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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?

Qmechanic
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James Miller
  • 103

3 Answers3

4

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.

Thomas Fritsch
  • 42,352
4

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.

Muphrid
  • 7,329
2

Hints:

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

Qmechanic
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