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Averaging a function in spherical polar coordinates?

Hello, I've come across this formula in some notes but I can't seem to find where it has come from.

The notes read:

We can find the average of any quantity f f from

⟨f⟩=∫f(θ,ϕ)sin⁡θ dθ dϕ∫sin⁡θ dθ dϕ.\langle f\rangle=\dfrac{\int f(\theta,\phi)\sin\theta\ \mathrm{d}\theta\ \mathrm{d}\phi}{\int\sin\theta\ \mathrm{d}\theta\ \mathrm{d}\phi}.

Can someone explain where this comes from? Or direct me to somewhere online I can find it? Google searches of "averaging a function in spherical polar coordinates" don't seem to get anywhere :frown:

Thanks.
Original post by KeyFingot
Hello, I've come across this formula in some notes but I can't seem to find where it has come from.

The notes read:

We can find the average of any quantity f f from

⟨f⟩=∫f(θ,ϕ)sin⁡θ dθ dϕ∫sin⁡θ dθ dϕ.\langle f\rangle=\dfrac{\int f(\theta,\phi)\sin\theta\ \mathrm{d}\theta\ \mathrm{d}\phi}{\int\sin\theta\ \mathrm{d}\theta\ \mathrm{d}\phi}.

Can someone explain where this comes from? Or direct me to somewhere online I can find it? Google searches of "averaging a function in spherical polar coordinates" don't seem to get anywhere :frown:

Thanks.


Your formula assumes that the radius is a constant.

An area element on a sphere in spherical polar co-ordinates is r2sin⁡θ dθ dϕr^2\sin\theta\ \mathrm{d}\theta\ \mathrm{d}\phi

Since r is a constant you can pull it out of the integral.

Your formula is the integral of f over the given area, divided by the given area, which is just the integral of the area element. The r^2 will cancel, leaving your formula.
Reply 2
Avatar for Mark85
Mark85
16
Well, normally, the average of a function ff on some manifold MM with volume form dvold_{\mathrm{vol}} is:

∫Mfdvol∫Mdvol=1volume(M)∫Mfdvol\frac{\int_M f d_{\mathrm{vol}}}{\int_M d_{\mathrm{vol}}} = \frac{1}{\mathrm{volume(M)}} \int_M f d_{\mathrm{vol}}

In this instance, I am assuming that your function is on the 2-sphere: sin⁡(θ)dθdϕ\sin(\theta)d\theta d\phi is the volume form for the 2-sphere and thus the average is just the normal one as mentioned above.

This just comes from the determinant of the Jacobian with respect to the change of coordinates from cartesian to spherical.
Reply 3
Avatar for KeyFingot
KeyFingot
OP
Original post by ghostwalker
Your formula assumes that the radius is a constant.

An area element on a sphere in spherical polar co-ordinates is r2sin⁡θ dθ dϕr^2\sin\theta\ \mathrm{d}\theta\ \mathrm{d}\phi

Since r is a constant you can pull it out of the integral.

Your formula is the integral of f over the given area, divided by the given area, which is just the integral of the area element. The r^2 will cancel, leaving your formula.


I thought it was that, but I was confused because, like you say, there were no r's in the formula. The case this is being applied to, r is constant

Thanks!

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