Metric Line Element Use: Do's & Don'ts for Accelerated Dummies?

[Lluis Olle]

TL;DR Summary

As it seems that I have some issues applying correctly the metric in the case of the Bob twin, which is the non-inertial guy in the twin paradox story, I write down here what I do and what conclusions I get, so hope I can find the mistake(s).

From Wikipedia article about Hyperbolic motion, I have the following coordinate equations of motion for Bob in his accelerated frame:

$t(T)=\frac{c}{g} \cdot \ln{(\sqrt{1+(\frac{g \cdot T}{c})^2}+\frac{g \cdot T}{c})} \quad (1)$

$x(T)=\frac{c^2}{g} \cdot (\sqrt{1+(\frac{g \cdot T}{c})^2}-1) \quad (2)$

where I use lowercase (t,x) to denote the Bob's NIRF coordinates, and uppercase (T,X) to denote Alice's IRF coordinates. Obviously, for Bob, that stays at rest at the origin of its accelerated frame, t is his proper time $\tau$.

The metric for Bob's accelerated frame is given by:

$ds^2=-(1+\frac{g \cdot x}{c^2})^2 \cdot d(ct)^2+(dx)^2 \quad (3)$

Then, the question is: can Bob use his metric "ruler" to measure or deduce the Alice's proper time? So, Bob in his accelerated laboratory, wants to deduce which is the proper time of Alice using his metric, or at least, check that all equations meet. Let's say, Bob takes about 2.5 years to travel from Earth to somewhere, and he wants to check how many years have elapsep for his twin sister Alice in Earth, as he needs to prepare in advance the birthday's gifts Alice will surely want to receive.

As I was quoted in another thread, it seems that Bob has to solve for $d\tau$ in the following equation:

$c^2 d\tau^2 = c^2 \, \left( 1+ \frac{g \cdot x}{c^2} \right)^2 \, dt^2 - dx^2 \quad (4)$

$\tau = \int_{0}^{t_2}\sqrt{\left( 1+ \frac{g \cdot x}{c^2} \right)^2 \, - \frac{v^2}{c^2}} \cdot dt \quad (5)$

where $t_2$ is the proper time for Bob to go from Earth to somewhere planet X.

But, here is the thing. If Bob uses his metric to measure the worldline path of Alice, I think that Bob will obtain... the Bob's proper time that Alice takes to follow that path, not the Alice's proper time! But seems that I'm wrong? If so, I don't understand why ?

So, lets check that, and do the numbers using equation (5) as starting point.

First thing to be cautious. The velocity in equation (5) is the coordinate velocity in Bob's accelerated frame. I attach one of the graphics I used in my thread, from where you can see that the coordinate velocity of Alice is... more that the speed of light. So, go figure.

In that post I wrote:

I attach both Alice and Bob's worldlines. For the Alice worldline, remark that the coordinate speed of Alice is faster than the coordinate speed of c, but I think that the "real" speed of light line should be drawn as $c \cdot (1+\frac{g \cdot x}{c^2})$, as for Bob in the origin of a non-inertial frame, the "speed of light" will be dependent on the position in which "the photon" is located relative to Bob. Someone could confirm this point?

Unfortunately, that was unanswered and unconfirmed, although it seemed to me and interesting question . Anyway, the v in equation (5) is the coordinate velocity, and is not equal to the velocity Alice observes from Earth.

Let's calculate first the coordinate velocity. Using the parametrized equations (1) and (2), the coordinate velocity will be the quotient of the derivatives of those equations with respect the parameter T, which is the proper time for Alice. And so:

$\frac{d x}{dT}=\frac{g \cdot T}{\sqrt{\left(\frac{g \cdot T}{c}\right)^2}+1} \quad (6)$

$\frac{d t}{dT}=\frac{c \cdot \left(\frac{g^2 \cdot T}{c^2 \cdot \sqrt{\left(\frac{g \cdot T}{c}\right)^2+1}}+\frac{g}{c}\right)}{g \cdot \left( \sqrt{\left(\frac{g \cdot T}{c}\right)^2+1} +\left(\frac{g \cdot T}{c}\right)\right)} \quad (7)$

and finally:

$v=g \cdot T \quad (7b)$

As said previously, the coordinate speed seems not limited by the c-limit in an accelerated frame.

From the Wikipedia article, we know that the relation between Alice's and Bob's time is given by:

$T(t)=\frac{c}{g} \cdot \sinh{\left(\frac{g \cdot t}{c}\right)} \quad (8)$

Plugging (2), (7b) and (8) into (5), we finally obtain:

$\tau = \int_{0}^{t_2}\sqrt{\left( 1+ \frac{g \cdot x}{c^2} \right)^2 \, - \frac{v^2}{c^2}} \cdot dt=t_2$

Big surprise, isn't it?

But as Saint Thomas the Apostle, lets plug the finger using another approach.

