Re: Train gedanken pitfall

On Jun 20, 8:31 pm, "Dono." <sa...@xxxxxxxxxxx> wrote:
On Jun 20, 5:10 pm, Bruce Richmond <bsr3...@xxxxxxxxxxx> wrote:





On Jun 20, 5:26 pm, "Dono." <sa...@xxxxxxxxxxx> wrote:

On Jun 20, 12:17 pm, Bruce Richmond <bsr3...@xxxxxxxxxxx> wrote:

On Jun 20, 11:04 am, "Dono." <sa...@xxxxxxxxxxx> wrote:

On Jun 18, 6:43 pm, Bruce Richmond <bsr3...@xxxxxxxxxxx> wrote:

BTW, your sync by slow transport assumes that moving the clock does
not affect its setting.  It is generally accepted that SR agrees with
that assumption, but I can show otherwise.  Put the clock on a moving
train Next to the clock at A' and sync it with that clock.  Slowly
transport it along the train and place it next to the clock at B'.
According to SR the transported clock must now be in sync with the
clock at B'.
From the tracks we see that the clock at B' is out of
sync with the clock at A', so the setting of the transported clock had
to change as it was transported to be in sync at its new location.
We are not talking about time dialation here.
Transport the clock back
to A' and the setting reverts back so that it is in sync with the
clock at A'.

No. It will not. See my post to Whoever- Hide quoted text -

- Show quoted text -

I can show that it does happen using the LT, but I'm not going to
waste my time for nothing.  How about if I show you and you can't pick
out any error you will admit that you didn't know what you were
talking about? :)

I'll pick the error , don't worry about it. So, let's see your
computations.- Hide quoted text -

- Show quoted text -

Ok, I'll trust you not to renege on your part of our agreement.

At t=0 A' is at x=0 and the train is traveling at .1c along the x axis
as measured in the track frame.

B' is also on the train at .1 light seconds along the x axis as
measured in the track frame.

The LT says t'=(t-vx/c^2)/sqrt(1-v^2/c^2)

I used units of c to simplify things.  With c=1 the c^2 terms go
away.  Plugging in the info for B' at t=0 we get

t'=(0-.01)/sqrt(1-.01)

t'=(-.01)/sqrt(.99)= -.010050378

At t=0 the clock at A' read

t'=(0-0)/sqrt(.99)=0

So in the track frame the clocks at A' and B' have different

readings at t=0.  They are not is sync in the track frame.

Are you still with me?  Any errors yet?

Please use symbolic calculations only, i.e. express the
desynchronization between the clocks in A' and B' in terms of the
variables x, \gamma, v, etc. It is not possible to follow your
calculations where you plug in numbers (I think that you have it
correct so far but you need to use symbolic calculations, this is the
proper way). Otherwise it will get a lot messier when you want to move
the clocks around.- Hide quoted text -

- Show quoted text -

Those symbols are used to represent real numbers. When you get around
to doing an actual calculation you plug in the real numbers that the
symbols represent. You do know how to do math with real numbers don't
you? For example if a=4 and b=5 then a+b=4+5=9. It's not that hard.
I think we covered it somewhere around seventh or eighth grade.

You do agree with t'=(t-vx/c^2)/sqrt(1-v^2/c^2) don't you? I know it
doesn't look the same with it all written in one line as it does here

http://en.wikipedia.org/wiki/Lorentz_transformation_equations

but it means the same thing. I don't write gamma because I don't want
any confusion over what it represents. Since gamma equals 1/sqrt(1-
v^2/c^2) multiplying by gamma is the same as dividing by sqrt(1-v^2/
c^2). That is all I did to get from t'=gamma(t-vx/c^2) to t'=(t-vx/
c^2)/sqrt(1-v^2/c^2).

The same formula t'=(t-vx/c^2)/sqrt(1-v^2/c^2) is used to determine
the clock readings for both A' and B' at t=0. The only difference is
that for A' you plug in 0 for x in (t-vx/c^2) while for B' you plug
in .1 for x. You can see that plugging in any non-zero number for x
will give you a non-zero answer can't you?
.

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