# Why is it easier to ignite a low density plasma?

1. Oct 10, 2014

### rogerk8

Hi!

I have kind of learned that a low density plasma is easier to ignite than a high density plasma.

Rearranging the Saha equation gives:

$$n_i^2=2.4*10^{21}n_nT^{3/2}\exp-(\frac{U_i}{kT})$$

Now, if the neutral particle density, nn is low, temperature needs to be even higher.

On the other hand the dominant part of the equation is the exponetial term.

So a high kT (or Ek) gives free ioized particles, ni.

While T (and thus Ek) is closely related to v2 you may consider temperature as particle velocity, v.

Now, for the neutral particles to be able to have high speed without hitting eachother all the time which would reduce the net speed and thus temperature, the neutral particle density needs to be low.

The last one is my guess.

How far from the truth am I?

Best regards, Roger.

2. Oct 10, 2014

### Staff: Mentor

If you want to get the same charged particle density, yes. If you want to get the same ionization fraction, the left side goes down faster than the right side if you reduce pressure.

I don't understand your arguments in the rest of the post.

3. Oct 10, 2014

### rogerk8

What do you mean by "ionization fraction"?

That term does not even excist in Wikipedia.

And what is it with the rest of my post you do not understand?

Best regards, Roger

4. Oct 10, 2014

### rogerk8

I see three types of pressure in a Tokamak.

$$p_n=n_nkT...[1]$$

$$p_i=n_ikT...[2]$$

$$p_e=n_ekT...[3]$$

If we consider pure Hydrogen, ni = ne = n.

Still they probably have different velocities.

They do have different velocities due to kT being equal while having different mass.

On the other hand, why should kT be equal?

Maybe I should write

$$p_i=nkT_i...[2']$$

$$p_e=nkT_e...[3']$$

But the first equation is valid as long as it isn't a plasma, which perhaps may defined as ni/nn<1%?

Best regards, Roger

5. Oct 10, 2014

### Staff: Mentor

Fraction of atoms that have been ionized.

Mainly this:
What does that mean? Neutral particles hitting each other do not lose speed - unless they lose energy by ionization.

Relating temperature to speed (even an average speed) is problematic for particles of different mass.
In equilibrium, the temperature is the same everywhere. Note that the material (does not matter which state) does not have to be in equilibrium...

6. Oct 11, 2014

### rogerk8

This is very interesting. Imagine a very dense gas under high pressure if you like. Increasing the temperature will make the particles move faster. But only if there is room for the particles to actually increase the speed, right? Or should I see it as temperature being vibrations in this case? Even though vibrations as a temperature source usually is associated with solid state materials.

The other part of your explanation is clear.

I don't see that problem because to me

$$kT=\frac{mv^2}{2}...[4]$$

is more or less valid at all times.

This is very interesting too. I have in another thread finally understood that there can be no temperature gradients for an extended time if there is no energy source. Also called the law of thermal equilibrium, yes?

The question is however if we may consider an ignited Tokamak plasma to have the same kT all over it because if we can we may write:

$$r_L=\frac{mv}{|q|B}$$

and using [4] gives the Langmor radius (the perpendiculate gyration radius around the lines of force, B) as

$$r_L=\frac{\sqrt{2kTm}}{|q|B}$$

Using the_wolfman's data

1) kT=10keV
2) B=5T

then gives

$$\frac{r_{Li}}{r_{Le}}=43==\frac{n_i}{n_e}=\frac{p_i}{p_e}$$

where it has been assumed that the ion/proton thermal energy is the same as the electron thermal energy.

How wrong is this?

Best regards, Roger
PS
The numerical value actually comes from sqrt(mp/me). But it is interesting to note that RL(mp)=2,9mm. Also, I find this small radius strange. Maybe there is some problem with my speed vs energy equation [4] after all?

7. Oct 12, 2014

### Staff: Mentor

There is no need for room. Increasing the speed at the same density makes them collide more often, giving a higher pressure. So what?

