# What are some ways to justify the Einstein field equations?

Since they are a postulate of general relativity, it is not really possible to "derive" the Einstein field equations $$R_{ab} + \left(\Lambda - \frac{1}{2}R\right)g_{ab} = -8\pi T_{ab}$$

in any very meaningful way. It is, however, possible to come up with derivation-like constructions that yield some insight. What are some of these?

First, we notice that the paths traced by particles through spacetime under the influence of a gravitational field seem to depend only on their positions and velocities, i.e. they are independent of any identifiable charge or composition of the particles. It is almost as if the particles were moving along tracks carved into some curved surface.

Although even Galileo noticed this, we happen to be passingly familiar with differential geometry to the point where we can actually make something of it. We realize that, if this observation really held, it would be possible to describe the curvature mathematically by assigning to spacetime a nonzero Riemann curvature tensor $R^a_{bcd}$. This object describes the rotation of a vector when dragged along a closed loop through the spacetime.

What should we couple $R^a_{bcd}$ to? Well, gravity is obviously sourced by mass somehow, but we know because we invented special relativity that mass and energy are intimately connected. We also know we can uniquely describe the energies and pressures of classical matter fields in a nicely coordinate-independent way by their stress-energy tensor $T_{ab}$. So, making the leap that perhaps pressure can source gravity as well, we write down

$$R^a_{bcd} = -\kappa T_{ab}.$$

Unfortunately this equation makes no sense because the RHS and LHS have a different number of indices. No matter - we can just contract some of the indices on the left hand side. This gives us the Ricci tensor $$R_{ab} \equiv R^j_{ajb}$$ so we write $$R_{ab} = -\kappa T_{ab}.$$

This equation predicts the perihelion precession of Mercury, so we are very happy. Unfortunately it has a rather serious flaw: $T_{ab}$ had better have identically zero divergence if energy is at least locally conserved - i.e. if energy does not simply appear out of nowhere in small regions of spacetime. But $R_{ab}$ unfortunately can have nonzero divergence.

Again, no problem - we can simply subtract the divergence: $$R_{ab} - \frac{1}{2}g_{ab}R = -\kappa T_{ab}$$ where $$R\equiv R^a_a$$ is the so-called Ricci scalar. Actually this is a bit restrictive: the divergence is determined only up to an arbitrary constant $\Lambda$. Thus: $$R_{ab} + \left(\Lambda - \frac{1}{2}R\right)g_{ab} = -\kappa T_{ab}.$$

What about the free parameter $\kappa$? This we can't fix with the theory, so it will have to be measured, and then set to enforce agreement with experiment. In particular, in the limit where the curvature is very weak, we should be able to reproduce Newtonian gravity. It turns out to do this we have to set $\kappa = 8\pi$, so that we end up with

$$R_{ab} + \left(\Lambda - \frac{1}{2}R\right)g_{ab} = -8\pi T_{ab}.$$

• Wait; you are saying the term $\frac{1}{2}g_{ab}R$ is equal to the divergence of $R_{ab}$?? Jun 25, 2022 at 0:52
• Yes, the Einstein tensor is the Ricci tensor with its divergence subtracted
– AGML
Jun 29, 2022 at 21:19

## Newtonian limit

As a starting point, let us take Newtonian gravitation. There, we know two things: $$- \Phi_{,i} = \frac{d^2 x^i}{d t}$$ $$\Delta \Phi = 4 \pi G \rho$$ We must first investigate how does $\Phi$ correspond to the metric $g_{\mu \nu}$. For that, we assume that Newtonian gravity is valid for slow particles $v/c \ll 1$ on an almost Minkowski background $$g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$$ where $h_{\mu \nu} \ll 1$ including its first and second derivatives. We start from the geodesic equation $$\frac{d^2 x^{\mu}}{d\tau^2} = \Gamma^\mu_{\; \nu \kappa} \frac{d x^{\nu}}{d\tau} \frac{d x^{\kappa}}{d\tau}$$ and considering only terms linear to $h_{\mu \nu}, v/c$ (and no $h_{\mu \nu} v/c$ cross-terms), we come to the conclusion that to lowest order of approximation the geodesic equation yields $$\frac{d^2 x^i}{d t} \approx \frac{d^2 x^i}{d \tau} \approx \frac{1}{2} h_{00,i}$$ That is, we conclude that $h_{00} \sim -2 \Phi$.

