Rutherford's $\alpha$-Particle Scattering Experiment

In Ernest Rutherford's $$\alpha$$-particle scattering experiment, it is well-known that the solid angle density of $$\alpha$$-particle flux

$$\frac{d\dot{N}}{d\Omega}$$

on the inner surface of a sphere of which the centre is the point of impingement of a beam of $$\alpha$$-particles on a (extremely) thin gold foil is proportional to

$$\csc^4\frac{\phi}{2}$$

where $$\phi$$ is the angle away from the pole of the apparatus - the point on the sphere upon which an undeflected beam would impinge. This proportionality relation incurs the problem of having no normalisation: integration of it over even the smallest non-zero region of the sphere comprising the pole will yield an infinite flux.

But there are various mechanisms operating to blur the distribution of flux over the sphere and 'wash-out' the singularity at the pole, such as the non-zero width of the beam and the non-zero spread of directions in the beam. Clearly, at the pole flux-density is not going to exceed the flux density of the beam, and

$$\frac{d\dot{N}}{d\Omega}\leq \frac{a^2\dot{N_0}}{\pi b^2}$$

where $$\dot{N_0}$$ is the flux in the beam, $$a$$ is the radius of the sphere, & $$b$$ the radius of the beam.

I do not find in Rutherford's paper that there is any attempt to hardwire this blurring mathematically into the proportionality relation - it probably wasn't necessary to do so in order sufficiently to evince that the nucleus is an extreme concentration of charge and mass; and besides, the detection medium at the point on the sphere antipodal to the $$\alpha$$-particle source was almost certainly utterly saturated: but I am curious anyway as to how it would be hardwired in. I fairly sure the greatest contribution to the blurring would be the finite width of the beam, followed by the spread of directions in the beam. I think the contribution from the finity of cross-section about a nucleus, in that more than a certain distance away another nucleus would be encountered, would be a small one.

So the question is primarily "how would the finite cross-sectional area of the beam be hardwired into the expression for solid-angle-density of flux (the $$\csc^4\frac{\phi}{2}$$ expression) so that it can be normalised?". It's almost certainly going to be an integration over a disc centred on point at polar-angle $$\phi$$ (I know $$\phi$$ normally denotes azimuth these days, but Rutherford used it for polar-angle); but I can't quite catch the particular details of how to do it.

Postscript

It's also quite remarkable, I think, the sheer serendipity of how in those days they just barely had the resources to do this experiment: if you calculate what proportion of the flux must be collimated out to get a beam reasonably narrow and sufficiently close to thoroughly parallel, and factor in that $$\alpha$$-particles cannot escape from deep within a bulk of substance, it transpires that radium has just sufficient activity to make the experiment feasible within a reasonable time span. These days the experiment could be reproduced in a trice with a piece of polonium-210, or one of many other nuclides.

• I think one of the ways to make the total cross section finite is to use some sort of electron shielding/charge screening. Basically, the electron cloud in the gold foil does change the Coulomb force. The screened electric potential looks like the Yukawa potential (en.wikipedia.org/wiki/Yukawa_potential). However, I never did the calculations, and can't say for sure how well this works. – Batominovski Nov 14 '18 at 0:41
• Indeed, replacing the Coulomb potential by a Yukawa potential removes the pole at $\phi=0$. See farside.ph.utexas.edu/teaching/qmech/Quantum/node133.html. – Batominovski Nov 14 '18 at 0:47
• @Batominovski -- your comment inspired an attempt to address this problem by a similar method. I have cast it as an answer to my own question, though, as I could not possibly fit it into a comment. – AmbretteOrrisey Nov 14 '18 at 8:29

Let $$r$$ be radius within the $$\alpha$$-particle beam away from the axis of the beam; and let $$b$$ be the closest approach to the nucleus of an $$\alpha$$-particle travelling straight at it.

$$r=\frac{b}{2}\cot\frac{\phi}{2}$$

Introduce a 'unit' sigmoid function (ie has unit gradient at origin, is odd, and tends to ±1 as its argument tends to $$\pm\infty$$) $$\digamma$$ that 'confines' $$r$$ to $$r\leq a$$. This could be $$\arctan x$$ or $$\tanh x$$ or $$\dfrac{x}{\sqrt{1+x^2}}$$, or anything atall that fits the sigmoidity critærion; and set

$$r=a\digamma\left(\frac{b}{2a}\cot\frac{\phi}{2}\right).$$

This basically 'squashes' the whole arrangement into a circle of radius $$a$$, with $$a$$ acting as a parameter.

Also let $$\dot{N}$$ denote flux & $$\dot{n}$$ denote flux density & $$\dot{n}_0$$ be the $$\alpha$$-particle flux density in the beam, & $$\dot{N}_0$$ be the total $$\alpha$$-particle flux in the beam. We have then

$$\frac{d\dot{N}}{d\phi} = \frac{\dot{n}_0\pi ab}{2}\digamma\left(\frac{b}{2a}\cot\frac{\phi}{2}\right)\digamma'\left(\frac{b}{2a}\cot\frac{\phi}{2}\right)\csc^2\frac{\phi}{2}$$

It can be seen that if $$\digamma(x)$$ tends to the horizontal line $$y=\pm1$$ even so slow as $$x^{-\epsilon}$$ (epsilon an arbitrarily small positive number, then the index of the resulting function as $$\phi\rightarrow 0$$ will be $$-1+\epsilon$$, by reason of $$\digamma' ×\csc^2$$, whence the total function integrable.

