# Effect of photon gravity on another photon traveling in the opposite d

1. Aug 6, 2014

### Sting33

Imagine a theoretical universe which contained only two photons with the same energy. Imagine that each photon began at approximately the same point and traveled in exactly opposite directions. I believe that general relativity predicts that photons exert a gravitational force on each other. In this case, what happens to these two photons? Will they simply redshift to some slightly lower energy level due to the gravitational effect, however small? Or will the gravitational effects ever even reach the other photon? I would hypothesize that the photons would never feel the influence of each other's gravitational force because from the point of view of one photon, it's gravity would be traveling at C, chasing after the other photon also traveling at C, unless they were not traveling at exactly opposite directions.

Does anyone have any theories that explain how this system would behave from a relativistic perspective from the point of view of an observer at the origin? I'd love to know how this looks from the point of view of one of the photons, but I'm afraid that might not possible to predict.

I have a little trouble with a redshift occurring because it seems to me that in this system the relativistic conservation of energy would not hold in that case.

2. Aug 6, 2014

### Simon Bridge

I'd suspect so too.

The photon frame is meaningless. It is also meaningless to talk about how an observer at the origin sees the photons - an observer there wouldn't. The only way to see the photons is to intercept one. It would be valid to ask what an observer intercepting one photon would observe compared with the case where there is only one photon in the Universe.

Conservation of energy is a tricky concept in general relativity anyway. You could always have a go working the stress-energy tensor for the problem - or work the approximation for Newtonian gravity.

3. Aug 6, 2014

### Chronos

You may wish to examine Noether's theorem on the concept of conserved quantities.

4. Aug 6, 2014

### Ich

My two cents: after release, the photons outrun any change of the gravitational field that the release itself produced. Thus they effectively still "feel" the static field that was present before the breakup. From there on, you're einther deep in the problematic notion of a "pointlike particle" or you deal with generation of energy from nothing. Both leads nowhere, so I think the result depends strongly on what you imagine a "photon" to be and how exactly these "photons" are arranged prior to release.

5. Aug 6, 2014

### pervect

Staff Emeritus
I don't know the answer offhand, but I can tell you how GR would describe what you need to look for.

GR, as a classcal theory, doesn't have "photons". But it can model beams of light - or other radiation.

The solution for the gravitational field of a beam of light would be something called a pp-wave. See The Wiki Entry

In particular, assuming your beam of light is axissymmetric, you'd look for an axissymmetric pp-wave solution. (You could also find plane pp-wave solutions, but these would not be of as much relevance.)

For two colliding beams of light. the solution would involve a colliding pp-wave space-time. "Starting out at the same place" is really just a collision, if you go back in time a bit (before the collision).

Google in fact finds a paper on colliding axissymmetric pp-waves on arxiv, but it's not clear to me if it was published in a peer reviewed journal. The Wiki article also talks a bit about colliding pp wave, it'd be the best place to start, I think.

6. Aug 7, 2014

### Sting33

Thank you all for your very insightful posts. Based on your responses, it looks like I am way in over my head at this point on this one. Fascinating thought experiment, but to find a solution it appears that the math is way too advanced for me. I really respect you all for your great knowledge.

7. Aug 7, 2014

### Simon Bridge

I was kinda hopeing someone would do it - but looking over pervect's post, this is actually a situation for quantum field theory. He's right - the statement "gravity of a photon" involves mixing up models.
Although you could get further by considering a light pulse.

This sort of thing gets discussed a lot, i.e. see:

8. Aug 13, 2014

### Sting33

So, does anyone know if light traveling directly away from an object exerts any gravitational force at all on that object?

9. Aug 14, 2014

### Simon Bridge

By Newton's third law, if light experiences gravity from a massive body then the massive body experiences an equal and opposite force from the light.

In GR, gravity is a pseudoforce that depends on your reference frame.

10. Aug 14, 2014

### Sting33

So, assuming a massive body and a beam of light come into existence at the same time, and the beam of light is traveling directly away from the massive body, I would hypothesize that the beam of light would never exert a gravitational force on the massive body and vice versa because the gravitational force, traveling at C, would never catch up with the light, traveling at C.

In order for it to ever catch up, the beam of light would need to be traveling on a line that does not pass through the massive body. Does anyone have anything that would tend to reject or modify this hypothesis?

