Singularity in Rutherford cross section

[klingerdinger]

The Rutherford differential cross section $$\frac{d\sigma}{d\Omega}$$ goes like

cosec($$\vartheta$$)^4

which means at $$\vartheta$$=0 the differential cross section is infinite, which is ok.
My question is, given that the differential cross section is proportional to the probability per unit solid angle $$\frac{dP}{d\Omega}$$, which is proportional to the expected rate of scattered particles, why/how does the expected rate go to infinity at $$\vartheta$$=0?

I take I've got something a little/very wrong...

 

[Bob S]

You are correct. After integrating over solid angle for a deflection greater than θ, the scattering cross section is

σ(>θ) = (πb²/4) cot²(θ/2)

where b= "collision diameter" = zZe²/(½mv²)

For small scattered angles, the impact parameter x** (distance of closest approach) is very large. But the Coulomb field does not extend to infinity, due to screening of the nuclear Coulomb field by atomic electrons. Thus at small angles, the cross section cuts off, and the total cross section is finite.

** x = (b/2) cot(θ/2)

I hope this helps.

Bob S

[added] Very roughly, the maximum impact parameter x should be set to about the "radius" of the 1s atomic electrons, just as electron screening begins to shield the nucleus.

When you do a Rutherford scattering experiment using an alpha source and a gold foil, you will see the counting rate drop by several orders of magnitude as θ increases, following the Rutherford scattering formula. Then at about θ = ~90 degrees the differential cross section starts deviating remarkably from the Rutherford cross section (Coulomb scattering). You can use this to establish the impact parameter of the gold nucleus.

 

[klingerdinger]

Thanks that really helps

 

[Vanadium 50]

The problem with Bob's answer is that you still have a singularity in the scattering with a bare nucleon. This singularity is real.

What you have to remember is that the Rutherford cross-section is a theoretical calculation. Like all calculations, how well it matches the real world depends on how well the assumptions match the real world.

The assumption that matters is that the interaction between the scattered particle and the scattering particle is instantaneous and depends pretty much only on the closest point of approach of the particles. This is a good approximation for wide angle scatters: a free particle comes in close, has a sudden interaction, and then a free particles moves out.

For interactions where the two particles never get close, this is a poor approximation - and that's exactly where the Rutherford cross-section becomes singular. So the lesson you should take from the singularity is not that you can ignore it because it's difficult to prepare a system that's not under the influence of nearby electrons: it's that you should recognize that this is telling you that (at least) one of the assumptions in the derivation is no longer valid.

 

[Bob S]

Coulomb scattering (the Rutherford scattering model) is for a charged particle scattering in a 1/r² central force (i.e., the Coulomb field). The calculated trajectory is hyperbolic, and is exact for a simple scattering of a charged particle in a central Coulomb field. As you may realize, this is exactly the same as the scattering of an extra-solar object in the gravitational field of the Sun. The main differences are that for Coulomb scattering of a nucleus, 1) the force may either attractive (electron, negative muon) or repulsive (alpha, proton, positive muon), 2) the force at large distances is shielded (screened) by atomic electrons, and 3) the force is non-Coulomb at small distances (due to strong interactions, radiative interactions, vacuum polarization, etc.).

The infinite cross section for very small-angle scattering is correct for Coulomb scattering off a bare nucleus, because the Coulomb field extends to infinite distances. But for neutral atoms, the Coulomb field is cut off due to screening by atomic electrons. A good electron-screening model is the Thomas-Fermi model, which can be approximated by a cutoff maximum impact parameter of x = 0.529·Z^-1/3 x 10^-8 cm. Moliere scattering theory also addresses the small scattering angle, large impact parameter regime.

Bob S

 

[tupos]

I registered here only to tell that you guys wrong.

It ca't be singularity simply from mathematical point of view. Your Coulomb wave functions are finite for any angle. But you have singular cross section. What is it? Then try to calculate total cross section or transport, which is extremely important for calculating kinetic characteristics in plasma or semiconductors!

When you have a semiconductor, let's say extrinsic. There you have screening radius Rs. or in angles 1/(kRs), k is a wave vector of your particle(p=hk). then when kRs^2/r >1 then you don't have screening ( r=n^{1/3}, n-concentration)? What are you doing in this case?

You mathematically calculate your cross section incorrectly.

I know that cross section is finite(Not actually the differential cross section itself, but sigma(\te)*sin(\te) remains finite for any angle!), I calculated it.

Regards.

 

[Bob S]

I registered here only to tell that you guys wrong.

