Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kelvin-Stokes' theorem in the presence of singularities

  1. Oct 23, 2015 #1
    My concern regards solving a class of somewhat ill-defined surface integrals occurring in Mathematical Physics and EMF Theory. I'll be using a simplified, representative example.

    Consider the surface S given by
    (x/2)^2 + (y/2)^2 + (z/3)^2 = 1
    0 <= z.
    And the integral ∫F⋅ds where we have in cylindrical coordinates F = eρ/ρ.
    That is to say, we are integrating over half an ellipsoid, with the boundary of a radius 2 circle at z = 0 and the height of z = 3. Furthermore we have a singularity at z = 3 -> ρ = 0. The field amounts to, short of a constant, the electric field of an infinitely long, thin wire at the z-axis.

    There are various ways of evaluating this, for example by arguments invoking Gauss' law, or through a painstaking direct evaluation using iterated integrals, with a limit around the singularity. However, this is one of those idealized cases where the electric field has no divergence, and so we have a vector potential; F = ∇ × A.

    It is therefore appealing to invoke Stokes' Theorem. However, we clearly cannot use it directly - the singularity means that the surface is not simply connected. But going back to the idea of examining limiting behaviour around the singularity, we can make the following argument that gives us the correct result: Consider the boundary δS1 at z = 0, and a small closed loop δS2 close to z = 3. In the limit of a small δS2, by taking the difference of the line integrals of A over the two loops with the appropriate orientations, we should arrive at the original, desired integral over S. We can now observe that the integral over S would, by similar boundaries, be the same as an integral over a cylinder of radius 2 and height 3 with a closed top with a small hole in it. But we have no flux in the z-direction, and we can thus simply open the top, and do our surface integral over a cylinder with open tops and bottoms. This takes us back to the results known from EMFT, and the trivial computation gives us, in this case, the answer of +-6π depending on orientation.

    I understand that, despite the apparent problems, this argument is not entirely invalid. My question is, under what parameters is it valid? It appears to me in this case that the argument works because we don't have any flux in the z-direction, so we don't have to consider the line integral in the neighbourhood of the singularity. Are there other pitfalls? Geometrically, it appears to me that you can use the line integral around a hole to "connect" a region even if the integrand cannot be defined in the entire region to be excised, but I'm not entirely sure this works mathematically.

    I hope the above is clear enough!
  2. jcsd
  3. Oct 23, 2015 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The critical step is in ensuring that the surface integral over a dome with a hole in the top is equal to the line integral in one direction (say clockwise) around δS1 plus the line integral in the opposite direction around δS2. The rest is just a question of applying limits.

    Whether Stokes' Theorem authorises that step depends on which version we use. A simple presentation of it may make the requirement for the surface to be simply connected because otherwise one needs more detail to specify the directions of line integration around the different boundaries. But if I remember my homology theory correctly (it's been a while), the notion of a boundary of a non-simply connected surface is well-defined, including indications of direction for each loop and, if we integrate around each component loop in the boundary, in the correct direction and add them up, we should get the surface integral.

    I'm pretty sure that a more mathematical version of Stokes' Theorem would not require the surface to be simply connected, in which case your approach would be valid. One just has to be careful to do the line integrals in the correct directions, which in this case is opposite.
  4. Oct 24, 2015 #3
    That makes sense to me. So, a follow-up question: Provided I excise the singularity, can I invoke the zero divergence at all defined points to show the existence of a vector potential? I recall that this can be a problem with scalar potentials.

    I'm still a bit iffy about the limiting behaviour of the line integral around the singularity as the radius becomes small. But of course, all I do is show that it approaches another line integral, bounding a surface where the surface integral is not dependent on the radius of the line integral.

    Intuitively and geometrically, of course, I understand that the lack of divergence allows me to deform the surface in a manner appropriate with respect to symmetry (i.e. enclosing no new field lines). But as always I'm concerned about where this intuition breaks down in mathematical terms. I'm going to make for a very neurotic physicist...
    Last edited by a moderator: Oct 24, 2015
  5. Oct 24, 2015 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I see from this page that there are some technical conditions that need to be met before one can make the converse inference that zero divergence implies existence of a vector potential. The page links to a page on the Helmholtz Decomposition, which appears to explain the technical conditions, but I am not familiar with it and did not work through it, as it was rather long.

    Regarding your concern over the limiting behaviour of the line integral: If I'm reading your problem statement correctly, isn't it always zero, so that the limit as the radius approaches zero must be zero?

    BTW, this may not help with the general case, but it seems to me that the surface integral can be easily computed without Stokes by just expressing it as the sum of the integrals over two surfaces obtained by making a vertical slice through the dome. For the givenb vector field F, the integral over an infinitesimal area at one place on the dome would be the negative of the integral at the opposite side, so the two integrals must cancel exactly.
  6. Oct 24, 2015 #5
    Yeah, I don't think there should be a problem with the decomposition.

    Hmm, not the line integral of the vector potential. I can't seem to recall what it turns out to be at the moment, though. Anyway, my concern is not really with evaluating the limit (however appropriate that would be) but with the fact that to make it arbitrarily close to the line integral I want (in order to deform the surface), I have to make it arbitrarily close to the singularity. It's not immediately clear to me whether this might be problematic in the general case.

    There are a myriad ways of solving this particular problem, but I'm not exactly sure about your reasoning here. The field lines all enter the surface with the same orientation, so they don't cancel... The answer in this case is simply the area of an open cylinder enclosing the dome.
  7. Oct 25, 2015 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yeah I completely stuffed that up. I mixed up my line and surface integrals. Let's pretend I didn't write that para.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook