Re: SR fundamental contradiction
- From: Brian Kennelly <bwkennelly@xxxxxxx>
- Date: Tue, 24 Oct 2006 08:44:12 -0700
mluttgens@xxxxxxxxxx wrote:
Brian Kennelly wrote:Your equations put in length contraction and time dilation by hand, while SR shows that they follow from the light postulate and the principle of relativity.mluttgens@xxxxxxxxxx wrote:Brian Kennelly wrote:It seems rather ad-hoc and inconsistent. What is the speed if,mluttgens@xxxxxxxxxx wrote:I assume that y'=ct', simply because the vector v is thenBrian Kennelly wrote:You have <x-vt=(c-v)t> when x=ct. You previously derivedmluttgens@xxxxxxxxxx wrote:Yes, x'=(c-v)t *according* to S, but *in S'* -as S' is moving at vBrian Kennelly wrote:As far as I can see, you answered in the affirmative, thatmluttgens@xxxxxxxxxx wrote:The answer is given in mine (and your) almost simultaneous post.Brian Kennelly wrote:If you look at the text quoted above, you stated > in S', aftermluttgens@xxxxxxxxxx wrote:I thought that my sentence was clear enough: According to S, etc...Not at all, the factor (c-v) doesn't mean that light speedI don't follow.
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'.
You state that:
x'=(c-v)t
and
x'=(c-v)t'
From these two equations, we find that t'=t.
Thus, I stated that x = (c-v)t and x' = (c-v)t'.
a time t, light has travelled a distance
(c-v)t.Does that not say that x'=(c-v)t?
x'=(c-v)t. From that and your postulated x'=(c-v)t', we derive
t'=t when x=ct.
relative to S-, its time should be t'<>t, hence x'=(c-v)t'.
<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 ofActually, your equation requires that t'=at when x=ct, and, as I
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.
showed, t'=at for all transformations.
You had <y=sqrt(c^2-v^2)t> and <x=vt> for the light signal alongAs shown above, one has x'=(c-v)t', not x'=ct'.Then why do you reject x'=ct'? If c is independent of theYou 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'.
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.
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'.
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.
perpendicular to the vector c. This justify the velocity c
along the y' axis in S'.
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 isYour equations also rely on the t1'=t2' assumption, because you
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 ?
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.
Your transformations cannot be derived from any consistent assumptions, unless g=1, in which case they reduce to the Galilean.
Assuming g <>1 in your equations leads to transformation equations that are different for each observer, and which predict different effects for positive and negative velocities for most observers. IOW, they have a built in preferred reference system.
That is not an ad-hoc assumption. You must mark both ends of a moving object at the same time to measure its length.
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.
It is also possible to measure the length of a moving object, if you know its speed, by measuring the time it takes to pass a fixed point. Using this very natural method to measure length, the LT lead to contraction of moving objects without the need to invoke t1'=t2' or t1=t2.
If one choose t1=t2, t1'<>t2'You don't get anything directly meaningful. It is similar to measuring the time of a trip from LA to NY by using the local time of departure and arrival. There is a systematic error that skews the results. Each observer must consistently use his own time reference.
and one doesn't get length contraction.
Their predicted 'relativeI do not agree. It would be a defect if it was possible to demonstrate absolute simultaneity, but this has not been done.
simultaneity' is thus their rehibitory defect.
But SRists,The validity of SR must stand on its proven internal consistency, and on its agreement with observation. On those standards, it continues to be accepted.
unscientifically and illogically, will play down such lack of coherence
and continue to defend the validity of the LT.
If you have an alternative theory, you must enumerate its assumptions, and demonstrate that it can explain the observations as well as, or better than, SR.
It appears that your suggested transformations are inconsistent with the principle of relativity. This alone does not make them wrong, but you will need to provide some very strong arguments for such a move.
.
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