It ca't be singularity simply from mathematical point of view. Your Coulomb wave functions are finite for any angle. But you have singular cross section. What is it? Then try to calculate total cross section or transport, which is extremely important for calculating kinetic characteristics in plasma or semiconductors!

When you have a semiconductor, let's say extrinsic. There you have screening radius Rs. or in angles 1/(kRs), k is a wave vector of your particle(p=hk). then when kRs^2/r >1 then you don't have screening ( r=n^{1/3}, n-concentration)? What are you doing in this case?

You mathematically calculate your cross section incorrectly.

I know that cross section is finite(Not actually the differential cross section itself, but sigma(\te)*sin(\te) remains finite for any angle!), I calculated it.
.

Please read my previous post on electron screening by atomic electrons, and the Thomas Fermi model:

The infinite cross section for very small-angle scattering is correct for Coulomb scattering off a bare nucleus, because the Coulomb field extends to infinite distances. But for neutral atoms, the Coulomb field is cut off due to screening by atomic electrons. A good electron-screening model is the Thomas-Fermi model, which can be approximated by a cutoff maximum impact parameter of x = 0.529·Z^-1/3 x 10^-8 cm. Moliere scattering theory also addresses the small scattering angle, large impact parameter regime.

Bob S

 

[Vanadium 50]

but sigma(\te)*sin(\te) remains finite for any angle!), I calculated it.

Apparently, not correctly.

$$\frac{d\sigma}{d\Omega} = \frac{A}{\sin^4(\theta)}$$

$$\sigma = \int \frac{A}{\sin^4(\theta)} d\Omega$$

$$\sigma = \int \int \frac{A}{\sin^4(\theta)} \sin(\theta) d\theta d\phi$$

$$\sigma = \int \frac{2 \pi A}{\sin^3(\theta)} d\theta$$

$$\sigma = \frac{A \pi}{4} \left[ \sec^2(\theta) - \csc^2(\theta) + 4 \log(\tan(\theta/2)) \right]$$

which is clearly divergent.

The Rutherford cross-section has an honest to goodness singularity. However, it is an approximation to a physical cross-section, which does not have one.

 

[klingerdinger]

tupos, i think you've completely understood what it is that was asked.

I was curious to why the Rutherford cross section tends to infinity. I think we're all aware that the its not the complete picture, but i wanted to know why it broke down so catastrophically in the small angle region.

And as Vanadium 50 has very kindly shown, it does.

 

[tupos]

Hello guys,

The problem with vanadium calculation is that he use asymptotic calculation of differential cross section. If you open any book on quantum mechanics, you'll find out the integral equation for scattering theory. it is a sum of plain wave and integral of \int{g(r-r')\psi(r')V(r')dr'}, you know the green function exp[ik|r-r'|]/|r-r'|, then you expand your green function and obtain asymptotic representation exp[i\vec k \vec r] + f(\te) exp{ikr}/r, from here you make a conclusion that your d\sigma is proportional to |f(\te)|^2!

But if you carefully think for the Coulomb case you can't expand the green function, because it is mathematically incorrect. And your asymptotic representation can't be used. That is why your Rutherford cross section calculated wrong.

Except of this you should use exact solution of Schroedinger equation in the Coulomb field

exp{i \vec k \vec r}F(i\xi,1,ikr(1-cos(\te))), which is a sum of two functions U_1 and U_2,

U_1 has spherical wave asymptotic and U_2 has plain wave.(See Mors and Feshbah for more details). After this you should calculate flux in a usual way

\psi^*\nabla \psi - \psi\nabla\psi* using U_1 function as a \psi in this formula.

And only after this you should calculate the cross section using standard definition

\si = j_{sc} / j_0 * r^2 d\Omega.

If you do all this you find out the your cross section is proportional to function |G_1(\te)|^2*sin(\te). G_1 behaves on small angles like ln(x), (Mors,Feshbah). From here you can see that differential cross section is finite because |ln(x)|^2 * x = 0 when x->0.


From here you can calculate total cross section which is of course not infinite and has the following form

\sigma_total = 2\pi r /k * I(\xi),

\xi here and upper in F is the same and characterized your incident beam, \xi = \alpha /(hv), where v is a velocity v = hk/m, alpha = Ze^2. And I is some function which depends only from \xi.

Maybe now you understand it better. It is a problem of every book of quantum mechanics. People simply calculated this incorrectly for the whole angle region.


Regards.


P.S. asymptotic calculation are correct in angle region \te>\te_0

where \te_0 describe the "near" zone, a zone where you can't distinguish plain and spherical wave from each other.

