole and do not disagree with it. My statement was with regard to a black hole that may be moving relative to it own previous position.

We agree to disagree on the third point. It is a known fact from observations and experiments, as well as theory that light experiences a red-shift Doppler effect when it rises against a force of gravity. A red-shift of a photon is a certain indication that has less energy than before the shift. The frequency is lower, so the product of the Planck constant with the lower frequency show a loss of a definite amount of energy.

On the last point, if a photon is rising perpendicular to a perpetual force of gravity, then its frequency will be reduced until it has none. Then we have the product of Planck's constant with zero.

Still, I understand your points and they have my respect. Thanks for bringing them forth.

 

[Orodruin]

a photon whose trajectory is perpendicular to the event horizon will be there as long as it exists.

The event horizon is not a spatial surface, it is a null surface.

I understand your relativity of motion point with the unknown motion of the black hole and do not disagree with it. My statement was with regard to a black hole that may be moving relative to it own previous position.

In relativity, "its own position" does not have a well defined and universal meaning.

We agree to disagree on the third point.

This is not something you can agree to disagree on. Your claim is demonstrably false in GR. What determines gravitational redshift is difference in gravitational potential, not gravitational force.

On the last point, if a photon is rising perpendicular to a perpetual force of gravity, then its frequency will be reduced until it has none. Then we have the product of Planck's constant with zero.

Again, your statement is demonstrably false so it makes no sense to speculate further in this direction.

Sorry to come off as harsh, but unless you accept that your statements are false you cannot learn what the theory actually predicts.

 

[Nugatory]

a photon whose trajectory is perpendicular to the event horizon will be there as long as it exists.

There is no such thing as a photon whose trajectory is perpendicular to the event horizon - the horizon is a lightlike surface, not spacelike. This will be most clear if you plot the path of a flash of light on a Kruskal diagram.

My statement was with regard to a black hole that may be moving relative to it own previous position.

"Moving relative to its own previous position" just means that you're using coordinates in which an observer moving relative to the black hole is at rest. There is no physical difference between this situation and the when the black hole is "at rest" (meaning that we are using coordinates in which the position of the black hole does not change); this goes all the way back to classical Galilean relativity.

 

[Eytan Suchard]

Hi,
1) It definitely depends on either you are a Lagrangian or an Eulerian observer. Suppose you are a Lagrangian observer who freely falls through the gravitational field in an elevator whose cable was cut off by some malicious entity. Suppose you have a photon that bounces between two horizontal mirrors. You will see no change in the wavelength of the photon as you fall. On the other hand, an observer who stands on the ground of the Earth will see a blue-shift of the photon just as you almost hit the ground. The wavelength he will see is Lambda * sqrt( (1-2GM/(r2 * c^2)) / (1-2GM/(r1 * c^2)) where r2 is the final radius of the ground in the coordinates of a far Eulerian observer and r1 is the the radius of the horizontal level of the elevator as it started to fall, also in far Eulerian observer coordinates. G is Newton's gravity constant and c is the speed of light. In general, there is no energy conservation for an Eulerian observer in General Relativity.
2) In other cases such as the expansion of the cosmos, we may say that the gravitational energy that mass gains as the cosmos expands is on the expense of red-shifted photons that reach far observers from the photon source. That explanation sounds rational but it can't explain the properties of cosmic expansion as currently observed. Cosmic expansion is a global violation of the conservation of energy unless we assume the existence of negative energy.
3) It is also important to note that Noether's theorem heavily relies on symmetry and on dependence of an action Lagrangian on first order derivatives, otherwise conservation laws do exist but are more complex. Physics of higher order derivatives may be at play such as in Conformal Gravity. Quantum Field Theory also relies on the description of wave functions by first order derivatives, e.g. Dirac's wave function. Despite the accuracy of predictions by the Standard Model which derives from QFT, it may be wrong especially in Planck scale physics. Deviations from QFT at the Planck scale will lead to new conservation laws. The big picture of conservation cannot be understood from GR alone and physics may have to be updated. After all, physics is a predictive language and language is not reality but a human interpretation of reality.

 

[Hugh de Launay]

This is addressed to Orodruin's first post: I agree with you about the many forms of energy. The energy I was referring to is electromagnetic as is found in each photon. I'm not certain, but I think you are telling me the energy of a photon in one inertial frame of reference may be different in a different frame of reference that has a different velocity even though the speed of light is the same.

But I am not sure about your last sentence. Photons may not qualify as objects, but they do gain and lose electromagnetic energy.

 

[PeterDonis]

@Hugh de Launay there is no point in making a post that just quotes someone else's post without any content of your own.

Also, please do not quote an entire post by someone else. Only quote the specific things you are responding to.

Your post made after #8 has been deleted.

