Re: A twin contraction paradox


"tomtom" <Carmam1534@xxxxxxx> wrote in message news:1167914793.972501.61010@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx


Just point at what you think is the error and explain what
is wrong with it.

This I have done, and will now do again. You say that according to B,
dx' = L ( = 1 meter), so that is the length to use. You also say that
according to B, dx' = 0, so you should use that, but you don't.

You probably mean dt' = 0.

This is
not consistent. Consistency is using the contracted length at all
places in the relevant equation, or using the proper length - not
mixing them. As the prongs are 0.866m apart when they strike A (as
measured from A of course), that is the length to use. Please try not
to make disparaging personal remarks, keep to the subject at hand. Your
or my argument will stand or fall on its own merits, such comments as
being unable to read are to say the least unhelpful. Please tell me why
you use dx' = 1, but do not use dt' = 0 .

And I don't use dt' = 0?

Look carefully - for the seventh time:

How much does B's aft prong strike A before B's front prong?
You know that they strike A simultaneously according to B, so
dt' = 0 <=== LOOK HERE: dt' = 0
and you know that the distance according to B between them is L, so
dx' = L.
You can use the equation
dt = g ( dt' + v/c^2 dx' ) <=== LOOK HERE:
we will take dt' = 0 and dx' = L
= g v/c^2 L <=== LOOK HERE:
we took dt' = 0 and dx' = 0
= g L/(2c) (with your v = c/2)
That is the time the aft prong strikes before the front prong.
So the aft prong of B strikes A during this time.
Since it has a velocity v = c/2, the length of the scratch on
A according to A, is the time multiplied with the velocity.
We get
g L/(2c) c/2
or in your notation
(gamma * L / 2c ) * c/2

Just point at what you think is the error and explain what
is wrong with it.
And this time, try very hard to avoid writing that I don't
use dt' = 0. I know it seems difficult, but do your best.

Dirk Vdm

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