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I encounter a problem in fluid dynamics that requires the Laplacian of Green's function in spherical coordinate. I am wondering can we get an explicit form of laplacian of a second-order tensor in spherical coordinate? Here is the fact I know: the Laplacian of a scalar and a vector in spherical coordinates is already shown here. How about the second-order tensor?

Given a tensor: $$\mathbf{T}=T_{rr}\mathbf{e_{r}}\mathbf{e_{r}}+T\mathbf{e_{r}}\mathbf{e_{\theta}}+T\mathbf{e_{r}}\mathbf{e_{\phi}}+T\mathbf{e_{\theta}}\mathbf{e_{r}}+T\mathbf{e_{\theta}}\mathbf{e_{\theta}}+T\mathbf{e_{\theta}}\mathbf{e_{\phi}}+T\mathbf{e_{\phi}}\mathbf{e_{r}}+T\mathbf{e_{\phi}}\mathbf{e_{\theta}}+T\mathbf{e_{\phi}}\mathbf{e_{\phi}}$$ what should be $\nabla^{2}\mathbf{T}$ looks like explicitly in all the nine terms?

I know it may be long and tedious to show the explicit format but it is really useful. many thanks!

Arctic Char
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Wz S
  • 381
  • is there any explicit form rather than this compact form? – Wz S May 11 '21 at 20:38
  • What is keeping you from writing it out yourself given the tensor form? What roadblock do you hit? – vb628 May 11 '21 at 20:45
  • I am not sure about the derivative of the Christoffel symbol. – Wz S May 11 '21 at 22:39
  • It may be easier to write the tensor out as its tensor product for example $$ T = \dfrac{\partial}{\partial x^k} \otimes \dfrac{\partial}{\partial x^\ell} : T^{k\ell} \text{ .}$$ If you do this when $T$ depends on Christoffel symbols, it becomes clear how to take the correct derivative (you're just taking partial derivatives and adding in new Christoffel symbols). If you are doing more than taking simple partial derivatives of the Christoffel symbols then something is going wrong. – vb628 May 11 '21 at 23:28

1 Answers1

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As per my own answer to my question about the tensor Laplacian, the formula for the tensor Laplacian is

$$(\boldsymbol{\triangle}\mathbf{T})^{i_1\dots i_r}{}_{j_1\dots j_s}=g^{kl}(\nabla(\nabla\mathbf{T}))^{i_1\dots i_r}{}_{j_1\dots j_s~kl}$$

By second order tensor I assume you mean a $(0,2)$ tensor in which case we get

$$(\boldsymbol{\triangle}\mathbf{T})_{ij}=g^{kl}(\nabla(\nabla\mathbf{T}))_{ij~kl}$$

NB: In the following I am using the so called mathematics convention for spherical coordinates, i.e $x=r\cos\theta\sin\phi$, etc.

There are two relevent formulae you'll need to compute the tensor Laplacian. The first of course is the inverse metric (written in the proper $(2,0)$ contravariant form)

$$\mathbf{g}^{-1} =\begin{bmatrix} \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}\\ \begin{bmatrix} 0\\ (r^2\sin^2\phi)^{-1}\\ 0 \end{bmatrix}\\ \begin{bmatrix} 0\\ 0\\ 1/r^2 \end{bmatrix} \end{bmatrix}$$ And the covariant derivative, which is going to be applied twice, first for a $(0,2)$ tensor

$$(\nabla \mathbf{T})_{ij~k}=\partial_kT_{ij}-\Gamma^a_{ik}T_{a j}-\Gamma^{a}_{jk}T_{ia}$$

And then for a $(0,3)$ tensor:

$$(\nabla(\nabla\mathbf{T}))_{ijk~l}=\partial_lT_{ijk}-\Gamma^a_{il}T_{ajk}-\Gamma^{a}_{jl}T_{iak}-\Gamma^{a}_{kl}T_{ija}$$

The computations are pretty heavy, so I'm going to use Mathematica to assist me.