 

[Orodruin]

The energy I was referring to is electromagnetic as is found in each photon.

You are still not getting the point. Energy is not a substance that is possessed by an object. In GR there are many different ways of defining what energy is that are all compatible with our usual notion from special relativity. It is not a question of which "form" the energy is in, it is a question of what you mean when you say "energy".

I'm not certain, but I think you are telling me the energy of a photon in one inertial frame of reference may be different in a different frame of reference that has a different velocity even though the speed of light is the same.

Right. This also has nothing to do with relativity per se - it is true also for classical objects in classical Newtonian mechanics.

 

[Hugh de Launay]

This is addressed to Orodruin's second post: Thanks for persisting. I need to see the perspectives had by knowledgeable people.

I am aware the event horizon is a condition of space-time around the core of a black hole and not a thing.

I accept this criticism as valid since it is obvious I am out of touch the routine jargon of physicists.

I know photons have less energy after rising up through a field of gravity based on what I have read which was written by other physicists. A photon that experienced a red-shift is less energetic than it was before the shift took place, and that is a quantitative fact.

The energy is conserved in some way. If the photon has less energy than before, that energy went somewhere else.

I don't mind your being harsh, because I know you are frustrated by my attitude, so I don't blame you and I will accept all civil harshness because we are all human and need to get tough at times. Harshness scares me a little, but doesn't bother me because I understand it.

 

[PeterDonis]

I know photons have less energy after rising up through a field of gravity based on what I have read which was written by other physicists.

Please give some specific references. As has already been pointed out, the term "energy" does not have a single unique meaning. We need to see the specific references you have read in order to understand what you are thinking.

To give just one example illustrating the possible complexities involved: in stationary spacetimes, there is a quantity called "energy" which is conserved for geodesic motion. A photon rising in a gravitational field is moving in a stationary spacetime; so its "energy" in this sense is conserved. This is true even though static observers at increasing altitudes will observe this photon to have an increasingly redshifted frequency.

So you can't just make a blanket statement that "photons have less energy after rising up in a gravitational field". You need to specify what specifically you mean by "energy" (notice that I specifically described how the "redshift" of photons rising in a gravitational field is observed).

The energy is conserved in some way. If the photon has less energy than before, that energy went somewhere else.

You should read this article by Sean Carroll:

http://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/

 

[Orodruin]

You should not think of a photon as something that "has energy" since that is a frame dependent statement. Ultimately, you can observe any photon colocated with you to have any energy, depending on your own state of motion. When we talk about "the" energy of a photon in relation to gravitational redshift, we talk about the energy that a stationary observer would measure, which is a hidden assumption that is rarely mentioned. However, it is vital to keep this in mind to achieve any level of appropriate understanding.

The energy is conserved in some way. If the photon has less energy than before, that energy went somewhere else.

This is also not true in GR and it is not the case that there necessarily is a globally applicable measure of energy. The conservation laws in GR are local, meaning that they locally look like conservation laws (in small enough regions of spacetime).

 

[Hugh de Launay]

there is no point in making a post that just quotes someone else's post without any content of your own.

Sorry. I did not know you could type between the quotes by others.

Also, please do not quote an entire post by someone else. Only quote the specific things you are responding to.

I will figure this out next time.

Your post made after #8 has been deleted.

Your post made after #8 has been deleted.[/QUOTE]
Thanks for your notice.

 

[Hugh de Launay]

You should not think of a photon as something that "has energy" since that is a frame dependent statement. Ultimately, you can observe any photon colocated with you to have any energy, depending on your own state of motion. When we talk about "the" energy of a photon in relation to gravitational redshift, we talk about the energy that a stationary observer would measure, which is a hidden assumption that is rarely mentioned. However, it is vital to keep this in mind to achieve any level of appropriate understanding.

I can see that my perspective on a lot of physics needs to be updated. Once I can speak using your perspective and language I will post about this again.

This is also not true in GR and it is not the case that there necessarily is a globally applicable measure of energy. The conservation laws in GR are local, meaning that they locally look like conservation laws (in small enough regions of spacetime).

The general perspective today in GR has changed a lot since Einstein published his theory. I will need to track the changes while I try to catch up with you

 

[Hugh de Launay]

Please give some specific references. As has already been pointed out, the term "energy" does not have a single unique meaning. We need to see the specific references you have read in order to understand what you are thinking.

Thanks for advising me. I will bring references if needed next time.

To give just one example illustrating the possible complexities involved: in stationary spacetimes, there is a quantity called "energy" which is conserved for geodesic motion. A photon rising in a gravitational field is moving in a stationary spacetime; so its "energy" in this sense is conserved. This is true even though static observers at increasing altitudes will observe this photon to have an increasingly redshifted frequency.

