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Stream function/velocity potential in fluid mechanics

  1. Dec 11, 2013 #1
    Hi! When you are calculating ##\psi## or ##\phi## from the horizontal or vertical velocity components of a fluid's velocity, you end up with a constant ##C## carried along with the function. In my class, the professor just said something about them being arbitrary (I think..) and set them to zero.

    Could somebody please explain to me exactly WHY we can just ignore them? Just a random guess: Does it have something to do with us being able to define origo of our coordinate systems anywhere we like, including at a point which makes ##C=0##?
  2. jcsd
  3. Dec 11, 2013 #2

    Andy Resnick

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    If I understand your question, since the velocity field is set by spatial derivatives of the stream function, the 'arbitrary constant' doesn't affect the velocity field and so can be ignored.
  4. Dec 11, 2013 #3
    so because it's only the velocity field which has real-world applications, the constants can be ignored?
  5. Dec 11, 2013 #4


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    Not really. Think about how the stream function and velocity potential are related to the velocity in a two-dimensional flow:
    [tex]u = \dfrac{\partial \phi}{\partial x} = \dfrac{\partial \psi}{\partial y}[/tex]
    [tex]v = \dfrac{\partial \phi}{\partial y} = -\dfrac{\partial \psi}{\partial x}[/tex]
    If you have something along the lines of [itex]\phi = f(x,y) + C[/itex] or [itex]\psi = g(x,y) + C[/itex], then when you take its derivative and extract a velocity, the constant [itex]C[/itex] falls out anyway. In that sense, the constant is arbitrary and doesn't matter. You only need the general solution to the differential equation, not the specific solution for a given set of boundary conditions.

    If instead you are looking at the value of [itex]\psi[/itex] or [itex]\phi[/itex] directly (for instance, for plotting streamlines or equipotential lines), then the constant does mean something. In that case, if you look in a 2D plane, setting an actual value for [itex]C[/itex] and holding it constant while you vary the other parameters in your equations [itex]f(x,y)[/itex] and [itex]g(x,y)[/itex], the resulting paths will trace out the equipotential lines and streamlines in the [itex](x,y)[/itex]-plane for [itex]f[/itex] and [itex]g[/itex] respectively.

    For example, if you want to plot the stagnation streamline on a body in potential flow, you would use some point that you know on that streamline (typically the stagnation point itself) and use its coordinates in your function [itex]g(x,y)[/itex] to find the value of the constant. Then, in varying one coordinate to find the other, you will trace out the stagnation streamline. Or you could just pick random values of the constant to map out random streamlines to your heart's content.
  6. Dec 11, 2013 #5
    Sometimes the stream function is defined such that it bears a specific physical relationship to the volumetric flow rate. For example, if you have flow in a tube, the stream function can be taken to be zero at the centerline, and equal quantitatively to the axial volumetric flow rate between the centerline and an arbitrary stream surface at radial location r: ψ=ψ(r). If the constant were chosen other than zero (such that the stream function were not zero at the centerline), then ψ(r) would not be the volumetric flow rate between r = 0 and arbitrary r. The volumetric flow rate would in this case would then be ψ(r)-ψ(0). Often it is convenient to define a certain stream surface as ψ=0 so that the stream function on any stream surface quantitatively represents the volumetic flow rate between the stream surfaces.
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