# Plausibility arguments for noninteraction of parallel beams of light

1. Jul 15, 2011

### bcrowell

Staff Emeritus
There is a classic result by Tolman, Ehrenfest, and Podolsky that in classical GR, the interaction between pencil beams of light is zero in the case where the beams are parallel, nonzero otherwise. (The case where they're not parallel can always be reduced to a case where they're antiparallel by a change of frames.) The original paper is
Tolman, R.C., Ehrenfest, P., and Podolsky, B. Phys. Rev. (1931) 37, 602,
which I don't have access to, but there's a freely available presentation here:
Mitskievich, "Gravitational interaction for light-like motion in classical and quantum theory," http://arxiv.org/abs/1007.2589
This type of gravitational interaction of light with light was the dominant form of gravity in the early universe, which was radiation-dominated. In the lab, it's too small to measure, and the dominant mechanism for photon-photon interaction is represented by a Feynman diagram with a fermion box in it.

What do people here think of the following plausibility arguments for the vanishing of the interaction in the parallel case? If there are flaws in them, can the flaws be patched up?

1. If the interaction didn't vanish, then the two beams would oscillate back and forth across each other or twine in a double helix or something like that. That would be a clock made out of photons, which is impossible.

2. The three-force of a beam with energy E1 acting on a beam with energy E2 is proportional to E1E2 in the weak-field limit, where gravity is linear. By doing Lorentz boosts to chase the beams, you can Doppler shift them so as to make E1E2 shrink by a factor of D2, where D is the Doppler shift factor. The inertia of E2 only shrinks by D, so its acceleration shrinks by D, and the beams take longer to collide. The time to collision can be made as long as desired. Transforming back into the original frame, this becomes an even longer time. Since there is no limit on D, the time to collision must be infinite.

3. Since the system's total energy and momentum E and p are conserved, the equivalent rest-mass $m=\sqrt{E^2-p^2}$ of the whole system is conserved as well. Initially m=0. If the beams were deflected and became non-parallel, they would then have an equivalent rest-mass that was nonzero, violating conservation of energy-momentum.

2. Jul 15, 2011

### Staff: Mentor

I am not sure why you say that is impossible. It certainly isn't possible for a single photon, but why not for several?

I like this one. Also, you can always boost to a frame where the energy is arbitrarily low, so the interaction is arbitrarily small.

This wouldn't be a static spacetime would it? I am not sure that energy-momentum is conserved.

3. Jul 15, 2011

### bcrowell

Staff Emeritus

Hmm... well, I'll give you a strong argument that it's not possible for a fleet of photons flying in formation: their center of mass frame moves at c, so the time dilation factor is infinite. This is actually the only case that is needed here.

I think it also holds for any collection of photons, but I don't know how to prove it. Maybe someone here who is more comfortable with conformal invariance, etc., could explain it. The reason I'm pretty sure it's true for any collection of photons is that that's a central idea of Penrose's conformal cyclic cosmology.

Another good point. It looks like the Tolman-Ehrenfest-Podolsky metric is general enough to include beams with both finite and infinite lengths. I think my argument works in the finite-length case, since then the spacetime is asymptotically flat, and we have a conserved energy-momentum for asymptotically flat spacetimes.

4. Jul 15, 2011

### pervect

Staff Emeritus
Well, I may be a bit ambitious, but I'd like the audience to come away with the notion that there are electric and magnetic components of gravity, by analogy to electromagnetism. (I'm not ambitious enough to motivate the topological component, at least not for light. It mighyt be worth motivating this for a stationary body though).

You can motivate the electric component of the gravity by pointing out that light bends a gravitational field, and pointing out that the momentum can be defined (in a suitable asymptotically flat sense) and that it changes, suggesting that the light must change the momentum of the gravitating body, just as the gravitating body alters the path of the light.

This explains how light interacts with a stationary gravitating body.

