Re: Principle of equivalence

rbwinn wrote:
On Apr 16, 8:43�pm, Bryan Olson <fakeaddr...@xxxxxxxxxxx> wrote:
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'?

How do you know what the clock is doing relative to S? Only because I
told you.

I used your given to show what you asked me to show, yes.

The clock was thrown from S'. How do you figure the time
on the clock from that frame of reference when it hits the floor?

Dude, what you challenged me to show was that, in the case you
described, t'=t. That's why I showed that in the case you
described, t'=t. *You* defined the rest frame, from which we
state velocity. You accused me of flipping frames when what I
actually did was use the reference coordinate system you stated.

As for what you are asking now, the time on the clock when that
clock hits the floor is the same in all frames. Simultaneity of
displaced evens is frame-relative. The clock hits the floor at
one place.


[...]> >>> 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.

Well, here is how I work the problem.
w=velocity of light
x=wt
x'=wn'
x'=x-vt
wn'=wt-vt
n'=t(1-v/w)

v= velocity of S' relative to S
-v= velocity of clock relative to S'
0 = velocity of clock relative to S

n'=t(1-0/w)
n'=t

You've confused failing to comprehend the issue with knowing
how to solve the problem.


--
--Bryan
.

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