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I learned the proof of the fact that 3-D laplacian is invariant under all rigid motions in space in Strauss's Partial Differential Equations:

Any rotation in three dimensions is given by ${\bf x'}=B{\bf x},$ where $B$ is an orthogonal matrix. The laplacian is $\Delta u=\sum_{i=1}^3 u_{ii}=\sum_{i,j=1}^3\delta_{ij}u_{ij}$ where the subscripts on $u$ denote partial derivatives. Therefore, $ \Delta u=\sum_{k,l}\Big(\sum_{i,j}b_{ki}\delta_{ij}b_{lj}\Big)u_{k'l'} =\sum_{k,l}\delta_{kl}u_{k'l'} =\sum_{k}u_{k'k'}. $

Added: I believe the chain rule is needs here, but I don't see how it is applied here.

Here is my question:

How can I get the first equality, i.e. $\Delta u=\sum_{k,l}\Big(\sum_{i,j}b_{ki}\delta_{ij}b_{lj}\Big)u_{k'l'}?$

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    It looks to me that you are trying to show something like $B^{-1} \triangle ~ B = \triangle $.2011-08-01

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I strongly dislike the use of undecorated subscripts for partial derivatives, so I will write $\partial_j$ for the partial differentiation operator $\partial / \partial x_j$. For convenience I will use the summation convention for repeated indices.

Firstly, we have the change-of-coordinates equation \tilde{x}_{i'} = B_{i' j} x_j. The chain rule states \frac{\partial}{\partial x_j} = \frac{\partial \tilde{x}_{i'}}{\partial x_j} \frac{\partial}{\partial \tilde{x}_{i'}} so, writing \tilde{\partial}_{i'} for the operator \partial / \partial \tilde{x}_{i'}, putting these together, we have \partial_j = B_{i' j} \tilde{\partial}_{i'}. The Laplacian is defined by $\nabla^2 = \delta_{i j} \partial_i \partial_j$, so substituting the previous equation in we obtain \nabla^2 = \delta_{i j} B_{i k'} B_{j l'} \tilde{\partial}_{k'} \tilde{\partial}_{l'}. By orthogonality, \delta_{i j} B_{i k'} B_{j l'} = B_{j k'} B_{j l'} = \delta_{k' l'}, so we are done.

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    @Jack: It's the bit starting at ‘substituting the previous equation in’. It is literally just the chain rule.2011-08-02