# About EM waves creating gravitational field

1. Sep 24, 2014

### Ookke

If EM waves create gravitational field "around" them (as I understand is the case) which reference frame we should use?

We could imagine several observers, some moving in the same direction with EM wave with different speeds, others moving in the opposite direction. The different observers could measure very different energy levels for the EM wave, so maybe we could expect very different gravity fields depending on the observer. Are all fields equally correct or should we pick one?

Sorry if this is old question, but I didn't find anything exactly matching. Thanks.

2. Sep 24, 2014

### Staff: Mentor

All are correct. Gravitational fields for different observers look different. This shouldn't be surprising, as electromagnetic fields have the same effect.

3. Sep 24, 2014

### Staff: Mentor

Yes, it is.

The question you should ask before even asking that one is, how should we describe an EM field as a source of gravity? In GR, we describe sources of gravity with a stress-energy tensor; and the SET for an EM field is well known:

$$T_{\mu \nu} = F_{\mu \alpha} F_{\nu}{}^{\alpha} - \frac{1}{4} g_{\mu \nu} F_{\alpha \beta} F^{\alpha \beta}$$

where $F_{\mu \nu}$ is the EM field tensor.

The reason we use the SET is that it tells us how the EM field acts as a source of gravity in any frame we like; just compute the components of the SET in that frame. The particular components will change, because the components of the EM field tensor (which are related to the ordinary electric and magnetic field 3-vectors) change when you change frames; but the same equation (the Einstein Field Equation) will still determine what kind of gravity the EM field produces, given its SET.

Last edited by a moderator: Sep 25, 2014
4. Sep 25, 2014

### Ookke

Ok, thanks. But I guess that gravity field of EM wave is somewhat different than field created by object with rest mass.

We can imagine two EM waves traveling to same direction, side by side and very close each other. Observer "at rest" would see just two waves with nothing special, but another observer moving very fast to opposite direction would measure the wave energy much higher. In principle, this energy could be large enough to cause gravitational pull, making the two waves collide i.e. merge together.

I don't think this would really happen, just interesting to imagine this kind of experiment. The same with objects that have kinetic energy, maybe that somehow contributes the gravity field as well.

5. Sep 25, 2014

### Staff: Mentor

The stress-energy tensors are somewhat different, yes.

They would, but that wouldn't change the mutual gravity between the waves themselves. How could it? Observers moving at different speeds past the waves doesn't change the waves; it just changes the observers.

(Similar remarks apply to objects with rest mass; see below.)

Actually, it turns out that two EM waves moving in the same direction do not attract each other gravitationally; but they do if they are moving in opposite directions. This is one way in which EM waves are different from objects with rest mass.

It does, but perhaps not in the way you were thinking.

As I noted above, different observers moving at different speeds relative to a pair of gravitating objects will measure the objects to have different kinetic energies, but that won't change the mutual gravity between the objects, because it doesn't change the objects themselves; it just changes the observers.

However, suppose I have a gravitating object composed of bodies all at rest relative to each other, and another gravitating object composed of similar bodies, with the same rest mass, but moving relative to each other in a bound system (for example, a bunch of particles flying around in a box, compared to the same particles sitting at rest in the same box). The kinetic energies of the objects in the second case will contribute to the overall gravity of the object; it will be larger than it would be if the bodies inside the object were all at rest.

(Also, the motion of a body passing by a gravitating object does depend on its velocity relative to the object; bodies flying by a gravitating object at relativistic speeds, for example, "fall" faster than bodies flying by at slow speeds. This is because, in Newtonian terms, the "force" of gravity in General Relativity has a velocity-dependent component.)

6. Sep 26, 2014

### pervect

Staff Emeritus
It turns out that parallel propagating EM waves wind up not attracting each other at all. Anti-parallel propagating EM waves do, however attract each other.

See for instance:
https://www.physicsforums.com/threads/light-and-gravity.747824/#post-4720898

and the references therin. You can probably find other threads too, this issue has been discussed a number of times.

This behavior will undoubtedly be puzzling, until one drops the Newtonian idea that "mass" causes gravity and realizes that in GR it's the stress-energy tensor that causes gravity, as a few other posters have mentioned. The components of the stress energy tensor include energy density, momentum density, and pressure. If one ignores the pressure for the time being (it is present in the light beam, however), one can say that the energy and the momentum of the light beam both "cause gravity".

Analogies between electromagnatism and gravity may be helpful as well, and are drawn in a few of the papers via the formalism of gravitoelectromagnetism, GEM, a weak-field solution to GR.

One final point - there isn't any huge difference between rapidly moving matter approaching the sped of light and light itself. Ignoring the issue of pressure (which is a tiny effect), the limit of taking rapidly moving matter of fixed energy in the limit as the rest mass goes to zero gives the same limit as light. This is often called a "null dust" solution.

7. Sep 26, 2014

### Ookke

I was going to develop thought experiment just like that. No need to :)
This is indeed a subject that creates confusion in many readers I'm sure, but it's more clear now. Your answers have been most helpful.

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