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Consider the stationary (i.e, independent of time) Stokes equations $$\mathrm{div}~ \sigma = f$$ where $\sigma$ is the stress tensor, $f$ is the external force.

Denote by $M,L,T$ the mass, length, and time, respectively. Then the dimension of $\sigma$ is $ML^{-1}T^{-2}$, which leads to the dimension of $\mathrm{div}~\sigma$ is $ML^{-2}T^{-2}$. However, it contradicts to the dimension of force $f$, which is $MLT^{-2}$.

Can anyone explain me the discrepancy?

1 Answers1

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The units of $\text{div}(\sigma)$ (taken with respect to cartesian components) is $ML^{-2}T^{-2}$.

The units of force $F$ is $MLT^{-2}$, as you said. But note that the units of (volume) force density $f$ is $MLT^{-2} / L^3 = ML^{-2}T^{-2}$

The misunderstanding you're having is that in the equation $\text{div}(\sigma) = f$, the $f$ is not force, but rather the (volume) force density, which is consistent with the dimensional analysis above.

By the way, the same is true also in electrodynamics, where you have a similar equation $\text{div}(\sigma) = f$, (assuming the Poynting vector $S$ doesn't depend on time; i.e the "static" case), where $\sigma$ is the Maxwell stress tensor, and $f = \rho E + J \times B$ is the (volume) electromagnetic force density.

Again, both sides of this equation have units of $ML^{-2}T^{-2}$.

peek-a-boo
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    But usually, people often out of laziness simply refer to $f$ as the force, rather than force density. Another example is that in electrodynamics, the quantity $u = \dfrac{1}{2} \left( \varepsilon_0 |E|^2 + \dfrac{|B|^2}{\mu_0}\right)$ is strictly speaking, the energy density of the electromagnetic field, but people often refer to it as the energy of the electromagentic field. So, the message is: (assuming your algebra is right) trust your dimensional analysis more than the words used. – peek-a-boo Dec 23 '19 at 18:46