My question is, can the (classical) Lagrangian be thought of as a metric? That is, is there a meaningful sense in which we can think of the least-action path from the initial to the final configuration as being the shortest one? Then the equations of motion would be geodesics - not in physical space, but in phase space, with the metric defined by the Lagrangian. Below is a slightly expanded outline of this idea, outlining why I think it might make sense.

(Classical) Lagrangian mechanics is derived by assuming the system takes the path of minimum action from its initial state to its final one. In the formalism of classical mechanics we normally write time $t$ as a special type of coordinate, quite distinct from the other (generalised) coordinates $\mathbf{q}$.

However, if we want we can lump these together into a single set of "generalised space-time coordinates" $\mathbf{r} = (t, q_0, q_1, \dots)$, to be considered a vector space of dimension $1+n$, where $n$ is the number of degrees of freedom. Then the action is just a function of curves in $\mathbf{r}$-space, and the equations of motion are determined by choosing the least-action path between two points, $\mathbf{r_0}$ and $\mathbf{r_1}$. In this picture, $L$ depends upon the position and direction of the curve at each point along it, and the action is obtained by a line integral along the curve.

This has a pleasingly geometrical feel, and so my question is, is there a meaningful sense in which I can think of the least-action path as being the shortest path, with the Lagrangian playing the role of a metric on $\mathbf{r}$-space? Or is there some technical reason why $L$ can't be thought of as a metric in this sense?

Note that this isn't in general the same as a metric defined on space-time. For an $m$-particle system in $3+1$ space-time, $\mathbf{r}$ will have $3m+1$ dimensions, rather than 4. I'm interested in the notion of a metric in $\mathbf{r}$-space defined by $L$, rather than the more usual picture of $L$ being defined in terms of a metric on space-time.

  • 2
    $\begingroup$ No, not always. The kinetic energy part yes, is usually a Riemannian metric, but the potential energy has a different nature. One can always think of the least action path as extremizing a variational functional, but not every variational functional has to be quadratic in velocities/momenta, which is required for it to come from a metric. $\endgroup$
    – Conifold
    Commented Nov 26, 2016 at 2:34
  • $\begingroup$ @Conifold if you could cash that out a bit with an explicit counterexample, it would be an answer. $\endgroup$
    – N. Virgo
    Commented Nov 26, 2016 at 3:42
  • $\begingroup$ It is all the way down. $\endgroup$
    – Conifold
    Commented Nov 29, 2016 at 0:15
  • $\begingroup$ related: physics.stackexchange.com/q/702436/226902 $\endgroup$
    – Quillo
    Commented Apr 6, 2022 at 12:57

5 Answers 5


The dynamics of a large class of mechanical systems can be described as a geodesic motion in some ambient space. This is the essence of the Kaluza-Klein theory.

The basic and most elementary example is the case of a charged particle in $3D$ coupled to a magnetic field which can be described as a neutral particle geodesically moving in a background metric in $4D$, such that the fourth dimension has the shape of a circle. Please see Marsden and Ratiu (Introduction to mechanics and symmetry - page 200). The $4$-th component momentum of the particle along the circle becomes the electric charge.

Thus this theory can account for origin of charge. The circle $S^1$ spanning the $4$-th dimension that we started with ends up to be the internal space $U(1)$ of the electric charge. (Also, since, in quantum mechanics, momenta along compact spaces are quantized, this theory also explains the quantization of the electric charge).

Generalization of the Kaluza-Klein theory describing non-Abelian charges and intercations such as spin and color also exist. Please, see for example the following article by Harnad and Pare.(The generalization of the Kaluza Klein theory to include non-Abelian charges and interactions with Yang-Mills fields was initiated by Kerner).

The generalization to multiple particles interacting through gravitational, electromagnetic and Yang-Mills fields is straightforward. One only has to couple them to the same metric in the ambient space. Also, there is no difficulty to formulate the theory relativistically, since the relativistic rules for coupling to a metric are known.

