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Finding vector potential

  1. Sep 17, 2011 #1
    In problem of finding the vector potential of a vector F = yz i + xz k + xy j,
    the solution gives in Griffith's solution manual is

    http://img843.imageshack.us/img843/2725/vectorpotential.jpg [Broken]

    Uploaded with ImageShack.us

    But I don't understand how we can integrate

    [itex]\frac{\partial Az}{\partial y}[/itex] = yz + [itex]\frac{ \partial Ay}{\partial z}[/itex]

    and get only f (x,z), why can't the partial w.r.t z be a function a function of y?? eg A = xyz

    then the partial w.r.t z is xy, which is a function of y.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 17, 2011 #2

    vanhees71

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    I don't know, how Griffiths comes to his solution, but I'd use the following simpler idea. The equation

    [tex]\vec{F}=\vec{\nabla} \times \vec{A}[/tex]

    has a solution [itex]\vec{A}[/itex] for a given [itex]\vec{F}[/itex], if and only if

    [tex]\vec{\nabla} \cdot \vec{F}=0,[/tex]

    which is fulfilled for your example.

    This solution is not unique, but only determined up to a potential field, i.e., a gradient of a scalar field. Thus, we can impose one constraint. Here, I use the "axial gauge condition"

    [tex]A_z=0.[/tex]

    Then we have

    [tex]\vec{\nabla} \times \vec{A}=\begin{pmatrix}
    -\partial_z A_y \\
    \partial_z A_x \\
    \partial_y A_x - \partial_x A_y
    \end{pmatrix} \stackrel{!}{=} \begin{pmatrix}
    yz \\
    xz \\
    xy
    \end{pmatrix}
    [/tex]

    The first line leads to

    [tex]A_y=-\int_0^z \mathrm{d} z F_x + A_y'(x,y) = -\frac{1}{2} y z^2 + A_y'(x,y)[/tex]

    and the second line

    [tex]A_x=\int_0^z \mathrm{d} z F_y+A_x'(x,y)=\frac{1}{2}x z^2 + A_x'(x,y).[/tex]

    The last line now reads

    [tex]F_z=-\int_0^z \mathrm{d} z (\partial_x F_x + \partial_y F_y)+\partial_x A_y'-\partial_y A_x'.[/tex]

    Because of [itex]\vec{\nabla} \cdot \vec{F}=0[/itex], we have

    [tex]F_z=\int_0^z \mathrm{d} z \partial_z F_z+\partial_x A_y'-\partial_y A_x'
    =F_z(x,y,z)-F_z(x,y,0) + \partial_x A_y'(x,y) - \partial_y A_x'(x,y)
    .[/tex]

    Now we can again arbitrarily set [itex]A_x'(x,y)=0[/itex]. To fulfill the above equation, we just have to set

    [tex]\partial_x A_y'=F_z(x,y,0)=xy \; \Rightarrow A_y'=\frac{1}{2} x^2 y+A_y''(y)[/tex].

    Of course, [itex]A_y''=0[/itex] is good enough since it doesn't contribute to the curl at all. Plugging everything together leads to

    [tex]\vec{A}=\begin{pmatrix}
    x z^2/2 \\
    (x^2 y-y z^2)/2 \\
    0
    \end{pmatrix}.
    [/tex]

    Finally, it's good to check, whether everything is fine. Thus we take the curl

    [tex]\vec{\nabla} \times \vec{A}=\begin{pmatrix}
    -\partial_z A_y \\
    \partial_z A_x \\
    \partial_x A_y - \partial_y A_x
    \end{pmatrix}=\begin{pmatrix}
    yz \\ xz \\ xy
    \end{pmatrix}=\vec{F}.[/tex]

    Thus, we have found a vector potential for [itex]\vec{F}[/itex].
     
  4. Sep 18, 2011 #3
    Can you explain what the axial gauge condition is and how it makes Az = 0?
     
  5. Sep 18, 2011 #4

    vanhees71

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    The vector potential for a given solenoidal vector field is determined up to a gradient of a scalar field since [itex]\vec{\nabla} \times \vec{\nabla} \chi=0[/itex] for any scalar field, [itex]\chi[/itex].

