Special Relativity in Rotating Reference frame

[Jokar]

TL;DR Summary

I have two reference frames K, K' rotating with respect to each other. I want to relate the coordinates of these two reference frames

There are two coordinate frames K and K'. A particle is moving. The position of the particle in K frame in (t, x, y, z) and K' frame is (t', x', y', z'). the z axis of K and K' are aligned. and K' is rotating with respect to K with an angular velocity w. Relate the equation of motion in these two frames.

1. I want the relation between (t, x, y, z) and (t', x', y', z'). [ From one thread I got an answer that t=t', z=z' and other two coordinates are related by sin(wt') and cos(wt') type of factor. However that can not be correct. Let is imagine that the particle is fixed in K frame. In that case in K' frame its moving with velocity wr . Therefore, it should face time dilation and length contraction etc. So I want to know how those coordinates will be related.]
2. How can we relate the equation of motion in these two coordinates. I mean I want $\frac{d^2 t}{d\tau^2}, \frac{d^2 x}{d\tau^2}, ... $ in term of the dashed coordinates. How can I do that.

 

[Ibix]

There's nothing wrong with the coordinate system in your point 1, but the axes aren't orthogonal. That is why you don't see length contraction-like, time dilation-like, and relativity of simultaneity-like effects. You can construct frames with orthogonal axes so that they share the basis vectors of a locally co-moving inertial frame. I think that only works at one chosen radius, and you definitely have to have at least one discontinuity in your coordinates.

Rotating coordinate systems are much, much trickier to handle in relativity than they are in Newtonian mechanics. You basically need 90% of the maths you need for GR to use them with any degree of ease.

 

[Orodruin]

You can use rotating coordinates, just as you can use curvilinear coordinates. However, your metric will no longer be the standard Minkowski form related to an inertial frame and so conclusions as the one you are trying to draw in (1) need particular care. You cannot simply apply the conclusions reached from considering coordinates in an inertial frame.

 

[Sagittarius A-Star]

TL;DR Summary: I have two reference frames K, K' rotating with respect to each other. I want to relate the coordinates of these two reference frames

There are two coordinate frames K and K'. A particle is moving. The position of the particle in K frame in (t, x, y, z) and K' frame is (t', x', y', z'). the z axis of K and K' are aligned. and K' is rotating with respect to K with an angular velocity w. Relate the equation of motion in these two frames.

1. I want the relation between (t, x, y, z) and (t', x', y', z'). [ From one thread I got an answer that t=t', z=z' and other two coordinates are related by sin(wt') and cos(wt') type of factor. However that can not be correct. Let is imagine that the particle is fixed in K frame. In that case in K' frame its moving with velocity wr . Therefore, it should face time dilation and length contraction etc. So I want to know how those coordinates will be related.]
2. How can we relate the equation of motion in these two coordinates. I mean I want $\frac{d^2 t}{d\tau^2}, \frac{d^2 x}{d\tau^2}, ... $ in term of the dashed coordinates. How can I do that.

For an often used rotating coordinate system, see under "Langevin observers in the cylindrical chart":
https://en.wikipedia.org/wiki/Born_coordinates#Langevin_observers_in_the_cylindrical_chart

 

[Jokar]

There's nothing wrong with the coordinate system in your point 1, but the axes aren't orthogonal. That is why you don't see length contraction-like, time dilation-like, and relativity of simultaneity-like effects. You can construct frames with orthogonal axes so that they share the basis vectors of a locally co-moving inertial frame. I think that only works at one chosen radius, and you definitely have to have at least one discontinuity in your coordinates.

Rotating coordinate systems are much, much trickier to handle in relativity than they are in Newtonian mechanics. You basically need 90% of the maths you need for GR to use them with any degree of ease.

Sorry, I don't understand. Let's say all the coordinates are orthogonal in K. Now in K' frame the particle is moving with velocity wr. So, the particle's time will show some time dilation and t cannot be equal to t'.

Can you please elaborate? Maybe I have some problem in understanding.

 

[Orodruin]

Sorry, I don't understand. Let's say all the coordinates are orthogonal in K. Now in K' frame the particle is moving with velocity wr. So, the particle's time will show some time dilation and t cannot be equal to t'.

No it will not. While the spatial coordinates are still orthogonal in K', the time coordinate is not.

Can you please elaborate? Maybe I have some problem in understanding.

