Re: A twin contraction paradox
- From: "tomtom" <Carmam1534@xxxxxxx>
- Date: 9 Jan 2007 06:37:19 -0800
Dirk Van de moortel wrote:
Yet your first
2 calculations are calculating how far apart the scratch marks on B are
as seen from A.
That is not what I calculated.
Dirk Vdm
Maybe it is not what you intended to calculate, but that is what you
did.
You can use the constant L and leave it at L for as long as you like.
What you must not do is to lose sight of what it represents, or the
fact that it cannot be changed. In this thought experiment it
represents 1, and is a fixed value. You still cannot see your error in
your calculation (1)
You wrote :-
Calculation of how far B's prongs are apart according to A
dx' = g ( dx - v dt )
giving
L = g ( dx - v 0)
Giving
dx(actually dx') = L / g
Constants :- dx = L = 1 : v = 0.5 : dt = 0 : gamma =
1.155
Now put the values in
1. dx' = g ( dx - v dt ) = 1.155 * (1 - ( 0.5 * 0 )) = 1.155
giving
2. L = g ( dx - v 0) = 1.155 * (1 - 0.5 * 0)) = 1.155 Which
is 1 = 1.155 ??!!
giving (as dx' now has the value 1.155)
3. dx' = L / g = 1 / 1.155 = 0.866 Which is 1.155 = 0.866
??!!
The above three equations are clearly nonsense when put together.
Equation 1 on its own calculates the distance apart of the scratch
marks on B caused by A, and dx' does not represent the same thing as
dx' in equation 3. You are being confused by the fact that there are
two separate things to calculate, there is the distance apart of the
prongs and the distance apart of the start of the scratch marks. The
prongs could be represented by dx' and the scratches by dy' . What
you said we should be calculating is how far B's prongs are apart
according to A.
Equation 2 is nonsense in its own right as it has 1 = 1.155 as its
solution.
Equation 3 is also nonsense if it is used in conjunction with 1 and 2.
If it is not used in conjunction with 1 and 2, it correctly calculates
the distance apart of the prongs on B as seen from A.
4. dx' = L / g = 1 / 1.155 = 0.866
Now it is your turn to read what I have written, and reply sensibly.
By the way, why use both L and dx' in the same equation to refer to the
distance apart of the start of the scratch marks on A?
.
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