I understand what you are advising me. Thank you.

So you can't just make a blanket statement that "photons have less energy after rising up in a gravitational field". You need to specify what specifically you mean by "energy" (notice that I specifically described how the "redshift" of photons rising in a gravitational field is observed).

I will qualify what I am saying henceforth.

You should read this article by Sean Carroll:

http://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/

Thank you for this reference. I just read the article and see what the problem is with what I have said.

 

[Hugh de Launay]

Hi,
1) It definitely depends on either you are a Lagrangian or an Eulerian observer. Suppose you are a Lagrangian observer who freely falls through the gravitational field in an elevator whose cable was cut off by some malicious entity. Suppose you have a photon that bounces between two horizontal mirrors. You will see no change in the wavelength of the photon as you fall. On the other hand, an observer who stands on the ground of the Earth will see a blue-shift of the photon just as you almost hit the ground. The wavelength he will see is Lambda * sqrt( (1-2GM/(r2 * c^2)) / (1-2GM/(r1 * c^2)) where r2 is the final radius of the ground in the coordinates of a far Eulerian observer and r1 is the the radius of the horizontal level of the elevator as it started to fall, also in far Eulerian observer coordinates. G is Newton's gravity constant and c is the speed of light. In general, there is no energy conservation for an Eulerian observer in General Relativity.
2) In other cases such as the expansion of the cosmos, we may say that the gravitational energy that mass gains as the cosmos expands is on the expense of red-shifted photons that reach far observers from the photon source. That explanation sounds rational but it can't explain the properties of cosmic expansion as currently observed. Cosmic expansion is a global violation of the conservation of energy unless we assume the existence of negative energy.
3) It is also important to note that Noether's theorem heavily relies on symmetry and on dependence of an action Lagrangian on first order derivatives, otherwise conservation laws do exist but are more complex. Physics of higher order derivatives may be at play such as in Conformal Gravity. Quantum Field Theory also relies on the description of wave functions by first order derivatives, e.g. Dirac's wave function. Despite the accuracy of predictions by the Standard Model which derives from QFT, it may be wrong especially in Planck scale physics. Deviations from QFT at the Planck scale will lead to new conservation laws. The big picture of conservation cannot be understood from GR alone and physics may have to be updated. After all, physics is a predictive language and language is not reality but a human interpretation of reality.

It will take me a while to analyzed what you are trying to communicate to me. I will be back when I have figured it out, but don't hold your breath!

 

[PeterDonis]

@Hugh de Launay why do you keep quoting people's entire posts after you've been asked not to and acknowledged that request?

 

[Hugh de Launay]

Sorry, again. I need to learn your system. I was not aware I was messing things up. I will learn the system before I answer anyone else.

 

[PeterDonis]

I was not aware I was messing things up.

Then you need to look at your posts after you post them. Looking at them should make it obvious that they were messed up. You have a window of time where you can edit a post after you first make it, to allow you to correct errors like this. Use that.

I will learn the system before I answer anyone else.

You evidently didn't do this before making your post #20.

 

[PeterDonis]

@Hugh de Launay I have now edited some of your posts to correct the quotes--you had your own responses mixed in with quotes from other people, which gives the impression that other people said things they did not say.

Please learn how to use the quote system correctly.

 

[russ_watters]

Then you need to look at your posts after you post them. Looking at them should make it obvious that they were messed up. You have a window of time where you can edit a post after you first make it, to allow you to correct errors like this. Use that.

@Hugh de Launay also try using the "preview" button to see what your post looks like before you post it.

 

[Hugh de Launay]

russ_watters said: ...also try using the "preview" button...

Thanks. I will do that with this post.

 

[Hugh de Launay]

PeterDonis said: ... you need to look at your posts after you post them...

I will do this. Thanks.

PeterDonis said: You evidently didn't do this before making your post...

Correct. I thought I had solved it but still needed yours and russ-watters comments to pin things down. Thanks.

 

[russ_watters]

Correct. I thought I had solved it but still needed yours and russ-watters comments to pin things down. Thanks.

This is better, but it would help if you use the quote function, which encloses quotes in boxes for better visibility...as you can see with my post.

 

[Hugh de Launay]

russ_watters said: ...it would help if you use the quote function...

Thanks. I will do this as soon as I figure how to do it.

 

[jbriggs444]

One way to proceed is to use the Reply button. In the editing pane you will see something like:

[QUOTE="russ_watters, post: 6082659, member: 142"]This is better, but it would help if you use the quote function, which encloses quotes in boxes for better visibility...as you can see with my post.[/QUOTE]
​

Now suppose that you wanted to quote only "it would help if you use the quote function". You would strip out the "This is better..." and you strip out the ", which encloses...". That leaves you with

[QUOTE="russ_watters, post: 6082659, member: 142"]it would help if you use the quote function[/QUOTE]​

And you can then tack on your response.

