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My notes say that this is the equation for the "advection of a quantity A at speed v":

$$ \frac{\partial \mathbf{A}}{\partial t} = \nabla\times(\mathbf{v}\times\mathbf{A}) .$$

Is this true? How can I see (mathematically) that this equation corresponds to advection?

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Is there any special property that $\mathbf{A}$ has to satisfy? – Mr. Gentleman Apr 9 '14 at 15:05
no, $\mathbf{A}$ is a general vector property. In my example it is the magnetic field $\mathbf{B}$ – SuperCiocia Apr 9 '14 at 15:11
Well, B does have special properties, such as always-zero divergence. – Kvothe Apr 9 '14 at 15:13
up vote 5 down vote accepted

This term looks like Faraday's law that is used in Ideal magnetohydrodynamics (MHD). So yes, it is true.

In order to see the mathematical advection, you'll need to apply some vector calculus:

$$ \nabla\times\mathbf a\times\mathbf b = \mathbf a\left(\nabla\cdot\mathbf b\right) - \mathbf b\left(\nabla\cdot\mathbf a\right) + \left(\mathbf b\cdot\nabla\right)\mathbf a - \left(\mathbf a\cdot\nabla\right)\mathbf b $$ Then grouping the $\mathbf a$ and $\mathbf b$ terms, $$ \nabla\times\mathbf a\times\mathbf b = \left(\nabla\cdot\mathbf b+ \mathbf b\cdot\nabla\right)\mathbf a - \left(\nabla\cdot\mathbf a+\mathbf a\cdot\nabla\right)\mathbf b $$ which then reduces to give the advection equation in the form $$ \frac{\partial\mathbf A}{\partial t}=\nabla\times\mathbf v\times\mathbf A=\nabla\cdot\left(\mathbf A\mathbf v^T - \mathbf v\mathbf A^T\right)\tag{1} $$ This now gives us a time-derivative of vector $\mathbf A$ equal to the gradient of $\mathbf A$ times the vector velocity, $\mathbf v$, which is pretty much what your advection equation is supposed to be.

It's probably a little bit more easily seen in component form: $$ \frac{\partial}{\partial t}\left(\begin{array}{c}A_x \\ A_y \\ A_z\end{array}\right) = \left(\begin{array}{c}\frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z}\end{array}\right)\left(\begin{array}{ccc}A_xv_x -v_xA_x& A_xv_y-v_xA_y & A_xv_z-v_xA_z \\ A_yv_x-v_yA_x & A_yv_y-v_yA_y & A_yv_z-v_yA_z \\ A_zv_x-v_zA_x & A_zv_y-v_zA_y & A_zv_z-v_zA_z\end{array}\right) $$ which reduces to $$ \frac{\partial}{\partial t}\left(\begin{array}{c}A_x \\ A_y \\ A_z\end{array}\right) = \left(\begin{array}{c}\frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z}\end{array}\right)\left(\begin{array}{ccc}0 & A_xv_y-v_xA_y & A_xv_z-v_xA_z \\ A_yv_x-v_yA_x & 0 & A_yv_z-v_yA_z \\ A_zv_x-v_zA_x & A_zv_y-v_zA_y & 0\end{array}\right)\tag{2} $$ And, very neatly, you have yourself some divergence free terms in the zero's along the diagonal: $$ \frac{\partial A_x}{\partial t}=\frac{\partial}{\partial x}0+\frac{\partial}{\partial y}\left(A_xv_y-A_yv_x\right)+\frac{\partial}{\partial z}\left(A_xv_z-v_xA_z\right) $$

Now with regards to the actual question, how can we show that this is an advection equation. Advection simply is the transport of some material along a fluid flow. Usually this means that $$D_t\psi=\frac{\partial}{\partial t} \psi+\mathbf u\cdot\nabla\psi=0,$$ where $\mathbf u$ is the velocity of the fluid flow. In the particular case that $\nabla\cdot\mathbf u=0$, then we can use $$\frac{\partial}{\partial t}\psi+\nabla\cdot(\mathbf u\psi)=0\tag{3}$$ which is the more-familiar continuity equation.

But that is for scalar advection. In the case considered here, we have a vector to advect. This is easily done by replacing the scalar $\psi$ with the vector $\mathbf a$ in Equation (3) $$\frac{\partial}{\partial t}\mathbf a+\nabla\cdot(\mathbf u\mathbf a)=0$$ But we have this tricky $\mathbf{ua}$ term here. It can't be the dot product because divergence of a scalar is meaningless, so it must mean some other geometric product. In this case it makes a dyad: $$ \mathbf{ua}=\left(\begin{array}{ccc}u_xa_x & u_xa_y & u_xa_z \\ u_ya_x & u_ya_y & u_ya_z \\ u_za_x & u_za_y & u_za_z\end{array}\right) $$ Taking the derivative of this matrix (a poor way to think of a tensor) gives you a vector. Thus, we should write this as $$\frac{\partial}{\partial t}\mathbf a+\nabla\cdot(\mathbf u\mathbf a^T)=0$$ to specify that we mean a tensor product there.

This now gives the form of Equation (1) that I claimed was an advection equation. The extra minus term in Equations (1) & (2) comes from the fact that your equation from Faraday's law contains a vector product of two terms which leads to two tensor products, rather than just one.

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thanks, but how does this correspond to advection? I thought advection meant 'transport', how is that translated mathematically? – SuperCiocia Apr 9 '14 at 16:55
Advection is the transport of some variable along a flow: $\partial_tq+\nabla\cdot(\mathbf{v}q)=0$. Faraday's law as given is an advection equation with a particular constraint that $\nabla\cdot\mathbf{B}=0$. – Kyle Kanos Apr 9 '14 at 16:59
I know about the continuity equation, but $q$ is a scalar and $\mathbf{v}$ is the velocity at which the fluid is moving. Here $\mathbf{B}$ is a vector quantity, how do I adjust the continuity equation? – SuperCiocia Apr 9 '14 at 18:52
Well you don't adjust the advection equation, you derive a new one. In the case of the magnetic field, the temporal evolution of the $\mathbf B$ field depends on the electric field, you then can use the Lorentz force law (assuming no forces) to get that $\mathbf E=\mathbf v\times\mathbf B$, thus $\partial_t\mathbf B=\nabla\times\mathbf v\times\mathbf B$. – Kyle Kanos Apr 9 '14 at 19:00
The momentum conservation equation (another advection equation) takes the form $\partial_t(\rho\mathbf v)+\nabla\cdot(\rho\mathbf v\mathbf v^T)=\nabla p$ where $p$ is the pressure. – Kyle Kanos Apr 9 '14 at 19:01

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