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

Physics Asked by Spaceman Spiff on February 10, 2021

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?

One Answer

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:

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

$$dS = 8 pi r dr $$

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

Answered by user97261 on February 10, 2021

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