Re: A twin contraction paradox

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

(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.

dx = L / g (your 0.866m)

You asked me to spot the typo(s). There are many typos and errors in
the above. The first is the distance between the prongs which is not dx
but dx' . Then there is double use of the word "must", then comes dx' =
L : L in this thought experiment is a constant with a value of 1.
It is the distance apart of the prongs on A as seen from A, or on B as
seen from B. Then we come to the three equations. The first two have no
bearing on the distance apart of the prongs on B as seen from A, and
although it is irrelevant, the second uses L when it should use dx' .
The third is the correct equation, if you make it dx' instead of dx
(which presumably is the typo you refer to).

(2) Calculation of how much B's aft prong strikes A before
B's front prong strikes A:
You know that they strike A simultaneously according to B, so
dt' = 0 <=== LOOK HERE: dt' = 0
i.o.w. THE FACT THAT THEY STRIKE SIMULTANEOUSLY
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
i.o.w. THE FACT THAT THEY STRIKE SIMULTANEOUSLY
= g v/c^2 L <=== LOOK HERE:
we took dt' = 0 and dx' = 0
i.o.w. THE FACT THAT THEY STRIKE SIMULTANEOUSLY
= 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

Do you understand the difference between the two calculations?
Yes.
Can you understand the first without looking at the second?
Yes
Can you understand the second without looking at the first?
Yes
Do you realize that the dx, dt, dx' and dt' in the first represent
 *different* things than in the second?

dx, dt, and dt’ do not appear in either calculation – except of
course where you erroneously used them in the first two equations in
calculation 1. dx’ in the first calculation refers to the distance
apart of the prongs on B as seen from A. As it does not appear in the
second except where I have used it below, the question is rather
immaterial.

As I said above, I will jump through no more hoops for you. You can now
answer my question, which has gone unanswered for all this time. Why do
you insist on using L and then dx’ in the same equation to calculate
the total distance apart of the start of the scratch marks on A as seen
by A?

You use :- TL = (g * L / (2c) * c/2) + dx' = (1.155 * 1 / 2 * 0.5)
+ 0.866 = 1.155 This is inconsistent

All these calculations are from within IFR A, or if you prefer it, with
respect to IFR A. The prongs which strike A are 0.866m apart, so that
is the length to use, not 1m.

So I use :- TL = (g * dx' / (2c) * c/2) + dx' = (1.155 * 0.866 / 2
* 0.5) + 0.866 = 1.116 This is consistent

Do not answer by saying this is the way we do it, or this is the way
Einstein did it. Do not say that you have explained all this to me
already – you have not explained – you have dictated. You have said
that this is the way it is done. That is not an explanation. I have
asked a simple English language question, which should be answered by
an equally simple English language answer.

.

Relevant Pages

  • Re: A twin contraction paradox
    ... You know that they strike A simultaneously according to B, ... that is the distance apart of the prongs as seen from B. You ... proper length of 1 meter, and 0.866 refers to the measured length. ...
    (sci.physics.relativity)
  • Re: A twin contraction paradox
    ... dx' and dt' to represent distances and times measured in IFR B. ... A quantity dx is measured parallel to the x-axis ... We want to know the distance dx according to A of the ... prongs on B. ...
    (sci.physics.relativity)
  • Re: A twin contraction paradox
    ... talk about two events on A's prongs and another part talks ... I will repeat a part and add a new calculation at the end. ... A measures his own prongs to be a distance L apart. ... We translate this physical statement into coordinates language now: ...
    (sci.physics.relativity)
  • Re: A twin contraction paradox
    ... while the distance apart of the start of the scratches on A as ... prongs on B to get the total distance apart of the scratch marks on A. ... Now look at your earlier posting of Dec 10 ...
    (sci.physics.relativity)
  • Re: A twin contraction paradox
    ... while the distance apart of the start of the scratches on A as ... prongs on B to get the total distance apart of the scratch marks on A. ... Now look at your earlier posting of Dec 10 ...
    (sci.physics.relativity)
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