Now, I will do a change of variable in the (5) integral, and instead of $dt$, I'll use $dT$ instead, and the boundaries of the integration will go from 0 to T_2, being T_2 the proper time Alice's measures for the voyage. Aha, this surely will now work, because something was wrong in the before proceeding.

We previously have obtained the derivative $\frac{d t}{dT}$ in equation (7):

$\frac{d t}{dT}=\frac{c \cdot \left(\frac{g^2 \cdot T}{c^2 \cdot \sqrt{\left(\frac{g \cdot T}{c}\right)^2+1}}+\frac{g}{c}\right)}{g \cdot \left( \sqrt{\left(\frac{g \cdot T}{c}\right)^2+1} +\left(\frac{g \cdot T}{c}\right)\right)} \quad (7)$

and so:

$\tau = \int_{0}^{T_2}\sqrt{\left( 1+ \frac{g \cdot x(T)}{c^2} \right)^2 \, - \frac{v(T)^2}{c^2}} \cdot dT \quad (9)$

If you calculate $\sqrt{\left( 1+ \frac{g \cdot x(T)}{c^2} \right)^2 \, - \frac{v(T)^2}{c^2}}$, this gives 1. That's it, 1. So, equation (9) gets "simplified" to:

$\tau = \int_{0}^{T_2}\frac{c \cdot \left(\frac{g^2 \cdot T}{c^2 \cdot \sqrt{\left(\frac{g \cdot T}{c}\right)^2+1}}+\frac{g}{c}\right)}{g \cdot \left( \sqrt{\left(\frac{g \cdot T}{c}\right)^2+1} +\left(\frac{g \cdot T}{c}\right)\right)}\cdot dT \quad (10)$

And the result of the integral is:

$\frac{c}{g} \cdot \sinh^{-1}{\left( \frac{g \cdot T}{c} \right)} \quad (11)$

which for equation (8) is... $t_2$ again. So, no way, it's consistent.

Bob now quits, and breaks the metric "ruler". But, overnight, he decides to think again what's the issue here. He must be doing something wrong using his metric to obtain the Alice's proper time. But, what is it?

Time to go to his old notes when he was studying to become an astronaunt. Let's see, acceleration in Special Relativity, seems weird because SR only deals with constant velocities. Umm. But, how's it's done, what the Physicists did long ago to treat acceleration in SR?

So, how's its made? Then, Bob remembers and argues the following: As the Lorentz Transform goes from one IRF to another IRF, then is a piece of cake, we pull out from our magician relativistic top-hat an MCRF "rabbit", and then we consider that at every instant in time, there's a Momentarily Comoving Reference Frame which is, obviously, Inertial. That MCRF is at rest (coincidence?, God playing dice?...) in that instant with respect that "ugly" accelerated frame in which I rest, so I can Lorentz transform to that frame instead. Isn't that great?

But he also needs to believe in the "Clock hypothesis", so to say that what's really important for SR is the velocity, no how you get to that velocity... in SR, doesn't matter, is not its business. You get to that velocity, and that's it. Ok, let's take that for granted, because it seems plausible, and he thinks that's been experimentally probed up to high g values.

Then for the IRF and thanks to Mr. Lorentz, we can deduce that $A=\gamma^3\alpha$, and from there to the infinite and beyond. Ok, this is what my hand-written notes read anyway.

So, what I'm (Bob) measuring? What I want to measure? Take it easy, and let's read the fine print of all this stuff. How's it was made? I'm now in deep space, in an accelerated spaceship, but how all this formulas and equations where initially made?

Wait a minute, where is that MCRF that suddenly pops up for Alice, and she says is at rest in my frame? Because is that MCFR what underpins all the acceleration framework in SR! If Alice says there's such MCRF, then that same MCRF is in my frame also, and is... AT REST obviously, but in a x-coordinate that lies in Earth. I have to consider it as it is, has no velocity for me, is at rest in my accelerated frame.

Now, the light-bulb shines upon Bob's head, and he says: Eureka, that's it! The proper time of Alice traveling along her worldline has to be calculated with my "ruler" as the "tiny" time dilations from each of the infinite succession of MCRF's that magically pop up along the path. And then, I know how to apply my "ruler" metric, I must consider that the coordinate velocity of Alice is cero, because I consider the velocity of the MCRF's, and all are AT REST for me. Is not the "ruler" what is wrong, but I was measuring something different with it.

Let's try.

$\tau = \int_{0}^{t_2}\sqrt{\left( 1+ \frac{g \cdot x}{c^2} \right)^2 \, - \frac{v^2}{c^2}} \cdot dt \quad (5)$

$\tau = \int_{0}^{t_2}\sqrt{\left( 1+ \frac{g \cdot x}{c^2} \right)^2} \cdot dt \quad (12)$

Plugging equation (2) which gives the path of Alice in her wordline, and (8) into (12):

$\tau= \int_{0}^{t_2} \sqrt{\sinh^2{\left(\frac{g \cdot t}{c}\right)}+1} \cdot dt=\frac{c}{g} \cdot \sinh{\left(\frac{g \cdot t_2}{c}\right)}=T_2$

We got it!

