Re: SR fundamental contradiction


<mluttgens@xxxxxxxxxx> wrote in message news:1161509686.026227.152980@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

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'.

An expression like "its time should be" shows that you
have no idea what events and coordinates are.

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.

At no point in your derivation, the meanings of x, t, x' and t'
is understood by you :-)

Dirk Vdm


.

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