# Light speed relative to accelerating & const vel source

1. Aug 22, 2008

### brightonb

Two observers, A & B, are moving apart at constant velocity V. At distance D, B sends a pulse to A which arrives T seconds later. If B were instead accelerating at a rate such that he attains velocity V just when he is at distance D, and sends a pulse at this instant (while still accelerating but at velocity V), would the pulse still arrive T seconds later?

Thanks for any responses, Manny

2. Jul 16, 2009

### JCOX

Let me start by saying I am not a physics major and am starting to try and learn this stuff so I would appreciate any skepticism in my response.

You canâ€™t say T sec. later. You are using T as a time without any particular Frame of reference. T for the B will not be T for A. Mainly due to the fact that B is accelerating.

3. Jul 16, 2009

### vin300

Yes, definitely, the time depends only on the distance.

4. Jul 16, 2009

### JCOX

Ah ... are you taking into account G.R... I mean time is not the same for someone accelerating compared to someone not accelerating ... Am I correct??

5. Jul 16, 2009

### JCOX

What I mean is WHO is holding the clock ... you?

6. Jul 16, 2009

### JesseM

The question is not really well-defined unless you specify what frame (coordinate system) you're using to define time and distance. If you are using A's inertial rest frame, for example, then it would make no difference whether B was accelerating when he emitted the signal, for example. But if you use an accelerating frame in which B is at rest, the answer will probably be different than it would have been if B had been moving inertially and you used B's inertial rest frame.

7. Jul 16, 2009

### vin300

The time measured by A would be the same but to find that measured in any other frame in relative motion divide the time measured by A bygamma, that is 1/sqrt.(1-v^2/c^2) v is the relative velocity that gives a lesser value

8. Jul 16, 2009

### JesseM

That's not right, the time dilation formula only works for events which occur at the same location in frame where they are separated by a time T (for example, ticks of a clock which is at rest in that frame), then in another frame the same pair of events will be separated by a larger time T/sqrt(1 - v^2/c^2). In this example we are dealing with the time between the signal being emitted by B and the signal being received by A, which don't occur at the same location in A's frame; if the time between them in A's frame is T and the distance between them in A's frame is X then you would use the temporal part of the Lorentz contraction equation, T' = (T - vX/c^2)/sqrt(1 - v^2/c^2), to find the time T' between the events in some other inertial frame moving at speed v relative to A's frame.

9. Jul 19, 2009

### brightonb

To clarify this problem, what I am actually asking is whether a signal from 2 transmitters will arrive simultaneously at the observer given that one transmitter is accelerating & the other moving at constant velocity relative to the observer. The signals are sent as the 2 transmitters pass each other (i.e. from the same position).

10. Jul 19, 2009

### ZikZak

In that case, since the signals travel at c, and they are being sent from the same spacetime event, then they arrive simultaneously at the receiver. It makes no difference whether the emitter is accelerating, at rest, moving, running in circles, or wearing its underwear on its head. Photons travel at c.

11. Jul 19, 2009

### diazona

Man, I'm going to have some strange dreams tonight...