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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:

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 rr axis.

Deriving the metric

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

x=rcosθy=rsinθz=z \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:

ds2=dx2+dy2+dz2=(cos2θ+sin2θ)dr2+r2(sin2θ+cos2θ)dθ2+dz2=dr2+r2dθ2+r2dz2 \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

grr=1gθθ=r2gzz=1 \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 tt and set θθ+ωt\theta \rightarrow \theta + \omega t, i.e. make the frame rotating in time:

t=tx=rcos(φ)y=rsin(φ)y=z \begin{align*} t &= t \\ x &= r \cos(\varphi) \\ y &= r \sin(\varphi) \\ y &= z \end{align*}

with φ=θ+ωt\varphi = \theta + \omega t.

Now, the line element becomes

ds2=dt2+dx2+dy2+dz2=(1r2ω2cos2(φ)+r2ω2sin2(φ))dt2+dr2+r2dθ2+r2dz2+cross terms in t, θ=(1ω2r2)dt2+dr2+r2dθ2+r2dz2+cross terms in t, θ \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 ω\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:

d2xμdτ2=Γνρμdxνdτdxρdτ \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. tτt \approx \tau, dxμ/dτ=(1,v,0,0)\mathrm{d} x^\mu / \mathrm{d}\tau = (1, v, 0, 0) with v20v^2 \approx 0, the geodesic equation for the rr coordinate becomes:

a:=d2rdt2=Γttr a := \frac{\mathrm{d}^2 r}{\mathrm{d} t^2} = - \Gamma^r_{tt}

So we only need to derive a single Christoffel symbol:

Γttr=12grλ(2tgtλλgtt)\Gamma^r_{tt} = \frac{1}{2} g^{r\lambda} \left( 2 \partial_t g_{t\lambda} - \partial_\lambda g_{tt} \right)

The only λ\lambda for which grλ0g^{r \lambda} \neq 0 is λ=r\lambda = r, but tgtr=0\partial_t g_{tr} = 0. So we’re left with:

Γttr=12grrrgtt=ω2r\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=rω2a = r\omega^2

or F=ma=mrω2F = 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 rω2r\omega^2 along the rr coordinate in a rotating coordinate system. Pretty neat, isn’t it? :)