But, Bob gets and tachyon-pulsed telegram from the Planet Federation, which says what he argues is not correct, and the Klingon empire has issued a warning!

So, what's wrong? Where's the issue?

Any help would be appreciated.
Thank all.

 

[Dale]

can Bob use his metric "ruler" to measure or deduce the Alice's proper time?

Yes. Proper time is an invariant so any frame’s metric can be used to deduce it.

I am still looking at the bulk of the question

 

[anorlunda]

This video from Fermilab explains the twin paradox without acceleration. Instead of acceleration, it uses three parties, A, B and C.

 

[Ibix]

Finding this difficult to follow on a small screen, but the process to compute Alice's elapsed time is:

Express Alice's trajectory in $X,T$ coordinates (trivial: $X=0$, $T=T$).

Transform these into $x,t$ coordinates, which should give you functions $x(T)$ and $t(T)$.

Eliminate $T$ between those two, so you (hopefully) have $x(t)$ for Alice. Calculate $dx/dt$.

You want $\int ds$ for Alice. Substitute from your (3) to get $ds=\sqrt{Adx^2-Bdt^2}$, then rewrite it as $ds=\sqrt{A(dx/dt)^2-B}dt$. Plug in your expressions for Alice's $x$ and $dx/dt$. The limits are the coordinate times in Bob's frame for departure and return. The result should be Alice's elapsed time.

 

[Dale]

I have the following coordinate equations of motion for Bob in his accelerated frame:

$t(T)=\frac{c}{g} \cdot \ln{(\sqrt{1+(\frac{g \cdot T}{c})^2}+\frac{g \cdot T}{c})} \quad (1)$

$x(T)=\frac{c^2}{g} \cdot (\sqrt{1+(\frac{g \cdot T}{c})^2}-1) \quad (2)$

where I use lowercase (t,x) to denote the Bob's NIRF coordinates, and uppercase (T,X) to denote Alice's IRF coordinates. Obviously, for Bob, that stays at rest at the origin of its accelerated frame,

OK, so first issue is here. Since Bob "stays at rest at the origin" then in Bob's accelerated frame his worldline is $(t,x,y,z) = (\tau_B,0,0,0)$ (I am going to use units where $c=1$)

So I think that these are the coordinates of Alice in Bob's frame, but I am not certain that this worldline is actually inertial. If they are not inertial then that is fine, they can still be used as Alice's worldline in Bob's frame. It will just mean that this is not a standard twin's scenario with one inertial and one non-inertial twin.

t is his proper time $\tau$.

I am always opposed to this kind of statement without very careful wording. If $t$ is a coordinate time and $\tau$ is a proper time then it is not correct that $t=\tau$ in general. The issue is the domain. $\tau$ is only defined on some specific worldline while $t$ is defined on an entire coordinate chart. It can be the case that on the given worldline $\tau=t$ but the identification of the specific worldline must be part of the statement of the equality, and even then it should not be stated that they are the same thing, only that they are equal on the worldline.

So here I would say "On Bob's worldline his proper time $\tau_B$ is equal to his coordinate time $t$." Note that I am going to use $\tau_B$ to refer to proper time on Bob's worldline and $\tau_A$ to refer to proper time on Alice's worldline. Since each is defined only on their own worldline it makes sense to use two different variables for them. I will also use $\tau$, without any subscript, to refer simply to the metric in a timelike convention, i.e. $ds^2=-c^2 d\tau^2$

The metric for Bob's accelerated frame is given by:

$ds^2=-(1+\frac{g \cdot x}{c^2})^2 \cdot d(ct)^2+(dx)^2 \quad (3)$

OK, so it looks like you are using Kottler Moller coordinates described here:
https://en.wikipedia.org/wiki/Rindler_coordinates#Variants_of_transformation_formulas

Then, the question is: can Bob use his metric "ruler" to measure or deduce the Alice's proper time?

Yes. Proper time on any worldline is invariant so his metric can certainly be used.

it seems that Bob has to solve for $d\tau$ in the following equation:

$c^2 d\tau^2 = c^2 \, \left( 1+ \frac{g \cdot x}{c^2} \right)^2 \, dt^2 - dx^2 \quad (4)$

$\tau = \int_{0}^{t_2}\sqrt{\left( 1+ \frac{g \cdot x}{c^2} \right)^2 \, - \frac{v^2}{c^2}} \cdot dt \quad (5)$

where $t_2$ is the proper time for Bob to go from Earth to somewhere planet X.

Almost, $t_2$ is the coordinate time in Bob's frame, which is only his proper time on his worldline.

I think that Bob will obtain... the Bob's proper time that Alice takes to follow that path, not the Alice's proper time! But seems that I'm wrong? If so, I don't understand why ?

