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(I searched for an answer online already but I couldn't quite find what I was looking for...)

I thought about this for a long time now. If two Photons fly in the same direction, one behind the other one, for my understanding the one behind the other one should be pulled towards the photon in front of it due to it's gravity, and because it cant get faster it should increase it's frequency and therefore gain energy. The one in front cant be pulled backwards though because gravity travels with the speed of light itself(?) and therefore the gravity of the rear photon cant reach the one infront of it, which would therefore not lose energy.

But that would break the law of conservation of energy, wouldn't it? So I'm confused...

Am I thinking something wrong? Or how does it work/what would actually happen in this scenario?

Thanks for answers in advance!

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    $\begingroup$ @Qmechanic - This is not a duplicate of those questions. It is a thought experiment asking about the blue shift of light due to gravity, as explained in your second link. This question is "Since the rear photon is passing through the gravitational field of the forward photon, does the rear photon get blue shifted? If so, does the blue shift increase over time? If so, how is that possible, given conservation of energy?" $\endgroup$ – mbeckish Oct 13 '16 at 20:54
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    $\begingroup$ Instead of single photons, it may be easier to model a situation where a spherical mass emits two perfectly spherical shells of radiation in succession (becoming slightly lighter due to the energy loss), and then ask what happens to the distance between them, as measured by observers at rest in Schwarzchild coordinates at different large $r$ coordinates. This would have the advantage that the spacetime between the light fronts have to be Schwarzschild-shaped by symmetry, and the different regions of spacetime have to meet at matching $r$ coordinates. $\endgroup$ – hmakholm left over Monica Oct 13 '16 at 21:14
  • $\begingroup$ @HenningMakholm Nice simplification. Would you think the simplification might hide certain physics that is present in the OP's set up? $\endgroup$ – QuantumDot Oct 14 '16 at 2:21
  • $\begingroup$ Google search revealed this paper "The gravitational field of a light wave" - J.W. van Holten. Which you will find to be useful. Your problem is a special case of their analysis. Do you have a background in GR? I am leaving this as a comment because I could be wrong. If someone can check it I would be happy. Again This analysis is possible only if you consider classical waves and are not interested in studying the possible effect of quantum gravity. $\endgroup$ – Prathyush Oct 14 '16 at 2:27
  • $\begingroup$ You will also find this article to be useful iopscience.iop.org/article/10.1088/1367-2630/18/2/023009/meta I haven't checked it myself. $\endgroup$ – Prathyush Oct 14 '16 at 2:44
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Each photon will be at rest relative to the other, since they both travel in the same direction at the same speed. A photon at rest has zero frequency, hence zero energy. So, at least to first order, there should be no gravitational interaction between the two photons within their mutual rest frame.

I'm not sure what an outside observer would see. It seems to be accepted that gravitational waves interact with each other very very weakly, which suggests that the same should be true of electromagnetic waves.

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    $\begingroup$ I don't know the answer to the question, but I am pretty skeptical of any train of thought that invokes the rest frame of a photon. $\endgroup$ – WillO Dec 7 '19 at 16:27
  • $\begingroup$ @S. McGrew, Why you put equals sign between electromagnetic & gravitational fields ? The way gravitational waves interacts does not necessary means that same way electromagnetic fields self-interacts. $\endgroup$ – Agnius Vasiliauskas Dec 7 '19 at 16:33
  • $\begingroup$ The two are not equal, but are analogous in this case. @WillO, you're right that the "rest frame of a photon" is risky to invoke. But it's useful to consider it as a limit. Allegedly what led Einstein to Special Relativity was thinking about what a light wave would look like if you could run alongside it. $\endgroup$ – S. McGrew Dec 7 '19 at 21:00
  • $\begingroup$ I don't see analogy here at all. They roots are different, thus behavior can be too. So you claim is just unsupported by anything $\endgroup$ – Agnius Vasiliauskas Dec 9 '19 at 7:00
  • $\begingroup$ If you check relevant literature, you will find that there are analogies between electromagnetism and gravitation that have been explored extensively. Analogies, not equalities. A good place to start is en.wikipedia.org/wiki/Gravitoelectromagnetism. $\endgroup$ – S. McGrew Dec 9 '19 at 14:27
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The speed of gravity is also only c, so the photon in front never gets overtaken by the gravity of the photon in the back. The photon in the back travels through the gravitational field of the one in the front, but since the direction of pull is to the front while the back photon is already travelling with c in the direction on the pull it already has the maximum velocity which it can not exceed. Since the distance stays constant so does the energy.

