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Let's consider an arbitrary scalar field. If I act twice on the scalar field with a gradient operator, I will obtain second-order tensor. If I will take a trace of this tensor, I will obtain another scalar field. Is the resulting scalar field invariant under coordinate transformation? For example, if I will do the same operation in spherical coordinates, do I obtain the same result?

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Yes! The trace of a tensor is one of its invarients, so the trace will remain constant no matter which coordinate representation you use.

Here's a proof from the mathematics stack exchange that the characteristic polynomial of a matrix is independent from your choice of basis.

Hope this helps.

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  • $\begingroup$ Isn't it proofed just for linear transformations? transformation from cartesian to spherical coordinates is non-linear. $\endgroup$ – Tomek Aug 10 '19 at 23:08
  • $\begingroup$ You can't take the gradient of a gradient. You can however take the divergence of a gradient which gives you the Laplacian. A tensor can be viewed as a linear or multi-linear mapping. $\endgroup$ – Cinaed Simson Aug 11 '19 at 4:32
  • $\begingroup$ Thus what is that? math.stackexchange.com/questions/156880/… $\endgroup$ – Tomek Aug 11 '19 at 9:40

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