# Lorentz transformations and a time dilation don't return the same result [closed]

Lets deal with a problem like this:

Lets say we have a twin A in coordinate system $x'y'$ who does a round trip to a star which is $12$ light years away and he is travelling with a speed $u=0.6c$. Twin B stays on Earth with coordinate system $xy$. How many years will pass for the twin B for a twin A to return. How many years will it pass for the Twin B?

In this case i ll set myself in a coordinate system $xy$

I calculate the time $\Delta t$ (this is easy because $u$ is a constant):

$$u = \frac{\Delta x}{\Delta t} \longrightarrow \Delta t = \frac{\Delta x}{u} = \frac{2 \cdot 12~l.y.}{0.6c}=\frac{2\cdot 12\cdot c \cdot 1 y}{0.6c} = 40y$$

So for twin B it will pass 40 years. According to time dilation every clock that is not at rest in the chosen system doesn't measure proper time. In the curent chosen coordinate system $xy$ only clock of the observer B is at rest so we choose a proper time like this $\boxed{\Delta t \equiv \tau}$. From this it follows that:

\begin{align} \Delta t' = \gamma \Delta t = 1.25 \cdot 40y = 50y \end{align}

So the observer in coordinate system $xy$ would say that for his brother it has passed 50 years. Now i will try to calculate the same result using the Lorentz transformations for coordinate system $xy$:

\begin{align} \Delta t' &= \gamma \left(\Delta t - \Delta x \frac{u}{c^2}\right)\\ \Delta t' &= 1.25\left(40y - 2\cdot 12 \cdot c \cdot 1y \frac{0.6c}{c^2}\right)\\ \Delta t' &= 1.25\left(40y - 2\cdot 12 \cdot 1y\cdot 0.6\right)\\ \Delta t' &= 32y \end{align}

The result is not the same...

The results would be the same if i chose the proper time differently $\boxed{\Delta t' \equiv \tau}$ from this it would then folow:

$$\Delta t = \gamma \Delta t \longrightarrow \Delta t' = \frac{1}{\gamma} \Delta t = \frac{1}{1.25}40y = 32y$$

But i can't do this because the clock which measures time $\Delta t'$ is not at rest in the coordinate system $xy$.

So we have a problem. I realy don't know what i did wrong. Is my perception of a proper time wrong or is the Lorentz transformation wrong. What am i missing to fully understand this?

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Why do you guys vote down EVERY single homework problem? Why do this forum than even have a homework tag??? I did include my attempt and everything... –  71GA Jun 22 at 20:18
Though this problem looks like a specific homework problem, I do believe there is a specific conceptual question hidden in there. –  Alfred Centauri Jun 22 at 22:17
I think that your first step to a better understanding of this would be to learn how to draw and use spacetime diagrams. It seems likely that the problem here is that you're changing frames of reference in the middle of a calculation. –  Alfred Centauri Jun 22 at 22:26
@AlfredCentauri, I agree with you that there probably is a good conceptual question here. It's just not apparent to me at all what it is. –  Colin McFaul Jun 22 at 22:52
@71GA We are not homework help. We deal with concepts here. Try PhysicsForums if you want homework help –  Manishearth Jun 29 at 10:27

## closed as off-topic by user1504, Manishearth♦Jun 29 at 10:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

(1) $\Delta t' = \gamma \Delta t$

(2) $\Delta t' = \gamma \left( \Delta t - \Delta x \dfrac{u}{c^2} \right)$

Now, the 2nd equation is generally true. The 1st equation is true only if $\Delta x = 0$ (well, and also if $u = 0$ but that's not the case here).

But $\Delta x$ isn't zero!

I think that the root of your confusion is the failure to identify what the symbols in these equations actually represent.

Step by step:

(1) There are two coordinate systems, primed and unprimed, in uniform relative motion. In the unprimed frame, the primed frame has a velocity $u = 0.6c$.

(2) In the primed coordinate system, twin A is at rest at the spatial origin.

(3) In either system, we are considering the coordinates of twin A. Thus, $\Delta x$ is the displacement of twin A in the unprimed system and $\Delta x'$ is the displacement of twin A in the primed system.

So, for the outbound leg, $\Delta x = 12ly, \Delta t = 20y$ since the speed of twin A, in the unprimed system is $0.6c$.

Now, in the primed system, $\Delta x' = 0$ (again, twin A is at rest in the primed system).

By the invariance of the interval, we have:

$(c\Delta t)^2 - \Delta x^2 = (c\Delta t')^2 - \Delta x'^2$

With the values we have, we can solve for $\Delta t'$:

$\Delta t'^2 = \Delta t^2 - (\frac{\Delta x}{c})^2 = (20y)^2 - (\frac{12ly}{c})^2$

$\Delta t' = 16y$

But, there's another, equivalent, way to find this $\Delta t'$. Since twin A is at rest in the primed frame, twin A's proper time is just $\tau_A = \Delta t'$

So, we have:

$\gamma_u \tau_A = \Delta t = \gamma_u \Delta t'$

So, where you went wrong was writing $\Delta t' = \gamma \Delta t$ without checking to see if it made sense in this problem. This is why that equation gives the wrong answer.

Your conceptual issue seems to be with correctly identifying proper time.

The proper time you're interested in is the proper time of twin A (we already know the proper time for twin B) and, since twin A is at rest in the primed frame, the correct identification of twin A's proper time is:

$\tau_A = \Delta t'$.

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You said: "You don't dilate the time recorded by the stationary observer". But this makes no sense as he is stationary and only a stationary observer measures a time which is the shortest among all and we call it the proper time. This is why i set $\Delta t \equiv \tau \longrightarrow \Delta t' = \gamma \Delta t$. NOTICE: that $\gamma$ is always greater than one and therefore $\Delta t'$ will always be bigger than $\Delta t$ which is matching with a statement that a proper time is the shortest... I really don't understand or this theory is wrong... –  71GA Jun 22 at 21:57
Lets say a muon is in the coordinate system $x'y'$ and we are in $xy$. If you position yourself (an observer) in a muon frame where muon is stationary, his half time $\Delta t'$ is the proper time $\Delta t' \equiv \tau$... which is exactly what i stated above: "Only a stationary observer measures a time which is the shortest". If you want to calculate the time passed in a frame $xy$ you then only use $\Delta t = \gamma \Delta t'$... and you will always get the hundred times longer time $\Delta t$. –  71GA Jun 22 at 22:48