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.

 

[Sagittarius A-Star]

$$\frac{d\tau_{p_2}}{d\tau_{p_1}}=\sqrt{\frac{1-(\omega_1+\omega_2)^2r_2^2/c^2}{1-\omega_1^2r_1^2/c^2}}\quad,$$where $~\omega_2=\frac{d\phi}{dt}~$ is the relative angular velocity.

I think you can't use Galilean (angular) velocity addition for $\omega_1$ an $\omega_2$.

 

[JimWhoKnew]

I think you can't use Galilean (angular) velocity addition for $\omega_1$ an $\omega_2$.

Good point. We should be careful here. However, in this case the $t$ coordinate is the same for both coordinate systems, and since $~\phi_{p_1}=\phi-\omega_1 t~$ , $~\omega_2=d\phi_{p_1}/dt~$ is the difference in angular velocity between the two circular motions, as measured by the Lorentzian observer (p3). So I think that a simple (Galilean) addition is OK (please re-check me on this).

The obtained result can be verified easily: for the p3 observer $~d\tau=dt~$ , and we have$$d\tau_{p_i}=\sqrt{1-\omega_i^2 r_i^2/c^2}~d\tau \quad.$$Substituting$$\omega_{i=2}:=\omega_1+\frac{d\phi_{p_1}}{dt}~~(=\omega_1+\omega_2)$$(if my arguments above are correct), we obtain immediately the result of #16.

Edit:
Note that the result for $~d\tau_{p_2}/d\tau_{p_1}~$ in post #16, comes from the calculation when the straightforward substitution of #7 is used:$$d\tau_{p_2}=\sqrt{\left(1 - \frac{\omega^2 r^2}{c^2}\right) dt^2 - 2 \frac{\omega r^2}{c^2} d\phi dt - \frac{r^2}{c^2} d\phi^2}\quad.$$As long as you don't challenge this substitution, you end up with a Galilean addition.

 

[Sagittarius A-Star]

I think you are only calculating in the simple non-rotating frame and use the time-dilation factors of SR.

since $~\phi_{p_1}=\phi-\omega_1 t~$ , $~\omega_2=d\phi_{p_1}/dt~$

BTW.: From this I conclude: $\omega_2 = { d\over dt} (\phi-\omega_1 t~) = -\omega_1$

This doesn't answer the question of the OP for a calculation in the rotating frame.
The Born coordinates are not orthogonal and are defined only in the range $-\pi < \phi < +\pi$.
Source:
https://en.wikipedia.org/wiki/Born_coordinates#Transforming_to_the_Born_chart

 

[PeterDonis]

This doesn't answer the question of the OP for a calculation in the rotating frame.

The OP already did that. @JimWhoKnew was just checking to see that the answer the OP got in the rotating frame is consistent with the calculation in the inertial frame.

 

[Sagittarius A-Star]

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.

I think you cannot integrate over 100 revolutions, because the used Born coordinates are only defined in the range $-\pi < \phi < +\pi$.

Source:
https://en.wikipedia.org/wiki/Born_coordinates#Transforming_to_the_Born_chart

 

[PeterDonis]

I think you cannot integrate over 100 revolutions, because the used Born coordinates are only defined in the range $-\pi < \phi < +\pi$.

Source:
https://en.wikipedia.org/wiki/Born_coordinates#Transforming_to_the_Born_chart

You are misinterpreting what that coordinate restriction means. The Born $\phi$ coordinate of the rotating observers the OP is using (the Wikipedia page you reference calls them Langevin observers) is constant. So as long as both of those observers have constant $\phi$ coordinates within the allowed range, there's no problem integrating along their worldlines.

What you can't do is integrate directly along the non-rotating observer's worldline for 100 revolutions, because that observer's $\phi$ coordinate will end up outside the allowed range in the first revolution. But you don't need to do that to do the calculation the OP is trying to do, because the Born $T$ coordinate already keeps track of the "reference time" that the rotating observers' proper times are being compared to.

 

[Sagittarius A-Star]

The Born $\phi$ coordinate of the rotating observers the OP is using (the Wikipedia page you reference calls them Langevin observers) is constant.

To my understanding this is valid for oberserver $p_1$, but not for $p_2$ described in the same coordinates.

 

[PeterDonis]

To my understanding this is valid for oberserver $p_1$, but not for $p_2$ described in the same coordinates.

I'm not sure the OP is actually using the same Born coordinates for both observers. The calculation in post #7 looks like it's calculating p2's proper time in Born coordinates in which p2 is at rest.

If the OP's intent was to calculate p2's proper time in Born coordinates in which p1 is at rest, then I agree there will be a limit beyond which p2's worldline can't be described in p1's Born coordinates.

 

[JimWhoKnew]

I think you are only calculating in the simple non-rotating frame and use the time-dilation factors of SR.

This doesn't answer the question of the OP for a calculation in the rotating frame.

