Summary: Still More Tensor Identities
Three more identities with gradients, divergences, Laplacians, and cross products. Later today, the fun stuff, special relativity begins!
$\vec{A} \cdot \left(\vec{B} \times \vec{C}\right) = \vec{B} \cdot \left(\vec{C} \times \vec{A}\right)$
$= A_i \epsilon_{ijk} B_j C_k$
As long as we only cycle the indices in the Levi-Civita symbol, we won't cause a sign change, so the above is also equal to
$= A_k \epsilon_{ijk} B_i C_j$
Which we can commute to get
$= B_i \epsilon_{ijk} C_j A_k = \vec{B} \cdot \left(\vec{C} \times \vec{A}\right)$
Done!
$\vec{\nabla} \cdot \left(\vec{\nabla} \times \vec{A} \right) = 0$
$=\partial_i \epsilon_{ijk} \partial_j A_k$
$= 0$
If $i$ and $j$ are equal, then the Levi-Civita evaluates to zero. If they are not equal, then swapping the two indices produces the same mixed partial derivative result, but with a negative sign inserted by swapping indices in the Levi-Civita symbol. These equal but opposite terms all sum to zero giving the advertised result.
$\vec{\nabla} \times \vec{\nabla} \times \vec{A} = \vec{\nabla}\left(\vec{\nabla} \cdot \vec{A}\right) - \nabla^2 \vec{A}$
$= \epsilon_{lmi} \partial_m \epsilon_{ijk} \partial_j A_k$
Don't let all the dels, nablas, whatever you'd like to call them, throw you off. At the end of the day, this is just the same as the $\vec{A} \times \vec{B} \times \vec{C}$ example. Use the two Levi-Civitas to four deltas identity.
$= \epsilon_{lmi} \partial_m \epsilon_{ijk} \partial_j A_k = \epsilon_{ijk} \partial_j \epsilon_{lmk} \partial_l A_m = \left(\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}\right)\partial_j\partial_l A_m$
$=\delta_{il}\delta_{jm}\partial_j\partial_l A_m - \delta_{im}\delta_{jl}\partial_j\partial_l A_m$
$= \partial_i \partial_j A_j - \partial_j \partial_j A_i = \vec{\nabla}\left(\vec{\nabla}\cdot \vec{A}\right) - \nabla^2 \vec{A}$
Picture of the Day:
Hiking trail near Redwood City, CA
Three more identities with gradients, divergences, Laplacians, and cross products. Later today, the fun stuff, special relativity begins!
$\vec{A} \cdot \left(\vec{B} \times \vec{C}\right) = \vec{B} \cdot \left(\vec{C} \times \vec{A}\right)$
$= A_i \epsilon_{ijk} B_j C_k$
As long as we only cycle the indices in the Levi-Civita symbol, we won't cause a sign change, so the above is also equal to
$= A_k \epsilon_{ijk} B_i C_j$
Which we can commute to get
$= B_i \epsilon_{ijk} C_j A_k = \vec{B} \cdot \left(\vec{C} \times \vec{A}\right)$
Done!
$\vec{\nabla} \cdot \left(\vec{\nabla} \times \vec{A} \right) = 0$
$=\partial_i \epsilon_{ijk} \partial_j A_k$
$= 0$
If $i$ and $j$ are equal, then the Levi-Civita evaluates to zero. If they are not equal, then swapping the two indices produces the same mixed partial derivative result, but with a negative sign inserted by swapping indices in the Levi-Civita symbol. These equal but opposite terms all sum to zero giving the advertised result.
$\vec{\nabla} \times \vec{\nabla} \times \vec{A} = \vec{\nabla}\left(\vec{\nabla} \cdot \vec{A}\right) - \nabla^2 \vec{A}$
$= \epsilon_{lmi} \partial_m \epsilon_{ijk} \partial_j A_k$
Don't let all the dels, nablas, whatever you'd like to call them, throw you off. At the end of the day, this is just the same as the $\vec{A} \times \vec{B} \times \vec{C}$ example. Use the two Levi-Civitas to four deltas identity.
$= \epsilon_{lmi} \partial_m \epsilon_{ijk} \partial_j A_k = \epsilon_{ijk} \partial_j \epsilon_{lmk} \partial_l A_m = \left(\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}\right)\partial_j\partial_l A_m$
$=\delta_{il}\delta_{jm}\partial_j\partial_l A_m - \delta_{im}\delta_{jl}\partial_j\partial_l A_m$
$= \partial_i \partial_j A_j - \partial_j \partial_j A_i = \vec{\nabla}\left(\vec{\nabla}\cdot \vec{A}\right) - \nabla^2 \vec{A}$
Picture of the Day:
Hiking trail near Redwood City, CA
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