Another relativity check please.

[rede96]

Suppose there are three space ships in a line, A, B and C each at rest wrt each other. Ship A is 100 million km from ship C and ship B is roughly in the middle, 60 million km from ship A.

There is a really long fibre optic cable coiled up in ship A, which we'll say is 300 million km long.

We use a forth ship to feed one end of the cable through ship B, and then attach it to ship C.

So there is 100 million km of cable between ship A and ship C, which passes through ship B and 200 million km of cable left on ship A.

Ship A sends a light pulse through the cable. Assuming 300,000 km/s for c, and that the light pulse would travel at c in the cable, it would take 1,000 seconds for the light pulse to travel the length of the cable and reach ship C and approximately 866.6 seconds for it be detected by ship B, as there is 260 million km of cable between ship B and ship A.

Ship A also sends a light beam directly to ship B at the same time it sends the pulse through the cable (Same time from ship A’s frame.)

So ship B would detect the light beam after 200 seconds and would then detect the light pulse in the cable 666.6 seconds later.

Ok so far I hope.

Ship C, where the other end of the cable is attached, now instantly accelerates away to a constant velocity of 0.75c, pulling the cable with it from ship A.

Just at the point where there is exactly 60 million km of cable left between ship A and ship B (the exact distance between them.) ship A sends a final light pulse through the cable just before it is pulled loose. Also at the same time it sends a light beam to ship B. (Same time from ship A’s frame.)

As ship A and ship B are still at rest wrt each other, the light beam still takes 200 seconds to reach ship B.

However, as the light pulse is traveling in the cable, and the cable is now traveling at 0.75c wrt to ship A and ship B, does that mean that the light pulse is closing the gap at a rate of 1.75c wrt to ship B and thus wouldn’t the light pulse arrive at ship B before the light beam?
If so, then ship A has sent information to ship B quicker than the speed of light?

Obviously that can’t be so, but I can’t figure out why.

 

[Doc Al]

However, as the light pulse is traveling in the cable, and the cable is now traveling at 0.75c wrt to ship A and ship B, does that mean that the light pulse is closing the gap at a rate of 1.75c wrt to ship B and thus wouldn’t the light pulse arrive at ship B before the light beam?

No. You must add the velocities relativistically. If the speed of the light with respect to the cable is c, then its speed with respect to ship B is also c. (In a real cable, the light pulse will travel less than c, which makes things a bit more interesting.)

 

[rede96]

No. You must add the velocities relativistically. If the speed of the light with respect to the cable is c, then its speed with respect to ship B is also c.

Ok, so using this formula I take it... w = (u+v)/(1+uv/c²)

I thought because the light pulse was dependent on the cable that I couldn't do this.

So would I still need to add velocities relativistically if instead of a fibre optic cable, I had a really ridged wire and I sent a vibration through it that traveled at c (or as close to c as the physics will allow?)

As the vibration is dependent on the wire and the wire is moving towards ship B, logic would say that the vibration has to get there quicker than if the cable was at rest wrt ship B.

(In a real cable, the light pulse will travel less than c, which makes things a bit more interesting.)

Not thought about it much, but how so?

 

[Doc Al]

Ok, so using this formula I take it... w = (u+v)/(1+uv/c²)

I thought because the light pulse was dependent on the cable that I couldn't do this.

That's the correct formula to find the speed of the pulse.

So would I still need to add velocities relativistically if instead of a fibre optic cable, I had a really ridged wire and I sent a vibration through it that traveled at c (or as close to c as the physics will allow?)

Yes, you'd add the velocities relativistically. Realistically, a vibration would travel down the wire at the speed of sound--much less than the speed of light.

As the vibration is dependent on the wire and the wire is moving towards ship B, logic would say that the vibration has to get there quicker than if the cable was at rest wrt ship B.

Your reasoning is correct. To find out how much faster, add the two speeds using the above formula.

Not thought about it much, but how so?

My only point was that if one of the speeds you're adding is c, the resultant speed is also c. So you don't gain anything. Of course, realistically the speed through the fiber optic cable or the wire will be less than c, so the resultant speed will be greater when you add in the effect of the cable's motion. Slightly more interesting.

 

[rede96]

My only point was that if one of the speeds you're adding is c, the resultant speed is also c. So you don't gain anything. Of course, realistically the speed through the fiber optic cable or the wire will be less than c, so the resultant speed will be greater when you add in the effect of the cable's motion. Slightly more interesting.

Ok, thanks for that.

 

[Dale]

The seminal experiment that confirms what Doc Al is saying is called the Fizeau experiment, and is considered one of the earliest experiments supporting relativity:

http://en.wikipedia.org/wiki/Fizeau_experiment

 

[rede96]

The seminal experiment that confirms what Doc Al is saying is called the Fizeau experiment, and is considered one of the earliest experiments supporting relativity:

http://en.wikipedia.org/wiki/Fizeau_experiment

Thanks for the link.

I'm not sure I understand it fully. Does the experiment show, through the interference pattern, that the light has moved at different speeds in the different directions?

 

[Dale]

Yes, exactly.

 

[rede96]

Yes, exactly.

Apologies for the stupid question, but how do I know from the experiment that the reason for the difference in speeds wasn't due to light traveling faster than c in one direction and slower than c in the other?