# Speed of Light

1. Dec 10, 2011

### mananvpanchal

Suppose, A is moving with constant speed c/2 and B is stationary. A fires a beam in direction of his travel. A measures the speed of light as c. B also measures the speed of light as c obviously. But when we ask B to claim what is speed measured by A, then what should be B's answer?

I guess the B's answer is according to STR:

= (c - c/2)/(1 - (c*c/2)/(c^2))

= c

If I am right in above then how can B claim that A measure speed of light as c?

Summary:

A measures c.
B measures c.
But how can B claim that A measures c?

2. Dec 10, 2011

### Pengwuino

That's the point of special relativity. Light travels at c according to all observers.

3. Dec 10, 2011

### ghwellsjr

Just remember that in order to measure the speed of light, A needs a light source and a timer traveling with him and he needs a mirror that stays some measured distance in front of him. He starts his timer while he sets off a flash of light and stops the timer when the reflected like gets back to him and he calculates the speed of light by taking double the measured distance to the mirror divided by the time interval.

But B realizes that A is experiencing length contraction and time dilation (because you said he is moving) and so the smaller time interval he measures is exactly compensated by the shorter distance to the mirror and his calculation comes out equal to c.

4. Dec 10, 2011

### mananvpanchal

So, Length contraction and Time dilation force B to claim that A measure speed as c.
Then

Suppose another situation where A fires a beam in anti direction of his travel, now B have to claim that:

= (c + c/2)/(1 + (c*c/2)/(c^2))

= c

Is there Length dilated and Time contracted???

5. Dec 10, 2011

### Pengwuino

Yes. One thing to remember is that the Lorentz Transformations that give you these velocity addition formulas were created with the assumption that all reference frames will have the speed of light as an invariant quantity. So, of course, they will be consistent with that.

6. Dec 10, 2011

### ghwellsjr

Yes, it works in both directions. Consider that when A is firing his beam in the direction he is moving, it will take longer for the light to reach the mirror but the reflected light will take less time to get back to A. When he shines the light in the opposite direction than he is moving, the times for the two parts of the trip are interchanged. It will take less time for the light to reach the mirror (which is now behind him) but it will take longer for the reflected light to get back to him.

7. Dec 10, 2011

### mananvpanchal

Are you telling that mirror is in A' frame of reference?

8. Dec 10, 2011

### ghwellsjr

They each need their own mirror that stays a fixed measured distance away from themselves and they each need their own timer. This is so that whatever frame you want to view the observer's measurement in, they will always get the same answer for the speed of light, even though it is propagating at c with respect to your chosen frame and not with respect to the observer. This works because in your chosen frame, the observer's mirror is closer to him than his measurement indicates and his timer is running slower than it indicates, and so his calculation of the speed of light comes out to c.

Of course the observer is not aware of whatever frame you are using and he is free to use any frame of his own choosing, including one in which he is at rest, in which case his measurement of the distance to his mirror will not not be contracted and the rate that his timer ticks will not be dilated and the light will propagate at c in his frame.

9. Dec 12, 2011

### mananvpanchal

Can we not re imagine the whole situation by placing both mirrors in B's frame, and A fires beam in both direction?

10. Dec 12, 2011

### ghwellsjr

All mirrors are in all frames. It's difficult to understand what you are asking.

If you are asking if B, moving with respect to A, has a mirror and a timer that are both stationary with respect to B, and he measures the speed of a beam of light generated by A, then, yes, he will get the value of c.

If you are asking if A does not have a mirror that is stationary with respect to himself, but rather attempts to measure the speed of light using a mirror moving with respect to him, then he will not be able to perform the measurement. He needs to know the distance from himself to the mirror at the moment the light reflects off it. Since he cannot track the progress of the light, he cannot know when it hits the mirror, and if he doesn't know when it hits the mirror, he cannot know how far away the mirror is when the light reflects off the mirror and without that measurement of distance, he cannot perform the needed calculation.

If you are asking about something different, please describe it in terms of the relationship of the mirrors with respect to each observer.