Twin Paradox - difference between outbound and inbound

[Dale]

The non correct task is "solved" in the article. One "solution" is shown there.
The another "solution" is:
A= ((1-v/c)/(1+v/c))1/2,
D= ((1-v/c)/(1+v/c))1/2
B=0
C=0

That is indeed a linear transform, but it is not a boost of any sort and it is specifically not a Lorentz transform. As @Ibix mentioned, it is a rescaling transform. I cannot think of what this transformation would be useful for, since there doesn’t appear to be any natural meaning for $v$ in that context.

Please remember that all posts should be consistent with the professional scientific literature. If you believe this transform deserves further consideration, please post the peer reviewed scientific publication that derives and explains it.

 

[PeterDonis]

Lorentz formulas are one way ticket?

No, every Lorentz transform has an inverse.

What will you say travelling twin on the back way? What will show his watch?

In the stay-at-home twin's frame, the traveling twin is moving at speed $- v$ on the return leg. So just do the Lorentz transform with $-v$ instead of $v$.

 

[Enovik]

No, every Lorentz transform has an inverse.


In the stay-at-home twin's frame, the traveling twin is moving at speed $- v$ on the return leg. So just do the Lorentz transform with $-v$ instead of $v$.

OK. And Lorentz transform will give the speeding-up of the time $t'$. Is that so? And there is the derivation for the appropriate formulas?

 

[PeterDonis]

And Lorentz transform will give the speeding-up of the time $t'$. Is that so?

No.

And there is the derivation for the appropriate formulas?

Look in any relativity textbook. Or in any number of places online. This information is easy to get.

 

[Sagittarius A-Star]

OK. And Lorentz transform will give the speeding-up of the time $t'$. Is that so? And there is the derivation for the appropriate formulas?

Note down the Lorentz transformation for differences between the coordinates of 2 events:
(1) $\ \ \ \ \Delta x' = x_2' - x_1' = \gamma (x_2-vt_2) - \gamma (x_1-vt_1) = \gamma (\Delta x-v \Delta t)$
(2) $\ \ \ \ \Delta t' = t_2' - t_1' =\gamma(t_2 - \frac{v}{c^2}x_2) - \gamma(t_1 - \frac{v}{c^2}x_1) = \gamma(\Delta t - \frac{v}{c^2}\Delta x)$
with $\gamma=1/\sqrt{1-v^2/c^2}$.

Source:
http://www.scholarpedia.org/article...nematics#Galilean_and_Lorentz_transformations

If the 2 events are ticks of a clock at rest in the unprimed frame, then $\Delta x = 0$.
Time dilation with respect to the primed frame of the clock with $\Delta x = 0$ in equation (2):
$\Delta \tau/\Delta t' = \Delta t_{\text{clock}}/\Delta t' = 1/ \gamma$.

The time dilation-factor is independent of the direction, in which the clock moves. The $\gamma$ factor does not depend on the sign of $v$.

 

[PeterDonis]

Due to the OP's refusal to acknowledge correct responses and consequent unscheduled vacation, this thread is now closed.

 

[Nugatory]

You article can't to be named as Lorentz transform derivation. It is some formulas derivation without a task statement. Lorentz transform must to follow from two postulates. It must to be confirmed.

The Lorentz transformation is, by definition, that coordinate transformation which under which the speed of light is the same in all inertial frames, and that's what @Ibix derived in #27 above (Note that his derivation does rely on the two postulates: The choice of a linear map is justified by the first postulate and the second postulate is why we choose the $A=(1-Ev^2)^{-1/2}$ solution over the $A=1$ solution).

Note that the Lorentz transformations are coordinate transformations, not time transformations or distance transformations. That is, if an event (point in spacetime) is assigned coordinates $(t,x,y,z)$ when using one frame, it will be assigned coordinates $(t',x',y,z)$ when using a another frame whose origin is moving at a constant velocity $v$ in the positive x direction relative to the origin of the first frame.
To get length contraction or time dilation from these formulas we need two pairs of events: the endpoints of the length being contracted or the time interval being dilated. The coordinates of these events are transformed and then the dilated time and/or contracted length can be calculated more or less as @Sagittarius A-Star suggests in #36 above, although there are some additional subtleties in the length contraction case.
(I am repeating this point because some of your other comments in this thread suggest that you may not have understood it. You may even be confusing the time dilation formula with the Lorentz transformation of the $t$ coordinate.)

More than. The time transformation can't to depend on value $x$. There are common time in all fixed frames.

But the Lorentz transformation is a transformation of the time coordinate, not of time intervals. There is a sensible coordinate transformation in which the time coordinate is common to all frames: the Galilean transforms used by Galilean relativity, which were taken for granted by all physicists for centuries before Einstein. These do not honor Einstein's second postulate.