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Deriving Vector Potentials

  1. Oct 11, 2013 #1
    https://dl.dropboxusercontent.com/u/22024273/vectorpotential.png [Broken]

    In the above passage, can someone explain to me where (3.102) comes from?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 11, 2013 #2


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    Take the curl of A and see if it yields B ... as long as it works it proves their contention.

    How did they arrive at it? If I were attempting this I would take the curl of a general expression and then work out which parts are needed/not needed ... and this is apparently what is left!
  4. Oct 11, 2013 #3


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    Obviously a necessary condition for [itex]\vec{B}[/itex] to be the curl of a vector field is
    [tex]\vec{\nabla} \cdot \vec{B}=0.[/tex]
    Then the vector potential is only determined up to the gradient of a scalar field. Thus one can impose one additional condition ("choice of a gauge"). In this case an axial gauge is chosen, i.e.,
    [tex]\vec{\nabla} \times \vec{A}=\vec{B}, \quad A_x=0.[/tex]
    Then we have in Cartesian coordinates
    [tex]B_x=\partial_y A_z-\partial_z A_y, \quad B_y=-\partial_x A_z, \quad B_z=\partial_x A_y.[/tex]
    From the last equation we get
    [tex]A_y=\int_{x_0}^{x} \mathrm{d} x' B_z(x',y,z)+\tilde{A}_y(y,z)[/tex]
    and from the 2nd equation
    [tex]A_z=-\int_{x_0}^x \mathrm{d} x' B_y(x',y,z) + \tilde{A}_z(y,z).[/tex]
    The first equation now reads
    [tex]B_x(x,y,z)=-\int_{x_0}^{x} \mathrm{d} x' [\partial_y B_y(x',y,z)+\partial_z B_z(x',y,z)]+\partial_y \tilde{A}_z(y,z)-\partial_z \tilde{A}_y(y,z).[/tex]
    Now from [itex]\vec{\nabla} \cdot \vec{B}=0[/itex] we find
    [tex]B_x(x,y,z)=B_x(x,y,z)-B_x(x_0,y,z) +\partial_y \tilde{A}_z(y,z)-\partial_z \tilde{A}_y(y,z)[/tex]
    [tex]B_x(x_0,y,z)=\partial_y \tilde{A}_z(y,z)-\partial_z \tilde{A}_y(y,z).[/tex]
    This equation we can satisfy by setting
    [tex]\tilde{A}_y=0, \quad \tilde{A}_z(y,z)=\int_{y_0}^{y} \mathrm{d}y' B_x(x_0,y',z).[/tex]
    Putting all together we get
    [tex]A_x=0, \quad A_y=\int_{x_0}^{x} \mathrm{d} x' B_z(x',y,z), \quad A_z=\int_{y_0}^{y} \mathrm{d}y' B_x(x_0,y',z)-\int_{x_0}^x \mathrm{d} x' B_y(x',y,z).[/tex]
    This is what's claimed in the OP.
  5. Oct 11, 2013 #4
    Thank you for the replies.
    Makes sense now.
    Thanks for taking the time to go through all the working as well vanhees, was helpful in checking my own working through it.
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