Nikola Tesla didn't believe in relativity. More historical context here. He made the following argument against general relativity in a 1931 interview with Hugo Gernsback:

Tesla contradicts a part of the relativity theory emphatically, holding that mass is unalterable; otherwise, energy could be produced from nothing, since the kinetic energy acquired in the fall of a body would be greater than that necessary to lift it at a small velocity.

Filling in some of the steps in the logic, it seems that he is assuming that the passive gravitational mass equals the relativistic mass, and therefore the force of the earth's gravity during the fall would be greater than the force during the slow lifting. That means the work around a closed path would be nonzero.

Is there a simple explanation for this? It's easy to pick holes in the argument, since GR doesn't describe gravity as a force, and gravitational interactions depend on the stress-energy tensor, not the mass-energy. But that doesn't feel like a complete resolution of the issue to me.

The relation $W=\int F dx$ is exact in special relativity if $F$ is the three-force, since $dE/dx=(dE/dp)(dp/dt)(dt/dx)$, and $dE/dp=p/E$. However, this seems ambiguous in the context of GR, since, e.g., the tension in a hanging piece of rope is subject to a correction equal to the gravitational redshift factor evaluated between the ends.

A possible way to get at this would be to imagine a slightly different scenario that may be simpler to reason about. We shoot a test particle straight down into a planet's gravity well with relativistic energy $E_1$. At the bottom, we reduce its speed to exactly escape velocity, extracting energy $E_2$ from it, and then reflect it back up, so that it rises back up with zero total energy. Then Tesla's argument would seem to be that $E_2>E_1$, which seems unlikely since we have a conserved energy for the geodesic motion of a test particle in this field.

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    $\begingroup$ I think the modern argument against Tesla would be "the mass in indeed inalterable, you're right, but the momentum grows non-linearly with velocity." $\endgroup$
    – Asher
    Jul 7, 2017 at 23:28
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    $\begingroup$ @Asher: I don't think that works as an explanation. As noted in the question, $W=\int F dx$ is exact in special relativity, regardless of the fact that momentum is a nonlinear function of velocity. And although the modern convention is to define mass as a relativistic invariant, that doesn't mean that gravitational interactions don't depend on kinetic energy. They do. That's what the discussion of the stress-energy tensor in the question is about. $\endgroup$
    – user4552
    Jul 8, 2017 at 2:04
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    $\begingroup$ The effect of gravity felt by a test particle on, let's say, a Schwarzchild metrics, to be specific, does not depend on the kinetic energy of the test particle. So in that important case, Tesla's argument is falsified from the start. $\endgroup$
    – user154997
    Jul 8, 2017 at 4:31
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    $\begingroup$ @LucJ.Bourhis: The effect of gravity felt by a test particle on, let's say, a Schwarzchild metrics, to be specific, does not depend on the kinetic energy of the test particle. I don't know whether this statement is true or false, because it depends on what you mean by "effect." Tesla's argument is falsified from the start. His argument certainly lacks validity on the face of it, because he's trying to introduce relativity into Newtonian mechanics simply by replacing mass with mass-energy, and we know that doesn't work. If that worked, we wouldn't need GR. $\endgroup$
    – user4552
    Jul 8, 2017 at 15:20

1 Answer 1


For a stationary metric with an asymptotically timelike Killing field, the contraction of the Killing field and any geodesic four-velocity is a constant quantity $E$ that could reasonably be called the "mechanical energy", which is conserved in the absence of external forces. For simplicity, let's consider the outside of a spherically symmetric matter distribution, which is described by the Schwarzchild metric. A standard calculation gives that $E = (1 - 2 G M / r)\ dt/d\tau$ is conserved in this region.

When faced with an apparent violation of conservation of energy, the best way to see the problem is usually to try to come up with some kind of cycle that exploits the apparent violation to provide an infinite source of energy. If we shoot a particle in from infinity at zero velocity, the energy $E$ will equal the initial $\gamma := dt/d\tau$ factor $\gamma_0 = 1$, so that $\gamma \equiv 1/(1 - 2 G M / r) = r/(r - r_*)$, where $r_*$ is the Schwarzchild radius $2 G M$. As it falls inward, the $(1 - 2 G M / r)$ factor will decrease from $1$ and the $\gamma$ factor will increase accordingly. We can roughly think of $\gamma$ as being like the "rest plus kinetic energy" and $(1 - 2 G M / r)$ as being like the "potential energy" (except that their product rather than their sum is conserved). We see that, precisely as in the nonrelativistic situation, any energy we extract by bouncing the particle off a piston or something will decrease the maximum radius that it can bounce back to, so we will extract less and less energy with each bounce, with a finite limit on the total energy.

Carroll explains what's going on very well on pg. 208 of his textbook:

The energy of a particle with four-momentum $p^\mu$, as measured by an observer with four-velocity $U_\mu$, would be $-p_\mu U^\mu$. This not equal, or even proportional to [the conserved quantity E], even if the observer is taken to be static ($U_i = 0$) .... $-p_\mu U^\mu$ may be though of as the inertial/kinetic energy of the particle, while $[E =\, ] p_\mu K^\mu$ is the total conserved energy, including the potential energy due to the gravitational field. The notion of gravitational potential energy is not always well-defined, but the total energy is well-defined in the presence of a timelike Killing vector.

Tesla was only considering the "rest plus kinetic energy" $\gamma$, which is indeed not conserved, even when added to any possible potential $V(r)$. But he was failing to consider the "gravitational potential energy" $(1 - 2 G M / r)$, because in GR you have the extremely non-Newtonian property that the product rather than the sum of the "kinetic energy" and "potential energy" is conserved.

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    $\begingroup$ Nice answer, thanks. Hope you don't mind that I cut the first paragraph, which IMO was not relevant. $\endgroup$
    – user4552
    Aug 10, 2017 at 5:04

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