B Trying to calculate proper time of worldlines using rotating frames

[Karin Helene Elise]

TL;DR Summary

Trying to calculate a proper time of worldlines not using an inertial frame, but a rotating frame.

So, to calculate a proper time of a worldline in SR using an inertial frame is quite easy. But I struggled a bit using a "rotating frame metric" and now I'm not sure whether I'll do it right.

Couls someone point me in the right direction?

"What have you tried?"

Well, trying to help truly absolute layppl with some variation of a "Circular Twin Paradox" not using an inertial frame of reference for whatevere reason.

I thought it would be a bit of a challenge so I made a derivation or approach to do that. But I'm not a 100% sure I did it right.

The point is not some weird umpteenth variation of the twin paradox, but a calculation not using inertial frames.

Thanks in advance,
Well, me.

 

[PeterDonis]

Could someone point me in the right direction?

Look up Born Coordinates.

 

[PeterDonis]

I thought it would be a bit of a challenge so I made a derivation or approach to do that. But I'm not a 100% sure I did it right.

The only way for us to evaluate that would be for you to post your calculation. But it might be better for you to look up Born Coordinates first, and then try to do the calculation in those coordinates and see what you get. Then, if you're still not sure, you can post that calculation here.

 

[pervect]

TL;DR Summary: Trying to calculate a proper time of worldlines not using an inertial frame, but a rotating frame.

So, to calculate a proper time of a worldline in SR using an inertial frame is quite easy. But I struggled a bit using a "rotating frame metric" and now I'm not sure whether I'll do it right.

Couls someone point me in the right direction?

"What have you tried?"

Well, trying to help truly absolute layppl with some variation of a "Circular Twin Paradox" not using an inertial frame of reference for whatevere reason.

I thought it would be a bit of a challenge so I made a derivation or approach to do that. But I'm not a 100% sure I did it right.

The point is not some weird umpteenth variation of the twin paradox, but a calculation not using inertial frames.

Thanks in advance,
Well, me.

If you were to use tensor methods, you would calculate ds from the formula
$$ds^2 = g_{\mu\nu} dx^{\mu}dx^{\nu}$$

(So, to get ds, you need to take the square root of the right hand side).

Use of the Einstein summation convention is assumed, which means you sum over all 16 value pairs of $\mu$ and $\nu$, For example , 00, 01, 02, 03, 10, 11, 12, 13, 21, 22, 23, 30,31,32,33 if $\mu$ and $\nu$ each have the range of {0,1,2,3}.


Here $g_{\mu\nu}$ is the metric tensor. For an inertial frame, it's diagonal. Sign conventions do vary, the one I use sets $g_{00}= -1$ and $g_{11} = g_{22} = g_{33} = +1$.

If you identify dt=$dx^0$, dx=$dx^1$, dy=$dx^2$ and dz=$dx^3$, you get the famliar relation
$$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$$

This brings up another note on conventions. I am using, by default, geometric units which set c=1 by a units choice. This is just a difference in the metric tensor, we replace -dt^2 with -c^2 dt^2 with standard units.

To do this in a rotating frame, you just use the metric tensor for the rotating frame. To get the metric tensor in the rotating frame, start with the metric tensor in cylindrical coordinates and then make the identity
$$\phi' = \phi - \omega t$$

I believe the metric tensor in cylindrical coordintes is just
$$ds^2 = -dt^2 + r^2 d\phi^2 + dr^2 + dz^2$$, I don't have a referernce. You can use the same method to compute this yourself as I suggested above to go from cartesian coordinates to cylindrical coordiantes.

You should get something similar to the metric tensor (more formally, the line element) in the wiki article on Born coordinates, which is however, firmly in the A-level category as to its exposition.

$$ds^2 = -(1 - \omega^2 r^2) dt^2 + 2 \omega r^2 dt d\phi + dr^2 + dz^2 + r^2d\phi^2$$

In short, this is just standard tensor methods for changing coordinates. But I'm guessing that's not the approach you were trying to take. As I write this, I realize there is a fair amount of background needed, but I hope this may be motivational enough for you to pursue learning more about tensor methods.

