High School Light Beams Attraction: e=mc^2 & Asymmetry

[c704710]

Appearently, two beams of light in a vacuum are attracted to each other. Photons have no Newtonian mass, but their relativistic mass (e=mc^2 or mass-energy equivalence) causes this (as is my understanding). So will a 10^4 Hz beam bend more towards a 10^20 Hz beam than the 10^20 Hz bends towards the 10^4 Hz beam? After all, e=mc^2 indicates the 10^20 Hz beam has more relativistic mass doesn't it? However, if they equally bend, why?

 

[PAllen]

You are applying special relativity concepts to GR. In GR, the problem you pose is not simple. For example, are you aware that two beams of light traveling in the same direction do not attract, while if traveling in opposite directions they do? Note that relativistic mass is a dubious construct in SR, and has no relevance whatsoever for gravity (GR).

Published research I've seen using GR uses identical beams. However, one thing to note is that there is always a frame of reference where the beams have equal energy density. One may then state that deflection is equal such coordinates.

 

[PeterDonis]

two beams of light in a vacuum are attracted to each other

Only if they are not moving parallel to each other. See below.

Photons have no Newtonian mass, but their relativistic mass (e=mc^2 or mass-energy equivalence) causes this (as is my understanding)

The source of gravity in GR is the stress-energy tensor. This includes energy (what you are calling "relativistic mass", but it's better to just call it energy), but also pressure (and other stresses, but pressure is the relevant one here).

It turns out that, for light beams moving parallel to each other, the effects of energy and pressure exactly cancel, so the beams don't attract each other at all. For light beams moving antiparallel, the effects of energy and pressure add, and the actual attraction between them turns out to be four times what you would expect based on the energy alone.

will a 10^4 Hz beam bend more towards a 10^20 Hz beam than the 10^20 Hz bends towards the 10^4 Hz beam?

The attraction (for beams not parallel to each other--see above) is proportional to the product of the beam energies (and pressures, see above), so it's not really a question of which one "bends more".

 

[c704710]

Thank you for the information concerning the parallelism of the beams. I was unaware of that and the 4 times effect as well. So I _think_ my question was answered. The result is a shared effect of the beams rather the each beam effecting the other? Therefore symmetrical regardless of asymmetric frequencies?

This means a 14 TeV laser beam can be diverted with a pocket laser!?

 

[PeterDonis]

Therefore symmetrical

What does "symmetrical" mean? I think you will find that you are implicitly adopting a particular coordinate system.

 

[PAllen]

Thank you for the information concerning the parallelism of the beams. I was unaware of that and the 4 times effect as well. So I _think_ my question was answered. The result is a shared effect of the beams rather the each beam effecting the other? Therefore symmetrical regardless of asymmetric frequencies?

This means a 14 TeV laser beam can be diverted with a pocket laser!?

Note that in a frame moving near c in the direction of powerful laser, the two beams have equal energy due to Doppler. The description of deflection in these coordinates is symmetric. Then, a Lorentz transform gets you the description in any other frame.

Specifically, if the power ratio in one frame is r > 1, then the fraction of lightspeed needed to equalize the power is (r-1)/(r+1) in the direction of the more powerful beam.