Airsteve0 said:
Homework Statement
In my general relativity class my professor mentioned that in dimension 4 there are only six statements in the Einstein equations and that this is exactly the number needed. Homework Equations
G_{\alpha\beta}+\Lambdag_{\alpha\beta}=8{\pi}T_{\alpha\beta}The Attempt at a Solution
Really I just want to understand why 6 statements are all that is needed (apparently exactly what is needed) in 4 dimensions. Any clarification would be greatly appreciated, thanks!
In D dimensional space-time there are (1/2)D(D + 1) field equations
G_{\mu \nu}= k T_{\mu \nu}. \ \ \ \ (1)
But we also have the identities,
\nabla^{\mu}G_{\mu \nu} \equiv 0. \ \ \ (2)
These D identities hold whether or not G_{\mu \nu} themselves satisfy the field equations. Therefore, covariantly, the dynamics is described by only (1/2)D(D - 1) field equations (in 4 dimensions, only six out of ten equations are dynamical ones, and the remaining four ones are constraints). Indeed, the Einstein equation involves \ddot{g}_{ij} but not \ddot{g}_{0 \mu}. Thus, the invariance of the action under general coordinates transformation shows that the action principle and with it the field equations can not determined the metric tensor unless the coordinate system is specified in some non-covariant way.
Of course, this failure is not dangerous. Similar situation arises in Maxwell’s equations
\partial^{2}A^{\nu} - \partial^{\nu} \partial \cdot A = J^{\nu}. \ \ (3)
These four equations (for the four unknowns A_{\mu}) do not determine A_{\mu} uniquely, because the left-hand-sides of these equations are related by a single divergence identity analogous to eq(2),
<br />
\partial_{\nu}( \partial^{\sigma}\partial_{\sigma}A^{\nu} - \partial^{\nu}\partial_{\sigma} A^{\sigma}) \equiv 0. \ \ (4)<br />
This identity reduces the number of functionally independent equations to three. In Maxwell’s case, the ambiguity in the solutions is removed by choosing a particular gauge. For example, given any solution B_{\mu}, one can always construct a new solution, A_{\mu}, such that
\partial^{\mu}A_{\mu} = 0, \ \ \ (5)
by setting
A_{\mu} = B_{\mu} + \partial_{\mu}\Omega ,
where \Omega is given by
\partial^{2}\Omega = - \partial \cdot B .
Thus eq(5) (regarded as a field equation) and eq(3), with appropriate boundary conditions, could generally determine A_{\mu} uniquely.
In similar way, in order to determine g_{\mu \nu}, it is necessary to introduce a coordinate condition, explicitly violating the invariance of the theory under general coordinates transformation. This procedure is often called “gauge fixing”. Of course this is an abuse of language and one must be careful about the difference between the choice of a coordinate system and gauge fixing. The former has physical meaning: inertial force (i.e. observable effects) actually depends on it. In fact, the properties of the solution to eq(1) depend on the choice of a coordinate system: in order to make definite predictions, one needs to add D coordinate conditions (treated like field equations) to the Einstein equations.
Mathematically, the most natural choice of the coordinate condition (invariant under GL(n)) is that of de Donder
g^{\mu \nu}\Gamma^{\sigma}{}_{\mu \nu} = 0. \ \ \ (6)
This is also called Harmonic Coordinate Condition. To understand why the term Harmonic is used, you need to use the fact
\frac{1}{\sqrt{-g}}\partial_{\sigma}\left( \sqrt{-g} g^{\sigma \rho}\right) = - g^{\mu \nu}\Gamma^{\rho}{}_{\nu \mu}.
Now, if you insert \delta^{\rho}_{\mu} = \partial_{\mu}x^{\rho} in the left hand side, you can rewrite the de Donder condition as
\frac{1}{\sqrt{-g}}\partial_{\sigma}\left( \sqrt{-g}\ g^{\sigma \mu}\ \partial_{\mu} x^{\rho}\right) = 0.
Thus the term Harmonic coordinate condition makes sense because the operator (1/ \sqrt{-g}) \partial_{\sigma}( \sqrt{-g}g^{\sigma \mu}\partial_{\mu}) is nothing but the curve space generalization of the d’Alembertian \partial^{2}.
Ok, now I leave you with the following little exercise: Prove that one can always find a coordinate system such that g^{\mu \nu}\Gamma^{\rho}{}_{\mu \nu}=0. Generalize the de Donder condition and show that there exists a coordinate system (called non-rotating) in which the energy-momentum tensor (of the matter field) is conserved globally.
Sam
I feel a little more confident commenting. To elaborate on what I said above, I would have expected 42 separate equations, not 20. Wikipedia then points out that these tensors are symmetric, and that each one therefore has 10 independent components, The article states that the EFEs are therefore a set of 10 equations, which is exactly what I would expect if you represent each tensor as a 4x4 matrix: the 4 diagonal elements are independent, and then due to symmetry, each of the 12 off-diagonal elements is equal to its mirror element reflected across the diagonal, so you would halve this number and end up with 4 + 6 independent components.
