Re: A twin contraction paradox


"tomtom" <Carmam1534@xxxxxxx> wrote in message news:1167764949.829123.194270@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Dirk Van de moortel wrote:
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
and you know that the distance according to B between them is L, so
dx' = L.

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

Hello everybody, I trust you had a merry Christmas and a happy new
year.

Now back to the point at hand. We don't seem to be getting anywhere do
we?

No, you don't seem to get anywhere indeed.

I am now going to point at the error (again) and tell you what is
wrong. You (being third person plural means all relativists) insist on
using dx' = L (ie 1 meter) in the first part of the formula, because
that is the distance apart of the prongs as seen from B. You
conveniently ignore the fact that as seen from B, dt' = 0. If you
insist on using dx' = 1, then to be consistent, you must use dt' = 0.
This you do not do.

You seem to be unable to read, so I will repeat (for the 6th
time, if I count correctly):

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
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' )
= g v/c^2 L
= 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.

Dirk Vdm


.

Relevant Pages

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