Theoretical experiment on relativity

[drme1981]

Consider a train , its proper length = 375000 km , traveling at a speed of 180000 km/sec (0.6 of speed of light) relative to person (name him B)

Consider a beam of light that travels in the same direction of the train

The train traveler (name him A) will measure the train length = 375000 km , & will measure time required for light beam to travel from the back of the train to the front of the train = 1.25 second (as light travels in a speed of 300000 km/second)

B will measure the length of the train = 300000 km (as length contraction occurs) , & will measure time required for light beam to travel from the back of the train to the front of the train = 1.5625 second (as time dilation occurs for time of A)

B will measure the distance that light beam traveled = 180000 * 1.5625 (the distance that train traveled) + 300000 (the length of train as B measured it) = 281250 + 300000 = 581250 km (B will measure that light travels this distance at 1.5625 second) i.e. B will measure the speed of light = 372000 km/second

Does this prove that light speed is not constant for all observers?

Thanks

 

[Doc Al]

B will measure the length of the train = 300000 km (as length contraction occurs) ,

Right.

& will measure time required for light beam to travel from the back of the train to the front of the train = 1.5625 second (as time dilation occurs for time of A)

Wrong. You can't apply the simple time dilation formula to the time of flight of the light beam as measured by A, since it does not represent a time measured by a single moving clock.

Does this prove that light speed is not constant for all observers?

Realize that the time dilation and length contraction formulas that you are appealing to in your argument are derived from the fact that the speed of light is constant for all observers.

 

[JesseM]

Hi drme1981, Doc Al gave you the correct answer here but I want to elaborate on this point of his:

Wrong. (You can't apply the simple time dilation formula to the time of flight of the light beam as measured by A, since it does not represent a time measured by a single moving clock.)

As he says, one clock is not enough to measure the speed of a light beam, but each observer can measure the speed of a light beam with two clocks which are at rest and synchronized in their own frame--first note the time the light passes the first clock, then subtract that from the time the light passes the second clock, and divide the distance between the clocks by this time interval. But a key thing to understand is that two clocks which are synchronized in one frame will not by synchronized in another, assuming they are both synchronized according to the convention Einstein described in his original 1905 paper in SR, where each observer synchronizes his clocks using the assumption that light travels at the same speed in all directions in his own rest frame. Using this assumption, each observer would define two of his own clocks to be synchronized if he could set off a light flash at the midpoint of the two clocks and both read the same time at the moment the light reached them. But a little thought shows that this automatically implies that clocks which are defined as "synchronized" in one frame will not appear synchronized in another. For example, if you see me traveling in a train in your frame and I set off a flash at the midpoint of the train, using it to synchronize clocks at the front and back of the train, then if you assume the light travels at the same speed in both directions in your frame, then since from your point of view the front of the train is moving away from the point where the flash was set off while the back is moving towards it, in your frame the light must reach the back clock before it reaches the front clock, meaning that my "synchronization" procedure leaves the two clocks out-of-sync in your frame. So if I use these two clocks to measure how much time a light beam takes to get from one end of the train to another, to figure out what I will measure you have to take into account both time dilation (which makes my two clocks appear slowed down by the same amount in your frame) and the fact that one of my clocks will be ahead of the other in your frame.

If you're interested, a while ago I made up a little example to show how the fact that each observer sees light travel at the same speed in their rest frame can be understood in terms of their different definitions of clock synchronization--see my first post on this thread.

 

[dextercioby]

"Theoretical experiment" sounds awful to most physicists, fortunately Einstein came up with "Gedankenexperiment"...

Daniel.