Dear Dipanker Das,

As I thought about your question for a few more minutes, and I wrote down the integral of the Rutherford (Coulomb) scattering cross-section over the full solid angle, I saw that you were indeed correct: the integral of the standard Rutherford scattering cross section formula over the full solid angle does indeed diverge to an infinite flux. However as we noted earlier in this thread, this cannot be physically correct. So I began to review the assumptions of the mathematical model carefully, and I realized that there is an approximation in the model that is a reasonably good approximation for most angles but not for very low angles or very high angles (near total-backscatter, angle = pi or angle = 180 degrees.)

The approximation in the model is that the incident particle is much less massive than the target particle (for example in Rutherford's experiment the helium nucleus is much less massive than the gold nucleus) and therefore the conservation of momentum results in very little energy transfer from incident particle to target particle. In the simple Rutherford cross-section, the assumption is that ZERO energy is transferred to the target particle (the target particle remains stationary during the whole interaction). If you eliminate this assumption and formulate the interaction correctly, you should get a result that still has zero in the denominator at deflection angle = 0 (the physical meaning here is that the area of space far away from the target particle is "infinitely large" compared to the area near the particle) but its integral over the full solid angle must converge to probability P = 1, that is, flux in = flux out.

Unfortunately the mathematical expression for the correct trajectory (including the resultant motion of the heavier particle and the energy transfer) is going to be horrendously complex and not integrable using symbolic mathematics: you would have to integrate the resulting expression numerically using a computer and accept an approximate numerical solution as the result.

As an example of how the correct treatment, including energy transfer and the resultant motion of the target particle, can transform the divergent (infinite) flux of the Rutherford approximation into a convergent (finite) flux in a physically real scattering system, see the discussion on this web page:

http://www.plasmaphysics.org.uk/coulomb.htm

Even the case discussed there is an oversimplification. The physicist who wrote that article, realizing that a full two-body treatment of the Coulomb scatter interaction would be unsolvable symbolically, chose a simpler energy-transfer model. He chose to model the case where there is a Coulomb interaction at a distance, and then at a certain finite (very small) radius there is an effective energy transfer as the incident particle changes direction near a spherical "surface", using an equation from the well-known mechanics textbook by Landau and Lifschitz. Using the resultant energy transfer to derive an approximate correction to the Rutherford scattering formula, he shows that this correction removes the divergence and allows the total cross-section to be normalized to a finite flux.

Well-spotted. Your intuition was correct that there was something wrong with the standard "Rutherford scatter cross-section" formula for the Coulomb interaction for small deflection angles.

That is an academic way of saying that most of the stuff is missing the target

Google optical theorem

and no, you will not collect an infinite no. of particles in the forward direction. The DIFFERENTIAL cross sec. is infinite, the integral is finite

It means your model has a problem.

(my point of view) In the Electromagnetic case, elastic scattering in the forward direction blows up because infinite range of EM interaction. Even though the particle doesn't actually scatter, it does interact (because infinite range). Thus the divergence in the diff cross section.

In case of forces of infinite range (electromagnetic, gravitational forces) the total cross sections are infinite... of course.(force goes to zero at infinite distance only).

So, in agreement with the optical theorem, the scattering amplitude at angle zero is also infinite. Mathematically speaking as in practice it is not a problem, there is something like an effective cutoff effect.

Take a narrow beam, with widening angle < \alpha, and place a detector downstream at an angle \beta. If

\beta < \alpha, the detector will be swamped by non-scattered particles, but if \beta >> \alpha, you will only catch the scattered ones. And now let \beta \to 0, always keeping \alpha

I just would like to remember that the assertion "diff cross section blows up in the forward direction" is also valid, for example, for a particle crossing a medium, not only within collider physics. This "blowing up" means the particle is interacting with all other particles in the medium, even though not actually scattering.

This question is not well done, since the medium, or environment is not clear.

