Re: Principle of equivalence

In sci.physics.relativity, rbwinn
<rbwinn3@xxxxxxxx>
wrote
on Tue, 15 Apr 2008 20:55:23 -0700 (PDT)
<a05cbec9-6ad2-4f0e-a2c6-c2ca3815b00b@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
The basic problem with current interpretation of relativity can be
seen from considering two frames of reference S and S' and how they
relate to the principle of equivalence.
S is a set of coordinates at rest and S' is a set of coordinates in
motion in the x direction relative to S with a velocity of v. Let S'
represent a train car in which we will drop a cesium clock at t'=t=0
from the roof. In S we construct an apparatus on a floor level with
the train car floor at x=0 from which we will drop an identical cesium
clock from a height equal to the height of the train car roof.
Now the train car comes by at velocity v and at t'=t=0 both
cesium clocks are dropped. If we measure the time on the cesium clock
in S when it hits the floor, scientists tell us that the cesium clock
in S' will still be in the air at the time the S clock hits and will
register a time of
t'=(t-vx/c^2)gamma at that moment. Scientists tell us that the
experiment comes out the same in S and S' because when the S' clock
does hit, it registers the same time that the S clock registered when
it hit. Then if you looked at the experiment from the frame of
reference of the train car, the S' clock would hit first, and the S
clock would still be in the air and would register less time than the
S' clock.
Now we look at the experiment from the perspective of Galileo,
who discovered the principle of equivalence. We will make one change
in the experiment. When the S' clock is released at the roof of the
train it will be sent opposite the direction of the motion of the
train with a speed relative to S' that is equal to the speed of the
train. The clock will therefore drop straight down relative to S,
falling alongside the clock in S. According to Galileo, both clocks
will hit the floor at the same time, and both will read t'=t.
The Lorentz equations disagree with this. There are two events for
the S' clock. It is released at t'=t=0, and it hits at x=0 at a time
of t'2. According to the Lorentz equations, the clock hits at x=0 at
a time of
t'2=(t-v(0)/c^2)gamma =t gamma. The Lorentz equations show the S'
clock still in the air when the S clock hits even though it is falling
right beside it, and Galileo sees the two clocks hit the floor at the
same time.
Scientists say, If this is a problem, it is just a tiny
problem.
Well, if you are figuring times and distances in millions of
miles such as trying to land a Mars probe, you might have problems.
Did anyone notice any problems with Mars probe landings?
Fortunately for us, Galileo had some transformation equations
that explain this phenomenon.

x'=x-vt
y'=y
z'=z
t'=t

Now with regard to cesium clocks, scientists tell us that a
cesium clock in S' will show light to be traveling at c=300,000 km/sec
the same as a cesium clock in S. How does this relate to Galileo's
equations shown above?
You will notice that the equation shows t'=t. That means
that a cesium clock in S' cannot be called t' because t' is already
defined to be t'=t. A clock sitting on the floor in S at x=0 shows
t'. The time on a cesium clock in S' has to be shown by another
variable, n'.

So n' != t'? What is n, the counterpart to n' in the non-prime
reference frame? How does it relate to t'? You've given a
function below relating n' in terms of t, which is a start.

If light is traveling at c,

Not proven to some.


x=ct
x'=cn'

x'=x-vt
cn'=ct-vt
n'=t(1-v/c)

If you're going to define a transformation T(v)(P)
that is usable in the physical realm, you'd better have
inverse(T(v)) = T(-v) or things get ridiculous.

We have the transform:

x' = x-vt
t' = t
n' = t(1-v/c)

We now ask the logical question as to how n relates to t'.
We have part of the inverse already:

x = x'+vt'
t = t'

but where's the rest of it?

One can assume

n = t'(1+v/c)

but this doesn't do much for me.


Disregarding time differences resulting from the y axis, which will be
negligible, both clocks in Galileo's part of the experiment will hit
the floor at the same time and will both read the same trime, t'=t. A
clock dropped in S' which has the same momentum as S' will hit the
floor at t'=t, but will read

n'=t(1-v/c)
when it hits the floor.
S is a preferred frame of reference with regard to time, in this
case because the train is controlled by the gravitation of the earth,
not the other way around. If all clocks in S are synchronized, they
will continue to read the same. If all clocks in S' are synchronized,
they will continue to read n'=t(1-v/c) unless they are moved relative
to S'. A clock that is moved relative to S' has had its velocity
changed relative to S and will then read differently from the rest of
the clocks in S'. We have considered one such clock which was moved
relative to S' to make it conform with the special case, t'=t.
In any event, these equations show the principle of equivalence to
be in effect, regardless of what the Lorentz equations say, and since
that is true, Isaac Newton's equations for gravitation should agree
with these equations, but I have not yet made a study of that.
Robert B. Winn


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