# (Why) is it necessary to invoke Newtonian Gravity to fix normalization constants in General Relativity?

I'm studying GR on Carroll's book. In Chapter 4 the constant appearing in Einstein's equation is fixed by requiring to obtain Poisson's equation in a Newtonian framework.

In chapter 5 the constant of the Schwarzschild metric is fixed invoking the weak field metric.

I understand that Einstein's equation is built, not derived or proved, and hence we somehow do not have a real ground to build on, but we have to work the other way round. Still, one would expect GR to be self contained, so what is going on here? Is it actually necessary to use Newtonian results?

• – Qmechanic Jul 8 '17 at 19:59

(First of all, point of terminology: these days "classical" is usually used to mean "not quantum," and "classical gravity" and "general relativity" are often used interchangeably. It would be more standard to refer to the nonrelativistic limit that you're referring to as "Newtonian gravity.")

The Einstein field equations (without a cosmological constant) say that the Einstein and stress-energy tensor are proportional - no more and no less. What we want the proportional constant to be is entirely up to us.

If we'd known about GR when we developed our system of physics units, then we would have chosen the proportionality constant to be $1$. Instead, by historical accident, we've chosen to use units defined in terms of arbitrary things like a particular platinum bar sitting in vault in Paris, and the time in takes for the Earth to rotate. (More accurately, within the past few decades we've redefined those units in terms of true physical quantities like the speed of light and cesium atoms, but with bizarre proportionality constants chosen to match up with the old bar of platinum and Earth's rotation.)

If we decide to keep mass, length, and time as separate base units (which prevents us from setting the proportionality constant to $1$), then the proportionality constant is simply a free parameter which can certainly be measured directly with no reference to Newton's law of universal gravitation. Call it $\kappa$.

It's a useful exercise to show that in the nonrelativistic limit (a term which of course needs to be carefully defined), the Einstein equation implies that a nonrelativistic particle of mass $M$ attracts all other nonrelavistic particles toward it with an acceleration

$$|{\bf a}| = \frac{\kappa}{8 \pi} \frac{M}{r^2},$$

where $r$ is the distance to the particle being attracted. This is just Newton's law of universal gravitation if we let $G = \kappa / (8 \pi)$. This is a useful exercise in GR (and of course was a crucial sanity check before we had direct experimental tests of GR), but the fact that the prefactor ends up being $\kappa / (8 \pi)$ is not terribly interesting or significant. (For example, it's not true in other numbers of dimensions.) So in fact it's Einstein's equation that determines the form of Newton's law of universal gravitation, not the other way around.

The reason we usually set the proportionality constant equal to $8 \pi G$ instead of just calling it $\kappa$ or something again boils down to historical accident - we already had a precisely-measured constant with the right units that was related to gravity, so we just reused it. Nothing deep there.

It is not. You can 100% fix the gravitational constant using experiments. A simple way to see this is to pick a classical free-fall experiment. Here's the version with only some unnamed constant $\kappa$ where

$$G_{\mu\nu} = \kappa T_{\mu\nu}$$

If you consider the earth as a Kerr metric source, then, neglecting angular momentum (which is fairly small), we'll end up with the Schwarzschild metric

$$ds^2 = -(1 - \frac{r_S}{r}) c^2 dt^2 + (1 - \frac{r_s}{r})^{-1} dr^2 + r^2 d\Omega^2$$

$r_s$ is the Schwarzschild radius, some parameter of the metric. To compute it, we perform the analysis of the Komar mass. If we go 100% without any definite constants from previous theories, let's just write the Komar mass as proportional to some constant.

$$M = -\alpha \int_S dA n_\mu \sigma_\nu \nabla^\mu \xi^\nu$$

with $\xi$ the normalized Killing timelike vector (we'll just pick $c\partial_t$), $\sigma$ the vector normal to the Cauchy surface, and $n$ the vector normal to the surface of integration. The constant $\alpha$ is related to the constant $\kappa$ since the Komar integral is alternatively written in term of the stress-energy tensor

$$M = -\alpha \int_\Sigma R_{\mu\nu} n^\mu \xi^\nu = -\alpha \kappa \int_\Sigma (T_{\mu\nu} + \frac 12 T g_{\mu\nu}) n^\mu \xi^\nu dV$$

We will really want $M$ to be of the same dimension as the mass. Since $T$ has the dimension of an energy density, we want $\alpha = \beta / c^2 \kappa$ for some constant $\beta$.

Then, picking the surface of constant $t$ and $r$ as $S$, we get

$$M = -\alpha \int_S r^2 \sin \theta d\theta d\varphi \nabla^r\xi^t$$

$\nabla^r\xi^t$ has for only non-zero term $c g^{rr} \Gamma^t_{rt}$, which is equal to $c g^{rr} g^{tt} g_{tt,r}/2 = -c r_S/2r^2$, so

$$M = -\alpha \int_S \sin \theta d\theta d\varphi c r_s = 8\pi c \alpha r_S$$

We may then write $r_S = M / 8\pi c \alpha = M \kappa / 8\pi \beta c$.