As @PeterDonis said in #22, the OP calculated in the Born coordinates (see the "edit" note at the end of #20). I just "tidied up" the result a little, and observed that it matches the non-rotating result.

BTW.: From this I conclude: $\omega_2 = { d\over dt} (\phi-\omega_1 t~) = -\omega_1$

In hindsight, my definition for $\omega_2$ in #16 is awkward. That's why I used the odd $\omega_{i=2}$ in #20. Therefore the quoted equation means $~\omega_2=\omega_{i=2}-\omega_1$ .

The Born coordinates are not orthogonal and are defined only in the range $-\pi < \phi < +\pi$.

If we limit ourselves to a patch where the use of Born coordinates is locally valid (small $r$ differences, small angles...), we can obtain some meaningful results like the expression for the constant $~d\tau_{p_2}/d\tau_{p_1}$ .

Edit: added clarification.

 

[Karin Helene Elise]

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.

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

Thanks for your reply.

I know that when we mention things like the Earth’s rotation it quickly turns into a GR problem, but that had already been made clear in the forum I posted this in.

Just to clarify: in my derivation I wasn’t using Born coordinates.

I only set it up this way because I wanted to see if I could calculate the proper time in a rotating frame myself, without relying on the standard textbook formulas. It’s not my daily work, I just find relativity very intriguing. So I wasn't completely sure that I did it all correctly. That’s why I posted it here, to check my reasoning and see where it might go wrong.

Now, I cannot fully understand everything being said after your reply (without delving into the literature), but I believe I did not make mistakes for this specific problem, namely working out the proper times from the perspective of a rotating frame. And I think I can conclude I did it correctly. (Haven't I?)

 

[Karin Helene Elise]

Just to clarify for anyone reading: in my derivation I did not use Born coordinates. I simply took the Minkowski metric in polar coordinates and transformed it into a rotating frame to calculate proper times. While the resulting metric has a cross-term similar to what appears in Born coordinates, I did not impose rigid rotation or the Born coordinate conditions.

Any conclusion that my calculation was done in Born coordinates is likely due to my limited explanation and/or elaboration of the, well, strange and very unusual setup or thought experiment of the TS in the forum I posted it on. I hope this clears up misunderstanding (I was afraid of) about that.

 

[PeterDonis]

in my derivation I wasn’t using Born coordinates.

Then I'm confused. The metric you gave in post #7 is for Born coordinates.

 

[PeterDonis]

I simply took the Minkowski metric in polar coordinates and transformed it into a rotating frame

That is Born coordinates.

While the resulting metric has a cross-term similar to what appears in Born coordinates, I did not impose rigid rotation or the Born coordinate conditions.

I don't know what you mean by this. A number $\omega$ appears in your metric; that appears to be a fixed number (whose physical interpretation is the angular velocity of rotation of the coordinates relative to an inertial frame). That is "rigid rotation".

 

[Karin Helene Elise]

Oh, I see. Thanks!

I realize now that by using a fixed $\omega$ for all points and transforming the Minkowski metric via $\theta' = \theta - \omega t$, I unwittingly ended up using Born coordinates, and so hadn’t explicitly framed it that way or emphasized the rigid-rotation interpretation.

 

[Karin Helene Elise]

Every time I learn something new about relativity, it’s pretty confusing at first. Once I get it, though, it often doesn’t seem all that hard. But if I don’t work with it regularly, coming back to it can be confusing again. I find that very typical for relativity.

 

[Karin Helene Elise]

I'm not sure the OP is actually using the same Born coordinates for both observers. The calculation in post #7 looks like it's calculating p2's proper time in Born coordinates in which p2 is at rest.

If the OP's intent was to calculate p2's proper time in Born coordinates in which p1 is at rest, then I agree there will be a limit beyond which p2's worldline can't be described in p1's Born coordinates.

Well, my intent was to calculate p2’s proper time (in Born coordinates) in which p1 is at rest, but with $\omega r \ll c$, so the velocity limit isn’t an issue.

 

[PeterDonis]

my intent was to calculate p2’s proper time (in Born coordinates) in which p1 is at rest, but with $\omega r \ll c$, so the velocity limit isn’t an issue.

The velocity limit won't be, but you won't be able to integrate over p2's worldline past a certain point, because, as has been pointed out, only a limited portion of p2's worldline can be covered by Born coordinates in which p1 is at rest.

However, you can take the approach that @JimWhoKnew suggested in post #27, and calculate the ratio of the rates at which the proper times of p1 and p2 "tick" relative to coordinate time in Born coordinates in which p1 is at rest. Since that coordinate time is the same as coordinate time in an inertial frame, that calculation will end up giving you the information you need.

 

[Karin Helene Elise]

Allright. I don't really understand the why, but I'll figure it out, reading about it.

Good to know where I got it incorrect.

My thanks to everyone involved.