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Solid angle question

  1. Jul 9, 2010 #1
    it is often written in books that the solid angle [itex]\Omega[/itex] subtended by an oriented surface patch can be computed with a surface integral:

    [tex]\Omega = \int\int_S \frac{\mathbf{r}\cdot \mathbf{\hat{n}} }{|\mathbf{r}|^3}dS[/tex]

    where r is the position vector for the patch dS and n its normal (see also wikipedia).
    However I would like to know how to derive this formula from the definition of solid angle, that is: the area of the the projection of a surface on the unit sphere.

    I can already see that:

    [tex]\frac{\mathbf{r}}{|\mathbf{r}|} \cdot \mathbf{\hat{n}} dS = cos(\theta)dS[/tex]

    where [itex]\theta[/itex] is the angle between the position (unit)-vector for dS and the normal vector for dS

    Unfortunately I don't understand where that [itex]|\mathbf{r}|^{-2}[/itex] comes from.
    Last edited: Jul 9, 2010
  2. jcsd
  3. Jul 9, 2010 #2


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    I think the 1/|r|2 factor just scales the area down to its projection on a unit sphere. Your expression

    \frac{\mathbf{r}}{|\mathbf{r}|} \cdot \mathbf{\hat{n}} dS = cos(\theta)dS

    gives the projection of the surface patch on the sphere of radius |r|. Since area is proportional to the square of the radius, you need the 1/|r|2 to scale it down to the unit sphere.
  4. Jul 9, 2010 #3
    Ok, thanks.
    Now I see how it works.

    I was just wondering how to sketch a rigorous proof that the surface area of an infinitesimal "disk" dA is projected onto an infinitesimal spherical cap [itex]d\Omega[/itex] having area [itex]|\mathbf{r}|^{-2}dA[/itex].
  5. Jul 10, 2010 #4


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    In order to have a rigorous proof of anything involving "infinitesmals" you will need to to "nonstandard analysis" where infinitesmals themselves are rigorously defined! Otherwise you will need to be content with limit proofs. What does the "[itex]\mathbf{r}[/itex]" represent in [itex]|\mathbf{r}|^{-2}dA[/itex]?
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