# Moving emitter

1. Jul 7, 2009

### keepitmoving

is it safe to say that the lead photon emitted from a moving emitter will always be at a distance from the emitter equal to c x time (from when the first photon was emitted) no matter what the emitter does?
Also, will the photon emitted at the end of the first second always be at a distance equal to c x time minus one second (from when the first photon was emitted)?

2. Jul 7, 2009

Depends on the reference frame. Let's say you're at rest relative to the emitter when the photon is fired, and then the emitter starts moving in the opposite direction. From your frame of reference, the distance between them will be greater than c x time. From the emitter's frame of reference, however, the distance will be c x time no matter what.

Again, depends on the reference frame.

Last edited: Jul 7, 2009
3. Jul 7, 2009

### pesto

Would it be proper to say then that "From my frame of reference, the photon is travelling faster than C relative to the emitter" while the emitter is moving opposite the direction of the photon, or is the distance is greater than c x time as far as we can take it?

I know it's a semantics question but that seems to be where I run into the most trouble.

Thanks.

4. Jul 7, 2009

### JesseM

There's a phrase physicists like to use for the type of "relative speed" you're talking about, it's "closing speed" (or 'closing velocity' if you want to treat it as a vector), meaning the speed that the distance between one thing and another is closing from the perspective of a frame of reference where neither object is at rest. So yes, in your frame the closing speed between the photon and the emitter can be faster than c, even though the speed of the photon in the emitter's own rest frame is exactly c, and of course the speed of the photon in your frame is c as well.

Last edited: Jul 7, 2009
5. Jul 7, 2009

### pesto

Thank you very much.

6. Jul 7, 2009

### keepitmoving

Jesse - i hope youre right. Thats howit seems to me but i have heard that the photon moves relative to every observer at C, which doesnt make a bit of sense to me, the photon would have to be in more than one place at one time. Have you heard this "speed c relative to every observer"?
Also, what does the photon do if the emitter stops or disintegrates ? does the photon put on its retro rockets?

7. Jul 7, 2009

### pesto

Hi keepitmoving. Mind if I take a crack?

You said you heard that "the photon moves relative to every observer at c". It does. The key phrase is "relative to the observer." The photon, relative to the observer, from the frame of reference of the observer, is c.

The photon, from the frame of reference of the observer, relative to the photon gun as it moves away from the photon, is > c.

If that's wrong I hope someone will correct me.

pesto

8. Jul 7, 2009

### Staff: Mentor

After the photon has been emitted, its travels are unaffected by what happens to the emitter, just like when you throw a baseball, it keeps on going even if you drop dead after the ball leaves your hand.

9. Jul 7, 2009

### JesseM

Yes, but when they say that it means they are talking about the speed of the photon in each observer's rest frame, not the closing speed between one observer and the photon as seen by a third observer. Note that at the end of my post I said:
No, I should have made clear that this discussion only applies to the frames of inertial observers who don't accelerate.

10. Jul 8, 2009

### keepitmoving

if you make a chart or a graph of a moving emitter and the photons that it emits second by second such that the photons always are 186,000 x time away from the emitter, there is no bunching of waves (blueshift). How would a stationary observer see a blueshift that isnt there?

11. Jul 8, 2009

### sylas

If the photons are always c x time away from the emitter, then you are in a frame where the emitter ISN'T moving.

Observed in a frame where the emitter is moving, the photons are always c x time away from where the emitter was when that photon was emitted... but the emitter is moving, and so the distance between photons and emitter at a given instant in this frame not c x time.

12. Jul 8, 2009

### keepitmoving

if what you say is true, then an emitter moving at .99c would see his light moving away at only .01 c and ive heard that is not true.

13. Jul 8, 2009

### ZikZak

No. In the frame in which the emitter is moving at 0.99c, the photon travels at c, and so the separation between them increases at 0.01c. In the rest frame of the emitter, on the other hand, the emitted photon travels at c so the separation increases at c.

The emitter sees the photon travel at c. The receiver sees the photon travel at c. Everyone sees the photon travel at c.

14. Jul 8, 2009

### keepitmoving

i guess i misunderstand what you say -"in the moving frame of the emitter the separation increases at c" and you also said that the photon moves at c relative to the rest frame.
That sounds to me like the photon moves away from the moving emitter at .01 c. Ive heard that is not true.

15. Jul 8, 2009

### sylas

From the perspective of the emitter, the emitter is at rest, by definition. In this frame, as I said previously, the light is moving away at c.

For another frame, in which the emitter is moving, the rate of separation between the emitter and the light is not a velocity of an object, but a difference of two velocities.

So. The emitter ALWAYS sees the photons they emit moving away from the emitter at c. This is the frame in which the emitter is at rest.

For another observer, the photons are still moving at c. But if the emitter is moving, then the distance between the emitter and the photon is changing at less than c (if the emitter is chasing the photon in this frame) and at more than c (if the emitter is moving away from the photon in this frame).

It just depends on the perspective of the observer.

Cheers -- sylas