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Scalar Potentials Conceptual Confusion

  1. Mar 8, 2008 #1
    I'm kind of confused as to how to determine when a vector field is conservative. For example, if we consider the following scalar field:

    [tex]\varphi[/tex] = arctan([tex]\frac{y}{x}[/tex])

    We see that the gradient is:

    F = [tex]\nabla[/tex][tex]\varphi[/tex] = [tex]\frac{-y i + x j}{x^{2} + y^{2}}[/tex]

    However, F is not a conservative vector field. Can someone please tell me why this is so? In class we've been discussing topology topics including simply connected domains, and other relevant topics.

    I don't think its because F is undefined at the origin because when i look at the problem in the context of electrodynamics, we know that the scalar potential associated with the Electric Vector Field goes off as 1/r. This would therefore make the Electric field non-conservative, which isn't true since we can simply take the end points when computing potential differences.

    So, is it because the field is not simply connected (i.e. the domain has a hole which is the z axis)?

    If someone could please help me understand this or point me to a place which could help that would be great.
     
  2. jcsd
  3. Mar 8, 2008 #2

    Dick

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    Yes, it's not necessarily conservative because the domain is not simply connected. But just because the domain is NOT simply connected doesn't mean that field is NOT conservative. The E field is conservative, your F field isn't. Integrate it around a closed loop circling the hole.
     
  4. Mar 8, 2008 #3

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    The proof of the theorem that a force derived from a potential is conservative relies on the fact that the underlying space is simply connected. Specifically, we need to turn the integral around a loop into a surface integral over an area bounded by the loop (using stoke's theorem), but the existence of such an area is equivalent to the loop being contractible. A simply connected space is one in which every loop is contractible, so the proof goes ahead fine in such spaces, but in general there may be non-contractible loops, and the integral around these may not vanish.

    n-dimensional euclidean space ([itex]\mathbb{R}^n[/itex]) is simply connected, but in 2 dimensions, if you remove a point you get a non-simply connected space (if you do this in 1-dimension, you get a non-connected space, and there are also higher dimensional analogues of these properties, studied in the theory of homotopy groups). Since your function doesn't have a well-defined value at 0, you can only define it on [itex]\mathbb{R}^2-\{0\}[/itex], which isn't simply connected, so the theorem doesn't apply.

    Actually, you can't even defined it here, at least not continuously: you have to pick an angle where it jumps by 2[itex]\pi[/itex]. The proper domain for the function is a riemann surface, specifically the one for ln(z), shown here. But you can still define it locally (ie, on a simply connected subset), and the theorem still works for an contractible loop.
     
    Last edited: Mar 8, 2008
  5. Mar 11, 2008 #4
    Clearly, if the domain isn't simply connected, F can be un-conservative. Is there some systematic way to determine if a vector field in such a domain is conservative or not? I have been trying to do the following, but I'm not sure if its enough:

    First i check that [tex]\nabla[/tex] [tex]\times[/tex] F = 0

    Then is it enough to simply compare 2 line integrals taken about different paths between any two points? I am concerned that I might run into a situation where the following happens:

    [tex]\int[/tex][tex]_{C1}[/tex]FdR = [tex]\int[/tex][tex]_{C2}[/tex]FdR

    even though the field is really un-conservative. Is this even possible? Are there other ways of analyzing a vector field in a non-simply connected domain to determine if it is conservative or not? Can I tell somehow from the potential, [tex]\varphi[/tex] alone if the vector field, F = [tex]\nabla[/tex][tex]\varphi[/tex] is conservative or not (once again, in a domain that isn't simply connected). Thank you so much for all the help! I appreciate it greatly.
     
  6. Mar 12, 2008 #5

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    By definition, a vector field is conservative if its line integral around any closed loop is zero. Two equivalent conditions are:

    1. Its integral between any two points is independent of the path between the points.
    2. It is the gradient of a scalar function.

    These equivalences are not hard to prove. By scalar function, I mean one that is continuous and unambiguously defined everywhere, so arctan(y/x) doesn't count. Since we've already seen F isn't conservative, we can deduce there is no scalar function it is the gradient of.

    The rest of the post discusses some of the deeper theory underlying this stuff. Sorry for the length. Anyway, this is all more naturally expressed in the language of differential forms, where there is a single operation d that, when applied to a scalar field (0-form) gives its gradient, and when applied to a vector field (1-form) gives its curl.

    (More precisely, a 1-form is something different than a vector, but this distinction disappears when the manifold in question is a subset of euclidean space. Also, curl is only a vector (or more precisely, a pseudo-vector) in 3 dimensions. In general, it is a 2-form, and, for example, in 2D it is a (pseudo) scalar (which can be taken as the z-component of the curl if you add an imaginary 3rd dimension), and in 1D it doesn't exist (since all vectors are parallel). But we'll ignore this here.)

    A form f for which df=0 is called closed, and one for which there exists a g such that f=dg is called exact. It turns out ddf=0 for all forms, so every exact form is closed. The question then arises whether every closed form is exact.

    In this case, an exact 1-form is a vector field which is the gradient of a scalar, ie, the exact 1-forms are precisely the conservative vector fields. A closed 1-form is one whose curl is zero. As we've seen, F is closed but not exact, so the answer to the question at the end of the last paragraph is no, in general. However, it is possible to define a function locally that it is the gradient of (just take arctan on a region that doesn't go all the way around the hole). This is true in general for closed forms, ie, we can locally construct forms g such that f=dg, but it isn't always possible to extend the definition of g to the entire manifold.

    But in euclidean space, stokes theorem says that for any curlless vector field we can unambiguouly construct a (globally defined) scalar function which it is the gradient of. So there's some difference between euclidean space and [itex]\mathbb{R}^2-\{0\}[/itex] that means all closed forms are exact in the former but not in the latter.

    This idea is generalized and studied in the theory of de Rham cohomology, which is one of many examples (along with the homotopy groups I mentioned earlier) of an algebraic object assigned to a topological space which encodes topological information about the space. It basically counts the number of closed p-forms that aren't exact, and for each one of these we can say there is a "p dimensional hole" in the space. For example, the letter A has a 1D hole (you could trap a line in it) while lower case i has a 0D hole, ie, it is disconnected (you could trap a plane in it) and a hollow ball has a 2D hole (you could trap a point in it. I'm just sketching the absolute basics of the really rich and interesting field of algebraic topology.
     
    Last edited: Mar 12, 2008
  7. Mar 12, 2008 #6

    Dick

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    If the curl is zero then the field is 'locally' the gradient of a function. That means the integral is zero around any loop that does not contain one of the holes in the domain. If you check that in addition the integral around each of the holes is also zero, then you can conclude that the integral around any closed contour is zero. You can prove this is true just by looking at the ways you can deform a contour without changing its integral.
     
  8. Mar 12, 2008 #7
    Thanks for all the help! So, the way to go about determining whether such a field is conservative or not is to do line integrals around the the holes in the domain
     
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