The motion of a point particle in classical mechanics is given by Newton's equation, $\mathbf{F}=m\mathbf{a}$. Suppose all forces considered are conservative and we have a constant total energy $h$. Let $M$ be the configuration space of our system, $T^*M$ its cotangent bundle, $(\mathbf{q},\mathbf{p})$ the natural coordinates on $T^*M$ and $\gamma$ a curve connecting the start- and end-points of our particle's motion. Then $\int_\gamma \mathbf{p}\,\mathrm{d}\mathbf{q}$ is a viable action integral. The Maupertuis theorem states that $$\int_\gamma\mathbf{p}\,\mathrm{d}\mathbf{q}=\sqrt{2}\int_\gamma\mathrm{d}\rho,$$ where $\mathrm{d}\rho$ is the Riemannian metric given by $$\mathrm{d}\rho=\sqrt{h-U(\mathbf{q})}\,\mathrm{d}s,$$ $\mathrm{d}s$ is the standard metric on $M$ and $U(\mathbf{q})$ the potential energy. This implies that

$$\text{Newtonian mechanics $\Longleftrightarrow$ geodesic problem of some pair $(M,\mathrm{d}\rho)$}$$

The motion of a point particle in general relativity is given by the equation $$F=a\tag{1}$$ where $a:=\nabla_{\dot\gamma}\dot\gamma$, $\gamma$ is the path of the particle, $\nabla$ is the Levi-Civita connection of the spacetime $(\mathcal{M},g)$, and $F$ is some "force" 4-vector. $F$ can be seen as an obstruction to the geodesy of $\gamma$, since $a=0$ is just the geodesic equation. (In the same way, $\mathbf{F}$ is an obstruction to geodesy on $\mathbb{R}^n$ since $\mathbf{a}=0$ $\Leftrightarrow$ $\mathbf{x}$ is a straight line and $U(\mathbf{q})$ in the case of constrained Lagrangian mechanics.) (Proofs of the above statements of classical mechanics can be found in Arnold, V.I. Mathematical Methods of Classical Mechanics. Springer, 1989.)

Is there some pair $(\mathcal{M}',g')$ for which the geodesic problem is (1), i.e. a suitable generalization of the Maupertuis principle to relativistic mechanics in curved spacetime?

  • $\begingroup$ What would be the meaning of F (in the relativistic case) is conservative? $\endgroup$ – MBN Dec 19 '15 at 10:24
  • $\begingroup$ @MBN I do not know. Perhaps $F_\mu=-\partial_\mu U$ for some function $U$. (That might be another question right there.) $\endgroup$ – Ryan Unger Dec 19 '15 at 10:40
  • $\begingroup$ The currently-unchallenged Wikipedia item "Principle of least action" claims that there is, through the "Einstein-Hilbert action". The actual remarks by Maupertuis that the Wiki cites are much more general, but I believe those made earlier by Liebniz, to whom the article traces that principle, are more specific. $\endgroup$ – Edouard Aug 1 at 2:47

I) We assume OP's question concerns a massive point particle of rest mass $m_0>0$ in a Lorentzian spacetime manifold $(M,g)$ [of signature $(+,-,\ldots, -)$] between an initial and a final spacetime point $p_i, p_f\in M$, which should be causally connected. Let us work in units where the speed of light $c=1$ and rest mass $m_0=1$ are both one.

II) Before discussing Maupertuis's principle, we should first have a stationary action principle (SAP). OP's question mentions an unspecified $4$-force. To have a variational principle, we must demand that that $4$-force comes from a potential $U$. The action is then

$$ S[x;\lambda_i,\lambda_f] ~=~\int_{\lambda_i}^{\lambda_f}\! d\lambda~L, \qquad L~=~T-U, $$ $$\tag{1} T~:=~-\sqrt{T_0},\qquad T_0~:=~ g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}, $$

where $\lambda$ is a parameter, and dot means differentiation wrt. $\lambda$.

III) In the SAP, we impose Dirichlet boundary conditions (BC)

$$\tag{2} x^{\mu}(\lambda_i)~=~x_i^{\mu}\qquad\text{and}\qquad x^{\mu}(\lambda_f)~=~x_f^{\mu}, $$

and keep $\lambda_i,\lambda_f$ fixed.

IV) We'll assume that the action (1) is reparametrization invariant

$$\tag{3}\lambda\quad\longrightarrow\quad \tilde{\lambda}~=~f(\lambda), $$

since physics should be geometric.

V) The Lagrangian $4$-momentum and energy function become

$$\tag{4} p_{\mu}~:=~\frac{\partial L}{\partial\dot{x}^{\mu}}~=~ -\frac{g_{\mu\nu}\dot{x}^{\nu}}{\sqrt{T_0}}-\frac{\partial U}{\partial\dot{x}^{\mu}},$$


$$\tag{5} h~:=~ p_{\mu}\dot{x}^{\mu} - L~=~\left(1 - \dot{x}^{\mu}\frac{\partial }{\partial\dot{x}^{\mu}}\right)U, $$ respectively.

VI) Usually when discussing the abbreviated action principle, it is assumed that the Lagrangian (1) has no explicit $\lambda$-dependence, so that the energy (5) is conserved on-shell. No explicit $\lambda$-dependence may sound natural & innocent, but together with the reparametrization invariance (3), it severely restricts the possible potentials $U$. The Lorentz potential $U\propto A_{\mu}\dot{x}^{\mu}$ is still allowed, of course.

VII) In fact, no explicit $\lambda$-dependence & reparametrization invariance (3) basically imply that the energy function (5) vanishes identically.

VIII) The abbreviated action becomes

$$\tag{6} A[x;E,\lambda_i,\lambda_f] ~=~\int_{\lambda_i}^{\lambda_f}\! d\lambda~ p_{\mu}\dot{x}^{\mu}$$

for virtual paths of constant and the same energy

$$ \tag{7} h~=~E, $$

satisfying Dirichlet BC (2), but free $\lambda_i$ and $\lambda_f$. From Section VII, we know that the energy $E=0$ drops out of the abbreviated action (6).

IX) Returning to OP's question, there seems little hope to achieve a Jacobi square root form of the abbreviated action principle (6) without further assumptions.

The natural next step is to assume that the potential $U$ does not depend on the $4$-velocity $\dot{x}^{\mu}$. Then the energy function $h\equiv U$ becomes just the potential energy, and $p_{\mu}\dot{x}^{\mu}\equiv T\equiv -\sqrt{T_0}$.

However, with all the other above requirements, this basically implies that the potential $U=0$ is zero!

Of course, without a potential $U=0$, we unsurprisingly achieve a Jacobi square root form of the abbreviated action principle

$$\tag{8} A[x;E,\lambda_i,\lambda_f] ~=~-\int_{\lambda_i}^{\lambda_f}\! d\lambda~\sqrt{T_0} .$$

The abbreviated action (8) is identical to the SAP (1) that we started from, essentially due to reparametrization invariance.

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  • $\begingroup$ The action integral you gave is the standard integral for geodesic motion on a spacetime. In the OP I'm asking for an action integral for nongeodesic motion on $(\mathcal{M},g)$ posed as geodesic motion on some other $(\mathcal{M}',g')$. As such, this does not (appear to) answer the question. $\endgroup$ – Ryan Unger Dec 19 '15 at 8:00
  • $\begingroup$ Oh, I didn't notice the $4$-force. I updated the answer. $\endgroup$ – Qmechanic Dec 19 '15 at 8:29

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