1
$\begingroup$

I looked at a proof of Gauss's law, as seen here:

Theorem - The flux of a point charge. Given a point charge $q$ and a surface surrounding it, the flux through the surface is just $\frac{q}{\epsilon_{0}}$. To prove this proposition we first consider a spheircal surface of radius $r$. We get $$\Phi=\int_{S} \boldsymbol{E} \cdot d \boldsymbol{a}=\int \frac{q \hat{\boldsymbol{r}}}{4 \pi \varepsilon_{0} r^{2}} \cdot r^{2} \sin \theta d \theta d \varphi \hat{\boldsymbol{r}}=\frac{q}{\varepsilon_{0}}$$ As we wanted. Moving to a general closed surface $G$ containing $q$, we first find a small spherical surface $S$ of radius $r$ around $q$ that is contained in $G$. The rays starting from the point charge $q$ passing through the surface element $\boldsymbol{\Delta a}$ of $S$ project in onto a surface element $\boldsymbol{\Delta A}$ of $G$. If $\boldsymbol{R}$ is the vector distance between $q$ and $\Delta A$, $\cos \theta$ is the angle between $\boldsymbol{R}$ and $\boldsymbol{\Delta A}$, i.e. $\cos \theta = \hat{R}\cdot \boldsymbol{\Delta A} /\Delta A$, then $$\Delta A=\Delta a\left(\frac{R}{r}\right)^{2} \frac{1}{\cos \theta}$$the electric field at $\boldsymbol{\Delta a}$ is $$\boldsymbol{E}(\boldsymbol{r})=\frac{q \hat{\boldsymbol{r}}}{4 \pi \varepsilon_{0} r^{2}}$$ ad at $\boldsymbol{\Delta A}$ is $$\boldsymbol{E}(\boldsymbol{R})=\frac{q \hat{\boldsymbol{R}}}{4 \pi \varepsilon_{0} R^{2}}$$ consequently the flux through $\boldsymbol{\Delta A}$ is $$\begin{aligned} \Phi(\Delta \boldsymbol{A}) &=\boldsymbol{E}(\boldsymbol{R}) \cdot \Delta \boldsymbol{A}=\frac{q}{4 \pi \varepsilon_{0} R^{2}} \hat{\boldsymbol{R}} \cdot \Delta \boldsymbol{A} \\ &=\frac{q}{4 \pi \varepsilon_{0} R^{2}} \cos \theta \Delta A \\ &=\frac{q}{4 \pi \varepsilon_{0} R^{2}} \cos \theta \Delta a\left(\frac{R}{r}\right)^{2} \frac{1}{\cos \theta} \\ &=\frac{q}{4 \pi \varepsilon_{0} r^{2}} \Delta a \\ &=\Phi(\Delta \boldsymbol{a}) \end{aligned}$$ enter image description here

In the proof of the general case, we use some strange geometry: $$\Delta A=\Delta a\left(\frac{R}{r}\right)^{2}\frac{1}{\cos\theta}$$ I'm having a hard time understanding why this is correct. I know it has something to do with the term "Solid Angle", but I'm unsure what it is exactly (I saw the the definition but I don't understand where it appears here).

So I'd also appreciate it if you could explain a bit about solid angles and perhaps provide some intuition.

I've visited posts such as this one, that show $\Omega = \frac{A}{R^2}$ as the definition of a solid angle, but I don't see why that's useful and where it appears in my case.

$\endgroup$
4
  • $\begingroup$ Hello! It is preferable to type out screenshots; for formulae, one can use MathJax. Thanks! $\endgroup$
    – jng224
    Commented Jun 20, 2021 at 13:04
  • $\begingroup$ @Jonas Is that ok? $\endgroup$
    – Ariel Yael
    Commented Jun 20, 2021 at 13:26
  • $\begingroup$ Does the book provide any picture? If so, it would be much clearer to show it to us $\endgroup$
    – FGSUZ
    Commented Jun 20, 2021 at 20:18
  • $\begingroup$ @FGSUZ added :) $\endgroup$
    – Ariel Yael
    Commented Jun 20, 2021 at 20:24

2 Answers 2

2
$\begingroup$

In order to understand the solid angle, you need to understand the plain angle first. Of course, you know how angles work... but have you really thought about them? How can you DEFINE an angle? How can you define a 75,32º angle? It is not as easy as it might seem

It can be shown mathematically that angles keep a linear relation with the length covered by the angle in a fixed radius circumference. This means that a double angle implies double length, and so on...

