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If F central field then its potential V is radial

  1. Dec 3, 2009 #1
    1. The problem statement, all variables and given/known data

    "Let F be a central vector field, that is, [tex]F(x,y,z) = h(x,y,z)(x,y,z)[/tex] for some smooth function [tex]h[/tex].
    Prove that its potential [tex]V[/tex] is radial, i.e., [tex]V(x,y,z) = g(\sqrt{x^2 + y^2 + z^2})[/tex] for some smooth function [tex]g[/tex]"

    2. Relevant equations

    Since we know that a central vector field must be conservative, we have that [tex]F(x,y,z)=-\nabla{V(x,y,z)}[/tex]

    3. The attempt at a solution

    The original problem statement is slightly different, in fact, I have to prove that given a central vector field F, there exists a radial function [tex]g[/tex] such that
    [tex]F=g(\sqrt{x^2+y^2+z^2})(x,y,z)[/tex], but the existence of this function follows directly once the first problem statement is proved.
    My first attempt to prove it has been using Taylor, using the [tex]V(x,y,z)[/tex] smoothness and the fact that [tex]\nabla{V(x,y,z)} = -h(x,y,z)(x,y,z)[/tex].
    By Taylor polynomial we have for any point [tex]a = (a_1,a_2,a_3)[/tex] in the domain of [tex]V[/tex],

    [tex]V(x,y,z) = V(a_1,a_2,a_3) + \nabla{V(x-a_1,y-a_2,z-a_3)}^t(x-a_1,y-a_2,z-a_3) + R_{2,a} (x,y,z) [/tex]


    [tex]V(x,y,z) = V(a_1,a_2,a_3) - h(x-a_1,y-a_2,z-a_3)(x-a_1,y-a_2,z-a_3)^t(x-a_1,y-a_2,z-a_3) + R_{2,a} (x,y,z) = [/tex]
    [tex]V(a_1,a_2,a_3) - h(x-a_1,y-a_2,z-a_3)||(x-a_1,y-a_2,z-a_3)||^2 + R_{2,a} (x,y,z) [/tex]

    But that does not show anything.
    It would be interesting to have a hint of how should I proceed to prove this statement.
    Thanks for your help.
    Last edited: Dec 3, 2009
  2. jcsd
  3. Dec 3, 2009 #2
    Why not just integrate [itex]F(x,y,z)[/itex] and show that

    V(x,y,z)=-\int F(x,y,z)\cdot d\mathbf{r}
  4. Dec 8, 2009 #3
    First of all, thanks for your attention, JDWood983.
    What I want to prove is that given F a central vector field, then its potential V scalar function, defined as

    [tex] -\nabla V(x,y,z) = F(x,y,z) = - (\frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z}) [/tex]

    is a radial function, that is

    [tex] V(x,y,z) = g(\sqrt{x^2+y^2+z^2}) [/tex]

    for some scalar function [tex]g[/tex].
    Therefore, integrating the way you say does not show anything.
    In fact, the result would be another vector field, wouldn't it? Anyway, thanks for your attention!
    Does anyone know how to proceed? Any hint?
  5. Dec 8, 2009 #4


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    I have a hunch that you're not using the fact that the vector field is conservative to help you. What does that fact imply? I suspect that thinking about this will help you on your way.
  6. Dec 8, 2009 #5
    Actually, if you notice on integral I gave you, it is the dot product of the force vector with the infinitesimal change in position vector that you are integrating over. The dot product of two vectors is a scalar.

    I would take AEM's advice and look at what it means to be conservative, then use the integral I gave you in terms of ([itex]r,\theta,\phi[/itex]) rather than ([itex]x,y,z[/itex]).
  7. Dec 8, 2009 #6


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    I've been pondering your problem a little. first, can you explain this notation

    [tex]F(x,y,z) = h(x,y,z)(x,y,z)[/tex] for some smooth function [tex]h[/tex]

    a little more. I'm from an older generation and we'd probably write this out in terms of unit vectors i,j,k. As I read your notation, there is absolutely nothing to indicate to me that you have prescribed a central vector field. However, this expression


    immediately tells me that the function g depends on the distance from the origin of your coordinate system. That being the case, with the other hints you've been given, the demonstration is almost immediate.
    Last edited: Dec 8, 2009
  8. Dec 8, 2009 #7
    It looks to me that he means this as


    where [itex]h(x,y,z)[/itex] is just some function. If this is the case, I agree the calculation should be nearly immediate.
  9. Dec 8, 2009 #8


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    Homework Helper

    I thought it was
    since [itex]x\hat{x}+y\hat{y}+z\hat{z}[/itex] is a radial vector. So the original problem talks about a function h which is a function of x, y, and z multiplied by that radial vector.
  10. Dec 8, 2009 #9
    Correct. I started writing it as [itex](\hat{x},\hat{y},\hat{z})[/itex] then decided I should write it as the vector instead and forgot my components! Big boo-boo there!
  11. Dec 9, 2009 #10
    First of all, thanks a lot to JDWood983, AEM and Diazona.

