How does the propagation of gravity work for photons?

Question

As explained in the answers to this post, photons apparently exert a gravitational pull on other objects. It has also been explained on this site, that gravity propagates at the speed of light.

I'm wondering, though, how do you reconcile these two facts? I'm trying to imagine the gravitational field made by a photon and it seems like there are some paradoxes. For example, how can gravity propagate ahead of the photon at the speed of light, if the photon is also travelling at the speed of light? My guess is the solution is probably found in relativity, but I certainly can't figure it out.

Right now, the best I can do is to think about gravity as sound, and a photon as an object travelling at the speed of sound. Can anyone help me out?

Answer 1

Gravity is well described with general relativity, which is a classical theory dealing with four dimensional space time, where the effect we call gravity is a distortion of this space time. This distortion depends on the energy momentum vector of the item under study, and it holds true also in the case of photons, which though of zero mass, have energy =h*nu, nu the frequency and h Planck's constant, and a large number of them build up the classical electromagnetic wave. That photons do interact with gravity, is seen in gravitational lensing.

The exact way is still a research project. If you have access to the publication through a library

The gravitational field of photons

W. B. Bonnor

The gravitational field of a photon on an infinite straight path is a single sheet of plane-fronted gravitational wave accompanying the photon and perpendicular to its track. This field cannot arise from a retarded potential generated by the photon, and I suggest that it arises in the process of emission. The near-field depends on the energy of the photon, but the far-field does not. The field of a steady beam of photons is compared with that of a static material rod, and the differences discussed.

The photon itself has no electric or magnetic field when measured, but in the mathematical expression that builds up the electromagnetic wave there are phases in complex amplitudes that generate the classical electromagnetic field, light.

A complete description for the photon's gravitational identity has to wait for the quantization of gravity. At the moment these are effective theories . One point should be clear that both the photon and the gravitational wave will be travelling with velocity c.

see also answers here.

Answer 2

Assuming speed of propagation of gravity same as speed of the photon, the contribution of the photon towards gravity (towards curving the space), can be considered a single shape (a circle, sphere or another appropriately curved shape). The single shape (say, smallest circle in your figure), moves with the photon, and there are no other circles (they just fade away). Now, what is difficult to figure, is given the same speed, the photon may be in the center of the shape, or, it may be towards the front end of the shape. There would not be any shock wave as photon does not have external source of energy to keep creating new waves. It is kind of field that is moving with it.

However, I am not aware of any experiment where speed of propagation of gravity (or of curving of space) has been measured to be equal to c. Gravitational waves have been known to propagate at c, but they are not same as gravity itself. Gravity is always attached to the source while gravitational waves are detached from the source and move on their own once created.

Answer 3

Photons travel along a gravitational compression wave of gravitational lines traveling in the same direction caused when the energy was released from an electron during a shell jump. Photons travel at the same speed as the wave. The reason they may not reach the same destination simultaneously is that photons collide with atoms along the way and are re-emitted taking some time in the process.

Answer 4

A possible way to arrive at an answer is by the following procedure:
∙ start with the background field G = classical Minkowski space-time.
∙ repeat {
∙ calculate the stress tensor for the background G, call it T.
∙ write out the electromagnetic field as a quantum field over classical curved space-time background G.
∙ write out the single photon state ψ for the field over the background G.
∙ calculate the vev of the stress tensor for the field in the state ψ with respect to G, call it δT.
∙ solve the Einstein field equations for T + δT, call the solution G'.
∙ replace G by G'.
∙ } until convergence

I'm not entirely sure making it the vev (vacuum expectation value) will reset the "add photon" part of the procedure for the background. So, you may be just piling on photons on top of each other with this. We want to add in the kick-back of the gravity without adding in the actual object with each update.

It should be something like: vev photon + stress tensor of photon's gravity + stress tensor of photon's gravity's gravity + ..., but gravity doesn't have a stress tensor, per se. So, I'm not totally sure if this is going to work, to add in just the gravity to the background at each stage, so as to bring about this kind of sequence.

Answer 5

It is important to note that in special relativity, all objects that travel at the speed do so in every inertial reference frame, including other inertial frames that are themselves traveling at the speed of light. The photon "observes" gravitons radiating away from it (if indeed that picture is accurate) at the speed of light, because in it's reference frame it is at rest. Likewise the gravitons that affects it coming from other objects, are "seen" to approach it at the speed of light. And so it's response to gravity in its reference frame is identical to an object experience gravity at rest. It is only when observing the photon interacting with the graviton from an outside reference frame that any peculiarity is noticed.