I'll paste in the Mathematica code below.

coords = {r, \[Theta], \[Phi]};
n = Length[coords];
p = {r*Cos[\[Theta]] Sin[\[Phi]], r*Sin[\[Theta]] Sin[\[Phi]], 
   r*Cos[\[Phi]]};
g = Simplify[
   Table[D[p, coords[[i]]].D[p, coords[[j]]], {i, 1, n}, {j, 1, n}]];
inverseg = Simplify[Inverse[g]];
\[CapitalGamma] = 
  Simplify[Table[
    inverseg[[k, k]]*
     D[p, coords[[k]]].D[D[p, coords[[i]]], coords[[j]]], {k, 1, 
     n}, {i, 1, n}, {j, 1, n}]];
T = {{Trr[r, \[Theta], \[Phi]], Tr\[Theta][r, \[Theta], \[Phi]], 
    Tr\[Phi][r, \[Theta], \[Phi]]}, {T\[Theta]r[r, \[Theta], \[Phi]], 
    T\[Theta]\[Theta][r, \[Theta], \[Phi]], 
    T\[Theta]\[Phi][r, \[Theta], \[Phi]]}, {T\[Phi]r[
     r, \[Theta], \[Phi]], T\[Phi]\[Theta][r, \[Theta], \[Phi]], 
    T\[Phi]\[Phi][r, \[Theta], \[Phi]]}};
Clear[CovariantDerivative]
CovariantDerivativeSecondOrder[T_] := 
 Table[D[T[[i, j]], coords[[k]]] - 
   Sum[\[CapitalGamma][[a, i, k]] T[[a, j]], {a, 1, n}] - 
   Sum[\[CapitalGamma][[a, j, k]] T[[i, a]], {a, 1, n}], {i, 1, 
   n}, {j, 1, n}, {k, 1, n}]
CovariantDerivativeThirdOrder[T_] := 
 Table[D[T[[i, j, k]], coords[[l]]] - 
   Sum[\[CapitalGamma][[a, i, l]] T[[a, j, k]], {a, 1, n}] - 
   Sum[\[CapitalGamma][[a, j, l]] T[[i, a, k]], {a, 1, n}] - 
   Sum[\[CapitalGamma][[a, k, l]] T[[i, j, a]], {a, 1, n}], {i, 1, 
   n}, {j, 1, n}, {k, 1, n}, {l, 1, n}]
Simplify[Table[
  Sum[inverseg[[k, l]]*
    CovariantDerivativeThirdOrder[
      CovariantDerivativeSecondOrder[T]][[i, j, k, l]], {k, 1, n}, {l,
     1, n}], {i, 1, n}, {j, 1, n}]]

Here is a screenshot. I've removed the redundant [r,θ,ϕ] all over the place. Note that in Mathematica, the (a,b,c) superscript means an a-fold partial derivative in the first argument of the function and so on.

Finally

This would take me quite a while to typeset, and I am unfortunately very busy at the moment so I really don't have time. I will get to it hopefully in the next couple of days.

K.defaoite
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  • thank you for your detailed answer which is really helpful. i am wondering why the final answer does not have the partial derivative of elements in T? for example, I expect there should be some term like \frac{\partial^{2}T_{rr}}{\partial r^{2}}, but I do not see it. – Wz S May 12 '21 at 19:42
  • @WzS Goodness me, good catch. I should have made the components of $T$ functions of $r,\theta,\phi$. Please disregard this answer as it is now. I will correct it. – K.defaoite May 13 '21 at 00:17
  • @Wzs I pasted the updated result from Mathematica. I will typeset it at some point in the next few days. – K.defaoite May 13 '21 at 00:25
  • really thank you. I know this is kind of tedious. so really thank you for your patience. But I still wondering there should be some second-order derivatives rather than just first-order shown in the result now, what do you think? – Wz S May 13 '21 at 01:18
  • @WzS You're right, that is... curious. I'll investigate. – K.defaoite May 13 '21 at 01:27
  • @WzS I found the error. I forgot the double square bracket around coords. – K.defaoite May 13 '21 at 01:35
  • @WzS I uploaded the new result. I think finally this is right! I'm really sorry about the several failed attempts. I'll typeset this sometime in the near future, but for now the screenshot is available. It contains second derivatives as expected. – K.defaoite May 13 '21 at 01:40
  • thanks, I think now it is in a good form. you do not have to type it, you have already done enough, I am going to type it myself and am glad share to you if you need it. Again, really appreciate it. just want to make sure the notation, the T_{rr}^(0,1,0) means \frac{\partcial{T_{rr}}}{\partcial\theta}, right? – Wz S May 13 '21 at 02:58
  • @WzS Yes, that's right. – K.defaoite May 13 '21 at 11:56