# Physical meaning of the Einstein tensor

The Einstein tensor is the tensor field $G$ on spacetime $M$ with components

$$G_{\mu\nu}=R_{\mu\nu}-\dfrac{1}{2}g_{\mu\nu}R$$

so that Einstein's field equations can be written as:

$$G_{\mu\nu}=\dfrac{8\pi G}{c^4}T_{\mu\nu}.$$

Now this tensor field, by definition, is a function $G:\Gamma(TM)\times\Gamma(TM)\to C^\infty(M)$, which takes two vector fields and outputs a function. In the chart $(U,x)$ we have

$$G(v,w)=G_{\mu\nu}v^\mu w^\nu=R_{\mu\nu}v^\mu w^\nu-\dfrac{1}{2}g_{\mu\nu}v^\mu w^\nu R.$$

The question here is: what is the significance of this tensor field? It esbalishes a multilinear relation between two vector fields and a number, but what is the physical significance of it?

What is the Einstein tensor from a physics point of view?

• Related/possible duplicate: physics.stackexchange.com/q/315941/50583 Mar 4, 2017 at 15:53
• Maybe check out arxiv.org/abs/gr-qc/0401099 Mar 4, 2017 at 15:56
• Einstein actually argued that $\kappa^{-1} G_{\mu \nu}$ can be seen as a (negative) stress-energy of the gravitational field. He proposed an interpretation of his equations $G_{\mu \nu} = \kappa T_{\mu \nu}$, which is based on the assertion that "total stress-energy of matter and gravity is always zero". That being said, I don't know all the details about this. I trust you can google-search more info if you are interested. Mar 6, 2017 at 13:44

Do you understand Jacobi fields (i.e., geodesic deviation)? They are probably the easiest way to explain what curvature tensors mean. Say I have a geodesic $\gamma$ and its tangent vector is $\xi$. Then using the Riemann tensor, I can define an operator

$$M^a{}_b \equiv R^a{}_{cbd} \, \xi^c \xi^d$$

which describes the behavior of vectors which are transported along $\gamma$ via the map $\zeta^a \to M^a{}_b \, \zeta^b$. If we lower its first index, then we can see that $M_{ab} \equiv R_{acbd} \, \xi^c \xi^d$ is a symmetric matrix, which means the deformations it describes will distort the transverse sphere $S^{n-1}_\bot$, defined by the set of vectors $\{ \zeta^a : g_{ab} \zeta^a \xi^b = 0, \; g_{ab} \zeta^a \zeta^b = 1 \}$, into an ellipsoid as one moves along $\gamma$. So, that is what the Riemann tensor describes: how the transverse sphere $S^{n-1}_\bot$ (orthogonal to our direction of travel) distorts into an ellipsoid as we move along a geodesic.

Now, the Ricci tensor is given by the trace $R_{cd} = R^a{}_{cad}$, so if we look along the same geodesic, our Ricci tensor just gives us the trace of the matrix $M^a{}_b$:

$$R_{cd} \, \xi^c \xi^d = M^a{}_a,$$

and the trace of the infinitesimal ellipsoidal deformation gives us the change in area (multipled by some constant) of $S^{n-1}_\bot$ as we move along $\gamma$. In a sense, the specific changes in shape of $S^{n-1}_\bot$ have been averaged out, and one is left only with the change in overall size.

To obtain the Ricci scalar, we then take the trace of $R_{cd}$, which means that we average over all directions $\xi^a$ for possible geodesics emanating from a given point. In each given direction, the Ricci tensor measures the change in area of $S^{n-1}_\bot$ along that geodesic; therefore, the Ricci scalar must measuer the total change in the area of an $S^n$ centered at our point. That is, the Ricci scalar gives the deficit solid angle (again multiplied by some constant).

Now, since we have not worked out exactly what the constants are that relate $R_{cd} \, \xi^c \xi^d$ and $R$ to the actual changes in area, it is difficult to provide a precise notion of what $G_{ab}$ means. But we can give a general idea: The Ricci tensor part of

$$G_{ab} \, \xi^a \xi^b \equiv R_{ab} \, \xi^a \xi^b - \frac12 R g_{ab} \, \xi^a \xi^b$$

is giving the directionally-dependent change in area as we move along the geodesic $\gamma$, but then we subtract an amount of the total change in area. The end result is an averaged change; the particular choice of average is made such that

$$\nabla_a G^{ab} = 0,$$

which is important in order to couple the curvature to a conserved current like $T_{ab}$.

