Re: Train gedanken pitfall

"Bruce Richmond" <bsr3997@xxxxxxxxxxx> wrote in message news:a5a0205c-cab0-4767-98b4-fa84722a1f7e@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
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?

The point that is in disagreement is your statement

"According to SR the transported clock must now be in sync with the clock at B'."

If you have a clocks A', B', and C' that are in sync, where A' and C' are at the same x' location, and B' is further away, and you move C' from being next to A' to being next to B', then C' and B' are no longer in sync. if you move C' back to A', they will no longer be in sync.

Slowness of transport just reduces the difference in sync. It doesn't eliminate it.

And that still does not address your claim that it is NOT the case that moving two clocks in opposite directions at the same speed and then bringing them to a stop, all relative to an inertial frame, results in the two clock being in sync as observed in the inertial frame. You very clearly disagree with me when I said that the would be in sync with each other.


.

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