Is This a Valid Simplification of Einstein's Equations?

[Physicist97]

Einstein's Equations are usually written $R_{\mu\nu}-(1/2)g_{\mu\nu}R={\kappa}T_{\mu\nu}$ , where $\kappa$ is a constant. If you were to multiply both sides by $g^{\mu\nu}$ , then this becomes $R=-{\kappa}g^{\mu\nu}T_{\mu\nu}$ . I have used the relations $g^{\mu\nu}g_{\mu\nu}=4$ and $R=g^{\mu\nu}R_{\mu\nu}$ . $g^{\mu\nu}T_{\mu\nu}$ is the trace of the Stress Energy Tensor, which now leaves you with $R=-{\kappa}T^{\mu}_{\mu}$ . It seems to me that writing the Einstein Field Equations this way simplifies them, but I've never seen them written this way. Have I done something wrong using this approach, or is it not practical to write them in this form? Thank you for any explanation :).

 

[bcrowell]

What you've done is to take the trace of both sides of the Einstein field equations. The result is valid, but it doesn't contain as much information as the original field equations. A metric could satisfy it without satisfying the field equations.

For example, if the stress-energy tensor is that of an electromagnetic field, then its trace is zero. Suppose that on the left-hand side we put the metric of Minkowski space. Then the trace equation is satisfied, but Minkowski space is not a solution of the field equations when there is an electromagnetic field present.

 

[Physicist97]

Would it then be better to write $R=-{\kappa}g^{\mu\nu}T_{\mu\nu}$ , $g^{\mu\nu}R_{\mu\nu}=-{\kappa}g^{\mu\nu}T_{\mu\nu}$ , $R_{\mu\nu}=-{\kappa}T_{\mu\nu}$ . I understand why the other equation didn't carry enough information. The Einstein Equations are actually 16 equations (10 due to symmetric tensors), and writing the trace takes away information, but this way you once again have 16 equations. Written as $R_{\mu\nu}=-{\kappa}T_{\mu\nu}$ would it carry the same information? Is it even right to write it like this?

 

[bcrowell]

Written as $R_{\mu\nu}=-{\kappa}T_{\mu\nu}$ would it carry the same information? Is it even right to write it like this?

Written this way it isn't true. This would only be true in the case of a vacuum solution (right-hand side = 0).

$R=-{\kappa}g^{\mu\nu}T_{\mu\nu}$

This could be more compactly written as $R=-T^\mu{}_\mu$. (Let's take $\kappa=1$.) The metric raises an index. This is a true statement about the trace, but doesn't carry as much information as the full field equations.

$g^{\mu\nu}R_{\mu\nu}=-{\kappa}g^{\mu\nu}T_{\mu\nu}$

This is just a longer way of writing the preceding equation.

 

[pervect]

Einstein's Equations are usually written $R_{\mu\nu}-(1/2)g_{\mu\nu}R={\kappa}T_{\mu\nu}$ , where $\kappa$ is a constant. If you were to multiply both sides by $g^{\mu\nu}$ , then this becomes $R=-{\kappa}g^{\mu\nu}T_{\mu\nu}$ . I have used the relations $g^{\mu\nu}g_{\mu\nu}=4$ and $R=g^{\mu\nu}R_{\mu\nu}$ .

It's certainly simpler, but you've lost some equations along the way - you are down to one equation after the contraction, wheras the full set has four equations. But Baez does basically this in some of his lecture notes on GR - for instance http://math.ucr.edu/home/baez/gr/outline2.html. I think there was another webpage by Baez which was en more similar than this, but I'm not quite sure which one.

The one equation you have left though does give a lot of physical insight, though, I think Baez discusses this as wll.

 

[m4r35n357]

I think there was another webpage by Baez which was en more similar than this, but I'm not quite sure which one.

The one equation you have left though does give a lot of physical insight, though, I think Baez discusses this as wll.

This looks like it might be the article you were referring to . . . ?