What do I mean?

There are two kind of equalities, or two ways to interpret an equality.

Take for example the ideal gas law $$PV = Nk_BT$$

We all know what this equation means: when you calculate both sides of the equation, you find the same physical quantity. This equation in other words is saying that temperature of an ideal gas is proportional to the pressure and the volume of the enclosing container, and inversely proportional to the number of molecules of the gas. There are so many of these equations in physics, but there is another, more subtle kind.

Take this other equation about from the statistics of ideal gas: $$\langle E\rangle = \frac{1}{2}mv_{rms}^2 = \frac{3}{2}k_bT$$ Now, this equation can also be taken as an expression of proportionality. However, this can also be taken as a definition for temperature. We can read this equation to mean that temperature (a macroscopic phenomenon) is the average kinetic energy of a gas particle (up to a multiplication).

Incidentally, one take $$F=ma$$ in a similar way. For example, when we are in a rotating or generally accelerated frame, the equation actually defines the fictional force in terms of the acceleration.

So is $$G = 8\pi T$$ an expression of proportionality, or a definitional identity? and why?

Follow up

The ideal gas law does not tell us why $lhs = rhs$, it expresses a law, it does not explain nature. On the other hand, the second equation informs us on the nature of temperature, it explains nature, it tells us: this is what temperature is. I find these kinds of equations very satisfying, and they are much rarer in physics.

If you disagree on any of this, please leave a comment, I am interested.

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    $\begingroup$ Is this question ontological, or is it mechanical? $\endgroup$ Oct 2, 2017 at 21:40
  • $\begingroup$ @can-ned_food It's definitely ontological. In the kinetic example, that equation to me answers the question: "what is temperature, really?". One hypothetical answer to my question about GR is, stress energy is, really an aspect/consequence of the geometry of spacetime. Not sure by what you mean by "mechanical" in this context. $\endgroup$
    – Andrea
    Oct 2, 2017 at 23:20

6 Answers 6


There is a pure geometric definition of the Einstein tensor $G_{\mu\nu}$ in terms of derivatives of the metric. Independent of any physics.

Likewise, given a field theory, you could in principle calculate the stress energy tensor.

GR is a physical theory which couples the geometry, through the Einstein tensor, to the matter content, via the stress energy tensor. There are other self-consistent theories which couple geometry to matter in different ways. In this sense, the equation $G=8\pi T$ is model dependent. Just like the ideal gas law $PV=NRT$, which only applies for ideal gases, not interacting gases.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Oct 5, 2017 at 10:50

In order to elucidate a potential answer to your question, it is helpful to consider the Einstein field equations in the context of field theory. The action,

$$S=\frac{1}{16\pi G}\int d^4 x \, \sqrt{|g|} \, R + \int d^4x \, \sqrt{|g|} \, \mathcal L_M$$

gives rise to the Einstein field equations through the principle of stationary action, with a right hand side corresponding to the usual symmetric definition of the stress-energy tensor of a field theory described by $\mathcal L_M$. In this context, we can think of,

$$R_{\mu\nu}-\frac12 g_{\mu\nu}R = 8\pi G \, T_{\mu\nu}$$

as being the equations of motion for the metric, and any fields in $\mathcal L_M$ coupled to gravity. So it's not so much that stress-energy is curvature but rather the fields (or other quantities) which contribute to the stress-energy are coupled to gravity.


Short answer:

  • If you like, you can say that the Einstein field equations define the active gravitational mass (or, more precisely, the active gravitational mass-energy-momentum-stress).

  • But active gravitational mass equals inertial and passive gravitational mass, so this makes the EFE more than a definition.

  • The EFE also contain all kinds of information that has nothing to do with sources. For example, they say that certain vacuum fields are not possible, and they predict the existence of gravitational waves.

  • There are some ambiguities that come into play in the case of dark energy.

Long answer:

The Einstein tensor $G$ is measurable. For example, when I drop a pencil and see how long it takes to hit the ground, I am finding out something about the Riemann tensor in a certain region of space. By doing enough measurements, I can measure the entire Riemann tensor and then determine $G$. This constitutes an operational definition of $G$. We could define $G$ in some other way, but we don't need to, it seems undesirable, and nobody does it.