(It looks kind of odd that the expression explicitly in terms of $$\digamma$$ & its derivative should have a value independent of the precise nature of $$\digamma$$ ... but that becomes clear by noting that the integral is simply of $$r\cdot dr$$; also that insofar as it itself is smaller it's derivative is greater, & vice versa, over most of its range.)

And then

\begin{align}\frac{d\dot{N}}{d\Omega}&=\frac{d\dot{n}}{2\pi\sin\phi d\phi}=\frac{d\dot{n}}{4\pi\sin\frac{\phi}{2}\cos\frac{\phi}{2} d\phi} \\&=\frac{\dot{n}_0\pi ab}{8}\digamma\left(\frac{b}{2a}\cot\frac{\phi}{2}\right)\digamma'\left(\frac{b}{2a}\cot\frac{\phi}{2}\right)\sec\frac{\phi}{2}\csc^3\frac{\phi}{2} \\&= \frac{\dot{N}_0\pi b}{8a}\digamma\left(\frac{b}{2a}\cot\frac{\phi}{2}\right)\digamma'\left(\frac{b}{2a}\cot\frac{\phi}{2}\right)\sec\frac{\phi}{2}\csc^3\frac{\phi}{2}\end{align}

applying the normalisation, which is now an elementary matter - no integration needs to be done because we have effectively done the derivation by reversing the normalisation integral.

By expressing it using the total $$\alpha$$-particle flux density, it covers the scenario of there being many nuclei instead of just one - the many 'discs' each of radius $$a$$ behaving like just one 'disc with all the flux going through it ... except insofar as many take up space laterally! The matter of how the non-zero radius - indeed many atoms radius - acts to blurr the beam. I still haven't sorted that ... but what I have done here is to introduce a roughly plausibly physical tweak to the distribution - plausible in that an $$\alpha$$-particle with its trajectory departing laterally from a nucleus will be caught up eventually in the field of another at a distance from the first that corresponds to the parameter $$a$$ - the radius of the disc within which, in the above derivation, the field is slightly distorted to fall to zero at the edge. If $$a$$ be larger by a substantial ratio than $$b$$, which is physically realistic, as in reality $$b$$ is much less than the semi-interatomic separation ... so the distribution will be pretty much the same as before - just now normalised. Speaking qualitatively, the reason the distribution in its untwoken form has a singularity (and a ^4 one at that!) is that there is in (growing) infinitude of lateral displacement corresponding to $$\phi\rightarrow 0$$.

@Batominovski

I haven't as yet tried (to completion) that method with the Yukawa potential. I found that when I did try it, and commenced the derivation for $$r$$ in terms of $$\phi$$, I got an integral I couldn't solve. Even for the case of a simple Coulomb potential, I had to wheel-out Gradstein & Ryzhik! The integral necessary for the solution using Yukawa potential might be buried in there somewhere ... but I couldn't find it in the time. The Wolfram Integrator gave up the ghost! But then ... I might be using an unnecessarily hard method for solving for the deflection in terms of perpendicular distance of nucleus from undeviated path of $$\alpha$$-particle (isn't "impact parameter" the correct name for that?) - I do that sort of thing sometimes. If anyone knows a simple method, please tell me!

But the Yukawa-potential method looks very similar to mine in many respects, with its single best-fit parameter.

And that

$$(\frac{2mV_0}{\hbar^2\mu})^2\frac{1}{(2k^2(1-\cos\theta)+\mu^2)^2}$$

is exactly the kind of expression I was looking for. But I haven't as yet figured-out the precise mathematical machinery for actually obtaining it! Also, I notice it removes the pole in the density (solid angle density $$d\dot{N}/d\Omega$$) distribution itself. My solution for that is still singular (or rather can still be - this does depend on the choice of $$\digamma$$), and depends on the factor $$2\pi\sin\phi$$ (still clinging to Rutherford's notation!) that arises in integration over a sphere for integrability.

Actually, looking it up, they don't even do it, the classical way atall. Doing it the classical way, conserving momentum & energy about the nucleus, the integral is gotten

$$\int\frac{dr}{r\sqrt{r^2 -br\exp(-kr)-a^2}}$$

, with $$a$$ the impact parameter, is obtained. I cannot find this in Gradsteyn & Ryzhyk (how many "i"s & "y"s are there in the correct spelling of that!?), & the Wolfram Online Integrator reports a "Computation time exceeded!" error. It's fine - though a tad awkward - with $$k=0 \therefore$$ the exponential factor $$=1$$ in it.

Also the $$b$$ in this is not the same $$b$$ as in the non-Yukawa-ised form of this this problem & simply brought over from it: rather $$b$$ will now be given by a lambertw function ... but that's not so very bad atall. Infact, it'll just be

$$\frac{w(kb_0)}{k} ,$$

with $$b_0$$ being the $$b$$ in the non-Yukawa-ised form. The closest approach of the α-particle being well within the shielding corresponds to $$kb_0$$ being a small fraction of 1, & therefore $$b$$ being not much less than $$b_0$$ ... $$b≈b_0(1-kb_0(1-\frac{3}{2}kb_0(1-\frac{16}{9}kb_0)))$$, infact.

Does anyone by any chance know how to do the special "f" for "function" in LateX? It's character 133 in ASCII.