# .5 c

1. Jul 13, 2009

### keepitmoving

a moving emitter passes point A traveling at .5 c and emits one photon along the direction of travel. At that same instant a freight train traveling at .5 c passes point A in the opposite direction.
Question 1 - two seconds later where is the moving emitter with respect to point A?
Question 2 - two seconds later where is the freight train with respect to point A?
Question 3 - where is the photon at that 2 second point with respect to point A?
Question 4 - where is the photon at that 2 second point with respect to the moving emitter?
Question 5 - where is the photon at that 2 second point with respect to the freight train?
Question 6 - how far apart are the freight train and the moving emitter at that 2 second point?
Question 7 - have the photon and the freight train separated at faster than c ?
This isnt homework, im just trying to understand this stuff.

2. Jul 13, 2009

### JesseM

I assume "two seconds later" refers to the frame in which each of them are traveling at 0.5c, and where point A is at rest? In this case each of them would be 1 light-second from point A, although of course the answer would be different in other frames.
Again, if you're talking about the frame where both are moving at 0.5c in opposite directions, then the photon is 2 light-seconds away from the point A where it was emitted, which means it's 1 light-second from the emitter and 3 light-seconds from the train.
2 light-seconds apart.
In this frame the "closing speed" between them is indeed faster than c, although if you were to switch to the train's own rest frame the photon would be moving away from it at exactly c.

3. Jul 13, 2009

### ZikZak

You continue to ask questions like this, which is good, but you seem to be missing the critical point that observers in different reference frames will disagree on the distance between two bodies and the duration between two events.

There is no absolute answer to the question. It will depend on the reference frame in which the measurement is made, for both "where" and "two seconds later" mean different things to different observers.

In the rest frame of point A ("the embankment"), the answer is 1 light second in the +x direction from A.

In the rest frame of the freight train, the answer is different, for in that frame, A travels at speed 0.5c, but the emitter travels at speed 0.5c + 0.5c = 0.8c, yielding a separation rate of 0.3c. Thus two seconds later, as measured by the freight train, the emitter is 0.6 light-seconds from A.

In other frames, the answer will be still different.

There is no absolute answer to the question. It will depend on the reference frame in which the measurement is made, for both "where" and "two seconds later" mean different things to different observers.

There is no absolute answer to the question. It will depend on the reference frame in which the measurement is made, for both "where" and "two seconds later" mean different things to different observers.

In the rest frame of A, the photon travels at c, so after 2 seconds, the photon is 2 light-seconds from A in the +x direction.

In the rest frame of the emitter, the photon travels at c and A travels backwards at 0.5c. Thus they separate at 1.5c and after 2 seconds the photon is 3 light-seconds from A.

In the rest frame of the train, the photon travels at c and A follows behind it at 0.5c. Thus they separate at 0.5c and after 2 seconds the photon is 1 light-second from A.

In other frames, the answer will be different.

There is no absolute answer to the question. It will depend on the reference frame in which the measurement is made, for both "where" and "two seconds later" mean different things to different observers.

There is no absolute answer to the question. It will depend on the reference frame in which the measurement is made, for both "where" and "two seconds later" mean different things to different observers.

There is no absolute answer to the question. It will depend on the reference frame in which the measurement is made, for both "where" and "two seconds later" mean different things to different observers.

There is no absolute answer to the question. It will depend on the reference frame in which the measurement is made, for both "where" and "two seconds later" mean different things to different observers.

In some frames, the train and photon have separated at more than c. In others, less than c. In still others, c.

4. Jul 13, 2009

### keepitmoving

what does rest frame mean?

5. Jul 13, 2009

### keepitmoving

yes. im referring to the frame where each is moving at .5 c frompoint A
Now, if the photon is at 600km from point A after 2 seconds and the photon is at 300km from the emitter, then the light didnt move away from the emitter at light speed and this would mean that if the emitter were moving at .9999 c, the light emitted would move away from the emitter at less than .01 c. Is that right?

6. Jul 13, 2009

### ZikZak

In the frame you have specified, the light separates from the emitter at that rate, correct. In the rest frame of the emitter (the reference frame in which the emitter is at rest), then that is not correct. In the rest frame of the emitter, the light travels at c and thus separates from the emitter at c. Light travels at c in all reference frames.

7. Jul 13, 2009

### keepitmoving

Zik Zak
Did you say that the light separates from the moving emitter (moving relative to point A) at c minus the velocity of the emitter? Or did i state the question right?

8. Jul 13, 2009

### ZikZak

Yes, in the frame of the embankment, the photon travels at c and the emitter follows it at 0.5c, so they indeed separate at 0.5c.

Note that this does not mean that they separate at 0.5c in an absolute sense: other observers measure a different rate of separation. In particular the emitter itself observes the separation at a rate of c.

9. Jul 13, 2009

### keepitmoving

im not familiar with the embankment?
Also, ive heard many times that the emitter measures the light moving at c from him (her) but that doesn`t make sense to me if light is moving from A at c as well.
Do i have to just believe or can it make sense to me if i know more?

10. Jul 13, 2009

### A.T.

No, physics is not about believing, but about measuring and accepting the reproducible results, which are that the same light moves at c for every inertial observer.
There is no more to know. You just have to get used to the idea, that "your sense" is based on experience of everyday life, and therefore of limited applicability.

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