Re: SR fundamental contradiction


Brian Kennelly wrote:
mluttgens@xxxxxxxxxx wrote:
Brian Kennelly wrote:
mluttgens@xxxxxxxxxx wrote:
Brian Kennelly wrote:
mluttgens@xxxxxxxxxx wrote:
Brian Kennelly wrote:
mluttgens@xxxxxxxxxx wrote:
Brian Kennelly wrote:
I reject the POR.
On what basis? The principle of relativity has been a
fundamental part of physics since at least the time of Newton.
It is consistent with all observations.

I don't reject Galilean relativity, what I reject is Einsteinian
relativity, based on
the hypothesis of the constancy of light speed.
Your proposed equations are inconsistent with Galilean relativity.

I didn't claim that they were.
You can't have it both ways. Either you reject Galilean
relativity or you don't. Above, you claimed that you do not
reject it, and now you acknowledge that your equations are
inconsistent with it. Of what use are your equations?


Starting from Galimean relativity, I got equations that can be
considered
as an extension of it. Like most SRists, you like to quibble.


Your equations can be used, but they single out one system as
preferred, and you lose the POR.

It is also a problem that your equations lead to the prediction
of *dilation* of moving objects.

Assume a stick of length L' at rest in S'. Let the left end be
at the S' origin. Then your equations give:
0=(x1-vt)/g
:: x1=vt
L'=(x2-vt)/g
:: x2=gL'+vt

Subtracting, we find the that length in S is L=gL'. The length
is expanded compared to its length at rest.

What my transformation predicts is that a moving stick is contracted
wrt a
stick at rest, i.e. L'=L/g.
Knowing the length L' of the moving stick, it is clear that the length
of the
stick at rest is L = gL', Iow, it is dilated in S.
Calling the moving stick a stick at rest, because it is at rest in the
moving frame S',
but forgetting that the rest frame remains S, is the type of logical
error systematically
made by SRists.
I am afraid you just lost a lot of ground. The rest length is
the length measured from a reference frame moving with the stick
(IOW, in which the stick is at rest).
This is exactly what I said: a stick comoving with S' is at rest in S'.

But, let's do it your way. A stick of length L in S is moving
with speed v to the right. The left end is x1=vt, and the right
end is x2=L+vt
The length of the moving stick is x2-x1=L. There is no change
in length for a moving stick.
Sure, in S, its length is L, in S', it is L'=L/g.
So you have the remarkable prediction that a moving stick has
no change of length when measured from the system in which it
has a non-zero velocity, but is contracted when measured from a
system in which it has a zero velocity.

Of course, the measuring instruments are also moving with S', so
they will be contracted as well, and S' will not measure any
change of length.

You misunderstood, I didn't refer to the constancy of the length of
the stick at rest in S' and measured in S'. I clearly said " in S, its
length is L, in S', it is L'=L/g", meaning that the length of the stick
measured in S' is contracted wrt its length measured in S.
Using your equations, we conclude that the length of the stick
is independent of is speed for any observer. We do find that
its length depends on the speed of the observer, but that the
observer will not be able to measure the change. Of what use
are your equations?


Quibbling again, unless you didn't understand what I said.



Not until you explain how you calculated these numbers.
One has g=1/sqrt(1-v^2/c^2) and t1'=t1/g, thus g corresponds to t1/t1'.
Similarly, g(LT) corresponds to t2/t2', with t2'= T/g.
We can calculate g(LT) from t2 = T(1 + v/c - v^2/c^2)
g(LT) = t2/(T/g) = g(1 + v/c - v^2/c^2)
Now we can compare the ratio g(LT) for different values of v.
Your g(LT) cannot be compared to g, because it was computed from
a different S time interval.
I compared g(LT) to g, simply because t2 = g(TL)*t2', by analogy with
t1 = gt1'.
Interval comparisons require two times. t1=gt1' is just a
convenient shorthand for (t1-0)=g(t1'-0), when the clocks were
synchronized at zero. You need to identify the starting and
ending times in both systems for your g(LT).


It is implicit that the clocks have to be beforehand synchronized.

You computed g from the interval from 0 to T, but you computed
g(LT) from the S interval from 0 to T(1+v/c-v^2/c^2), then
divided it by T. Your numbers are meaningless.

They are not meaningless, as t2 = g(TL)*t2'.
g(TL) should at least have a relation with g, that whould have a
physical meaning.
Using o2 and o2' to designate the starting times, the correct
equation is
(t2-o2)=g(t2'-o2')

Identify o2 and o2', and you will find that g(LT)=g

g(LT) is a ratio, which is a function of v:
g(LT) = t2/(T/g) = g(1 + v/c - v^2/c^2)
And g(LT)/g = 1 + v/c - v^2/c^2.
Could you show how you get g(LT)/g = 1, when v<>0 ?
You did not identify the starting times of the intervals.
Because your analysis is based on t1'=t2', you must also use
o1'=o2' (as you wrote above, "the clocks have to be beforehand
synchronized"). The length of the interval is L, and the left
end is located at the origin of S':
x1'=0
x2'=L

At o1'=0, we find:
o1=g(o1'+v0/c^2)
=0
o2=g(o2'+vL/c^2)
=g(vL/c^2)

At t1'=T', we find, as before:
t1=gT'
t2=g(T'+vL/c^2)

It is obvious that t2-o2=g(t2'-o2') and t1-01=g(t1'-01'), with
the same value of g at both ends.