That the time coordinate is not orthogonal to the spatial coordinates imply you cannot just lift some of the results that appeared in regular Minkowski coordinates (the coordinates of an inertial frame). In particular, to compute the proper time passed by the particle, you cannot simply apply a factor $\sqrt{1-v^2/c^2}$. You need to compute the metric and apply it appropriately.

 

[Jokar]

No it will not. While the spatial coordinates are still orthogonal in K', the time coordinate is not.


That the time coordinate is not orthogonal to the spatial coordinates imply you cannot just lift some of the results that appeared in regular Minkowski coordinates (the coordinates of an inertial frame). In particular, to compute the proper time passed by the particle, you cannot simply apply a factor $\sqrt{1-v^2/c^2}$. You need to compute the metric and apply it appropriately.

Thanks for your reply.

Thats what I said. In K frame all are orthogonal. But not in K' frame.

So, what will be the metric in the K' frame. Do you have any idea or any reference? It will be helpful.

 

[Orodruin]

So, what will be the metric in the K' frame. Do you have any idea or any reference? It will be helpful.

It is easily computable. Just do the coordinate transformation. (It is even easier if you use polar coordinates in the plane of rotation.)

 

[Jokar]

It is easily computable. Just do the coordinate transformation. (It is even easier if you use polar coordinates in the plane of rotation.)

What kind of coordinate transform? Thats what is the question. One thing you can do is

$t = t'$
$r = r'$
$\theta = \theta' + w * t'$
$z = z'$

But there is the problem. In this coordinate transform you are assuming that $t = t'$. How can that be true? Because in the $K'$ frame the particle is moving with velocity $r w$.

So please explain properly what the correct answer will be and what I am doing wrong. Please explain with full mathematical details not with 1 line comments.

 

[Orodruin]

Yes, $t = t’$ is fine. That just means the time coordinates have the same values everywhere. However, $t’$ has different coordinate lines from $t$ because the spatial coordinates are different.

 

[thomsj4]

Begin by letting the standard Minkowski metric in $(t,x,y,z)$ be
$$
ds^2 = c^2\,dt^2 \;-\; dx^2 \;-\; dy^2 \;-\; dz^2.
$$

Define a vector field $X$ on $\mathbb{R}^4$ by
$$
X = \omega\,\bigl(y\,\partial_x \;-\; x\,\partial_y\bigr),
$$
where $\omega$ is a constant. Observe that $X$ generates rotations in the $(x,y)$-plane at angular speed $\omega$. For each fixed $t$ and $z$, let $\bigl(x(\sigma), y(\sigma)\bigr)$ satisfy the flow equation
$$
\frac{d}{d\sigma}
\begin{pmatrix} x(\sigma) \\[4pt] y(\sigma) \end{pmatrix}
=
\omega
\begin{pmatrix} y(\sigma) \\[2pt] -\;x(\sigma) \end{pmatrix},
\quad
\bigl(x(0), y(0)\bigr) = (x_0, y_0).
$$

Solving this gives
$$
\begin{pmatrix} x(\sigma) \\[3pt] y(\sigma) \end{pmatrix}
=
\begin{pmatrix}
x_0 \cos(\omega \sigma) \;+\; y_0 \sin(\omega \sigma)\\[4pt]
-\;x_0 \sin(\omega \sigma) \;+\; y_0 \cos(\omega \sigma)
\end{pmatrix}.
$$

Define a diffeomorphism $\Phi:\mathbb{R}^4\to\mathbb{R}^4$ by letting $\Phi(t,x,y,z)$ be the point
$$
\bigl(t, x(t), y(t), z\bigr)
\quad
\text{where}
\quad
\begin{pmatrix}x(t)\[2pt]y(t)\end{pmatrix}
=
\begin{pmatrix}x\[2pt]y\end{pmatrix}
\ast
\text{the above flow from } \sigma=0 \text{ to } \sigma=t.
$$

Hence, for each fixed $(x,y)$, we follow the integral curve of $X$ up to the “time” $\sigma = t$. Explicitly, if $\Phi(t,x,y,z) = (t', x', y', z')$, then
$$
t' = t,
\quad
\begin{pmatrix}x'\[3pt]y'\end{pmatrix}
=
\begin{pmatrix}
x\cos(\omega t) \;+\; y\sin(\omega t)\[3pt]
-\;x\sin(\omega t) \;+\; y\cos(\omega t)
\end{pmatrix},
\quad
z' = z.
$$