The Einstein field equations relate the Einstein tensor to the stress-energy tensor. In the nonrelativistic limit, this is simply equivalent to the Newtonian equation $g=Gm_a/r^2$, which relates the active gravitational mass $m_a$ to the gravitational field. This can be taken as the definition of active gravitational mass in Newtonian physics, and there is no other way to define it. However, this does not make Newton's law of gravity a tautology or a definition, because in Newtonian gravity the active gravitational mass is strictly equal to both the passive gravitational mass and the inertial mass. Since we have other types of experiments that can measure inertial mass, there is no circularity involved. Furthermore, Newton's law of gravity specifies the distance dependence of the field, which is not a matter of definition. This $1/r^2$ form of the force law results, for example, in the prediction of elliptical orbits.

Similarly, in GR, the active gravitational mass (or, more precisely, active gravitational mass-energy-momentum-stress) is defined as $G/8\pi$, and there is no other way to define it. However, this does not make the Einstein field equations tautological or a matter of definition, for the same reasons as in the Newtonian theory.

Note that in GR, the equality of inertial, active, and passive gravitational masses is not just an optional feature as in Newtonian gravity. If any of these equalities fails, then GR is falsified and cannot be fixed by tinkering. (E.g., it's a theorem in GR that test particles follow geodesics.)

One place where I think it gets a little trickier to make proper operational definitions is in the case of dark energy. We have no way to measure the inertia or passive gravitational mass of dark energy. This is basically because our model of dark energy is a cosmological constant, and the Einstein field equations do not allow us to simply make solutions in which the cosmological constant varies from point to point. Such solutions always violate the field equations. Therefore the cosmological constant is ordinarily modeled as a constant -- it has no dynamics. (You can have a dynamical dark energy, but doing so requires something more elaborate than just letting $\Lambda$ vary.)

This lack of dynamics in $\Lambda$ prevents us from measuring dark energy's inertial or passive gravitational mass. For this reason, it's not uncommon to see different people making different choices about whether or not to include the dark energy piece as part of the stress-energy tensor.


I am not sure on the distinction you draw between 'proportionality' and 'definitional identity' specially in the examples you use.

For example, one could argue that $PV = nk_BT$ is just a definition of pressure (why not?), in the same way that $F = ma$ is used to define force. The interesting thing is that these concepts are consistent with other elements in the theory, for example, you can also see that if you place the gas in chamber of area $A$ then the force a side of the box experiences is $F = PA$. According to your argument this last expression would automatically change the label of $PV = nk_BT$ from definition to proportionality.

With this in mind, if you define both $G$ and $T$ someway and show that $G = 8\pi T$ then in light of your argument this would be a mere proportionality. If, on the other hand, you never defined one of them, you could take this expression as a definition. Turns out this terms are separately defined and Einstein's equations are a result of the theory. But again, you could argue that this is a definition and then show that the other instances where $G$ appears in the theory are proportionalities.

Finally I would like to comment on your last statement. I certainly believe the word law is not as black-and-white as you assume. To give you an example,

$$ {\bf F} = -\frac{GMm}{r^2}\hat{\bf r} $$

is known as the law of gravity, whereas

$$ G_{\mu\nu} = 8\pi T_{\mu\nu} $$

is labeled as the theory of general relativity, which is essentially gravity, but somehow we dediced not use the word law anymore. Why? I think it is just a historical reason which reflects the fact that at some point we realized laws may be wrong

