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I was wondering how can one change from Cartesian coordinate system to some other like polar coordinates or spherical coordinates, in the context of special relativity. For example, with the four-velocity, $$V^{\mu}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau},\tag{1}$$ where $\mu=0,1,2,3$, $x^0=ct$, $x^1=x$, $x^2=y$ and $x^3=z$, if I want to change to cylindrical coordinates, can I simply use the transformations $x=r\cos (\theta)$, $y=r\sin(\theta)$, $z=z$, $t=t$ and then use the chain rule with partial derivatives to work out the four velocity in spherical coordinate, or I am missing something?

Another example, in the relativistic Lagrangian, $$\mathcal{L}=-mc^2\sqrt{1-\frac{|\mathbf{v}|^2}{c^2}}-U(\mathbf{r}),\tag{2}$$ it is correct to choose $r$ and $\theta$ as generalised coordinates, substitute $|\mathbf{v}|^2=\dot{r}^2+r^2\dot{\theta}^2$, to have $$\mathcal{L}=-mc^2\sqrt{1-\frac{\dot{r}^2+r^2\dot{\theta}^2}{c^2}}-U(r),\tag{3}$$ to create a "relativistic central force problem"?

Another question that I have is how this will affect the metric tensor? And the line element?

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  • $\begingroup$ "or am I missing something?" Please explain why you think you're missing something here. It seems you're doing it right. $\endgroup$ – Ari Mar 28 at 2:53
  • $\begingroup$ @Ari It is because I don't know how it will affect the Minkowski metric, or even if it still counts as Special and not general relativity. $\endgroup$ – Don Al Mar 28 at 2:57
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You are doing it correctly.

In reference to your comment, to find the minkowski metric in cylindrical coordinates, it is easiest to transform the line element so: $$ ds^2=g_{\mu\nu}dx^\alpha dx^\beta \tag{1}, $$ Where usually the minkowski metric is denoted by $\eta_{\mu\nu}$

Which is $$ ds^2 = -c^2dt^2+dx^2+dy^2+dz^2 \tag{2} $$ In Cartesian.

If you use the transformation equations and chain rule as you said, you get: $$ ds^2=-c^2dt^2 + dr^2 + r^2d\theta^2+dz^2 \tag{3} $$

It's easy to see that $\eta_{\mu\nu}\neq g_{\mu\nu}$ from looking at the first equation.

To summarize:

$$\eta_{\mu\nu} = Diag(-1, 1, 1, 1) \tag{4}$$ $$g_{\mu\nu}= Diag(-1, 1, r^2, 1) \tag{5}$$

With $g_{\mu\nu}$ being the minkowski metric in cylindrical coordinates.

And this usually falls under General relativity rather than special as it's concerned with the curvature of space time.

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    $\begingroup$ Yes, technically this is special relativity as spacetime is flat, but the coordinates are curvilinear, requiring maths normally associated with general relativity. $\endgroup$ – Charles Francis Mar 28 at 6:59
  • $\begingroup$ The curved surface is embedded in Minkowski space - just like the curved surface of the Earth is embedded in Euclidean space. In both cases the the Riemann tensor $R=0$, i.e., both spaces are flat. $\endgroup$ – Cinaed Simson Mar 28 at 8:08

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