## Basic form of field equations

Now we proceed to the derivation of the full covariant field equations. As a starting point, we use the Laplace equation and the observation that it says "gravity (geometry)"="matter(energy)". Furthermore, the "left hand side" should somehow involve the second derivative of the metric because $\Delta g_{00} \sim \Delta\Phi$, and the natural candidate for the "right hand side" is some function of the stress-energy tensor because $\rho \sim T_{00}$. (Only $T^{\mu \nu}$ and derived tensors include energy density.)

But covariant derivatives of the metric vanish, and coordinate derivatives are not covariant. The foundational covariant tensor based on the geometry from which any other geometric tensor is built (with help of the metric contractions), is the Riemann tensor $R^\mu_{\;\nu \kappa \sigma}$. The elegance of metric gravity shows itself in the fact that this "most basic geometric tensor" also involves second-order derivatives of the metric. Hence, we conclude that "the left hand side" must somehow be a function of the Riemann tensor.

Now let us investigate the tensor rank of the desired equation. If we don't want to do anything ugly with the stress-energy tensor which would produce more indices such as $T_{\mu \nu; \kappa \sigma}$, we are left with three possibilities: a 0-index (scalar), 1-index (vector/form) or 2-index (bilinear, 2-vector, 2-form) equation. Since we are interested to construct a theory based only on metric quantities, we have no tensor with an odd number of indices and we thus cannot form any 1-index equation.

## Scalar equation

The 0-index equation leads us to the only possibility $$R + ... = \lambda T +... \; \; \; \; (\rm S)$$ where the $...$ signify higher order quantities either in the stress-energy or curvature.

The main problem of this theory is that the degrees of freedom of the geometry are too unconstrained. The metric has ten components, and the equation involves its first and second derivatives, the whole being governed by a single equation. In this form, the equations cannot be solved from any boundary-value data nor from an initial value problem on a space-like hypersurface. The only solution is to constrain the Riemann tensor by some other condition.

One of the possible conditions is to set the traceless part of the Ricci tensor to zero (i.e. specify the components of the Ricci tensor independent of $R$): $$S_{\mu \nu} \equiv R_{\mu \nu} - \frac{1}{4}R g_{\mu \nu} =0$$ But when you substitute for $R$ in the above equation from equation $(\rm S)$, you get a unified 2-index equation (with a total of 10 independent components) $$R_{\mu \nu} = \frac{\lambda}{4}T g_{\mu \nu}$$

Another possibility of constraining the Riemann tensor (20 independent components) is to fix the part which is independent of the Ricci tensor (10 independent components), the so-called Weyl tensor (10 independent components): $$C^\mu_{\;\nu \kappa \lambda} = 0$$ This theory has some historical importance because it is equivalent to Nordström's (second) theory of gravity, the only scalar gravity fulfilling the weak equivalence principle. Manifolds in which the Weyl tensor vanishes are conformally flat which, amongst other thing, means that light-rays do not bend. (But we know they do, so Nordström gravity is incorrect.)

There are of course other ways to constrain the Riemann tensor, but if we do not want to combine about 9 random curvature invariants or introduce new natural constants (to set some of the tensors equal to a constant times the metric, or so), this is about what one can do with a scalar equation.

Nevertheless, the main problem with all the mentioned possibilities is the fact that they do not conserve the stress-energy tensor and thus violate the Einstein equivalence principle. If we were to put $T^{\mu \nu}_{\; \; \; ; \nu}=0$ along with equation $(\rm S)$ with $\lambda=-8\pi G$, we would get Einstein equations (see below).

## Why do we need $T^{\mu \nu}_{\; \;\; ; \nu}=0$

The point of Einstein gravity is that locally, a free-falling observer has no idea about whether they are falling or not. This shows itself e.g. in the fact that around any point, a set of normal coordinates can be found, in which the metric looks like $$g_{\mu \nu} = \eta_{\mu \nu} - \frac{1}{2} R_{\alpha \mu \beta \nu} x^\alpha x^\beta$$ This is how free-falling observers at $x^\mu=0$ actually sees the metric and geometry.

One other property of normal coordinates and the point of view of a free-falling observer is that Christoffel symbols vanish and thus the covariant derivative is just a gradient $A_{\mu ; \nu} = A_{\mu,\nu}$ (in most courses/constructions of relativity, this is the defining property of a covariant derivative). This means that a free-falling observer understands $T^{\mu \nu}_{\; \; \;;\nu}$ as $T^{\mu \nu}_{\; \; \;,\nu}$.

