Re: time dilation

On Apr 17, 2:58 pm, rbwinn <rbwi...@xxxxxxxx> wrote:
On Apr 17, 10:42 am, PD <TheDraperFam...@xxxxxxxxx> wrote:

On Apr 15, 5:22 pm, rbwinn <rbwi...@xxxxxxxx> wrote:

On Apr 15, 2:40�pm, PD <TheDraperFam...@xxxxxxxxx> wrote:

No, they do NOT say this. You are subject to a gross misapprehension.
The position of scientists is that the laws of physics are identical
in S and S'. This does NOT mean that the results must be the same.

PD

Well, all right , PD, but what is the distinction made with regard to
this experiment?
� Scientists are saying that in S, a clock will drop in a time of
t=1. �A clock in S' drops in a time of t'=1,

No, that is NOT what scientists are saying. That would be counter to
the laws of physics, which have to be of the same form for both S and
S'.

Well, it is what scientists have been telling me in this topic I
started.  They say as seen from S, the S clock hits the floor at t=1,
and when that happens, the S' clock is still in the air as seen from S
and registers a time of t'=(t-vx/c^2)gamma and will hit the floor at a
time of t'=1 in S'.

but as seen from S, a
clock does not hit the floor until after the S clock hits, and a clock
in S' will show less time than a clock in S.

No, that is not what scientists are saying either. Maybe it would do
some good to find out what scientists would really say, rather than
just making up what scientists would say.

Well, you may be a scientist who would say something different.
However, at least three scientists have given this exact scenario in
responding to this topic.

Since you seem to be having difficulty with this, let's simplify
things and drop only a single object, a rock, and let that one rock be
released at an event we'll label HERE and let that one rock land on
the ground at an event we'll label THERE. The rock is only dropped
once, so there are only the two events HERE and THERE. Now you can ask
what we know about the relationship between HERE and THERE, as viewed
from two reference frames, S and S'.

What we know is that the time at event one when the rock was released
was t'=t=0.  Event two takes place at x,t.  In the case of a clock in
S' thrown such that it will land at x=0, according to Galileo, the
clock will hit at t=1, the same as the S clock.

                   t'=(t-vx/c^2)gamma = 1 gamma

  You scientists still have the clock in the air when the S clock
hits.

Now, it may be observed in S that the time measured on (S)'s clock
between HERE and THERE is t. And it may be observed in S' that the
time measured on (S')'s clock is t'. There is no law of physics on the
planet that says t = t'. And in fact it is counter to the laws of
physics that t=t'. And in fact, nature observes those laws of physics
and real measured t and t' will not quite be the same.

Well, these other scientists were willing to concede that in their S'
laboratory, the t' clock would hit the floor at t'=1.  However, they
all said that the t' clock would only read t'=(t-vx/c^2)gamma when the
t clock hit the floor, and would still be in the air.
    It does not run counter to the laws of physics for t'=t.  All it
means is that a cesium clock in S' does not show t' except in
instances like this special case we are considering. If the velocity
of a cesium clock in S' is exactly the same as the velocity of a
cesium clock in S relative to S, then the cesium clock in S' will show
t'=t.  However, usually a cesium clock in S' has the same velocity as
S', therefore, it is usually not going to show t'=t.  It is going to
show n'=t(1-v/c).
     I am sorry, but my equations are right and yours are wrong.

That isn't determined by looking at equations. That's determined by
comparing whether, when you plug in *measured* values for those
quantities in the variables, whether the left side of the equation and
the right-side of the equation are the same. This is where experiment
becomes important, Bobby.

Now, you can say, "But the problem is you're trying to stick in
*measured* quantities in for t and t'. The equation is right but t'
isn't what you measure. What you measure is some other number." That's
fine if you want to invent comic-book heroes and other mythological
things, but it's not physics.

Well, yes, it is physics, PD.  I even got one scientist to show with
the Lorentz equations that t'=t for this particular problem.  He had
to flip frames of reference to do it, but that is standard procedure
with the Lorentz equations.
    At any rate, there is no doubt in physics that Galileo was right
about the principle of equivalence.

About the principle of relativity, yes. The principle of equivalence
means something else.

 My equations agree with the
principle of equivalence.

No, they don't. They make the laws of physics violate the principle of
relativity, which is what the principle of relativity is all about.
Under your equations that describe a possible transformation between
coordinates between reference frames, the laws of physics and in
particular the laws of electrodynamics are no longer of the same form
in both reference frames. That is explicitly a violation of the
principle of relativity.

Of course, you'd never know that by looking at just the set of
equations you're considering. That's part of the problem.

 Now since you claim to have found a mistake
in my mathematics, just go ahead and show everyone the mistake.

No, I didn't say there was a mistake in the mathematics. That's the
point. There can be a perfectly well-formed and mathematically
consistent set of equations that nevertheless does not match reality.
The determination of reality is NOT based on inspection of equations
to find out if there is a math error somewhere. The determination of
reality is based on whether, when you put actual *measured* values in
for the variables in the equation, the left-hand side of the equation
agrees with the right-hand side of the equation.

In your case, you wrote t=t'. This equation doesn't describe reality
because when you put *measured* values of t and t' in, as taken from a
real experiment, you find that the left side and right side don't
agree. This does NOT mean that there is a mathematical mistake in the
derivation of the equations someplace. It just doesn't describe
reality.

The development of physics is littered with theories (since abandoned)
that are completely logical, completely mathematically consistent, and
completely plausible -- and dead wrong. And the fact that they are
wrong is determined when confronted with *measured* values in
experiment.

Robert B. Winn

.

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