There are several resources, including this question as well as for example problem 5-1 in Kovetz "The Principles of Electromagnetic Theory" which state that the time derivative of a flux can be given:
$$
\frac{d}{dt} \int_{S(t)} \vec{A}\cdot \vec{n}ds = \int_{S(t)} (\frac{\partial \vec{A}}{\partial t} + (\nabla \cdot \vec{A})\vec{v} – \nabla \times (\vec{v} \times \vec{A})) \cdot \vec{n}ds.
$$
And I have attempted to try and reproduce this result using a method similar to that of the Reynold's Transport Theorem Wikipedia Article.
But I end up with some extra "shear" terms, and I would love to understand where I am going wrong.
The derivation goes as follows. I want to examine the integral of the flux, and so we can rewrite our integral:
$$
\int_{S(t)} \vec{A}\cdot\vec{n}ds = \int_{S(t)} \vec{A}(x, y, z, t) \cdot \vec{n} ds = \int_{S(t)} A_x(x(y, z, t), y, z, t) dydz – A_y(x, y(x, z, t), z, t) dxdz + A_z(x, y, z(x, y, t), t) dxdy
$$
We can look at this sum of three integrals as three separate integrals, and so we can begin by examining the first one. We first reparameterize from the time-dependent surface of y,z to r,s to get: $$\int_{S(t)} A_x(x(y, z, t), y, z) dydz = \int_{S_0} A_x(x(r, s, t), y(r, s, t), z(r, s, t), t) \lvert\frac{\partial(y, z)}{\partial(r, s)}\rvert drds$$ where $\frac{\partial(y, z)}{\partial(r, s)}$ is the Jacobian of the transformation from y, z to r, s and is a function of time.
We can now differentiate this first term with respect to time to get (employing the product rule and the chain rule for determinants of matrices:
$$\frac{d}{dt}\int_{S(t)} A_x(x(r, s, t), y(r, s, t), z(r, s, t), t)\lvert\frac{\partial(y, z)}{\partial(r, s)}\rvert drds =\\ \int_{S_0} \frac{\partial A_x}{\partial t} + \nabla A_x \cdot \vec{v} + A_xtr(\frac{\partial(r, s)}{\partial(y, z)}\frac{\partial(v_y, v_z)}{\partial(r, s)})\lvert\frac{\partial(y, z)}{ \partial(r, s)}\rvert drds$$
Now at a given time, $v_y, v_z$ are only functions of r and s, so we can simplify the trace of the determinant to be just $tr(\frac{\partial(r, s)}{\partial(y, z)}\frac{\partial(v_y, v_z)}{\partial (r, s)}) = \frac{\partial v_y }{\partial y} + \frac{\partial v_z}{\partial z}$
Then, bringing everything together we have $$\frac{d}{dt}\int_{S_0} A_x(x(r, s, t), y(r, s, t), z(r, s, t), t)drds = \int_{S_0} (\frac{\partial A_x}{\partial t} + \nabla A_x \cdot \vec{v} + A_x\frac{\partial v_y }{\partial y} + A_x \frac{\partial v_z}{\partial z})\lvert\frac{\partial(y, z)}{\partial(r, s)}\rvert drds$$
Now, we can identify some like terms, and adopting the notation $\frac{\partial}{\partial x} = \partial_x$, we have by adding and subtracting $\partial_y (A_yv_x)$ and $\partial_z (A_zv_x)$:
$$\frac{d}{dt}\int_{S(t)} A_x(x(y, z, t), y, z, t)dydz = \int_{S(t)} \frac{\partial \vec{A_x}}{\partial t} + (\nabla \cdot \vec{A})\vec{v}_x – (\nabla \times (\vec{v} \times \vec{A})))_x + \color{red}{A_y \partial_yv_x + A_z \partial_zv_x}dydz$$
Basically, I am curious if I am doing something wrong, or is it necessary the cumulative shear terms cancel out? (It doesn't seem like that is the case, but open to it). In their derivation Kovetz does not take velocity shear into account, but I imagine these effects would be taken into account in other derivations. Best and thanks in advance!
