We have a time and even "position" invariant vector field and a surface. If the surface is moving with constant velocity, is the flux through the surface should constant in time? Also, is there an easy to follow proof for the formula $$\frac{d}{dt} \iint_{S(t)}\overline{V}(\overline{r},t) \cdot d\overline{A} $$

  • $\begingroup$ What do you mean a "position" invariant vector field and which vector field? The vector field you want to compute the flux or the velocity field? $\endgroup$ – achille hui Jun 15 '14 at 18:01
  • $\begingroup$ The vector whose flux has to be computed is constant in time and is the same everywhere. $\endgroup$ – user42768 Jun 15 '14 at 18:29
  • $\begingroup$ If $\vec{V}$ is constant in time and space, then $\int_{S(t)} \vec{V} \cdot d\vec{A}$ is identically zero by divergence theorem. $\endgroup$ – achille hui Jun 15 '14 at 18:32
  • $\begingroup$ I am terribly sorry, I don't know why I said closed surface. The surface is not closed. I'm again very sorry. $\endgroup$ – user42768 Jun 15 '14 at 18:48
  • $\begingroup$ Your formula would seem to be incomplete. Or, do you mean to ask, "is there some way to simplify this derivative..." Fwiw, I do think the answer is affirmative, the flux is constant if a rigid surface moves at constant velocity through a constant vector field (constant in both space and time here). $\endgroup$ – James S. Cook Jun 15 '14 at 19:16

Here is a proof, though I'm not sure it is easy to follow.

In below derivation, we will use the adjective "nice" for anything sufficiently regular to make the argument works. If I'm not mistaken, continuous differentiable up to second order is sufficient.


  • $\vec{v}(\vec{x},t)$ be a "nice" velocity field.
  • $\vec{B}(\vec{x},t)$ be a "nice" vector field which we want to compute the flux.
  • $S$ be a compact "nice" surface with piecewise "nice" boundary in $\mathbb{R}^3$.
  • $S(t)$ be a family of surfaces generated by $S$ using the velocity field $\vec{v}$.
    More precisely, for each $\vec{x} \in S$, consider the initial value problem: $$\frac{d}{dt} \gamma_{\vec{x}}(t) = \vec{v}(t,\gamma_{\vec{x}}(t)),\quad t \in (-\epsilon, \epsilon ) \quad\text{ with initial condition }\quad \gamma_{\vec{x}}(0) = \vec{x}. $$ Define a function by $$\Gamma : S \times (-\epsilon, \epsilon ) \quad\mapsto\quad \Gamma(\vec{x},t) = \gamma_{\vec{x}}(t)\in\mathbb{R}^3$$ Standard theory of ODE tell us $\Gamma$ is a "nice" function (as least for small enough $\epsilon$).

    $S(t)$ is simply the image $\Gamma(S \times \{t\})$. As a geometric object, I'm not 100% sure $S(t)$ is "nice".
    For the sake to get a result, let's assume they are.

The number we want to compute is following time derivative $$\mathscr{F}_{t} = \frac{d}{dt}\int_{S(t)} \vec{B}\cdot d\vec{A}$$ where $d\vec{A}$ is the area element associated with $S(t)$. To carry out the computation, let us assume $S$ is small enough to fit into a single coordinate chart $$\varphi : \mathbb{R}^2 \supset \mathcal{O} \ni (r,s) \quad\mapsto\quad \varphi(r,s) \in S \subset \mathbb{R^3}$$ We will further assume the domain $\mathcal{O}$ has piecewise "nice" boundary${}^{\color{blue}{[1]}}$. Using $\Gamma$, we can extend it to a function

$$\vec{X} : \mathcal{O} \times (-\epsilon, \epsilon ) \ni (r, s, t) \quad\mapsto\quad \vec{X}(r,s,t) = \Gamma(\varphi(r,s),t)$$

For any function (or expression) $\psi$ that depends on $(r,s,t)$, we will adopt the shorthand $$\psi_r = \frac{\partial\psi}{\partial r},\quad \psi_s = \frac{\partial\psi}{\partial s},\quad\text{ and }\quad \psi_t = \frac{\partial\psi}{\partial t}$$

