# Rotating Observers in Special Relativity: Coriolis-like effect?

Do any non-inertial "forces" [terms in the metric] (like Coriolis in Newtonian mechanics) appear to a rotating observer (reference frame) in special relativity? Is the resulting spacetime after performing a change of coordinates in Cartesian Minkowski metric still FLAT, i.e., is it still special (and not general) relativity?

• See en.wikipedia.org/wiki/Ehrenfest_paradox . SR reduces to Newtonian mechanics in the limit $v\ll c$, so certainly you get effects like the Coriolis force.
– user4552
Feb 25, 2019 at 21:08

Do any non-inertial "forces" [terms in the metric] (like Coriolis in Newtonian mechanics) appear to a rotating observer (reference frame) in special relativity?

Yes, fictitious forces will appear. To see this let's proceed step by step.

In an inertial reference frame $$(t,x,y,z)$$ we have the Minkowski metric $$ds^2=-c^2dt^2+dx^2+dy^2+dz^2. \tag{1}$$

Now let's define a reference frame $$(t',x',y',z')$$ which is rotating by an angular velocity $$\omega$$ around the $$z$$-axis. The coordinate transform between the two reference frames is \begin{align} t&=t' \\ x&=x'\cos(\omega t')-y'\sin(\omega t') \\ y&=x'\sin(\omega t')+y'\cos(\omega t') \\ z&=z' \end{align} \tag{2}

By inserting transformation (2) into metric (1) we can get the metric in the rotating frame $$(t',x',y',z')$$. After a lengthy but simple calculation we arrive at $$ds^2=(-c^2\bbox[yellow]{+\omega^2(x'^2+y'^2)})dt'^2 +dx'^2+dy'^2+dz'^2 \bbox[yellow]{+2\omega(x'dy'-y'dx')dt'}. \tag{3}$$

The resulting metric (3) is the Minkowski metric augmented with two additional terms. The one additional term $$(\propto \omega^2 dt'^2$$) gives rise to the centrifugal force, the other ($$\propto \omega\ dt'$$) to the Coriolis force, as will be sketched now.

From the metric (3) you can calculate the Christoffel symbols $$\Gamma^\mu{}_{\alpha\beta}$$, and some of them turn out to be non-zero: \begin{align} \Gamma^x{}_{tt}&=-\omega^2x' \\ \Gamma^x{}_{ty}=\Gamma^x{}_{yt}&=-\omega \\ \Gamma^y{}_{tt}&=-\omega^2y' \\ \Gamma^y{}_{tx}=\Gamma^y{}_{xt}&=\omega \\ \text{all others} &=0. \end{align} Using these you can write down the geodesic equations and get the following differential equations for the inertial motion (with $$\tau$$ being the proper time). \begin{align} \frac{d^2t'}{d\tau^2}&=0 \\ \frac{d^2x'}{d\tau^2}&= \omega^2x'\left(\frac{dt'}{d\tau}\right)^2 +2\omega\frac{dy'}{d\tau}\frac{dt'}{d\tau}\\ \frac{d^2y'}{d\tau^2}&= \omega^2y'\left(\frac{dt'}{d\tau}\right)^2 -2\omega\frac{dx'}{d\tau}\frac{dt'}{d\tau}\\ \frac{d^2z'}{d\tau^2}&=0 \end{align} \tag{4}

The terms appearing on the right side can straightforwardly be interpreted as centrifugal and Coriolis acceleration.

In the non-relativistic limit ($$v\ll c$$) the approximation $$t'\approx\tau$$ (and thus $$\frac{dt'}{d\tau}\approx 1$$) holds, and equations (4) reduce to the ones as known from Newtonian mechanics.

Is the resulting spacetime after performing a change of coordinates in Cartesian Minkowski metric still FLAT, i.e., is it still special (and not general) relativity?

From the above mentioned Christoffel symbols you can calculate the Riemann curvature tensor $$R^\rho{}_{\sigma\mu\nu}$$, and find all of its components to be zero. (I omit this extremely tedious but simple calculation here.) Hence the rotating reference frame is still flat. Of course this result was to be expected. Because the Riemann curvature tensor is zero in the inertial reference frame $$(t,x,y,z)$$, it is necessarily zero in all other transformed reference frames, especially in the rotating reference frame $$(t',x',y',z')$$.

So we are still in the realm of special relativity (i.e. no gravity, no curvature), although we use the differential geometry calculus which is more often used in general relativity.

• You might wish to clarify that only the Riemann components are zero. The Christoffel components are non-zero, and in fact if you write out the geodesic equation it can be straightforwardly interpreted in terms of centrifugal and Coriolis forces/accelerations when $v\ll c$. Mar 13, 2020 at 11:23
• But the Wikipedia article and other answers here to similar questions say that the spatial geometry has been curved and not the whole space time. So if I try to calculate the Riemann tensor for the 4d space time I should get 0 like you said, but if I try to calculate the Riemann tensor for the 3d spacelike manifold I should get a non vanishing Reimann tensor because spacelike slice is curved. If I put dt=0 to get a spacelike slice the induced metric is just the Euclidean and so I will get a vanishing Riemann tensor showing it’s not curved. Could you please suggest what is that I am wrong. Apr 23, 2021 at 5:07
• @Shashaank That seems like a good question. But I can't give a simple answer (at least not in the scarce space available here). May be it is better to post this as a new question (instead of a comment here) to address a wider audience. Apr 23, 2021 at 11:18