Can we say if the covariant derivative is zero, the partial derivative is also zero because locally covariant derivative reduces to partial derivatives (since locally spacetime is flat)? Because, in the right hand side of EFE we prove that the divergence (sum of PD along 4 axises) = zero, but in the left hand side we prove that the covariant derivative of Einstein Tensor = 0.
$\begingroup$
$\endgroup$
3
-
$\begingroup$ the divergence When you take divergence of $T^{\mu\nu}$ to show local energy-momentum conservation, you take the covariant divergence. This is the sum of four covariant derivatives, not the sum of four partial derivatives. See Wikipedia. So your question is based on a false assumption. $\endgroup$– GhosterMay 28 at 3:44
-
$\begingroup$ It's based on the assumption that Energy momentum is only conserved locally - so the covariant divergence of stress energy tensor is zero only locally, right? But the covariant divergence of Einstein Tensor is zero everywhere, not just locally. Does that mean EFE should be valid only locally? $\endgroup$– Nayeem1May 28 at 4:14
-
$\begingroup$ The covariant divergence of the stress energy tensor vanishes everywhere. But local conservation everywhere is not the same as global conservation. In general, you can’t define the total/global energy or total/global momentum in curved spacetime and show that they stay constant. For example, dark energy in an expanding universe violates global energy conservation, and physicists don’t consider this a problem. $\endgroup$– GhosterMay 28 at 4:38
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
First, note that for scalar functions, the covariant derivative reduces to the partial derivative. So for scalar functions, it is true that if the covariant derivative is zero at a point, then the partial derivatives are also zero at that point.
So, we move to a more general situation. If the covariant derivative of a (non-scalar) function is zero at a point, then there exists a coordinate system where the partial derivative is zero at that point. However, in a general coordinate system the partial derivative will not be zero at that point.
An easy example is provided by the metric tensor. Assuming a metric-compatible connection, the covariant derivative of the metric is zero at every point in spacetime. However, the partial derivative of an individual metric component is usually not zero. Consider the metric for a flat spacetime in spherical coordinates $$ g_{ij} dx^i dx^j = ds^2 = dr^2 + r^2 d\theta^2 + r^2\sin^2\theta d\phi^2 $$ Clearly $\partial_r g_{\theta\theta} = \partial_r r^2\neq 0$ in general, even though $\nabla_r g_{\theta\theta} = 0$.
-
-
1$\begingroup$ @Nayeem1 what’s the issue? A certain tensor field on the LHS (Einstein tensor) is set equal to another tensor field of the same valence (a multiple of stress energy momentum tensor). If you’re getting confused by tensors and differentiations (of all sorts) then you need to review the definitions (including the definition of divergence), and understand what they say vs what they’re not saying. $\endgroup$ May 28 at 5:51
-
$\begingroup$ @Nayeem1 By asserting they are equal and noting there is no contradiction (the equations don't reduce to something like $1=0$. $\endgroup$– AndrewMay 28 at 14:15