\te_0 = \sqrt{2/(kr)}

 

[sirapwm]

The indefinite integral effected in this PF post (from 2010) is not the correct expression for the Rutherford differential cross section.

Here is the correct expression in a slightly different notation (so I'll try to be careful to define all the symbols).

The expression for the differential cross section for charged point particles of charge $Z_1$ and $Z_2$ interacting by the Coulomb interaction $V_c(r)=\frac{Z_1 Z_2 e^2}{r}$ as typically derived (see p. 438 of Sakurai, Modern Quantum Mechanics) is:
$$\frac{d\sigma}{d\Omega} = \frac{\gamma^2}{k^2\, 4\sin^4(\theta/2)},$$ where $\theta$ is the scattering angle in the center-of-mass of the interacting particles. Here, $\gamma k = Z_1 Z_2 e^2 m, m^{-1} = m_1^{-1} + m_2^{-1}, k = \sqrt{ 2 m E}$, where $E$ is the center-of-mass energy and we recognize $m$ as the reduced mass of the interacting particles. (We are employing natural units where $\hbar = 1 = c$.)

Note that the expression in the post linked above has $\sin\theta$ compared to $\sin(\theta/2)$ here.

We define the following integral, indefinite with respect to the limits on the scattering angle $\theta$ but definite with respect to the azimuthal angle $\phi$:
$$\sigma_{\text{int}}(\theta, E) \equiv \int d\theta\sin\theta \, \int_0^{2\pi} d\phi\, \frac{d\sigma}{d\Omega}.$$ (Note that $d\Omega = d\theta\sin\theta\,d\phi$.) Substitution of the expression for the differential cross section above into this definition and effecting the trivial integral over $\phi$ gives the expression:
$$\sigma_{\text{int}}(\theta, E) = \frac{\pi \gamma^2}{2 k^2} \int d\theta\, \frac{\sin\theta}{\sin^4(\theta/2)}.$$ Recognizing $\sin x = 2 \sin(x/2)\cos(x/2)$, making a substitution $u = \sin(\theta/2)$ gives the elementary integral
$$\int d\theta\, \frac{\sin\theta}{\sin^4(\theta/2)} = 4 \int du\, \frac{1}{u^3} = -2 u^{-2} + C = \frac{-2}{\sin^2(\theta/2)} + C,$$ where $C$ is a constant that depends on the (neglected) limits of integration.

Clearly, this quantity is divergent in the limit $\theta \to 0$, which the post linked above correctly states.

Note, however, that user tupos points out -- also correctly -- that the exact wave function for the Coulomb Schrödinger equation is finite at all finite ranges $r$ between the scattering particles. The origin of the singularity in the Rutherford expression for the differential cross section is the application of the asymptotic form for the confluent hypergeometric functions $_1F_1(-i/k,1,ik \eta)$ where $\eta \equiv r - z$ (see p. 566 of Landau & Lifshitz Quantum Mechanics; the asymptotic form is given op. cit. p.662, Eq.(d.16).)

 

[mfb]

@sirapwm I merged your post with this thread and @Vanadium 50 can have a look.

 

[sirapwm]

@sirapwm I merged your post with this thread and @Vanadium 50 can have a look.

Many thanks. -S

 

[vanhees71]

Some careful analysis of the problem, using the exact solution using the confluent hypergeometric functions can be found here:

https://arxiv.org/abs/quant-ph/0403050

 

[sirapwm]

Some careful analysis of the problem, using the exact solution using the confluent hypergeometric functions can be found here:

https://arxiv.org/abs/quant-ph/0403050

I'm aware of this paper. It agrees with my statement above, regarding tupos' comment, that the exact (not the asymptotic expression for $\eta = r-z \to \infty$) wave function is finite at all finite $r$. And yes, as you point out, the exact solution is a linear superposition of two linearly independent functions -- a regular term and an irregular (singular as $r\to 0$) term -- both of which satisfy the confluent hypergeometric equation. (See Landau & Lifshitz, Quantum Mechanics, Appendix d, p.659ff.) I haven't found the asymptotic expression for $r\to\infty$ (not $\eta\to\infty$) written anywhere. I think it should be possible to employ the addition theorem for Kummer's function to do this. I will post it here if I come up with anything.

 

[vanhees71]

It's really hard to find a thorough discussion of the exact solution. All textbooks I know either derive the asymptotic expression which is singular for forward scattering and discuss the apparent singularity at $\vartheta \rightarrow 0$, for which of course $\eta \rightarrow 0$ no matter how large $r$. So the above mentioned paper was the best I could come up with, but I think one just can use the asymptotics of the said hypergeometric functions for $\eta \rightarrow 0$.