Re: Principle of equivalence

rbwinn wrote:
Bryan Olson wrote:
rbwinn wrote:
Show the math, Bryan. The S' clock hits at x=0.
t'=(t-vx/c^2)gamma = t gamma
In order to agree with Galileo's result, t' would have to equal t. If
you can get t' to equal t, let's see you do it.

Ah, a challenge. That clock, as you said "will be sent opposite the
direction of the motion of the train with a speed relative to S'
that is equal to the speed of the train." Thus in S, the velocity of
that clock is v=0.

gamma = c / (c**2 - v**2)**(1/2)
gamma = c / (c**2 - 0)**(1/2)
gamma = c / (c**2)**(1/2)
gamma = c / c
gamma = 1

t'=(t- v x/c^2) * gamma
t'=(t- 0 x/c^2) * 1
t'= t - 0
t'= t

You challenged me to show "t' would have to equal t". Challenge met.

Well, you did exactly what I said you would do. You flipped frames of
reference and said the clock was in S.

You claim I "said the clock was in S"? I try not to say things
quite that vacuous (though I'll admit the showing t'=t was
pretty trivial). All objects are in all frames. One of us says
silly thinks like "the cesium clock in S," but it ain't me.

I did no flipping of frames of reference. Robert, you defined S
as the rest frame.

Figure it from S'.
If you can get it to work from S', I will have to say you have done
something.

In what sense is it not working from S'?

[...]

Well, as I said, if you are correcting papers, just show the correct
math as you see it.
Done.

I have to give you an incomplete on this one.

Gotta at least pass the course before you can do the grading.


--
--Bryan

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