Label the new coordinates by $(t', x', y', z')$. In these coordinates, write the differentials
$$
dt' = dt,
\quad
dx' = \frac{\partial x'}{\partial t}\,dt + \frac{\partial x'}{\partial x}\,dx + \frac{\partial x'}{\partial y}\,dy,
$$
$$
dy' = \frac{\partial y'}{\partial t}\,dt + \frac{\partial y'}{\partial x}\,dx + \frac{\partial y'}{\partial y}\,dy,
\quad
dz' = dz.
$$

Substitute into
$$
ds^2 = c^2\,dt^2 \;-\; dx^2 \;-\; dy^2 \;-\; dz^2.
$$
For instance,
$$
dx = \bigl[\cos(\omega t)\,dx' \;-\;\sin(\omega t)\,dy'\bigr]
$$
plus an extra piece from the partial derivatives with respect to $t$. Carefully expand:
$$
dx = \cos(\omega t)\,dx' + \sin(\omega t)\,dy'
\quad
\text{only if } dt=0,
$$
but here $t' = t$ implies there is also
$$
\frac{\partial x}{\partial t} = -\,\omega\,y(\sigma)\ldots
$$

Writing each of $dx$ and $dy$ in terms of $dt', dx', dy'$ leads to cross-terms of the form $dt'\,d\phi'$. After all substitutions, one finds that the metric in $(t',x',y',z')$ picks up off-diagonal terms and changes in the coefficient of $dt'^2$. In cylindrical-like form (setting $r' = \sqrt{x'^2 + y'^2}$, $\phi'$ its polar angle), the resulting line element becomes
$$
ds^2
=
\bigl(c^2 - \omega^2\,r'^2\bigr)\,dt'^2
\;-\;2\,\omega\,r'^2\,dt'\,d\phi'
\;-\;dr'^2
\;-\;r'^2\,d\phi'^2
\;-\;dz'^2.
$$

Symbolically,
$$
ds^2
=
c^2\,dt'^2
\;-\;
(\text{some expression})\,dt'\,d\phi'
\;-\;\dots
$$
where $\phi'$ is the azimuthal angle around the $(x',y')$-plane. The key observation is that $\partial/\partial t'$ in the primed system is not globally orthogonal to the spatial directions, which reveals that a particle’s proper time $\tau$ is *not* simply $\tau = t'\sqrt{1 - v^2/c^2}$. Instead, we obtain
$$
d\tau^2
=
-\frac{1}{c^2}\,g_{\mu\nu}\,dx'^\mu\,dx'^\nu,
$$
where $g_{\mu\nu}$ is the induced metric from the above transformation. If the particle remains “at rest” with $(x',y')$ fixed (so $dr'=0$ and $d\phi'=0$), then
$$
ds^2
=
\bigl(c^2 - \omega^2\,r'^2\bigr)\,dt'^2,
\quad
d\tau
=
\sqrt{1 - \frac{\omega^2\,r'^2}{c^2}}\;dt',
$$
proving there is a time-dilation factor at radius $r'$. In general, a trajectory at rest in the rotating chart is actually undergoing centripetal acceleration in the original inertial frame, and the integral
$$
\tau
=
\int
\sqrt{
-\,g_{\mu\nu}\,
\frac{dx'^\mu}{d\lambda}\,
\frac{dx'^\nu}{d\lambda}
}
\,d\lambda
$$
will differ from
$\int dt'$, showing that time coordinate labels in a rotating chart are not identical to the proper time of rotating observers. The non-orthogonality of the time coordinate to spatial directions is made explicit by the off-diagonal components $g_{t'\phi'} \neq 0$. This geometric fact ensures that we cannot simply take the usual inertial-frame time-dilation factor $\sqrt{1 - v^2/c^2}$ to compute elapsed proper time. Instead, all such relativistic effects emerge from the full metric tensor, which carries the imprint of rotation through these mixing terms, even though we began in flat Minkowski space.

 

[Orodruin]

Begin by letting the standard Minkowski metric in $(t,x,y,z)$ be
$$ ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2. $$

Define a vector field $X$ on $\mathbb{R}^4$ by
$$
X = \omega\,\bigl(y\,\partial_x - x\,\partial_y\bigr),
$$
where $\omega$ is a constant. Observe that $X$ generates rotations in the $(x,y)$-plane at angular speed $\omega$. For each fixed $t$ and $z$, let $(x(\sigma), y(\sigma))$ satisfy the flow equation
$$ \frac{d}{d\sigma}
\begin{pmatrix} x(\sigma) \\ y(\sigma) \end{pmatrix}
=
\omega
\begin{pmatrix} y(\sigma) \\ -x(\sigma) \end{pmatrix},
\quad
(x(0), y(0))=(x_0,y_0). $$

