1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Complex Potential

  1. May 30, 2009 #1

    What I've done is worked backwards

    Relevant Equations

    I know ∂ψ/∂y = ∂φ/∂x = u

    and -∂ψ/∂x = ∂φ/∂y = v (Cauchy Riemann equations)

    For the complex potential ω(z) = φ(x,y) + iψ(x,y)

    u = ui + vj

    and dω/dz = u - iv

    Solution working backwards

    So I took his v = 2kx(x2+y2-a2)/D

    And I realised D = (x2+y2-a2)2

    So that means v = 2kx/(x2+y2-a2)

    So what I did was equate them all

    v = -∂ψ/∂x = 2kx/(x2+y2-a2)

    => ∂ψ = -2kx/(x2+y2-a2)∂x

    => ψ = -k log(x2+y2-a2)

    => ψ = -k log(((x-a) + iy)((x+a) - iy))

    And that's how I should have started. I see Z0 = ±a from the question, and so a is real.

    So I would have two equations -ik log((x-a) + iy) and -ik log((x+a) + iy)

    What do I do from here to get to -ik log(((x-a) + iy)((x+a) - iy)) like he did? :/

  2. jcsd
  3. May 30, 2009 #2


    User Avatar

    The first thing to note that since this is fluids which are described by complex potentials, the real and complex components of the potential separately satisfy Laplace's (and Poisson's) equation.

    As the Poisson equation is linear, we may superimpose solutions pertaining to a particular line vortex distribution to obtain the desired, more complicated line vortex distribution.

    We are given that for a single line vortex of strength k at position [tex]z_0[/tex]:

    [tex]\phi(z,z_0) = - i k log(z - z_0)[/tex]

    We want the potential for a line vortex of strength k at [tex]z_0 = a[/tex]:

    [tex]\phi(z,a) = - i k log(z - a)[/tex]

    and a line vortex of strength k at [tex]z_0 = -a[/tex]:

    [tex]\phi(z,-a) = - i k log(z + a)[/tex]

    As the Possion equation is linear, we may superimpose the two:

    [tex]\phi(z) = \phi(z,a) + \phi(z,-a) = - i k log(z - a) - i k log(z + a)[/tex]

    As these are logarithms, adding them gives:

    [tex]\phi(z) = - i k log(z - a) - i k log(z + a) = -i k log[(z-a)(z+a)][/tex]

    As z = x + i y, we get the expected result.
  4. May 30, 2009 #3
    Ah ok thankyou I see perfectly now :)
  5. May 30, 2009 #4
    For the second bit am I supposed to draw this on an argand diagram? How am I supposed to plot it?

    Is a line vortex a point vortex?
  6. May 30, 2009 #5


    User Avatar

    A line vortex is indeed a point vortex, but only in 2D. In 3D, a point vortex and a line vortex represent different things.

    We know that this is a 2D problem as complex potentials are by their nature, two dimensional.

    However the technicality of quoting 'line vortex' in this question is to ensure that this represents the same solution as a line vortex in 3D.

    If you work out the differentials, u and v represent fluid moving in tangential directions.

    Sketching a complex function is pretty difficult. I believe they actually meant you to sketch the flow lines, which would be the velocities.

    Here is a good example of a plot of a complex function:
    Last edited: May 30, 2009
  7. May 30, 2009 #6
    Woah lol, this is a past exam paper question, I doubt he wants me to break out my crayons like that :)

    Could you mildly explain how to sketch these flow lines, I have 3 examples only on this, each one he found the components ur and uθ at z0 = 0

    If I do this I get (saving the working out)

    at z0 = 0

    ur = ∂φ/∂r = 0

    uθ = (1/r)∂φ/∂θ = -k/r

    Then (for the example similar to this question) all he did was draw a point vortex with circular flow lines about z0 = 0 going anticlockwise

    So I draw the same thing but at z0 = -a and z0 = a, and... then I don't know what to do from here!
  8. May 30, 2009 #7


    User Avatar

    Do you have any experience with electric and magnetic field?

    They satisfy Laplace's equation, so if you know how to sketch those, you can apply it directly to fluids.

    Other than that you can either rely on intuition, or look at the equations.

    For intuition, we note than an irrotational point source emits fluid moving in radial directions, and an irrotational point sink sucks in fluid in radial directions. If we superimpose the two, we have lines moving from the source to sink like this http://en.wikipedia.org/wiki/File:Camposcargas.PNG.

    For a mathematical approach, you can only slog through by looking an u and v, and taking them to be the x and y component of velocity at every point.

    You could also look at equipotential surfaces. For fluid problems, the respective real and complex components of complex potentials separate into stream lines and velocity lines. Your flow lines will be the equipotential surfaces of one of them.

    How difficult this is depends on the coordinate system you pick. For a point vortex, we get a function representing equipotentials as circles, and a function representing equipotentials as radial lines. Given that our flow is tangential, we know that velocities flow along equipotenials of one, and perpendicular to the surface of equipotentials of the other.

    Note that we could have easily said that the radial lines represent flow lines, and we would have gotten the same scenario, except for a point source.

    In general, only the cartesian system is useful.

    It helps to identify points of symmetry. For the example above, if one represents clockwise flow, and the other anticlockwise flow, superimposing the velocities mean that the line equidistant to both of them is a flow line, and anything off this flow line forms circles around their respective source. Like this:

    This reference is useful
    http://www.up.ac.za/academic/www.me.up.ac.za/cpc/lectures/english/12/12.html [Broken]
    Last edited by a moderator: May 4, 2017
  9. May 30, 2009 #8
    Oh I see, so we know the fluid is irrotational, so z0 = a and z0 = -a are either a soure or a sink, how do I identify which?

    I'm assuming theyre sources since the question wants me to show it's a strength 2k

    my attempt:


    So because they superimpose on each other, thats 2 times the strength?
    Last edited by a moderator: May 4, 2017
  10. May 30, 2009 #9


    User Avatar

    Exactly, but only at large distances, where it 'appears' as one source with a strength of 2k.

    At distances that are large relative to the separation of the sources, you essentially get equally spaced radial lines pointing out from the center.

    At small distances the variation in the flow lines is sufficient to deduce that there are actually two sources.

    Source or sink for a positive or negative sign is a matter of convention. I believe that the standard convention is to treat positive signs as sources, as well as clockwise flow if rotational.
  11. May 30, 2009 #10
    Great thankyou, thanks alot for the help.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook