About a year ago, I took a course on general relativity. It isn't until now that I realize that, given a metric, I am unsure how to find a particle's dynamics. What I mean by that is, normally I would set up either a Lagrangian and Hamiltonian and try to find dynamics the usual ways from these methods. But how do these methods change when I am in curved space time? During the relativity course, we spent a very large amount of time studying tensor analysis, covariant derivatives, geodesics, etc, but not finding the dynamics of a physical system. I am sure we learned this somewhere, maybe I just forgot.

I thought to myself that maybe I could use the geodesic equation, but I don't think this equation would be valid in the presence of a potential. Can I just use the normal methods but just replace the normal derivative with the covariant derivative?


Thanks for all the answers and the bounty put up. I will wait a couple more days to pick the best answer. I thought I would add this for future readers of this question (and possibly for more discussion).

I found a section in my relativity book (Spacetime and Geometry by Sean Carroll) in the beginning of Chapter 4, he says in order to do the "physics of space time" do the following:

  1. Take a law of physics, valid in inertial coordinates in flat spacetime
  2. Write it in a coordinate invariant (tensorial) form.
  3. Assert that the resulting law remains true in curved spacetime.

His example is then to do these steps with a free particle to obtain the geodesic equation, so according to this, these steps could be followed for any dynamical theory. The second step being the most complicated part as (for example) second derivatives need modifications to be considered a tensorial object (which he covers in the book).

  • $\begingroup$ What I guess is that uncharged particles just follow the geodesics and for charge particles, you'll need to set up the Lagrangian which involves the potential. I guess wait till someone more knowledgeable comes up to answer. $\endgroup$ Commented May 7, 2017 at 16:14
  • $\begingroup$ The word "particle" invokes quantum mechanics. General relativity is not quantized , it is a classical theory. One would have to go to something like a string theory model, which allows for quantization of gravity , imo. For flat spaces there is no problem to have an effective quantization but I imagine that for curved ones a particular model has to be specified.. $\endgroup$
    – anna v
    Commented May 7, 2017 at 17:32
  • 5
    $\begingroup$ Anna, by particle I don't mean quantum mechanics, I mean just a "point particle." Such particles are common in any kind of classical analysis, for example any kinematics problem you treat the object as a point particle. $\endgroup$
    – user41178
    Commented May 7, 2017 at 20:06

4 Answers 4


As in classical mechanics, there are several approaches to determining the dynamics of a particle in general relativity. Crucially we should observe that, in general relativity, the effect of a gravitational field is entirely captured by the behaviour of the metric. Providing the particle in question experiences no other external influences, we can determine its motion using the geodesic equation as you suppose. This is one of the central axioms of general relativity:

A particle moving only under the influence of gravity follows geodesics in curved spacetime. These geodesics are timelike if the particle is massive and null if the particle is massless.

Dynamics à la Newton:

Knowing the above, we can directly write down the equation governing particle motion. In Riemannian geometry, it is natural to define a geodesic as a path that extremises the length between two fixed points. For us, this definition comes with some subtleties, due to the fact that two distinct points can be zero distance apart. For our purposes, there is a better definition:

A geodesic is a path whose tangent vector is parallel transported along it. In wordier terms, if you take the tangent vector at one point, and then carry it along the path without changing the direction it points in, the vector remains tangent to the path at the destination.

In less wordy terms, geodesics are straight lines.

Mathematically this is expressed as

$$ \nabla_U U = 0 \qquad \mathrm{where} \qquad U^\mu = \frac{\mathrm{d}x^\mu}{\mathrm{d}\lambda}\,.$$ Here $\lambda$ is a parameter along the curve and $U$ is tangent to it. If we expand out this equation using the definition of the covariant derivative, we arrive at the famous geodesic equation: $$ \frac{\mathrm{d}^2 x^\mu}{\mathrm{d}\lambda^2} + \Gamma^\mu{}_{\nu \rho}\frac{\mathrm{d}x^\nu}{\mathrm{d}\lambda}\frac{\mathrm{d}x^\rho}{\mathrm{d}\lambda} = 0 \,.$$ This equation is the analogue of Newton's first law: particles under the influence of no external force move in straight lines at constant speed. It is nothing other than

