## The Transport Theorem

Theorem
The rate of change of flux from a vector field
$\mathbf{F}$
through a surface
$S$
both time dependent i
$\frac{d \Phi}{dt} = \int \int_{S_t}( \frac{\partial \mathbf{F}}{\partial t} + \mathbf{v} \mathbf{\nabla} \cdot \mathbf{F}) \cdot d \mathbf{n} dS + \oint_{C_1} (\mathbf{F} \times \mathbf{v}) \cdot d \mathbf{r}$

Proof
At any time
$t_1$
the vector field is
$\mathbf{F} (\mathbf{r}, t_1)$
. Let the surface at
$t=t_1$
be
$S_{t_1}$
and let the boundary of
$S_{t_1}$
be
$C_{t_1}$
.
The flux out of
$S_{t_1}$
is
$\Phi = \int_{S_{t_1}} \mathbf{F} (\mathbf{r}, t_1) \cdot \mathbf{n}_{t_1} dS_{t_1}$
.
$\frac{d \Phi}{dt} = lim_{t_2 \rightarrow t_1} \frac{\Phi_{t_2} - \Phi_{t_1}}{t_2 -t_1} =lim_{t_2 \rightarrow t_1} \frac{\int \int_{S_{t_2}} \mathbf{F} (\mathbf{r}, t_2) \cdot \mathbf{n}_{t_2} dS_{t_2} - \int \int_{S_{t_1}} \mathbf{F} (\mathbf{r}, t_1) \cdot \mathbf{n}_{t_1} dS_{t_1}}{t_2 -t_1}$

Apply the Divergence Theorem to the volume traced out by
$S$
between
$t_2$
and
$t_1$
gives
\begin{aligned} \int \int \int_{V_{t_1}} \mathbf{\nabla} \cdot \mathbf{F}(\mathbf{r}, t_{t_1}) dV_{t_1} &= lim_{t_2 \rightarrow t_1} ( \int \int_{S_{t_2}} \mathbf{F} (\mathbf{r},t_2) \cdot \mathbf{n}_{t_2} dS_{t_2} \\ &- \int \int_{S_{t_1}} \mathbf{F} (\mathbf{r},t_1) \cdot \mathbf{n}_{t_1} dS_{t_1} + \int \int_{Sides} \mathbf{F} (\mathbf{r},t_2) \cdot \mathbf{n}_{t_2} dS_{t_2}) \end{aligned}

$\mathbf{F}(\mathbf{r}, t_2) \simeq \mathbf{F}(\mathbf{r}, t_1) + \frac{\partial \mathbf{F}}{\partial t} dt$

so
where
$dt=t_2 -t_1$
.
\begin{aligned} \int \int_{S_{t_2}} \mathbf{F} (\mathbf{r},t_2) \cdot \mathbf{n}_{t_2} dS_{t_2} - \int \int_{S_{t_1}} \mathbf{F} (\mathbf{r},t_1) \cdot \mathbf{n}_{t_1} dS_{t_1} &\simeq= \int \int_{S_{t_2}} \mathbf{F}(\mathbf{r}, t_1) dS_{t_2} \\ &+ \int \int_{S_{t_2}} \frac{\partial \mathbf{F}}{\partial t} dt dS_{t_2} - \int \int_{S_{t_1}} \mathbf{F} (\mathbf{r},t_1) \cdot \mathbf{n}_{t_1} dS_{t_1} \\ &+ \int \int \int_{V_{t_1}} \mathbf{\nabla} \cdot \mathbf{F}(\mathbf{r}, t_{t_1}) dV_{t_1} - \int \int_{Sides} \mathbf{F} (\mathbf{r},t_2) \cdot \mathbf{n}_{t_2} dS_{t_2} \end{aligned}

$dV \simeq v dS dt$

Hence
\begin{aligned} \int \int_{S_{t_2}} \mathbf{F} (\mathbf{r},t_2) \cdot \mathbf{n}_{t_2} dS_{t_2} - \int \int_{S_{t_1}} \mathbf{F} (\mathbf{r},t_1) \cdot \mathbf{n}_{t_1} dS_{t_1} &\simeq= \int \int_{S_{t_2}} \mathbf{F}(\mathbf{r}, t_1) dS_{t_2} \\ &+ \int \int_{S_{t_2}} \frac{\partial \mathbf{F}}{\partial t} dt dS_{t_2} - \int \int_{S_{t_1}} \mathbf{F} (\mathbf{r},t_1) \cdot \mathbf{n}_{t_1} dS_{t_1} \\ &+ \int \int \int_{V_{t_1}} \mathbf{\nabla} \cdot \mathbf{F}(\mathbf{r}, t_{t_1}) v dS_{t_1} dt - \int \int_{Sides} \mathbf{F} (\mathbf{r},t_2) \cdot \mathbf{n}_{t_2} dS_{t_2} \end{aligned}

Now divide by
$dt=t_2 -t_1$
to get
$\frac{d \Phi}{dt} = \int \int_{S_t}( \frac{\partial \mathbf{F}}{\partial t} + \mathbf{v} \mathbf{\nabla} \cdot \mathbf{F}) \cdot d \mathbf{n} dS + \oint_{C_1} (\mathbf{F} \times \mathbf{v}) \cdot d \mathbf{r}$