Let $(t,x,y,z)$ be the standard coordinates on $\mathbb{R}^4$ and consider the Minkowski metric

$$ds^2 = -dt^2+dx^2+dy^2+dz^2.$$

I am trying to compute the metric coefficients under the change of coordinates given by:

$t' = t$

$x' = \sqrt{x^2+y^2}\cos(\omega t + \phi)$

$y' = \sqrt{x^2+y^2}\sin(\omega t + \phi)$

$z' = z$

where $\phi = \tan \frac{y}{x}$. I know one way to do this is to compute write each of the original coordinate maps as a composition of these new coordinate maps and then take derivatives to express $(\partial_{t'},\partial_{x'},\partial_{y'},\partial_{z'})$ in terms of $(\partial_t,\partial_x,\partial_y,\partial_z)$. This is rather cumbersome though and it seems like there should be a more elegant way to do this. Can anyone suggest a cleaner way to transform the metric?


3 Answers 3


Edit edit: as has been pointed out, I was incorrect to say $\partial_t = \partial_{t'}$ and so on. Serves me right for trying to look at it by inspection instead of being rigorous.

Nevertheless, I do think cylindrical coordinates simplifies the problem somewhat. Recall the cylindrical line element:

$$ds^2 = -dt^2 + dr^2 + r^2 \, d\phi^2 + dz^2$$

Now, there are several ways you can compute $\partial_{\mu'}$. In this case, it's simple enough to invert the coordinate system transformation, which naturally expresses the Jacobian terms in the correct manner. Note that $r' = r$ and $\phi' = \phi + \omega t$. We can then start reading off the transformations of the partial derivatives.

$$\begin{align*} \partial_{t'} &= \frac{\partial t}{\partial t'} \partial_t + \frac{\partial \phi}{\partial t'} \partial_\phi = \partial_t - \omega \partial_\phi \\ \partial_{r'} &= \frac{\partial r}{\partial r'} \partial_r = \partial_r \\ \partial_{\phi'} &= \frac{\partial \phi}{\partial \phi'} \partial_\phi = \partial_\phi \\ \partial_{z'} &= \frac{\partial z}{\partial z'} \partial_z = \partial_z\end{align*}$$

These give as a metric,

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

There is only one derivative of any consequence to calculate. Moreover, in cylindrical coordinates, we can see clearly that there is some strange stuff going on at $r' = 1/\omega$.

Convert this back into your primed cartesian coordinates, and you're done.

  • $\begingroup$ Alright. Well I suppose it's time to take out a pen and paper and run through the calculations. Thanks! $\endgroup$ Feb 12, 2013 at 20:28
  • $\begingroup$ @Muphrid, Just some thoughts: I don't see any point in answering "I can't find any answer" (as long as you don't give a mathematical proof that it is impossible - that would be a quite different case). It happens sometimes with some questions, and I can't understand it. Leave it open, perhaps other users can find some mathematical or at least notational workaround, as it is indeed the case in the answer of Qmechanic with the DeMoivre formula. By answering that and becoming (I don't know why) the green point, it is much less likely that other users think about the problem... $\endgroup$ Feb 12, 2013 at 21:58
  • $\begingroup$ as Qmechanic writes, $\partial_t\neq\partial_{t'}$ since $\partial_{\mu} = \frac{\partial x^{\mu'}}{\partial x^\mu}\partial_{\mu'}$ and ($\partial_t x'\neq0$ and $\partial_t y'\neq 0$) $\endgroup$ Feb 13, 2013 at 1:24
  • 2
    $\begingroup$ Muphrid, you forgot the off-diagonal term in the transformed metric: -(r')^2\omega d\phi' dt' $\endgroup$
    – Fizzerman
    Oct 17, 2017 at 14:42
  • 2
    $\begingroup$ @Fizzerman Almost: the off-diagonal term is $+2 \omega { r^{\prime} }^2 dt^{\prime} d \omega^{\prime}$. $\endgroup$
    – apdnu
    Apr 2, 2018 at 0:02

This answer is essentially a combination of @Muphrid's answer, the comments given there, together with some significant simplifications of my own.

