Twins Paradox: Solving for Total Time Lapse

[Vrbic]

Homework Statement

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:
$$T_M=\frac{4}{g}\sinh{(\frac{gT_F}{4})}$$

Homework 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

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?

 

[TSny]

I tried to integrate (1) with $v=gt$

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?

 

[Vrbic]

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?

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 $\beta=gt$?

 

[TSny]

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?

Yes, that's right.

And $v=gt$, may I use only becouse $gt<<c$ is it true?

No, there is nothing in the problem statement that would allow you to make that approximation.

But still there is $\beta$ in transformation of $a$, but now $\beta$ is not constant. May I write $\beta=gt$?

With $v'=0$, you have

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

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

 

[Vrbic]

No, there is nothing in the problem statement that would allow you to make that approximation.

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

 

[TSny]

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

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?

 

[Vrbic]

With $v'=0$, you have

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

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

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

 

[Vrbic]

Where did you get this formula? I believe there is a sign error in the denominator.

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

the acceleration of Florence according to Florence

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?

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?

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

 

[TSny]

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

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

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

 

[TSny]

Yes...there should be + sign.

So, that would change your equation to

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

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

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.

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?

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.

 

[Vrbic]

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

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

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...

 

[TSny]

I have $v=\frac{gt}{\sqrt{1+g^2t^2}}$...nice.
Now I put it in formula from #1 post and integrate...

Yes. Good.

 

[Vrbic]

Yes. Good.

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?

 

[TSny]

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?

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.

 

[Vrbic]

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.

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 $\int\sqrt{1-v^2}$.

 

[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 $v = 0$.

 

[Vrbic]

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

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

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.

 

[TSny]

OK. Good.