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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

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  • $\begingroup$ I think you should post it over at physics.stackexchange.com. This website is for math, while the other one is for physics. $\endgroup$ – BadAtGeometry Jan 13 at 22:54
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    $\begingroup$ 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. $\endgroup$ – lightxbulb Jan 13 at 23:03
  • $\begingroup$ @Michael ... which is why this question, which is exclusively about mathematics and has no physics content, is off-topic at Physics SE. $\endgroup$ – E.P. Jan 14 at 3:06
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    $\begingroup$ Cross-posted to Physics SE. $\endgroup$ – E.P. Jan 14 at 3:08
  • $\begingroup$ @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. $\endgroup$ – E.P. Jan 14 at 3:10
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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

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

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  • $\begingroup$ Thank you for the answer. The last part I assume can also be derived from the Jacobian or the first fundamental form? $\endgroup$ – lightxbulb Jan 14 at 9:31
  • $\begingroup$ I guess. I dont know how to calculate from those. $\endgroup$ – Reign Jan 14 at 9:35
  • $\begingroup$ Similar to here:math.stackexchange.com/questions/131735/… '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. $\endgroup$ – lightxbulb Jan 14 at 9:39
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    $\begingroup$ 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 $y$ $\endgroup$ – Reign Jan 14 at 19:46
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    $\begingroup$ Thank you for the help on this problem. $\endgroup$ – lightxbulb Jan 14 at 20:06

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