I Potential vector's formulas

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1. Jan 15, 2017

enerieire

I'm having some trouble with vector potential formulas. Ihave always used this one :
A=(μ/4pi )*(J/r) dV,
where r is a distance.

I don't understand where this formulas comes from:
1) A=(1/cr)*∫JdV
where c is the speed of light;

2) H=(1/cr) Ȧ
where H is the magnetic field's vector.

Thanks

2. Jan 15, 2017

Delta²

Equation 1) is approximately(read further on) the same formula as the one you use to know it is just in another system of units (your original equation is in SI units while 1) is probably in Heaviside units) is . Also the term 1/r has gone outside of the integral, an approximation that is valid for distances far away from the source of current density.

I cant help you though with equation 2) because from what I know $\dot{A}$ is the time derivative of A and that's the non conservative constituent of the electric field vector, not of the magnetic field vector.

3. Jan 15, 2017

enerieire

And if it was a space derivative of A? Would it have any sense?

S= c/4pi H^2 n

4. Jan 15, 2017

Delta²

This is the Poynting vector for a linearly polarized wave. In the case of linearly polarized wave you can prove that the electric field has essentially the same equation as the magnetic field but its direction is always perpendicular to that of the magnetic field. So for example if the propagation of the wave is in the z direction, and if the magnetic field vector is on the x direction $B=B(r,t)\hat{x}$ then the electric field vector is in the y-direction and $E=cB(r,t)\hat{y}$.

Last edited: Jan 16, 2017
5. Jan 16, 2017

enerieire

Thank you very much for the answers!!
I'm going to post another problem about Poynting's flux in a new question.

6. Jan 16, 2017

vanhees71

This is the Biot-Savart Law for the vector potential for magnetostatics. The derivation starts from the static Maxwell equations for the magnetic field (here written in Heaviside-Lorentz units)
$$\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{B}=\frac{\mu}{c} \vec{j}.$$
From the first equation we can write
$$\vec{B}=\vec{\nabla} \times \vec{A},$$
but $\vec{A}$ is defined only up to a gradient field (gauge invariance for the special case of magnetostatics). This can be used to impose one additional condition on $\vec{A}$. As we shall see in a moment, the following Coulomb-gauge condition is particularly convenient in this case:
$$\vec{\nabla} \cdot \vec{A}=0.$$
Now we use the inhomogeneous equation (Ampere's Law):
$$\vec{\nabla} \times \vec{B}=\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla} (\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}=\frac{\mu}{c} \vec{j}.$$
Now using the Coulomb-gauge condition this simplifies finally to
$$\Delta \vec{A}=-\frac{\mu}{c} \vec{j}.$$
This is the same equation as for the electrostatic potential, just for every Cartesian component of $\vec{A}$. Thus we know the solution via the superposition of Coulomb fields (more formally it's the use of the Green's function of the Laplace operator):
$$\vec{A}(\vec{x})=\frac{\mu}{4 \pi c} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\vec{j}(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
Note that this is only consistent, if
$$\vec{\nabla} \cdot \vec{j}=0,$$
which is charge conservation for the static case. If this integrability condition (which follows also from the Maxwell equation (Ampere's Law)) is fulfilled, then the found solution also fulfills the Coulomb-gauge condition, as it must be for consistency.

7. Jan 16, 2017

enerieire

Thank you,

A=(1/cr)*∫JdV

8. Jan 16, 2017

vanhees71

This formula is obviously not fully correct. It's only valid for $r=|\vec{x}| \gg R$ (where $R$ is the radius within which you have a non-vanishing $\vec{j}$). It's the first term in the magnetostatic multipole expansion.

9. Jan 16, 2017

enerieire

okay, now it's clear.
Could you help me with this other connected problem?