Now comes Einstein's postulate:

The outcome of any local non-gravitational experiment in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime.

But if $T^{\mu \nu}_{\; \; \;,\nu} = f(R,R_{\mu \nu},...)$, then it is possible to detect the space-time geometry by local experiment, e.g. by observing the non-conservation of energy, momentum and angular momentum.

There is also a deeper, conceptual issue with Einstein strong equivalence: the mere existence of the term "inertial frame". If gravity is unshieldable and even a freely-falling laboratory is affected by gravity, there is no physical realisation of an "isolated" or "free" particle. Amongst other things, this would mean that even special relativity is built on physically non-existent terms and must thus be necessarily invalid!

Simply said, Einstein's equivalence postulate is not quite "an extra empirical requirement", Einstein equivalence is a consistency check of the whole body of relativity. $T^{\mu \nu}_{\; \; \;;\nu}$ is thus a necessity.

## Finale

We are thus looking for a set of equations which is a 2-index equation and involves second derivatives of the metric. There is a single arbitrary or "theoretical prejudice" criterion in the presented construction and it is coming right now: we will also require that the equations involve only up to second derivatives of the metric. We can understand most of the contemporary modified gravities as theories taking advantage of this caveat by using higher-derivative field equations.

Nevertheless, we now arrive at this (second-derivative) form of the field equations $$R_{\mu \nu} + \alpha R g_{\mu \nu} + \beta g_{\mu \nu} = \gamma T_{\mu \nu} + \delta T g_{\mu \nu}$$ By taking a trace of these equation we come to $$(1+ 4\alpha) R + 4\beta = (\gamma+4 \delta) T$$ I.e. we can solve for $T$ and get rid of the $T$ term by substitution into the original equation. Hence, we can restrict ourselves to the form
$$R_{\mu \nu} + \alpha R g_{\mu \nu} + \beta g_{\mu \nu} = \gamma T_{\mu \nu}$$ The term $\beta g_{\mu \nu}$ is automatically divergenceless, because $g_{\mu \nu;\kappa}=0$. A slightly technical derivation based on the Bianchi identities yields $$R_{\mu \nu ;}^{\;\;\; \, \nu} = \frac{1}{2} R_{,\mu}$$ I.e. we can render the left-hand side of the equations divergenceless by setting $\alpha=-1/2$. This enforces the divergence-lessness of $T^{\mu \nu}$.

The constant $\beta$, commonly written as $\Lambda$ is free and undetermined by the Newtonian limit, apart from the fact that $\Lambda \ll \gamma \rho$. But since it should be smaller than even the relatively small densities we encounter in our Solar system, it could as well be zero, right? It is a really ugly term which would be very natural to set to zero, but Nature thought different.

The constant $\gamma$ can be obtained by solving the linearized equations around Minkowski $$g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$$ The procedure is not so simple; we have derived that $h_{00} \sim -2 \Phi$ but it is important to also derive in linearized theory that e.g. in spherical symmetry we have $h_{ii} \sim -2 \Phi$. This leads to $R_{00} \sim \Delta \Phi$ and $R \sim 2 \Delta \Phi$. As a result the $00$ component of Einstein equations reads $$2 \Delta \phi \sim \gamma T_{00} \sim \gamma \rho$$ From this we are able to see that $\gamma$ has to be $8 \pi G$.

Finally, we obtain the Einstein equations with a single free parameter $\Lambda$ $$R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} + \Lambda g_{\mu \nu} = 8 \pi G T_{\mu \nu}$$

Sidenotes: I have omitted higher curvature terms $\sim R^2,R^3,...$ on the left-hand side. It can be easily shown that these terms cannot generate a divergenceless tensor (it has mainly to do with the fact that $R_{\mu \nu}$ is the "only two-index tensor in shop" apart from the divergenceless $g_{\mu \nu}$).

I have not given an account of the derivation of Einstein equations from an action principle. The construction via an action (with the usual coupling to matter!) gives divergenceless left-hand sides automatically. Furthermore, David Lovelock showed that in dimension 4, the Einstein equations are the unique second-order field equations generated by an action. I.e., you could derive Einstein equations by requiring that they are generated by an action and that they are of second order in the derivatives of the metric.

I find the above presented approach more physical and direct. Furthermore, for the justifiability of the requirement of an action principle one would have to prove the following implication: if $T^{\mu \nu}_{\; \; \;; \nu}=0$, then the field equations can be generated by an action principle. (I am not sure if that implication is or is not proven.) Otherwise, there could be a lurking non-action-generated set of field equations satisfying all our physical criteria.

