Consider a general curved space-time with torsion. In the standard Einstein-Cartan-Kibble-Sciama theory (ECKS or ECSK), torsion is non-dynamical and doesn't propagates in free space. But a more general theory could allow torsion to be dynamical and propagates. In general, geodesics (the shortest or extremal length curves) and auto-parallels (the straightest curves) are different curves in space-time.

In classical general relativity (without torsion), test-particles without spin should follow geodesics. This is a statement of inertia motion.

But with torsion, what curve should a test-particle follow ? A geodesic or a auto-parallel curve ?

It can be shown that if torsion is totally antisymetric (which is just a special case), geodesics and auto-parallel curves are the same. But in general they aren't.

I feel that Newton's inertia principle is really about the straightest curves, and not the shortest (or extremal) curves.

Is there any indication, clue or argument, that the spineless test-particles should follow auto-parallel curves in space-time, instead of geodesics ? Would it be more natural in some way ?

If the auto-parallel curves are more fundamental (from an inertia point of view), then would it imply that the lagrangian method for fields is falling apart as a general principle, since it's (i.e. was) motivated by the inertia principle ?

  • $\begingroup$ Isn't the variational principle related to the shortest proper time, assuming the curve on the space-time manifold is parametrized by proper time? This variational principle should replace whatever principle occurring in Newtonian mechanics (Newton's inertial principle holds for motion in a 0 vector total field of forces, and this is unrelated to notions of "straightest" or "shortest" curves). $\endgroup$
    – DanielC
    Oct 11, 2017 at 13:50
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    $\begingroup$ I'm voting to close this question as off-topic because this identical question has been asked here: physics.stackexchange.com/q/318200 $\endgroup$
    – user87745
    Oct 11, 2017 at 15:32
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    $\begingroup$ @Dvij: that's not what off-topic means: it could be marked as a duplicate, but even that is problematic as the question it is a duplicate of has no anwers $\endgroup$
    – Christoph
    Oct 11, 2017 at 15:39
  • $\begingroup$ @Christoph I agree. But there is a bug in the StackExchange network which prevents one to do so in certain cases. See: physics.meta.stackexchange.com/q/9949 $\endgroup$
    – user87745
    Oct 11, 2017 at 16:19

2 Answers 2


My vote would go to the auto-parallels as well, which seems to me the correct generalization of the idea that bodies should persist in their state of motion.

According to Kleinert, Pelster (doi, arxiv), due to the closure failure of parallelograms in spaces with torsion, the action principle needs to be modified, leading to the appearance of an additional torsion term in the Euler-Lagrange equations $$ \frac{\partial L}{\partial q^\lambda(\tau)} - \frac{d}{dt} \frac{\partial L}{\partial \dot q^\lambda(\tau)} = 2S_{\lambda\mu}{}^\nu(q(\tau)) \dot q^\mu(\tau) \frac{\partial L}{\partial \dot q^\nu(\tau)} $$

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    $\begingroup$ The arguments from the paper cited in your answer are pretty strong and convincing : inertia and locality imply that a spinless particle should follow an auto-parallel curve, instead of a geodesic. From the paper : "Because of its inertia, a particle will change its direction in a minimal way at each instant of time, which makes its trajectory as straight as possible. If it were to choose a path which minimizes the length of the orbit it would have possessed some global information of the geometry." $\endgroup$
    – Cham
    Oct 11, 2017 at 17:38

Without prejudicing the issue on either side, we can follow the Gandalf / Toucan Sam Strategy and follow our nose.

The matter can be addressed by applying the continuity equation to the stress tensor density for the particle: $$𝔗^μ_ν(x) = \int_{-∞}^{+∞} p_ν(s) v^μ(s) δ(x - x(s)) ds, \hspace 1em v^ρ(s) = \frac{dx^ρ}{ds},$$ with the momentum vector being given by: $$p_ν(s) = m g_{μν}(x(s)) v^μ(s),$$ where $m > 0$ is the mass of the particle, and $s ↦ x(s)$ its worldline with parameterization, $g_{μν} v^μ v^ν = 1$; i.e. $s$ is the proper time, if the convention is chosen of setting $g_{00} = 1$ in the Minkowski limit.

