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Homework Help: LHS to RHS integral problem

  1. Mar 25, 2006 #1
    I have a problem with a integral. Could someone tell me how to get from the LHS to RHS? It must be very easy, but as so often I don't get it.

    Since I have not mastered Latex yet, I attached a pdf file.


    Attached Files:

  2. jcsd
  3. Mar 25, 2006 #2


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    In LaTex what you have is
    [tex]\frac{1}{k^2+ m^2}\int e^{i\vec{k}\dot\vec{r}}d\vec{k}= \int k d\phi \intk sin(\theta)d\theta \int \frac{e^{ik|r|cos(/theta)}}{k^2+ m^2}dk[/tex]
    (Click on that for a pop-up box showing the code.)

    I had to stare at that a while until I realized- those blasted engineers have mixed up [itex]\theta[/itex] and [/itex]\phi[/itex] again! Also, it is using "k" for the length of [itex]\vec{k}[/itex]. I take it that "[itex]\vec{k}[/itex]" is the position vector of a point in the space being integrated over (here all of R3).

    The right hand side is in spherical coordinates, using k instead of the more standard [itex]\rho[/itex]. In spherical coordinates (mathematics version!)
    [tex]dV= \rho^2 sin(\phi)d\rho d\phi d\theta[/tex]
    In "engineer speak", as here, it is
    [tex]dV= \k^2 sin(\theta)dk d\theta d\phi[/tex]

    That has been separated into
    [tex](k d\phi)(k sin(\theta)d\theta)(dk)[/tex]
    Of course, since [itex]\vec{u}\dot\vec{v}= |u||v|cos(\theta)[/itex], that
    [itex]i\vec{k}\dot\vec{r}= ik|r|cos(\theta)[/tex]

    It's not clear to me that breaking it up that way is valid, because clearly k is a variable (dk in one integral) and so putting k into the [itex]d\theta[/itex] and [itex]d\phi[/itex] integrals doesn't make sense.
  4. Mar 25, 2006 #3


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    Basically, all that's happened is writing the integral over all space in spherical coordinates [itex]d^3\vec k=k^2\sin \theta dk d\theta d\phi[/itex] and using [itex]\vec k \cdot \vec r=|r|k\cos \theta[/itex].
    But the way it's written it looks like a product of three integrals, which it is not!
    And unless [itex]\vec r[/itex] points along the z-axis you must not confuse the angle between [itex]\vec k[/itex] and [itex]\vec r[/itex] with the angle [itex]\theta[/itex] over which you are integrating.
    Last edited: Mar 25, 2006
  5. Mar 25, 2006 #4
    Thanks HallsofIvy, thanks Galileo!

    Good to know that this time it was not all my fault.
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