# I Problem understanding crosssection

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1. Mar 13, 2017

### chandrahas

I understand that the nuclear cross section determines the probability of fusion taking place. When I was talking to my friend he said that the cross section doesn't depend on anything but the particles and their energies. But I fail to understand this. For example, let us consider ionization. I have a hot cathode, and an anode placed at distance from each other. And between them, I have a gas. So the electrons would ionize the gas and move to the anode. But this is rare as most of them miss. So if I have a magnetic field, the electrons spiral and there is a higher probability of them colliding with neutral atoms (This is used in ion thrusters).

• Now, since the probability increased, shoudn't the cross section increase too? So the cross section depends on a lot of factors, right? But the confusion arises when almost all websites say that the collision cross section is basically pi(2r)^2 . But isn't it dependent on other aspects? like the magnetic field we just talked about? Then why is it pi(2r)^2?

• Doesn't the same principle apply to fusion? For a certain energy, the gamows factor determines the probability of fusion. Indirectly, the more collisions we have, the higher the fusion cross section. Since, the collisions or collision cross section are higher the probability of fusion increases. And because the collision cross section depends on external factors like magnetic fields, nuclear cross section does too, doesn't it?

• The D-T fusion graphs only work for plasmas that are confined and follow a maxwellian distribution don't they? In other words, they only work for devices that similar to the tokamak.

• So, is the cross section really dependent on various factors like magnetic fields? electric fields? Am i right?

2. Mar 13, 2017

### John Park

Look up "mean free path". Yes, the cross section is the effective area of a body involved in a possible collision, but it's not the only factor that determines the collision probability. With something like an ion moving through a gas, the chances of a collision depend on how far the ion has to travel. You'll find that the mean free path (mfp) is the mean distance between collisions (it depends on the cross section and on the density of the gas). When an ion travels less than one mfp, it's unlikely to hit anything; if it travels several mfp, it's very likely to have at least one collision. The magnetic fields you refer to don't change cross sections; they force the electrons to take a longer route from the cathode to the anode.

3. Mar 13, 2017

### chandrahas

How exactly is the cross section defined as? I am kind of confused about this. I mean, do we have any mathematical definition? Because apparently the nuclear cross section is defined this way. Number of reactions per second = Number density of deuterium ions * Number density of tritium ions* Cross section. I have seen the article you are referring to and it says that the cross section = 1/( number density * mean free path). How do we relate the cross section to the probability of collision? I understand what you said, the larger distance and ion travels, the higher the probability, but can we get a number?

4. Mar 13, 2017

### John Park

In general, it's complicated, because what's causing the scattering is a force (or a potential energy, depending on how you look at it), and that force varies with distance. In the classic case of Rutherford scattering, moving charged particles are scattered by charged nuclei, so the force between them varies as the inverse square of the distance. One can work through the mechanics of this (without quantum theory in the simple case) and find the probability that a particle is scattered through a given angle, depending on how close it comes to the nucleus. The cross-section is a parameter constructed to summarise the results of these calculations. It's an effective area. I can't recall all the details, but a description of Rutherford scattering will give you an idea of what's involved. The scattering forces might not be inverse-square or even electromagnetic, and might have to be treated quantum mechanically, so each case is different in detail. For gas molecules a "hard-sphere" ("billiard ball") model is often adequate, and then the cross section does roughly correspond to the physical area.

5. Mar 13, 2017

### chandrahas

Since the probability of fusion is determined by the Gamows factor, Is it safe to assume that out of all the collisions that take place, the particles that do not fuse participate in elastic collisions? So, what I mean is there are 2 ways in which particles can interact with each other Elastic (coulomb collsions) collisions and fusion. So say that at a particle energy E , the probability of fusion is 0.5, 50% of them fuse while the other 50% of them participate in coulomb interactions and scattering? I know particles would resort to the maxwell boltzmann distribution, but for simplicity, they all have energy E. Then where does the nuclear cross section come into play? If I can use QM to calculate the tunneling probabilities (Gamows factor), and know the energy distribution of the particles, do I even need cross section to figure the probability out?

