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Unit of the Green's function

  1. Mar 5, 2010 #1
    Is there a physical unit related to the Green's function of the wave equation?

    In particular, let
    [tex]\nabla^2 P -\frac{1}{c^2}\frac{\partial^2 P}{\partial t^2} = f(t)[/tex]
    where P is pressure in Pa. Since the Green's function solves the PDE when f(t) is the delta function, the Green's function G has a unit of Pa.

    The solution to the inhomogeneous PDE is

    P = G*f(t),

    where * denotes convolution. This leads to contradiction since LHS is in Pa and RHS is in Pa * Pa/m^2 * (sec m^3), where (Pa/m^2) is the unit of f(t) and (sec m^3) denotes the integration in both space and time.

  2. jcsd
  3. Jul 20, 2011 #2

    Recall the sifting property of the delta function:
    [tex]\int_{-\infty}^{\infty} \delta(r) dr = 1. [/tex] With this in mind, we see that the units of RHS (actually both sides) change from the original wave equation once it is replaced with a delta function. This is because the delta function must have units which are the inverse of its argument's units in order to make the sifting property true. So in the example above r has units of distance, and the delta function must therefore have units of 1/distance. The delta function returns a unitless quantity after integration.
    Hence, the RHS of [tex]\nabla^2 G -\frac{1}{c^2}\frac{\partial^2 G}{\partial t^2} = \delta(t)\delta(r),[/tex] where G is the Green's function and the RHS contains the necessary spatial and temporal delta functions, has units 1/(m^3*s). To figure out the units of G, we demand that this equation has dimensional consistency among all its terms. Thus, matching units between the second term of the LHS (an arbitrary choice) and the RHS demands that
    [tex][-\frac{1}{c^2}]*[\frac{\partial^2 G}{\partial t^2}] = [\delta(t)\delta(r)],[/tex] dimensionally (and only dimensionally!). Thus, [tex][\frac{s^2}{m^2}]*[\frac{G}{s^2}] = [\frac{1}{m^3*s}],[/tex] and (for this example only) G has the following units: [tex][G] = [\frac{1}{m*s}]. [/tex] Finally, we check this result by performing the convolution with f(t) from the original wave equation. Remember that f(t) has units Pa/m^2 due to dimensional consistency of the wave equation. Thus, convolving G and f(t) in space and time, we get
    [tex]\int [G]*[f(t)]*[dr dt] = [\frac{1}{m*s}]*[\frac{Pa}{m^2}]*[m^3*s] = Pa,[/tex] which are the correct units for pressure.
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