Re: IRT: A New Theory of Relativity

From: Jesse Mazer (vze2ztqw_at_mail.verizon.net)
Date: 02/04/05 

Date: Fri, 04 Feb 2005 03:46:23 GMT

kenseto wrote:

>
>
>>If that were true, it would mean the track observer would see the light
>>moving at different speeds relative to himself in different directions.
>>Suppose that in the track-observer's frame, at t=0 seconds the train's
>>center is at x=0 light-seconds, the front is at x=2 light-second, and
>>the back is at x=-2 light second. At this moment a light pulse is
>>emitted from the center. If the train is moving at 0.5c, and the pulse
>>of light moving to the right is moving at exactly c in the
>>track-observer's frame, then the light would hit the front of the train
>>at t=4 seconds, at position x=4 light seconds. Are you saying the light
>>beam moving to the left would also hit the back of the train at t=4
>>seconds in the track observer's frame? Well, since the back of the train
>>started out at x=-2 ls and is moving at 0.5c, after 4 seconds it will be
>>at position x=0 ls. So if that's when the light hits the back of the
>>train, the light beam must have had a velocity of 0 in the
>>track-observer's frame! Obviously this doesn't match with the second
>>postulate of relativity...for the second postulate to be satisfied, the
>>light pulse moving to the left must also have a velocity of c in the
>>track observer's frame, so it must hit the back end at t=4/3 seconds, at
>>which point the back of the train will have moved to the right 2/3
>>light-seconds, so its position will be x=-2+2/3=-4/3 ls.
>>
>>
>
>Your calculations are pure nonsense. At the time the light pulses are
>emitted simultaneously the track observer will say that both ends of the
>train are of the same length L from him. Therefore he will say that the
>light pulses will hit the ends of the train simul;taneously at time L/c
>seconds.
>

By the way, regardless of whether we're talking SR or your own "IRT", it
seems clear to me that this argument is inconsistent with the claim that
light travels at the same speed in both directions in the track
observer's frame. If the front end of the train is initially at position
x=L, then if it's moving at velocity v, the position of the front of the
train as a function of time will be x(t) = L + vt in the
track-observer's coordinate system, will it not? So if the time for the
light to reach it is t=L/c, then at that point its position will be
x(L/c) = L + vL/c = (vL + cL)/c = L(v + c)/c. But if the light beam hit
the front of the train at position x = L(v + c)/c, and it took time
t=L/c to get there, then if you calculate the speed that the track
observer saw the light beam move in terms of distance/time, you conclude
that he saw the light beam move with a velocity of v + c.

If you think this is wrong, which part do you disagree with? Do you
disagree that the position of the front of the train as a function of
time in the track-observer's frame is x(t) = L + vt? Or do you disagree
that the track-observer should define the speed of the light ray in
terms of the distance he saw it travel in a given amount of time?

Jesse

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