In terms of $\vec{X}$, the area element $d\vec{A}$ is simply $\vec{X}_r \times \vec{X}_s dr ds$ and

$$\begin{align} \mathcal{F}_t &= \frac{d}{dt} \int_{\mathcal{O}} \vec{B}\cdot (\vec{X}_r \times \vec{X}_s) dr ds\\ &= \int_{\mathcal{O}}\left[ \frac{d\vec{B}}{dt} \cdot (\vec{X}_r \times \vec{X}_s) + \vec{B} \cdot \left( (\vec{X}_r)_t \times \vec{X}_s + \vec{X}_r \times (\vec{X}_s)_t \right) \right] dr ds\\ &= \int_{\mathcal{O}}\left[ \left(\vec{B}_t + \color{firebrick}{(\vec{v}\cdot\vec{\nabla})\vec{B}}\right) \cdot \color{firebrick}{(\vec{X}_r \times \vec{X}_s)} + \vec{B} \cdot\left( \vec{v}_r \times \vec{X}_s + \vec{X}_r \times \vec{v}_s\right) \right] dr ds \end{align} $$ Notice the second term in the integrand $$ \vec{B} \cdot\left( \vec{v}_r \times \vec{X}_s + \vec{X}_r \times \vec{v}_s\right) = \vec{B} \cdot \left( ( \vec{v} \times \vec{X}_s)_r - ( \vec{v} \times \vec{X}_r)_s \right)$$ can be rewritten as $$ ( \vec{B}\cdot( \vec{v} \times \vec{X}_s) )_r - ( \vec{B}\cdot( \vec{v} \times \vec{X}_r) )_s + \color{red}{((\vec{X}_r\cdot\vec{\nabla})B) \cdot (\vec{X}_s \times \vec{v})} + \color{red}{((\vec{X}_s\cdot\vec{\nabla})B) \cdot (\vec{v}\times\vec{X}_r)} $$ Now for any five vectors $\vec{a},\vec{b},\vec{c}, \vec{p}, \vec{q} \in \mathbb{R}^3$, we have an identity${}^{\color{blue}{[2]}}$: $$ (\vec{p}\cdot\vec{q})(\vec{a}\cdot(\vec{b}\times\vec{c})) = (\vec{p}\cdot\vec{a})(\vec{q}\cdot(\vec{b}\times\vec{c})) + (\vec{p}\cdot\vec{b})(\vec{q}\cdot(\vec{c}\times\vec{a})) + (\vec{p}\cdot\vec{c})(\vec{q}\cdot(\vec{a}\times\vec{b})) $$ $\vec{\nabla} \otimes \vec{B}$ is a rank 2 tensor, we can decompose it as a sum of outer product of vectors $$\vec{\nabla} \otimes \vec{B} = \vec{p}_1 \otimes \vec{q}_1 + \vec{p}_2 \otimes \vec{q}_2 + \cdots$$ If we substitute $\vec{a},\vec{b},\vec{c}$ in above identity by $\vec{v}$, $\vec{X}_r$ and $\vec{X}_s$ respectively, it is not hard to deduce:

$$\color{firebrick}{((\vec{v}\cdot\vec{\nabla})\vec{B}) \cdot (\vec{X}_r \times \vec{X}_s)} + \color{red}{((\vec{X}_r\cdot\vec{\nabla})B) \cdot (\vec{X}_s \times \vec{v}) + ((\vec{X}_s\cdot\vec{\nabla})B) \cdot (\vec{v}\times\vec{X}_r)} = \color{green}{( \vec{\nabla}\cdot \vec{B}) (\vec{v}\cdot(\vec{X}_r \times \vec{X}_s)}.$$ Substitute this back into the integrand for $\mathcal{F}_t$, we get

$$ \mathcal{F}_t = \int_{\mathcal{O}} \left[ \left((\vec{B}_t + \color{green}{(\vec{\nabla}\cdot \vec{B}) \vec{v}} )\cdot \color{green}{( \vec{X}_r \times \vec{X}_s )}\right)_{\color{blue}{\verb/I/}} + \left( ( \vec{B}\cdot( \vec{v} \times \vec{X}_s) )_r - ( \vec{B}\cdot( \vec{v} \times \vec{X}_r) )_s \right)_{\color{blue}{\verb/II/}} \right] dr ds\\ $$ The integrand split into two pieces. It is obvious what Piece $\color{blue}{\verb/I/}$ is. For piece $\color{blue}{\verb/II/}$, we can transform it first to a line integral along $\partial\mathcal{O}$ using the classical Green's theorem in $\mathbb{R}^2$. We then re-express it as a line integral along $\partial S(t)$ in $\mathbb{R}^3$. Finally, we convert it back to a surface integral over $S(t)$ using Kevin-Stokes theorem:

