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What is the physical interpretaion of the vector potential.

  1. Jun 15, 2011 #1
    I was reading the text of electricity and magnetism by griffiths. Here I read a term called magnetic potential but I did not completely understood the physical essence of the term, neither it is explained in the book. It should have some physical interpretation as it is named a potential. In what sense it is a potential??
  2. jcsd
  3. Jun 15, 2011 #2


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    As far as I know, it's simply a mathematical tool with no concrete physical meaning. I can't imagine it has physical meaning since for any configuration, the vector and scalar potentials are not uniquely defined.
  4. Jun 15, 2011 #3
    Hello girsimran,

    There are many vector potential functions in Physics.

    Basically if we can assign a scalar value to every point in some region of space, the vector potential is the gradient of this scalar as we pass from one point to another.

    Mathematically if we have a scalar function of position, [itex]\varphi[/itex] at some point

    then the vector a = [itex]\nabla\varphi[/itex] is the vector value and direction of the vector potential at this point.

    All the scalars form a scalar field and all the vectors field over the region in question.

    Examples are Gravitational potential, Electrostatic potential, Magnetostatic potential, Fluid velocity potential.

    Very often (as with the fluid velocity field) we have the vector potential and infer the existance of a scalar from it.

    go well
  5. Jun 15, 2011 #4
    Hey first of thanks for the answer but what you wrote about is the scalar potential. I'm asking about vector potential. This one http://en.wikipedia.org/wiki/Vector_potential
  6. Jun 15, 2011 #5


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    The idea is the same. Potentials have no significant meaning in classical e/m since they're just mathematical tools that are not unique to any given situation. Things get a little clouded when you start looking at quantum mechanical effects, however. You may want to look up the Aharonov-Bohm effect.
  7. Jun 15, 2011 #6
    Because the magnetic field is solenoidal ([itex]\mathrm{div} \vec{B} = 0[/itex]), it can be expressed as the curl of a vector field:
    \vec{B} =\mathrm{curl} \, \vec{A}
    Since the curl of a gradient of any scalar function is zero ([itex]\mathrm{curl} \, \mathrm{grad} \, \phi = 0[/itex]), the vector potential is determined only up to a gradient of a scalar function:
    \vec{A} = \vec{A'} + \mathrm{grad} \, \Lambda
    The flux of the magnetic field through a closed contour is:
    \Phi = \int_{S}{\vec{B} \cdot \hat{n} \, da} = \int_{S}{\mathrm{curl} \vec{A} \cdot \hat{n} \, da} = \oint_{C}{\vec{A} \cdot d\vec{l}}
    is given by the circulation of the vector potential around its boundary. We see that the arbitrariness of the definition of the vector potential disappears since the circulation of a gradient around a closed contour is always zero:
    \oint_{C}{\mathrm{grad} \, \Lambda \cdot d\vec{l}} = 0
    Last edited: Jun 15, 2011
  8. Jun 16, 2011 #7
    Thanks for answering. But I do not agree that potentials are just mathematical constructs. Gravitaional potentail or electric potential has significant well understood physical interpretation which is in accordance with its name, potential (ie potential to do work.. due to its location in space)
  9. Jun 16, 2011 #8


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    That's why I specifically noted classical e/m potentials. I should have further noted this actually only applies to the vector potential as well.
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