And what 'g' do you get in between?


(You can also do this directly from the equation t2=t1+vL/c^2.
When o1=0, o2=vL/c^2. If you insist on using L=(c-v)T, get
o2=v^2T/c^2-vT/c. Either way, the L drops out when comparing
time intervals.)



As far as I can see, g depends only on the velocity, and is the
same at every point.
Both g and g(LT) depend only on velocity.
No, you demonstrated a difference due to position. Your error
introduced a dependence on position.
One can explain the variation with v of the ratio g(LT)/g by referring
to the LT,
which is evident, as g(LT) has been derived from them, but you can't
give
it a coherent physical meaning. This proves that the LT are physically
incoherent.
You are correct. Your g(LT) has no coherent physical meaning,
but, because it is a symptom of your confusion, and nothing to
do with the LT, you have proven nothing.


g(LT) has nothing to do with the LT? The confused one is you.
No, because you used clocks that were not synchronized at the
beginning of the interval, but are synchronized at the end, it
is obvious that they are running at different rates, and the
comparison is meaningless.

If you use synchronized clocks, you find that g has the same
value everywhere.


To calculate the time dilation factor, you need to compare time
intervals between the systems.

Let us look at a stick of length L', at rest in S', and
calculate the S times when t' is 0 and T'.

You make again the same logical mistake, because S' remains the moving
frame.
Iow, the stick continue to move wrt the S-frame, even if one considers
it at rest
in the S'-frame.
I fully agree that a stick at rest in S' is moving in S. If I
made a logical mistake, then we made it together.
Then what is the purpose of stressing that the stick can be considered
at rest in the moving frame? Jumping from one frame to another, make
logical mistakes and draw wrong conclusions?

Any frame can be considered as the rest frame; it is a feature
of Galilean relativity. The stick is at rest in the frame in
which it has zero velocity.

No, physically, not any frame can be considered as the rest frame.
What, then, is the physical content of Galilean relativity? (As
Newton put it, "The motions of bodies included in a given space
are the same among themselves, whether that space is at rest, or
moves uniformly forwards in a right line without any circular
motion.")

How about if I more clearly state that the frames must be
inertial?

Yes, but those SRists who consider that the Earth can be considered
as orbiting around an Earth satellite forget that in SR, the frames
must be inertial.



When you tried to derive your equations, you had to put in the g
factor by hand. You made an ad-hoc assumption about light speed
along y', without physical justification.

It is now the nth time that I claim that the physical justification for
y'=ct' is that
the vector v is perpendicular to the vector c, hence, vectorially,
c+v=c.
From that equation, then the light speed in the original frame
is sqrt(c^2+v^2), but you stated that it was c. (Remember that
you affirmed that relative velocity was invariant).

What is the original frame? I presume that you mean the rest frame.
Yes, in S, the velocity ot the light signal following the y'-axis is
sqrt(c^2+v^2), but it remains c in S'.

Previously, you stated that the speed of the light signal in the
rest system was 'c'. Which is it?

Then I was inattentive, sorry.



You obtained a result that is inconsistent with the results we
can obtain from considering the x direction alone (which
requires that a=1).
There must be some flaw in your assertion, as I obtain physically
meaningful
transformations.
Only when g=1. I demonstrated earlier that, from the x
direction alone, we can conclude a(u-v)*a(v)=a(u). That
requires that a=exp(bv), with b constant. From a(v)=a(-v), we
find that b=0, so a=1.


Did you try your demonstration in the case where x'=ct' instead
of x'=(c-v)t' ?
I used x=ut and x'=(u-v)t' to derive the relation for any value
of u.

So, it doesn't apply to x'=ct'. Anyhow, it is wrong:

For the moment let us ignore the y direction. We agree that,
with the assumption of an invariant relative velocity we have
reached:

x'=a(v)(x-vt)
t'=a(v)t

'a' is a function of v, a(0)=1, and from symmetry a(v)=a(-v).

When v = 0, a(0)=1, x'=x and t'=t. End of the story!

From these equations, with x=ut, we get

x'=a(ut-vt)
=(u-v)t'

Yes.

So, for any velocity in the x direction, the relative velocity
between that point and the S' origin is invariant.

The transformation from S' to S" is
x"=a(u-v)(x'-(u-v)t')
t"=a(u-v)t'

What is S"?
What is the velocity S" relative to S'? It seems to be u-v.

Combining these, we get the transformation:
x"=a(u-v)a(v)(x-vt-(u-v)t)
=a(u-v)a(v)(x-ut)
t"=a(u-v)a(v)t

????

The direct equations give:
x"=a(u)(x-ut)
t"=a(u)t

????????

Consistency requires that a(u-v)*a(v)=a(u).

Setting u=0, we find that a(v)*a(-v)=1. This gives a(v)^2=1, or
a(v)=1 (the negative root is excluded, because a(0)=1).

Setting u=0, we have
x'= (u-v)t'= -vt'
t'= t/g

We do not need to examine the y direction to conclude that your
transformation must reduce to:
x'=x-vt
t'=t .

Marcel Luttgens

.

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