The Minkowski metric rewritten in cylindrical cooordinates is $$ds^2=-dt^2+dr^2+r^2d\phi^2+dz^2. \tag{1}$$

Your coordinate transformation expressed in cylindrical coordinates simply is $$\begin{align} t'&=t \\ r'&=r \\ \phi'&=\phi+\omega t \\ z'&=z \end{align} \tag{2}$$

Combining (1) and (2) you can immediately write down the metric in primed cylindrical coordinates $(t',r',\phi',z')$. There is no need to fiddle with the derivatives $(\partial_{t'},\partial_{r'},\partial_{\phi'},\partial_{z'})$ and $(\partial_{t},\partial_{r},\partial_{\phi},\partial_{z})$.

$$\begin{align} ds^2&=-dt'^2+dr'^2+r'^2(d\phi'-\omega\ dt')^2+dz'^2 \\ &=(-1+r'^2\omega^2)dt'^2+dr'^2+r'^2d\phi'^2+dz'^2-2r'^2\omega\ d\phi'dt' \end{align} \tag{3}$$

Transforming from cylindrical coordinates $(t',r',\phi',z')$ back to cartesian coordinates $(t',x',y',z')$ you finally get (I leave out the calculational details here) $$ds^2=(-1+\omega^2(x'^2+y'^2))dt'^2+dx'^2+dy'^2+ dz'^2 + 2\omega(y'dx'-x'dy')dt'. \tag{4}$$

The resulting metric (4) is the Minkowski metric augmented with two additional terms. The one additional term ($\propto dt'^2$) gives rise to the centrifugal force, the other ($\propto dt'$) to the Coriolis force.



  1. The coordinate $z^{\prime}=z$ is a passive spectator variable, so one may consider the reduced $2+1$ dimensional problem.

  2. View the remaining two spatial coordinates as one complex coordinate, i.e., $$u~:=~x+iy, \qquad u^{\prime}~:=~x^{\prime}+iy^{\prime}. $$

  3. The rotational transformation then simplifies to $$t^{\prime}=t, \qquad u^{\prime}~=~ u e^{i\omega t}. $$

  4. Recall the chain rules $$ \frac{\partial}{\partial t} ~=~\frac{\partial t^{\prime}}{\partial t}\frac{\partial}{\partial t^{\prime}} +\frac{\partial x^{\prime}}{\partial t}\frac{\partial}{\partial x^{\prime}} +\frac{\partial y^{\prime}}{\partial t}\frac{\partial}{\partial y^{\prime}}, $$ $$ \frac{\partial}{\partial u} ~=~\frac{\partial t^{\prime}}{\partial u}\frac{\partial}{\partial t^{\prime}} +\frac{\partial u^{\prime}}{\partial u}\frac{\partial}{\partial u^{\prime}} +\frac{\partial \bar{u}^{\prime}}{\partial u}\frac{\partial}{\partial \bar{u}^{\prime}}. $$

  5. Derive via the chain rule $$ \frac{\partial}{\partial t}~=~\frac{\partial}{\partial t^{\prime}} + \omega\left( x^{\prime}\frac{\partial}{\partial y^{\prime}} -y^{\prime}\frac{\partial}{\partial x^{\prime}}\right), \qquad \frac{\partial}{\partial u} ~=~e^{i\omega t}\frac{\partial}{\partial u^{\prime}}, $$ or conversely, $$ \frac{\partial}{\partial t^{\prime}}~=~\frac{\partial}{\partial t} + \omega\left( y\frac{\partial}{\partial x} -x\frac{\partial}{\partial y}\right), \qquad \frac{\partial}{\partial u^{\prime}} ~=~e^{-i\omega t}\frac{\partial}{\partial u}. $$

  • $\begingroup$ I get the trick you used with the exponential, although the $\phi$ term seems to complicate things a little bit, but shouldn't $\partial_t = \partial_{t'}$? $\endgroup$ Feb 12, 2013 at 22:23
  • $\begingroup$ No, $\partial_t$ and $\partial_{t'}$ are not equal. There are transport-like terms coming from the chain rule, see pt. 4. $\endgroup$
    – Qmechanic
    Feb 12, 2013 at 22:34
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    $\begingroup$ @JamesMiller Qmechanic's comment is correct, but if that doesn't help you understand what's going on try this: Remember the partial derivative is the derivative keeping other things constant. Normally it's left implicit what "other things" means, but in this case it's important. $\partial_t$ keeps $x,y$ constant, whereas $\partial_{t'}$ keeps $x',y'$ constant. Since $x',y'$ differ from $x,y$ in a time dependent way the derivatives differ, even though $t'=t$. $\endgroup$
    – Michael
    Feb 13, 2013 at 2:21
  • $\begingroup$ How, does one get from the first line in 5. to the second line? I can see that it is basically switching primed and unprimed and $\omega \rightarrow -\omega$ but why is this possible mathematically? Can you please point that out to me $\endgroup$
    – Marsl
    Oct 29, 2018 at 17:56

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