The above expression is in terms of $\tau$ which is just the metric. If you integrate it along Alice's worldline then it is $\tau_A$, Alice's proper time. If you integrate it along Bob's worldline then it is $\tau_B$, Bob's proper time.

Using the parametrized equations (1) and (2), the coordinate velocity will be the quotient of the derivatives of those equations with respect the parameter T,

So this is Alice's worldline expressed in Bob's coordinates. Since it is Alice's worldline the proper time you find will be $\tau_A$.

which is the proper time for Alice.

Again, same caveat as above. $T$ is the coordinate time in Alice's frame and is equal to her proper time $\tau_A$ only on her worldline.

However, since you are evaluating it on her worldline indeed

and finally:

$v=g \cdot T \quad (7)$

Yes, that is what I get also.

Plugging (2), (7) and (8) into (5), we finally obtain:

$\tau = \int_{0}^{t_2}\sqrt{\left( 1+ \frac{g \cdot x}{c^2} \right)^2 \, - \frac{v^2}{c^2}} \cdot dt=t_2$

I don't get this. I get $$\tau_A = \int_{0}^{t_2} \sqrt{1-\sinh(g t) + \sinh(g t)^2} \ dt \ne t_2$$ The integral doesn't have an analytical solution according to mathematica, but to get what you got the integrand would have to be equal to 1, which it isn't.

So, no way, it's consistent.

It should be consistent, but I didn't check this math since I didn't get what you got above.

So, what I'm (Bob) measuring? What I want to measure? Take it easy, and let's read the fine print of all this stuff. How's it was made? I'm now in deep space, in an accelerated spaceship, but how all this formulas and equations where initially made?

Wait a minute, where is that MCRF that suddenly pops up for Alice, and she says is at rest in my frame? Because is that MCFR what underpins all the acceleration framework in SR!

You can always make a MCRF, but it is never necessary and often not even helpful. We can simply directly use the metric expressed in a given set of coordinates.

I must consider that the coordinate velocity of Alice is cero,

This is not correct. Alice's coordinate velocity is not zero.

 

[PeterDonis]

If Bob uses his metric to measure the worldline path of Alice, I think that Bob will obtain... the Bob's proper time that Alice takes to follow that path, not the Alice's proper time! But seems that I'm wrong?

Yes.

If so, I don't understand why ?

The metric measures arc length along a curve. Arc length along a timelike curve is the proper time of the observer who has that curve as their worldline. So arc length along Alice's worldline is Alice's proper time, not Bob's.

Note that this is true regardless of which metric you use. You can use Alice's metric (i.e., her inertial rest frame) to measure her worldline path and you will get her proper time. Or you can use either Bob's or Alice's metric to measure Bob's worldline path and you will get Bob's proper time.

In short, whose proper time you get by measuring arc length along a worldline depends on whose worldline you are measuring, not on whose metric you are using. Any valid metric must give the same arc length along a given curve.

 

[Lluis Olle]

This video from Fermilab explains the twin paradox without acceleration. Instead of acceleration, it uses three parties, A, B and C.

Thanks, will watch it.

 

[Lluis Olle]

Yes.The metric measures arc length along a curve. Arc length along a timelike curve is the proper time of the observer who has that curve as their worldline. So arc length along Alice's worldline is Alice's proper time, not Bob's.

Note that this is true regardless of which metric you use. You can use Alice's metric (i.e., her inertial rest frame) to measure her worldline path and you will get her proper time. Or you can use either Bob's or Alice's metric to measure Bob's worldline path and you will get Bob's proper time.

In short, whose proper time you get by measuring arc length along a worldline depends on whose worldline you are measuring, not on whose metric you are using. Any valid metric must give the same arc length along a given curve.

Yes, I think I understand this.

I will try to explain what happens more briefly. Let's say that the accelerated voyage from Earth to planet X takes 5 years for Bob's proper time (Alice's proper time will be greater, we know).

If Bob uses the metric for the accelerated frame, and integrates over the Alice worldline using the equation (5), the result is... 5 years. You only get Alice's proper time (for example, 10 years) if you consider that the coordinate speed in the equation is zero. You can check it by yourself if you want.

In other words, the result I obtain is that the arclength of Alice worldline - Alice is who moves from Bob's perspective - using the metric is... the Bob's time, not the Alice's time.

Bobs's time$= \int_{0}^{t_2}\sqrt{\left( 1+ \frac{g \cdot x}{c^2} \right)^2 \, - \frac{v^2}{c^2}} \cdot dt \quad (5)$

I think I messed up a little the use of the times translating to LaTex in the post, but originally it's done using wxMaxima. I can attach the wxMaxima file if you wish.

 

[Lluis Olle]

OK, so first issue is here. Since Bob "stays at rest at the origin" then in Bob's accelerated frame his worldline is $(t,x,y,z) = (\tau_B,0,0,0)$ (I am going to use units where $c=1$)

...

This is not correct. Alice's coordinate velocity is not zero.

