1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Potential vector's formulas

  1. Jan 15, 2017 #1
    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.

    Can someone please help me?
  2. jcsd
  3. Jan 15, 2017 #2


    User Avatar
    Homework Helper
    Gold Member

    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.
  4. Jan 15, 2017 #3
    And if it was a space derivative of A? Would it have any sense?

    Do you know something about this form of the Poynting vector?
    S= c/4pi H^2 n
  5. Jan 15, 2017 #4


    User Avatar
    Homework Helper
    Gold Member

    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
  6. Jan 16, 2017 #5
    Thank you very much for the answers!!
    I'm going to post another problem about Poynting's flux in a new question.
  7. Jan 16, 2017 #6


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    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.
  8. Jan 16, 2017 #7
    Thank you,
    even if my doubt was about this form of the formula:

  9. Jan 16, 2017 #8


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    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.
  10. Jan 16, 2017 #9

    okay, now it's clear.
    Could you help me with this other connected problem?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Potential vector's formulas
  1. Vector Potential (Replies: 5)

  2. Vector potential (Replies: 7)

  3. Vector potential (Replies: 3)

  4. Vector Potential (Replies: 1)

  5. Vector potential (Replies: 5)