In mathematics, this theory is called "SubRiemannian geometry", Please see the following review by I. Markina. The most general construction is to allow the particle to start from any point in the ambient space, and move along geodesic lines, but to constrain its velocity to lie on certain subspaces of its tangent bundle. Thus this theory in its full generality describes nonholonomic constraints and has many applications in geometric mechanics.

Due to all the possible interactions that can be explained by the Kaluza-Klein approach, it seemed very attractive for the unification of all the fundamental forces. But, there is a strong argument by Witten that we cannot obtain massless fermions chirally coupled to gauge fields in this approach. (We can, of course, introduce this interaction by hand, but then we would loose the unification principle). Since, this type of interaction Lies in the basis of the standard model, this approach was virtually abandoned.


You can derive the equations of motion (equations of geodesics) for a particle in curved spacetime by using the Lagrangian $$L = \frac{1}{2} \sum_{\mu,\nu} g_{\mu\nu} \frac{dx^\mu}{dt} \frac{dx^\nu}{dt},$$ so the answer is yes. You could regard the configuration manifold as the manifold, it need not be the physical spacetime.

I would like to clarify that $L$ is the metric of the manifold under consideration, not $g_{\mu\nu}$, which are its components in the specific basis.


The motion of a particle on a surface of revolution $r = r(z)$ in $\mathbb{R}^3$ is described in cylindrical coordinates $(r,\theta,z)$ by $L = \frac{1}{2}(\dot{r}^2 + r^2\dot{\theta}^2 + \dot{z}^2)$, $g_{\mu\nu} = \text{diag}(1,r^2,1)$. This has cyclic coordinate $\theta$ and two degrees of freedom, which means that the problem can be solved. Indeed, since the hamiltonian is time-independent, energy is conserved and $|\dot{\vec{x}}|=\text{const}$. If the angle between the meridian and $\dot{\vec{x}}$ is $a$, then $r\dot{\theta} = |\dot{\vec{x}}| \sin a$, and, since $p_\theta = r^2\dot{\theta} = \text{const}$, we obtain $r\sin a = \text{const}$, the equation of the geodesics.

  • $\begingroup$ I'm interested in whether this carries over to the case of a general Lagrangian - multiple particles interacting under multiple forces, bound by multiple constraints, and not necessarily in a relativistic context. The fact that it works for a single relativistic particle moving inertially is suggestive, but by itself it isn't enough to demonstrate that the general case will hold. (See the final paragraph of my question.) But many thanks for the answer! $\endgroup$
    – N. Virgo
    Commented Mar 30, 2014 at 9:44
  • 2
    $\begingroup$ I suppose another way to put the question is, can we always write the Lagrangian in this form, even if $x$ refers to the "generalised space-time coordinates" for many particles (as defined in the question), and $\mu$ and $\nu$ range over a different number of dimensions than the dimensionality of space-time? $\endgroup$
    – N. Virgo
    Commented Mar 30, 2014 at 9:50
  • $\begingroup$ Thanks for your second example, but it seems to be showing something different. The metric you describe is only for the spatial coordinates, and doesn't include time. Still, it's again suggestive of a "deep" relationship between Lagrangian mechanics and geodesics (at least in the case where there are no forces besides the constraints), and so I appreciate it. $\endgroup$
    – N. Virgo
    Commented Mar 31, 2014 at 2:20
  • $\begingroup$ By the way, I don't get notified when you update your answer, so if you make any changes you'd like me to see, it's a good idea to leave a comment to let me know. $\endgroup$
    – N. Virgo
    Commented Mar 31, 2014 at 2:20
  • $\begingroup$ @Nathaniel I wasn't certain what $(t)\times \bf{q}$ means, now I see that you meant $\mathbb{R} \times \mathbb{R}^3$. If we treat time as a coordinate, we obtain the Euler equations for fields. This can be generalized to many particles with constraints, but I cannot think of something specific at the moment. $\endgroup$
    – auxsvr
    Commented Mar 31, 2014 at 8:26

Turning time into an extra dimension does not really help because "time" in this context is not a coordinate, it is a parameter used to represent the path. Even if we add an extra dimension the paths would still have to be parametrized to take "line integrals", so a parameter will be reintroduced anyway. But even if that was made sense of (maybe by fixing the parametrization somehow, I don't really see how) such "lifted" Lagrangian will most certainly not be in the special form of geodesic functionals, because of how the "time" enters it.

