Re: time dilation

On Apr 11, 10:24�am, Darwin123 <drosen0...@xxxxxxxxx> wrote:
� � Okay, I see here that I made a mistake. The acceleration g is the
almost the same in both frames. �There is a small difference in g and
g' due to what I described before, but that doesn't completely explain
the difference in clocks. Sorry, sorry, sorry. The problem is with
simultaneity. I will assume in this post that g=g'. Sorry, sorry,
sorry.

On Apr 11, 11:32 am, rbwinn <rbwi...@xxxxxxxx> wrote:> On Apr 11, 8:04 am, Darwin123 <drosen0...@xxxxxxxxx> wrote:
Darwin,
� � � I like your idea of having the weight be a clock. �I came to the
same idea, but was just using the hitting of the floor as a tick. �I
like your idea better. �Use a cesium clock as the object to be
dropped.
� � � Now we set up the experiment this way. �Put an identical
railroad car, or an apparatus that drops a clock the same distance in
S. �Then we hire a photographer to take pictures of the two clocks as
they drop. �Ok, the train comes by at a high rate of speed, and the
two clocks are dropped at t'=t=0.

� � The clock that is started stationary with respect to the tracks
will hit the floor earlier than the clock that started out traveling
with the train. The first clock is being acted on by a larger
gravitational force than the second clock, as measured in S. Neither S
nor S' sees the two clocks hit the floor at the same time.
� �Why don't you address the statement? The two clocks don't hit the
floor at the same time in either frame.

I just don't believe it, Darwin. I don't see any difference in the
fall of the clock in the railway car and the fall of the clock in S.
I think they are going to hit the floor at the same time. As I said,
y'=y. They both fall the same distance. Why wouldn't it take the
same amount of time?

Our photographer takes a bunch of
pictures as the clocks drop, and the last picture shows the two clocks
hitting the floor at the same time.

� � Can't happen. Any statement based on this false statement is
likely to be wrong.>
Well, what about Galileo's two weights dropped from the leaning tower
of Pisa? Why would it be different for two weights, one dropped in a
railroad car and one dropped the same distance beside the car?
�Say the train is going at the
speed of the planet Mercury, 30 miles per second, then if the clock in
S reads 1 sec when it hits the floor, the clock in S' will read .99984
sec when it hits the floor.

� �I plugged in your numbers. When I plugged the numbers in, I caught
a mistake in my previous analysis. The previous analysis is correct
only when the train is going very close to the speed of light. So
thank you for introducing numbers.
� � �When the clock hit the floor, the train has traveled a
significant distance in frame S. The coordinate x does not equal x'.
In the S' system, x'=0. However, the train has travelled a distance of
0.00016x30=0.00048 miles. So x=0.00048 miles.
� �t'=gamma(t-vx/c^2)
where
gamma=1/sqrt(1-v^2/c^2)^0.5
� �v=30 miles per second, c=186000 miles per second, x=0.00048 miles.
� �The answer is t'=1 second. The xv/c^2 term exactly cancels out the
gamma term.
I am getting something different here. Why don't we kick the
velocity up to 1/2 the speed of light and figure it there?

� � Therefore, when the clock hits the floor both clocks read 1
second. Both S and S' agree on this. However, the two observers
"disagree" as to where the clock hit the floor. Observer S thinks the
clock hit the floor at position x=0.0048 miles, and observer S' thinks
the clock hit the floor at x=0.00048 miles. The observers also
disagree as to which clock was clicking slower. Both observers think
the other observer had a clock clicking slower.
30 miles per second does not show much difference. Let's do it at 1/2
the speed of light.

� � �There is another disagreement. When S' looks at the S clock, he
thinks the S clock said 1 s at a value of x'=-0.00048 miles. So he
thinks the S clock said 1 s after the object hit the floor. In other
words, there is a disagreement as to the simultaneity of the S clock
saying 1 s and the S' clock hitting the floor.
� � This is a major exception to the rule, "two wrongs don't make a
right." I will modify this proverb by saying that "sometimes, two
disagreements make a right." Both observers are in disagreement
because they think their clock clicked faster. Both observers are in
disagreement as to the distance the train moved by the time the clock
hit the floor. The observers are in agreement that their clock (not
the other guys clock) saind 1 s when the clock hit the floor.
� � � �Your major error is this. Both observers agree that the other
observers clock said 0.99984 s when the object hit the floor. However,
if they were ignorant of relativity they would say that the other
clock was wrong. The other clock was running slowly. Their clock is
correct, and their clock said the other clock hit the ground at 1.0 s.
Each observer agrees that his clock said 1 sec when the object hit the
floor.
The problem I have with the reasoning of scientists with regard to
this is, how does one clock end up reading less when they are brought
back together? Why wouldn't both observers continue to think the
other's clock was slower? Scientists all agree that a clock orbiting
earth is slower and shows less time when two identical clocks are put
back together.

� � Okay, I see here that I made a mistake. The acceleration g is the
same in both frames. Sorry, sorry, sorry. The difference in the
gravitational term is insignificant at these speeds. I was thinking
what happens when the train is going very close to the speed of light.> � � �Now we put the photographer on the train and run the experiment
again. �This time the pictures show the two clocks hitting the floor
at the same time, but the clock in S' reads 1 second, and the clock in
S reads .99984 sec.

� � � The photographer (S') sees the S clock read 0.99984 sec when the
object hits the floor, but he also thinks that the S clock said that
at a distance -0.00048 miles behind the object when the object hits
the floor. The photographers clock (S') says 1 second when the object
hits the floor, but the distance of the object is 0 miles. The
photographer only agrees with the clock that is near him. The other
clock is too far away.> � � �Do you agree with this result, and if so, does anything about >it seem contradictory to you?

� � �I disagree with your result, that the other clock says 1 sec when
the clock hits the floor. Each observer sees the other clock say
0.99984 sec when the clock hits the floor. I agree with your result
that their own clock says 1 sec when it hits the floor. When the clock
hits the floor the two observers are a long distance apart. That
resolves the contradiction. It leaves an ambiguity, as the two
observers can't agree as too which clock is working properly.
� � � � It no longer seems contradictory to me. I now refer to it as
an unavoidable ambiguity. The physical events are unambiguous, but the
order in which they occur is ambiguous. An ambiguity is not the same
as a contradiction.

OK, in your scenario, the other clock is still in the air and reads .
99984 when an observer in either frame of reference sees his own clock
hit the floor. I say, all things being equal, the clocks hit the
ground at the same time and the clock in the train car reads .99987
when it hits the ground from either frame of reference, and the clock
in S reads 1 from either frame of reference.

Robert B. Winn
.

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