The divergence can be tabulated as follows: $$\begin{align} ∂_μ𝔗^μ_ν(x) &= \int_{-∞}^{+∞} p_ν(s) v^μ(s) ∂_μ δ(x - x(s)) ds \\ &= \int_{-∞}^{+∞} p_ν(s) \left(-\frac{d}{ds} δ(x - x(s))\right) ds \\ &= -\left[p_ν(s) δ(x - x(s))\right]_{-∞}^{+∞} + \int_{-∞}^{+∞} \frac{dp_ν}{ds} δ(x - x(s)) ds \\ &= \int_{-∞}^{+∞} \frac{dp_ν}{ds} δ(x - x(s)) ds \end{align}$$

Now ... the continuity equation is for the stress tensor: $$∇_μT^μ_ν ≡ ∂_μT^μ_ν - Γ_{μν}^ρ T^μ_ρ + Γ_{μρ}^μT^ρ_ν = 0.$$ The Levi-Civita connection is $Γ_{μν}^ρ - C_{μν}^ρ$, where $C_{μν}^ρ$ is the contorsion, or just $Γ-C$ for short. A well-known (to General Relativists) identity for it is given by: $$(Γ-C)_{μρ}^μ = \frac{1}{2} g^{μν} ∂_ρ g_{μν} = \frac{∂_ρ \sqrt{|g|}}{\sqrt{|g|}}.$$ Applying this to the continuity equation: $$∂_μT^μ_ν + \frac{∂_ρ \sqrt{|g|}}{\sqrt{|g|}}T^ρ_ν - Γ_{μν}^ρ T^μ_ρ + C_{μρ}^μT^ρ_ν = 0,$$ multiplying by $\sqrt{|g|}$, and noting that the stress tensor density, itself, is given by $𝔗^ρ_ν = \sqrt{|g|}T^ρ_ν$, we can write: we get: $$∂_μ𝔗^μ_ν - Γ_{μν}^ρ𝔗^μ_ρ + C_{μρ}^μ𝔗^ρ_ν = 0.$$

Upon substitution, we get: $$ 0 = \int_{-∞}^{+∞} \left(\frac{dp_ν}{ds} - Γ_{μν}^ρ(x) p_ρ(s) v^μ(s) + C_{μρ}^μ(x) p_ν(s) v^ρ(s)\right) δ(x - x(s)) ds \\ \int_{-∞}^{+∞} \left(\frac{dp_ν}{ds} - Γ_{μν}^ρ(x(s)) p_ρ(s) v^μ(s) + C_{μρ}^μ(x(s)) p_ν(s) v^ρ(s)\right) δ(x - x(s)) ds. $$

We assume that there is a foliation of the space-time manifold - at least in the vicinity of the worldline - into spacelike 3-surfaces $Σ_s$, such that (1) $x(s) ∈ Σ_s$ for each $s ∈ (-∞,+∞)$ and (ii) we can decompose the delta function into $δ(x_s - x(s')) = δ_{Σ_s}(x_s - x(s))δ(s - s')$. Then an integral of the following form may be worked out as follows: $$\begin{align} \int_{Σ_s} \left(\int_{-∞}^{+∞} F(s') δ(x_s - x(s')) ds'\right) dx_s &= \int_{Σ_s} \left(\int_{-∞}^{+∞} F(s') δ_{Σ_s}(x_s - x(s))δ(s - s') ds'\right) dx_s \\ &= \int_{-∞}^{+∞} F(s') \left(\int_{Σ_s} δ_{Σ_s}(x_s - x(s)) dx_s\right) δ(s - s') ds' \\ &= \int_{-∞}^{+∞} F(s') δ(s - s') ds' \\ &= F(s). \end{align}$$

