I'm reading through Classical Electrodynamics by Jackson and he makes use of a particular identity that I've not actually seen before. I can't seem to find a proof for it anywhere. It is just as the title says:

$\cos(\theta)da=r^2 d\Omega$

where $d\Omega$ is $\sin\theta d\theta d\phi$, the solid angle.

As an example for context consider Gauss' law. If the electric field E at a point on a Gaussian surface due to the charge q within the surface makes an angle $\theta$ with the unit normal, then the normal component of E times the area element is:

$\vec{E} \cdot \hat{n} da=\frac{q}{4\pi\epsilon_{0}}d\Omega$

Since E is directed along the line from the surface element to the charge q, $\cos\theta da=r^2 d\Omega$, where $d\Omega$ is the element of solid angle sutended by da at the position of the charge. Therefore,

$\vec{E} \cdot \hat{n} da=\frac{q}{4\pi\epsilon_0}d\Omega$.

A simple geometric proof is all I need, I just want to know where this comes from.

  • $\begingroup$ I believe this would be more suitable for a physics forum. That said, how do your variables-$\theta, r, \Omega $-relate? $\endgroup$ Jan 10, 2017 at 21:08
  • $\begingroup$ Contextualize the question. Can you show us a an equation/step where he uses this equality? $\endgroup$
    – b00n heT
    Jan 10, 2017 at 21:08
  • $\begingroup$ also what is da? $\endgroup$ Jan 10, 2017 at 21:11
  • $\begingroup$ And now there is also a $\phi$ :-).. $\endgroup$ Jan 10, 2017 at 21:12
  • $\begingroup$ edited with all that in mind. da is not specified in the book, that's part of the problem. $\endgroup$
    – Karl
    Jan 10, 2017 at 21:24

1 Answer 1



I suppose that you found the formula in the context of Gauss Theorem. If so then this figure (from here) shows how to find the formula.

enter image description here

  • $\begingroup$ Bless you, I was losing my mind because of this for about an hour and I found your answer $\endgroup$ Apr 20, 2021 at 10:22

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