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?


1 Answer 1


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):

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


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

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

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

$$dS = 8 \pi r dr $$

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

  • $\begingroup$ $dS$ 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. $\endgroup$
    – Cham
    Jul 10, 2019 at 14:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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