# 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 '19 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 $$ds^2=(-c^2+\omega^2(x'^2+y'^2))dt'^2+dx'^2+dy'^2+dz'^2 +2\omega(x'dy'-y'dx')dt'. \tag{3}$$

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 ($$\Gamma^x{}_{tt}=-\omega^2x'$$, $$\Gamma^x{}_{ty}=\Gamma^x{}_{yt}=-\omega$$, $$\Gamma^y{}_{tt}=-\omega^2y'$$, $$\Gamma^y{}_{tx}=\Gamma^y{}_{xt}=\omega$$, all others $$=0$$). 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 $v\ll c$. – Michael Seifert Mar 13 at 11:23