Derivation of $\frac{\cos(\theta)dA}{r^2} = d\omega$

Question

I've been looking for a (formal) derivation of the following equation $\frac{\cos(\theta)dA}{r^2} = d\omega$. Where $d\omega = \sin(\theta_x)d\theta d\phi$ is the differential solid angle, and $dA$ is some oriented surface differential area element. $r$ is the distance to this element, and $\theta$ is the angle between the radius vector and the normal of the area element. There was already a similar question on here: Proof of $\cos(\theta) da=r^2 d\Omega$

Please refer to the image provided there for a clarification. However the thread linked above was marked as 'answered' even though no formal proof was provided. I have the intuition why this works, I just want to see the formal way one would prove this. For one thing all computer graphics books that use this fact always ignore the derivation, so I have been very interested in how one can prove it.

Edit: Found a formal proof: http://www.stewartcalculus.com/data/ESSENTIAL%20CALCULUS%20Early%20Transcendentals/upfiles/challenge/ess_cp_13_stu.pdf

I think you should post it over at https://physics.stackexchange.com/. This website is for math, while the other one is for physics. — BadAtAlgebra, Jan 13 '19 at 22:54
This is not specifically about physics though I think. It's multivariable calculus/differential geometry as far as I can tell. Thank you anyway, I'll ask there too if I get no answers. — lightxbulb, Jan 13 '19 at 23:03
@Michael ... which is why this question, which is exclusively about mathematics and has no physics content, is off-topic at Physics SE. — E.P., Jan 14 '19 at 03:06
@lightxbulb "ask there too if I get no answers" generally means that you should wait 24 - 48 hours to cross-post. And if you do cross-post, then you should always include links on all versions to all other versions. — E.P., Jan 14 '19 at 03:10
I literally have a disclaimer up above the post on the physics forum. And to be fair to Michael the problem does seem to arise in proving the Gauss Law, not to mention that the field I got it from (CG, specifically light transport) is based on geometric optics which is once again physics. All references avoid proving it though, e.g. http://www.pbr-book.org/3ed-2018/Color_and_Radiometry/Working_with_Radiometric_Integrals.html "We will not derive this result here, but it can be understood intuitively" — lightxbulb, Jan 14 '19 at 03:16
@lightxbulb A disclaimer is not the same as a link, which should go on both versions. You did well, but the site standards are a bit higher - I'm not here to berate you, just some quick info for next time. As for whether this is physics or not - this is indeed maths that comes up with some frequency in physics contexts, much like, say, 1+1, but that doesn't make it physics. — E.P., Jan 14 '19 at 06:56

score 2 · Accepted Answer · answered Jan 14 '19 at 06:24

2

We can start by thinking a differential surface ($dA$), which is oriented with an angle $\beta$ with respect to the normal of the surface of the sphere and this surface has area vector $d\vec{A}=dA\hat {n}_A$

And the normal of the sphere is $\hat {r}$

Hence, we can take the projection of $dA$ on to the sphere by $d\vec {A} \cdot \hat {r}$ since length of $\hat{r}$ is $1$ we can write, $$ d\vec{A} \cdot \hat {r} =dAcos(\beta)$$ So this is the area element that falls on the surface of the sphere. Lets call this area element $dA'$

Then from here we can write $$dA'/r^2=dw$$

The general derivation of $dA'/r^2=dw$ is fairly simple. The area element on the sphere can be calculated from the cross products of other two elements, so the area element $ds_r$ can be written as, $$d\vec{s}_r=d\vec{s}_{\theta} \times d\vec{s}_{\phi}$$ where $0<\theta<\pi$ and $0<\phi<2\pi$.

