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Finding the Vector Potential

  1. Nov 20, 2005 #1
    [tex]\mathbf{F}(x,y,z)=(x^2+yz,y^2+zx,-2z(x+y))[/tex] Find the vector potential.

    A vector potential [tex]\mathbf{V}[/tex] would have to satisfy



    [tex]\Rightarrow \mathbf{V}_y=zx+M_y(y)[/tex]
    [tex]\Rightarrow M_y(y)=y^2[/tex]
    [tex]\Rightarrow \mathbf{V}=\frac{x^3}{3}+xyz+\frac{y^3}{3}+N(z)[/tex]
    [tex]\Rightarrow \mathbf{V}_z=xy+N_z(z)[/tex]

    However, here I can't find [tex]N_z(z)[/tex].
  2. jcsd
  3. Nov 20, 2005 #2

    F is a vector field, right?,,, if so, a SCALAR Potential function V satisfies the relations you have written: F=(-) grad V, ( the minus sign is just a convention used in certain physical problems and may be removed).

    A vectorial potential function A of F, (as it is understood in physics) must satisfy F= rot A, where rot is the rotational operator.

    In general, You may try to express F as

    F= (-)grad V + rot A

    If you really mean a Vector Potential then it must be the rot part, i think.
    Last edited: Nov 20, 2005
  4. Nov 20, 2005 #3

    I think is the opposite, V is the potential function, and scalar function V=V(x,y,z) and F is the Vector Field F=f1(x,y,z) i + f2(x,y,z) j + f3(x,y,z) k , where i,j,k are unit orthogonal vectors in the x,y,z directions.

    If V were a vector (potential) function, you won't be able to apply to it the del operator as impliying 'gradient'.
  5. Nov 20, 2005 #4
    I don't think so

    Nop, F is a vector function!

    del=nabla= grad operator is a vector operator

    del= (d/dx) i + (d/dy) j + (d/dz) k

    hence V, NEEDS to be and scalar function in order to apply to it the del operator and hence the nomenclature scalar potential for V.

    When you apply del to V, you get a VECTOR function

    F= del V= dV/dx i + dV/dy j + dV/dz k= [dV/dx, dV/dy, dV/dz], and hence the nomenclature vector field for F.

    If V is a vector function (potential) the application of the operator del to it makes any sense, at least in the 'traditional' sense. (if you think in diads or something like that then maybe).

    a Vector potential field for a vector field F may be obtained, for instance, by the cross product of del operater and another vector field V1

    so that

    F= del x V1

    And example from physics.

    The Electric Field may be expresed as

    E= - del (phi) with phi being the SCALAR electric potencial

    and the Magnetic Field as

    B= del x A with A being the VECTOR magnetic potential
  6. Nov 20, 2005 #5
    http://en.wikipedia.org/wiki/Potential, I am a math guy but I have taken a fair bit of physics. Arguments such as these seem to occur frequently between the two opposing camps. I go by what I read in my books. I am aware that there is an underlying physical interpretation which you seem to be commenting on. I have been referencing a couple of standard Calculus books, but I found a link at wikipedia which references my claim. The book which has the defintion I read, and more or less regurgated in my last post, is Larson, Hostetler Edwards, Mulitivariable Calculus 6th edition page 988.
  7. Nov 20, 2005 #6
  8. Nov 20, 2005 #7
    Notice that in the links you post they refer to F as a vector field and V as the scalar potential. That Is what i'm saying.

    I know that there may be ambiguities in nomenclature between different sciences, but notice also that in the original post he/she says: 'find the vector potential', so, either he/she made a mistake and was trying to say scalar potential or he/she REALLY meaned vector potential, that is what motivated my post.
  9. Nov 20, 2005 #8
    I mean the potential of a conservative vector field. The choice of V is completely arbitrary. Sorry about the confusion, I am in math and didn't realize notations could be ambiguous to the physics people. I've since then solved it. Thanks for the help.
  10. Nov 20, 2005 #9

    Ok, lats post about this, because he/she has already solved the problem.

    If F=del(V) , F is the conservative vector field and V the scalar potential.

    If V=del(F), V is the conservative vector field and F the scalar potential.

    What I've posted is ok for both cases. No confusion.

    F in the original post has components, it's a vector field, and hence
    F=del(V) is the one that applies.



    See also, Stweart, "Calculus" & Marsden & Tromba "Vector Calculus", or your favorite text ;).
  11. Nov 21, 2005 #10
    Now I am confused. To many references to F and V. What I wrote above is wrong. F and V should be transposed. I was referencing to the F and V in my book. The problem is just really familiar to the ones I encountered while being a reader for multi calc so the solution was more or less profunctory. Unfortunately my inability to correctly delineate my solution probably did more harm than good. Sorry.
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