Re: Help with SR time dilation

From: Harry (harald.vanlintel_at_epfl.ch)
Date: 02/07/05 

Date: Mon, 7 Feb 2005 13:41:07 +0100

<dseppala@austin.rr.com> wrote in message
news:4206fe71.290891242@news-server.austin.rr.com...
> On Mon, 07 Feb 2005 03:11:43 GMT, "Todd" <abc@zzz.com> wrote:
>
> >
> ><dseppala@austin.rr.com> wrote in message
> >news:4204cc44.146952872@news-server.austin.rr.com...
> >> Can anyone please explain the following, supposedly simple, SR problem
> >> to me.
> >>
> >> Let there be two identical water troughs parallel to the x-axis. Let
> >> them move in opposite directions along the x-axis with a velocity
> >> V=0.866c. Let each water trough have a pump attached to it that
> >> pumps 10 liters of water per second, as measured in the frame of the
> >> trough that its attached to. The pumps are initially off, and the
> >> water troughs are nearly filled with water. At time t0, lets say the
> >> the two pumps are at the same x coordinate. At that point in space
> >> and time, both pumps are turned on. The pump on trough A pumps water
> >> from trough A into trough B, and the the pump on trough B pumps water
> >> from trough B into trough A.
> >>
> >> As viewed from the trough A frame, according to SR, pump B is pumping
> >> at half the rate that pump A is pumping. The question I have, is how
> >> do observers in frame A explain why trough B never overflows?
> >> (We can have a series of pumps, and turn them on as each pump passes a
> >> pump on the other trough. This will hopefully minimize posting that
> >> have to do with the length of the troughs).
> >> Thanks,
> >> David Seppala
> >
> >David,
SNIP

Todd wrote:
> >What do observers in frame A predict? The mistake that can easily be
made
> >here is to assume that from the perspective of frame A, 'time dilation'
> >requires that the pumps in B are pumping water into the bucket at less
than
> >N gallons per minute and so the water level should go down! But, in
fact,
> >this is a wrong conclusion based on a hasty application of the concept of
> >'time dilation'.

OK, now I understand the question!
And Todd already replied for the case that there is no symmetry.

I have too little time, but I can say the following for the symmetrical
case:

The clocks in the "moving" frame seem to run slow by factor gamma, and on
top of that their distance seems contracted by gamma as well. Thus when they
pass by, the clocks would be gamma~2 behind on yours if not for
synchronization. For they also seem out of sync by factor gamma^2: each next
clock appears gamma^2 advanced on the preceding clock. The net effect is
that all clocks indicate the same as yours when passing each other.
In other words, their synchronization errors compensate for time dilation
and length contraction.
Just make a few sketches and you'll understand. :-)

Cheers,
Harald

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