# Is work done by gravity equal to zero in general relativity?

I've heard it said that, according to Einstein's equations, gravity isn't actually a force but is instead an effect of the curvature of spacetime. In other words, a body in a gravitational field follows its geodesic without the need for a force dragging it along.

Given that work done on a body equals zero when either the displacement or net force on the body is zero, this non-force view of gravity implies work done by gravity is also (technically) zero.

I don't see any practical reason to treat gravitational work this way, but I'm curious definitionally.

• "Force" and "Work" are really only concepts that have meaning in Newtonian physics Apr 19, 2021 at 0:49

gravity isn't actually a force ... I'm curious definitionally.

Ok, so “isn’t actually a force” is more of a pop-sci summary than something definitionally correct in General Relativity.

In Newtonian mechanics the gravitational force is mediated by the gravitational field which gives the gravitational acceleration at each event.

In general relativity the mathematical quantity that gives the gravitational acceleration at each event is the Christoffel symbols. This is the same quantity that produces inertial forces (also known as fictitious forces) like the centrifugal force and the Coriolis force.

This makes some sense because gravity, like all inertial forces, cannot be detected by an accelerometer, is always proportional to the mass, and can be made to appear or disappear through a change of coordinates. However, gravity is somewhat different because when spacetime is curved there is no coordinate system that can make gravity disappear everywhere.

Because the force of gravity is like an inertial force, some people will describe it as not actually a force at all. Such people typically dismiss all inertial forces as not being forces, and in particular the centrifugal force is often described as not a force. However, in the coordinates where an inertial force exists it does work, produces acceleration, and (depending on the details) can have an associated potential energy. So in coordinates where the force of gravity exists, it does work just like any inertial force

• I strongly disagree with your assessment that the distinction is a pop-sci summary rather than definitionally true in GR. In the model, there is no invariant sense in which the trajectory of a test mass under the influence of only gravity is accelerating. These trajectories are explicitly defined in GR as non-accelerating curves, using the only invariant notion of acceleration available. They are the closest thing to a straight line achievable on a curved manifold. Apr 18, 2021 at 19:57
• @jawheele while you are certainly free to disagree, I stand by my statement. The actual math is more nuanced than the simple statement "gravity isn't actually a force" can convey. So I am comfortable calling that un-nuanced statement a pop-sci summary. You won't see such a statement in a standard GR textbook without several paragraphs or pages of support to convey the nuances. I tried to provide a taste of that here.
– Dale
Apr 18, 2021 at 20:04
• @jawheele. In Lagrangian mechanics the equation of motion reads $$\frac{d}{dt}\frac{\partial L}{\partial \dot q_i} = \frac{\partial L}{\partial q_i}.$$ The right hand side is the Lagrangian definition of a (generalized) force, and that can vanish in one choice of coordinates and not in another. Apr 19, 2021 at 9:31
• @md2perpe The object's 4-momentum tangent vector. That is, its velocity tangent 4-vector multiplied by mass. One can derive the geodesic equation from an action principle, wherein this agrees (perhaps up to a dualization) with the "generalized momentum". The difference between the relativistic notion of force and the "generalized force" of the action description is a covariant versus coordinate time derivative. So one can be said to be intrinsically associated to the manifold while the other cannot. The geodesic equation is the statement that the momentum's covariant time derivative is zero. Apr 19, 2021 at 19:15
• I wouldn’t say that the Christoffel symbols give rise to inertial forces or should be interpreted as their cause. Stick to Minkowski flat space time. Pick an inertial frame. Now just change coordinates ( non linearly) to polar coordinates. You are sticking in the same inertial frame just changing from Cartesian to polar coordinates not going in a rotating frame. There won’t be any fictitious forces yet the Christoffel symbols will not vanish.....Let me know if you do not agree... Apr 22, 2021 at 5:26

When discussing things in general relativity, one generally strives to describe physical phenomena in a way independent of arbitrary coordinate choices, so that one can be sure they're extracting real physical information about the model rather than coordinate artifacts. Forces on point masses can be described in a relativistic setting in keeping with this philosophy through the concept of four-force, the covariant derivative along the mass $$m$$'s trajectory of its four-momentum $$P^\mu = m U^\mu$$, where $$U^\mu$$ is its four-velocity:

$$f^\mu = U^\nu \nabla_\nu P^\mu.$$

Covariant descriptions of classical electromagnetism describe the Lorentz force on a charged particle as a four-force, for example:

$$f^\mu_\text{Lorentz} = q F^{\mu \nu}U_\nu,$$

where $$F^{\mu \nu}$$ is the electromagnetic tensor and $$q$$ the particle's charge. The magnitude of the net four-force is physical in the sense that it is the force the particle "feels" in its rest frame, i.e. its rest mass times its acceleration according to a momentarily co-moving inertial observer.