Applying this to the reworked continuity equation, we get: $$\frac{dp_ν}{ds} - Γ_{μν}^ρ p_ρ v^μ + C_{μρ}^μ p_ν v^ρ = 0,\label{1}\tag{1}$$ where all $x$-dependent quantities are evaluated on the worldline at $x(s)$. The first two terms are the momentum-space version of the terms that appear in the geodesic and autoparallel equations: $$\frac{dp_ν}{ds} - Γ_{μν}^ρ p_ρ v^μ = m g_{μν} \left(\frac{dv^μ}{ds} + Γ_{νρ}^μ v^ν v^ρ\right). \label{2}\tag{2}$$ This requires on that the connection $Γ$ be metrical and involves the chain rule on $g_{μν}(x(s))$ in $p_ν = g_{μν}v^μ$, along with the metrical property of the connection: $$ \frac{dp_ν}{ds} = m\left(\frac{dg_{μν}}{ds} v^μ + g_{μν} \frac{dv^μ}{ds}\right), \hspace 1em \frac{dg_{μν}}{ds} = ∂_ρ g_{μν} v^ρ = \left(g_{σν} Γ_{ρμ}^σ + g_{μσ} Γ_{ρν}^σ\right) v^ρ \\ ⇒ \frac{dp_ν}{ds} = m\left(g_{σν} Γ_{ρμ}^σ + g_{μσ} Γ_{ρν}^σ\right) v^ρ v^μ + m g_{μν} \frac{dv^μ}{ds}. $$ We can rewrite $$Γ_{μν}^ρ p_ρ v^μ = Γ_{ρν}^σ p_σ v^ρ = m g_{μσ} Γ_{ρν}^σ v^μ v^ρ.$$ Therefore, upon cancellation, we get: $$\begin{align} \frac{dp_ν}{ds} - Γ_{μν}^ρ p_ρ v^μ &= m g_{σν} Γ_{ρμ}^σ v^ρ v^μ + m g_{μν} \frac{dv^μ}{ds} \\ &= m g_{μν} \frac{dv^μ}{ds} + m g_{μν} Γ_{ρσ}^μ v^ρ v^σ \\ &= m g_{μν} \left(\frac{dv^μ}{ds} + Γ_{ρσ}^μ v^ρ v^σ\right). \end{align}$$

Rewriting the contorsion term in ($\ref{1}$), $$C_{μρ}^μ p_ν v^ρ = -p_ν N = -m g_{μν} N v^μ, \hspace 1em N = -C_{σρ}^σ v^ρ,$$ divide out the mass $m > 0$: $$ g_{μν} \left(\frac{dv^μ}{ds} + Γ_{ρσ}^μ v^ρ v^σ - N v^μ\right) = 0 $$ and invert the metric in $g_{μν}$ to get: $$\frac{dx^μ}{ds} = v^μ, \hspace 1em \frac{dv^μ}{ds} + Γ_{ρσ}^μ v^ρ v^σ = N v^μ. \tag{3}$$

That's a variation of the auto-parallel equation, but with the need to reparametrize $s$. If you reset the parameter to $S$, such that $$\frac{d}{ds}ln\left(\frac{dS}{ds}\right) = N = -C_{σρ}^σ v^ρ, \tag{4}$$ then you'll get: $$\frac{dx^μ}{dS} = v^μ, \hspace 1em \frac{dv^μ}{dS} + Γ_{ρσ}^μ v^ρ v^σ = 0. \tag{5}$$

There's a warping of some sort on the proper time $s$ into $S$.

The contorsion is warping the time.

I did these calculations extemporaneously. You should check it closely. It needs a once-over. I might try it again at a later time. I'm pretty sure of ($\ref{1}$) and ($\ref{2}$) is fairly standard and well-known.


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