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Deriving the centrifugal force with general relativity differential geometry

Note: Since writing this article, I’ve adapted it into a full article for my uni physics student journal which I never got around to properly publish unfortunately.

The idea of this post is to show how the math behind general relativity can be used to derive the apparent forces in a rotating coordinate system. This is something I came up with a couple of years ago while studying GR for the first time, and I think it’s a really beautiful example of how there are many possible ways in physics to mathematically describe a phenomenon.

For this post, I’ll assume basic knowledge of general relativity, i.e. in particular:

• arbitrary spacetime metrics $g_{\mu \nu}$
• covariant derivatives
• the Christoffel connection
• the geodesic equation

I’m going to use the $(-, +, +, +)$ sign convention here.

The problem

In any classical mechanics class, you’ll learn that Newton’s laws don’t apply in accelerated or rotated coordinate systems. This is only partially true - differential geometry provides the mathematical framework to construct the relativistic version of Newton’s laws in such a way that you can choose arbitrary coordinate systems, including accelerated ones.

We’re first going to derive the metric of a rotating coordinate system, use it to calculate the relevant Christoffel symbol, plug it into the geodesic equation and see that once you take the non-relativistic limit, the geodesic equation predicts an acceleration along the $r$ axis.

Deriving the metric

Let’s consider a reference frame rotating around the $z$ axis. We’ll start by writing down the curvilinear coordinates for a spatial cylindrical coordinate system:

$$\begin{align*} x &= r \cos \theta \\ y &= r \sin \theta \\ z &= z \end{align*}$$

The spacetime interval is given by the pullback of the Euclidean metric along this coordinate transformation:

$$\begin{align*} \mathrm{d}s^2 &= \mathrm{d} x^2 + \mathrm{d} y^2 + \mathrm{d} z^2 \\ &= ( \cos^2 \theta + \sin^2 \theta ) \mathrm{d} r^2 + r^2 (\sin^2 \theta + \cos^2 \theta) \mathrm{d}\theta^2 + \mathrm{d}z^2 \\ &= \mathrm{d}r^2 + r^2 \mathrm{d}\theta^2 + r^2 \mathrm{d}z^2 \end{align*}$$

so we have

$$\begin{align*} g_{rr} &= 1 \\ g_{\theta \theta} &= r^2 \\ g_{zz} &= 1 \end{align*}$$

So far so good, but we could’ve already guessed that. But it starts getting really interesting once we add in a time coordinate $t$ and set $\theta \rightarrow \theta + \omega t$, i.e. make the frame rotating in time:

$$\begin{align*} t &= t \\ x &= r \cos(\varphi) \\ y &= r \sin(\varphi) \\ y &= z \end{align*}$$

with $\varphi = \theta + \omega t$.

Now, the line element becomes

$$\begin{align*} \mathrm{d}s^2 &= -\mathrm{d}t^2 + \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2 \\ &= -(1 - r^2 \omega^2 \cos^2(\varphi) + r^2 \omega^2 \sin^2(\varphi)) \mathrm{d}t^2 + \mathrm{d}r^2 + r^2 \mathrm{d}\theta^2 + r^2 \mathrm{d}z^2 + \text{cross terms in t, $\theta$}\\ &= -(1 - \omega^2 r^2) \mathrm{d}t^2 + \mathrm{d}r^2 + r^2 \mathrm{d}\theta^2 + r^2 \mathrm{d}z^2 + \text{cross terms in t, $\theta$} \end{align*}$$

We can directly read off this expression that for high $\omega$, time will pass more slowly for an observer far away from the center of rotation, analogously to the gravitational time dilation effect.

Geodesic equation

Let’s take the geodesic equation:

$$\frac{\mathrm{d}^2 x^\mu}{\mathrm{d}\tau^2} = - \Gamma^\mu_{\nu \rho} \frac{\mathrm{d} x^\nu}{\mathrm{d} \tau} \frac{\mathrm{d} x^\rho}{\mathrm{d} \tau}$$

At the non-relativistic limit, i.e. $t \approx \tau$, $\mathrm{d} x^\mu / \mathrm{d}\tau = (1, v, 0, 0)$ with $v^2 \approx 0$, the geodesic equation for the $r$ coordinate becomes:

$$a := \frac{\mathrm{d}^2 r}{\mathrm{d} t^2} = - \Gamma^r_{tt}$$

So we only need to derive a single Christoffel symbol:

$$\Gamma^r_{tt} = \frac{1}{2} g^{r\lambda} \left( 2 \partial_t g_{t\lambda} - \partial_\lambda g_{tt} \right)$$

The only $\lambda$ for which $g^{r \lambda} \neq 0$ is $\lambda = r$, but $\partial_t g_{tr} = 0$. So we’re left with:

$$\Gamma^r_{tt} = -\frac{1}{2} g^{rr} \partial_r g_{tt} = - \omega^2 r$$

Reinserting this back into the geodesic equation, we get:

$$a = r\omega^2$$

or $F = ma = mr\omega^2$, which is precisely the classical result for the centrifugal force.

To summarize, the geodesic equation predicts that in the non-relativistic limit, a geodesic (i.e. straight line in spacetime) will have an acceleration of $r\omega^2$ along the $r$ coordinate in a rotating coordinate system. Pretty neat, isn’t it? :)