Re: Help with SR time dilation

dseppala_at_austin.rr.com
Date: 02/07/05 

Date: Mon, 07 Feb 2005 14:28:34 GMT

On Mon, 7 Feb 2005 13:41:07 +0100, "Harry" <harald.vanlintel@epfl.ch>
wrote:

>
><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
>
Thanks for the reply. A while back I posted that problem using clocks
and you are correct that was the answer given - mainly their
synchronization errors compensate for time dilation and length
contraction - which I could follow. However, when I was looking into
how an electrical circuit might work with SR, things didn't work out
with that circuit. I recognized that the water trough problem is
analogous to the problem I was having with understanding this
particular electrical circut. And I then recognized that the pump
problem was analogous to my previously posted clock problem (where the
pumps are simply clocks). However, the synchronization explanation
doesn't seem to work in the pump problem.
   In the water trough problem posted in this thread, with a single
pump in Frame A, and a single pump in Frame B, there is no
synchronization issue. The pumps are turned on at the same point in
space and time, and they are not turned off. Over time neither trough
gains or loses water (due to the symmetry of the problem). So at
first I thought the problem could have some explanation due to the
pumps being at ever increasing distances apart. But I could not make
that explanation work. There were two problems I had there. One is
if the relative velocity is V between the two frames, I can make the
water velocity within a given trough extremely high and in any
direction. Doing that doesn't have any effect on the rate pump A or
pump B removes water from their respective troughs, but it eliminates
the argument about how fast the water reaches a steady-state level.
    The other problem with the explanation related to every increasing
distances is that from the point of view of observers in Frame A,
where I turned on both pump A and B when they meet, I could have a
series of pumps, labeled A0,A1,A2,A3,A4...AN, instead of just a single
pump A. In this scenario, pump B is turned on when it meets pump A0
and then pump B stays on. Whereas A0 is turned on momementariy when B
is in its vicinity, and then turned off, and then pump A1 is turned on
when B is in its vicinity and then turned off. From the point of view
of observers in Frame A, the series of A pumps can be made to sequence
such that on average they pump 10 liters per second into trough B just
as the single A pump does in the one pump variation, as measured in
Frame A. Pump B is always pumping at half the rate of the A pumps
according to the Frame A observers and from the point of view of
observers in Frame A, the series of A pumps always spray the water
into trough B in the vicinity of the B pump, yet the B trough never
overflows (since this is identical to the symmetrical problem in terms
of flow rates).
   In the pump problem, there is no synchronization issue (as far as I
can see) with the pumps, so the explanation given if the pumps were
merely clocks doesn't seem to work.
   David Smith posted the usual answer, mainly that pump A and pump B
catch up on the amount of water pumped because based on their
separation they are turned off at different points in time and space.
But in the problem I posted pump B is never shut off, and even if it
was shut off at some time, he didn't explain why trough B doesn't
overflow while pump B is on.
David

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