Re: time dilation

On Apr 10, 6:04�pm, xxein <xx...@xxxxxxxxxxx> wrote:
On Apr 10, 7:40�pm, rbwinn <rbwi...@xxxxxxxx> wrote:





The work of famous scientist Galileo Galilei provides us with a
question about time dilation and Dr. Albert Einstein's statement that
the laws of physics must remain the same in all frames of reference.
Galileo carried two lead weights of unequal sizes to the top of the
leaning tower of Pisa and dropped them at the same time, disproving
the idea of scientists of his time that the heavier of the two weights
would strike the ground first. �Of course, it took some time before
scientists accepted the results of his experiment. �They did not all
believe in the principle of equivalence the moment the two lead
weights hit the ground.
� � � � This brings us to another question about falling objects which
arises from the idea of dropping an object in a moving train car,
which writers of textbooks about relativity often use to show how the
Lorentz equations work. �If a weight is dropped from the top of a
train car to the floor, it falls a distance of y'. �In any
transformation equations this is always expressed as y'=y. � The
object travels the same distance vertically in S' as it does in S.
In Galileo's equations, it takes the same amount of time for the
object to travel from the roof of the train car to the floor in either
frame of reference. �t'=t.
� � �In the Lorentz equations, a clock in S', the frame of reference
of the train car, is slower than a clock in S, the frame of reference
of the train tracks.
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?
� � � 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 am sure that some of our scientific friends who believe in a
distance contraction will be anxious to explain this phenomenon.
Robert B. Winn

xxein: �This is not a case of any distance contraction. �In a train,
however, the speed is non-relativistic.

So let me explain it to you. �If you posed the question of how fast
light traveled in a traincar from the front to back, the answer is not
contained in any particular measurement. �You know that your clock to
measure by is also affected. (I hope you do anyway). �But it is
mathematically realised as c because that is what you will measure.

Here's the problem with your question. �Any notion of length
contraction (real or supposed) is based upon the 'direction' of
travel. �While any moving clock will beat slower, a distance will
remain the same distance. �It is only when a physical object like a
barn vs. pole is relativised that a contraction has to occur in some
fashion or another. �Why? �Because that's what we can measure.

But xxein, the question was not about distance contraction. I just
said that some of the people who believe in distance contraction
should want to answer this question. As I said, in anyone's
equations, y'=y. The problem I see is that a clock in S ticks once
while the object is falling from the ceiling to the floor of the train
car. A clock in the train car does not tick until after the object
hits the floor. The object is falling faster in the train car than in
S.

Is it real? �I'll let you make a stew for a while.
Well, the distance contraction is not real. As we can see from the
Galilean transformation equations, a clock that shows t' in those
equations ticks at the same time the clock in S ticks, when the object
hits the floor. The problem that scientists have with this is that
they have put clocks in satellites, and they came back reading less
time than an identical clock kept on earth. So let's look at what
scientists say a clock would read in the train car.

t'=(x-vt)/sqrt(1-v^2/c^2)

That clock would definitely tick after the object hits the floor
if a clock in S ticks when it hits the floor. This t' is less than
t. So we use two clocks in S'. One clock ticks exactly when the t
clock in S ticks, the other ticks when this t' clock from the Lorentz
equations ticks. If you measure the velocity of the falling object
with the Lorentz equation t' clock, the object is falling faster in S'
than the t clock in S shows it falling.
Now consider how Einstein came to the conclusion that the Lorentz
equation t' clock was running slower. He said that the Lorentz
equations satisfied the results of the Michelson-Morley experiment
because light was traveling at c=300,000 km per second in both frames
of reference. Therefore, these two equations would be true with
regard to the distance traveled by light.
x=ct
x'=ct'
That is fine for the Lorentz equations with their distance
contraction, but we are going to use the Galilean transformation
equations because we already have a clock running at a rate such that
t'=t as measured by that clock. If we are going to express the above
relationship with regard to the Lorentz equation clock using the
Galilean transformation equations, we have to use a different variable
than t' for the time on that clock, so we use n'. Calculating from
the Galilean transformation equations, we find that this n' clock has
a rate of n'=t(1-v/c). If the train were traveling at the speed of
the planet Mercury, 30 miles per second, n' would agree with t' from
the Lorentz equations to several decimal places. So this n' clock,
which we can say represents a cesium clock on the train, shows a time
contraction in terms of what clocks will show. However, we still have
the problem of this falling object which hits the floor before the n'
clock ticks. How fast is it falling? To keep the laws of physics the
same in all frames of reference, do we use the t'=t clock, or do we
use the n' clock?
It seems to me that if we are measuring the speed of light, we use
the n' clock. If we are measuring the velocity of the falling object,
we use the t'=t clock. That is just my opinion.

OK. �But a length contraction is only functional in a single
dimension. �You got that?
I already had that. y'=y.

�A clock will slow regardless of dimension.
Velocity only. �Velocity wrt what?
The velocity of the object is relative to the top of the train car and
the floor of the train car as seen from either frame of reference.
y'=y

We can measure pretty good. �We developed a math to cover it. �Good
for us. �But does it mean we understand the logic of a physic?
Apparently not.

Well, the distance from the top of the train car to the floor is the
same from either frame of reference. What we are concerned about is
the time dilation. A cesium clock in the train car shows the object
falling faster than an identical clock on the ground.

Missed you for a while. �Now bug off before you become the pest you
were in the past.

Well, I just can't believe you would think I am a pest since I have
the correct equations for the time dilation.
Robert B. Winn
.

Relevant Pages

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