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The Riemann curvature tensor $R_{\mu \nu \rho \sigma}$ has the geometric interpretation of giving how much parallel transport fails to close around tiny loops. The Ricci tensor $R_{\mu \nu}$ the Riemann curvature averaged over all directions, as in, if there is negative curvature in some direction there must be positive curvature in another if $R_{\mu \nu} = 0$.

What is the geometric interpretation of the Einstein tensor $$ G_{\mu \nu} = R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R? $$ Is there a way to understand $$ \nabla^\mu G_{\mu \nu} = 0 $$ Intuitively?

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  • $\begingroup$ See ch. 15 of Misner, Thorne, and Wheeler, and the brief summary at mathoverflow.net/a/238551/21349 . $\endgroup$
    – user4552
    Nov 9, 2018 at 21:36

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TL;DR:

$\nabla^\mu G_{\mu\nu} = 0$ is implied by the following statement - There are no gravitational magnetic charges.

Long Answer

It always helps when trying to intuit something to compare it to other things for which you already have intuition!

Let's start with electromagnetism. This is described in terms of a field strength $F_{\mu\nu}$ which satisfies the Bianchi identity $$ dF = 0 \implies \partial_{[\mu} F_{\nu\alpha]} = \partial_\mu F_{\nu\alpha} + \partial_\alpha F_{\mu\nu} + \partial_\nu F_{\alpha\mu} = 0 . $$ This condition is often understood as the absence of magnetic charges. In the presence of magnetic charge current $j_M^\mu$ this equation gets modified to $$ dF = \ast j_M \implies \partial_\mu F_{\nu\alpha} + \partial_\alpha F_{\mu\nu} + \partial_\nu F_{\alpha\mu} = \epsilon_{\mu\nu\alpha\beta} j_M^\beta . $$ This is the story with abelian gauge theories.

We can also extend this next to non-abelian gauge theories, where the Bianchi identity takes a slightly more complicated form $$ dF + A \wedge F = 0 \implies D_{[\mu} F_{\nu\alpha]} = D_\mu F_{\nu\alpha} + D_\alpha F_{\mu\nu} + D_\nu F_{\alpha\mu} = 0 , \qquad D_\mu F_{\nu\alpha} = \partial_\mu + [ A_\mu , F_{\nu\alpha} ] . $$ The intuition is the same as before - no non-abelian magnetic charges!

ASIDE - If you're unfamiliar with non-abelian gauge theories, simply think of $A_\mu$ as a matrix with elements $(A_\mu)^a{}_b$ and $[A,B] = AB-BA$ denotes matrix commutation. The field strength is defined in terms of $A$ as $$\tag{1} F = dA + A \wedge A \implies F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu + [ A_\mu , A_\nu ] . $$ Again, think of $F_{\mu\nu}$ as a matrix with elements $(F_{\mu\nu})^a{}_b$.

Having understood this intuition, lets go to the gravitational case! Roughly, the Christoffel symbol $\Gamma^\lambda_{\mu\nu}$ is our non-abelian gauge field (think of it as $(A_\mu)^\lambda{}_\nu = \Gamma^\lambda_{\mu\nu}$ and the Riemann tensor is our field strength, \begin{align} ( F_{\sigma\nu} )^\lambda{}_\mu = R^\lambda{}_{\mu\sigma\nu} &= \partial_\sigma \Gamma^\lambda_{\nu\mu} - \partial_\nu \Gamma^\lambda_{\sigma\mu} + \Gamma^\lambda_{\sigma\tau} \Gamma^\tau_{\nu\mu} - \Gamma^\lambda_{\nu\tau} \Gamma^\tau_{\nu\mu} \\ &= \partial_\sigma (A_\nu)^\lambda{}_\mu - \partial_\nu (A_\sigma )^\lambda{}_\mu + (A_\sigma)^\lambda{}_\tau (A_\nu)^\tau{}_\mu - (A_\nu)^\lambda{}_\tau (A_\sigma)^\tau{}_\mu \end{align} In matrix notation this is the same as equation (1).

