Timeline for The momentum constraints in the ADM formulation of general relativity
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7 events
| when toggle format | what | by | license | comment | |
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| Aug 18, 2015 at 19:08 | comment | added | o0BlueBeast0o | @BenCrowell yes you are absolutely correct. My mistake. | |
| Aug 17, 2015 at 21:05 | comment | added | user21349 | There is a standard way of extending the definition of the covariant derivative to include tensor densities; see the end of this section of the WP article: en.wikipedia.org/wiki/… . They actually address the derivative we're talking about as an example of a tensor density. Maybe another way of getting at it is the following. I assume that this generalization obeys the product rule, obeys the chain rule, and is metric compatible. So the derivative we're talking about can be evaluated down to derivatives of components of g, and these vanish. | |
| Aug 17, 2015 at 20:54 | comment | added | user21349 | I think the answer is right, but the comment is not. In the case of a (0,0) tensor, the covariant derivative is the same as an ordinary partial derivative. Therefore if the square root of the determinant of g were a (0,0) tensor, then its covariant derivative would be the same as its ordinary partial derivative, which does not in general vanish. | |
| Aug 17, 2015 at 15:44 | comment | added | o0BlueBeast0o | It is a function on the manifold though, a covariant derivative of which is defined. A function will just be a rank (0,0) tensor if you will. The covariant derivative of functions are incorporated in the definition of a connections as can be found in most text books on (pseudo)Riemannian geometry. | |
| Aug 17, 2015 at 15:14 | comment | added | Pun Huo | The square root of the determinant of g is not a tensor. So what do you mean by the covariant derivative of the square root of the determinant of g? Would you clarify that? | |
| Aug 17, 2015 at 14:04 | review | First posts | |||
| Aug 17, 2015 at 14:16 | |||||
| Aug 17, 2015 at 13:53 | history | answered | o0BlueBeast0o | CC BY-SA 3.0 |