Dale, it will take some time to answer you, because there's many things, and now I've to cook the dinner.

Thanks for you patience.

 

[Sagittarius A-Star]

First thing to be cautious. The velocity in equation (5) is the coordinate velocity in Bob's accelerated frame. I attach one of the graphics I used in my thread, from where you can see that the coordinate velocity of Alice is... more that the speed of light. So, go figure.

There is generally no issue, if a coordinate-velocity is greater than the speed of light in a non-inertial frame. SR postulate #2 refers to inertial frames.

For Alice's velocity with reference to the accelerated rest-frame of Bob, it is the case because the velocity is effected by "gravitational" time dilation. Maybe, my calculation in the accelerated frame helps:
https://www.physicsforums.com/threads/twin-paradox-with-accelerated-motion.1047089/post-6819344

 

[Lluis Olle]

This is not correct. Alice's coordinate velocity is not zero.

Just one quick answer. I said " I must consider that the coordinate velocity of Alice is cero, because I consider the velocity of the MCRF's, and all are AT REST for me".

So, to get the Alice's proper time, of she following the worldline that Bob observes, and using the Bob's metric, Bob's has to consider zero as the coordinate speed in that equation, because Bob has to consider the MCFR that at every instant is at rest for him, but at the location of Alice.

I leave now, or my wife will kill me.

 

[PeterDonis]

If Bob uses the metric for the accelerated frame, and integrates over the Alice worldline using the equation (5), the result is... 5 years. You only get Alice's proper time (for example, 10 years) if you consider that the coordinate speed in the equation is zero. You can check it by yourself if you want.

Sure, let's check it and see why you're wrong.

[Note: Some corrections have been made based on Sagittarius-A-Star's posts.]

Your equation (5) is (in units where $c = 1$):

$$
\tau = \int_{0}^{t_2} \sqrt{\left( 1+ g x \right)^2 - v^2}\, dt
$$

We know $t_2$ is 5 years. How? Because we know that Bob's worldline is characterized by $x = 0$ and $v = 0$, since Bob is at rest at the origin of his accelerated frame. And if we plug $x = 0$ and $v = 0$ into the above equation, we get the result $\tau = t_2$. And since we know Bob's $\tau$ is 5 years, we get that $t_2$ must be 5 years.

To evaluate the above integral for Alice, we need the functions $x(t)$ and $v(t)$ for Alice. Since $v(t) = dx / dt$, once we have $x(t)$, we can just differentiate it to obtain $v(t)$. So we need to find $x(t)$ for Alice.

We do that by noting that Alice's $X$ coordinate (in her inertial frame) is constant. (Note that with the equations you have written down for the accelerated frame, Alice's $X$ is not zero.) Call this constant $X_A$. Then we have for Alice

$$
x(t) = \sqrt{ \left( X_A + \frac{1}{g} \right)^2 - T(t)^2} - \frac{1}{g}
$$

where $T(t)$ is the coordinate time $T$ in the inertial frame, as a function of the coordinate time $t$ in the accelerated frame. From the transformation equations, we find that

$$
T(t) = \left( X(t) + \frac{1}{g} \right) \tanh \left( g t \right)
$$

For Alice, we have $X(t) = X_A$ (i.e., a constant), and plugging all this into the equation for $x(t)$ above gives

$$
x(t) = \left( X_A + \frac{1}{g} \right) \sqrt{ 1 - \tanh^2 \left( g t \right) } - \frac{1}{g} = \left( X_A + \frac{1}{g} \right) \frac{1}{\cosh \left( g t \right)} - \frac{1}{g}
$$

(Note, once again, that we cannot have $X_A = 0$.)

Differentiating this with respect to $t$ gives

$$
v(t) = - \left( g X_A + 1 \right) \frac{\sinh \left( g t \right)}{\cosh^2 \left( g t \right)}
$$

Plugging all this into the integral gives

$$
\tau = \int_0^{t_2} \sqrt{ \left( \frac{g X_A + 1}{\cosh \left( g t \right)} \right)^2 - \frac{ \left( g X_A + 1 \right)^2 \sinh^2 \left( g t \right)}{\cosh^4 \left( g t \right)} } \, dt
$$

This simplifies to

$$
\tau = \left( 1 + g X_A \right) \int_0^{t_2} \frac{1}{\cosh^2 \left( g t \right)} \, dt
$$

Which in turn evaluates to

$$
\tau = \left( \frac{1}{g} + X_A \right) \tanh \left( g t_2 \right)
$$

Which will always be greater than $t_2$ for the values we are considering.

 

[PeterDonis]

Just one quick answer. I said " I must consider that the coordinate velocity of Alice is cero, because I consider the velocity of the MCRF's, and all are AT REST for me".

So, to get the Alice's proper time, of she following the worldline that Bob observes, and using the Bob's metric, Bob's has to consider zero as the coordinate speed in that equation, because Bob has to consider the MCFR that at every instant is at rest for him, but at the location of Alice.

