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?
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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 $$\begin{align} ds^2&=-c^2dt'^2+dx'^2+dy'^2+dz'^2 \\ &+\bbox[yellow]{\omega^2(x'^2+y'^2)dt'^2} \\ &+\bbox[yellow]{2\omega(x'dy'-y'dx')dt'}. \end{align} \tag{3}$$
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.
Thomas Fritsch
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The simple answer is to point out that the Riemann curvature tensor and its various contractions are invariants i.e. they do not depend on the coordinates that you are using. For a non-rotating observer in flat spacetime the Riemann tensor vanishes everywhere, and therefore for the rotating observer the Riemann tensor must also vanish everywhere. Spacetime is flat for both observers.
Calculating the Minkowski metric in rotating polar coordinates has proved a bit much for me on a Friday evening, but can I suggest a similar comparison? The metric of a uniformly accelerating observer is the Rindler metric, and this metric has horizons like a Schwarzschild black hole (and in fact a Schwarzschild white hole). Despite this the spacetime is still flat and the Riemann tensor vanishes everywhere, as you can show with a simple coordinate transformation.
John Rennie
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Special relativity applies to inertial reference frames, not rotating ones. For example, lets imagine your eye is the reference frame, you look at a star in the sky, then in a fraction of a second you rotate your eye to look away from the star. From your eye reference frame, the star moved faster than the speed of light. [EDIT]
Manu de Hanoi
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