[QUOTE="russ_watters, post: 6082659, member: 142"]it would help if you use the quote function[/QUOTE]
OK, I'll try harder this time. But what is the quote function?
​

Now suppose that you also want to quote some additional text without attribution, you can simply stick it between [QUOTE] and [/QUOTE] tags. For instance:

[QUOTE]... as you can see with my post.[/QUOTE]
Oh, you are talking about those square bracketted [QUOTE] and [/QUOTE] thingies?
​

You can also browse to the Info => How To => BBCODES page on this site. Here it is: https://www.physicsforums.com/help/bb-codes.

I'm not terribly competent, but this post at least demonstrates [PLAIN], [QUOTE] and [INDENT]. I had to make heavy use of the [PLAIN] tag here so you could actually see the other tags I was trying to describe. I also had to edit this post more times than I care to remember in order to get it looking decent.

There is an easier way to do piece-meal quoting. Drag the mouse over a section you wish to quote and hit either "reply" or "quote" from the contextual pop-up. I usually prefer the main Reply button that quotes the whole post because it does a better job preserving embedded LaTeX equations.

 

[Hugh de Launay]

I also had to edit this post more times than I care to...

I do appreciate the work you did in pointing me in the right direction. It helped a lot.

 

[PeterDonis]

@Hugh de Launay I just edited your post #29. It should still be within the edit window for you at the time of this posting; you can click "Edit" to see how the QUOTE tags look now.

 

[Hugh de Launay]

you can click "Edit" to see how the QUOTE tags look now.

I did this. Thanks for your efforts to help me.

 

[PeterDonis]

@Hugh de Launay look at your post #31, which I am purposely not editing. Then look at your post #29, which I edited. Can you see the difference between them? Can you edit post #31 so it looks like post #29?

I don't know why you are having such trouble in properly using the quote system. But you're never going to learn to use it properly if you don't look at what you post and see what's wrong with it, compared to everyone else's posts, and try to fix it.

 

[PeterDonis]

I don't know why you are having such trouble in properly using the quote system.

And, btw, the reason I am making a big deal out of this is that if you don't learn to properly use the quote feature, your posts will be unreadable and nobody will want to respond to you.

 

[Hugh de Launay]

I don't know why you are having such trouble in properly using the quote system

I don't have a good excuse. I am somewhat ignorant in using this type of computer system. Of course you are right about turning people off from reading my input. I think I am close now enough to do a correct post. There may still be some mistakes, but I will preview them and look for ways to correct them with the BBcode guides before I post. Where I screwed up in post #29 and #31, I just had no clue at that time on how to do it right. Now I see what I missed after taking your advise in post #32. Thanks again for your help. You mentors are something special -- being all volunteers.

 

[PeterDonis]

I think I am close now enough to do a correct post.

Your post #34 is fine.

 

[Hugh de Launay]

It definitely depends on either you are a Lagrangian or an Eulerian observer.

Even though it is since Nov. 2, 2018 you posted your comments, I am now responding in case you are still interested. At first I did not understand most of your statements, but I think I understand them now after a little studying, so I will restate what I think you said in my own words.
(1) The Lagrangian observer sees the photon maintain the same wavelength when it is reflected between two mirrors as the mirrors drop down toward a source of gravity. This is because the Lagrangian observer travels alongside the mirrors as they descend. But the immobilized Eulerian observer measures just a short segment of the photon's path on one level and again at a lower level. The observer sees a blue-shift in the photon at the lower level. Without the GR conservation of energy requirement included, the photon has gained energy according to QM.

In other cases such as the expansion of the cosmos, we may say that the gravitational energy that mass gains as the cosmos expands is on the expense of red-shifted photons that reach far observers from the photon source.

In the expansion of the cosmos, the loss of energy during the photon's red-shift is transferred to mass -- an inadequate explanation of dark energy's force which is accelerating the expansion of the universe.

The big picture of conservation cannot be understood from GR alone and physics may have to be updated. After all, physics is a predictive language and language is not reality but a human interpretation of reality.

Conservation laws use higher order derivatives than Noether's first order derivatives in her Lagrangian equations. Conformal gravity mathematics use yet higher order derivatives. QFT uses first order derivatives and makes accurate predictions, through the standard model, that may nevertheless be wrong at Planck scales. New conservation laws will be found at Planck scales. GR is incomplete with respect to the full explanation of energy conservation laws, and physics may have to be developed further to gain a complete explanation of its subject matter.

I have had the good fortune to have heard that all knowledge, including physics and its models, is a virtual reality which can never be the same as what is physically real. For example a five ton tree is not the same as its virtual reality visualization in the three pound human brain.