Measuring the Speed of Light at Relativistic Velocities

[Drakkith]

Let's say you have a spaceship with an observer on it who has an experiment set up on the ship to measure the speed of light from a laser beam that is directed at the ship from behind. This experiment consists of two mirrors, one half silvered to only reflect part of the beam, that reflect the light down into two detectors. The half silvered mirror is place in the aft end of the ship while the other is placed in the forward end. The half-silvered mirror serves to reflect part of the light into the aft detector while letting the rest pass through to the forward detector. The laser sits in a stationary frame and the ship moves at 0.5c with respect to it.

Once the ship is accelerated up to 0.5c away from the laser the laser is then turned on and directed at the ship to hit both mirrors. The observer on board the ship measures when each photodetector goes off. Repeat the experiment as necessary to average out the shot noise and other effects.

What does the observer on the ship measure as the speed of the laser beam by measuring the time between the aft detector first detecting light and the forward detector first detecting light? I WANT to say c, but it doesn't make sense to me. At 0.5c you relativistic effects are only at a factor of 1.154, or about 15%, meaning that you would experience 1.154 seconds for every second of the stationary frame. It is my understanding that the light in the laser beam would take twice as long to catch up to the moving ship as it would if it were stationary since it is traveling at 0.5c. But if that is true, what is the observer measuring on the ship and does this particular experiment say anything about the speed of light or am I misunderstanding something, as accounting for time dilation and length contraction doesn't seem to be able to account for everything. I'm feel as if I've missed something horribly obvious but I can't seem to put my finger on it.

Thanks.

 

[jollyredgiant]

I think you're forgetting that simultaneity is relative. According to the stationary observer, the light will hit the beam splitter, then hit the back of the ship, then the front. However, according to the ship's frame of reference, the laser will hit the beam splitter and then both ends at the same time. The guy on the ship will still measure the speed of light to be c.

The classic thought experiment about simultaneity is the ladder/barn paradox.
http://en.wikipedia.org/wiki/Ladder_paradox

Hopefully this is helpful.

 

[Drakkith]

I think you're forgetting that simultaneity is relative. According to the stationary observer, the light will hit the beam splitter, then hit the back of the ****, then the front. However, according to the ship's frame of reference, the laser will hit the beam splitter and then both ends at the same time. The guy on the ship will still measure the speed of light to be c.

I think you have my setup confused. The beam splitter is on the aft end and it reflects part of the light into the aft detector while letting the rest pass through to a fully reflective mirror on the front that reflects the rest into the forward detector. I have edited my above post to clarify.

 

[jollyredgiant]

Ah, ok. I think its still basically the same problem. The stationary observer will claim the time delay between detections is different than what the observer on the ship claims. However, they will also claim the light traveled different lengths. They will both disagree on the time interval and the distance the light traveled, but they will agree that it moved at the speed of light. If you use the Lorentz transformations, being careful to label when/where each event occurs in both reference frames, you should be able to show that x/t = x'/t' = c.

 

[Mentz114]

In the ship, they measure the speed of light to be c. If I've got your setup correctly, then it's obvious from the diagrams. The first one is from the emitters Pov, the second the ship Pov. In the ship, the distance traveled by the light is equal to the time taken, so they measure c=1.

 

[ghwellsjr]

Let's say you have a spaceship with an observer on it who has an experiment set up on the ship to measure the speed of light from a laser beam that is directed at the ship from behind. This experiment consists of two mirrors, one half silvered to only reflect part of the beam, that reflect the light down into two detectors. The half silvered mirror is place in the aft end of the ship while the other is placed in the forward end. The half-silvered mirror serves to reflect part of the light into the aft detector while letting the rest pass through to the forward detector. The laser sits in a stationary frame and the ship moves at 0.5c with respect to it.

Once the ship is accelerated up to 0.5c away from the laser the laser is then turned on and directed at the ship to hit both mirrors. The observer on board the ship measures when each photodetector goes off. Repeat the experiment as necessary to average out the shot noise and other effects.

