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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?

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up vote 1 down vote accepted


  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, $$ \frac{\partial}{\partial t^{\prime}}~=~\frac{\partial}{\partial t} + \omega\left( y\frac{\partial}{\partial x} -x\frac{\partial}{\partial y}\right), \qquad \frac{\partial}{\partial u^{\prime}} ~=~e^{-i\omega t}\frac{\partial}{\partial u}. $$

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I get the trick you used with the exponential, although the $\phi$ term seems to complicate things a little bit, but shouldn't $\partial_t = \partial_{t'}$? – James Miller Feb 12 '13 at 22:23
No, $\partial_t$ and $\partial_{t'}$ are not equal. There are transport-like terms coming from the chain rule, see pt. 4. – Qmechanic Feb 12 '13 at 22:34
@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 $t'=t$. – Michael Brown Feb 13 '13 at 2:21

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.

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Alright. Well I suppose it's time to take out a pen and paper and run through the calculations. Thanks! – James Miller Feb 12 '13 at 20:28
@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... – Eduardo Guerras Valera Feb 12 '13 at 21:58
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 $\partial_t y'\neq 0$) – joshphysics Feb 13 '13 at 1:24
@JamesMiller My apologies for being misleading. I've blanked the old post and expanded on the idea of using cylindrical coordinates. – Muphrid Feb 13 '13 at 6:03
@Muphrid, thanks for considering my comment. I have undone the downvote and given an additional +1 – Eduardo Guerras Valera Feb 13 '13 at 12:54

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