Best Answer
I figured out the problem with my derivation! This will benefit from using index notation. We begin with a slight formalization of the problem to be able to apply ideas from continuum mechanics. Consider a vector field $\mathbf{A}(x, y, z, t)$ which is a function of space and time, and a surface $\mathcal{S}$ such that there exists some reference configuration $S_0$ and a differentiable, invertible map $\varphi : S_0 \times \mathbb{R} \rightarrow \mathcal{S} \times \mathbb{R}$ such that $\varphi(X, Y, Z, t) \in \mathcal{S}$ $\forall X, Y, Z \in S_0$ and our surface $\mathcal{S} \times \{t\} = \varphi(S_0, t)$. In English, this is just a surface that has a smooth enough trajectory over time. With this definition, we are also able to define what we mean by $\mathbf{v}_i = \frac{\partial\varphi_i}{\partial t}$. Here Roman letters correspond to spatial indices and derivatives. One final piece of convention is that capital letters will be used for coordinates and indices in our reference $S_0$ surface, and lower case will correspond to our spatial configuration or our $\mathcal{S}$ surface.
So armed we are ready to look at our integral again. We want to find $$\frac{d}{dt}\int_\mathcal{S} \mathbf{A} \cdot \mathbf{n} da$$ where $\mathbf{n}$ is the normal to the surface. In order to make sense of a time derivative of an integral over a surface moving in time, we first change coordinates to our reference configuration. To do this, we first rewrite our integral using index notation. We have then for some parameterization of $\mathcal{S}$ by coordinates $r$ and $s$, and use the differential transformation rule $dr^i = \frac{\partial \phi^i}{\partial X^A}dR^A$. \begin{align*} \frac{d}{dt}\int_\mathcal{S} \mathbf{A} \cdot \mathbf{n} da &= \frac{d}{dt}\int_\mathcal{S} \epsilon_{ijk} \mathbf{A}^k dr^ids^j \\ &= \frac{d}{dt}\int_{S_0} \epsilon_{ijk} \mathbf{A}^k \frac{\partial \phi^i}{\partial X^A}\frac{\partial \phi^j}{\partial X^B}dR^AdS^B \end{align*} This is where the derivation differs from that above. I am still not sure exactly where the error is, but likely comes from the fact that when we write our transformation, $J\frac{\partial (\varphi^{-1})^C}{\partial x^I}$ is not the same as the 2-d Jacobian, but captures additional information about changes in the direction of the normal. First, notice that our integrand $\epsilon_{ijk} \mathbf{A}^k \frac{\partial \phi^i}{\partial X^A}\frac{\partial \phi^j}{\partial X^B}$ is antisymmetric in the indices $A$ and $B$. Using the $\epsilon-\delta$ identity, we can rewrite this as $$\epsilon_{ijk} \mathbf{A}^k \frac{\partial \phi^i}{\partial X^A}\frac{\partial \phi^j}{\partial X^B} = \frac{1}{2}\epsilon_{ijk} \epsilon_{ABC} \epsilon_{CDE} \mathbf{A}^k \frac{\partial \phi^i}{\partial X^D}\frac{\partial \phi^j}{\partial X^E}$$ where I have abandoned hope of appropriate upper and lower indices here (to be improved) Now we can identify this term as a partial derivative of the Jacobian matrix of the transformation function. Fact: the Jacobian of our transformation matrix $\frac{\partial\varphi^i}{\partial X^A}$ can be written $$J = \det \frac{\partial\varphi^i}{\partial X^A} = \frac{1}{6}\epsilon_{ijk}\epsilon_{ABC}\frac{\partial\varphi^i}{\partial X^A}\frac{\partial\varphi^j}{\partial X^B}\frac{\partial\varphi^k}{\partial X^C}$$ Now notice that our integrand looks like $A^i\epsilon_{ABC}\frac{\partial J}{\partial (\partial_C\partial\varphi^i)}$ We finally use the fact that $\frac{\partial J}{\partial (\partial_C\partial\varphi^i)} = J\frac{\partial (\varphi^{-1})^C}{\partial x^i}$ (believable fact, just looks like multiplying by an inverse and is provable). We now have all of the pieces to solve this problem.
