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:
mluttgens@xxxxxxxxxx wrote:
Not at all, the factor (c-v) doesn't mean that light speed
is not c anymore. According to S, light is at ct and S' at vt,
hence in S', after a time t, light has travelled a distance
(c-v)t. To express such distance in S', one has to use t', thus
x' = (c-v)t'.
I don't follow.
You state that:
x'=(c-v)t
and
x'=(c-v)t'

From these two equations, we find that t'=t.

I thought that my sentence was clear enough: According to S, etc...
Thus, I stated that x = (c-v)t and x' = (c-v)t'.
If you look at the text quoted above, you stated > in S', after
a time t, light has travelled a distance
(c-v)t.
Does that not say that x'=(c-v)t?
The answer is given in mine (and your) almost simultaneous post.
As far as I can see, you answered in the affirmative, that
x'=(c-v)t. From that and your postulated x'=(c-v)t', we derive
t'=t when x=ct.
Yes, x'=(c-v)t *according* to S, but *in S'* -as S' is moving at v
relative to S-, its time should be t'<>t, hence x'=(c-v)t'.
You have <x-vt=(c-v)t> when x=ct. You previously derived
<x'=a(x-vt)> From these, we find <x'=a(c-v)t> when x=ct.

You then assert that <x'=(c-v)t'> when x=ct. I assume that this
is based on keeping the relative velocity invariant for the
light signal.

This yields <t'=at> when x=ct.

*******************************************************
Note that if the origin of S is moving to the left at v,
according to S', we can derive:
(-vt')=a(-vt)
t'=at when x=0
Using your time equation, we get
at=kt
So, a=k

From that result, and the previous equation, we find
at=ect+at
e=0
*******************************************************

Generally, if the relative velocity is invariant for three
different velocities, the transformation equations become:
x'=a(x-vt)
t'=at

We can see that all relative velocities in the x direction are
invariant.

In your original derivation, you required this condition for v,
c and -c. I derived it from 0, v, and c. (There is nothing
special about c in this case.)

Notice that x' represents the difference between the endpoint of
the light signal and the origin of S', the same endpoint having
the coordinate x=ct in S.
At this point of my derivation, the link between t' and t
is not yet known.
Actually, your equation requires that t'=at when x=ct, and, as I
showed, t'=at for all transformations.

You have not explained how you derived y'=ct'.

y' = ct' is a direct consequence of the hypothesis that c is
independent
of the motion of S'.
Then why do you reject x'=ct'? If c is independent of the
motion of S', it follows as directly as y'=ct'.

If you disagree, please explain how y'=ct' is a direct
consequence, but x'=ct' is not.
As shown above, one has x'=(c-v)t', not x'=ct'.
The case of y' is wholly different, because the light signal
follows the y' axis, which is moving at v wrt the y axis.
Iow, the vector v is perpendicular to the vector c, not
parallel as in the case of x,x'.
You had <y=sqrt(c^2-v^2)t> and <x=vt> for the light signal along
the y' axis.

(Assuming y'=y, from symmetry arguments, and using A for the
sqrt to shorten the equations)
y'=At
x'=0
t'=at

y'=At'/a

Now the relative velocity between the signal and the S' origin
is A. You assert that in S', it is c, so relative velocity is
invariant along the x axis, but not in other directions? Why is
the speed 'c' along the y' axis in S'? At what angle does the
relative velocity cease to be invariant, and the light signal
move with speed 'c'?

If we assume that relative velocity is invariant, we get
<y'=At'>, which results in a=1.

I assume that y'=ct', simply because the vector v is then
perpendicular to the vector c. This justify the velocity c
along the y' axis in S'.
It seems rather ad-hoc and inconsistent. What is the speed if,
in S', v and c form an angle of 89.9 degrees, or 45 degrees?

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).

From these equations, with x=ut, we get
x'=a(ut-vt)
=(u-v)t'

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'

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).

