# Speed of light in acclerated frame

1. Mar 28, 2013

### DiracPool

I've been hunting around online but have found only long-winded explanations of the speed of light in non-intertial or accelerated reference frames, which mostly relate to gravity. I'm looking for just a straightforword treatment dealing more specifically with traveling acceleration and not necessarily gravity.

My understanding is that if I'm traveling at a constant velocity .3c relative to my friend Al and I shine a flashlight, I will see the light move away from me at c. Al will also see the light travel at c, and will see me traveling at .3c in the direction of the light.

Here's my question...If I suddenly accelerate from .3c to .5c over a ten second period, and then proceed from thereon at .5c, what is the experience of me and Al? Does either of us measure the speed of light to change in this situation, that is, during the 10 second period of acceleration? If so, why and how? If not, why not and how not?

2. Mar 28, 2013

### Staff: Mentor

How are you doing the measuring while under acceleration?

3. Mar 28, 2013

### DiracPool

Hmm, I thought this question had already been thought out somewhere and there was some precedent manner in which to meaure this, so I'm not exactly sure.

However, for argument's sake, let's say I have a Michelson and Morley interferometer with me and point it in the direction of the acceleration for the first 5 seconds, and then point it at 45 degrees to the acceleration for the next 3 seconds, and then 90 degrees to the direction of acceleration for the last 2 seconds. How would the interference patterns look relative to those in my steady (constant velocity) rate of motion at .3c and .5c.

Also, say I had a light clock with me that oscillates at 1 cycle per second. Would that change during the acceleration epoch?

4. Mar 29, 2013

### 1977ub

Event A emission and event B detection of light of light. In IRF you measure the times and distances of events and end up with c. I take it that under acceleration there is ambiguity for you regarding times / distances of A & B since you don't have an IRF.

5. Mar 29, 2013

### Alain2.7183

6. Mar 29, 2013

### 1977ub

Is this in any way problematic?

7. Mar 29, 2013

### PAllen

Apparent faster than c motion of distant objects or light for an accelerating observer is not surprising at all if you look at it in a certain way. Consider a different type of non-inertial motion: turning your head suddenly. In your 'head' rotating frame, distant mountains appear to move faster than c. In spacetime, acceleration in one direction is represented as hyperbolic rotation - with similar arbitrary apparent speed effects as ordinary rotation.

Note that 'faster than c' should not be taken as superluminal. Superluminal would be a material body passing light. Though both bodies and light may appear to move > c for an accelerating observer, a body overtaking light is never seen. So all motion of matter is still observed to be subluminal, but could be > c.

8. Mar 30, 2013

### Staff: Mentor

OK, I didn't know the answer to this question so I didn't respond until I could work it out for myself. I will use units such that c=1 and such that the proper acceleration of the place where the two arms of the accelerometer meet is a=1. The proper length of each arm of the interferometer is L, and the interferometer is undergoing Born rigid acceleration (http://www.mathpages.com/home/kmath422/kmath422.htm)

In those units, the worldline of the center of the accelerometer is given by $(\sinh(\tau_c),\cosh(\tau_c),0,0)$, the worldline of the forward mirror is $((1+L)\sinh(\tau_f/(1+L)),(1+L)\cosh(\tau_f/(1+L)),0,0)$, and the worldline of the side mirror is $(\sinh(\tau_s),\cosh(\tau_s),L,0)$.

First, I set $\tau_c=0$ and solved for $\tau_s$ and $\tau_f$ such that the spactime interval was 0. Each had one negative and one positive root, so I discarded the negative root. Then I used those roots and solved twice for $\tau_c$ such that the spacetime interval was 0. Each one had one 0 root and one positive root, so I discarded the 0 root.

The light pulse which left the center at $\tau_c=0$ and traveled along the forward arm returned back at $$\tau_c=\ln\left( (1+L)^2 \right)$$

The light pulse which left the center at $\tau_c=0$ and traveled along the sideways arm returned back at $$\tau_c=\ln\left( \frac{2+L^2+L\sqrt{4+L^2}}{2+L^2-L\sqrt{4+L^2}} \right)$$

These times are not the same, so there are interference fringes. Furthermore, dividing these times by 2L gives us the measured speed of light. This is smaller than c for both the forward and sideways arms.

EDIT: please see below. The measured speed of light is greater than c. I mistakenly calculated τ/(2L) instead of 2L/τ here.

Last edited: Mar 31, 2013
9. Mar 31, 2013

### DiracPool

Thank you so much for taking the time to work this out. So, are you saying that measured speeds of light in the arm facing the forward direction and side direction are different, and each one shows a speed less than c? Which arm had the slower speed of the two? (Unfortunately I just recently lost my hyperbolic slide rule, so my calculating ability is currently compromised)

10. Mar 31, 2013

### pervect

Staff Emeritus
If we look at
τc=ln((1+L)2)

in a series expansion that's 2L - L^2 + (2/3)L^3

So for small L, tau is less than 2L, meaning a speed faster than C in the vertical direction.

My understanding (which might be flawed) is that all Born rigid flows are generated by Killing vector flows

http://arxiv.org/pdf/1004.1935.pdf
This immediately suggests that Fermi coordinates which have a time like Killing field would be the "natural" coordinates for Born rigid motion.

So in particular I'd expect that since rindler coordinates with the metric have a time-like Killing vector (as is obvious from the lack of any dependence of coordinates on t)

(1+gz)^2 dt^2 - dx^2 - dy^2 - dz^2

that they would represent Born rigid coordinates.

This would lead to the expectation hat the speed of light speeds up as you get "higher" in the accelerating frame, in particular at gz=2 the speed of light is doubled for where it is at gz=1, with the Rindler horizion being at z=0

It would also yield a prediction that the speed of light horizontally would be approximately constant for small L

Indeed, for the second expression, I get a formula like

2L - (1/12)L^3 for the second expression, which is also faster than C but the second order correction term is zero here.

11. Mar 31, 2013

### Staff: Mentor

D'oh

You are correct. I mistakenly calculated the speed of light as τ/(2L) instead of 2L/τ.

12. Mar 31, 2013

### Staff: Mentor

The forward arm was more affected than the sideways arm, although I got the direction wrong above.