Re: Help with SR time dilation
From: jem (xxx_at_xxx.xxx)
Date: 02/07/05
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Date: Mon, 07 Feb 2005 08:55:58 -0500
dseppala@austin.rr.com wrote:
> On Sun, 06 Feb 2005 17:56:20 -0500, jem <xxx@xxx.xxx> wrote:
>
>
>>dseppala@austin.rr.com wrote:
>>
>>
>>>On Sun, 06 Feb 2005 10:18:04 -0500, jem <xxx@xxx.xxx> wrote:
>>>
>>>
>>>
>>>>dseppala@austin.rr.com wrote:
>>>>
>>>>
>>>>>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?
>>>>
>>>>Measured from A, what are the densities (molecules/liter) of the water
>>>>being pumped out of/into B?
>>>
>>>Instead of talking about densities, let's just say that pump A pumps N
>>>atoms of water per second as measured by observers in Frame A, and
>>>that an identical pump in Frame B pumps N atoms of water per second as
>>>measured by observers in Frame B.
>>>David
>>
>>It doesn't make any difference whether molecules/sec or liters/sec are
>>considered. E.g. let Psec be a second in the pump frame, and Tsec be a
>>second in the trough frame, let g = 1/sqrt(1 - v^2) with c=1. Then the
>>pumps release water at N units/Psec, and the troughs receive it at
>>(N/g)units/Tsec, but from the trough perspective 1 Tsec = 1/g Psec, so
>>the release/receive rates are the same.
>>
>>P 31 NE 28
>
> I wasn't able to follow what you are saying. In Frame A, the pump
> releases water at N units/Psec. Ok. Now as measured in this same
> frame (Frame A), at what rate is water being pumped from the B frame?
> It is being pumped at N/g units per Psecs. Is that right?
It would be confusing to use the Psec/Tsec distinction when considering
the pump and trough in the same reference frame, so drop the prefixes
and assume time is measured in the A frame.
From the A frame perspective, pump A releases N water units over 1 sec,
which are spread over a proper length v*g of trough B, and pump B
releases N/g water units over 1 sec, which are spread over a proper
length v of trough A. In each case, N/(v*g) water units are added per
unit volume of trough by a pump moving along the trough at speed v (i.e
neither trough accumulates more water than the other).
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