11. Aug 15, 2014

### Simon Bridge

But a "beam" of light is not "travelling at c". It is either there or it is not. It is the waves that make up the beam that travel at c.

12. Aug 16, 2014

### David Horgan

Photons interact gravitationally but this gravitational scattering of light by light has been studied only rarely. Studies go back to Tolman, Ehrenfest and Podolsky in 1931 and to Wheeler in 1955 who analysed the gravitational field of light beams and the corresponding geodesics in the linear approximation of Einstein equations. They discovered that null rays behave differently according to whether they are parallel or anti-parallel to a steady, long, straight beam of light. Those rays moving parallel experience no attraction whilst those moving anti-parallel experience an attraction.

See: On the Gravitational field produced by light, Phys. Rev., 37, p. 602-615(1931).

This paper can be found here:
http://authors.library.caltech.edu/1544/1/TOLpr31a.pdf

Some recent papers that are relevant to this discussion are:

K. S. Ng and C. H. Raymond Ooi, “Gravitational force of a Bessel Light beam in a slow Light medium,” Laser Physics, vol. 23, no. 3, Article ID 035003 , (2013).

and

The Uncertainty Principle in Einstein Gravity, Vilasi G 2007 J.Phys.Conference Series 87012017

13. Aug 16, 2014

### Simon Bridge

Tolman et al did not study the gravity of photons did they, but of classical light beams? Isn't there an important distinction to be made here?

Since photons come from quantum mechanics, to answer questions about photon gravity, don't we really need a theory of quantum gravity?

All authors are talking about light beams or rays. OK the world-line of a photon is a light ray in space-time.

Probably closer.

However, I suspect OPs main question is about how the gravity of an object moving at the speed of light could possibly catch up to another such travelling away from it.

14. Aug 17, 2014

### Chronos

If you assume a gravitational field is continuously generated by a photon during its journey, there should be some kind of interaction between passing photons, however weak they may be. The stress energy tensor does not recognize any difference between a massless and massive particle.

15. Aug 17, 2014

### Simon Bridge

If you treat a photon as a classical particle, yes. Though, is "generated" how gravitational fields are usually thought of in GR? What does the gravity of a rapidly moving particle look like in GR?

The original question may be best answered by considering a short wave-packet or pulse rather than a photon.
I think Dave's linked article does that.

16. Aug 17, 2014

### pervect

Staff Emeritus
Last edited: Aug 17, 2014
17. Aug 17, 2014

### WhatIsGravity

Aren't photons, by definition, kinda off the grid so far as calculations go involving time?

18. Aug 17, 2014

### Staff: Mentor

In my more curmudgeonly moments, I've been tempted to suggest that the word "photon" should be banned from the relativity forum. Whenever anyone poses a question about "a passing photon", they could be gently corrected into speaking in terms of a flash of light, or as Simon suggests, a pulse or a short wave packet.

19. Aug 17, 2014

### Sting33

I think this is what I was hypothesizing David. Thank you for that paper at caltech.

Based on my limited understanding, I think what I understand is that parallel pulses of light will not exert gravitational influence on each other. In basic terms I think that this is because the gravitational wave, if I may call it that, never can reach the light pulse because it just never can catch up. But if they are not parallel, mathematically at some point some part of the gravitational wave will intersect with the other light pulse's position, at least assuming the pulses have been traveling forever. (Edit: I just thought some more about this and i don't think this explanation mathematically makes sense in many cases)

I would think though that parallel light pulses traveling towards each other would exert a gravitational force on each other.

The article made me think of another question though, and maybe someone can set me on the right path. It comes from a modification of the old question about what happens if you have a spaceship traveling at near the speed of light and you turn on the headlights.

Correct me if I am wrong, but if such a spaceship were coming directly towards you, the observer would "see" the light as gamma rays due to a significant blueshift.

Now my thought, which probably is just going to further reveal my ignorance of GR, is:

If a light pulse is traveling directly at you, what is the gravitational force from that pulse like when it gets to you? Would it create some sort of "shock wave" gravitational field from the accumulation of the gravitational fields during its travel with a corresponding gravitational vacuum behind it? This doesn't seem right to me, but it can't think of any other way to think about it.

Thank you all for considering my questions even when I show my ignorance and mix up models. I really appreciate you all indulging my curiosity. You have taught me something even though I realized soon after posting I was in over my head.

Last edited: Aug 17, 2014