The short answer is that the law of the gravitational field $$R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R = - \frac{8 \pi G}{c^4} T_{\mu\nu} (= G_{\mu\nu})$$

is the equivalent of the Newtonian theory of the gravitational field which is

$$\nabla^2\phi = 4 \pi G \rho$$

and thus

1. It may contain no differential coefficients of the $g_{\mu\nu}$ higher than the second.
2. It must be linear in these second differential coefficients.
3. Its divergence must vanish identically.

The $R_{\mu\nu}$ is the contraction of the $R^\sigma_{\mu\nu\tau}$ (by $\sigma$ and $\tau$) - the Riemann tensor, which can tell you how much will a vector change if it is moved by parallel displacement along a geodesic (moved along the curve from point A to point B).

$R^\sigma_{\mu\nu\tau}$ is obtained in the following way

$$\Delta A^\mu = - \oint\Gamma^\mu_{\alpha\beta}A^\alpha dx_\beta$$

which [equation of parallel displacement of a vector], with sufficient accuracy, is transformed algebraically to be

$$2 \Delta A^\mu = - R^\mu_{\sigma\alpha\beta}A^\sigma f^{\alpha\beta}$$

where $f^{\alpha\beta}$ is formed by integral along the curve of $\xi^\mu = (x_\mu)_B - (x_\mu)_A$ $$f^{\alpha\beta} = \frac{1}{2} \oint(\xi^\alpha d \xi^\beta - \xi^\beta d \xi^\alpha)$$ with $R^\sigma_{\mu\nu\tau}$ being

$$R^\sigma_{\mu\nu\tau} = - \frac{\partial \Gamma^\sigma_{\mu\nu}}{\partial x_\tau} + \frac{\partial \Gamma^\sigma_{\nu\tau}}{\partial x_\nu} + \Gamma^\sigma_{\rho\nu}\Gamma^\rho_{\mu\tau} - \Gamma^\sigma_{\rho\tau}\Gamma^\rho_{\mu\nu}$$

The scalar R is formed from $g^{\mu\nu} R_{\mu\nu}$.

The field equations (aka the law of gravitational field) furnish the energy principle of matter $$0 = \frac{\partial\mathfrak{W}^\alpha_\sigma}{\partial x_\alpha} - \Gamma^\alpha_{\sigma\beta} \mathfrak{W}^\beta_\alpha$$ $$\mathfrak{W}^\alpha_\sigma = T_{\sigma\tau}g^{\tau\alpha}\sqrt{-g}$$

which practically says that

the gravitational field transfers energy and momentum to the matter

The second term ($\Gamma^\alpha_{\sigma\beta} \mathfrak{W}^\beta_\alpha$) is the energy density of the gravitational field while the first ($\frac{\partial\mathfrak{W}^\alpha_\sigma}{\partial x_\alpha}$) expresses the energy density of matter.

As an application we can try to approximate the motion of N body singularities each located at position $\overset{k}{\xi}$, each surrounded by a closed surface :

$$\int^k{(\Phi_{\mu\nu} + 2\Lambda_{\mu\nu}) n_{k}dS} = 0$$

We define $g_{\mu\nu} = h_{\mu\nu} + \eta_{\mu\nu}$ with $\gamma_{\mu\nu} = h_{\mu\nu} - \eta_{\mu\nu}\eta^{\sigma\rho}h_{\sigma\rho}$ .
Additionally let $\tau = x^0 \lambda$ with $\lambda$ being the approximation parameter obtained from developing our $x^0$ into power series (as a sum of powers of $\lambda$).
The distance from k-th singularity to $x^s$ is $\overset{k}{r} = \sqrt{ \left[ (x^s - \overset{k}{\xi^s})(x^s - \overset{k}{\xi^s}) \right] }$.

