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I would like to compute the average distance travelled by particle points at constant speed $v>0$ with uniformly-distributed directions and placed uniformly at random inside a hollow sphere of radius R until it hits the sphere.

My guess would be to initiate the computation as

$$\frac{1}{V}\int_0^R f(r)4\pi r^2dr$$

where $f(r)$ would be the mean distance to the sphere, with $r \leq R$ (as this function is obviously radial).

My problem comes from this question :

enter image description here

To get $f(a)$, should I integrate over $\color{blue}{\theta}$ or $\color{green}{\phi}$?

My first instinct was to average over $\color{blue}{\theta}$, can somebody confirm that I am doing well? The instinct comes from my assumption that the direction of the particle is uniformly distributed around itself, not around the origin O, but perhaps due to integration/infinitesimal trickery the correct answer is actually $\color{green}{\phi}$? I am befuddled as the answers seem to point at the fact that averaging over an inner sphere amounts to integrating over $\color{green}{\phi}$, which I guess is what I want to do if I were to place this before $4\pi r^2 dr$, but this contradicts my basest intuition which points at the uniformity of directions of $\vec v$. So basically, what should prevail here, uniformity of directions ($\color{blue}{\theta}$) or averageness over an inner sphere ($\color{green}{\phi}$)?

Edit: actually it seems like $\color{green}{\phi}$ provides an isotropic averaging over the sphere, while $\color{blue}{\theta}$ provides a uniform averaging over directions. But uniform averaging over directions does not translate into isotropic averaging over the sphere because of asymmetry, so $\color{blue}{\theta}$ still seems to be the way to go. However, I would like some feedback anyway please.

Sorry if this sounds elementary, I've not practiced physics in a long time, so I am full of self-doubt. If the quantitative answer matters to passive viewers, both answers are to be found in the question linked above. In particular, I am not asking for the computation as it is already available and would be redundant.

Thank you for your answers.

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  • $\begingroup$ Note that blue-green color blindness is a thing, and about 1 in 500 people have it, which makes it likely that possible answerers here might have it. Consider changing one of the $\theta$'s to $\phi$ in the diagram and the text rather than distinguishing them by color. $\endgroup$
    – march
    Commented Apr 29 at 23:13
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    $\begingroup$ @march Done, thanks for the tip! $\endgroup$
    – Evariste
    Commented Apr 29 at 23:20

1 Answer 1

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This one, you just gotta break down to all the degrees-of-freedom, and slowly reject the trivial ones. There's too many to wing it.

Also: when on/in the sphere, don't deal with polar angle, deal with its cosine. Also, bring azimuth into the mix with:

$$ d^2\Omega=\sin\theta d\theta d\phi = d(\cos\theta)d\phi = dz d\phi $$

where the "dz" thing is not common, but I am going to use it--it's not unheard of. What it does, as you can tell, is: it turns your uniform sphere into a uniform rectangle, and that is convenient.

Also:

$$ d^3\vec r = r^2drd^2\Omega \equiv d^3V $$

OK, for colors, though they are nice, I don't have the skills. Unprimed is spacial angle, and primed are in velocity space

$$ d^3\vec v = v^2dv d\Omega'$$

(save that part for Maxwell's Distribution).

Though, I think, you have $|v| =v $ fixed? If so, we need to deal only with $d\Omega'$.

Now consider the distance as a function of all the coordinates--hang on, using "d" is dumb, since it's also the differential...I will call it $x$, no, $l$..no, looks like $1$...$L$:

$$ L(\vec r, \Omega') = {\rm ?} $$

(It would have been nice had you supplied the $?$", but well go with that abstract form and figure it out):

$$ L(\vec r, \Omega') = {\rm ?} $$

So, assuming:

$$ \int_V\int_{\Omega'} d^3Vd^2\Omega'= N$$

(if not, adjust accordingly), we can write the full integral for the expectation value of $L$:

$$ \langle L \rangle = \frac 1 N \int_V\int_{\Omega'}L(\vec r, \Omega')d^3Vd^2\Omega'$$

Now we can expand that into coordinates:

$$ \bar{L} = \frac 1 N\int_{r=0}^R\int_{z=1}^1 dz\int_{\phi=0}^{2\pi}d\phi \int_{z'=1}^{-1}dz'\int_{\phi'=-}^{2\pi} L(r, z,\phi,z',\phi')d\phi' $$

Now we can start chucking coordinates:

$$ L(r, z,\phi,z',\phi') = L(r, z') $$

(by symmetry), so we can do the constant integrals:

$$ \bar{L} = \frac 1 N 8\pi^2\int_{r=0}^Rr^2dr\int_{z'=1}^{-1} dz'L(r, z') $$

So now you need to use the formula for $L(r, z')$ and verify the $N=\langle 1\rangle$.

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