Here $d\vec{s}_{\theta} =rd\theta\hat {\theta}$ and $d\vec{s}_{\phi}=rsin(\theta)d\phi\vec {\phi}$ so we have,

$$d\vec{s}_r=rsin(\theta)d\phi\vec {\phi} \times rd\theta\hat {\theta}$$ $$d\vec{A'}=d\vec{s}_r=r^2sin(\theta) d\theta d\phi \hat {r}$$

or in magnitude, $$d{A'}=r^2sin(\theta) d\theta d\phi$$ and lets call $dw=sin(\theta) d\theta d\phi$ so we have $$dA'=r^2dw$$

but $dA'=dAcos(\beta)$ so we have

$dAcos(\beta)/r^2=dw$

answered Jan 14 '19 at 06:24

seVenVo1d

452

Thank you for the answer. The last part I assume can also be derived from the Jacobian or the first fundamental form? – lightxbulb Jan 14 '19 at 09:31
I guess. I dont know how to calculate from those. – seVenVo1d Jan 14 '19 at 09:35
Similar to here:https://math.stackexchange.com/questions/131735/surface-element-in-spherical-coordinates/131811#131811 'We can take the projection of the surface onto the sphere', I assume you define the mapping here $f(\vec{r}) = \frac{\vec{r}}{r}$? I am also assuming that this can be derived backwards from integrating the vector field $f$ and differentiating on both sides. Can you clarify why we have to integrate a vector field function as opposed to a scalar function? Also any reference (preferably for engineers) would be appreciated where I can look up more on differential area elements. – lightxbulb Jan 14 '19 at 09:39
I'll leave this open for a while longer, just to maybe get more and different derivations. – lightxbulb Jan 14 '19 at 09:46
I think you are making things more complicated by mapping and etc. The idea is much more simple, consider this: You have an area on the xy plane with an area of S with side length a and b.We can write the $dS$ instantly, $dS=ab\vec{k}$. However it can be written as $dS=a\vec {i} \times \ b\vec {j}=ab\vec {k}$. You can expend the idea for other coordinate systems. All area elements can be found the cross product of the other two elements. Because the normal vector must be perpendicular to the surface. – seVenVo1d Jan 14 '19 at 10:09
I never used f(r) or integrals to derive the surface area element. You should look the to curvilinear coordinates maybe to understand the idea. – seVenVo1d Jan 14 '19 at 10:10
The projection of the initial surface $S$ onto the unit sphere is precisely the mapping above though? More specifically you took $\hat{r}$ without formalising why. I am just arguing that the above mapping maps $S$ to $S^2$ and that's where $\hat{r}$ comes from. – lightxbulb Jan 14 '19 at 10:19
I didnt quite understand what you mean. I put r-hat because thats the normal vector to the surface? I am not saying you are wrong or anything like that. I am physics sophomore student my math knowledge is not really good in that sense. – seVenVo1d Jan 14 '19 at 19:21
If $\hat{r}$ is the normal vector, how do you get $\vec{dA}\cdot \hat{r} = dA \cos\beta$? Clearly $\vec{dA} = \vec{n}dA$ where $n$ is the normal vector and $\hat{r}$ is actually the normalized radius vector to the differential element. – lightxbulb Jan 14 '19 at 19:24
Let me clarify then with an image: https://i.imgur.com/w4TF096.png In our case $\vec{r} = \vec{x}-\vec{y}$, and $\vec{n}_y$ is the normal. – lightxbulb Jan 14 '19 at 19:31
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The normal vector for $dA'$ is $\vec {n_{A'}}$. It can have $r$,$\theta$ or $\phi$ componenets. But as you said, when we make a dot product with $r$ we get $dAcos(\beta)$. The surface element on the surface of the sphere has a normal vector of $r$ and only $r$ it doesnt not have any components. As you know $r$ is one of the unit vectors for that coordinate system. You dont have to write $r$ in terms of $x$ and $y$ – seVenVo1d Jan 14 '19 at 19:46
Oh, you are talking about the normal of the sphere and not the surface we are integrating! I think we are basically thinking the same thing, just from a different perspective. – lightxbulb Jan 14 '19 at 19:56
Maybe we are, I am not sure. – seVenVo1d Jan 14 '19 at 20:03
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Thank you for the help on this problem. – lightxbulb Jan 14 '19 at 20:06
Your welcome, I hope I didnt get you more confused about the topic. – seVenVo1d Jan 14 '19 at 20:09
I found a formal proof of the fact above: http://www.stewartcalculus.com/data/ESSENTIAL%20CALCULUS%20Early%20Transcendentals/upfiles/challenge/ess_cp_13_stu.pdf – lightxbulb Jan 20 '19 at 13:36
I ll check later – seVenVo1d Jan 20 '19 at 14:21

Derivation of $\frac{\cos(\theta)dA}{r^2} = d\omega$

1 Answers1