The geodesic equation for a purely radial motion will then be $$c^{-2} \dot r^2 + (1 - \frac{r_S}{r}) = E^2$$

Expanding the $r^{-1}$ term around $R_\oplus$, the radius of the earth, we get

$$\frac{1}{r} = \frac{1}{R_\oplus} - \frac{r - R_\oplus}{R_\oplus^2} + \mathcal O(R_\oplus^{-3})$$

This will give you the equation, cutting off the smallest term,

$$c^{-2} \dot r^2 + \frac{r_S}{R_\oplus^2} r = E^2 - 1 + \frac{2r_S}{R_\oplus}$$

If you take the proper time derivative,

$$c^{-3} \ddot r \dot r + 2 \frac{r_S}{cR_\oplus^2} \dot r= 0$$

or in other words,

$$\ddot r = - 2 c^2 \frac{r_S}{ R_\oplus^2}$$

The classic equation of free fall, which you can check experimentally. If you wish for less classical results, of course, you'll need to solve the geodesic equation properly, which I don't think can be done analytically (numerical simulations will do there).

You'll notice that this equation has the correct dimension and sign. We can then input back the expression of $r_S$

$$\ddot r = - 2 c^3 \frac{M \kappa}{8\pi \beta R_\oplus^2}$$

I think some $c$'s might have gotten lost in the shuffle, but you can see that globally, we get back roughly what we were looking for : the free fall of an object depends on some constants which, if you were to sort out a bit all the various constants, would turn out to be $G$.

• Re "It is not": If it's not, what's that $G$ doing in your first equation? – David Hammen Jul 8 '17 at 17:50
• This is, I think, the answer that best addresses OP's question. We invoke the classical limit because people have already gone to the trouble of fixing Newton's $G$, so we take advantage of that. But Newton's $G$ can't be derived from some limit, you have to do experiments. – Javier Jul 8 '17 at 17:52
• @DavidHammen: If $k$ is the unknown constant in the RHS of Einstein's equations, simply define $G = c^4 k / 8\pi$ and proceed to find the Schwarzschild solution. It is just a convenient redefinition of constants. – Javier Jul 8 '17 at 17:53
• Well, this was done kind of quickly, but if done properly, you can find that if you take this equation in terms of $\kappa$, then, given the fact that a $4$-vector will be of the form $(ct, x, y, z)$, you get in the end that $$\ddot r = - \kappa c^4 M/8\pi R_\oplus^2$$, the $c$ come from the derivatives with respect to time (if you want a reason for this, assume for instance we are measuring distances with lasers) and the $M$ and $8\pi$ come from the Komar mass of the metric – Slereah Jul 8 '17 at 18:05
• This is not reducing GR to a previous theory. You do not need to know that Newtonian gravity exists to make this prediction, you just need to perform the experiment and note the value of the acceleration (although you'll need to know the mass and radius of earth), but of course the experiments will be similar in the end, since they are both theories dealing with the same topic. – Slereah Jul 9 '17 at 7:44

When we develop a theory, it is necessary that it reduces to the theories we observe in certain regimes. In this case, we need to make sure the gravitational theory derived from the Einstein equations match whatever theory we have in the limit of slow bodies and weak gravitational fields (ie Newtonian gravity). This is where the factor of $8\pi G/c^4$ comes from in the Einstein equations.

To answer the meat of your question, we don't expect GR to be self-contained. In any theory coupling spacetime to matter, we need to indicate how strong this coupling is. The different coupling constants (ie different values of G) will lead to different quantitative results of measurements in the Newtonian limit. To fix these, we just compare the results we get at low energies to the low-energy limit of our theory. This is an incredibly common theme in physics. We use "hints" from low energy physics to fix the requirements in our high energy theory.

It is useful to note that GR is relatively unique. It is derived from the postulate that $G_{\mu\nu}\propto T_{\mu\nu}$ for some tensor $G$. Then, we require that $G$ contains at most second order derivatives of $g$, and that it is covariantly conserved (since the energy momentum tensor). This effectively fixes the Einstein equations to a proportionality constant, which we again derive from our low energy experiments.

I hope this helped!

• There's also the cosmological constant, which is a free parameter in GR. – tparker Jul 9 '17 at 18:29
• The cosmoligical constant doesn't really play a role in the typical gravitational scenarios that Newtonian gravity describes. Moreover, the cosmological constant is often best thought of as a part of the stress-energy tensor. That is my personal preference, so that it's not a free parameter, but a feature of the energy distribution of the universe. – Bob Knighton Jul 9 '17 at 20:01

Is it actually necessary to use classical results?

Of course. Any new theory in physics must replicate current theories in those areas where those current theories already do a very good job of predicting behaviors. This is true in quantum mechanics as well.

Newtonian gravity predicts behaviors quite nicely in the limit of small densities, large distances, and small velocities, with a very small discrepancy from observations with regard to the orbit of Mercury. General relativity has to explain the near-Newtonian behavior of the outer planets and explain that small discrepancy in Mercury's orbit. It does both. The explanation of Mercury's anomalous precession is a natural consequence of general relativity. But matching the near-Newtonian behavior of the outer planets was done by design.

• -1: GR must confirm with Newtonian Gravity. It doesn't mean we must use Newtonian results to determine something in GR. For example, as in Sleerah's answer, we can very well determine everything unknown in GR by experiments without actually referring to Newtonian Gravity. Of course, all this will be consistent with Newtonian results if done correctly but it doesn't mean we really invoked Newtonian results to derive ours. – Dvij Mankad Jul 8 '17 at 20:23

## protected by Qmechanic♦Jul 9 '17 at 4:29

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