The phenomenon you try to explain may be referent to atmosphere and movement and...

You may be asking something about 'Doppler effect'!

Pls search about it!

Best!

Murilo

In the scattering formula of diff. cross section you see sin(t) in divisor, so blow up in t=0 grad.

It only means that the particles will surely get scattered in the forward direction which makes sense because they will "collide" when impact parameter is zero. What is the problem?

In Compton scattering as the scattering angle tends to zero, the Klein-Nishina cross section formula blows up. That is what is referred here. For more details see the weblink below. You can also find a very good discussion on Compton scattering, Raleigh scattering and KN cross section limiting approaches with graphical plot in the text book "Nuclear Physics" by "Irving Kaplan"

" .......Compton scattering, photoelectric absorption and pair production are the three main energy transfer mechanisms in photon interactions with matter. Each of these processes can dominate under specific conditions determined chiefly by the incident photon energy hν and atomic number Z of the absorber.

1.1.1 Compton Scattering

• Klein-Nishina differential cross section for Compton scattering measures probabilty of photon re-emission into solid angle dΩ = d cos θ dφ as a result of a Compton interaction between an incident photon and a free electron

• exact expression is derived from quantum electrodynamics:

(1)

• Klein-Nishina form factor:

FKN = [1+ε(1−cosθ)]−2 (3) 􏰌 ε2(1−cosθ)2 􏰍 "



The above is part of text quoted from the lecture notes available at: http://www.physics.usyd.edu.au/~kuncic/lectures/RP3.pdf

Vaidejhi Venkatakrishnan, Ph.D., DABR

Remember that the number of particles scattered into a solid angle equals the product of diff cross section and incremental solid angle- this product remains finite, despite the diff cross section going to infinity.

@ Jindal,

Sir, I didn't get your first comment. What do you mean by 'collide' and 'surely scattered' when impact parameter is zero? Does it also mean, when the impact parameter is non-zero, it will 'not collide' and 'not surely scattered'? When the impact parameter is zero, do you always expect scattering amplitude to be infinity (Which is of course not always the case)??

In your second comment you said something that I want. But its not very precise. See, in coulomb scattering there is sin^4(\theta/2) at the denominator and there is a sin\theta in expression of d\omega. So, overall the product diverges as 1/\theta^3 for small \theta. Doesn't it? Correct me if I am wrong..

@Caio:

I don't understand the difference between being scattered and interacting. Please elaborate. And in the second comment, if I get it right, you are talking about multiple scattering which are neglected in the analysis and inspite of that the divergence occur.

@Adam,

Sir, please explain the 'effective cutt-off effect' or, suggest some basic references. I agree that the optical theorem suggests such a divergence, but isn't that a problem? In such cases, how would you expand the amplitudes in terms of partial waves?

In the forward direction the momentum transfer is zero which means according to the uncertainty principle that the particle is interacting with many particles of the target system. The scattered waves from that particles add up coherently in the forward direction independently from their local arrangement in the target system.

As I understand, we can also understand DCS as "probability of finding a scattered particle within a given solid angle" so it means in forward direction the "probability" of scattered particle is blow up....

In the simplest and most concrete interpretation, it means this:

"Scattering" is when a moving particle is deflected by a target particle into a different angle (vector) of motion.

"Differential Cross-section" is the probability (or equivalent area) where an incident particle is scattered by the second particle into a certain range of scattering angle.

Since "forward cross-section" means zero angular deflection, the simple physical interpretation is that the area of all of space far away from the target particle is infinitely larger when compared to the cross-section or area of the target particle.

Another way of saying this is, if you aim randomly through all of space, the odds are infinitely more probable that you will miss and not hit (deflect from) the target particle.