All of this is nonsense. Where are you getting all this from? Not from any textbook on SR that I'm aware of.

I think you are either trolling, or extremely overconfident in your understanding of SR.

 

[Sagittarius A-Star]

Call this constant $X_A$. Then we have for Alice$$
x(t) = \sqrt{X_A^2 - T(t)^2}
$$where $T(t)$ is the coordinate time $T$ in the inertial frame, as a function of the coordinate time $t$ in the accelerated frame.

Please notice, that from OP equation (5) follows, that in the OP not Rindler-coordinates, but Kottler–Møller coordinates are used, which have an x-offset of $\frac{1}{\alpha}$ compared to Rindler-coordinates. Then the equation must be:
$x(t) = \sqrt{(X_A+\frac{1}{\alpha})^2 - T(t)^2}-\frac{1}{\alpha}$.
Source:
https://en.wikipedia.org/wiki/Rindler_coordinates#Variants_of_transformation_formulas

 

[PeterDonis]

Please notice, that from OP equation (5) follows, that in the OP not Rindler-coordinates, but Kottler–Møller coordinates are used, which have an x-offset of $\frac{1}{\alpha}$ compared to Rindler-coordinates.

Yes. That's why I said Bob is at $x = 0$ instead of $x = 1 / \alpha$.

Then the equation must be:
$x(t) = \sqrt{(X_A+\frac{1}{\alpha})^2 - T(t)^2}-\frac{1}{\alpha}$.

This actually makes no difference in what I wrote because $X_A$ is a constant for Alice, so we can just absorb the $1 / \alpha$ offset into its definition, i.e., we can define $X_A$ to be Alice's $X$ coordinate (which still can't be zero) plus the $1 / \alpha$ offset. But yes, it is good to point out exactly what the correct definitions are.

 

[Dale]

So, to get the Alice's proper time, of she following the worldline that Bob observes, and using the Bob's metric, Bob's has to consider zero as the coordinate speed in that equation, because Bob has to consider the MCFR that at every instant is at rest for him, but at the location of Alice.

Again, this is not correct. I am not sure why you believe it is correct, but it is simply wrong. Bob does not need to consider Alice's speed to be zero when using Bob's metric. Bob does not need to consider any MCRF.

 

[Sagittarius A-Star]

This actually makes no difference in what I wrote because $X_A$ is a constant for Alice, so we can just absorb the $1 / \alpha$ offset into its definition

I didn't plausi-check your integral numerically, but I get one, which can be calculated symbolically.

The location of Alice in her frame is not relevant for a calculation in her frame of Bob's elapsed proper time, but for a calculation in Bob's accelerated frame of Alice's elapsed proper time.

Kottler–Møller:
$x(t) = \sqrt{(X_A+\frac{1}{g})^2 - T(t)^2}-\frac{1}{g}$
$T(t) = (X(t)+\frac{1}{g}) \tanh \left( g t \right) = (X_A+\frac{1}{g}) \tanh \left( g t \right)$
$\Rightarrow$
$x(t) = (X_A+\frac{1}{g}) \sqrt{1-\tanh^2 \left( g t \right)} -\frac{1}{g}= (X_A+\frac{1}{g}) \ \text{sech} \left( g t \right) -\frac{1}{g}$

Source:
https://www.wolframalpha.com/input?i2d=true&i=Sqrt[1-Power[\(40)tanh\(40)ax\(41)\(41),2]]

Differentiating this with respect to $t$ gives
$v(t)=-g(X_A+\frac{1}{g}) \tanh \left( g t \right)\ \text{sech}\ \left( g t \right)$
$v(t)=-g \tanh \left( g t \right)\ (x(t)+ \frac{1}{g})$
$v(t)=-\tanh \left( g t \right)\ (g \ x(t)+ 1)$

Source:
https://www.wolframalpha.com/input?i2d=true&i=D[bSqrt[1-Power[\(40)tanh\(40)ax\(41)\(41),2]],x]

Plugging this into
$\tau = \int_{0}^{t_2} \sqrt{\left( 1+ g x \right)^2 - v^2}\, dt$
gives:
$\tau = \int_{0}^{t_2} \sqrt{(g \ X_A+1)^2 (1-\tanh^2 \left( g t \right) ) - v^2}\, dt$
$\tau = \int_{0}^{t_2} \sqrt{(g \ X_A+1)^2 (1-\tanh^2 \left( g t \right) ) (1-\tanh^2 \left( g t \right) )}\, dt$
$\tau = (g \ X_A+1)\int_{0}^{t_2}(1-\tanh^2 \left( g t \right) )\, dt$
$\tau = (g \ X_A+1)\frac{\ tanh \left( g \ t_2 \right)}{g}$
$\tau = (X_A+\frac{1}{g}) \tanh \left( g \ t_2 \right)$

Source:
https://www.wolframalpha.com/input?i2d=true&i=Integrate[\(40)1-Power[\(40)tanh\(40)ax\(41)\(41),2]\(41),x]

 

[Lluis Olle]

Sure, let's check it and see why you're wrong.