The general form of a classical Lagrangian least action functional in generalized coordinates is $L=\int g_\gamma(\dot{\gamma},\dot{\gamma})-V(\gamma)\,dt$, where $g$ is the Riemannian metric for kinetic energy, $\gamma=\gamma(t)$ is a path, and $V$ is the potential term. The functional whose extremals are geodesics is only $\int g_\gamma(\dot{\gamma},\dot{\gamma})\,dt$ (for length we have to put square root under the integral, but square root is a monotone function, so minimizers will be the same). There is no reason why minimizing a least action integral should be equivalent to minimizing a geodesic one, perhaps with a different $g$, except in some very special cases (such as constant $V$). In coordinates the first term will be a sum of some functions of generalized positions multiplied by second degree monomials in generalized velocities, the potential term has no velocities at all.

So already Galileo's problem of falling bodies in constant gravity, with $L=\int\frac12(\dot{x}^2+\dot{y}^2)-\text{g}y\,dt$ that produces the usual assortment of parabolas, is not a geodesic problem. One may still wonder if there is some super-clever way to come up with a metric that produces an equivalent problem, for which the least action parabolas are the geodesics. But there is one special property that geodesics have which precludes it: there is only one of them (locally) through every point in every direction. This is not true of Galileo's parabolas: if you throw a ball in the same direction, say horizontally, but with different strengths (initial speeds), the parabolas it follows down to the ground will be different. This said, much of the geometric theory of geodesics, often encountered in modern textbooks only in their context, the eikonal, the conjugate points, the Jacobi fields, etc., does generalize to more general functionals, much more general even than the classical least action.

In contrast, the related brachistochrone problem is a geodesic problem, its Lagrangian is $L=\displaystyle{\int\frac{\dot{x}^2+\dot{y}^2}{2\text{g}y}\,dt}$, the corresponding "length" is the travel time, and this is why Bernoulli was able to solve it the way he did, with the optico-mechanical analogy. In particular, the aforementioned parametrization invariance property of the geodesics means that no matter the initial speed the trajectory of fastest descent from any point will be the same, as long as it is in the same direction.

Thus, Bernoulli's cycloids are geodesics but Galileo's parabolas are not. By the way, it was Galileo who first considered the brachistochrone problem, although he thought that the solutions were circular rather than cycloidal arcs (or at least, what he wrote is commonly interpreted that way). Galileo was also the one who first introduced the cycloid. History is funny sometimes.


I) OP talks about minimizing curves (rather than higher dimensional objects) so let us concentrate on point mechanics (as opposed to field theory) with Lagrangian $L$ (rather than Lagrangian density ${\cal L}$). We conventionally call the curve parameter time $t$, although it doesn't have to correspond to any physical time variable.

Let us for simplicity impose Dirichlet boundary conditions on the curve

$$ \tag{1} q(t_i)~=~q_i \qquad \text{and}\qquad q(t_f)~=~q_f. $$

To have the stationary action principle working as a minimum action principle, the potential term should be small, i.e. we should either study the sudden approximation $\Delta t:=t_f-t_i \ll \tau$ small ("so that the potential doesn't have time to act"), or study theories without potential terms, i.e.

$$\tag{2} L(q,\dot{q})~=~T(q,\dot{q})~=~ g_{jk}(q)\dot{q}^{j}\dot{q}^{k}.$$

To have a minimum principle, the metric $g_{jk}$ should be (semi)positive definite. Zero-modes should be gauge-fixed.