However this does not lead to infinite flux of particles or other non-physical result. Remember the "differential cross-section" is not a total probability, it is a probability PER UNIT SOLID ANGLE. When you integrate over the full solid angle, as Vijay Jindal posted above, you will get a finite result. If your probability distribution is correctly normalized, the integral over the full solid angle gives Probability = 1, that is, 1 particle in = 1 particle out. This is because the "forward cross-section" refers to angle=0 where the solid angle --> 0 in the limit. So the improper integral converges to a finite and physically reasonable result, even though there is a zero in the denominator of the fraction at angle=0.

If your integral over the solid angle is diverging to infinity (infinite flux out for finite flux in) you need to revisit the assumptions in your mathematical model. I agree with your assertion that a physically correct model must yield (flux in) = (flux out) when integrated over the full solid angle. However this conservation law may not be preserved in some approximations, which break down in certain limits.

Dear Dipankar,

by a cut-off I meant that in practice you are not interested in particles scattered by infinitesimaly small angle, but by cases in which a particle scatters by, lets say, an angle bigger than a microradian. Then you isolate the apparent singularity in the scattering amplitude.

Dear Dipanker Das,

As I thought about your question for a few more minutes, and I wrote down the integral of the Rutherford (Coulomb) scattering cross-section over the full solid angle, I saw that you were indeed correct: the integral of the standard Rutherford scattering cross section formula over the full solid angle does indeed diverge to an infinite flux. However as we noted earlier in this thread, this cannot be physically correct. So I began to review the assumptions of the mathematical model carefully, and I realized that there is an approximation in the model that is a reasonably good approximation for most angles but not for very low angles or very high angles (near total-backscatter, angle = pi or angle = 180 degrees.)

The approximation in the model is that the incident particle is much less massive than the target particle (for example in Rutherford's experiment the helium nucleus is much less massive than the gold nucleus) and therefore the conservation of momentum results in very little energy transfer from incident particle to target particle. In the simple Rutherford cross-section, the assumption is that ZERO energy is transferred to the target particle (the target particle remains stationary during the whole interaction). If you eliminate this assumption and formulate the interaction correctly, you should get a result that still has zero in the denominator at deflection angle = 0 (the physical meaning here is that the area of space far away from the target particle is "infinitely large" compared to the area near the particle) but its integral over the full solid angle must converge to probability P = 1, that is, flux in = flux out.

Unfortunately the mathematical expression for the correct trajectory (including the resultant motion of the heavier particle and the energy transfer) is going to be horrendously complex and not integrable using symbolic mathematics: you would have to integrate the resulting expression numerically using a computer and accept an approximate numerical solution as the result.

As an example of how the correct treatment, including energy transfer and the resultant motion of the target particle, can transform the divergent (infinite) flux of the Rutherford approximation into a convergent (finite) flux in a physically real scattering system, see the discussion on this web page:

http://www.plasmaphysics.org.uk/coulomb.htm

Even the case discussed there is an oversimplification. The physicist who wrote that article, realizing that a full two-body treatment of the Coulomb scatter interaction would be unsolvable symbolically, chose a simpler energy-transfer model. He chose to model the case where there is a Coulomb interaction at a distance, and then at a certain finite (very small) radius there is an effective energy transfer as the incident particle changes direction near a spherical "surface", using an equation from the well-known mechanics textbook by Landau and Lifschitz. Using the resultant energy transfer to derive an approximate correction to the Rutherford scattering formula, he shows that this correction removes the divergence and allows the total cross-section to be normalized to a finite flux.

Well-spotted. Your intuition was correct that there was something wrong with the standard "Rutherford scatter cross-section" formula for the Coulomb interaction for small deflection angles.

