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

Starting point not defined for contour integrals?

  1. Apr 6, 2006 #1
    I was asked to use antiderivative to evaluate the integral of z^(1/2) dz , over contour C1. for which the intergrand is the branch z^(1/2) = r^(1/2)exp(ix/2) , (r>0, 0<x<2*pi) and C1 is the contour from z=-3 to z=3.

    The books says the integrand is piecewise continuous on C1 and integral exist there but the branch of z^(1/2) is not defined on the ray x=0 (or pt z=3). And so we have to integrate it over another branch
    f(z) = r^(1/2)exp(ix/2) , (r>0, -pi/2<x<3*pi/2)

    I don't really understand why the branch is not defined at x=0.
    and also, why do we choose -pi/2 to 3pi/2 as our new range for the branch? How do we determine that?

    thanks a lot!
     
  2. jcsd
  3. Apr 7, 2006 #2
    As you know, the square root has two branches in the real numbers: the positive roots and the negative roots. Positive roots and negative roots are separated by the square root at x=0, which is the ramification point. You can prolongate any branch of the square root to x=0, just defining [tex]\sqrt{0} = 0[/tex], which is consistent, but that prolongation won't be differentiable at x=0.

    In the complex plane, something similar happens. If you take any branch of the square root and prolongate it as much as you can, you'll eventually find that you can define the branch everywhere except in a ray starting at x=0. This ray determines which branch are you taking. For example, you used two branches in your post: z^(1/2) = r^(1/2)exp(ix/2) , (r>0, 0<x<2*pi), which ray is [tex]\mathrm{arg}(z) = 0[/tex], and f(z) = r^(1/2)exp(ix/2) , (r>0, -pi/2<x<3*pi/2), which ray is [tex]\mathrm{arg}(z) = -\pi/2[/tex].

    As in the real case, you may extend any branch to z=0, defining [tex]\sqrt{0} = 0[/tex], but the problem is that in z=0 the square root is not differentiable. That means that in z=0 the square root is not holomorphic, hence you can't take much advantage of path integrals containing z=0.

    So the book is somewhat wrong: you can indeed take the square root of z=0, but you can't integrate over it using Cauchy theorem (for example). Your paths must turn around z=0.

    The second branch is chosen using two conditions:
    1) The new branch must define que square root of z=3.
    2) The new branch must be compatible with the first one in a nonempty open set.

    This may seem complicated. The idea is that any branch of the square root that you take is good, save the branch with range -pi to pi, which does not verify 2).

    The problem is that C1 has as ending point z=3, which is in the undefined ray of your branch with range (0,2pi). So you need another branch of the square root to complete the integral. And your new branch must be coincident with the old one somewhere, so you can compute the transition from one to the other. If you are familiar with manifold theory, the problem can be translated as the need of another chart to carry out the calculation.


    I hope this explanations helps.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Starting point not defined for contour integrals?
  1. Contour integration (Replies: 2)

  2. Contour integrals (Replies: 2)

  3. Contour Integral (Replies: 2)

Loading...