II) Assume that the configuration space $(M,g)$ is endowed with a metric $g_{jk}$. It seems natural in this context to mention Synge's world function $\sigma(q_f,q_i)$, which is half the square of the length of the (shortest) geodesic between the two generalized positions $q_i$ and $q_f$, cf. e.g. Ref. 1. Synge's world function $\sigma(q_f,q_i)$ may not be well-defined globally. Besides general relativity, the Synge's world function is used by various authors, such as, e.g., Refs. 2 and 3, in geometrically covariant approaches to field theory.


  1. E. Poisson, The Motion of Point Particles in Curved Spacetime, Living Rev. Relativity 7 (2004) 6; Section 2.1.

  2. Bryce DeWitt, Supermanifolds, Cambridge Univ. Press, 1992; Starting from around eq. (6.4.13).

  3. G.A. Vilkovisky, The unique effective action in quantum field theory, Nucl. Phys. B234 (1984) 125; Starting from around eq. (12).

  • $\begingroup$ Thanks for the answer - I somehow managed to miss it until today. Having defined Synge's world function, what does one then do with it? (The Wikipedia page doesn't say either.) $\endgroup$
    – N. Virgo
    Commented Apr 3, 2014 at 1:09
  • $\begingroup$ The metric as you write it seems to be defined on configuration space rather than generalised space-time. I'm wondering whether including time as a coordinate (and parameterising the curve with some other variable) would allow potentials to be considered. $\endgroup$
    – N. Virgo
    Commented Apr 3, 2014 at 1:18
  • $\begingroup$ @Nathaniel: Well, I tried to keep the answer quite general. As in string theory, one of the target space dimensions could in principle be a time coordinate. I'm not sure it would be constructive for me to elaborate since I'm not entirely sure what you would be interested in. Instead I have updated the answer with some references. $\endgroup$
    – Qmechanic
    Commented Apr 3, 2014 at 11:30

You don't need to assume that the path of least action is the path taken. You can show it from Newton's laws. See http://www.damtp.cam.ac.uk/user/tong/dynamics/two.pdf

The path of least action is the path for which $F = ma$ holds at each point. This is the geodesic. This is the shortest path through space-time. You get this path from the Lagrangian.

You can do this for many particle systems in general coordinates. It works for all the fundamental laws of physics.

But I am not sure this makes the Lagrangian a metric. A vector space is a metric space if it has a metric. A metric is a function that takes any vector and returns a length for that vector.

In this case, the vector space is space-time. The usual metric returns either a proper length or a proper time.

Are you saying that a valid replacement for length is given by $L = T-V$? This has energy units. It provides information about forces, not just space-time. Maybe this is OK. Geodesics contain information about forces. But more energy = bigger distance/time?

This doesn't look right, unless you are talking about a different vector space than I think you mean.

  • 2
    $\begingroup$ The vector space I'm talking about is not space-time. For a single-particle system it has the same dimensionality but a different metric. (The Lagrangian not only has dimensions if energy but also depends on the particle's mass.) For an $n$-particle system it has $3n+1$ dimensions. $\endgroup$
    – N. Virgo
    Commented Mar 31, 2014 at 0:17
  • $\begingroup$ OK. But I am still not sure about L as a metric. If you have 2 interacting particles, you have information about where both are. This tells you $V$. How do you get $T$ from just these coordinates? Do you need to add $\dot{q}$ coordinates? $\endgroup$
    – mmesser314
    Commented Mar 31, 2014 at 9:44
  • 2
    $\begingroup$ When calculating distance from a metric tensor, the integrand depends upon not only the position of each point along a curve, but also upon the direction of the curve at that point. The Lagrangian has the same property -- if you plot $\mathbf{q}$ against $t$ it's easy to see that $\mathbf{\dot q}$ tells you the direction of the curve. It's because of this similarity that I think $L$ might formally be a metric in $\mathbf{r}$ space, if expressed in the right way. $\endgroup$
    – N. Virgo
    Commented Mar 31, 2014 at 10:03

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