What does the observer on the ship measure as the speed of the laser beam by measuring the time between the aft detector first detecting light and the forward detector first detecting light? I WANT to say c, but it doesn't make sense to me. At 0.5c you relativistic effects are only at a factor of 1.154, or about 15%, meaning that you would experience 1.154 seconds for every second of the stationary frame. It is my understanding that the light in the laser beam would take twice as long to catch up to the moving ship as it would if it were stationary since it is traveling at 0.5c. But if that is true, what is the observer measuring on the ship and does this particular experiment say anything about the speed of light or am I misunderstanding something, as accounting for time dilation and length contraction doesn't seem to be able to account for everything. I'm feel as if I've missed something horribly obvious but I can't seem to put my finger on it.

Thanks.

Before you concern yourself with doing the experiment on a ship moving at 0.5c, you should think about how you would do the experiment on a stationary ship or in a laboratory.

Specifically, how does the observer "measure when each photodetector goes off"? This will either require two timing devices colocated with each detector or it will require one timing device that starts with the aft detector and stops with the forward detector.

If the latter scheme is used, what better way to send the start and stop signals to the timer than by using light signals? Then you can see that if the timer is placed near the aft detector, the measured time will be the same as the round-trip light travel time. And if you put the timer in the forward position, it will measure no time at all.

So maybe we better use the first scheme. Now we have the problem of knowing whether the two clocks are set to the same time. Well that's easy, we think, we just look at them both and see if they have the same time on them. But if we are positioned near the aft clock and we see the same time on them, then we should conclude that they really aren't at the same time because it takes time for the image of the forward clock to reach the rear. We have the same problem if we are at the front.

This is in fact the issue that Einstein discussed in his 1905 paper introducing Special Relativity and he said there is no way to resolve the problem apart from making an arbitrary definition of what the timing relationship should be between the two clocks. His solution is to look at the two clocks from the midpoint or put the single timing device at the midpoint.

Then it turns out that if we do this for the ship while it is stationary, we are defining the one-way speed of light to be equal to the two-way speed of light and then when we carry out the same experiment in the traveling ship with the same definition of time, we have to conclude that the length of the ship contracts and the timing devices run slower than normal so that when we repeat the exact same experiment, the measurement will yield the same answer.

 

[Drakkith]

Then it turns out that if we do this for the ship while it is stationary, we are defining the one-way speed of light to be equal to the two-way speed of light and then when we carry out the same experiment in the traveling ship with the same definition of time, we have to conclude that the length of the ship contracts and the timing devices run slower than normal so that when we repeat the exact same experiment, the measurement will yield the same answer.

How does a 15% time dilation and length contraction counteract a velocity relative to the laser at 0.5c? Or have I done my math wrong?

So maybe we better use the first scheme. Now we have the problem of knowing whether the two clocks are set to the same time. Well that's easy, we think, we just look at them both and see if they have the same time on them. But if we are positioned near the aft clock and we see the same time on them, then we should conclude that they really aren't at the same time because it takes time for the image of the forward clock to reach the rear. We have the same problem if we are at the front.

What issues are involved if you have the detectors record the times using their own clocks and then compare the recorded times afterwards, either while you are still moving or after you slow down?

 

[ghwellsjr]

Have you read the first couple of sections of Einstein's paper?

 

[Drakkith]

Have you read the first couple of sections of Einstein's paper?

I've read a couple of books, including one that had some of his papers/articles. I guess I may need to go re-read them.

 

[Nugatory]

i WANT to say c... I'm feel as if I've missed something horribly obvious but I can't seem to put my finger on it.

This seems so obvious that I'm reluctant to post it... But it's the exact same problem as if the ship were at rest and the laser were receding at .5c... And viewed that way, c really is no-kidding obvious. Now Lorentz-transform the two detection events into the laser frame (choose the origin of the laser frame to be the rear detector trigger firing - you don't care where the laser is) and it will all hang together.