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:
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.
Apparently, you don't understand your equations.

It is clear that I meant that the length L' of a moving stick is
contracted by g
relative to a stick of length L at rest. In S', the stick is at rest,
so its length is L.


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 get the same value of g at any point you choose. The
position, represented by <L>, drops out of the result.


(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.)


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

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.
Are you now saying that, in S, you have <y=ct>?

Can't you read? I just said that in S, the velocity ot the light signal

following the y'-axis is sqrt(c^2+v^2), but it remains c in S'.
Hence (cf. my derivation), y = sqrt(c^2 - v^2) * t and y'=ct'.



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:
By choosing u=c+v, you get the case of x'=ct'. It applies for
any velocity.


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"?
S" is the frame moving with the object with speed u.

What is the velocity S" relative to S'? It seems to be u-v.
Yes, that follows from the equations we already have.


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

????
OK. I will do it again slowly.
The transformation from S' to S", with speed <u-v> is
x"=a(u-v)(x'-(u-v))t
t"=a(u-v)t'

You don't seem to understand the meaning of the transformation
x'=(x-vt)/g
t' = t/g,
where x is the position of some object in frame S,
v is the velocity of a frame S' relative to S,
x-vt is the difference in S between x and the origin of frame S'
after a time t,
x'=(x-vt)/g is the coordinate of the object in S'.

In the former discussions, we used x=ct, but you can of course
choose x=ut. Then x'=(u-v)t/g. But notice that x' is the coordinate
of the object in S'.

Now you consider the frame S" of the object, and claim that
the transformation from S' to S", with speed <u-v> is
x"=a(u-v)(x'-(u-v))t"
t"=a(u-v)t'

This is plainly wrong, because the coordinate of the object
in S" is *zero*, thus x"=0.

Your demonstration must be a hoax!

Marcel Luttgens


Substituting the equations we already have for the
transformation from S to S':

x"=a(u-v)(a(v)(x-vt)-(u-v)a(v)t)
=a(u-v)a(v)(x-vt-(u-v)t)
=a(u-v)a(v)(x-vt-ut+vt)
x"=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

????????
These are simply your equations for the case v=u, giving the
transformation from S to S".



Consistency requires that a(u-v)*a(v)=a(u).
This is a functional equation, that must be satisfied by a, for
all values of u and v.


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'
I don't know if you were trying to advance your objection, or
just posting an equation. Was there a point?

t'= t/g
You ignored the functional equation, and substituted a=1/g,
without any justification. (Of course, the functional equation
can be expressed in terms of g, and, after inverting, gives
g(u-v)g(v)=g(u), leading to g=1.)


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

.

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