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


The resulting transform is

x' = (x-vt)/g
t' = t/g,

the inverse transform being

x = g(x'+vt')
t = gt'

Allow me to get back to your 'long way' derivation of length
contraction:

BEGIN
x1'=g(vT-vt1)
x2'=g(cT-vt2)
t1'=g(t1-v^2T/c^2) (x1=vT)
t2'=g(t2-vT/c) (x2=cT)

Setting t1'=t2', we get
t2=t1-v^2T/c^2+vT/c

Choosing t1=T, so that x1'=0
t2=T(1-v^2/c^2+v/c)
So:
x2'=g(cT-vT(1-v^2/c^+v/c))
=g(cT-v^2T/c-vT(1-v^2/c^2))
= g(cT(1-v^2/c^2)-vT(1-v^2/c^2))
=g((c-v)T(1/g^2)
=(c-v)T/g
END

You got x1'=0
To get x2'=(c-v)T/g, you had to assume that t1'=t2'.
Then x2'-x1'=(c-v)T/g
As (c-v)T represents x2-x1, which is the length L of the
stick in S, you demonstrated length contraction. Indeed,
L' = L/g

The 'short way' use the inverse LT

x1 = g(x1'+vt1')
x2 = g(x2'+vt2')

x2-x1 = g(x2'-x1') + gv(t2'-t1'), or
L = gL' + gv(t2'-t1')

To get the correct result L = gL', or L' = L/g, you
*have* to assume that t1' = t2'!

Otoh, length contraction is straightforwardly obtained with
my transformation x'=(x-vt)/g:

x1' = (vT-vT)/g = 0
x2' = (cT-vT)/g
x2'-x1' = (cT-vT)/g = (x2-x1)/g
L' = L/g

So, why should one prefer the Einsteinian LT, who *needs* an
assumption (t2'=t1') whose only justification is to obtain the correct
result, to my transformation ?

Your equations also rely on the t1'=t2' assumption, because you
have absolute simultaneity. t1=t2 implies t1'=t2'.

The real problem with your equations is that they do not satisfy
the principle of relativity.

I will simply remind that the proof of the pudding is in the eating.
Iow, my equations, even if you quibbled about their derivation, lead
straightforwardly to two physically well attested results, i.e.
length contraction and time dilation. And luckily, they have absolute
simultaneity.

Otoh, the Einstein transformation (the LT), precisely because they
predict the so-called relativity of simultaneity, a phenomenon that
by the way has never been proved, need the ad-hoc assumption that
t1'=t2' to derive length contraction. If one choose t1=t2, t1'<>t2'
and one doesn't get length contraction. Their predicted 'relative
simultaneity' is thus their rehibitory defect. But SRists,
unscientifically and illogically, will play down such lack of coherence
and continue to defend the validity of the LT.

Marcel Luttgens

.

Relevant Pages

  • Re: SR fundamental contradiction
    ... if the relative velocity is invariant for three ... different velocities, the transformation equations become: ... stick in S, you demonstrated length contraction. ...
    (sci.physics.relativity)
  • Re: SR fundamental contradiction
    ... if the relative velocity is invariant for three ... different velocities, the transformation equations become: ... follows the y' axis, which is moving at v wrt the y axis. ...
    (sci.physics.relativity)
  • Re: SR fundamental contradiction
    ... if the relative velocity is invariant for three ... different velocities, the transformation equations become: ... follows the y' axis, which is moving at v wrt the y axis. ...
    (sci.physics.relativity)
  • Re: SR fundamental contradiction
    ... To express such distance in S', one has to use t', thus ... I assume that this is based on keeping the relative velocity invariant for the light signal. ... follows the y' axis, which is moving at v wrt the y axis. ...
    (sci.physics.relativity)
  • Re: Deriving space-time interval
    ... On the other hand, to accept the Lorentz transformation, you must ... relative velocity was along your new axis. ... axis, I merely just have to find another axis that does align. ...
    (sci.physics.relativity)
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