The Bianchi identity for the Riemann tensor is $$ \nabla_{[\alpha} R_{\mu\nu]\rho\sigma} = \nabla_{\alpha} R_{\mu\nu\rho\sigma} + \nabla_{\nu} R_{\alpha\mu\rho\sigma} + \nabla_{\mu} R_{\nu\alpha\rho\sigma} = 0 $$ Using our previous intuition we can immediately conclude that the above implies no gravitational magnetic charges!

The conservation of the Einstein tensor is just one contraction of the above Bianchi identity (prove it by contracting the identity with $g^{\alpha\rho} g^{\mu\sigma}$).

PS - This answer seems to suggest that gravity is just a particular type of non-abelian gauge theory. In fact, this is not the case due to other issues. However, the presentation here is good for intuition.

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TL;DR- General relativity (GR) is based upon general coordinate invariance; this means that physics is invariant under a general coordinate transformation (GCT). This invariance implies the contracted Bianchi identities ($\nabla_\nu G^{\mu \nu} =0$), which in turn give us constraints on the equations of motion. We needed some constraints anyway since we allowed for arbitrary GCT in the beginning. That Einstein's field equations do not determine $g_{\mu \nu}$ uniquely, but only up to $4$ arbitrary coordinate transformations, is explained by the contracted Bianchi identities.

Ignoring factors of $16$ and $\pi$ throughout the answer, the Einstein-Hilbert action is $$S = \int_Vd^4x \sqrt{-g} R,$$ where $V$ is the spacetime region of integration. Taking the variation of $S$ (with respect to $g^{\mu \nu}$) gives $$\delta S = \int_V d^4x \sqrt{-g} G_{\mu \nu} \delta g^{\mu \nu} \tag{1}.$$

Under arbitrary variations of the metric, $\delta g^{\mu \nu}$, the least action principle $\delta S=0$ gives us Einstein's equations of motion in vacuum: $G_{\mu \nu}=0$. (You can repeat the procedure by adding a matter Lagrangian). So $G_{\mu \nu}$ is the source-free part of the equations of motion for the metric field $g_{\mu \nu}$. In $3+1$ dimensions, $G_{\mu \nu}$ is the unique tensor (apart from the metric tensor itself) that is constructed from $g_{\mu \nu}$ and its first and second derivatives, is symmetric in its two indices and is divergence-free (Lovelock's theorem). In higher dimensions, $G_{\mu \nu}$ is no longer unique, if you allow non-linear functions of 2nd derivatives of the metric. But if you allow only linear functions of 2nd derivatives of the metric, $G_{\mu \nu}$ remains unique.

But without hurrying up to obtain the equations of motion, you can gain useful information just from the form of the action variation in $(1)$.

GR must be invariant under GCT, so $S$ must be invariant under GCT $\Rightarrow \delta S = 0$ under GCT: $x \rightarrow x'$. What could this tell us? It is sufficient to consider infinitesimal GCT. So, suppose $$x^\mu \rightarrow x'^\mu = x^\mu + \epsilon^\mu,$$ where $\epsilon^\mu$ is arbitrary inside $V$ but constrained to vanish on the boundary of $V$: the hypersurface $\partial V$.

Under this infinitesimal GCT, evaluate the variation of the metric tensor field, $$\delta g_{\mu \nu} = g'_{\mu \nu}(x) - g_{\mu \nu}(x),$$ up to first order in $\epsilon$, ignoring $O(\epsilon^2)$ and higher terms. Raising the indices gives you $$\delta g^{\mu \nu} = \nabla^\mu \epsilon^\nu + \nabla^\nu \epsilon^\mu. \tag{2}$$

Note the symmetry of indices $\mu, \nu$. Substitute $(2)$ in $(1)$ to get $$\delta S = \int_V d^4x \sqrt{-g} \ G^{\mu \nu} \nabla_\nu \epsilon_\mu.$$