Rewriting our integral, we can rewrite it as an integral in initial-configuration space, move the derivative inside, differentiate, and then recast in terms of the current configuration. We then have:
\begin{align} \frac{d}{dt}\int_\mathcal{S} \mathbf{A} \cdot \mathbf{n} da &= \int_{S_0} \frac{d}{dt} (A^i \epsilon_{ABC} \frac{\partial (\varphi^{-1})^C}{\partial x^i})J dR^A dS^B\\ &= \int_{S_0} (\dot{A}^i \frac{\partial (\varphi^{-1})^C}{\partial x^i}J + A^i\frac{\partial J}{\partial t}\frac{\partial (\varphi^{-1})^C}{\partial x^i} + A^iJ \frac{\partial^2 (\varphi^{-1})^C}{\partial t\partial x^i})dR^AdS^B \end{align} Where $\dot{A}^i = \frac{\partial A^i}{\partial t} + \partial_j A^i v^j$ is the material time derivative of $A$ (really just applying the chain rule to $A(\varphi(X,Y,Z,t))$). We then have $\frac{\partial J}{\partial t} = J \nabla \cdot \mathbf{v}$, and are left with the last term $\frac{(\varphi^{-1})^C}{\partial t\partial x^i})$. We can find this term: \begin{align*} \frac{(\partial^2\varphi^{-1})^C}{\partial t\partial x^i} &= -\frac{\partial (\varphi^{-1})^C}{\partial x^j} \frac{\partial v^j}{\partial X^B} \frac{\partial (\varphi^{-1})^B}{\partial x^i}\\ &= -\frac{\partial (\varphi^{-1})^C}{\partial x^j} \frac{\partial v^j}{\partial x^i} \end{align*} Where we exchange the order of partial derivatives, find the derivative of the inverse of a matrix, and then use the chain rule to get the gradient of $v$. Now, writing using vector notation and relabeling dummy indices, we have (with $\nabla$ referring to spatial coordinates): \begin{align*} \frac{d}{dt}\int_\mathcal{S} \mathbf{A} \cdot \mathbf{n} da &= \int_{S_0} (\dot{\mathbf{A}} + \mathbf{A}\nabla \cdot \mathbf{v} + (\nabla \mathbf{v}) \mathbf{A})^i J \epsilon_{ABC}\frac{\partial (\varphi^{-1})^C}{\partial x^i}dR^AdS^B\\ &= \int_\mathcal{S} \left(\frac{\partial A}{\partial t} + (\nabla \mathbf{A})\mathbf{v} + \mathbf{A} \nabla \cdot \mathbf{v} - (\nabla \mathbf{v})\mathbf{A}\right)da \end{align*} Finally, we need an identity. This is not too bad to see with index notation, but it is that $$\nabla \times (\mathbf{v} \times \mathbf{A}) = (\nabla \mathbf{v})\mathbf{A} - (\nabla \mathbf{A})\mathbf{v} + \mathbf{A} \nabla \cdot \mathbf{v} - \mathbf{v} \nabla \cdot \mathbf{A}$$ Simply matching terms in our integrand, we are left with the statement: \begin{align*} \frac{d}{dt}\int_\mathcal{S} \mathbf{A} \cdot \mathbf{n} da = \int_\mathcal{S} \frac{\partial \mathbf{A}}{\partial t} + \mathbf{v} \nabla \cdot \mathbf{A} - \nabla \times (\mathbf{v} \times \mathbf{A}) \end{align*} as desired.