$$\vec{a} = 0\,.$$

This comparison informs us how to modify the geodesic equation when external forces are present. For definiteness, suppose that the particle is moving under the influence of an electromagnetic field, in addition to gravity. We simply take geodesic equation and replace the right hand side with force over mass. Writing the Lorentz force in the covariant form used in special relativity, and choosing $\lambda$ to be proper time $\tau$, we find that a charged particle obeys the equation: $$ \frac{\mathrm{d}^2 x^\mu}{\mathrm{d}\tau^2} + \Gamma^\mu{}_{\nu \rho}\frac{\mathrm{d}x^\nu}{\mathrm{d}\tau}\frac{\mathrm{d}x^\rho}{\mathrm{d}\tau} = \frac{q}{m} F^\mu{}_\nu \frac{\mathrm{d}x^\nu}{\mathrm{d}\tau} \,.$$ Note that if we turn off the gravitational field by setting the connection components to zero, we recover the special relativistic form of Newton's second law for a particle moving in an electromagnetic field, as we must do.

Dynamics à la Lagrange:

One of the advantages of Lagrangian over Newtonian mechanics is that it's often easier to write down some scalar 'energy' that characterises our system than it is to specify the vectorial forces acting on it. With no external influences present, we want an action that, when extremised, gives us geodesics. For massive particles, we can employ the definition of geodesic familiar from Riemannian geometry – action equals length:

$$S = m\int\mathrm{d}\tau = m\int \mathrm{d}\lambda \sqrt{-g_{\mu \nu}\frac{\mathrm{d}x^\mu}{\mathrm{d}\lambda}\frac{\mathrm{d}x^\nu}{\mathrm{d}\lambda}} \,.$$

It's a worthwhile exercise to check that this indeed yields the geodesic equation when extremised (note that one should be careful about the limits on the integral when performing the variation!).

With this free-particle action in place, it's a simple matter to add extra terms that capture the effect of interactions of the particle with external fields or other particles. If an electromagnetic field is present, the action becomes

$$ S = m\int \mathrm{d} \tau - q \int \mathrm{d} \tau A_\mu \frac{\mathrm{d}x^\mu}{\mathrm{d}\tau} \,. $$

For massless particles this picture is complicated by the fact that every null path has zero length. To talk about the Lagrangian mechanics of massless particles, we need to introduce a new action which contains a Lagrange multiplier $e(\lambda)$ defined along the worldline. The following action is valid for particles of any mass:

$$ S = \frac{1}{2} \int \mathrm{d} \lambda \left(\frac{1}{e(\lambda)} g_{\mu \nu}\frac{\mathrm{d}x^\mu}{\mathrm{d}\lambda}\frac{\mathrm{d}x^\nu}{\mathrm{d}\lambda} - m^2 e(\lambda)\right)\,.$$

This is a technical point – more information can be found in this and this stackexchange post. At the end of the day, we're defining these actions so that they return the geodesic equation when extremised. Adding extra terms to the action to account for interactions is then often easier than adding extra terms directly to the geodesic equation.

Dynamics à la Hamilton:

Perhaps some other time; this answer is already long enough. Let me just note here that it is entirely possible to formulate a Hamiltonian approach to particle dynamics in a curved spacetime, but a little tricky. There are additional subtleties associated with the fact that the conjugate momenta are not independent – they satisfy a mass-shell constraint – which compels us to include a Lagrange multiplier as above. For more information see these lecture notes by Paul Townsend.

  • $\begingroup$ Regarding your final section: see my answer for a simple, covariant Hamiltonian description. $\endgroup$ Commented May 17, 2017 at 17:04

In general relativity, in the absence of other fields, point particles move along geodesics. Note that a Newtonian potential is replaced in general relativity by a component of the metric. In the presence of other fields, the motion need no longer be geodetic.

More specifically, for a particle of constant mass $m>0$ and charge $e$, there is a simple covariant Hamiltonian description: In the presence of a gravitational field (metric tensor) $G(x)$ and an electromagnetic field $A(x)$, the Hamiltonian is given by $$H=\frac{1}{2m}(p+eA(q))^TG(q)(p+eA(q)).$$ This Hamiltonian generates the trajectory parameterized by the particle's eigentime. For an uncharged particle (or in the absence of an electromagnetic field) this reduced to $H=p^TG(q)p/2m$, which gives the geodetic motion.

