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.