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From classical mechanics and general relativity, we know that the natural motion of a particle in general curvilinear coordinates, assuming the Levi-Civita connection, is given by the geodesic equation. $$ \frac{d^2x^\mu}{d\tau^2} = \Gamma^\mu_{\nu \sigma} \frac{d x^\nu}{d\tau}\frac{dx^\sigma}{d\tau} $$

Where $x$ are the coordinates of the particle and $\tau$ is an affine parameter (usually proper time).

I have never seen it discussed, but is it possible to derive the usual non-inertial/fictitious forces in a rotating reference frame, just from say, replacing the derivatives in Newton's second law with a covariant derivative?

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Yes it is possible already in classical physics.

In classical physics, spacetime is a 4-dimensional affine space $\mathbb{A}^4$ equipped with a surjective affine map $T: \mathbb{A}^4 \to \mathbb{R}$ defined up to additive constants corresponding to absolute time. The 3-planes $\Sigma_t := \{p \in \mathbb{A}^4 \:|\: T(e)=t\}$ are the absolute spaces at time $t$. They are equipped with a metric structure (a positive scalar product in the 3-space of translations of each $\Sigma_t$) smoothly depending on the global affine structure.

In this picture, inertial reference frames are defined by a subclass of 4-dimensional Cartesian coordinates such that, in particular, one coordinate is $T$ itself (*) and the remaining three coordinates are tangent to the planes $\Sigma_t$ and form a 3-dimensional Cartesian orthonormal system of coordinates in each $\Sigma_t$.

As a consequence, the relations between a pair of these Cartesian coordinates is a generic Galileian transformation

$$x'^0= x^0+c$$

$$x'^k = c^k+ tv^k + \sum_{h=1}^3{R^k}_h x^h\quad k=1,2,3$$

where $c, c^k, v^k \in \mathbb{R}$ and $[{R^h}_k]_{h,k=1,2,3} \in O(3)$.

Then, it is possible to define a 4-dimensional affine connection $\nabla$ in $\mathbb{A}^4$ just by declaring the connection coefficients of $\nabla$ vanish in these (inertial) reference frames. Evidently a curve $\gamma$, given by $$t \mapsto (t, x^1(t),x^2(t), x^3(t))$$ in inertial coordinates, is a geodesic with respect to $\nabla$ if and only if it is an inertial motion $$x^k(t) = u^k t + c^k\quad k=1,2,3\:.$$
When using other reference frames associated with coordinates $y^\alpha$, the equation reads in full generality but keeping the parametrization in terms of absolute time, $$\frac{d^2y^\mu}{dt^2} = -\Gamma^\mu_{\alpha \beta} \frac{dy^\alpha}{dt} \frac{dy^\beta}{dt}\:.\tag{1}$$

Above, directly from the transformation law of the connection coefficients $$\Gamma^\mu_{\alpha \beta} = \sum_{\sigma,\delta=0}^3\frac{\partial^2 y^\mu }{\partial x^\sigma \partial x^\delta} \frac{\partial x^\sigma}{\partial y^\alpha} \frac{\partial x^\delta}{\partial y^\beta} \tag{2}$$ with $x^\mu$ being Cartesian coordinates of any inertial reference frame.

A generic reference frame of the classical physics with coordinates $y^\alpha$ is related to an inertial reference frame with coordinates $x^\beta$ through the general relation

$$y^0=x^0 + c \quad (=t+c')\:,$$

$$y^k = c^k(x^0) + \sum_{h=1}^n {R^k}_h(x^0) x^k \quad k=1,2,3\:.$$

Above, the maps $\mathbb{R} \ni t\mapsto [{R^k}_h(x^0)] \in O(3)$ and $\mathbb{R} \ni t\mapsto c^k(x^0) \in \mathbb{R}$ are smooth.

If we use there relations to compute the coefficients $\Gamma^\mu_{\alpha \beta}$ according to (2), in particular defining $$\omega^p := -\frac{1}{2} \sum_{k,q,s=1}^3 \epsilon_{pqs} {R^q}_k \frac{d{R^s}_k}{dt}\quad p=1,2,3$$ Eq.(1) turns out to become, where we also multiply with the mass $m$ of the point, $$m\frac{d^2 \vec{y}}{d t^2} =- m \vec{a}(t)- 2 m\vec{\omega} \wedge \vec{v} - m\vec{\omega}\wedge (\vec{\omega} \wedge \vec{y})-m \dot{\vec{\omega}} \wedge \vec{y}$$ where $v^k = \frac{dy^k}{dt}$, and $\vec{a}(t)$ is the acceleration of the origin of the coordinates $y^\alpha$ computed in the inertial coordinates $x^\mu$ and finally expressed in the basis of the reference frame $y^\alpha$.

It is now clear that, within this formalism, the connection coefficients account for the inertial forces.


(*) This means that the vector $e_0$ defining that coordinate satisfies $\langle e_0, dT \rangle =1$.

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    $\begingroup$ @DanielC As usual, thanks for correcting my huge number of typos... $\endgroup$ Nov 9, 2023 at 16:59
  • $\begingroup$ Reading what you write, professor, was, is an always be my pleasure. -:) $\endgroup$
    – DanielC
    Nov 10, 2023 at 8:02
  • $\begingroup$ :-( I spotted another typo, corrected... $\endgroup$ Nov 10, 2023 at 8:44
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Of course, the Euler-Langrange mechanism to derive equations of motion implies terms, quadratic in the velocities by the kinetic energy term $$\frac{d}{dt}\nabla_{\dot x(t)}(\dot x\cdot g(x(t)) \cdot \dot x)-\nabla_{x(t)}(\dot x\cdot g(x(t)) \cdot \dot x)$$

By the chain rule, the time derivative of the first term and the grad of the second together yield the virtual forces.

Since the euclidean square in curvilinear coordinates is the metric tensor applied to the coordinate velocities, the kinetic energy is identical to the line element as a representation of the metrics.

For a 2D coordinate system rotating at constant frequency $\omega$, with $$T= \frac{m}{2}(\dot r(t)^2 + r(t)^2 (\dot \phi(t) +\omega)^2 )$$ the Christoffel symbols are identified as coefficents of the quadratic forms in the equations of motion, yielding centrifugal force for $\ddot r$ and the Coriolis force for $\ddot \phi$

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