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| Hermann Weyl |
\nabla_\mu F^\mu=\frac{1}{\sqrt{\left|g\right|}}\frac{\partial}{\partial x^\mu}\left(\sqrt{\left|g\right|}F^\mu\right)
$$It immediately solved an outstanding problem from a year ago that I had with the Laplacian. While proving it I find three other interesting formulas:
Jacobi's formula is$$
\partial_kQ={\widetilde{Q}}^{ji}\partial_kQ_{ij}
$$where ##Q_{ij}## is a tensor, ##Q## is its determinant and what might be new is ##{\widetilde{Q}}^{ij}## which is the cofactor of ##Q_{ij}## and thus as a matrix, ##{\widetilde{Q}}^{ji}## is the transpose of the cofactors of ##Q_{ij}##. The cofactor of a component ##Q_{ij}## is the determinant of the matrix made by removing row ##i##, column ##j## then multiplying by ##\left(-1\right)^{i+j}##. The important thing is that the transpose of the cofactors is called the adjugate, so ##{\widetilde{Q}}^{ji}## is the adjugate and the inverse of a matrix is the adjugate divided by its determinant. We write that:$$
Q^{ij}=\frac{1}{Q}{\widetilde{Q}}^{ji}
$$Finally on the way to proving the Voss-Weyl formula we find the marvellously symmetrical $$
Q^{-1}\partial_kQ=Q^{ij}\partial_kQ_{ij}
$$where ##Q^{-1}## is the determinant of the inverse of a rank 2 tensor ##Q_{ij}## which has determinant ##Q##. It is also well known that ##Q^{-1}=\frac{1}{Q}##.
\partial_kQ={\widetilde{Q}}^{ji}\partial_kQ_{ij}
$$where ##Q_{ij}## is a tensor, ##Q## is its determinant and what might be new is ##{\widetilde{Q}}^{ij}## which is the cofactor of ##Q_{ij}## and thus as a matrix, ##{\widetilde{Q}}^{ji}## is the transpose of the cofactors of ##Q_{ij}##. The cofactor of a component ##Q_{ij}## is the determinant of the matrix made by removing row ##i##, column ##j## then multiplying by ##\left(-1\right)^{i+j}##. The important thing is that the transpose of the cofactors is called the adjugate, so ##{\widetilde{Q}}^{ji}## is the adjugate and the inverse of a matrix is the adjugate divided by its determinant. We write that:$$
Q^{ij}=\frac{1}{Q}{\widetilde{Q}}^{ji}
$$Finally on the way to proving the Voss-Weyl formula we find the marvellously symmetrical $$
Q^{-1}\partial_kQ=Q^{ij}\partial_kQ_{ij}
$$where ##Q^{-1}## is the determinant of the inverse of a rank 2 tensor ##Q_{ij}## which has determinant ##Q##. It is also well known that ##Q^{-1}=\frac{1}{Q}##.
I must thank David A. Clarke of Saint Mary’s University, Halifax NS, Canada for his guidance.
Proof here: Commentary 4.1 Voss-Weyl formula and more.pdf (2 pages)
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