Solving this gives
$$ \begin{pmatrix} x(\sigma) \\ y(\sigma) \end{pmatrix}
=
\begin{pmatrix}
x_0 \cos(\omega \sigma) + y_0 \sin(\omega \sigma)\\
-x_0 \sin(\omega \sigma) + y_0 \cos(\omega \sigma)
\end{pmatrix}. $$

Define a diffeomorphism $\Phi:\mathbb{R}^4\to\mathbb{R}^4$ by letting $\Phi(t,x,y,z)$ be the point
$$ (t, x(t), y(t), z)
\quad
\text{where}
\quad
\begin{pmatrix}x(t)\\y(t)\end{pmatrix}
=
\begin{pmatrix}x\\y\end{pmatrix}
\ast
\text{the above flow from } \sigma=0 \text{ to } \sigma=t. $$

Hence, for each fixed $(x,y)$, we follow the integral curve of $X$ up to the "time" $\sigma = t$. Explicitly, if $\Phi(t,x,y,z)=(t',x',y',z')$, then
$$ t' = t,
\quad
\begin{pmatrix}x'\\y'\end{pmatrix}
=
\begin{pmatrix}
x\cos(\omega t)+y\sin(\omega t)\\
-x\sin(\omega t)+y\cos(\omega t)
\end{pmatrix},
\quad
z' = z. $$

Label the new coordinates by $(t',x',y',z')$. In these coordinates, write the differentials
$$ dt' = dt,
\quad
dx' = \frac{\partial x'}{\partial t} dt + \frac{\partial x'}{\partial x} dx + \frac{\partial x'}{\partial y} dy, $$
$$ dy' = \frac{\partial y'}{\partial t} dt + \frac{\partial y'}{\partial x} dx + \frac{\partial y'}{\partial y} dy,
\quad
dz' = dz. $$

Substitute into $ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2$. For instance,
$$ dx = (\cos(\omega t) dx' - \sin(\omega t) dy') $$
plus an extra piece from the partial derivatives with respect to $t$. Carefully expand:
$$ dx = \cos(\omega t) dx' + \sin(\omega t) dy' \quad \text{only if } dt=0, $$

but here $t' = t$ implies there is also
$$ \frac{\partial x}{\partial t} = -\omega y(\sigma)\ldots $$

Writing each of $dx$ and $dy$ in terms of $dt',dx',dy'$ leads to cross-terms of the form $dt' d\phi'$. After all substitutions, we find that the metric in $(t',x',y',z')$ picks up off-diagonal terms. Symbolically,
$$ ds^2 = c^2 dt'^2 - (\text{some expression}) dt' d\phi' -\dots $$

where $\phi'$ is an azimuthal angle around the $(x',y')$-plane. The key observation is that $\partial/\partial t'$ in the primed system is not globally orthogonal to the spatial directions, which reveals that a particle's proper time $\tau$ is not simply $\tau = t'\sqrt{1-v^2/c^2}$. Instead, we obtain
$$ d\tau^2 = -\frac{1}{c^2} g_{\mu\nu} dx'^\mu dx'^\nu $$

where $g_{\mu\nu}$ is the induced metric from the above transformation. If the particle remains "at rest" with $(x',y')$ fixed, it must actually follow an accelerated trajectory in the original inertial frame, and the integral
$$ \tau = \int \sqrt{- g_{\mu\nu} \frac{dx'^\mu}{d\lambda} \frac{dx'^\nu}{d\lambda}} d\lambda $$

will differ from $\int dt'$, proving that time coordinate labels in a rotating chart are not identical to the proper time of rotating observers. The non-orthogonality of the time coordinate to spatial directions in the new metric is made explicit by the off-diagonal components $g_{t'\phi'}\neq 0$. This geometric fact ensures that we cannot simply take the usual inertial-frame time-dilation factor $\sqrt{1-v^2/c^2}$ to compute elapsed proper time. Instead, all such relativistic effects emerge from the full metric tensor, which carries the imprint of rotation through these mixing terms, even though we began in flat Minkowski space.

The factor in front of $dt’^2$ is not $c^2$. It needs to be $c^2 - \omega^2 r^2$ to reproduce the time dilation for an observer at rest in the rotating system for which $v = \omega r$ in the inertial frame.