Details are discussed, e.g., in Thirring's course in mathematical Physics, Part I, Chapter 5.


The equation for a point particle in a curved space time (if you assume the particle doesn't change the spacetime metric) is obtained by extremizing the action (see here):

$$ S = \int_{\xi_0}^{\xi_1} d\xi \sqrt{g_{\mu\nu}(x)\frac{\partial x^\mu(\xi)}{\partial \xi}\frac{\partial x^\nu(\xi)}{\partial \xi}} = \int ds,$$ the Euler-Lagrange equations for this metric are just the geodesic equations.

If you want to include a potential, the condition you have to satisfy is that the equations are covariant, but it is much easier to work with the action. Notice that the action above has the same value no matter how you parametrize the curve $x^\mu(\xi)$, i.e., if you change variables and apply the chain rule, the integral is the same. The covariance of the corresponding equations of motion follows from this property of the action. Since any function that is invariant with respect to reparametrizations of the curve is a function of the length, the most general action you can write is just

$$ S = \int f(s)ds. $$

Every potential you can couple corresponds to a choice of $f(s)$.


In Lagrange mechanics the covariant generalized momenta and forces are $$p_{i} =\frac{\partial L}{\partial u^{i} },$$ $$F_{i} =\frac{\partial L}{\partial x^{i} } $$ with appropriate Lagrangian $L$ and four-velocities $u^i$. The Euler-Lagrange equations $$\frac{d}{d\lambda } \frac{\partial L }{\partial u^{i } } -\frac{\partial L }{\partial x^{i } } =0$$ are rewritten in form $$\frac{dp_{i } }{d\lambda } -F_{i} =0.$$ The contravariant momenta bind to the physical energy and momentum of the particle $p^j=g^{ji}p_i$ and the gravitational force acting on it is mapped to associated vector $F^j=g^{ji}F_i$. When passing to these values Prespacetime Journal, Vol 13, No 1 (2022), 32 , the Euler-Lagrange equations take the form $$\frac{dp^{k} }{d\lambda } + g^{kl } \frac{\partial g_{li} }{\partial x^{j} } u^{j} p^{i} =F^{k}.$$ The presence of the second term on the left side reflects that in the gravitational field not only the 4-momentum of matter, but the 4-momentum of matter together with the gravitational field is stored (see L.D. Landau & E.M. Lifshitz, Classical Theory of Fields, Fourth Edition, § 96). The Lagrangian of a material particle with a rest mass is as follows $$L_m=cm\sqrt{g_{ij}u^iu^j}.$$ For material particles, the parameter $\lambda $ coincides with the interval: $\lambda=s$. The energy-momentum vector takes the form $$p^i=cmu^i.$$ The first component determines the energy $E=cp^1$. The gravitational force acting on a material particle is determined by the formula $$Q^k=cF^k=\frac{1}{2}c^2mg^{kl}\frac{\partial g_{ji} }{\partial x^{l} }u^iu^j.$$ This expression is non-covariant and therefore can be used in the limit of weak gravity, where its values converge asymptotically, or it can be considered in a dedicated isotropic reference frame. The Lagrangian corresponding to the principle of stationary integral of the photon energy Wikipedia gives the energy-momentum vector $$p^{i } =h\nu _0\frac{u^{i} }{u^{1} u_{1} },$$ where $\nu _0$ is some fixed value of the photon frequency. If we allow a thought experiment consisting in the direction of a photon with the help of a mirror to a point with which the reference frame is associated, then this will be its frequency at that point. This is possible if the observer and the photon's trajectory are not separated by the event horizon. The vector of gravitational force acting on the photon is $$ Q^{k } = h\nu _0\frac{g^{kl}}{2u^{1} u_{1} } \frac{\partial g_{ij} }{\partial x^{l} } u^{i} u^{j}.$$ Examples of this force for various metrics are in J. Phys.: Conf. Ser. 1251 012048.

  • $\begingroup$ User41178 suggested edit, that is deviated from the original intent of the post, but makes sense in its own right. "Expression $Q^k$ is non-covariant in the general case, but will be covariant for linearized metrics." $\endgroup$
    – Bel Vl
    Commented Aug 30, 2023 at 12:45

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