Deriving Solid Angle Formulation from Definition

[mnb96]

Hello,
it is often written in books that the solid angle $\Omega$ subtended by an oriented surface patch can be computed with a surface integral:

$$\Omega = \int\int_S \frac{\mathbf{r}\cdot \mathbf{\hat{n}} }{|\mathbf{r}|^3}dS$$

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:

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

where $\theta$ is the angle between the position (unit)-vector for dS and the normal vector for dS

Unfortunately I don't understand where that $|\mathbf{r}|^{-2}$ comes from.

 

[LCKurtz]

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.

 

[mnb96]

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 $d\Omega$ having area $|\mathbf{r}|^{-2}dA$.

 

[HallsofIvy]

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 "$\mathbf{r}$" represent in $|\mathbf{r}|^{-2}dA$?

 

[CellCoree]

Hello,

Thank you for your question. The solid angle is defined as the ratio of the area of a surface projected onto a unit sphere to the square of the radius of the sphere. In other words, it is a measure of how much of the unit sphere is covered by the surface.

To understand where the \frac{1}{|\mathbf{r}|^3} term comes from in the surface integral formula, we need to consider the geometry involved. The position vector \mathbf{r} represents the distance from the center of the sphere to the surface patch dS. Since we are working with a unit sphere, the radius is equal to 1. Therefore, the magnitude of the position vector, |\mathbf{r}|, is also equal to 1.

Now, let's consider a small element of the surface, dS. This element can be represented by a small circle on the surface of the unit sphere. The area of this circle is given by dA = \frac{|\mathbf{r}|^2}{2}d\Omega, where d\Omega is the solid angle subtended by this element. Since we are working with a unit sphere, dA = \frac{1}{2}d\Omega.

Now, if we consider the projection of this element onto the unit sphere, we can see that the area of the projection is equal to the area of the element itself. Therefore, dA = dS.

Putting all of this together, we can see that dS = \frac{1}{2}d\Omega = \frac{|\mathbf{r}|^2}{2}d\Omega = \frac{1}{2|\mathbf{r}|^2}d\Omega. This is where the \frac{1}{|\mathbf{r}|^2} term comes from in the solid angle formula.

To get the full solid angle, we need to integrate over the entire surface, which is represented by the double integral in the formula you provided. I hope this explanation helps to clarify where the \frac{1}{|\mathbf{r}|^3} term comes from in the solid angle formula. Let me know if you have any further questions.