Re: SR fundamental contradiction


Dirk Van de moortel wrote:
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Dirk Van de moortel wrote:
<mluttgens@xxxxxxxxxx> wrote in message news:1159879051.169882.3140@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Dirk Van de moortel wrote:
<mluttgens@xxxxxxxxxx> wrote in message news:1159791347.852389.224720@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Dirk Van de moortel wrote:
<mluttgens@xxxxxxxxxx> wrote in message news:1159737966.456569.228880@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Dirk Van de moortel wrote:
<mluttgens@xxxxxxxxxx> wrote in message news:1159692403.878602.225890@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Dirk Van de moortel wrote:
<mluttgens@xxxxxxxxxx> wrote in message news:1159612637.044922.196540@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Dirk Van de moortel wrote:
<mluttgens@xxxxxxxxxx> wrote in message news:1159535570.216186.311940@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

[snip repetitive demonstrations of your imbecility]

Where does the length of the stick come from, if not from
the LT x' = g(c-v)t ?

Hey, retard, when I tell you to imagine a stick of length 5, do
you ask where 5 comes from?
Yes, that figures. Okay, I'll tell you a secret: it comes out of thin air.
How is that?

You said:

"For this event E, as seen in S', the light signal has covered the
distance
x' = c t' = g (c - v) t
This is a distance of the event E in the S' frame.

Now imagine a stick with this particular length
x' = g (c-v) t
at rest in the S' frame."

And now, you claim that its length comes out of thin air!

Again, you forgot my opening line:
| Consider the event E on the light signal with x = c t for some
| chosen value of t.
"For some chosen value of t"... that's your thin air.


You are a stupid liar!

I'm sorry, but you are too stupid to be qualified to know whether
someone is lying to you or not.
That is quite Amusing :-)

Of course, one can choose any value for t. What counts is the
formula x' = g (c-v) t, which means that the distance x = (c-v)t
measured in the S-frame is *dilated* by g in the S'-frame, whereas
it should be *contracted* by 1/g.

No, Marcel, it does not.
the formula x' = g (c-v) t is not what counts.
What counts is the meanings of the variables.
What counts is that you never understood them and you never
will. You invested too heavily in failing to understand, remember?
http://perso.orange.fr/mluttgens/

The meaning of the variables is clear to everybody, and should be
clear, even to you.

Alas, clearly not clear to you ;-)
http://perso.orange.fr/mluttgens/LTfalse.htm
http://perso.orange.fr/mluttgens/twinpdx1.htm
http://perso.orange.fr/mluttgens/mmx.htm
and
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LuttRel.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/DidntUseSR.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SpeedV.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NegativeCrap.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/ApplyDerivation.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SRSymbols.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/CorrectRelations.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SRLuttgens.html

Dirk Vdm

Van de Moortel wrote in
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Fumblamental.html

"Consider the event E on the light signal with x = c t for some
chosen value of t.
Then c t - v t is the distance between the origin of S' (the 'moving
observer') and the light signal, as seen at time t in the S-frame
(the 'stationary frame'), and, by the way, so c - v is by definition
the closing velocity between the two.

For this event E, as seen in S', the light signal has covered the
distance
x' = c t' = g (c - v) t
This is a distance of the event E in the S' frame."


Let's imagine a stick with length
x = (c-v) t = ct - vt
at rest in the S frame.

Logically, the length of the stick corresponds to the distance
between two points fixed in S, which are occupied by the ends
of the stick simultaneously, i.e. at the same time t.

The coordinates of those two points in the S-frame are:

x2 = c t (the light signal, as seen at time t in the S-frame) and
x1 = v t (the origin of S', as seen at time t).

In S', the corresponding coordinates are, according to the LT:

x2' = c t' = g (c - v) t and
x1' = 0.

Hence the length of the stick in S' is given by
x2' - x1' = g (c - v) t.

No. Length of a moving stick must be measured by taking the
distances to the end points simultaneously.
The events (t,x1) and (t,x2) are not simultanous in frame S':

{ x1' = g ( x1 - v t ) = g ( v t - v t ) = 0
{ t1' = g ( t - v x1 /c^2) = g ( t - v v t / c^2 ) = t / g

{ x2' = g ( x2 - v t ) = g ( v t - c t ) = g (c-v) t
{ t2' = g ( t - v x2 /c^2) = g (t - v c t / c^2) = t sqrt(1-v/c) / sqrt(1+v/c)

Noone (in his right mind) would call x2' - x1' the length of the stick,
since x1' and x2' are distances at *different* times in the S'-frame,
as you can see.

In your right mind, what is the length of the stick in the S'-frame, if
not
g (c - v) t ?

(c-v) t / g

You should of course demonstrate your solution.
Notice that if you find any value different from (c-v)t/g, the
Lt is false. And don't try to escape by telling me that SR has no
solution.

Sigh.
So your stick has length in the S-frame = dx = (c-v) t, with some
chosen value for t. You want t = 5? You get t = 5.

Since the stick is at rest in S, the end-points can be measured
at any time, so dt for the measuring events doesn't matter.
Since the stick is moving in S', the end-points must be taken
simultaneously in S, so the measuring events must have dt' = 0.
Transformation:
{ dx' = g ( dx - v dt ) [1]
{ dt' = g ( dt - v dx / c^2 ) [2]
or
{ dx = g ( dx' + v dt' ) [3]
{ dt = g ( dt' + v dx' / c^2 ) [4]

You want a connection between dx' and dx, where dt' is known
to be 0, so the simplest way to go about is with equation [3], giving
dx = g dx'
and thus
dx' = dx / g

So the length in S' is (c-v) t / g.
So I don't find a value different from (c-v) t / g.

