Re: A twin contraction paradox


"tomtom" <Carmam1534@xxxxxxx> wrote in message news:1168208899.123948.5180@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Dirk Van de moortel wrote:
> Apart from answers to the questions at the end, I expect you to
> find and show a little typo I made in my other reply.
> We don't "stay in IFRs".
> 10 December:
> | No, Tom, we are at our desk or table or whatever.
> | We *oversee* the frames K and K', and use coordinates x and t
> | and x' and t' to talk about events as measured in those frames.
> | I already told you this. Please try to keep this in mind.
> You seem to be unable to keep it in mind.

I am perfectly able to keep in mind that we are conducting a thought
experiment, and the spaceships are not moving any more than we are. To
avoid confusion, I use the phrase stay in one IFR as a shorthand, and
you know what it means as well as I do - so please stop bantering
semantics again.

Ah, but this is not about semantics.
This is about you failing to understand the activity of doing a
thought experiment and even more failing to understand the
very basics of the theory in which these experiments are
conducted.
You don't understand that we calculate what would or could
be measured in one frame when the measurements made in
another frame are known.

I will now answer your questions but I will jump
through no more hoops for you.

I'm not asking you to jump through hoops for me. I'm asking
you to do a little effort to understand the very root of your
gross mistakes. For some reason (that I perfectly understand,
by the way), you seem to be not interested.

>
> (1) Calculation of how far B's prongs are apart according to A:
> If A wants to measure the distance dx between the prongs, he
> must must measure the coordinates of the prongs simultaneously
> (dt = 0), and we know the distance between the prongs
> according to B, namely dx' = L.
> So we use an equation that contains dx, dt and dx' and we don't
> care about dt' because if B wants to measure the distance between
> *his* prongs, he can measure the x'-coordinates at *any* time.
> So we use
> dx' = g ( dx - v dt )
> giving
> L = g ( dx - v 0 )

These two equations are totally wrong :- dx’ [or L] = 1.155 * (1
– (0.5 * 0)) = 1.155 . This is the distance apart of the scratch
marks on B as seen by A, not the distance apart of the prongs on B as
seen by A. Perhaps this error is a consequence of your continual frame
jumping.

Listen, I just EXPLICITLY DEFINED the distance between B's
prongs according to B as dx' = L where L is an abstraction of
your numerical 1 meter.

Now I know where the error is. You look at an equation without
even *trying* to understand the meanings of the variables, and
then of course you go in a panic. You refuse to even read what
I write.

When you are prepared to read what I write *before* looking
at the equations, feel free to reply again to my previous post.

Dirk Vdm

.

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