6. Mar 13, 2017

### John Park

Fusion isn't an area I'm familiar with, so I can't give you a really good answer. It looks as though the Gamow factor gives the probability of fusion after a collision occurs, but I think the cross section would give you actual the probability of collision, for a given density and velocity distribution. So you'd need to know all of these factors to calculate the fusion rate. (For your example, maybe 50% of the particles fuse--but how long does that take? You need to know the collision frequency.)

A side note: elastic collisions are not the only possibility. Internal energy states can be altered, if translational kinetic energy is taken or given up by the colliding bodles. I suspect in air this effect is small and mostly averages out, but Raman scattering (q,v,) is the inelastic scattering of photons, when vibrational states of a molecule are excited by light of much higher frequency.

7. Mar 13, 2017

### Khashishi

I think you left out the speed. It should be
Number of reactions per second = Number density of deuterium ions * Number density of tritium ions * relative speed between deuterium and tritium * cross section.
Or
$r = n_D n_T v_r \sigma_{DT}$
Typically, we have a distribution of speeds, so we have to integrate over the distributions. Taking a distribution average over (relative speed * cross section) gives the rate coefficient (usually denoted q).
$q = \int {f_D(v_D) f_T(v_T) |(v_D-v_T)| \sigma_{DT} d^3v_D d^3v_T}$
Note that it has nothing to due with the physical size of the ions.

We don't really ever talk about the probability of a collision resulting in fusion, because ions don't collide like billiard balls. When we talk about collision frequency of ions, what we really mean is the inverse of the time it takes for a bunch of electromagnetic interactions between ions to accumulate into a deviation in the direction of an ion of about 90 degrees. There isn't any connection between "colliding" in this manner and a fusion reaction.

8. Mar 13, 2017

### Khashishi

The rate coefficient $q$ is usually plotted for Maxwellian distributions, but you can easily calculate the rate coefficients for some other distribution function if you know the cross section. For a monoenergetic beam colliding with a cold target, $q$ is just $v_\text{beam}\sigma$

I don't think there's a significant dependence on magnetic field or electric field. The nucleus is much smaller than the gyroradius for achievable fields, and the electric potential across one radius of distance is very small for an external field.

9. Mar 13, 2017

### chandrahas

.
Then how do collisions occur? and how does fusion occur at all if there are no collisions? And how often do particles get close enough, so that gamows factor can describe the probability?

But what if the deuterium and tritium are spiraling in opposite directions with the same gyroradius? I mean, there is a dense plasma in the center, and also as per the example that I gave, Doesn't that change the fusion cross section? Because, they are taking longer paths through the plasma, as John park said. But in the equations above, (Rate of fusion=Number density of deuterium*Number density of Tritium *Velocity* Cross section), I don't see any term accounting for the distance the ions traveled through the plasma (due to magnetic fields) , Hence, shouldn't the cross section include that factor? Where am I wrong?

10. Mar 13, 2017

### John Park

"Collision" is a convenient term but it probably is misleading. What happens when two bodies get close depends on the forces between them. Even in the case of Rutherford scattering it is the electrostatic inverse-square law that causes the repulsion, it's not really like a billiard-ball collision; the two particles approach until the electrostatic repulsive potential balances the kinetic energy of the incoming particle. It's more like rolling up an increasingly steep hill than running into a wall.

In the case of fusion, the important forces involved in the scattering are I believe electrostatic and nuclear. The electrostatic forces provide a strong repulsion, but with enough kinetic energy the particles can get close enough for the nuclear forces to become dominant. At this point quantum mechanics takes over and my understanding starts to decay exponentially.

As I understand it, the cross section hasn't changed--it's a description of the interaction of two particles; what has changed is the velocity distribution of the particles. (What does "velocity" in that equation mean exactly? Presumably it's some sort of effective value.) It sounds as if it could be a matter of convention, though, how the components of the "collision" frequency are divided up. Really, it's that frequency that is important. As long as all the relevant factors are included, it ultimately doesn't matter how they're grouped together.

11. Mar 14, 2017

### Khashishi

I don't know too much about calculating the fusion cross section. But you can look here for an approach that uses the Gamow factor:
http://zuserver2.star.ucl.ac.uk/~idh/PHAS2112/Lectures/Current/Part7.pdf
It uses $\pi \lambda^2$ where $\lambda$ is de Broglie wavelength as an effective size for a quantum collision

There's no reas
All that matters is the relative speed of the two particles when they are close to each other. The gyro motion will tend to randomize the collision angle, but if the initial distribution functions are isotropic (such as a thermal distribution), then the distribution with magnetic field is still isotropic, so it doesn't matter. If the initial distribution isn't isotropic (e.g. you are colliding two beams together) then, yes, the magnetic field will tend to scramble the beam directions such that the relative velocity may be affected.

