Ngô Quốc Anh

January 20, 2010

RG: Divergence of an (p,s)-tensor

Filed under: Nghiên Cứu Khoa Học, Riemannian geometry — Ngô Quốc Anh @ 14:24

Followed by this topic where we define the divergence of a vector field X as follows

\displaystyle {\rm div} X = dx^i \left( \nabla_{\frac{\partial}{\partial x^i}} X\right)

that is, in local coordinates,

\displaystyle {\rm div} X = \left\langle {\frac{{\partial {X^j}}}{{\partial {x^i}}}\frac{\partial }{{\partial {x^j}}} + {X^j}\Gamma _{ij}^k\frac{\partial }{{\partial {x^k}}},d{x^i}} \right\rangle = \frac{{\partial {X^i}}}{{\partial {x^i}}} + {X^j}\Gamma _{ij}^i

it’s time to take opportunity to define the divergence of an (p,s)-tensor where p\geq 1. We assume \alpha is an (p,0)-tensor, that is,

\displaystyle \alpha= {\alpha _{{i_1}{i_2} \cdots {i_p}}}d{x^{{i_1}}} \otimes d{x^{{i_2}}} \otimes\cdots\otimes d{x^{{i_p}}}.

We then defined the divergence of \alpha to be an (p-1,0)-tensor given as the following

\displaystyle {\rm div}{(\alpha)_{{i_1}{i_2} \cdots {i_{p - 1}}}} = {g^{\alpha \beta }}{\nabla _\alpha }{\alpha _{\beta {i_1}{i_2} \cdots {i_{p - 1}}}} = {\nabla _\alpha }{\alpha _{\alpha {i_1}{i_2} \cdots {i_{p - 1}}}}.

In particular, for a 1-form X, that is an (1,0)-tensor with the form X=X_idx^i in local coordinates, one has

\displaystyle {\rm div}(X) = {g^{\alpha \beta }}{\nabla _\alpha }{X_\beta }.

In general, if \alpha is an (p,s)-tensor with p\geq 1, then the divergence of \alpha to be an (p-1,s)-tensor given as the following

\displaystyle {\rm div}{(\alpha)_{{i_1}{i_2} \cdots {i_{p - 1}}}}^{j_1j_2\cdots j_s} = {g^{\alpha \beta }}{\nabla _\alpha }{\alpha _{\beta {i_1}{i_2} \cdots {i_{p - 1}}}}^{j_1j_2\cdots j_s} = {\nabla _\alpha }{\alpha _{\alpha {i_1}{i_2} \cdots {i_{p - 1}}}}^{j_1j_2\cdots j_s}.

Note that, for a vector field X=X^i\frac{\partial}{\partial x^i} which is an (0,1)-tensor, we need to change to an (1,0)-tensor, an 1-form. To this purpose, we need to change \frac{\partial}{\partial x^i} to g_{ij}dx^j. Therefore,

\displaystyle X = {X^i}\frac{\partial }{{\partial {x^i}}} = {X^i}{g_{ij}}d{x^j}.

Since

\displaystyle {\nabla _\alpha }({X^i}{g_{i\beta }}d{x^\beta }) = \frac{{\partial {X^i}}}{{\partial {x^\alpha }}}{g_{i\beta }}d{x^\beta } + {X^i}{g_{i\beta }}{\nabla _\alpha }d{x^\beta } = \frac{{\partial {X^i}}}{{\partial {x^\alpha }}}{g_{i\beta }}d{x^\beta } - {X^i}{g_{i\beta }}\Gamma _{\alpha i}^\beta d{x^i}

then

\displaystyle {\nabla _\alpha }{X_\beta } = \frac{{\partial {X^i}}}{{\partial {x^\alpha }}}{g_{i\beta }} - {X^i}{g_{ij}}\Gamma _{\alpha \beta }^j.

Thus

\displaystyle {\rm div}(X) = {g^{\alpha \beta }}{\nabla _\alpha }{X_\beta } = {g^{\alpha \beta }}\left( {\frac{{\partial {X^i}}}{{\partial {x^\alpha }}}{g_{i\beta }} - {X^i}{g_{ij}}\Gamma _{\alpha \beta }^j} \right) = \frac{{\partial {X^\alpha }}}{{\partial {x^\alpha }}} - {X^\alpha }\Gamma _{\alpha \beta }^\beta .

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