The opening sentence of section 2.8 starts "Tensors possess a compelling beauty and simplicity". That fills me with fear.
We are told that the Levi-Civita symbol, which is not a tensor, is defined as$${\widetilde{\epsilon }}_{{\mu }_1{\mu }_2\dots {\mu }_n}=\left\{ \begin{array}{ll}
+1 & \mathrm{if}\mathrm{\ }{\mu }_1{\mu }_2\dots {\mu }_n\mathrm{\ is\ an\ even\ permitation\ of}\ 01..(n-1)\ \\
-1 & \mathrm{if\ }{\mu }_1{\mu }_2\dots {\mu }_n\mathrm{\ is\ an\ odd\ permitation\ of}\ 01..\left(n-1\right) \\
0 & \mathrm{otherwise} \end{array}
\right.$$and (Carroll's (2.66)) that given any ##n\times n## matrix ##M^{\mu }_{\ \ \ \mu '\ }##, the determinant ##\left|M\right|## obeys $${\widetilde{\epsilon }}_{{\mu '}_1{\mu '}_2\dots {\mu '}_n}\left|M\right|={\widetilde{\epsilon }}_{{\mu }_1{\mu }_2\dots {\mu }_n}M^{{\mu }_1}_{\ \ \ \ \ {\mu '}_1}M^{{\mu }_2}_{\ \ \ \ \ {\mu '}_2}\dots M^{{\mu }_n}_{\ \ \ \ \ {\mu '}_n}$$We are invited to check this for 2×2 and 3×3 matrices which we do and do discover some beauty comparing traditional methods for calculating the determinant of a matrix using cofactors or using the Levi-Civita symbol.
Setting ##{\mu '}_1{\mu '}_2\dots {\mu '}_n=01\dots (n-1)## we get the even simpler$$\left|M\right|={\widetilde{\epsilon }}_{{\mu }_1{\mu }_2\dots {\mu }_n}M^{{\mu }_1}_{\ \ \ \ \ 0}M^{{\mu }_2}_1\dots M^{{\mu }_n}_{\ \ \ \ \ (n-1)}$$Other combinations of ##{\mu '}_1{\mu '}_2\dots {\mu '}_n## either give ##0=0## or the same as ##01\dots (n-1)## or other cofactor expansions of the determinant.
However if the equation had been simplified the next equation (2.67) would not work and (2.67) is the punch line.
We also prove that the determinant of the metric under a coordinate transformation is given by
$$g\left(x^{\mu '}\right)={\left|\frac{\partial x^{\mu '}}{\partial x^{\mu }}\right|}^{-2}g\left(x^{\mu }\right)$$ See the details at
We are told that the Levi-Civita symbol, which is not a tensor, is defined as$${\widetilde{\epsilon }}_{{\mu }_1{\mu }_2\dots {\mu }_n}=\left\{ \begin{array}{ll}
+1 & \mathrm{if}\mathrm{\ }{\mu }_1{\mu }_2\dots {\mu }_n\mathrm{\ is\ an\ even\ permitation\ of}\ 01..(n-1)\ \\
-1 & \mathrm{if\ }{\mu }_1{\mu }_2\dots {\mu }_n\mathrm{\ is\ an\ odd\ permitation\ of}\ 01..\left(n-1\right) \\
0 & \mathrm{otherwise} \end{array}
\right.$$and (Carroll's (2.66)) that given any ##n\times n## matrix ##M^{\mu }_{\ \ \ \mu '\ }##, the determinant ##\left|M\right|## obeys $${\widetilde{\epsilon }}_{{\mu '}_1{\mu '}_2\dots {\mu '}_n}\left|M\right|={\widetilde{\epsilon }}_{{\mu }_1{\mu }_2\dots {\mu }_n}M^{{\mu }_1}_{\ \ \ \ \ {\mu '}_1}M^{{\mu }_2}_{\ \ \ \ \ {\mu '}_2}\dots M^{{\mu }_n}_{\ \ \ \ \ {\mu '}_n}$$We are invited to check this for 2×2 and 3×3 matrices which we do and do discover some beauty comparing traditional methods for calculating the determinant of a matrix using cofactors or using the Levi-Civita symbol.
Setting ##{\mu '}_1{\mu '}_2\dots {\mu '}_n=01\dots (n-1)## we get the even simpler$$\left|M\right|={\widetilde{\epsilon }}_{{\mu }_1{\mu }_2\dots {\mu }_n}M^{{\mu }_1}_{\ \ \ \ \ 0}M^{{\mu }_2}_1\dots M^{{\mu }_n}_{\ \ \ \ \ (n-1)}$$Other combinations of ##{\mu '}_1{\mu '}_2\dots {\mu '}_n## either give ##0=0## or the same as ##01\dots (n-1)## or other cofactor expansions of the determinant.
However if the equation had been simplified the next equation (2.67) would not work and (2.67) is the punch line.
We also prove that the determinant of the metric under a coordinate transformation is given by
$$g\left(x^{\mu '}\right)={\left|\frac{\partial x^{\mu '}}{\partial x^{\mu }}\right|}^{-2}g\left(x^{\mu }\right)$$ See the details at
Commentary 2.8 Tensor Densities.pdf (6 pages)
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