# Div G_ab = 0

1. Jun 10, 2008

### snoopies622

What physical meaning can be ascribed to the non-divergence of the Einstein tensor? I find it counterintuitive since I associate divergence with field sources (like the electrical field of a proton) and obviously a gravitational field has a source. Is there a parallel with Newton's formulation of gravity that might be instructive?

Last edited: Jun 10, 2008
2. Jun 10, 2008

### pmb_phy

Since the Einstein tensor is proportional to the stress-energy-momentum tensor, T it means that energy and momentum is conserved since div T = 0. This holds true even when the cosmological constant is non-zero.

Pete

3. Jun 10, 2008

### haushofer

You can also see them as constraints on the equations of motion. Intuïtively you can understand them as follows: in general relativity one wants to solve for the metric tensor, which is symmetric and thus can have 10 independent entries ( n*(n+1)/2 ). However, one is free to choose the coordinates, and this gives some freedom in your equations ( which can be seen as a gauge-freedom ). The consequences of this can be calculated to be the Bianchi-identities, which give that the Einstein tensor is divergence-free. Note that this is an identity rather than a symmetry; the description of physics doesn't depend on your coordinates.

A parallel with Newton's formulation is a little tricky; but you have to be aware of the fact that the Einstein tensor already contains second order derivatives of the metric, just like Poisson's equation is a second order differential equation of the classical gravitational field ! A sensible parallel would appear to me that due to constraints on the gravitational field the third order divergence of the classical gravitational field would be zero.

4. Jun 11, 2008

### snoopies622

So Einstein created the stress-energy tensor first, then made $$G_{ab}$$ to match it?