Your equation (5) is (in units where $c = 1$):

$$
\tau = \int_{0}^{t_2} \sqrt{\left( 1+ g x \right)^2 - v^2}\, dt
$$

Ok,

Lets also consider g=1, so we get simpler formulas. I should have done that from the beginning.

$\tau = \int_{0}^{t_2} \sqrt{\left( 1+ x \right)^2 - v^2}\, dt \quad$

Now, without any physical evidence at the moment, consider that v(t)=0, just to see what happens. The equation is much simpler:

$\tau = \int_{0}^{t_2} \left(\, 1+ x(t) \, \right) \, dt$

The x(t) for Alice, comes from equation (2) in post#1, and are directly taken from the article of Wikipedia that I also linked in post#1, and that I repeat here.

Of course, such equations are for x(T), so we also have to use equation (8), to change them to x(t). So we have:

$x(T)=(\sqrt{1+T^2}-1) \quad (2)$
$T(t)=\sinh{(t)} \quad (8)$
$d\tau=(1+\sqrt{1+\sinh^2{(t}})-1)\,dt$
$\int_{0}^{t_2}\sqrt{1+\sinh^2{(t})}\,dt=\sinh{(t_2)}$

As you can check elsewhere, if proper time of Bob twin is $t_2$, then proper time of Alice twin is just $\sinh{(t2)}$.

Now, try to imagine a function of v(t), which is not always zero, square it and subtract it from the "time" component of the "metric" equation. It seems to me impossible that you can get the same result, which is, by the way, the correct result.

I didn't analyze how you obtain x() for Alice, because it would take me a lot of time... will do later, because I'm trying to answer everyone, and I'm not that fast.

Thanks

 

[PeterDonis]

without any physical evidence at the moment, consider that v(t)=0, just to see what happens

This is valid for Bob. It is not valid for Alice.

The equation (1) is much simpler:

For Bob, it's very simple, since for Bob $x = 0$.

For Alice, the equation, as above, is not even valid. So the rest of your post is simply wrong.

If you continue to repeat things you have already been told are wrong, multiple times, this thread will be closed and you will receive a warning. Doing that is wasting the time of other posters for no good reason.

 

[PeterDonis]

try to imagine a function of v(t), which is not always zero, square it and subtract it from the "time" component of the "metric" equation. It seems to me impossible that you can get the same result

In terms of the accelerated frame metric, the reason why Alice's proper time is greater than Bob's is simple: Alice is at a value of $x$ that is greater than zero, and the effect of that is to increase her proper time compared to Bob because of the $g x$ term in the $dt^2$ component of the metric. And that effect is larger than the effect of her nonzero coordinate speed in the non-inertial frame, which reduces her proper time compared to Bob because of the minus sign in the $dx^2$ component of the metric. So the net effect is that her proper time is increased.

 

[Lluis Olle]

For Bob, it's very simple, since for Bob x=0.

The same way you do, x(t) is the worldline of Alice, is not the position of Bob.

I just did the exercise of considering v(t)=0 in the equation, just for the sake to see what's the outcome. The outcome is the correct result. This is what I say that's strange.

 

[Lluis Olle]

I would ask you to wait until I can, at least, answer to the people just posted until now. Then, do what you consider.

 

[PeterDonis]

The same way you do, x(t) is the worldline of Alice, is not the position of Bob.

The equation for $x(t)$ for Alice that you get, as far as I can tell, is not the same as mine.

For Bob, $x(t) = 0$ for all $t$.

I just did the exercise of considering v(t)=0 in the equation, just for the sake to see what's the outcome. The outcome is the correct result. This is what I say that's strange.

The reason you are finding "strange" things in your results is that you are doing things wrong. You need to stop doing things wrong and start doing things right. You have already been given good guidance on how to do things right in multiple posts. Indeed, my post #12 did the correct analysis explicitly for you. I'm not sure what more we can do at this point.

 

[Lluis Olle]

Ok, let me answer Dale, and that's it. I'll not answer anybody except you and Dale.

 

[PeterDonis]

I would ask you to wait until I can, at least, answer to the people just posted until now. Then, do what you consider.

I didn't say I was going to close the thread now. I said I would close the thread if you continued to repeat statements that you have already been told, multiple times, were wrong. So if you would like the thread to remain open, just don't do that and you'll be fine.

It is true that doing that might require you to think a lot more carefully about what you post before you post it, than I suspect you have been doing up to now.

 

[Lluis Olle]

I didn't say I was going to close the thread now. I said I would close the thread if you continued to repeat statements that you have already been told, multiple times, were wrong. So if you would like the thread to remain open, just don't do that and you'll be fine.

It is true that doing that might require you to think a lot more carefully about what you post before you post it, than I suspect you have been doing up to now.

