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Helmholtz Decomposition theorem stats:

"Let $\vec F$ be a vector field on a bounded domain $V$ in $\mathbb R^3$, which is twice continuously differentiable, and let $S$ be the surface that encloses the domain $V$. Then $\vec F$ can be decomposed into a curl-free component and a divergence-free component"

$$ \vec F = -\nabla \phi + \nabla \times \vec A$$

It is easy to see that $\phi$ is equal to $V$ in electrostatic where $V$ is the electric potential. Can someone tell me what $\vec A$ would represent?

Is there a similiar analogy for $\vec B$?

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    $\begingroup$ The term you're looking for is "vector potential" $\endgroup$ – Kyle Kanos Nov 27 '14 at 3:57
  • $\begingroup$ Interesting, why would this term appear in the equation for electric field? I wonder. And why is it zero in electrostatics? $\endgroup$ – Bajie Nov 27 '14 at 4:18
  • $\begingroup$ It appears due to Maxwell's equations. Electrostatics implicitly assume that the magnetic field just does not exist, hence $\mathbf A$ not appearing. $\endgroup$ – Kyle Kanos Nov 27 '14 at 4:21
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$\nabla\times\vec{A}$ is the solenoidal component of the vector field: it is the divergenceless component.

A good way to intuitively visualise the Helmholtz theorem is to think in Fourier space, so that all fields become their Fourier transforms. In this visualisation, $\nabla\times \vec{E}$ is the component of the Fourier transform that is at right angles to the wavevector. Then the differential operations become dot and cross products with the position in Fourier space i.e. the wavevector of plane wave component. So, instead of $\vec{E}(\vec{r})$, we have now a field $\tilde{\vec{E}}(\vec{k})$ specifying the superposition weight (amplitude) of a plane wave component $\exp(i\,\vec{k}\cdot\,\vec{r})$ in the Fourier decomposition of the field. The Fourier transforms of the following $\nabla$ operations on vector and scalar fields $\vec{E}(\vec{r})$ and $\phi(\vec{r})$ are:

$$\nabla\phi(\vec{r}) \leftrightarrow i\,\vec{k}\,\tilde{\phi}(\vec{k})$$ $$\nabla\cdot\vec{E}(\vec{r}) \leftrightarrow i\,\vec{k}\,\cdot\tilde{\vec{E}}(\vec{k})$$ $$\nabla\times\vec{E}(\vec{r}) \leftrightarrow i\,\vec{k}\,\times\tilde{\vec{E}}(\vec{k})$$

where the tilde quantities are the Fourier transforms. So, at each point in space, $\tilde{\vec{E}}(\vec{k})$ resolve the vector into unique components parallel to and orthogonal to the wavevector (the position vector in Fourier space):

$$\begin{array}{lcl}\tilde{\vec{E}}(\vec{k}) &=& \tilde{E}_\parallel(\vec{k})\,\frac{\vec{k}}{|\vec{k}|}+\tilde{E}_\perp(\vec{k})\,\frac{\vec{k}\times \tilde{\vec{E}}(\vec{k})}{|\vec{k}|\,|\tilde{\vec{E}}(\vec{k})|}\\&=&-i\,\left(\mathfrak{F}(\nabla.\vec{E}) \,\hat{k} + \mathfrak{F}(\nabla\times\vec{E})\right)\end{array}$$

where $\mathfrak{F}$ is the Fourier transform. The curl is the component of the field at right angles to the unit radius $\hat{k}$ (modulo the scaling constant $i$), since $\hat{k}\cdot(\vec{k}\times \tilde{\vec{E}})=0$ and the divergence is the component along the unit radius $\hat{k}$

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  • $\begingroup$ Hmm I thought the fourier transform had no effect on spatial operators i.e. $d/d_z$ is unchanged $\endgroup$ – Bajie Nov 27 '14 at 4:56
  • $\begingroup$ @IllegalImmigrant No, that's only true if the derivative is with respect to a non-transform variable: otherwise you have the fundamental rule $\mathrm{d}_x\,\int_{-\infty}^\infty e^{i\,k\,x}\, f(x)\,\mathrm{d}x = i\,k\,F(k)$. $\endgroup$ – Selene Routley Nov 27 '14 at 6:59
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$\vec{A}$ is the vector potential. $\vec{E}$ is equal to $-\nabla\phi$ only in the electrostatic case. If the electric field varies with time, we have \begin{equation} \nabla\times\vec{E} = -\frac{\partial\vec{B}}{\partial t} \end{equation} Since $\nabla\cdot\vec{B}$ is always zero, $\vec{B}$ is written as a curl of another field, called the vector potential. That is $\vec{B} = \nabla\times\vec{A}$. Putting this in the above equation, we get \begin{equation} \nabla\times\left(\vec{E} + \frac{\partial\vec{A}}{\partial t}\right) = 0 \end{equation} Curl of a field is zero means that it can be expressed as a gradient of a scalar field. In this case, it is $-\phi$. That is, \begin{equation} \vec{E} + \frac{\partial\vec{A}}{\partial t} = -\nabla{\phi} \end{equation}

One can surely use Helmholtz theorem for magnetic field as well and write it as $\vec{B} = -\nabla\psi + \nabla\times\vec{A}$, where $\psi$ is the magnetic scalar potential. It is non-zero only if $\nabla\times\vec{B} = 0$. (Please note that $\psi$ is usually defined using the $\vec{H}$ field. However, the idea is applicable to $\vec{B}$ field as well. You can refer to http://en.wikipedia.org/wiki/Magnetic_potential#Magnetic_scalar_potential for additional information.)

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  • $\begingroup$ Hi is the vector potential a physical or mathematical quantity? $\endgroup$ – Bajie Nov 27 '14 at 4:18
  • $\begingroup$ I do not know if the vector potential is truly a physical quantity. I used to think that the Aharanov-Bohm effect indicates that it is indeed a physical quantity and not just a mathematical convenience. However, it appears that not every one agrees with this view. See, for example, en.wikipedia.org/wiki/…. $\endgroup$ – Amey Joshi Nov 27 '14 at 5:24

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