Re: Water Trough - non-sensical SR result?

Thank you Sal for the posting. I really appreciate it. One aspect of
this motion I don't understand. I posted a very simple question in a
new thread titled, "SR and mass - simple question". Perhaps you can
answer that very simple question so I can try to apply it to the
situation I posted in this thread.
Thanks again,
David

On Mon, 15 Jan 2007 11:09:19 -0500, sal <pragmatist@xxxxxxxxxx> wrote:

On Mon, 15 Jan 2007 07:14:39 -0600, David wrote:

On Sat, 13 Jan 2007 21:37:35 -0500, sal <pragmatist@xxxxxxxxxx>
wrote:

On Sat, 13 Jan 2007 14:16:13 -0600, David wrote:

On Thu, 11 Jan 2007 13:59:30 -0500, sal <pragmatist@xxxxxxxxxx>
wrote:

Oh, wow, one of _these_ problems!

Here, David, I'll give you the key, but it's in the form of a
question:

When Al gets to pump B, what time is it at pump A?

When Al gets to pump B, as my original post stated, the time shown
on the clock at pump A is 250 seconds.

Wrong.

Now, using what you know of Lorentz transforms and simultaneity in
special relativity, tell me why I could tell you it was wrong
without even checking your arithmetic.

Here's a hint: No matter what _single_ answer you gave, it would be
wrong. Tell me why.
Oh, I thought you were asking what the moving observer thought the time
shown on the clock at pump A was when he got to pump B. He thinks the
clock shows 250 seconds, the rest frame observers think it shows 1000
seconds, and observers in other frames have different views.

Thank you. Correct.

But, enough with the red herrings. This is an interesting problem and
my first guess wasn't quite right. Here's the real scoop, first in words,
then in numbers:

The "still volume" of water in the trough (the volume the water would
occupy if it were not moving) is _different_ in Al's frame and the
stationary frame after the run. That's obvious; that's what you computed,
and that's what you objected to.

But, as you observed, the depth of the water in the trough can't be
different in the two frames, and in fact it's not. Here's why:

The water is _MOVING_ in frame S (the stationary frame) because of the
action of the pumps. So, it's Lorentz contracted along its
length. The result is that the depth of the water is reduced by a
factor of gamma(water) -- the gamma for the water's velocity. When
the water is flowing, its depth in the trough is less than when it's
still.

The water is also moving in Al's frame, and it's _not_ moving at the
same rate as frame S is relative to Al, because, again, the water is
flowing (in both frames). Al sees a different total "still volume" of
water, but he also sees a different value for gamma(water) in his
frame; the water is contracted to a different degree, and the result
is that the depth he measures is the same as the depth measured in
frame S.

I ran this calculation through using a spread*** -- it gets messy --
and it did work out. I'll give the highlights here. (I didn't work
it out for the general case; I just ran it through with numbers. Feel
free to work out the general case if you like, I'm not going to!)

Rather than 1 liter/sec for the flow rate, which leads to a very ugly
calculation, I used 0.1 cubic light-second/second of pump rate. I
used a depth and width of 1 light-second each for the trough. Note
that the amount of water Al added in my version of the scenario is 25
ls^3, rather than 250 l; the amount pumped out during the run is 100 ls^3
rather than 1000 l. (If you use a 1x1xN trough, in ls, and a pump rate of
1 ls^3/s, the velocity of the water works out to C, which is not so good
-- that's why I divided the pump rate by 10.)

Frame "S" = the frame in which the trough is stationary.

Let

W = "still volume" of water (frame S) = 866 ls^3

L = length of trough in frame S = 866 ls

P = pump rate in frame S = 0.1 ls^3/s

Velocity of the water in frame S is the "still volume", divided by the
pump rate, divided by the length of the trough. If both pumps started
up simultaneously in frame S, the water velocity and gamma would be:

V(water) = W*P/L = 0.1

gamma(water) = 1.005

But that didn't happen. What actually happened was that Al started
pump A first, and while A pumped out 100 ls^3, Al added 25 ls^3. So the
"still volume" of water in the trough after the run, as seen in frame
S (where I've used a single prime for values after the run in frame
S), is

W' = Net volume after run = 866-100+25 = 791 ls^3

V'(water) = W'*P/L = 0.1095

gamma'(water) = 1.0060

Now the _depth_ of the water is the "still volume", divided by gamma,
divided by the length of the trough, or

D' = (W'/gamma'(water))/L = 0.9079 light-seconds

In Al's frame, I found the velocity of the water using composition of
velocities; using two primes for Al's frame after the run, we have

V''(water) = (V'(water) + V(Al))/(1 - V'(water)*V(Al))

= 0.8910

gamma''(water) = 2.2027

The "still volume" of water in Al's frame, after the run, is

W'' = Al's net volume after run = 866-25+25 = 866 ls^3

It didn't change (of course, that's why Al added 25 ls^3 of water).
The depth of the water in Al's frame, after the run, is

D'' = (W''/gamma'')/(trough-length in Al's frame)

= (866/2.2027)/433 = 0.9079 light-seconds

And the depths match.

Feel free to do it over using your original value of 1 liter/second
instead of the "simpler" value, or using symbols instead of numbers to
show it for an arbitrary set of velocities and pump rates; this is as far
as I'm going on calculations for this one.

It was entertaining.




David


David


On Thu, 11 Jan 2007 07:57:14 -0600, David wrote:

A couple of days ago I posted a problem but it included an
acceleration, and some reponders mentioned "reseting"
clocks. I've setup a similar problem with no acceleration - just
a very basic SR problem and the results seem non-sensical to me.
Can any of you experts point out the error.

Let there be a water trough 866 light-seconds long as measured in the
rest frame of the water trough. Let there be a pump, labeled pump A,
at one end of the water trough that pumps water out of the trough at
one liter per second. At the other end of the trough let there be a
pump labeled pump B that pumps water into the trough at one liter
per second. At the start of this experiment let both pumps be turned
off. And let the trough be filled with water to a constant level mark
labeled C.
Now let a person, who I'll call Al, travel along the this trough
with a velocity V = 0.866c. As measured in the rest frame of the
water trough, it takes 1000 seconds for Al to travel from one end of
the trough to the other end. As measured by Al's watch in his moving
frame, it only takes 500 seconds.
When Al is at Pump A, he turns Pump A on. Pump A starts pumping
water out of the trough. When he arrives at Pump B he turns Pump B
on. Pump B starts pumping water into the trough. Al's boss wants the
pump to remain at the constant level mark C. So as Al traverses
along the trough he adds water to the trough since he turned Pump A
on first. When he arrives at Pump B, he looks at his watch. It shows
that 500 seconds have elapsed since Pump A was turned on. Al says,
according to my calculations, only 250 seconds have elapsed at the
rest frame clock located at Pump A. Therefore, 250 liters of water
have been pumped out of the trough. Now that I'm turning on Pump B,
and Pump B and Pump A pump at the same rate, if 250 liters of water
are added to the trough, the water will return to the constant level
mark C. So Al adds a total of 250 liters of water to the trough.
In the rest frame of the trough, Pump A was turned on 1000
seconds
before pump B. Therefore 1000 liters was pumped out of the trough
before Pump B was turned on. Al only added 250 liters of water.
Therefore, the water in the trough is 750 liters below the constant
level mark. How does Al explain the missing water? Thanks, David
Seppala
.

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