A stick has a length L in its rest frame.
When measured from a moving frame, that stick has length L / g.
What can be so difficult about that?

Your demonstration leads to the correct result, i.e.
the length in S' is (c-v) t / g, but from the *ad hoc* postulate that
"as dt for the measuring events doesn't matter, the measuring
events must have dt' = 0."

Remember that you wrote yesterday: "x1' and x2' are distances
at *different* times in the S'-frame":

Yes, because YOU TOOK THEM at the same time t in S:
>> >> > x2 = c t (the light signal, as seen at time t in the S-frame) and
>> >> > x1 = v t (the origin of S', as seen at time t).

If two events are simultaneous in S then they are not so in S'.
If two events are simultaneous in S' then they are not so in S.
Big deal.

If you want to measure the length of a moving stick, you must
measure the distances to the end-points at the same time.
If you want to measure the length of a non-moving stick, you
can measure the distances to the end-points at any time, since
the end-points of a non-moving stick aren't going anywhere.
Big deal.
Too bad that you don't understand this :;-)


"No. Length of a moving stick must be measured by taking the
distances to the end points simultaneously.
The events (t,x1) and (t,x2) are not simultanous in frame S':

{ x1' = g ( x1 - v t ) = g ( v t - v t ) = 0
{ t1' = g ( t - v x1 /c^2) = g ( t - v v t / c^2 ) = t / g

{ x2' = g ( x2 - v t ) = g ( v t - c t ) = g (c-v) t
{ t2' = g ( t - v x2 /c^2) = g (t - v c t / c^2)
{ = t sqrt(1-v/c) / sqrt(1+v/c)

Noone (in his right mind) would call x2' - x1' the length of the stick,

since x1' and x2' are distances at *different* times in the S'-frame,
as you can see."

You should know (after how many years?) that the value of x' is
always zero. It is wholly independent of any value of t or t'.


yes, that was what I wrote.
This on the other hand, I didn't write:

On the other side, x2' = g(c-v)t at t2' = gt(1 - v/c).
But to t2' = gt(1 - v/c) corresponds t = t2' / g(1 - v/c), and
x1' is also independent of that specific value of t.

You are babbling.

I gave it on a platter:
For your chosen value of t and two events simultaneous in S
with resp. x2 = c t and x1 = v t you get
{ x1' = g ( x1 - v t ) = g ( v t - v t ) = 0
{ t1' = g ( t - v x1 /c^2) = g ( t - v v t / c^2 ) = t / g
and
{ x2' = g ( x2 - v t ) = g ( v t - c t ) = g (c-v) t
{ t2' = g ( t - v x2 /c^2) = g (t - v c t / c^2) = t sqrt(1-v/c) / sqrt(1+v/c)

So x2' is *not* independent on the chosen value of t, so x1'
cannot be like you say "*also* independent" of t. Stop babbling.
You haven't got a clue, Marcel.


Iow, the value of x1' remains 0 at t2', hence the fact that
the events (t,x1) and (t,x2) are not simultaneous in frame S'
is irrelevant, and x2' - x1' = g (c - v) t is the length of
the stick in the S'-frame.

You have NO IDEA about events.
You have NO IDEA about the meaning of the variables.
You just don't know what you are babbling about.


Thus, according to the Einsteinian LT, the stick is *dilated*,
instead of *contracted*, in the S'-frame.

No. I proved to you that the length of a moving stick is
its proper lenght divided by gamma, i.e. *contracted*.
This has been shown to school kids since a *century*,
but I guess you are too stupid to understand that.

If you decide to measure the front and the rear of a
moving train at different times and then call the difference
of those distances the *dilated length* of the train, by all
means be my guest and entertain us some more.


But if you use the correct LT
x' = (x-vt) / g
t' = t /g
you get the expected length contraction.

Marcel, you must be the Ultimate Imbecile ;-)

Dirk Vdm

You are so brainwashed by SR that you cannot think logically
anymore. In fact, you have become a crackpot.

Yes, by applying the LT's, one get

x1' = g ( x1 - v t ) = g ( v t - v t ) = 0
t1' = g ( t - v x1 /c^2) = g ( t - v v t / c^2 ) = t / g

which can be written

x1' = gt(v-v) = gt * 0
t1' = t/g

or

x1' = g^2 t1' * 0

Only a crackpot would deny that the value of x1' (=0) is independent
of t, or t1', or any other time at which one of the end of the stick
is measured in S'.

Otoh,

x2' = g ( x2 - v t ) = g ( v t - c t ) = g (c-v) t
t2' = g ( t - v x2 /c^2) = g (t - v c t / c^2) = t sqrt(1-v/c)

Notice that you made a stupid error, t2' is not t sqrt(1-v/c),
but t sqrt [(1-v/c)/(1+v/c)] !

One can also write

t2' = gt (c-v) / c, thus
t = c t2' / g (c-v)

Hence, x2' = g (c-v) t = c t2', the second end of the stick being
measured at a time t2'.

But at t2', x1', the first end of the stick, still measure 0.

If you deny this, you are a crackpot squared.

So, the length of the stick in the S'-frame is given by
x2' - x1' = g (c-v) t - 0 = g (c-v) t, or
= c t2' - 0 = c t2'

The length is thus *dilated* according to the Einsteinian LT.

You wrote:

"If you decide to measure the front and the rear of a
moving train at different times and then call the difference
of those distances the *dilated length* of the train, by all
means be my guest and entertain us some more."

This shows that you don't even understand the problem.
You are merely babbling.
The correct analogy is that of a car keeping the same
position after any time interval and another car moving away from
the first. After some time interval, the distance between the
two cars is of course increased, not reduced.
I guess you are too stupid to understand that.

Marcel Luttgens

.

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