In Feynman's Lectures on Physics (volume 2 chapter 42) he states that the field equation is equivalent to the following statement:

For all local inertial observers, the scalar curvature of space at a point is proportional to the energy density at that point.

Simple, right? By requiring the correct Newtonian limit the constant of proportionality can be found to be $16\pi G/c^4$.

Feynman doesn't prove that you can derive the Field equation from this, but it can be done -- there is a sketch of a proof in my question and the answer to it here.

Note that this motivates the Field equation with zero cosmological constant. To get the cosmological constant term you have to subtract an energy density of $\Lambda c^2/8\pi G$ from each point in space.

• I cannot find this quote in Feynman's Lectures on Physics (vol. 2 ch. 42 is here: feynmanlectures.caltech.edu/II_42.html). Is this a quote (as represented), or a summary? And if the latter, what is the material it summarizes? Oct 21, 2023 at 21:25

Here's a few methods to find the Einstein Field Equations :

1) The classical route

The classical method is to note the similarity between the geodesic equation

$$\ddot x^\sigma + {\Gamma^\sigma }_{\mu\nu} \dot x^\mu \dot x^\nu = 0$$

And the classical equation of motion for particles in a gravitational field :

$$\ddot {\vec x} = -\nabla \Phi (x)$$

(in the classical limit, only the terms $\Gamma^a_{00}$ remains)

meaning that we can associate the metric with the gravitational potential $g \approx \Phi$.

The Einstein equation will then be something akin to the Poisson equation $\Delta \Phi = \rho$. The simplest covariant tensor made of second derivatives of the metric tensor is the Riemann tensor. And, if we suppose the source of the gravitational field is the stress energy tensor (it can't be something simpler like its trace due to various theoretical reasons), then we have something generally of the form

$$F_{\mu\nu}(g, \partial g, \partial^2 g) \propto T_{\mu\nu}$$

We would also like this to have (local) conservation of energy built into it, which would be something of the form

$$\nabla^\mu F_{\mu\nu}(g, \partial g, \partial^2 g) \propto \nabla^\mu T_{\mu\nu} = 0$$

Fortunately, thanks to the Bianchi identity, there is a very simple form for such a thing, which is the Einstein tensor

$$\nabla^\mu (R_{\mu\nu} + \frac{1}{2} Rg_{\mu\nu}) = 0$$

That is the "usual" method of showing the EFE, which is a bit historical and a bit hand wavy.

2) The Fierz-Pauli method

The first tensor theory of gravity was the Fierz-Pauli theory, which is the simplest free theory of symmetric tensor fields, and corresponds today to linearized gravity

$$\frac 1 2 \square h_{\mu\nu} + \partial_\mu \partial_\nu h - \partial_{\{\nu}\partial^\sigma h_{\mu\}\sigma} + \eta_{\mu\nu} (\partial^\alpha\partial^\beta h_{\alpha\beta} - \square h) = \frac \kappa 2 T_{\mu\nu}$$

A rough analysis can show that if we want gravity to be a field theory, it has to be of this form (the dropoff in $r^{-2}$ means it's massless, the fact that it is always attractive means it is of even spin, and the deflection of light forbids it to be of spin 0).

This theory is invariant under some $SO(3,1)$ gauge, $h_{\mu\nu} \rightarrow h_{\mu\nu} + \partial_\mu \chi_\nu + \partial_\nu \chi_\mu$. If we use the equivalent of the Lorentz gauge, that is

$$\square h_{\mu\nu} = \kappa T_{\mu\nu}$$

If we try to find out the stress energy tensor of the full theory (the stress energy tensor of matter and of the gravitational field itself) with this basic theory, we find that energy is not conserved. To fix this, we have to add counterterms

$$\square h_{\mu\nu} = \kappa (T_{\mu\nu}(\phi) + T_{\mu\nu}(h))$$

With $T_{\mu\nu}(h)$ some non-linear term of the field. This is still not consistent, unfortunately. If we perform the same process at all orders, and perform the change of variable $g_{\mu\nu} = \eta_{\mu\nu} + \kappa h_{\mu\nu}$, we end up with the Einstein field equations.

