How many independent components are there in the Einstein equations?

[maxverywell]

I read that the number of Einstein equations is 10 and because of Bianchi identity, it reduces to 6. This is true because we have totally 16 components, but they are symmetric, so there are 10 independent components. However in four dimensional space-time, the number of second order partial differential equations for the metric tensor is 40, because, each component $g_{\mu\nu}$ is a function of $x^{\alpha}, \alpha=0,1,2,3$. So finally we have a system of 24 independent second order partial differential equations for the metric tensor. Is that correct?

 

[dextercioby]

Why do you multiply by 4? It makes no sense. The correct number is 6 independent components of the metric tensor in each spacetime point.

 

[Naty1]

I was just reading here

http://en.wikipedia.org/wiki/Einstein_field_equations#Solutions


where the prior post response [6] is explained in a bit more detail...

 

[PeterDonis]

I read that the number of Einstein equations is 10 and because of Bianchi identity, it reduces to 6.

The Bianchi identities don't "reduce" the number of independent components; they just let you choose which specific equations you want to solve--do you want to solve all ten components of the EFE, or solve only six of them and four Bianchi identities instead?

However, the more important point is that the Einstein Field Equations are not equations for the metric tensor; they are equations for the Einstein tensor, which is composed of second derivatives of the metric tensor.

The full accounting, IIRC, goes like this (MTW goes into this, I don't have my copy handy to look up an exact reference):

The metric tensor is a symmetric 2nd-rank tensor, so it has 10 independent components. However, we can fix all 10 of them at a given event by an appropriate choice of coordinates for a local inertial frame, because a local Lorentz transformation has ten parameters (four translations to fix the origin, three spatial rotations to fix the orientation, and three boosts to fix the particular state of local inertial motion that is "at rest").

There are forty first derivatives of the metric tensor. We can eliminate all forty of them at a given event by an appropriate choice of coordinates--basically we choose Riemann normal coordinates to set all of the connection coefficients to zero at our chosen event.

There are one hundred second derivatives of the metric tensor. We can eliminate eighty of them at a given event by using symmetries of the Riemann tensor, leaving twenty independent components. Ten of these are the ten independent components of the Einstein tensor, which can be found by solving the EFE. The other ten are the components of the Weyl tensor, which are not constrained at a given event by the EFE. (But they *are* constrained if you have a solution of the EFE not just at a single event but over an entire spacetime.)

 

[WannabeNewton]

See here for a discussion involving what Peter has just explained: https://www.physicsforums.com/showthread.php?t=677888

Also, if you can get access, check out chapter 10 of Wald. In this chapter, it is made quite clear that the constraint $\nabla^{a}G_{ab} = 0$ that results in 6 evolution equations for 10 independent metric components and what ostensibly seems like an under-determined system, is not physical but rather comes out of the diffeomorphism freedom which, evidently, is a type of gauge freedom for GR. By choosing a convenient gauge, or equivalently a convenient set of coordinates, we can eliminate the under-determination. Wald in particular chooses the harmonic gauge which is a set of coordinates $x^{\mu}$ solving $\nabla^{a}\nabla_{a}x^{\mu} = 0$, for the purposes of the referenced chapter.

It is akin to how for the source-free Maxwell's equations $\nabla^{a}F_{ab} = 0,\nabla_{[a} F_{bc]} = 0$ we have the constraint $D_{a}E^{a} = D_{a}B^{a} = 0$ on the initial value space-like hypersurface, which make Maxwell's equations seem under-determined but this can be eliminated by fixing a gauge, in particular the Lorenz gauge $\nabla^{a}A_{a} = 0$ for the 4-potential $A_{a}$.

EDIT: I forgot to mention, for the sake of completeness, that $D_{a}$ is the induced derivative operator on the initial value surface.

 

[PeterDonis]

See here for a discussion involving what Peter has just explained: https://www.physicsforums.com/showthread.php?t=677888

I knew we'd had a thread on this topic fairly recently. Thanks for finding it!

 

[PeterDonis]

The metric tensor is a symmetric 2nd-rank tensor, so it has 10 independent components. However, we can fix all 10 of them at a given event by an appropriate choice of coordinates for a local inertial frame, because a local Lorentz transformation has ten parameters (four translations to fix the origin, three spatial rotations to fix the orientation, and three boosts to fix the particular state of local inertial motion that is "at rest").

On re-reading this, I should note that I mis-stated it; the correct statement is in this post in the thread WannabeNewton linked to. Briefly, the correct accounting for how the metric components are fixed is not 10 = 4 + 6, but 10 = 16 - 6; there are 16 first-order coefficients in an expansion of a coordinate transformation at a given event, but 6 of them describe the 6 parameters of a Lorentz transformation (excluding translations, since choosing a particular event for the expansion fixes the origin), so there are 10 left over to describe the metric tensor itself. In other words, 10 of the 16 coefficients describe how to transform an arbitrary metric tensor at a given event to the Minkowski metric; the other 6 describe the family of all possible local inertial frames in which the metric is Minkowski at the given event.