Calculating the Solid Angle of a Nebula from Earth

[agnimusayoti]

Homework Statement

Great Nebula in Andromeda called M-31. The nearest of the large regular galaxies it is still 2 500 000 lightyears from solar system. Its diameter is about 125 000 light-years and it contains more than $10^{11}$ stars. a) Determine the angle subtended by the diameter of the Great Nebula M-31 when observed from the earth. Express it in radians and in degree of arc. (b) Find the solid angle subtended by the nebula. (FUndamental University Physics I, Chapter 2, Prob 2.18; Finn and Alonso)

Relevant Equations

Plane angle $\theta$ is defined by:
$$\theta = \frac{l}{R}$$
where l is arc of circle with radius R.

Solid angle $\Omega$ is defined by
$$\Omega = \frac{S}{R^2}$$
where S is the area of spherical cap intercepted by the solid angle.
Or,
$$d\Omega = \frac{dS}{R^2}$$

If I assume the nebula is a circle, than the length of arc viewed from Earth is a half of the circumference. So, here
$$l = \frac{1}{2} \pi D$$
From the problem, $D = 125 000 ly$.
Because the distance of nebula is much larger than the diameter; I try to approximate R with the distance of nebula from earth. Therefore, $R = 2 500 000 ly$
a) From the equation, I get
$$\theta = \frac{\pi}{20} radian$$
$$$\theta = 9^o$$

b) Actually, I don't really understand what solid angle is, so I try to find the problem to get solid angle. Here is what I try.
First I use the infinitesimal form of the equation. Using the spherical coordinate,
$$dS = R^2 \sin {\theta} d\theta d\phi$$

So,
$$d\Omega = \int \int \sin {\theta} {d\theta} {d\phi}$$

My domains of integration are:
$$0 <= \theta <= \frac{\pi}{20}$$
$$0 <= \phi <= 2\pi$$
Is it right? Thankss

 

[Charles Link]

$\theta=l/R$. The $l$ is essentially the same as $D$, the diameter of the galaxy. ( $\theta= \pi/20$ is incorrect).

For the solid angle, $\Omega= (surface \, area)/R^2 \approx \pi r^2/R^2=\pi (D^2/4)/R^2=\pi (\frac{D}{R})^2/4=\pi \theta^2/4$. I derived it for you, because it's kind of simple, but perhaps not obvious. To summarize, solid angle $\Omega=\pi \theta^2/4$. (Very similar to area of a circle $A=\pi D^2/4$).

(Your attempt with $dS=R^2 \sin{\theta} \, d \theta \, d \phi$ is correct, and can also get you there, with integrating $\Delta \phi=2 \pi$, and $\sin{\theta} \approx \theta$, so it integrates to $\theta_o^2/2$, with $\theta_o=\theta/2$ where $\theta=D/R$, but doing it this way by integrating in spherical coordinates is a lot more cumbersome than what I did with the first method. Notice also the $\theta$ limits of integration go from $0$ to $\theta_o$, where $\theta_o=\theta/2$.).

 

[agnimusayoti]

Oh I see. Because the nebula is far away from Earth then the diameter is approximated the same as the length of arc l right?

Nah, what kind of surface area in formula of solid angle?
Why the domain of integration is $0 <= \theta<=\theta_0/2$?

 

[Charles Link]

$\theta$ in $\sin{\theta} \, d \theta \, d \phi$ is the polar angle in spherical coordinates. When you cover the whole range of $\phi$ from $0$ to $2 \pi$, it makes a circle/cone around the north pole with the polar angle of $\theta_o=\theta/2$, where $\theta$ is the full cone angle. (The $\theta=D/R$ =full-cone angle, is basically a constant, and should really have a subscript, e.g. call it $\theta_f$, to distinguish it from the $\theta$ that we are integrating over. More correctly, with less confusion, $\theta_o=\theta_f/2$.)

Meanwhile, the surface area in the solid angle is that of the spherical cap on the cone, but for relatively small angles, it can be approximated as having the same area as the flat circle which forms the base of the cone, which is $A =\pi r^2=\pi R^2 \sin^2(\theta_f/2) \approx \pi R^2 \theta_f^2/4$.
[If you want an exact calculation, you can integrate $\int\limits_{0}^{\theta_f/2} \sin{\theta} \, d \theta =1-\cos(\theta_f/2)$, instead of integrating the approximate $\int\limits_{0}^{\theta_f/2} \theta \, d \theta=\frac{\theta_f^2}{8}$ like I did in post 2. If you have had Taylor series, you can use $\cos{\theta} \approx 1-\frac{\theta^2}{2}$, for small $\theta$, which then converts the exact result of $\Omega=2 \pi (1-\cos(\theta_f/2))$ to the simpler, (approximate, but close to being exact), result. The simpler result, (with $\Delta \phi=2 \pi$, so that the solid angle $\Omega \approx \frac{\pi}{4} \theta_f^2$), is perhaps the better one to use, unless the calculation involves large angles].

 

[Charles Link]

a follow-on: Putting the north pole of the spherical coordinate system at the center of the object of interest (per posts 2 and 4) may be a little clumsy, because it is somewhat necessary to have the object of interest to be shaped like a circle. It perhaps is also very useful to be able to compute the solid angle if the galaxy is elliptical, which this one really is. This one needs to be worked a little differently: For an ellipse, $A=\pi a b$, so that the solid angle would be $\Omega=\frac{A}{R^2}=(\frac{\pi}{4})( \frac{2a}{R})(\frac{2b}{R})=(\frac{\pi}{4})(\Delta \theta )(\Delta \phi)$, where we have located the center of the galaxy of interest on the equator of a spherical coordinate system, (where $\theta=\pi/2$). This way we can treat $\theta$ and $\phi$ on equal footing, as elevation and azimuth angles.
(Here $dA = R^2 d \theta \, d \phi$ because $\sin(\theta) \approx 1$. The north pole and south pole can be placed either above and below, or left and right, with the galaxy of interest straight ahead).

It's a little extra detail, but to really treat the solid angle subject properly, (as it applies to very practical problems), the extra detail can be important.

It may also be worth noting, by this same method, a rectangular shape of $w$ x $l$ will have area $A=wl$, and the solid angle will be $\Omega= \frac{A}{R^2}=(\frac{w}{R})(\frac{l}{R})=\Delta \theta \, \Delta \phi$.

 

[Charles Link]

One additional note: a google shows this galaxy has a ellipsoidal shape of 3.167 degrees x 1.0 degrees. These would need to be converted to radians, and the solid angle would be measured in steradians. Considering it to be ellipsoidal, and using the method of post 5 would give a better and more accurate answer, than to consider the galaxy as circular, and to do the computation per post 2.