Re: TIME DILATION


"Stamenin" <tasko.s@xxxxxxxxxxx> wrote in message news:1155690017.627763.167180@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

[snip]

I wrote the relations of the Lorentz transformation noted with 1) 2) 3)
and 4). I consider that these relations are valid and for the
description of the motion of a material body with a mass (m) and the
coordinates x in the coordinate system let say K atached to the
railways and the coordinate x' in the coordinate system K' atached to
the train. The t and t' are zero when the two coordinate systems are
superposed. When we know the coordinates x and t we can calculate x'
and t' with the relations 1) and 2). The invers passing we can do with
the relations 3) and 4). So we can conclude tha the relations 1) and 2)
are the inverse relations to the relations 3) and 4) and not 1) with 3)
or 2) with 4).
That is why we can't do the changes with Dt and Dt'.

By the way, while I wait for your reply to my yesterday's
message (the one with Dt = Dt' = 0, remember), ... you say
we can't combine (1) with (3) or (2) with (4).

Look, from your numbered equations below we have:
x' = (x-v t) / sqrt(1-v^2/c^2) (1).
t' = (t-v x/c^2) / sqrt(1-v^2/c^2).......(2)
x = (x'+v t') / sqrt(1-v^2/c^2) ...........(3)
t = (t'+v x'/c^2) / sqrt(1-v^2/c^2)........(4)


Write (1) and (3) as
g = 1/sqrt(1-v^2/c^2)
x' = g ( x - v t ) (1).
x = g ( x' + v t' ) (3).
put (3) in (1) and you get:
x' = g ( g ( x' + v t' ) - v t )
<==> x' = g^2 x' + g^2 v t' - g v t
<==> x' ( 1 - g^2 ) = g v ( g t' - t )
<==> x' ( 1 - 1/(1-v^2/c^2) ) = g v ( g t' - t )
<==> x' ( - v^2/c^2 / (1-v^2/c^2) ) = g v ( g t' - t )
<==> - x' v^2/c^2 g^2 = g v ( g t' - t )
<==> - g v x' / c^2 = g t' - t
<==> t = g t' + g v /c^2 x'
<==> t = g ( t' + v x' / c^2 )
<==> t = ( t' + v x' / c^2 ) / sqrt(1-v^2/c^2)
which is your equation (4)
Magic? No, just algebra. Okay?
Exercises below.


Of course they can be deduced from the transformation.
Look, this is the transformation:
{ x' = ( x - v t ) / sqrt(1-v^2/c^2)
{ t' = ( t - v x /c^2) / sqrt(1-v^2/c^2)
from which you get
{ Dx' = ( Dx - v Dt ) / sqrt(1-v^2/c^2)
{ Dt' = ( Dt - v Dx /c^2) / sqrt(1-v^2/c^2) .
For events satisfying Dx = 0 the last equation gives
Dt = Dt' sqrt(1-v^2/c^2)

I don't thik so, let take the basic relations of the Lorentz
transformation.
x'=(x-vt)/sqrt(1-v^2/c^2) (1).
t'=(t-vx/c^2)/sqrt(1-v^2/c^2).......(2)
which are the relations for obtaining the x' and t' on the train when
we know t and x on the railway, and the inverse relations then will be:
x=(x'+vt')/sqrt(i-v^2/c^2) ...........(3)
t=(t'+vx')/c^2)/sqrt(1-v^2/c^2)........(4)

Likewise, you can take *any* two equations from the system
(1), (2), (3), (4), and simply deduce the remaining two equations.

Exercise_1: what happens when you combine (2) and (4)?
Exercise_2: what happens when you combine (1) and (4)?

Dirk Vdm


.

Relevant Pages

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