Getting the result that the magnetic force exactly cancels the electric force when the beams are moving in the same direction is tricker. Directly addressing how light interacts with a velocity moving with some velocity 'v' might be possible, we'd expect the result that it should bend less if the body moves in the same direction as the light beam, and more if it moves in the opposite direction. I haven't calculated this, and I'm not sure if it really makes a good motivator, assuming it works like I think it should (though I don't see how it could avoid it, occasionally I"m surprised).

Perhaps the angle is the same, and it's just the red-shifiting of the light that's significant - it would have less momentum. Something interesting to calculate.....

It might be best to refer them to the literature first.

Another possibility is to take the limiting case of two low-rest mass particles being boosted up to 'c', I suppose.

Last edited: Jul 15, 2011
5. Jul 15, 2011

### Staff: Mentor

Sure, the clock doesn't have a rest frame, but it seems to me that by your logic you could say that a pulse of light can't have a frequency because then it could be used as a clock. The electric field cannot oscillate, etc. As long as the photon "clock" beats with a frequency proportional to its energy I don't think it violates anything.

At least, it is not obvious enough for me to use it as a plausability argument.

6. Jul 15, 2011

### PAllen

I find the following interesting:

The invariant mass of two or more parallel photons is zero. The invariant mass of photons moving any other way is nonzero.

Another observation: in the limit as rest mass goes to zero, for parallel particles, the energy in the center of momentum frame goes to zero, thus curvature goes to zero. In the limit as mass goes to zero for anti-parallel particles (keeping energy constant in some starting frame), the energy in the center of momentum frame does not go to zero, and curvature produced by the system does not change in this limiting process.

7. Jul 15, 2011

### bcrowell

Staff Emeritus
You can use a light wave's oscillation as a frequency in a particular frame of reference. What you can't do is go into the light wave's frame of reference (which doesn't exist) and use the wave's oscillation as a clock. A couple of other ways of putting this: (1) a clock should be an object that can measure time in its own rest frame, i.e., proper time; (2) a light wave isn't a system whose internal state can change over time.

8. Jul 15, 2011

### Staff: Mentor

By that definition plausibility argument 1 is not correct. The helical or oscillating light beams do not measure time in their rest frame since it does not exist and therefore they are not a clock. The fact that they twist in our frame doesn't imply that they do so in their frame.

I would stick with argument 2.

9. Jul 16, 2011

### pervect

Staff Emeritus
I think I've got a bit better handle on my argument, but I'm not quite sure where it's leading yet. It might show if the approach of using a conserved momentum and "backreaction" is a good one, or leads to confusion, though. At the moment, I think it's leading towards confusion. Which is dissapointing, but maybe we can learn something from the confusion, or someone can point out what I did wrong.

Consider a light beam, and it's effect on a stationary large mass first. We know that the light beam deflects, and our plausibilty argument says there is a conserved momentum (we'd need asymptotic flatness or somesuch to make it rigorous), so we expect an equal and opposite reaction on the large mass.

Now, what happens if our large mass is moving in the same direction as the beam? We transform to a coordinate system co-moving with the planet (a rather suspicious step since space is really curved), and the only difference is that the beam is redshifted, by the factor sqrt(1-beta)/sqrt(1+beta). The beam is deflected by the exact same amount, it's the same geometry, just a weaker beam.

We therefore conclude that dp/dtau , the four-force, is lower on the planet moving with the beam, and approaches zero as beta goes to infinity. So far, so good.

However, dp/dt appears to approach a steady limit, dp/dtau * dtau/dt = dp/dtau * gamma = sqrt(1-beta)/sqrt(1+beta)* 1/sqrt(1-beta^2) = 1 / 1 + beta

This isn't quite where we wanted to be, it seems to suggest that gravity caused by light dp/dt doesn't approach zero.

So, I suspect there's something fundamentally wrong, but I can't say exactly what, though I suspect the issue is one of ignoring the contribution of spatial curvature, which wouldn't transform like a force does, and that the notion of gravity as being a force that transforms like a four-vector is supsect.