    Yes, you are right, it is a line integral.

    Yes, that would be an approach, AEM, but I was given at class a proof using spherical coordinates and I was supposed to give one that does not involve sphercial conversion.
    Anyway, the professor only gave an sketch of the proof using spherical coordinates, so your effort on trying to lead me to prove this way has been rather useful too. Thanks.

    Yes, sorry for using that notation, it can be a little fuzzy.
    It means just what diazona says,

    Thanks a lot for your help. Well, finally I have been given a sketch of a proof that does not involve spherical coordinates, I will post here the argument, but I do not see the conclusion. I will see it in two coordinates, that is, [tex] \mathbb{R}^2 [/tex].

    We have a central vector field given by
    [tex] F = f(x,y)(x,y) [/tex] for some smooth function [tex] f(x,y) [/tex],
    that is,
    [tex] F_x = f(x,y)x [/tex]
    [tex] F_y = f(x,y)y [/tex]

    Since all central vector fields are conservative, we have that curl of F is zero, and from the cartesian equation of F curl, we have:

    [tex] \frac{\partial F_x}{\partial y} = \frac{\partial F_y}{\partial x} (1) [/tex]

    If we calculate these derivatives, we have

    [tex] \frac{\partial F_x}{\partial y} = \frac{\partial f}{\partial y}x [/tex]
    [tex] \frac{\partial F_y}{\partial x} = \frac{\partial f}{\partial x}y [/tex]

    And from equation (1),

    [tex] \frac{\partial f}{\partial y}x = \frac{\partial f}{\partial x}y [/tex]

    which implies

    [tex] \frac{\partial f}{y\partial y} = \frac{\partial f}{x\partial x} [/tex]

    Until here, everything is correct. But here comes the first dubious reasoning:


    [tex] \int y\partial y = y^2/2 [/tex]


    [tex] \int x\partial x = x^2/2 [/tex]

    We can conclude that

    [tex] y\partial y = d y^2/2 [/tex]


    [tex] x\partial x = d x^2/2 [/tex]

    So, from here we can conclude

    [tex] \frac{\partial f}{\partial y^2} = \frac{\partial f}{\partial x^2} (2) [/tex]

    I am not really confident with this argument, but that is what I was given as an sketch of an alternative proof.
    Once here, the conclusion was that the only (smooth) function that fulfilled equation (2) was the sphere, in this case 2-sphere or circumference, for geometers. (or 1-sphere for topologists).
    I can see that indeed the sphere fulfills that equation, but I did not believe it was the only one, and the professor gave me a proof of this, consisting of a certain substitution of variables that, from my point of view, was not rigurous at all.
    So, summarizing:
    -I am not sure if the steps taken to go from equation (1) to equation (2) are correct.
    -Is Equation (2) a definitive result, that is, can we interpret it as a characterization of the sphere function? If so, is there some easy and rigurous way to prove it, i.e., a proof that any (smooth) function that fulfills equation (2) is the sphere function? (here smooth means that the function [tex] f \in C^{\infty} [/tex]).

    Thanks a lot for your attention, for your patience and for your help!

  12. Dec 9, 2009 #11


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    Homework Helper

    Actually that is a perfectly valid argument (well... I'm not sure what mathematicians might think of it, but it usually works fine in physics, which is a physicist's working definition of "perfectly valid" ;-)

    You can also think of it as an application of the chain rule, if this makes any more sense. Let [itex]u = x^2[/itex], then

    [tex]\frac{\partial f}{x \partial x} = \frac{1}{x}\frac{\partial f}{\partial u}\frac{\partial u}{\partial x} = \frac{1}{x}\frac{\partial f}{\partial x^2}\frac{\partial x^2}{\partial x} = \frac{1}{x}\frac{\partial f}{\partial x^2}2x = 2\frac{\partial f}{\partial x^2}[/tex]

    Are you familiar with techniques for solving partial differential equations? Specifically, separation of variables? You might be able to use it here to show that a radially symmetric solution is the only solution to this equation, but I haven't worked through the math so I'm not 100% sure. If you're not familiar with that technique... maybe someone else can offer a clear explanation.
  13. Dec 14, 2009 #12
    Thanks a lot for your help, Diazona.

    Actually, I am not familiar with methods to solve PDEs.
    Now, the problem gets reduced to prove the following result:

    A smooth function [tex] f(x,y) [/tex] fulfills the following equation:

    [tex] \frac{\partial f}{\partial x^2} = \frac{\partial f}{\partial y^2} [/tex]

    if, and only if, [tex] f(x,y) [/tex] is a [tex] \emph{radial}[/tex] function, that is, [tex] f(x,y) = h(\sqrt{x^2 + y^2}) [/tex] for some smooth function [tex] h [/tex].

    I have been reading some theory about the separation of variables for solving PDEs, but I am not quite sure how to apply it here to prove that result.
    I will try to prove it by a different method and if I get any result I will post it here.
    If someone could help me with a hint on where to start, it would be very useful.
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