3) The Lovelock theorem

Lovelock's theorem states that if we have a differential equation of the type

$$D_{\mu\nu}[g] = T_{\mu\nu}$$

with $D$ some differential operator acting on the metric, that is only at most of order 2 (second derivatives of the metric), and such that $\partial_\mu T^{\mu\nu} = 0$, and the spacetime considered is four dimensional, then

$$D_{\mu\nu}[g] = a G_{\mu\nu} + b g_{\mu\nu}$$

With $G$ the Einstein tensor and the second term being a cosmological constant term. It is a rather lengthy theorem, the demonstration for which you can find in Straumann's "General Relativity"

4) Gauge and gauge-like methods for gravity

It is possible to describe gravity as a (sort of) gauge theory. The theory has to be invariant under the two following transformations :

\begin{eqnarray} x &\rightarrow& f(x)\\ \varphi(x) &\rightarrow& \Lambda(x) \varphi(x) \end{eqnarray}

The first is a diffeomorphism applied to the coordinates (a coordinate change), of the group $GA(4,\Bbb R)$, the general affine group (which can be reduced to $O(3,1) \ltimes \Bbb R^4$, or, if the spacetime is well behaved, $SO^\uparrow(3,1)\ltimes \Bbb R^4$, the special orthochronous Poincaré group), with the associated "gauge field" being the tetrad field, and the second is a $SO^\uparrow(3,1)$ gauge on the fields, with the associated gauge field being the spin connection. The Ricci tensor of the Lagrangian is then just the field strength of the connection.

There's a few other methods, but that's about the ones I can think of

• To 4): What, exactly, do you mean by "of the group $\mathrm{GA}(4,\mathbb{R})$"? The group of diffeomorphisms is far larger than the general affine group. Nov 24, 2015 at 0:02

Mindful of your bounty comment, I'm going to pick through your answer to try to offer some insight:

First, we notice that the paths traced by particles through spacetime under the influence of a gravitational field seem to depend only on their positions and velocities, i.e. they are independent of any identifiable charge or composition of the particles. It is almost as if the particles were moving along tracks carved into some curved surface.

Note that spacetime is the "block" universe that models space at all times. See Ben Crowell making the point here: "Objects don't move through spacetime. Objects move through space". Also note that objects don't really move along curved tracks carved into some curved surface. Their curved paths depend on their speed and direction, and on how strong the "force" of gravity is at that location. So for simplicity let's consider light only, and focus its motion through space as opposed to spacetime.

Although even Galileo noticed this, we happen to be passingly familiar with differential geometry to the point where we can actually make something of it. We realize that, if this observation really held, it would be possible to describe the curvature mathematically by assigning to spacetime a nonzero Riemann curvature tensor $R^a_{bcd}$.

We can describe a curvature mathematically, but this does not directly dictate the curved path of a light beam. See the picture below taken from the Wikipedia Riemann curvature tensor article.

CCASA image by Johnstone, see Wikipedia, amended by me

The curvature of the light beam depends on the slope of the plot, the first derivative of potential, the "force" of gravity. The Riemann curvature relates to the second derivative of potential, the tidal force. It's the "defining feature" of a gravitational field because without this curvature your whole plot is flat and level. You need this curvature to have a slope, but it's the degree of slope that that then dictates the curvature of the light path.

This object describes the rotation of a vector when dragged along a closed loop through the spacetime.

This is a somewhat abstract definition. IMHO it's better to say this object describes a gravitational field. As to how, drop a dimension and imagine you place optical clocks throughout "an equatorial slice through the Earth and the surrounding space". These clock run at different rates. When you plot these clock rates, depict lower slower clocks as lower down in a 3D image, and higher faster clock rates higher up. What your plot looks like, is the picture above. It's a picture of curved spacetime. And note this: your clocks don't run slower when they're lower because your plot of clock rates is curved.

What should we couple $R^a_{bcd}$ to? Well, gravity is obviously sourced by mass somehow, but we know because we invented special relativity that mass and energy are intimately connected. We also know we can uniquely describe the energies and pressures of classical matter fields in a nicely coordinate-independent way by their stress-energy tensor $T_{ab}$.

A photon has a non-zero "active gravitational mass". This is a measure of energy, and it's a concentration of energy that causes gravity. A massive body causes gravity because it is a concentration of energy, so let's consider light only. A photon causes gravity. Note how Void referred to action in his answer? For a photon we say energy E=hf, where h is action. Also note that one way of expressing the dimensionality of energy is pressure x volume.

So, making the leap that perhaps pressure can source gravity as well, we write down $R^a_{bcd} = -\kappa T_{ab}$. Unfortunately this equation makes no sense because the RHS and LHS have a different number of indices.