Huh? I am completely confused by what you mean by (due to magnetic fields). Distance traveled is proportional to the velocity.

12. Mar 14, 2017

### chandrahas

What I mean by that is that when there is no magnetic field, the particles just go straight through the plasma and hence, the probability of collision is small. But if you have a magnetic field, for example, A helical field, the particles spiral around the field the lines and drift along it.

Something like the image, but the magnetic field is helical too. So, the particles will have to spiral ALONG the direction of the field lines multiple times while simultaneously spiraling AROUND the field lines, before it can get to the other end. Sorry, I couldn't find a picture of a helical magnetic field. Anyway, so the probability of collision goes higher

13. Mar 14, 2017

### Khashishi

Yeah, but the arc distance along the helical path is just vt, just like for a straight path.

14. Mar 14, 2017

### the_wolfman

You are confusing the collision frequency with the probability of undergoing a collision before a particle escapes confinement.

In a uniform plasma the collision frequency is path independent. If the collision frequency is $1 s^{-1}$ then it is $1 s^{-1}$ if the particle is gyrating in a circle and it is $1 s^{-1}$ if the particle is moving in a straight line. The collision frequency is telling us that on average the particle will undergo 1 collision every second it stays in the plasma.

The probability of a particle undergoing a collision before escaping depends on the collision frequency and it's individual confinement time.

Adding a magnetic field will increase the particle's individual confinement time, but it does not change the collision frequency (assuming the density and temperature of the background plasma don't change).

In a magnetically confined plasma the collision frequency is low and the "fusion frequency" is even lower. To produce a net amount of energy we need to maximize the number of ions that fuse before they escape confinement. To do this we need long confinement times.

15. Mar 15, 2017

### chandrahas

I've been reading about cross sections and then I kind of got confused. What exactly is the difference between elastic scattering cross section, Differential cross section, Rutherford scattering cross section and coulomb collision cross section? Aren't these all elastic? Than why do we have so many terms? I don't understand the difference.

16. Mar 15, 2017

### the_wolfman

As the name implies the differential cross section is the derivative of the cross-section with respect to the solid angle. It is mathematically different from a cross-section.

There are many different elastic scattering processes. For example you can have elastic scattering due to the electrostatic interactions between particles, but you can also have elastic scattering due to nuclear forces. The term elastic scattering is more general than Rutherford scattering.

The difference between Coulomb collisions and Rutherford scattering is more nuanced. The Rutherford problem considers 1 collision between 2 charged particles and calculates a differential cross-section for this interaction. The Coulomb collision problem takes then takes the Rutherford problem a step further and ask what happens if you have distribution of such particles, like a plasma. It uses the result of the Rutherford problem to calculate an effective collision frequency. This can then be used to calculate a diffusion coefficient and other transport coefficients.

17. Mar 15, 2017

### chandrahas

Oh. Got it thanks. I finally understand.

18. Mar 16, 2017

### chandrahas

Many people have told me that beam target fusion wouldn't work because the particles are more likely to scatter and lose all their energy than fuse. But I am kind of confused, is the scattering cross section smaller for tokamaks than for particles passing through a relatively cold plasma? If so, why is this? and how much power do tokamaks lose through braking radiation per unit volume?Thanks for your help...

Last edited: Mar 16, 2017
19. Mar 16, 2017

### Staff: Mentor

In a tokamak, the particles don't lose their energy due to scattering, because all particles have a large kinetic energy.

There are some radiative losses, but those are a small effect in this context.

20. Mar 16, 2017

### the_wolfman

The problem with beam fusion is that on average scattering will knock particles out of a beam before they can fuse.

Scattering is a process that coverts a beam of particles into a thermal distribution of particles. Scattering does nothing to a uniform thermal distribution of particles. If you want to build a power plant that generates a net amount of fusion, then you have to account for scattering and build a devices that confines a thermal distribution of particles. This is the idea behind confinement.