# How to get an integral formula for the flux time derivative

$$\frac{d}{dt}\int \limits_{A} \mathbf B d \mathbf A = \int \limits_{A} \left( \frac{\partial \mathbf B}{\partial t} + \mathbf v (\nabla \cdot \mathbf B ) + [\nabla \times [\mathbf v \times \mathbf B ]\right)d\mathbf A,$$ where $d \mathbf A$ - infinitesimal element of the vector of a surface $\Sigma$ bounded by circuit $d\Sigma$ (look at the picture).

Who can help me to explain the derivation of this expression? It's not homework.

$d\mathbf A$, as I understand, is equivalent to the "allocation" surface of segment $d \mathbf l$ of the curve $d \Sigma$ fo the time $dt$ for a motion with the speed of $\mathbf v$: $$d \mathbf A = [d \mathbf l \times \mathbf v dt].$$

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The only sense I can give to the right hand side is if the surface is moving through the field, and you are asking for the change in flux as the surface moves. – Ron Maimon Sep 15 '12 at 5:04
Please provide a reference to this formula. – Qmechanic Sep 18 '12 at 6:38
Unfortunately, I saw it only in the "Faradey's law" article in russian Wikipedia. Also I saw a reference to the book "Theoretische Elektrotechnik, 5th edition" by K. Simonyi. – user8817 Sep 18 '12 at 14:17
There is a sign mistake in the last term(v6) as compared to the Russian version of this Wikipedia page. – Qmechanic Sep 18 '12 at 19:36

I don't have time for a detailed derivation, so the following can contain errors, so take it for what it's worth...In the following I assume that $\mathbf{B}$ is constant in time. If not, the difference will just give (in the first approximation) the first term in the integral in the right-hand side. Let us consider the volume formed by $\Sigma(t_0)$ and $\Sigma(t_0+dt)$. The flux of $\mathbf{B}$ over the surface of this volume will be approximately $dt \frac{d}{dt}\int_{A}\mathbf{B}d\mathbf{A}$. On the other hand, this flux equals the integral of the divergence $\nabla \cdot \mathbf{B}$. This gives the second term of the right-hand side, as $\mathbf{v}dt d\mathbf{A}$ is the elementary volume. The last term in the right-hand side seems to vanish, as it equals a flux of rotor through $\Sigma(t_0)$, which equals the circulation of vector $\mathbf{v} \times \mathbf{B}$ over $d\Sigma(t_0)$. As circuit $d\Sigma$ is constant, $\mathbf{v}$ should be directed along the circuit in the points of the circuit, so $\mathbf{v} \times \mathbf{B}$ should be orthogonal to the circuit in the points of the circuit, so its circulation will vanish.

EDIT (09/18): As the author of the question asked for details, please find below an explanation of some points of the original answer. Again, there may be some errors, especially with signs, so please take this for what it's worth. I suspect $v$ is the field of velocity of the points of the surface $\Sigma$. Let us consider two surfaces: $\Sigma(t_0)$ and $\Sigma(t_0+dt)$. Together, they limit a certain volume $V$ between them. Let us consider the following expression: 1) $\int_{\Sigma(t_0+dt)} B dA$-$\int_{\Sigma(t_0+dt)} B dA$ (I assume here for the sake of simplicity that $B$ does not depend on time; furthermore, $v$, $B$ and $A$ are everywhere vector values and should be written in bold font). This expression equals $\int_{\Sigma_V} B dA$ (where $\Sigma_V$ is the total surface of volume $V$), because $\Sigma(t_0)$ and $\Sigma(t_0+dt)$ enter in $\Sigma_V$ with opposite signs due to their different position with respect to the normal of volume $V$. On the other hand, expression 1) approximately equals the following expression: 2) $dt\frac{d}{dt}\int_{\Sigma(t_0+dt)} B dA$. As 1) is an integral of $B$ over the surface $\Sigma_V$, it is actually the flux of $B$ through the surface of $V$. According to the Gauss theorem, the flux of a vector field through the surface of a volume equals the integral of the divergence of the vector field over the volume. Therefore, 1) equals 3) $\int_V (\nabla\cdot B) dV$. On the other hand, if $Q$ is some scalar field, $\int Q dV$ approximately equals $dt \int_A Q (v\cdot dA)$ (remember that volume $V$ is very small if $dt$ is very small.) Therefore, 3) approximately equals $dt\int_A (\nabla\cdot B) (v\cdot dA)$. As 2)=3), you can divide both sides of this equality by $dt$ and get the second term.

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I understand it by the next time. $$\frac{d}{dt}\frac{1}{c}\int \mathbf B d \mathbf S = \frac{1}{c}\int_{S_{0}} \frac{\partial \mathbf B}{\partial t}d \mathbf S_{0} + \frac{1}{c}\frac{d}{dt}\int \mathbf B_{0} d\mathbf S,$$ where indexes "0" means values of $\mathbf B , d\mathbf S$ in the moment $t_{0}$. The first summand gives $$- \int \mathbf E_{0} d \mathbf l,$$ the second gives $$-\int \frac{1}{c}[\mathbf v_{0} \times \mathbf B_{0} ] d \mathbf l.$$ I don't understand apppearance of the summand $\mathbf v (\nabla \cdot \mathbf B)$. Can you help me? – user8817 Sep 16 '12 at 8:27
@Maxim_Ovchinnikov: You keep changing the notation - there was no $c$ or $S$ in your question. Please make some choice and stick to it. The second term in the right-hand side of your first formula in the comment is treated as described in my answer - expressed via a flux through Σ(t0) and Σ(t0+dt), which equals an integral of divergence over the volume formed by Σ(t0) and Σ(t0+dt). – akhmeteli Sep 16 '12 at 9:37
Sorry. $d \mathbf S -> d\mathbf A$. – user8817 Sep 16 '12 at 10:54

To build upon @akhmeti's excellent answer, let us relax the assumption that $\partial \Sigma$ is constant in time. We let the boundary of the surface move along with the fluid. Then we must correct the divergence formula for the flux coming out of this edge strip:

$\int_{\partial \Sigma} \mathbf{B} \cdot [d\mathbf{l} \times \mathbf{v} dt] = \int_{\partial \Sigma} (\mathbf{B}\times \mathbf{v}dt) \cdot d\mathbf{l} = \int_{\Sigma} d\mathbf{A} \cdot (\nabla \times (\mathbf{B} \times \mathbf{v}) )dt$

Which is the origin of the last term. I have not been careful with signs.

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In the left hand side, both B and the surface A are time dependant. In the right hand side, the first term is due to B changing (even if A is fixed), the second and third term come from the fact that A(t) is not constant.

To take into account the variation of A with time, you need to use the convective derivative: D/Dt=d/dt+(V.grad)

The integrand on the right hand side is: