# Average distance travelled by particle points placed uniformly at random in a sphere with speed $||v||$ and direction uniformly random?

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 :

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.

• 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. Commented Apr 29 at 23:13
• @march Done, thanks for the tip! Commented Apr 29 at 23:20

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 $$N=\langle 1\rangle$$.