But what does make sense is the stress-energy-momentum tensor. Take a look at the Wikipedia article. Note the energy-pressure diagonal. And note that the stress–energy tensor "describes the density and flux of energy and momentum in spacetime". A photon is the simplest instance of that I can think of. Let's come back to that.

No matter - we can just contract some of the indices on the left hand side. This gives us the Ricci tensor $R_{ab} \equiv R^j_{ajb}$ so we write $R_{ab} = -\kappa T_{ab}$.

Ricci curvature represents "the amount by which the volume of a geodesic ball in a curved Riemannian manifold deviates from that of the standard ball in Euclidean space". Can we get some intuition on that? I think so. Imagine we can fly towards the Earth on the above plot. Let's fly very close and effectively "zoom in" to avoid getting confused by the curvature of the spherical Earth. Now let's redraw the plot, flipping round the rubber-sheet tension to yield a stress, stress being directional pressure. Our depiction, with a suitable light beam, now looks like this:

We have cuboids instead of geodesic balls, but no matter. Plot the height of each cell and you're plotting Ricci curvature.

This equation predicts the perihelion precession of Mercury, so we are very happy. Unfortunately it has a rather serious flaw: $T_{ab}$ had better have identically zero divergence if energy is at least locally conserved

IMHO energy is conserved full stop. I know some people say it isn't conserved in GR, but I beg to differ. As for the intuitive reason, look at divergence. It's to do with sources and sinks. And despite the "waterfall" analogy derived from Gullstrand–Painlevé coordinates, space is not falling towards the Earth. You do not live in some Chicken-Little world where the sky is falling in. Instead that light beam curves because "the speed of light is spatially variable". That's why those optical clocks go slower when they're lower, not because your plot of clock rates is curved. And they do this because a concentration of energy in the guise of a massive star "conditions" the surrounding space altering its metrical properties, making it "neither homogeneous nor isotropic, compelling us to describe its state by ten functions (the gravitation potentials gμν)". Space is inhomogeneous, this inhomogeneity is non-linear, and we model it as curved spacetime.

Again, no problem - we can simply subtract the divergence

Because there is no sink, not for a static gravitational field like the Earth's. Note the shear stress term in the stress-energy-momentum tensor? This is telling you that space is modelled as a gin-clear-ghostly elastic continuum. Google on Einstein elastic space. OK, for an analogy, imagine you have a great big gedanken block of crystal-clear elastic jelly. Now put grid lines in it so you can see what's going on. Then insert a hypodermic needle into the centre of the block, and inject more jelly. See how this creates a pressure gradient in the surrounding jelly? Such that the grid lines are pushed outward? Such that cells that used to be square are now cuboid? OK, now let's take another look at the EFE:

Actually this is a bit restrictive: the divergence is determined only up to an arbitrary constant $\Lambda$. Thus: $R_{ab} + \left(\Lambda - \frac{1}{2}R\right)g_{ab} = -\kappa T_{ab}$.

On the left we have the Ricci curvature tensor $R_{ab}$. That describes how the block of jelly has been subjected to an outward directional-pressure or stress gradient. We then have Λ which is "the energy density of the vacuum of space". That's saying something about the density of your original block of jelly. Then $R$ is scalar curvature which is related to $R_{ab}$, maybe we can say that's a relative density. The $g_{ab}$ metric tensor is telling you how optical clock rates vary at difference locations, wherein the "coordinate" speed of light c = √(1/ε₀μ₀) is akin to the mechanics expression v = √(G/ρ). This is different to the locally-measured speed of light c in the kappa term. The stress-energy tensor $T_{ab}$ is in essence saying how much jelly you injected.

What about the free parameter $\kappa$?

I don't think's it's particularly important myself. It feels like a conversion factor, like big G. I think what is important is why there is an equation at all. Why does one side equal the other? Look again at the analogy. The elastic jelly represents space. You injected jelly. But that jelly represented energy. And IMHO the insight is that at some deep fundamental level, space and energy are but two aspects of the same thing.

PS : I mentioned the E=hf photon, the picture below depicts a photon, a singleton sinusoidal electromagnetic field variation wherein E and B are the spatial and time-derivatives of four-potential A. It's as if light is energy, a pressure pulse in space propagating through space. And look above the centre, above the A. The squares are slightly flattened, something like the depiction above. I am reminded that for every action, there is an equal and opposite reaction. It's as if the gravitational field is the reaction to action h.