# Traveling parallel to a ray of light through a gravitational field

1. Jan 18, 2012

### mrspeedybob

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. Jan 18, 2012

### RiddlerA

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

3. Jan 18, 2012

### Drakkith

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

4. Jan 18, 2012

### Bill_K

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. Jan 18, 2012

### Drakkith

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

6. Jan 18, 2012

### Staff: Mentor

The spacetime describing this situation is the Aichelburg-Sexl ultraboost:
http://en.wikipedia.org/wiki/Aichelburg–Sexl_ultraboost

I don't know the answer, but this is how you would find it.

7. Jan 18, 2012

### elfmotat

The four-force is given by $F^\mu = m\frac{DU^\mu }{d\tau}=m(\frac{d^2x^\mu }{d\tau ^2}+\Gamma^\mu_{~\alpha \beta }U^\alpha U^\beta)$.

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

8. Jan 18, 2012

### Drakkith

Staff Emeritus
Ah ok, got it. Thanks!

9. Jan 19, 2012

### pervect

Staff Emeritus
To a stationary observer, as your velocity approaches 'c', your path will approach the path of a light beam.

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 Aichelburg-Sexl 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 Aichelberg-Sexyl 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 parallel-moving 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".