Maybe this is also concluding too much, but it seems like I always run into these sorts of issues if I start thinking of gravity as a "force" too seriously. And if this is true, we don't want to encourage students to think of it that way - they'll probably continue to do it anyway, but we don't have to encourage them if we know it'll lead to problems further down the road.

Last edited: Jul 16, 2011
10. Jul 16, 2011

### bcrowell

Staff Emeritus
If the beams were initially parallel and then became non-parallel, then they would have no rest frame while parallel, but would have a rest frame later, while non-parallel.

Obviously I've made a botch of presenting argument 1, but I'm pretty sure it's correct. Let me see if I can get my conformal-invariance ducks in a row and come back with a better version of that argument.

11. Jul 16, 2011

### Staff: Mentor

Yes, this gets back to your 3rd argument, which I buy now.

Sure, I will be glad to re-visit it, but how many plausibility arguments do you need? You already have two good ones (IMO). I would think all you really need is one, and I would recommend you use the 2nd.

12. Jul 16, 2011

### bcrowell

Staff Emeritus
I just like to have more than one way of understanding something.

13. Jul 16, 2011

### martinbn

14. Jul 16, 2011

### pervect

Staff Emeritus
You can find Tollman's expression for the pencil of light in his book, "Relativity, Cosmology, and Thermodynamics", which might be easier to get a hold of than the original paper.

It's a weak field expression for a finite pencil, and it's pretty simple

ds^2 = (1+ h11) dt^2 - (1-h11)dx^2 - dy^2 - dz^2 - 2 h11 dx dt

where h11 is a function of x,y,z, and is expressed in terms of an intergal of the relativistic mass (!) of the light beam.

$$h11 = 4 \int \frac{\rho dV}{r}$$

I could copy down the exact form of h11(x,y,z) if there's enough interest, though I suspect it'll be easy enough to get a copy of the book, and you can do a lot of analysis without knowing more than the fact that the metric is of the above form.

It's not a function of time because while Tollman assumed a pencil of light of finite length, he assumed a steady pencil. So he's assuming the length is long enough that l doesn't matter to the field, while also assuming that l is finite to keep the integral from diverging.

I'll probably bang this around and look at the Riemann a bit. Tollman works out the acceleration of a co-moving pencil of light just using the metric and algebra, by the way, just by solving for ds=0, and uses this algebraic argument to show that the light moving near the pencil isn't attracted.

15. Jul 16, 2011

### bcrowell

Staff Emeritus
16. Jul 16, 2011

### bcrowell

Staff Emeritus
Ah, very cool -- thanks!

One thing about the paper doesn't make sense to me. Case #1:
Case #2:
When they talk about the factors of 2 being satisfactory, what they mean, as they explain later on, is that they're comparing with the result for deflection of light in a Schwarzschild field such as the sun's. Although they treat the two cases in the same way in the abstract, later they only emphasize the comparison in case #2. To me, the comparison in case #1 seems wrong. The integrated deflection angle in case #2 is unambiguously defined, because the spacetime is asymptotically flat; this is the same as in deflection of light rays by the sun. But in case #1, they're talking about an instantaneous acceleration, which can have any value you like depending on your coordinates. In the Schwarzschild case, you can talk about the acceleration of a test particle as measured by a static observer. But in the case they're talking about, the spacetime isn't static, so there is no way to define static observers.

17. Jul 16, 2011

### pervect

Staff Emeritus
We can get considerable simplification in the metric if we rewrite it as:

$$\left( \left(1 + \frac{h11}{2} \right) dt - \frac{h11}{2} dx \right)^2 - \left( \left(1 - \frac{h11}{2} \right) dx + \frac{h11}{2} dt \right)^2 - dy^2 - dz^2$$

Using the obvious choice of orthonormal basis vectors, we can then find

$$R_{\hat{y}\hat{t}\hat{y}\hat{t}} = R_{\hat{y}\hat{t}\hat{y}\hat{x}} = R_{\hat{y}\hat{x}\hat{y}\hat{x}} = \frac{1}{2} \frac{\partial^{2} h11}{\partial y^{2}}$$