In general relativity, it is posited that test masses under the influence of gravity alone follow geodesics, for which the equation of motion is

$$U^\nu \nabla_\nu U^\mu = 0.$$

Up to a factor of the constant mass, this is the statement that the covariant derivative of the particle's four-momentum is zero along its trajectory, i.e. that the net four-force is zero. This is what is meant when people say "gravity is not actually a force"-- the action of gravity alone is that of zero four-force. Though test particles following this motion can be said to have varying coordinate accelerations in different coordinate systems, they are not accelerating in the only invariant sense of acceleration available. They are following the straightest possible paths through the curved spacetime manifold.

So, what about work? Can we achieve a similarly natural relativistically invariant analog? Well, in a sense, though the concept doesn't seem to be discussed or utilized very much. Let's relax the assumption that our particle has constant mass, and recall that the four-velocity is the tangent vector to the trajectory when parameterized by proper time $$\tau$$, so that $$U^\mu U_\mu = -1$$ (I like signature $$(- \, + \, + \, +)$$ and units with $$c = 1$$), and we have $$-2m \frac{dm}{d \tau} =\frac{d}{d \tau}( -m^2) = \frac{d}{d \tau} (P^\mu P_\mu) = U^\nu \nabla_\nu (P^\mu P_\mu) = 2 P^\mu U^\nu \nabla_\nu P_\mu = 2 P^\mu f_\mu$$

$$\frac{dm}{ d \tau} = -U^\mu f_\mu.$$
The left hand side is the rate of change of the particle's rest mass with proper time, while the right hand side is what one would obtain by taking the Newtonian expression for work done per unit time (or power), $$P = \vec F \cdot \vec v$$ , and minimally substituting in the invariant relativistic analogs (the sign is an artifact of the signature-- if the vectors are parallel, the quantity is positive). This says something like "the relativistic work done by the net four-force is the change in rest mass". This might be what you'd guess, considering that the rest mass is the only coordinate-invariant measure of energy available (kinetic energy is right out, of course), and work is supposed to describe a change in energy.
In this sense, gravity of course does no work, as it contributes no four force. The Lorentz four force listed above also does no such work, since the tensor $$F^{\mu \nu}$$ is antisymmetric and hence $$F^{\mu \nu} U_\mu U_\nu = 0$$. This is saying, e.g., that electrons don't gain rest mass as they accelerate under the electromagnetic field.
In a general coordinate system, though, there's not really a meaningful sense in which this is the case. There's difficulty defining what the terms should mean, so it depends on what you want to mean. When the coordinates are structured well enough that one can meaningfully consider a notion of coordinate kinetic energy, it will generally change under geodesic motion (so gravity does "work" in the sense that it impacts the coordinate kinetic energy), but it is not nicely related to any of the natural candidates for what one might mean by the coordinate force doing work as described through power: $$\frac{d}{dt}(P^i)v^i$$, $$\frac{d}{d \tau}(P^i)v^i$$, $$\frac{d}{dt}(P^i)v^j g_{ij}$$, $$\frac{d}{dt}(g_{ij} P^i)v^j$$, $$m\frac{d}{dt}(v^i)v^i$$, $$m\frac{d}{dt}(v^i)v^j g_{ij}$$, etc. (here $$\vec v$$ is the coordinate velocity and $$P^\mu$$ the four-momentum). In general, of course, there need not even be a nice break down into $$3$$ spatial coordinates and a time coordinate, so that none of the prior expressions even make sense. The best one can do with these expressions is say that, in coordinates where they make sense, these coordinate quantities that look like work are nonzero but have no consistent relation to actual energy.