Dear Kenneth,

That was a very nice comment. I will look into the link you posted, then I shall make any comment. But before that, I want to state that this is not the problem of classical Rutherford formula only. This problem exists in QFT also. As far as I know, till now, QFT gives most accurate description of nature. But if you look at the tree level Bhabha scattering amplitude, the problem is also there. This divergence comes from the t- or u- channel photon propagator. Now, what is the approximation here? One may think that the divergence may be removed from higher order corrections. This is very unlikely, as long as perturbation expansion is valid. The higher order corrections involve different powers of the coupling constant. I have not seen a QFT book, which tries to calculate total scattering cross-section of Bhabha scattering- the all stop after calculating the tree level amplitude. There may be other similar examples. I suppose it is due to the photon propagator. I don't know how people justifies it theoritically. (By the way, the experimental meaning of this is mentioned in Strauss's comment in this thread).

Some might say, "No wonder! Photon is massles, it has infine range, so the total cross-section is expected to blow up." Is that good enough? I don't know. I mean, look at e^+e^- ----> \mu^+\mu^- . This is also photon mediated (mainly). But the cross-section is finite!

Actually collision is the physical contact phenomenon but scattering is a energy share process between charges. I am taking an example of scattering of electron from atom. In forward direction impact parameter is zero. In this case the electron will experience an attractive force rather than repulsive force. It will be accelerated in field of nucleus. So they are not deflected by repulsive force field. Thus flux of scattering electrons are more in this direction. As impact parameter is increased , there are competition between both forces( repulsive by electron cloud ) hence their direction will changed . If we look at the scattering flux in forward direction it gets maximum value. So in other word we can say that diffraction cross section is blow up in forward direction.

Small deflections, in general, are the consequence of larger collision are impact parameters. Larger the collision parameter weaker will be the interacting force between the colliding parteners. And we know long range forces like coulomb interaction become weaker and weaker at larger collision parameters which will certainly influence the scattering crossection in the forward direction. For coulomb interaction the differential scattering crossection, for small scattering angle, is given by a simple expression q(x) dx = 2.pai.b .(db/dx)dx = 2pai(z1.z2)2 /E2 . x2. Here x is the scattering angle, and E is the centre of mass energy of the colliding charged particles of charges z1 and z2. b is the impact parameter which is a function of x. For a better understanding you should consult an old but good book in Quantum Theory written by David Bohm. I am sure this book is still in some self of the SINP library.

Dear Dipankar,

I came across this page today because I was wondering about the exact same question as you. I went through the answers on this page and I think I can now probably summarize an answer to your question. Bear with me if this seems a little lengthy...

The divergence in the DCS is not due to the point-like nature of the target in the Rutherford model. Take for instance, the QM treatment of hard-sphere scattering. The total cross-section here is proportional to \pi R^2, where R is the radius of the sphere. In the limit that the target shrinks to a point, the total cross-section vanishes. Indeed, if the divergence in the Rutherford model was due to the point-like nature of the target, then it would most likely vanish when integrated, because the probability of an alpha particle lining up exactly with the point-like nucleus is zero. Thus we may eliminate one source of confusion: The divergence is not due to the point-like nature of the target because it's not just the DCS but also the total cross section (TCS) that diverges, as you too pointed out.

On the other hand, the comparison with hard-sphere scattering makes it obvious where the difference lies: The hard-sphere potential is short-ranged whereas the Coulomb force is not. What this means is that the beam particles will always be deflected no matter how far away they are from the target (in the transverse direction). However, these large-distance particles will also be deflected only slightly, because the Coulomb force is weak at these distances.

What does this add up to? Imagine the whole xy-plane filled with incoming particles meant to scatter off a target sitting at the origin. If one draws a circle with center at the origin and radius b, then all the hard-scattering, that is, scattering through angles larger than some \theta_min, happens for particles within this circle. All the other particles, from b out to infinity, scatter through angles less than \theta_min. These are the particles that fill up the small solid angle around \theta=0, and this is why the Rutherford DCS diverges in the forward direction.

Lastly, in reply to your specific question, note that the divergence comes from particles that are very far away from the target in the transverse direction. Therefore, any detector placed directly behind the target would not register an infinite flux (More precisely, to "see the infinity" in the DCS, the downstream detector would not only have to be directly behind the target but also at an infinite distance from it).