Edit: You also cannot ignore the $dt’$ terms in $dx$ and $dy$. This is precisely where the off-diagonal terms arise!

Edit 2: The comments regarding $dt’$ went off-screen on my mobile …

 

[Jokar]

OK. Let' rephrase my original question based on all the answers.

The line element in the K frame

$d\tau^2 = -dt^2 + dx^2 + dy^2 + dz^2$

According to the above proposition in K' frame

$d\tau^2 = - dt'^2 + dx'^2 + dy'^2 + dz'^2 + \omega^2 (x'^2 + y'^2) dt'^2 + 2\omega (x' \, dy' - y' \, dx') dt'$

The Christoffel symbols for this metric are given by

\begin{eqnarray}
& \Gamma^x_{tt} = -\omega^2 x' \qquad\qquad
\Gamma^x_{ty} = \Gamma^x_{yt} = -\omega \nonumber\\
& \Gamma^y_{tt} = -\omega^2 y' \qquad\qquad
\Gamma^y_{tx} = \Gamma^y_{xt} = \omega \nonumber\\
& \text{all others} = 0 \,.
\nonumber
\end{eqnarray}

Therefore, we can calculate the equation of motion as
\begin{eqnarray}
\frac{d^2 t}{d\tau^2} &=&\frac{d^2 t'}{d\tau^2} \nonumber \\
\frac{d^2 x}{d\tau^2} &=&\frac{d^2 x'}{d\tau^2} - \omega^2 x' \left( \frac{dt'}{d\tau} \right)^2 - 2 \omega \frac{dy'}{d\tau} \frac{dt'}{d\tau} \nonumber \\
\frac{d^2 y}{d\tau^2} &=&\frac{d^2 y'}{d\tau^2} - \omega^2 y' \left( \frac{dt'}{d\tau} \right)^2 + 2 \omega \frac{dx'}{d\tau} \frac{dt'}{d\tau} \nonumber \\
\frac{d^2 z}{d\tau^2} &=&\frac{d^2 z'}{d\tau^2} \,.
\nonumber
\end{eqnarray}

So we should have $\frac{d t}{d\tau} =\frac{d t'}{d\tau}$. Now we have two reference frames, $K$ and $K'$. In one reference frame particle has a velocity $w r$. So how can we have $\frac{d t}{d\tau} =\frac{d t'}{d\tau}$?

Lets say $w r = 1$. In such case we should expect $\frac{d t'}{d\tau} = 0$. Isn't? So what am I doing wrong?

In case math is correct, please explain me what is wrong in my conceptual understanding.

 

[Sagittarius A-Star]

But there is the problem. In this coordinate transform you are assuming that $t = t'$. How can that be true?

See the following animation:
https://upload.wikimedia.org/wikipedia/commons/7/7d/Simultaneity.webm

If $A$ and $B$ are modified clocks, which show (related to radius $R$) $\gamma$ times their proper time, they can run synchronous to the inertial clock $O$ in the center. But the observer $L$ at radius $R$ on the rotating disk in the middle between $A$ and $B$ will receive light flashes from $A$ and $B$ (sent out at the same displayed "time") not synchronously.

Via German Wikipedia:
https://de.wikipedia.org/wiki/Born-Koordinaten#Uhrensynchronisation_und_Abstandsmessung

 

[Orodruin]

OK. Let' rephrase my original question based on all the answers.

The line element in the K frame

$d\tau^2 = -dt^2 + dx^2 + dy^2 + dz^2$

According to the above proposition in K' frame

$d\tau^2 = - dt'^2 + dx'^2 + dy'^2 + dz'^2 + \omega^2 (x'^2 + y'^2) dt'^2 + 2\omega (x' \, dy' - y' \, dx') dt'$

The Christoffel symbols for this metric are given by

\begin{eqnarray}
& \Gamma^x_{tt} = -\omega^2 x' \qquad\qquad
\Gamma^x_{ty} = \Gamma^x_{yt} = -\omega \nonumber\\
& \Gamma^y_{tt} = -\omega^2 y' \qquad\qquad
\Gamma^y_{tx} = \Gamma^y_{xt} = \omega \nonumber\\
& \text{all others} = 0 \,.
\nonumber
\end{eqnarray}

