Re: time dilation

On Apr 16, 2:18�am, "harry" <harald.vanlintelButNotT...@xxxxxxx>
wrote:
"rbwinn" <rbwi...@xxxxxxxx> wrote in message


- Going back to your issue with SRT: if you had read the article on
simultaneity you might have noticed that according to measurements in S,
calibrated clocks that are at rest in the train car (S') are erroneously
synchronized as it takes longer for light to reach the front of the car
than
to reach the back of the car. Thus, by comparing two distant out-of-sync
clocks, or by "measuring" time-at-a-distance with a picture, the clocks
seem
to fall down in less time for people in the train car than according to
measurements in S. But according to measurements in S', all clocks are
well
synchronized at the start and the falling clocks are slow by the Lorentz
factor due to their high speed.
Thus everyone agrees about what is observed by everyone; the only
difference
is in the perception of causes. I don't expect you to understand that
immediately, but I hope that if you study the Wikipedia article it will
become clearer! :-)
Success.

Harald,
� �The position of scientists is that, taking the train car as a
laboratory, experiments run in this laboratory S' must have the same
results as experiments run outside the laboratory in S. �So by their
interpretation of the Lorentz equations, they get that result by
saying that events that are simultaneous in S' are not simultaneous in
S.

Almost, but not quite!Your fast reply as well as your confusions tell me
that you did NOT read that article.

I estimate that it will take you a whole day in order to more or less
understand my above comments, and even then only with the help of such
articles as the one I referred you to - assuming that you are about as
intelligent as I am. But without such understanding, further discussion is
meaningless: it's like trying to explain integration to a kid who cannot
correctly multiply yet.

According to scientists who have responded in this topic, when the
clock hits the floor in S, a clock at x'=0 in S' is still in the air
and reads less time than the clock in S.

I doubt that anyone made such a fuzzy statement - I for sure didn't do so.
You seem to have forgotten that you changed the description at least once,
and that that change affects the outcome. Changing the question can change
the answer! :-)

The corect statement by you (except for a little glitch in the equation) and
to which I reacted was:

t'=(t-vx/c^)/sqrt(1-v2/c^2). �According to this equation, it takes
less time for the object to fall from the roof of the train car to the
floor in S' than it does in S. �So how are the laws of physics the
same in both frames of reference?

OK, Harald, then we seem to have two groups of scientists. This was
my original statement, which scientists immediately jumped all over
and corrected, saying that the current interpretation of scientists
was that when the S clock hits the floor, the S' clock is still in the
air and reads a time of t'=(t-vx/c^2)gamma. You appear to be agreeing
with my interpretation of the problem except that I say the S' clock
hits the floor in a time of
n'=t(1-v/c), which agrees with the Lorentz equation value for t' to
several decimal places at 30 miles per second , the speed of the
planet Mercury.

� � �If a clock in S ticks once while an object is falling in the
train car, it will not tick in S' until after the object has hit the
floor. �This means that the object is falling with a faster velocity
in S' than in S.

I explained to you here above how that works, complete with a reference to a
full article about simultaneity which is essential for understanding. That
should be sufficient, but only if you study it!

Well, I agree with these other scientists about what the Lorentz
equations say. The S' clock is still in the air when the S clock hits
as seen by the S observer. So there is relativity of simultaneity.
Then from the other frame of reference, the S clock is still in the
air when the S' clock hits as seen by the S' observer. That is what
the Lorentz equations are saying with Einstein's interpretation. I am
not completely sure, but I believe that if you have the clocks hitting
simultaneously with the S' clock reading
t'=(t-vx/c^2)gamma, then you are agreeing with Lorentz's
interpretation.

This is all in keeping with S' being a laboratory where experiments are
the same as in S. �In order to accomplish this, a distance contraction is
necessary.
The Galilean transformation equations are showing something different.
They show S as being a preferred frame of reference, and it is not the
velocity of S' as a moving
laboratory that determines the times on clocks, it is the velocity of
individual clocks relative to S.

It may indeed be useful for you to explore the reasons why Galilean
transformations don't correspond to observation. As you correctly stated,
when you do so one frame becomes a preferred frame according to
observations. This is not the case in reality.

Well, it is the case in reality. I would challenge you to show
something in reality where time of events is not controlled by a
preferred frame of reference. In the case we are considering here,
time of events in the frame of reference of a railroad track is
obviously not controlled by time shown on a clock on a moving train.





For instance, if we drop two cesium clocks, the clocks are
moving relative to S and S' and will show less time when they hit the
floor than identical clocks on the floor in S and S'. �With regard to
clocks in S', a clock at x'=0 in S' has a velocity of v relative to S
and will read t(1-v/c) when a clock in S hits the floor.
A clock sent backwards in S' such that it falls straight down in S has
a velocity of v=0 as far as the x-axis of S is concerned and will read
exactly the same as a clock dropped vertically in S when it hits the
floor. �This means that if all clocks in S read the same, and S is a
preferred frame of reference with regard to time, clocks in S' will
read according to their velocity relative to S, and not all the same
in S' as they would do if S' were a laboratory such as scientists
imagine it to be.
� � Even this description is not exactly correct because not all
differences in time are being taken into account, since most would be
so small as to be almost non-existent. �In fact, for this example, the
differences in results from the Lorentz equations and the Galilean
transformation equations would be so small as to be almost non-
existent. �The major difference is the existence of a preferred
reference frame for time, which the Galilean transformation equations
show, and the Lorentz equations do not. �For instance, in the case of
a satellite in orbit around earth, the clock on earth represents the
preferred frame of reference, and the clock on the satellite
represents the clock in motion.
� �Now it may be centuries before this concept will ever be
considered by scientists.

Robert, I already pointed out to you that that concept has been considered
and rejected by scientists over one century ago, because it is in conflict-

Well, a few scientists have admitted that they have never seen what I
am presenting. Scientists have always taken the position that time on
a cesium clock in S' has to be represented by t'. That automatically
eliminated the Galilean transformation equations because t' is defined
in those equations to be t. What I do, to the horror of scientists,
is put a cesium clock on the ground in S and say, That clock
represents t'=t. Then I use a different variable n' for time on a
cesium clock in S'.

Robert B. Winn
.

Relevant Pages

  • Re: time dilation
    ... train car to the floor, it falls a distance of y'. ... In the Lorentz equations, a clock in S', the frame of reference ... "If you accept the fact that all objects are in constant motion within ... NOT TOO FASCINATING HOW YOU CALLED THIS A FACT. ...
    (sci.physics.relativity)
  • Re: time dilation
    ... train car to the floor, it falls a distance of y'. ... In the Lorentz equations, a clock in S', the frame of reference ... "If you accept the fact that all objects are in constant motion within ... NOT TOO FASCINATING HOW YOU CALLED THIS A FACT. ...
    (sci.physics.relativity)
  • Re: time dilation
    ... train car to the floor, it falls a distance of y'. ... In the Lorentz equations, a clock in S', the frame of reference ... than the clock in S if the picture is taken from S, ...
    (sci.physics.relativity)
  • Re: time dilation
    ... the laws of physics must remain the same in all frames of reference. ... train car to the floor, it falls a distance of y'. ... frame of reference. ... of the train car, is slower than a clock in S, the frame of reference ...
    (sci.physics.relativity)
  • Re: time dilation
    ... train car to the floor, it falls a distance of y'. ... In the Lorentz equations, a clock in S', the frame of reference ... We are using cesium clocks as weights. ...
    (sci.physics.relativity)
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