Suppose that the space-time has a time function. Let $g_{ij}$ be the Riemannian metrics of the time slices, and $K_{ij}$ be the second fundamental forms. It is by Codazzi equation that $$ D^{i}(K_{ij}-Hg_{ij})=0 $$ where $H=\text{tr}K$ and $D$ is the covariant derivative of $g$. It seems that, in the ADM formulation of general relativity, this equation is often re-expressed as the form $$ D^{i}p_{ij}=0 $$ where $p_{ij}=(\det g)^{1/2}(K_{ij}-Hg_{ij})$, and called the momentum constraint. But it seems to me that the above ``momentum constraint'' does not coincides with the Codazzi equation due to the extra factor $(\det g)^{1/2}$. How to explain the momentum constraint in a right way? Another question: It is obvious that $p_{ij}$ is not a tensor. Is it legal to apply the covariant derivative to $p_{ij}$?
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$\begingroup$ When I check both of the formulas on computer, I have found that the first is right and the second is wrong, according to my computation it is not constraint of GR. $\endgroup$– user95673Commented Jul 30, 2016 at 11:24
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2 Answers
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The connection is metric compatible. In particular the covariant derivative of the square root of the determinant of g is zero. So the above equation does coincide.
answered Aug 17, 2015 at 13:53
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1$\begingroup$ 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? $\endgroup$– Pun HuoCommented Aug 17, 2015 at 15:14
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1$\begingroup$ 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. $\endgroup$ Commented Aug 17, 2015 at 15:44
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1$\begingroup$ 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. $\endgroup$– user21349Commented Aug 17, 2015 at 20:54
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2$\begingroup$ 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. $\endgroup$– user21349Commented Aug 17, 2015 at 21:05
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$\begingroup$ @BenCrowell yes you are absolutely correct. My mistake. $\endgroup$ Commented Aug 18, 2015 at 19:08
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$D_i\gamma\ne\partial_i \gamma$, because $\gamma$ is not a scalar, but density of a scalar. Thus, all the expressions for momentum constraint are equivalent.
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$\begingroup$ Please: don't use answer boxes to make comments. (And please use proper LaTeX.) $\endgroup$ Commented Jul 30, 2016 at 19:33