So, I just went through the intuition behind the Divergence Theorem, and also the proof, so I understand why it is true. Still, though, I have some problems understanding the units involved. To be clear, I am thinking of this Divergence Theorem:
$\iiint_v (\nabla \cdot \vec{v})d\tau = \iint_S \vec{v} \cdot d\vec{a}$,
where $\nabla \cdot \vec{v} = \text{div } \vec{v}$, $d\tau = dxdydz$ in Cartesian coordinates, and $d\vec{a}$ is an infnitesimal area like $dxdy$, $dydz$, etc. multiplied by the normal unit vector to the surface, $\hat{n}$.
Let's assume that the components of $\vec{v}$ are just numbers, and have no units, and $dx$, $dy$, $dz$ all have units of length. In that case, it would seem to be that the triple integral on the leftside would have units of volume, while the right hand side would have units of area.
Again, I've seen the proof of the Divergence Theorem, so I know something in my thought process here must be wrong. Any idea where I'm going wrong? Thanks a lot in advance!
Best Answer
Note that \begin{align} \nabla \cdot v = \dfrac{\partial v_x}{\partial x} + \dfrac{\partial v_y}{\partial y} + \dfrac{\partial v_z}{\partial z} \end{align} And in each partial derivative, since $x,y,z$ have units of length, the units of the divergence of $v$ is $\dfrac{1}{\text{length}}$. Multiplying by the volume element $d\tau$ implies that $(\nabla \cdot v)\, d\tau$ has units of $(\text{length})^2 = \text{area}$. So, there is no contradiction anywhere, even the LHS has units of area.
So, the key part you were missing is that you didn't take into account the units of divergence. You incorrectly assumed that just because $v$ is dimensionless, $\nabla \cdot v$ is also dimensionless. This is incorrect, because the differentiation process of $\dfrac{\partial}{\partial x}$ etc. introduces an extra factor of $\dfrac{1}{\text{length}}$.
Now you should convince yourself that in general, both sides of the equation \begin{align} \iiint_{\Omega} (\nabla \cdot v) \, d\tau = \iint_{\partial \Omega} (v \cdot n) \, da \end{align} have units equal to $(\text{(units of $v$)} \cdot \text{area})$.