If we calculate the relative acceleration of the geodesics in the y direction, assuming that the 4-velocity is gamma, 0, -beta gamma, 0 , (and I'm not quite sure where the minus sign came from, we get

$$R_{\hat{y}\hat{t}\hat{y}\hat{t}} \gamma^2 + -(R_{\hat{y}\hat{t}\hat{y}\hat{x}} + R_{\hat{y}\hat{x}\hat{y}\hat{t}} ) \beta \gamma^2 +R_{\hat{y}\hat{x}\hat{y}\hat{x}} \beta^2 \gamma^2$$

This cancels out in one direction as expected, and adds in the other.

If we think of the geodesic deviations as being due to forces, we see we have a velocity-independent force, which we can think of as static gravity, which should be attractive, a velocity dependent force, which switches sign depending on our velocity, attractive in one direction and repulsive in the other, and another attractive force, due to the spatial curvature, proportional to velocity^2. And the net sum is zero when the magnetic force opposes the other two.

Because the terms are all equal, in one direction everything cancels, in the other direction it all adds together,the magnetic term is the only one that flips sign, and it's equal to the sum of the other two.

So the sum turns out to be zero as we expect, but we need to include the spatial curvature "forces".

To get something that more closely represents the usual notion of force, we need to integrate the geodesic deviation over some y-directed path to infinity, the geodesic deviation is really a "force gradient". I can't guarantee path independence,but if we choose the "obvious" path we'll get something like $\partial h11 / \partial y$,the metric coefficeint for dy is always unity. if we do this, reinforcing the notion of h as a sort of "potential".

One more thing, the expression for h11 (I put it into Maple, which latexified it for me, though sometimes it does an ugly job.

$$\frac{h11}{4 \rho} = \ln \left( {\frac {\sqrt { \left( l-x \right) ^{2}+{y}^{2}+{z}^{2}}+l -x}{\sqrt {{x}^{2}+{y}^{2}+{z}^{2}}-x}} \right)$$

also , at x=0 and z=0 and rho=1/4
$$\frac{\partial h11}{\partial y} = \frac{l}{\sqrt{l^2 + y^2} y}$$

So as a candidate for something proportional to the force (after we integrate the tidal force over some path), it's reasonable, it falls off as 1/y close to the pencil, and 1/y^2 far away. But the region of interest will be y<<l. Because it's essentially the Newtonian line-mass integral, we would expect 1/y for the force.

The second derivative was messy, but the series expansion was reasonable. Again, at x=z=0 and rho=1/4

$$\frac{\partial^2 h11}{\partial y^2} \approx ({y}^{-2}+1/2\,{l}^{-2}-{\frac {9}{8}}\,{l}^{-4}{y}^{2}+O \left( {y}^{ 4} \right) )$$

Last edited: Jul 16, 2011
18. Jul 16, 2011

### pervect

Staff Emeritus
Aha - this is the problem. There isn't any square root, the stress energy tensor is multiplied by a factor of (1-beta)^2/(1-beta^2) = (1-beta) / (1+ beta)

The doppler shift is only part of the energy density reduction, there's also a reduction in particle flux.

Now this argument does show zero dp/dt on a body comoving with the beam, the beam gets redshifted by a factor of (1-beta)/(1+beta) in the frame of the moving body

we multiply by gamma to convert dp/dtau into dp/dt, which is 1/sqrt(1-beta^2) and we get a limit of zero as beta->1.

Last edited: Jul 16, 2011
19. Jul 17, 2011

### yuiop

I think this third option is closest to the most plausible argument at least from a layperson's point of view. A system of photons that are all moving parallel to each other has no rest mass and hence no gravitational field. A system of photons that are not moving exactly parallel to each other has a centre of momentum frame and a collective rest mass giving rise to gravitational interactions.