 

[Sagittarius A-Star]

Minkowski ("West Coast" convention, unprimed coordinates):
$ds^2=c^2(dt)^2 -(dx)^2-(dy)^2-(dz)^2$
$ds^2=c^2(dt)^2 -(d(r \cos(\theta)))^2-(d(r \sin(\theta)))^2-(dz)^2$

Using rotating, primed coordinates for the special case of $r$ and $z$ being constant:
$\require{color} ds^2=c^2dt'^2 -(\color{red} dt' \color{black} {d \over \color{red} dt'\color{black} }(r\cos(\theta' + \omega t')))^2-(\color{red} dt' \color{black} {d \over \color{red} dt'\color{black} } (r\sin(\theta' + \omega t')))^2$
$ds^2=c^2dt'^2 -(-dt'(\omega+{d\theta' \over dt'})(r\sin(\theta' + \omega t')))^2-(dt'(\omega+{d\theta' \over dt'})(r\cos(\theta' + \omega t')))^2$
$ds^2=c^2dt'^2 -r^2(dt'(\omega+{d\theta' \over dt'}))^2$
$$ds^2=(c^2-\omega^2r^2)dt'^2 -2\omega r^2 d\theta'dt' - r^2 (d\theta')^2$$

 

[cianfa72]

Yes, as others said you can use Born chart for Minkowski spacetime in which the Langevin observers are at rest. Their worldlines form the Langevin congruence.

Take a look also to this recent thread.

 

[Karin Helene Elise]

Thanks for the replies and sorry I didn't reply earlier. (I was a bit sick and got really sick after posting this question, feeling a bit better right now.)

I'm afraid that if I post what I did exactly, it will be quite confusing, since it was an awnser to a pretty weird "Circular Twin/Triplet Paradox". Which described a situation in flat Minkowski space, where observers p1 and p2 revolved around a spherical object (resembling the Earth) a 100 times before crossing the inertial observer p3 again. But not using the inertial frame of p3, which would be easy.

So after using $\tau_1=T \times 100$ for p1's proper time (stationary in a rotating frame), it would be pointless to calculate p2's proper time (interval) when they are exactly the same but rotating in opposite directions. So for any not pointless calculation made from p1's frame the circular motions of p1 and p2 have to be different in radius or angular velocity.

But anyway I used this "rotating frame metric" or line element for it:

$$ds^2 = -\left(1 - \frac{\omega^2 r^2}{c^2}\right) c^2 dt^2 + 2\omega r^2 d\phi dt + dr^2 + r^2 d\phi^2 + dz^2$$

With $\frac{d\phi}{dt}=$ relative angular velocity of p2.

To calculate p2's proper time then using:

$$\tau_{p2} = \int \sqrt{ - \frac{ds^2}{c^2} } = \int \sqrt{\left(1 - \frac{\omega^2 r^2}{c^2}\right) dt^2 - 2 \frac{\omega r^2}{c^2} d\phi dt - \frac{r^2}{c^2} d\phi^2}$$

Than integrate $d\tau_{p2}$ over total time $\tau_1=T \times 100$

First I added a Sagnac time delay on top of that but realised (I think) isn't needed.

Hope it makes sense.

 

[PeterDonis]

I guess I have to learn how to use LaTeX here first.

You had two bugs, opening "itex" tags and closing "tex" tags. I just used magic moderator powers to fix both of them. Your equations should display OK now.

 

[PeterDonis]

I used this "rotating frame metric"

Your $d \phi d t$ cross term isn't quite right; the factor in front should be $2 \omega r$, not $2 \omega r^2$. (When doing these "rotating frame" metrics, basically each factor of $\omega$ should have a matching factor of $r$ so the coefficient as a whole is dimensionless.)

 

[Karin Helene Elise]

Ok, thank you.

I see I need to use ## and $$ instead of "itex" and "tex" tags. Sorry about that.

 

[PeterDonis]

I see I need to use ## and $$ instead of "itex" and "tex" tags. Sorry about that.

Both work, but the double pound signs and dollar signs are definitely easier and quicker to type, and easier to keep track of.

 

[Karin Helene Elise]

Your $d \phi d t$ cross term isn't quite right; the factor in front should be $2 \omega r$, not $2 \omega r^2$. (When doing these "rotating frame" metrics, basically each factor of $\omega$ should have a matching factor of $r$ so the coefficient as a whole is dimensionless.)

But I don't really understand.

I see your point about dimensions, but from the standard derivation in polar coordinates the cross term really does come out with $2 \omega r^2$.

Starting with flat spacetime in polar coordinates,

$$ds^2 = -c^2 dt^2 + dr^2 + r^2 d\theta^2,$$

$$ds^2 = -c^2 dt^2 + dr^2 + r^2 d\theta'^2 + 2 \omega r^2 dt\, d\theta' + \omega^2 r^2 dt^2.$$

So the cross term is $2 \omega r^2 dt d\theta'$. This also makes dimensional sense: $d\theta$ is dimensionless, so only with $r^2$ does the term have the correct dimensions of length squared.

The form $2 \omega r$ can appear if one rewrites things in terms of arc length $r d\theta$ instead of the angle $\theta$, which effectively shifts a factor of $r$.

But I'm no expert so I'm probably overlooking something.

 

[Karin Helene Elise]

Both work, but the double pound signs and dollar signs are definitely easier and quicker to type, and easier to keep track of.