    Now, suppose you have a solution to the equation
    [tex]\vec{\nabla} \times \vec{A}=\vec{F}.[/tex]
    Now, any field
    [tex]\vec{A}'=\vec{A}-\vec{\nabla} \chi[/tex]
    also fulfills this equation and is as good as the original [itex]\vec{A}[/itex]. Thus, to make our life easier, we can impose one additional constraint to our vector potential. Since it's easier to solve for two components rather than three components, we make one component vanishing. So, suppose for a moment, you have found a solution [itex]\vec{A}[/itex] and you like to find another representation such that the gauge transformed field obeys the axial-gauge condition
    [tex]A_z'=0.[/tex]
    Thus, I've to find a scalar field, [itex]\chi[/itex], such that
    [tex]A_z'=A_z-\partial_z \chi=0[/tex].
    It's very easy to see that one possible solution for this equation is
    [tex]\chi(x,y,z)=\int_{z_0}^z \mathrm{d} z' A_z(x,y,z').[/tex]
    Of course, [itex]\chi[/itex] is not completely determined by this condition. You can still add an arbitrary gradient of a gauge field that depends only on [itex]x[/itex] and [itex]y[/itex]. That's why we could choose arbitrarily [itex]A_x'(x,y)=0[/itex] in the solution presented yesterday.
     
  6. Sep 19, 2011 #5

    Bill_K

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    It's an ansatz. As a trial solution he splits each equation up into a pair of equations:

    ∂Az/∂y = ½ yz and ∂Ay/∂z = - ½ yz
    ∂Ax/∂z = ½ xz and ∂Az/∂x = - ½ xz
    ∂Ay/∂x = ½ xy and ∂Ax/∂y = - ½ xy

    from which

    Az = 1/4 z (y2 - x2)
    Ay = 1/4 y (x2 - z2)
    Ax = 1/4 x (z2 - y2)

    I suppose you could say there's no logic to it; but it works, and leads to a nice symmetrical solution.
     
  7. Sep 19, 2011 #6

    dynamicsolo

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    Another way of saying what vanhees71 has is that the given vector field is supposed to be the result of taking curl A = F . (So plainly this is not a conservative potential, or the curl would have given us zero.) The equations Griffiths shows are the components of this curl result, which are supposed to equal F. (And, as Bill K notes, it appears Griffiths, or the solver for the manual, has omitted factors of 1/2 somewhere...)

    We are doing the inverse problem of trying to figure out what the components of A must look like in order to have produced F. We have to integrate each of those components in two ways, since we have information only about the partial derivatives of the components. Integrating [itex]\frac{\partial Az}{\partial y} = \frac{1}{2}yz + \frac{ \partial Ay}{\partial z}[/itex] with respect to y gets us definitely only the [itex]\frac{1}{4} y^{2}z [/itex] term, but leaves an "arbitrary integration function" dependent on the two variables not involved in that integration (analogous to the "arbitrary constant" in single variable integration).

    There is also a second integration taking place by rearranging the curl component differential equation as [itex]\frac{\partial Ay}{\partial z} = \frac{ \partial Az}{\partial y} - \frac{1}{2}yz [/itex] and then integrating with respect to z. This gives Griffith's second result for this equation, and the "arbitrary function" now depends on x and y .

    We go through the same process for all three curl components, which gets us two pieces of information for each component of A. So, for example, on the Ax component, we have

    [tex]A_{x} = \frac{1}{4} xz^{2} + h( x , y ) and A_{x} = - \frac{1}{4} xy^{2} + l( x , z ) . [/tex]

    [And, incidentally, there is a typo for that arbitrary function l ("el" of x and z) . ]

    We see that Ax has a term with powers of x and z , a term with powers of x and y , an arbitrary function of x and y , and another arbitrary function of x and z . So the arbitrary function of one of the integrations is apparently the result for the integration with respect to the other variable. There's nothing left over, so we must have [itex]A_{x} = \frac{1}{4} xz^{2} - \frac{1}{4} xy^{2} [/itex] , and similar results for the other components of A, as Bill K lists.

    Also, if the expression you gave for F is correct, then all of these (1/4)'s should be (1/2)'s.

    (And I'll keep in mind that the solution manual for Griffiths is rife with typoes...)
     
    Last edited: Sep 19, 2011
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