1. Jul 23, 2017

### Vrbic

1. The problem statement, all variables and given/known data
In the twins paradox, suppose that Florence begins at rest beside Methuselah, then accelerates in Methuselah’s x-direction with an acceleration a equal to one Earth gravity, “1g”, for a time $T_F/4$ as measured by her, then accelerates in the −x-direction at 1g for a time $T_F/2$ thereby reversing her motion, and then accelerates in the +x-direction at 1g for a time $T_F/4$ thereby returning to rest beside Methuselah. Show that the total time lapse as measured by Methuselah is:
$$T_M=\frac{4}{g}\sinh{(\frac{gT_F}{4})}$$

2. Relevant equations
(1) $T_F=\int d\tau = \int\sqrt{dt^2-\delta_{ij}dx^idx^j}=\int_0^{T_M}\sqrt{1-v^2}dt$ where $d\tau$ is proper time of Florence

3. The attempt at a solution
I tried to integrate (1) with $v=gt$ but I have got some crazy result with ArcSin and square root. It is impossible to extract $T_M$.
Is my procedure generally wrong? Or it exists some better way how get result for this case?

2. Jul 23, 2017

### TSny

This is not the correct equation for the velocity $v$ of Florence relative to Methuselah. According to Methuselah, Florence does not have a constant acceleration during the first quarter of the trip. (Note that your equation would make $v$ become greater than the speed of light for large enough $t$.)

Florence "feels" a constant acceleration of g, but g is not the acceleration of Florence relative to Methuselah's frame. g is the acceleration of Florence relative to an inertial reference frame that is instantaneously co-moving with Florence.

Have you studied how accelerations transform between inertial frames?

3. Jul 24, 2017

### Vrbic

Ou I forgot it. Transformation of acceleration is $a=\frac{a'}{\gamma^3(1+v'\beta)^3}$. And velocity of Florence in her own frame is $v'=0$. Right?
And $v=gt$, may I use only becouse $gt<<c$ is it true?
But still there is $\beta$ in transformation of $a$, but now $\beta$ is not constant. May I write $\beta=gt$?

Last edited: Jul 24, 2017
4. Jul 24, 2017

### TSny

Yes, that's right.
No, there is nothing in the problem statement that would allow you to make that approximation.
With $v'=0$, you have

$a=\frac{a'}{\gamma^3}$.

Write this out explicitly in terms of $g$, $v$, and $dv/dt$.

5. Jul 24, 2017

### Vrbic

Hmm... I thought that for accelerated motion $u=\frac{at}{\sqrt{1-a^2t^2/c^2}}$.

6. Jul 24, 2017

### TSny

Where did you get this formula? I believe there is a sign error in the denominator.
What does $a$ represent here, the acceleration of Florence according to Florence, or the acceleration of Florence according to Methuselah? What does $u$ represent?

7. Jul 24, 2017

### Vrbic

$a=\frac{dv}{dt}=\frac{a'}{\gamma^3}=(1-\beta^2)^{3/2}\frac{dv'}{dt'}=(1-g^2t^2)^{3/2}g$ Right?

8. Jul 24, 2017

### Vrbic

Yes...there should be + sign. Sorry.

How would it looks like? I can't imagine this situation. How can be Florence accelerated with respect to herself? In her frame she will be in rest, no?

So I mean $a$ and $u$ are acceleration and velocity of Florence measurred by Methuselah.

9. Jul 24, 2017

### TSny

You cannot write $\beta$ as $gt$ because Florence does not have a constant acceleration $g$ relative to Methuselah.

Write $\beta$ as $v$ (since you are apparently letting $c = 1$). You then have

$\frac{dv}{dt}=(1-v^2)^{3/2}g$. This is a differential equation whose solution will give you $v(t)$.

It is helpful to rearrange it as $\frac{1}{(1-v^2)^{3/2}}\frac{dv}{dt} = g$. The trick is to note that

$\frac{1}{(1-v^2)^{3/2}}\frac{dv}{dt} = \frac{d}{dt} \left( \frac{v}{(1-v^2)^{1/2}} \right)$

10. Jul 24, 2017

### TSny

So, that would change your equation to

$u=\frac{at}{\sqrt{1+ a^2t^2/c^2}}$

So, $u$ is the same as $v$ that you used in your first post. OK.
But the equation $u=\frac{at}{\sqrt{1+ a^2t^2/c^2}}$ is still not correct if you are saying that $a$ is the acceleration of Florence relative to Methuselah. The correct formula is $u=\frac{a't}{\sqrt{1+ a'^2t^2/c^2}}$ where $a'$ is the acceleration that Florence feels.

Good point. By "the acceleration of Florence according to Florence" I meant "the acceleration of Florence relative to an instantaneously co-moving inertial frame". That is, it's the acceleration $g$ that Florence feels.

11. Jul 24, 2017

### Vrbic

Oooook :-) Now I understand, I have to be careful from what perspective are all variables taken.
Thank you for a hint...it is helpful. So than I have $v=\frac{gt}{\sqrt{1+g^2t^2}}$...nice.
Now I perfectly understand also the previous point, thanks this particular example. And also YOU ;-)
Now I put it in formula from #1 post and integrate...

12. Jul 24, 2017

### TSny

Yes. Good.

13. Jul 24, 2017

### Vrbic

Unfortunately I have another problem, with limits of integral in #1. How to interpret that it stops after some part of time and go back.
My result of integration is$$T_F=\frac{1}{g}(Arcsinh(gT2_M)-Arcsinh(gT1_M))$$

I suppose it doesn't matter of change of direction of velocity and also acceleration (it is seen from first formula where is $v^2$. And from result, if $g$ changes sign doesn't matter, because of sinh is an odd function.) So a time lapse in first part and third part is same and limits of the integral are from $0$ to $TM1$ resp. $TM3$ and it is $T_F/4$, but in second part it's different it travels half of Florence time $T_F/2$. Where $T_M=TM1+TM2+TM3$.
What am I missing?

14. Jul 24, 2017

### TSny

Break it up into four intervals $T_M=TM1+TM2+TM3 + TM4$ where each interval lasts $T_F/4$ of proper time. Argue that each quarter contributes the same.

15. Jul 24, 2017

### Vrbic

I understand, I thought about that, but I wasn't sure, wheather TM3 is same. Every of TMx starts with $v=0$ but TM3 starts with $v\neq 0$. Actually I would say it isn't same, in #1 there is $\int\sqrt{1-v^2}$.

16. Jul 24, 2017

### TSny

$TM1$ and $TM3$ start with $v = 0$ and end with $v = v_{max}$.

$TM2$ and $TM4$ start with $v = v_{max}$ and end with $v = 0$.

17. Jul 24, 2017

### Vrbic

I'm sorry I can't read :-) I've read it again and now I understand what the motion is.
Now it is clear, THANK YOU very much.

18. Jul 24, 2017

OK. Good.