Therefore, we can calculate the equation of motion as
\begin{eqnarray}
\frac{d^2 t}{d\tau^2} &=&\frac{d^2 t'}{d\tau^2} \nonumber \\
\frac{d^2 x}{d\tau^2} &=&\frac{d^2 x'}{d\tau^2} - \omega^2 x' \left( \frac{dt'}{d\tau} \right)^2 - 2 \omega \frac{dy'}{d\tau} \frac{dt'}{d\tau} \nonumber \\
\frac{d^2 y}{d\tau^2} &=&\frac{d^2 y'}{d\tau^2} - \omega^2 y' \left( \frac{dt'}{d\tau} \right)^2 + 2 \omega \frac{dx'}{d\tau} \frac{dt'}{d\tau} \nonumber \\
\frac{d^2 z}{d\tau^2} &=&\frac{d^2 z'}{d\tau^2} \,.
\nonumber
\end{eqnarray}

So we should have $\frac{d t}{d\tau} =\frac{d t'}{d\tau}$. Now we have two reference frames, $K$ and $K'$. In one reference frame particle has a velocity $w r$. So how can we have $\frac{d t}{d\tau} =\frac{d t'}{d\tau}$?

Lets say $w r = 1$. In such case we should expect $\frac{d t'}{d\tau} = 0$. Isn't? So what am I doing wrong?

In case math is correct, please explain me what is wrong in my conceptual understanding.

The velocity $\omega r$ is a coordinate velocity. It does not have a direct correspondence to physical measurements without the metric. I suggest you simply insert the solution to compute the proper time.

 

[A.T.]

In one reference frame particle has a velocity wr. So how can we have dtdτ=dt′dτ?

In a rotating frame you not only have kinetic time dilation, but also pseudo-gravitational time dilation from the centrifugal potential. For particles at rest in the inertial frame those two cancel each other.

As for the length contraction, it's even wierder. You can describe the space in a rotating frame as non-Eucludian, with more circumference than 2*pi*r. Here tangential length contraction manifests itself as occupying less of that expanded tangential space.

 

[Jokar]

The velocity $\omega r$ is a coordinate velocity. It does not have a direct correspondence to physical measurements without the metric. I suggest you simply insert the solution to compute the proper time.

Can you please elaborate.

The 4D Riemann Curvature tensors are $0$ as they won't change due to coordinate transform.

So, at a given instant $\tau$, we can assume two Minkowski coordinate frames moving with respect to each other. And therefore, $\frac{d t}{d\tau} \ne \frac{d t'}{d\tau}$ as per my understanding.

It will be really helpful if you can explain what's wrong.

Please explain in detail.

 

[Sagittarius A-Star]

One thing you can do is

$t = t'$
$r = r'$
$\theta = \theta' + w * t'$
$z = z'$

But there is the problem.

Solution: It is allowed to define a coordinate system, in which the time-coordinate does not represent the proper time of clocks at rest in this coordinate system.

However, you can use alternatively the Lorentz-transformation for a Langevin-observer at rest at constant radius $R$ on the rim of the rotating disk. In this case you would choose non-curvilinear coordinated and the time-coordinate would represent the proper time of clocks.

To use the LT, You must define the time-coordinate on the rim of the rotating disk locally by the Einstein synchronization between adjacent clocks of a clock chain. But this coordinate system would be limited to $-\pi \lt \theta \leq \pi$. Two clocks at the "same" location(s) $-\pi$ and $+\pi$ would have then locally an offset, which is half of the $\Delta T$, that results by an experiment testing the Sagnac-effect.

The LT treats $-\pi$ and $+\pi$ as different locations in the rotating system, which is relevant for the relativity of simultaneity.

 

[Jokar]

Solution: It is allowed to define a coordinate system, in which the time-coordinate does not represent the proper time of clocks at rest in this coordinate system.

However, you can use alternatively the Lorentz-transformation for a Langevin-observer at rest at constant radius $R$ on the rim of the rotating disk. In this case you would choose non-curvilinear coordinated and the time-coordinate would represent the proper time of clocks.

To use the LT, You must define the time-coordinate on the rim of the rotating disk locally by the Einstein synchronization between adjacent clocks of a clock chain. But this coordinate system would be limited to $-\pi \lt \theta \leq \pi$. Two clocks at the "same" location(s) $-\pi$ and $+\pi$ would have then locally an offset, which is half of the $\Delta T$, that results by an experiment testing the Sagnac-effect.

The LT treats $-\pi$ and $+\pi$ as different locations in the rotating system, which is relevant for the relativity of simultaneity.

Thank you very much. This was the coordinate that I was looking for.

 

[PeterDonis]

what will be the metric in the K' frame. Do you have any idea or any reference?

Look up Born coordinates.