So we define the radian as the angle whose subtended length equals the radius of the circle.

enter image description here

This property is essential because it lets us relate the circumference's length with the angle, via

$$L=r\cdot\varphi$$

This is a really relevant property of radians. Only radians fulfill this.

We can use this to measure angles via $\varphi=L/r$, that is, we can measure angles by just measuring lengths.

Now, imagine a segment (or a plain curve) in the middle of space. You want to know the angle subtended by that segment as seen by your origin O. How do we measure the angle with which O sees the segment?

One possibility is projecting the segment onto the circumference and then extracting the angle from there.

enter image description here

So you can say that the angle O sees the segment is just $\theta=s/r$, Like this, by just measuring lengths, you can determine the angle of the segment seen from O. The problem is how to determine $s$. Well, that's the projection of the segment onto the circumference. It can be done mathematically with integrals.


Solid angle

So, once you've understood this, you can easily understand solid angles. It is the same idea: we can determine the angles by measuring areas and dividing by $r^2$. the solid angle with which you see a surface is the area of the projection divided by $r^2$.

In other words, project the surface on a sphere of radius r, and then calculate the ratio between the projection and $r^2$.

enter image description here

So, the solid angle with which O sees surface S is the ratio of $A/r^2$, where $A$ is the projection of surface $S$ on a perfect sphere.


And what does this have to do with Gauss' law? Well, you do not need solid angles to prove it, but it can be useful to replace some formulas if you know that $A/r^2$ is the solid angle...

Because a double integral of the solid angle is really easy; $$\iint d\Omega = 4 \pi$$

If the total angle is $2\pi$ in a circle, the total solid angle is $4\pi$ on a Sphere. Obviously, the total area is $4\pi r^2$, if you do $A/r^2$ you get $4\pi$.

Consequently, the solid angle is a nice tool to calculate the total flux.

This is done for the point $r=0$, because that's the only problematic point.

$\endgroup$
2
  • $\begingroup$ Hi! Why do we divide by $r^2$? Just in order to keep the solid angle unit-less or is there another reason? Also, if the proof doesn't use solid angles, how is the relation \begin{equation}\frac{\Delta A \textrm{cos}\theta}{\Delta a}=\left(\frac{R}{r}\right)^2.\end{equation} Recieved? $\endgroup$
    – Ariel Yael
    Commented Jun 21, 2021 at 15:20
  • $\begingroup$ It is the definition of solid angle. It's a generalization of $\theta=s/r$, just $\Omega=A/r^2$. You're not measuring areas, you want another thing, which is related to them. As for the second question, you can compute the first integral directly, without reducing it to the sphere, but it is harder... $\endgroup$
    – FGSUZ
    Commented Jun 21, 2021 at 21:31
1
$\begingroup$

Applying the 'Areas of Similar Triangles Theorem', the ratio of the areas of two similar tetrahedrons is given by the square of the ratio between the two corresponding sides. Thus, \begin{equation}\frac{\Delta A \textrm{cos}\theta}{\Delta a}=\left(\frac{R}{r}\right)^2.\end{equation} Note that $\Delta \textbf{A}$ and $\Delta \textbf{a}$ are not parallel, and thus we have to compare two parallel areas $\Delta \textbf{A}\cdot \hat{R}$ and $\Delta \textbf{a}$.

$\endgroup$
1
  • $\begingroup$ Where are the tetrahedrons here? And how does all of this relate to solid angles? $\endgroup$
    – Ariel Yael
    Commented Jun 20, 2021 at 14:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.