Integrating by parts and using Gauss' theorem gives $$\delta S = -\int_V d^4x \sqrt{-g} \ (\nabla_\nu G^{\mu \nu}) \epsilon_\mu + \oint_{\partial V} d \Sigma_\nu G^{\mu \nu} \epsilon_\mu.$$

Recalling that $\epsilon_\mu$ is arbitrary in $V$ and that $\epsilon_\mu = 0$ on $\partial V$, we see that asking for action invariance, $\delta S=0$, under GCT implies $$\nabla_\nu G^{\mu \nu} = 0, \tag{3}$$ which are the contracted Bianchi identities. So we see that (3) can be seen as a consequence of imposing GCT on GR.

Einstein's equations $G_{\mu \nu} = T_{\mu \nu}$ seem like they imply that there are $10$ equations for $10$ unknowns in $g_{\mu \nu}$. But that's not the full story. Since we are allowed to do GCT, Einstein's equations do not uniquely determine $g_{\mu \nu}$, but only up to $4$ arbitrary coordinate transformations. The $4$ equations in $\nabla_\nu G^{\mu \nu} =0$ provide the missing link. Explicitly rewriting the Bianchi identities in terms of the timelike components of $G^{\mu \nu}$, we see $$\partial_t G^{t \nu} = \text{some expression of $G$ and $\Gamma$ that contains at most 2nd derivatives of $g_{\alpha \beta}$},$$ so that $G^{t \nu}$ on the LHS must contain at most 1st derivatives of $g_{\alpha \beta}$. But specifying the metric and its first derivative accounts for specifying the initial conditions, and it does not really account for dynamical equations of motion. We therefore are able to get $4$ constraint equations, and $10-4=6$ truly dynamical equations of motion.

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The divergence of the Ricci tensor is identically $\nabla^\mu R_{\mu \nu} = \frac{1}{2}g_{\mu \nu} R$, as a consequence of the Bianchi identities. So the Einstein tensor is by construction the Ricci tensor minus its divergence. In fact, though it's not obvious, Lovelock's theorem implies the Einstein tensor is the only divergence-free tensor that depends only upon $g_{ab}$ and its first two derivatives. I think that's the closest thing to an "intuitive" understanding that's really out there to be had.

This in itself is a geometry thing and has nothing to do with physics. Physics then motivates why we should care about the aforementioned properties: if we want to couple the metric to a divergence-free rank two tensor (i.e. the stress-energy tensor), and for physical initial value problems to require only knowledge of the ``position and velocity", and for physics to be manifestly coordinate-invariant, then the Einstein tensor is the only game in town.

It's worth pointing out that Einstein's original theory of gravity, the ``Entwurf" theory, in fact used the field equations

$$ R_{\mu \nu} = T_{\mu \nu}$$

which had to be coupled to a coordinate condition enforcing that $R_{\mu \nu}$ be divergence-free. This is identical to general relativity in vacuum and is in fact the theory Einstein used to predict the perihelion precession of Mercury. The Einstein tensor was introduced to restore full coordinate invariance by having $\nabla^\mu T_{\mu \nu} = 0$ enforced at the level of the field equations.

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At the risk of being downvoted, without any good arguments, I can tell you that you can understand these equations as a relativistic equations for a fluid dynamics, so the G equation can be viewed as sort of continuity equation. It indicates energy-momentum conservation. My advice: google relativistic fluid dynamics.

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  • $\begingroup$ I know that it leads to $\nabla_\mu T^{\mu \nu}$ via the Einstein equation. However, $\nabla_\mu G^{\mu \nu} = 0$ is tautological, and follows from the definition of $G^{\mu \nu}$. I am wondering if I can understand why it is a tautology independent of the Einstein Field Equation. $\endgroup$ Nov 9, 2018 at 21:14
  • $\begingroup$ Yes, you are right. Now I want to know that also.. $\endgroup$ Nov 10, 2018 at 0:18

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