Ok, I'll try to measure carefully everything I say... but I'm not very fast, so I ask for a little patience, because I'll go little by little.

 

[Lluis Olle]

The equation for $x(t)$ for Alice that you get, as far as I can tell, is not the same as mine.

I didn't derive that equations. I just use the set of equations that are in the Wikipedia article. Are the same equations I used in the original post about accelerated dummies, and are the equations used to draw the worldline of Alice in the Bob spacetime diagram.

If those equations from Wikipedia are incorrect, then all what I say is incorrect of course. I'll not enter into validating if the Wikipedia article is correct or not. I just take it for granted.

If we're using different motion equations for Alice, then this is a bit of a problem in the first instance.

Are the equations described in the Wikipedia article, which I reference as (1) and (2) in post#1, correct so to be used to "draw" the worldline of Alice in the spacetime diagram of Bob in his non-inertial frame?

$t(T)=\frac{c}{g} \cdot \ln{(\sqrt{1+(\frac{g \cdot T}{c})^2}+\frac{g \cdot T}{c})} \quad (1) \\$

$x(T)=\frac{c^2}{g} \cdot (\sqrt{1+(\frac{g \cdot T}{c})^2}-1) \quad (2)$

Because if this is not applicable, then there's nothing more to say.

 

[PeterDonis]

I didn't plausi-check your integral numerically, but I get one, which can be calculated symbolically.

Yes, I realized on reading your post that I left out some of the $1 / g$ correction terms in my post. I have now edited my post #12 to include those terms, and that gives me the same integral you got.

 

[PeterDonis]

I didn't derive that equations. I just use the set of equations that are in the Wikipedia article.

You can't just blindly read equations from Wikipedia. You need to actually understand what you are doing. Clearly you don't, and you apparently are either unable or unwilling to consider the feedback you have been given and correct what you are doing.

Are the equations described in the Wikipedia article, which I reference as (1) and (2) in post#1, correct so to be used to "draw" the worldline of Alice in the spacetime diagram of Bob in his non-inertial frame?

Not the way you are doing it. You have already been told that. Multiple times.

(Btw, you were also told in a previous thread, multiple times, that the graph you are posting cannot be described using a single non-inertial frame in the coordinates you are using.)

Because if this is not applicable, then there's nothing more to say.

Quite so.

This thread has become pointless and is now closed.

 

[Dale]

So I thought it might be helpful for me to post how I would work through this problem, step-by-step, if I were going to do it. I would define my two coordinate systems using units where $c=1$.

I would have an inertial frame with coordinates $(T,X,Y,Z)$ and the Minkowski metric $$ds^2=-d\tau^2=-dT^2 + dX^2 + dY^2 + dZ^2$$
I would have a Rindler frame with coordinates $(t,x,y,z)$ and use the Kottler–Møller form of the metric for compatibility with what you did above $$ds^2=-d\tau^2=-(1+gx)^2 dt^2 + dx^2 + dy^2 + dz^2$$
The transformations between the coordinate systems are $$T=\left( x+\frac{1}{g} \right) \sinh(gt)$$$$X=\left(x+\frac{1}{g}\right) \cosh(gt)-\frac{1}{g}$$$$Y=y$$$$Z=z$$$$t=\frac{1}{g} \tanh^{-1}\left( \frac{T}{X+\frac{1}{g}} \right) $$$$x=\sqrt{\left( X+\frac{1}{g} \right)^2-T^2}-\frac{1}{g}$$

Now, I will define Bob's worldline in the Rindler frame, $r_b$, and his worldline in the Minkowski frame, $r_B$, as functions of his proper time $\tau_B$ as $$r_b=\left(\tau_B,0,0,0\right)$$$$r_B=\left( \frac{1}{g} \sinh(g\tau_B), \frac{1}{g} \cosh(g\tau_B),0,0 \right)$$
Similarly I will define Alice's worldline in the Rindler frame, $r_a$, and her worldline in the Minkowski frame, $r_A$, as functions of her proper time $\tau_A$ as $$r_A=(\tau_A,X_0,0,0)$$$$r_a=\left( \frac{1}{g} \tanh^{-1} \left( \frac{\tau_A}{X_0+\frac{1}{g}} \right) , \sqrt{\left( X_0 + \frac{1}{g}\right)^2-\tau_A^2}-\frac{1}{g} ,0 ,0 \right)$$

Now, it is straightforward to set $r_b=r_a$ and solve for $\tau_A$ and $\tau_B$. $$\tau_A=\pm\sqrt{X_0\left(\frac{2}{g}+X_0 \right)}$$$$\tau_B=\pm\frac{1}{g}\ln\left( 1+g X_0+\sqrt{g X_0(2+g X_0)} \right)$$

At this point we are actually done. You can check to confirm that the same $\tau_A$ and $\tau_B$ also give $r_B=r_A$. You can also integrate $d\tau$ to show that you get the same results. I will probably try that later, but probably not today.