# Gravitational effects on photons

1. Oct 8, 2009

### jnorman

I was just re-reading some of Hawking's writings, and apparently he is fairly stong supporter of a quantum theory of gravity, as opposed to GR, and writes extensively about gravitons. this brought up a question for me which perhaps one of you can help clarify.

classical gravity indicates that only mass and distance control gravitational attraction, and treats it as a spooky force but does not describe any mechanism. GR indicates that gravity is not a force, but a reflection of the curvature of spacetime created by the presence of mass. quantum gravitational theory would indicate that gravity results from the interaction of two masses via particle exchange (gravitons), and is supported by the search for GUTs (i think).

however, the question that popped into my head is with regard to gravitational effects on photons. experiment has shown clearly that light is effected by gravity, as in lensing of light around massive stars. in GR this is accepted as light follows the geodesics created by the curved spacetime - no particle interaction required.

so, in a quantum theory of gravity, where the interaction is the result of graviton exchange between the two bodies/particles - how would gravitons interact with individual photons, which as i understand, have no specific location between the time they are emitted and absorbed? and photons, being massless, would not be emitters of gravitons themselves, and would not generate gravitational effects - which is not consistent with experiments which have demonstrated that beams of light do indeed affect each other via gravitational attraction (correct?).

i know i am a dunce here, so i hope my question is not too confused...

2. Oct 8, 2009

### Naty1

General relativity shows that gravity results from not only mass but energy and pressure. Since photons carry energy, they exhibit gravitational attraction....as far as is known, everything is affected by gravity....all particles, regardless of characteristics, even time and space.

don't know what that means...the speed of light is finite "c"....

3. Oct 9, 2009

### yuiop

This is a quote by Chris Hillman from an old thread. ".. it is possible to construct models showing that two parallel light beams will not attract each other, but two anti-parallel beams will attract one another." See https://www.physicsforums.com/showthread.php?t=154391&page=2

I have read similar statements on several occasions, so maybe there is something in it.

It is also frequently mentioned in cosmology that radiation has a gravitational effect which is greater (for a given energy density) than that of matter.

There is no generally accepted theory for gravity mediated by gravitons, so if you are having problems with how gravitons work, you are in good company as most physicists have problems with them too.

Another thing to consider is what would happen if anti matter was added to a black hole so that all the matter inside the black hole turned into pure radiation. I am pretty sure it would violate some rule of GR if the black hole was destroyed.

4. Oct 9, 2009

### FunkyDwarf

surely you mean energy and momentum?

5. Oct 9, 2009

### Naty1

Last edited: Oct 9, 2009
6. Oct 9, 2009

### granpa

if nothing else, the light will be bent by the gravitational time dilation.

7. Oct 10, 2009

### stevebd1

It's also interesting that gravity's effect on a photon moving tangential to an object of mass (as in the case of gravitational lensing) is twice as much as that of a photon moving radially towards the mass. For gravitational lensing, the angle of deflection is $\theta=4Gm/rc^2$ while the dilation on a radial path is half as much- $dt'=dt\ 2Gm/rc^2$.

Source-
https://www.worldscientific.com/phy_etextbook/6833/6833_02.pdf [Broken]

Last edited by a moderator: May 4, 2017
8. Oct 10, 2009

### Jonathan Scott

This does not sound like a meaningful comparison. The URL does not work for me, so I can't check the source.

In an isotropic metric around a static source in the weak field case, the simplest way to look at gravity is in terms of the effect it is has on the momentum of a test particle (massive or massless), Ev/c2, or, dividing by the total energy (which is constant for free fall in a static field), on v/c2. The field is represented as a Newtonian acceleration g:

$$\frac{d\mathbf{p}}{dt} = \frac{E}{c^2} \, \mathbf{g} \left ( 1 + \frac{v^2}{c^2} \right )$$

This expression holds regardless of the direction of the velocity. It says that the rate of change of momentum is in the direction of the Newtonian acceleration and that its magnitude depends on the speed, but not on the direction. For a photon, the rate of change of momentum is exactly twice that for an object at rest, regardless of its direction.

Last edited by a moderator: May 4, 2017
9. Oct 10, 2009

### stevebd1

Last edited by a moderator: Apr 24, 2017
10. Oct 10, 2009

### Jonathan Scott

Hmmm. I was able to access that one by overriding a security warning for a site name.

Lev Okun is normally better than that. He seems to have had a very odd way of thinking about mass.

I think that in this case, his description is inconsistent, or at least confusing; he talks about "force" (which is the rate of change of momentum, Ev/c2) but then about the deflection of light, which is related to velocity. As the coordinate value of c varies relative to any flat background metric, he needs to be more careful in distinguishing between quantities which differ in units by factors of c.

For the horizontal case at a tangential point, the potential is constant so c is constant, so the momentum is proportional to the velocity as in Newtonian theory and the only part of the momentum Ev/c2 which varies is the direction of v, which turns twice as fast as in the Newtonian case.

For the vertical fall case, the total energy is constant and the velocity is the coordinate value of c, so the momentum is simply E/c in the vertical direction, and varies with the coordinate value of 1/c. As the fractional variation of c is twice the Newtonian potential (because clock rates and ruler sizes both vary with potential), we again get the same result that the downwards momentum increases twice as fast as in Newtonian theory.

(Perhaps I should really use primes or something like that to denote coordinate values of c, because that's normally assumed to denote the standard local value, but since one doesn't usually use primes for the coordinate system, I think it's more consistent to use c for the coordinate value and some other notation for the standard local value).

I should point out that the concept of a "coordinate value of c" only applies in an isotropic coordinate system, and similarly the idea of a coordinate value of the mass also only applies in that case, in that it is given by the rest energy divided by c2. If the metric is not isotropic, the coordinate value of c has different values in different directions and is in general a tensor rather than a scalar.

Last edited by a moderator: Apr 24, 2017
11. Oct 10, 2009

### Austin0

Does this mean that in GR, where I assume that the metric is not isotropic that c would have different values radiating outward than inward, relative to the center of gravity?

12. Oct 10, 2009

### Jonathan Scott

If the metric is at least static, then the coordinate c is the same both ways along any particular line.