Calculating flux through a moving surface in a vector field that evolves with time

Question

Suppose we are given a vector field $\mathbf{F} \colon \mathbb{R}^4 \to \mathbb{R}^3$ that evolves with time and describes the way, say, liquid particles move in a tank. Also, we are given a parametric surface $\mathscr{S}$ that is parameterized by

$$\mathbf{r}(u, v, t) = x(u, v, t)\mathbf{i} + y(u, v, t)\mathbf{j} + z(u, v, t)\mathbf{k}.$$

(Note that the surfaces evolves with time too; it "moves".) Moreover, let the (spatial) domain of $\mathbf{r}$ be $A$.

Now the question I want to be able to answer is: how to compute flux through an area element taking motion of both the surface and the vector field into account?

See what I tried:

(1) First I need to compute the unit normal to the surface at a particular instant:

$$\mathbf{\hat{N}}(u, v, t) = \frac{\frac{\partial}{\partial u}\mathbf{r}(u, v, t) \times \frac{\partial}{\partial v}\mathbf{r}(u, v, t)}{\Big| \frac{\partial}{\partial u}\mathbf{r}(u, v, t) \times \frac{\partial}{\partial v}\mathbf{r}(u, v, t)\Big|}.$$

(2) Taking motion into account, we have that the surface element at time $t$ passes the space at rate $$\Big\langle \mathbf{\hat{N}}(u, v, t), \frac{\partial}{\partial t}\mathbf{r}(u, v, t) \Big\rangle.$$

(3) Now all we do at each area element is to take the difference of the two vectors: one vector from the vector field and the other one is from the motion of the surface. For example, if at some instant an area element moves with the same speed and direction as the vector implied by the vector field, the flux should be zero. Putting all together, I have the following:

$$ \begin{align} \Phi(t) &= \iint_{\mathscr{S}} \Bigg\langle \mathbf{F}, \mathbf{\hat{N}}\Bigg\rangle - \Bigg\langle \frac{\partial}{\partial t}\mathbf{r}, \mathbf{\hat{N}}\Bigg\rangle\, \mathrm{d}S \\ &= \iint_{\mathscr{S}} \Bigg\langle \mathbf{F} - \frac{\partial}{\partial t}\mathbf{r}, \mathbf{\hat{N}} \Bigg\rangle\, \mathrm{d}S \\ &= \iint_{A} \Bigg\langle \mathbf{F} - \frac{\partial}{\partial t}\mathbf{r}, \mathbf{\hat{N}} \Bigg\rangle \Bigg| \frac{\partial}{\partial u}\mathbf{r} \times \frac{\partial}{\partial v}\mathbf{r} \Bigg| \, \mathrm{d}u \, \mathrm{d}v \\ &= \iint_{A} \Bigg\langle \mathbf{F}(\mathbf{r}(u, v, t), t) - \frac{\partial}{\partial t} \mathbf{r}(u, v, t), \frac{\partial}{\partial u} \mathbf{r}(u, v, t) \times \frac{\partial}{\partial v}\mathbf{r}(u, v, t) \Bigg\rangle \, \mathrm{d}u \, \mathrm{d}v. \end{align} $$

Is the above calculation correct? If yes, than the volume through the surface $\mathscr{S}$ within time range $[a, b]$ is simply $\int_a^b \Phi(t) \, \mathrm{d}t$?

Answer 1

Yes, this the calculation is correct. In a constant density setting (such as a liquid), the flux you calculate is the (signed) amount of stuff that goes through the surface in an infinitesimal time interval.

Physical intuition dictates that these things must happen:

• If the surface does not move, this is just the usual flux.
• If the surface moves in a direction tangential to itself (like a sphere rotating but not moving), this should cause no flux.
• If the surface moves with the fluid flow ($\vec F=\partial_t\vec r$), then the flux should be zero. (Think of an impenetrable plastic bag moving in water. There is no flux through it.)
• If the surface is a disc of area $A$ that moves without any deformations at a constant speed, it wipes an area $A|\hat N\cdot\vec v|T$ in time $T$, as you can easily calculate from elementary geometry. If there is no flow ($\vec F\equiv0$), this should be the time integral of the flux, up to sign.
• Suppose the surface is $\mathscr S_t=\partial B(0,t)$ at any time $t>0$. This corresponds to $\partial_t\vec r=\hat N$. (See a remark below about spheres.) In the absence of any flow, the integral of the flux from $t_0$ to $t_1$ should be the difference of the volumes of the balls $B(0,t_1)$ and $B(0,t_0)$. The sign depends on which way you orient your surface. The integral can be evaluated explicitly in this case.

And indeed, these hold true. When doing such "physical calculus problems", I strongly suggest making sure that the solution exhibits physically correct behavior.

I did not only check these criteria. I also went through your reasoning, and you have explained your solution well.

Additional remarks:

• Sometimes you cannot parametrize a surface in the way you propose. For example, you cannot parametrize a sphere by a planar set. The parametrization can be done up to a small set (which doesn't matter for the integrals) or you can remove double counting if you parametrize some part many times, so you can still do as you did. You just need to be a bit more careful.
• If density is not constant, you need to include it in your integral. This might be relevant for a gas.