$$\begin{align} \text{Piece }\color{blue}{\verb/I/} &= \int_{S(t)}\left( \vec{B}_t + (\vec{\nabla}\cdot\vec{B})\vec{v}\right)\cdot d\vec{A}\\ \\ \text{Piece }\color{blue}{\verb/II/} &= \int_{\partial S(t)} ( \vec{B}\times\vec{v} ) \cdot d\vec{X} =\int_{S(t)} \vec{\nabla} \times (\vec{B} \times \vec{v}) \cdot d\vec{A} \end{align} $$ Combine this, we get

$$\mathcal{F}_t = \int_{S(t)} \left( \vec{B}_t + (\vec{\nabla}\cdot\vec{B})\vec{v} - \vec{\nabla} \times (\vec{v} \times \vec{B})\right)\cdot d\vec{A}\tag{*1}$$

When $S$ is too large to fit in a single coordinate chart, we can split $S$ into finite many pieces that fit. We then apply $(*1)$ to the individual piece and sum their contributions. At the end, $(*1)$ continue to work...


  • $\color{blue}{[1]}$ $\partial\mathcal{O}$ need to be nice enough to apply classical Green's theorem.
  • $\color{blue}{[2]}$ When $\vec{a},\vec{b},\vec{c}$ are linear independent, they form a basis of $\mathbb{R}^3$. The correspond dual basis consists of $\frac{\vec{b} \times \vec{c}}{\Delta}$, $\frac{\vec{c} \times \vec{a}}{\Delta}$, and $\frac{\vec{a} \times \vec{b}}{\Delta}$ where $\Delta = \vec{a}\cdot(\vec{b}\times\vec{c})$. The identity in main text is nothing special but a formula of the dot product for this particular pair of basis/dual basis.
  • $\begingroup$ Let me see if I understood correctly. First, we find a parametric representation of the initial surface S, $\mathcal{O}$. And we denote the body that is formed by the surface after it has moved for "t time" by S(t). Why is the derivative of the flux through S(t) equal to the difference between the flux through the surface "at moment t" and the surface "at moment 0" over t? Maybe I misunderstood something. Thank you! $\endgroup$ – user42768 Jun 16 '14 at 6:26
  • $\begingroup$ No, the $t = 0$ is nothing special. It is only used to pick an initial surface to generate parametrization for $S(t)$ at each $t \in (-\epsilon,\epsilon)$. There are no "differences" of any sort in the derivation. We don't compare flux at time $t$ and $0$ at all. Everything are derivatives and all expression I write down happens at a time $t$ not necessary equal to $0$. $\endgroup$ – achille hui Jun 16 '14 at 6:55
  • $\begingroup$ But this is how I understand this derivative: First we have a surface S and we now the flux through it. After a very short time, let's say dt, S has moved according to v. We now compute the flux through it. As a final step, we compute the flux difference over dt. In your proof, S(t) represents the volume "described by the moving surface", or the surface at moment t? Again, thank you for your time. $\endgroup$ – user42768 Jun 16 '14 at 7:08
  • $\begingroup$ $S(t_1)$ is the surface $\Gamma( \mathcal{O} \times \{t_1\} )$, not a body $V(t_1,t_2) = \Gamma( \mathcal{O} \times [t_1,t_2] )$. The way you described is the physicist's way of deriving the formula. In that case, the boundary of $V(t,t+dt)$ consists of 3 surfaces, $S(t)$, $S(t+dt)$ and $\Gamma(\partial\mathcal{O} \times [t,t+dt] )$. It is the last piece that will give you the curl piece $\vec{\nabla} \times ( \vec{v} \times \vec{B} )$. The physicists' way will be more easier to understand but has more logical loop holes. $\endgroup$ – achille hui Jun 16 '14 at 7:22
  • 1
    $\begingroup$ So, looking from a physicist's point of view, the first term in the right-hand side is caused by the change in the field, the second term is the divergence of the field "inside" that volume, and the last one considers the flux through the surface that "links" the boundaries of the surfaces at those two different moments in time? I am immensely appreciative for your understanding. $\endgroup$ – user42768 Jun 16 '14 at 7:28

${\color{#66f}{\large\tt\mbox{This is from $\tt\color{#c00000}{G\&R}$ Table,}\ 7^{\rm\underline{a}}\ \mbox{ed.}}}$

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