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A Relativity Question

Updated on March 1, 2008

This is a thought experiment to show how two different observers might measure the speed of light along a single path if they could get a cylinder to rotate at 107 revolutions per second. In figure 1 observer A has a hollow cylinder 3 meters long and not moving relative to the light source. Both ends are closed. In the top is a small hole near the circumference. In the bottom is a similar hole offset from the top hole by 36O or 1/10 the distance around the bottom. Light enters the top and travels the length of the cylinder. If the cylinder were rotating at 1.0 x 107 revolutions per second then the hole in the bottom and the light will arrive at the at the same point at the same time and the light will exit the cylinder. The light exits the bottom hole and traces a circle of light on a flat screen below the cylinder. The time for the light to travel 3 meters is 3/c = 10-8 second. The number of revolutions in that amount of time is 1.0 x 107 x10-8 = 0.1 revolution or 36O. If the cylinder were rotating at any other velocity the bottom hole and the light would not coincide and the light would not exit the cylinder.

Fig. 1  Observer A has a rotating cylinder that is not moving relative to the light source
Fig. 1 Observer A has a rotating cylinder that is not moving relative to the light source
Fig. 2 Observer B and his rotating cylinder are moving vertically toward the light source with a velocity of 0.6c
Fig. 2 Observer B and his rotating cylinder are moving vertically toward the light source with a velocity of 0.6c

In figure 2 observer B has the same kind of hollow cylinder and is moving past A with a speed of 0.6c toward the light source. However, observer A sees B's cylinder is contracted to 0.8 x 3 = 2.4 meters in length because it is moving with a velocity of 0.6c past him. Also since the cylinder is moving toward the light source the bottom of the cylinder will reach the light beam in half the time it took the light to travel the static cylinder. Since the light must travel 2.4(c+v)/c = 1.5 meters. This cylinder is still rotating at 107 revolutions per second by observer B's clock. But B's clock is moving 1.25 times slower than A's clock. This would mean that observer B should see the light reach the bottom of his cylinder in only [(107rev/s)/1.25][1.5/(3x108m/s)] = 0.04 revolution. In this amount of time his cylinder would have only rotated 0.04x360O = 14.4O. The hole and the light do not appear to line up. If both observers measure the velocity of light as the same, then both observers should see a circle of light on the screen below either cylinder. How is this possible?

If you can answer this, send me an E-mail?


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      8 years ago

      Yes I can answer it: it has been explained in the literature by Ives. I'll send you an email with details.


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