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    $\begingroup$ This is protty much what I wrote in my question except I dont understand the last sentence. What does the distance have to do with the frequency? $\endgroup$ – Rubydragon Oct 13 '16 at 20:11
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    $\begingroup$ The energy of a photon increases when the photon gets closer to the center of gravity. In our scenario the distance between the two photon stays constant, so the energy of our photons stays the same everywhere (assuming vacuum along the path) $\endgroup$ – Yukterez Oct 13 '16 at 20:35
  • $\begingroup$ If you conclude this based on an analysis of field equation I would be interested in the analysis. The analysis done by you does not seem to reflect any credible understanding of GR. Secondly Please don't use the term gravitation attraction of a photon so loosely. Quantum theory of Gravity is subject of lot of modern research. One must be very careful. $\endgroup$ – Prathyush Oct 14 '16 at 2:24
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    $\begingroup$ We don't have any theory of quantum gravity yet, therefore my answear is from the perspective of general relativity, where energy is equivalent to mass when it comes to gravity, and the propagation speed for gravity and light is locally c. If you take two energetic light beams instead of single photons and keep a certain minimum distance between them quantum gravity effects will average out. For example, you don't need quantum gravity to conclude that a box full of photons weighs more than an empty box! $\endgroup$ – Yukterez Oct 14 '16 at 17:06
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They do not interact because their rest masses are zero. You can not attach to the photon neither an inertial nor a gravitational "mass" as $$ m \neq\frac{\hbar \omega}{c^2}$$ See for example on this topic https://arxiv.org/abs/physics/9907017. This part is pure kinetic part and is related to the momentum of a photon in the energy–momentum relation (with $m_{rest}=0$ ) $$E=p c $$ As example, for a moving particle with non-zero rest mass, the momentum part $E=p c $ does not influence its gravitational interaction (i.e. the gravitational potential that it creates). Because you can always consider the case in a comoving reference frame where $p=0$.

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From my understanding:. Photons have a rest mass of zero, but the electromagnetic waves do carry energy which causes gravitational pull. Since light isn't affected by gravity because its rest mass is zero, but the space light is moving on gets curved. This is what causes light beams to curve around massive objects. The two light beams wont be pulled into one another, but will start 'bending' if they are not moving parrelel to each other. This could get them closer to each other without changing speed.

A better explanation can be found in this Physics.SE post

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I don't know enough relativistic E&M to answer Henning Markholm's salvage of the question...but, for the OP:

Photons are massless, so they exert no gravitational force. So neither will experience a force.

Furthermore, your notion of the "speed" of a force is fundamentally wrong. The "speed" of a force is the rate at which changes in the force field propagate. So even if two massive, yet lightspeed-traveling particles did exert a force on each other, the first particle would still feel the force exerted by the second particle. There are confusing pieces of physics associated here; see https://en.wikipedia.org/wiki/Wheeler%E2%80%93Feynman_absorber_theory (sorry no link; I'm writing this on a phone). If you really want to understand want the "speed" of a force means, look into getting yourself a textbook; any E&M textbook at the level of Purcell or better will do.

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    $\begingroup$ I think the OP is asking about the general relativistic effect, in which massless particles can gravitationally attract one another. $\endgroup$ – QuantumDot Oct 14 '16 at 2:20
  • $\begingroup$ Then I stand by the second part of my answer. Without knowing QED, I feel like, if the particles did exert forces on each other, that would violate the linearity of the theory. $\endgroup$ – Jacob Manaker Oct 14 '16 at 2:23
  • $\begingroup$ linearity of which theory? Quantum theory? $\endgroup$ – QuantumDot Oct 14 '16 at 2:25
  • $\begingroup$ Yes, QED. But, judging from your response, I'm gathering I don't know this stuff enough to post any further. $\endgroup$ – Jacob Manaker Oct 14 '16 at 2:27
  • $\begingroup$ As far as I know, the canonical formulation of quantum gravity coupling to photons does preserve linearity of the Hilbert space (at least as far as perturbation theory is concerned). $\endgroup$ – QuantumDot Oct 14 '16 at 2:29

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