1
$\begingroup$

The divergence of a vector field in Cartesian coordinate system (CCS) is defined as follows

$$ \mathrm{div}(\mathbf v) = \nabla \cdot \mathbf v = \begin{bmatrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{bmatrix} \cdot \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{bmatrix}^T \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} =\frac{\partial v_1}{\partial x}+ \frac{\partial v_2}{\partial y}+ \frac{\partial v_3}{\partial z} $$

where $\cdot$ denotes the dot product; it was changed to transposition with matrix multiplication. Now lets define some matrix

$$A = \begin{bmatrix} a_{11} && a_{12} && a_{13} \\ a_{21} && a_{22} && a_{23} \\ a_{31} && a_{32} && a_{33} \\ \end{bmatrix} $$

How to calculate divergence of matrix in CCS and how looks its 'dot' product and 'matrix multiplication' form?

$$\mathrm{div}(A) = ?$$

$\endgroup$
12
  • 1
    $\begingroup$ as far I know, the divergence of matrix should give vector as a result $\endgroup$ – Kamil Kiełczewski Dec 30 '19 at 17:04
  • 1
    $\begingroup$ This might help $\endgroup$ – A. Goodier Dec 30 '19 at 17:04
  • $\begingroup$ @mvpq thank for link, but someone told there in comments that forumla given in wiki is WRONG (idex-es are bad) $\endgroup$ – Kamil Kiełczewski Dec 30 '19 at 17:06
  • $\begingroup$ First calculate the gradient $\;g_{ij}=\partial_iv_j\;$ then calculate the divergence by contracting over the leading two indices, i.e. $\;\phi=\delta_{ij}g_{ij}=g_{ii}.\;$ This procedure can be extended to arbitrary tensors, just add more indices on the right, e.g. $$\eqalign{g_{ijklm}&=\partial_iv_{jklm}\\ \phi_{klm}&=\delta_{ij}g_{ijklm}=g_{iiklm}}$$ $\endgroup$ – greg Dec 30 '19 at 18:14
  • $\begingroup$ @greg in your approach (for 3x3 matrix) the div(A) will give scalar (if I read your formulas in good way) - but as far I know, the div(A) should give VECTOR as result $\endgroup$ – Kamil Kiełczewski Dec 30 '19 at 18:21
1
$\begingroup$

In this answer I use $x=x_1, y=x_2, z=x_3$ and Einstein notation. On wikipedia in this article I found following information (in article they use S instead A) for CCS:

$$ \nabla\cdot A = \cfrac{\partial A_{ki}}{\partial x_k}~\mathbf{e}_i = A_{ki,k}~\mathbf{e}_i = \begin{bmatrix} \frac{\partial a_{11}}{\partial x} + \frac{\partial a_{21}}{\partial y} + \frac{\partial a_{31}}{\partial z} \\ \frac{\partial a_{12}}{\partial x} + \frac{\partial a_{22}}{\partial y} + \frac{\partial a_{32}}{\partial z} \\ \frac{\partial a_{13}}{\partial x} + \frac{\partial a_{23}}{\partial y} + \frac{\partial a_{33}}{\partial z} \\ \end{bmatrix} $$

The result is contravariant (column) vector. But in this article is mention that $\mathrm{div}(A) \neq \nabla\cdot A$ and

$$ \mathrm{div}(A) = \nabla\cdot A^T = \cfrac{\partial A_{ik}}{\partial x_k}~\mathbf{e}_i = A_{ik,k}~\mathbf{e}_i = \begin{bmatrix} \frac{\partial a_{11}}{\partial x} + \frac{\partial a_{12}}{\partial y} + \frac{\partial a_{13}}{\partial z} \\ \frac{\partial a_{21}}{\partial x} + \frac{\partial a_{22}}{\partial y} + \frac{\partial a_{23}}{\partial z} \\ \frac{\partial a_{31}}{\partial x} + \frac{\partial a_{32}}{\partial y} + \frac{\partial a_{33}}{\partial z} \\ \end{bmatrix} $$

When A is symetric: $a_{ij}=a_{ji}$ then $\mathrm{div}(A) = \nabla\cdot A$

Wiki also mention that some authors use alternative definition: $\nabla\cdot A = \cfrac{\partial A_{ik}}{\partial x_k}~\mathbf{e}_i $ probably only for case when A is symmetric (for which that alternative definition is equal to original). However alternative definition is NOT compatible with general curvilinear definition which I found on wiki too:

$$ \nabla\cdot A = \left(\cfrac{\partial A_{ki}}{\partial x_k}- A_{li}~\Gamma_{kk}^l - A_{kl}~\Gamma_{ki}^l\right)~\mathbf{g}^i $$

$\endgroup$
1
  • $\begingroup$ currently I don't know what is exact definition of: $\mathbf{g}^i$ - probably it will give something like $\mathbf{e}_i$ (?) $\endgroup$ – Kamil Kiełczewski Dec 31 '19 at 10:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.