The field equations for our system are $$G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R$$

where we have the linear combinations of $R_{\mu\nu}$ (if we consider $\eta_{\mu\nu} >> h_{\mu\nu}$ ): $$\Phi_{\mu\nu} + 2 \Lambda_{\mu\nu} = -2(R_{\mu\nu} - \frac{1}{2}\eta_{\mu\nu}\eta^{\alpha\beta}R_{\alpha\beta} )$$

For each closed surface we have : $$\underset{2l}{\overset{k}{c_m}}(\tau) = \frac{1}{4 \pi} \int^{k}{ 2 \underset{2l}{\Lambda_{mn}} n_k dS}$$

So for all N bodies we will write $$\underset{2l}{\gamma_{mn,n}} = -\sum_{k=1}^{p} \left\{ \underset{2l}{\overset{k}{c_m}}(\tau) / \overset{k}{r} \right\}$$

Note: The $G_{\mu\nu}$

satisfies the Bianchi identity $$G^\mu_{\nu|\mu} + \Gamma^{\alpha}_{\alpha\beta}G_{\nu}^{\beta} - \Gamma^{\beta}_{\nu\alpha} G_{\beta}^{\alpha} = 0$$

References (and quotes):

1. The Meaning of Relativity - A.Einstein
2. Gravitational equations and problem of motion - A.Einstein, L. Infield, B. Hoffmann
3. On the motion of particles in general relativity theory - A.Einstein, L. Infield
• Maybe it is worth noting that the Poisson equation is actually the (00) component of the Einstein equation for the static weak field case. So the intuition for $G_{00}$ is the same as for $\Delta \phi$ in this case (though I don't have a good intuition for the Laplacian either...). Mar 4, 2017 at 20:49
• Sorry for the long post, but there is no simple explanation for the non linear equations of gravity such as the ones described by general relativity. Each field bends the space but each field follows the geodesic created by all the "bend-ations" of all the retarded fields. Mar 5, 2017 at 13:18

It arises as the variation of the Einstein-Hilbert action

S = ∫ (1/2κ) R √|g| d⁴x

with respect to the inverse metric

δS = ∫ (1/2κ) G_μν δg^μν √|g| d⁴x

As to what other geometric meaning the tensor has, apart from its application here and to the Einstein field equations: both it and its generalizations that include the cosmological coefficient, are used to define Einstein manifolds. These are manifolds in which the Ricci tensor has a fixed ratio to the metric tensor.

A couple notes on the coupling coefficient κ are in order here, to correct a widespread error that I see being repeated in some of the replies here.

Up to powers of light speed, c, the 8π factor in the expression κ = 8πG is specific only to 4 dimensions and is related to the 4π that appears in the formula for the surface area of a unit 2-sphere and the 4π/3 for the volume of the solid 2-sphere. For n > 3 dimensions it generalizes to (n-1)(n-2)V/(n-3), where V = √(πⁿ⁻¹)/((n-1)/2)! is the volume for the solid (n-2)-sphere, where the convention (-1/2)! = √π is used.

Second, the powers of c come directly out of the action formula by dimensional analysis. Using M, L and T respectively to denote dimensions of mass, length and time, the dimensions are given, for n-dimensional space-times, by

[S] = ML²/T = [h]
[R] = 1/L²,
[√|g| dⁿx] = Lⁿ,
[G'] = Lⁿ⁻¹/(MT²) = the n-dimensional version of Newton's constant G,
[c] = L/T,

where h is Planck's constant. Therefore,

[hκ] = [∫ R √|g| dⁿx] = Lⁿ⁻² = [hG'/c³]

the dimensions of the Planck n-2 area for n-dimensional space-times ... or (for n = 4) the Planck area. Therefore,

[κ] = [G'/c³], not [G'/c⁴]

and the correct expression for κ for dimensions n > 3 is

κ = (n-1)(n-2)/(n-3) √(πⁿ⁻¹)/((n-1)/2)! G'/c³

which for the case n = 4 (and G' = G) reduces to

κ = 8nG/c³.

The literature is interspersed with powers 2, 3 and 4 (e.g. Einstein used 2 in The Meaning of Relativity), the expression κ = 8nG/c⁴ comes by way of a faulty dimensional analysis which fails to properly distinguish tensors from tensor densities (e.g. the mass density ρ and pressure p are components of the stress tensor density, not the stress tensor) and/or to properly account for the dimensions of the stress tensor itself.

This is discussed in further detail in a recent thread under sci.physics.research - the standard forum for physics-related questions.

The issue of the numerical factor is also discussed here by Mansouri et al.: https://arxiv.org/abs/gr-qc/9609061