Traveling parallel to a ray of light through a gravitational field...by mrspeedybob Tags: field, gravitational, light, parallel, traveling 

#1
Jan1812, 05:12 AM

P: 686

I was thinking yesterday about a scenario where a ray of light passes a massive body and is deflected. If I were in a rocket moving at the speed of light along the same trajectory, I should follow the same path through the gravitational field and so should observe the ray of light to be traveling straight, except for the fact that traveling at C is a logical contradiction. So what happens in the limit as my velocity is set arbitrarily close to C?
One line of thought says that as V approaches C the trajectory of the light should approach a straight line in my reference frame. Another says that the trajectories will never be identical and that as V approaches C Length contraction will make the rays curve sharper. So, As V approaches C in this scenario do I observe the lights path to be straighter, more curved, or do the effects cancel and I see the same curvature at any speed? 



#2
Jan1812, 05:39 AM

P: 58

My guess is that the trajectory would approach a straight line......




#3
Jan1812, 05:47 AM

PF Gold
P: 11,044

If you are traveling an identical trajectory, would you only see it as it is normally except for length contraction?




#4
Jan1812, 05:57 AM

Sci Advisor
Thanks
P: 3,853

Traveling parallel to a ray of light through a gravitational field...
There are no global reference frames in a curved spacetime, only local ones. The light ray follows a geodesic, which is locally a straight line. The curvature of its path becomes apparent only when you consider the reference frame at infinity, and notice that it came out at a different angle than it went in. A freely falling particle traveling at v ≈ c will follow nearly the same path, and the same remarks apply: locally the trajectory appears straight, ultimately at infinity the path has been deflected.




#5
Jan1812, 05:08 PM

PF Gold
P: 11,044

Bill, quick question, are all inertial objects (those not under an acceleration) following geodesics?




#6
Jan1812, 05:44 PM

Mentor
P: 16,477

http://en.wikipedia.org/wiki/Aichelb...exl_ultraboost I don't know the answer, but this is how you would find it. 



#7
Jan1812, 05:53 PM

P: 260

It's easy to see that this reduces to the geodesic equation if F^{μ}=0 (i.e. there are no external forces): [itex]\frac{d^2x^\mu }{d\tau ^2}+\Gamma^\mu_{~\alpha \beta }\frac{dx^\alpha}{d\tau} \frac{dx^\beta}{d\tau}=0[/itex] 



#9
Jan1912, 12:58 AM

Emeritus
Sci Advisor
P: 7,439

Of course, no matter how fast you go, the light beam will appear to travel at 'c' relative to yourself. So while the static observer will see your trajectory as being pretty much the same as the light beam, you will still see the light beam as moving much faster. In your own frame, the metric is described (in the limit) by the AichelburgSexl ultraboost, as some posters have already mentioned. This is the metric of a gravitational plane wave. You can't easily measure gravitational forces, but you can easily measure the rate of change of gravitational forces  i.e. tidal forces. The geodesic deviation equation relates these tidal forces to change in separation from you to a nearby observer moving in a parallel trajectory. Because the AichelbergSexyl solution is a plane ave, most of the time you won't see any change in distance or relative forces (tidal forces) between you and any parallelmoving observer. But there will be a planar surface where you see an impulsive force, a very high magnitude and short duration force (like a bat hitting a baseball) , which will cause a sudden, impulse change between your trajectory and the trajectory of some observer who was previously "moving in parallel". 


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