# Calculating the Divergence of a Tensor

I am working through a fluid dynamics paper and came across this equation: $$\frac{\partial \vec{v}}{\partial t} + \vec{v}\cdot\nabla\vec{v}=\nabla\cdot T - \frac{1}{\rho}\nabla \phi\tag1$$ where T is a tensor defined by $$T_{ij} = -\frac{p}{\rho}\delta_{ij} + \nu\left(\frac{\partial v_i}{\partial x_j}+\frac{\partial v_j}{\partial x_i}\right) - \beta\gamma\frac{\partial c}{\partial x_i}\frac{\partial c}{\partial x_j}$$ The paper is concerned with 2D so $$\vec{x}=(x,z)$$ and $$\vec{v}=(u,w)$$. I started by writing out the individual components of the tensor $$T$$ and could pretty easily see that it is symmetric (not sure if this matters). I wanted to then write out the component-wise equations of $$(1)$$ but to do that I needed to expand $$\nabla\cdot T$$ but I honestly have no idea how to do that. A quick google search says that it should be: $$\nabla \cdot T = T_{ik,i}\cdot e_k$$ This doesn't really seem to make any sense to me though. For instance, in my case (for 2D) what are the values of $$i$$ and $$k$$ that I should be ranging over? Also what is the meaning of that comma in the index for the tensor - it is not anywhere included in the definition of $$T$$. An explanation of how to generally find the divergence of a tensor would be much appreciated. Also perhaps a simpler example worked out.

• The comma means partial derivative
– John
Commented Sep 7, 2021 at 21:34
• So $T_{ik,i}$ is $\frac{\partial T_{ik}}{\partial x_i}$? Commented Sep 7, 2021 at 22:07
• Yes, that is correct
– John
Commented Sep 7, 2021 at 23:14

The notation you found through google does not really make sense with the dot, but is ok without the dot. It means (in dimension $$3$$) $$\nabla\cdot T = T_{ik,i}\,e_k = \sum_{i=1}^3 \sum_{k=1}^3 \partial_{x_i}(T_{ik})\, e_k,$$ by the convention of summation over repeated indices. Hence, it is the vector with $$k$$-th coordinate $$(\nabla\cdot T)_k = \sum_{i=1}^3 \partial_{x_i}(T_{ik}).$$