  • 1
    $\begingroup$ I agree with you about the word "law". But most often "law" is used to refer to an equation that is empirical (drawn from experiments) and is not necessarily explained by other means. $\endgroup$
    – Andrea
    Oct 2, 2017 at 23:37
  • $\begingroup$ About your first bit, I disagree. When we see physics as a process of discovery of nature, we look for the why and the what of things, not just the how. The ideal gas law tell us how are those quantities related, and when it was discovered, we already had things that told us the what of these things. Pressure, volume and temperature were already defined. In contrast, the kinetic equation, relates temperature to something more fundamental (atoms), thus it can be used to replace the old definition of temperature in terms of thermometers and partial derivatives of potentials. $\endgroup$
    – Andrea
    Oct 2, 2017 at 23:46
  • $\begingroup$ @AndreaDiBiagio Correct, but my point still stands: you can use this relation to define pressure, the same way you use the equipartition law to define temperature. What gives the concept of temperature a privilege over pressure in this context? $\endgroup$
    – caverac
    Oct 3, 2017 at 1:37
  • $\begingroup$ In total absence of actual physical knowledge, yes, yes you could. I will try to reformulate. The ideal gas law was derived from experiments and it linked observations of directly measurable quantities. The reason it does not define anything in terms of anything is that, at that level, we already know what each term means. In contrast, kinetic theory provides explanation of macroscopic thermodynamic quantities in terms of mechanical quantities of atoms. In physics, atoms are taken as more fundamental, and so an equation linking a microscopic entity with a macroscopic one defines the latter. $\endgroup$
    – Andrea
    Oct 3, 2017 at 9:38
  • $\begingroup$ A few other examples from that theory. Pressure, the average force per unit aerea, is caused by the collisions of the atoms on the surface, the internal energy of the gas is the energy of the molecules and $S = k_B\log\Omega$. Remember that thermodynamics was developed before people accepted the existence of atoms, so these were not obvious relations when they were first written down. They explained what we see in terms of another layer of reality. $\endgroup$
    – Andrea
    Oct 3, 2017 at 9:42

No. For the simple reason that there is space-time curvature without any stress-energy.

The Schwarzschild solution is a vacuum solution. Everywhere that the solution holds, the stress-energy is zero yet spacetime is curved. The famous experiments showing star light "bending" when passing near the sun, or distant galaxies gravitational lensing, are approximated with vacuum equations outside of the sources.

Einstein's field equations only express the Ricci curvature. The full curvature is given by the Riemannian curvature.

  • 1
    $\begingroup$ I don't think this quite works. The OP didn't state which side of the Einstein field equations they wanted to use as a definition and which side as the thing being defined. There is nothing in your argument that prevents us from defining the stress-energy in terms of the curvature, which I would say is more reasonable than defining the curvature in terms of the stress-energy. $\endgroup$
    – user4552
    Oct 2, 2017 at 20:31
  • 1
    $\begingroup$ Also, although the OP used the word "curvature" in the title, when they actually stated the question in more mathematical detail they specified the Einstein tensor. It doesn't seem likely that you could define the Einstein tensor in terms of the stress-energy and then go on and define the whole Riemann tensor in some way that would fit logically -- but your argument doesn't address that. $\endgroup$
    – user4552
    Oct 2, 2017 at 20:32

I'd like to add the following notes to mainly Mr. Weather's good answer.

The Einstein tensor has definite geometric meaning independent of physics; there are two physics independent meanings we should take heed of:

  1. Its divergence vanishes, by dint of the first Bianci identity. This fact is an expression of the geometrical relationship $\partial^2 =\emptyset$ - the boundary of the boundary of a set is always the empty set. This is one intuitive reason why we can "wire up" the Einstein tensor to the stress energy $G\propto T$: the divergence of $G$ vanishes through geometry, and this forces the divergence of $T$ to be nought. In "wiring up" geometry to stress energy in this way, we force the theory to encode local conservation of energy-momentum through a fundamental, simple geometrical fact;

  2. To answer your title question: "definitely not!" The Einstein equation determines the Ricci tensor, and this tensor encodes only hypervolume distortion as I discuss further in my answer here. The full Riemann curvature tensor has further degrees of freedom not set by the Einstein equation; these further degrees of freedom encode shape distortion through the Weyl tensor. Boundary conditions are needed further to stress energy to fully define the Riemann - I discuss how the Riemann decomposes into Ricci and Weyl tensors, then to Ricci + Schouten tensor + Boundary conditions in my linked answer above. These degrees of freedom unspecified by the Einstein equation itself are what allow, amongst other things, gravitational waves to propagate in the vacuum, and also allow all kinds of interesting vacuum solutions to the Einstein equations, such as Calabi-Yau manifolds or even something as "mundane" as the Schwarzschild and other black hole solutions away from the singularity.


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