# Meaning of components of Maxwell's stress tensor $T_{ij}$

I am learning about Maxwell's stress tensor and what I understood is that the components, say $T_{ij}$ is something like a force parallel to the $j$th-direction acting on the surface with its normal in the $i$th-direction.

I was working on a problem which is to find the net force on the upper hemisphere of a uniformly-charged solid sphere of radius $R$ and charge $Q$.

Calculating the force using Maxwell's stress tensor and symmetry arguments(ignoring $F_x$ and $F_y$), I got

$$F = \int{T_{zz}da_{z} + T_{zx}da_{x} + T_{zy}da_{y}}$$

Then came the confusion. When calculating just the $\int{T_{zz}da_{z}}$ part, I got 0. Which meant $F_z$ arises only from shear forces $T_{zx}da_{x} + T_{zy}da_{y}$. I cannot visualize how this is possible given a $T_{zx}$ acting along $x$-direction give rise to a force in the $z$-direction and same for $T_{zy}$. What did I understand wrongly here?

To find the total force in the z axis you should sum over the z vector embedded in the field's matrix, which is the

The integral should be (for the net force in the z-axis):

$$F_{z} = \sum_{i = 1, j = 3}^{i=3} T_{i}^{j} \cdot \hat{n}dS$$

With

$$T_{ij} = \left( \begin{array}{ccc} xx & yx & zx \\ xy & yy & zy \\ xz & yz & zz \end{array} \right)$$

with n is a unit normal vector and dS is some area element, in the case of a sphere it would be:

$$S = 4 \pi r^{2}$$

$$dS = 8 \pi r dr$$

$$r = \sqrt{x^{2} + y^{2} + z^{2}}$$

• $dS$ isn't $8 \pi r \, dr$ on the surface of the sphere of radius $r$. It should be $dS = r^2 \, \sin{\vartheta} \, d\vartheta \, d\varphi$ instead.
– Cham
Jul 10, 2019 at 14:40