Yeah, definitely! You also have "math" "\math" tags, but such forums/platforms usually have a button to put them there. (Isn't there such a button here on PhysicsForums?)

 

[PeterDonis]

The form $2 \omega r$ can appear if one rewrites things in terms of arc length $r d\theta$ instead of the angle $\theta$, which effectively shifts a factor of $r$.

Ah, yes, that's right. I was implicitly thinking of it that way. Good catch!

 

[PeterDonis]

Yeah, definitely! You also have "math" "\math" tags, but such forums/platforms usually have a button to put them there. (Isn't there such a button here on PhysicsForums?)

Unfortunately PF doesn't have a way to automatically insert LaTeX tags or delimiters.

 

[JimWhoKnew]

I'm afraid that if I post what I did exactly, it will be quite confusing, since it was an awnser to a pretty weird "Circular Twin/Triplet Paradox". Which described a situation in flat Minkowski space, where observers p1 and p2 revolved around a spherical object (resembling the Earth) a 100 times before crossing the inertial observer p3 again. But not using the inertial frame of p3, which would be easy.

So after using $\tau_1=T \times 100$ for p1's proper time (stationary in a rotating frame), it would be pointless to calculate p2's proper time (interval) when they are exactly the same but rotating in opposite directions. So for any not pointless calculation made from p1's frame the circular motions of p1 and p2 have to be different in radius or angular velocity.

But anyway I used this "rotating frame metric" or line element for it:

$$ds^2 = -\left(1 - \frac{\omega^2 r^2}{c^2}\right) c^2 dt^2 + 2\omega r^2 d\phi dt + dr^2 + r^2 d\phi^2 + dz^2$$

With $\frac{d\phi}{dt}=$ relative angular velocity of p2.

To calculate p2's proper time then using:

$$\tau_{p2} = \int \sqrt{ - \frac{ds^2}{c^2} } = \int \sqrt{\left(1 - \frac{\omega^2 r^2}{c^2}\right) dt^2 - 2 \frac{\omega r^2}{c^2} d\phi dt - \frac{r^2}{c^2} d\phi^2}$$

Than integrate $d\tau_{p2}$ over total time $\tau_1=T \times 100$

First I added a Sagnac time delay on top of that but realised (I think) isn't needed.

Hope it makes sense.

Since you are interested in "circular twin paradox", the meaningful quantity is the constant$$\frac{d\tau_{p_2}}{d\tau_{p_1}}\quad,$$which from your calculation$$\frac{d\tau_{p_2}}{d\tau_{p_1}}=\sqrt{\frac{1-(\omega_1+\omega_2)^2r_2^2/c^2}{1-\omega_1^2r_1^2/c^2}}\quad,$$where $~\omega_2=\frac{d\phi}{dt}~$ is the relative angular velocity. Now you can multiply it by $~\tau_{p_1}~$ to get the corresponding $~\tau_{p_2}~$.

Which described a situation in flat Minkowski space, where observers p1 and p2 revolved around a spherical object (resembling the Earth)

In SR, it is better to say that p1 and p2 use their thrusters to achieve the circular motion.

 

[Karin Helene Elise]

Since you are interested in "circular twin paradox", the meaningful quantity is the constant$$\frac{d\tau_{p_2}}{d\tau_{p_1}}\quad,$$which from your calculation$$\frac{d\tau_{p_2}}{d\tau_{p_1}}=\sqrt{\frac{1-(\omega_1+\omega_2)^2r_2^2/c^2}{1-\omega_1^2r_1^2/c^2}}\quad,$$where $~\omega_2=\frac{d\phi}{dt}~$ is the relative angular velocity. Now you can multiply it by $~\tau_{p_1}~$ to get the corresponding $~\tau_{p_2}~$.


In SR, it is better to say that p1 and p2 use their thrusters to achieve the circular motion.

Well, I wasn’t really aiming at the twin paradox aspect itself, but more at working out the calculation from the perspective of a rotating frame, which I hadn’t done before.

Your explanation of the constant ratio is very clear, thanks!

As for the circular motion, whether you imagine thrusters or something else doesn’t really matter, and it wasn’t part of the question I was trying to answer on another forum.

 

[JimWhoKnew]

Well, I wasn’t really aiming at the twin paradox aspect itself, but more at working out the calculation from the perspective of a rotating frame, which I hadn’t done before.

You could have obtained the same result for $~\frac{d\tau_{p_2}}{d\tau_{p_1}}~$ by using p3 as a mediator, but you didn't. In that sense your goal was at least partially achieved.

As for the circular motion, whether you imagine thrusters or something else doesn’t really matter, and it wasn’t part of the question I was trying to answer on another forum.

Of course, but mentioning the Earth has a scent of GR.