The same subject is also discussued in sci.physics.relativity: SR is succesfully debunked, spread the word. Part of the reason of this questionary can be found in the last posting of this thread.
Nothing for nothing. There will be a lottery between the persons who answer all the questions in this questionary before 1 January 2003. Two boxes of Belgium Chocolate are at stake.
The answer on the second question is: Length Contraction and Time Dilation.
Based on the concept that the laws of physics are equivalent for all inertial observers, Observer C will not only always see the two flashes simultaneous, but also measure each time the same duration. For that being the case you need both Length Contraction and Time Dilation
A separate issue the interpretation of what you see and what actual is (i.e. a description of the "physical reality").
I. Newton L :
The Moving Train experiment in literature is used in order to explain relativity of simultaneity, with two observers as done in question 3.
An Observer can see (observe) two events (light signals) simultaneous. The question is than to decide if those events actual happend simultaneous.
You can also raise this question in a different way: Suppose two events happened simultaneous. Where is the plane from which you can see those two events simultaneous. To answer that question (experimental) is very difficult.
If two observers (A and C) are together in that plane at the moment when those two simultaneous events happen and observer C moves away from that plane then ofcourse A has a chance to see those two events simultaneous and C not. That is what the first to experiments demonstrate. Also Newton will agree with this.
If you measure the length of the Moving Train (in all experiments) as a function of v you get accordingly to SR a function like:
|l | ..... ... | ... .. | .. . | . | | ---------------0------------- V --> figure 6The problem is you can not test this curve in reality
/ / / .t3 / ./ / / . / / / . / / / . .t4 / / . / . / / . / . / / . / . / / . / .t5 / . / /| / . / / | /. / / | ----> v L1----------C-----------L2 | ------FD1------------A---------|--FD2------- track ---- | . | \ | x, t | . | \ | FD1= 0, 0 | . | \| A = 0.5*l0, 0 | . | .t5 FD2= l0, 0 | . | . | L1 = 0, 0 | . | . | C = 0.5*l, 0 | .t1 . | L2 = l, 0 | | . | | | . | | | . | | .t2 | figure 7
/ .t3 / / ./ . / / . / . / / . / .t2 / . / ./ / . / . / / . / . / / . / . / /. / . / .t1 / . / / . / . / / . /. / ----> v <--------C--------> -----------M1----X-----A----------M2---track ---- | . . | | | . . | | . . | | . | . | | . | . | | . | . | | . | . | | . | . | | .t4 . | | . | | | . | | | . | | | . | | | . | | .t5 | figure 8The top part shows the situation for the moving observer C.
IMO we can learn the most if Observer A measures the difference in arriving times of the two signals t5 and t4 (See figure 8) which is identical as the difference between t2 and t1
i.e. equal to 0.5*l/(c-v)-0.5*l/(c+v) = l*v/(c*c-v*v)
See :Example for more details. The difference is the same as between t4 and t3.
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