Compare the v you get for electrons and protons - they differ by a factor of 43. So which speed do you want to associate with your temperature?

Right.

The Larmor radius is different for different particles, right.

8. Oct 12, 2014

### the_wolfman

The Saha equation represents a thermodynamic balance between ionization and recombination. The rate at which ionization occurs in linearly proportional to neutral particle density. Recombination occurs when an electron an ion combine to form a neutral atom. The rate at which recombination occurs is proportional to the product of the electron and ion densities. In a quasineutral plasma with one ion species, the the electron density is proportional to the ion density. Therefore recombination is proportional to the ion density squared.

Decreasing the the total density (ion + neutral) decreases the recombination rate faster than it decreases the recombination rate. If you follow this train of logic, then ionization fraction will be higher at lower densities.

Also be careful with your terminology. In this context ignited has a very specific meaning. A ignited plasma is a plasma that is undergoing a self-sustaining fusion reaction.

9. Oct 12, 2014

### rogerk8

Thanks for teaching me. I see now what you mean which is that speed may be increased regardless of density. This is simply due to the fact that they will collide more often while they at same time do not loose speed by such "totally elastic" collisions.

Why do you even need a single speed to be associated with the temperature?

I know from the_wolfman that the poloidal diameter of the ITER Tokamak is some 2m.

Confining the 100MK hot plasma with the available 5T reduces the Langmor radius for the protons to 3mm only (and even less for the electrons).

That is an awful waste of space.

On the other hand, it is quite hot also :)

Best regards, Roger

10. Oct 12, 2014

### rogerk8

Understood but needed to be said.
Sounds reasonable but however not fully understood.
Why product of?
Understood.
Understood.

Should not bold say "ionization" rate?

From the above I get:
1) Ionization goes linearly with density
2) Recombination goes squared with density.

So the ionization fraction will be higher at lower density.

I will respect that!

Thank you very much for this explanation. I think I have read it 20 times before I think I understand it.

At least a rough part of it.

Best regards, Roger

11. Oct 12, 2014

### Staff: Mentor

I don't want to do that, you tried to do that in the first post if I understood that correctly.

Not all particles will make the same circle. It is still a plasma, with particle collisions, some broad density distribution and so on. A larger plasma volume is not necessary for the Larmor radius, but to reduce energy losses (surface to volume ratio).

The rate of "ion collides with electron" is proportional both to the density of ions and to the density of electrons.

Right.

12. Oct 12, 2014

### the_wolfman

Yep, that was a typo.

Break the problem down into 2 parts.
What is the the probability that one electron traveling through a sea of ions will recombine? This probability is proportional to the ion density.
Now if you have multiple electrons, just multiply the above probability by the number of electrons (per unit volume) to get the total recombination rate (per unit volume).

If you want a more detailed explanation go here:
http://chemwiki.ucdavis.edu/Physica...Kinetics/Collision_Theory/Collision_Frequency
The link is for simple chemical reactions. But the basic idea applies to atomic reaction.

The Larmor radius is not the radius of a magnetically confined plasma. It is usually the case that the Larmor radius is many times smaller than the size of the plasma.

13. Oct 13, 2014

### rogerk8

Thanks for explaining it so basically to me. I totally understand now.

It stroke me today that the Larmor radius is not 3mm for the protons all the time. It is

$$R_{Lfpp}=R_L*\sqrt{\frac{10Mev}{10keV}}$$

or some 92mm at ignition, yes?

Because you taught me that the fusion product protons (fpp) have an energy of some 10MeV.

But now you say there can be even larger radius than that?

Is it perhaps due to kinks and/or drifts in the plasma?

Best regards, Roger

14. Oct 14, 2014

### the_wolfman

Keep in mind that a plasma is a collection of many charged particles. If you want to determine the "size" and shape of a plasma you have to use a model deals with a collection of particles. The gyroradius partially describes the motion of a single individual particle. It does not account for the collective dynamics of many particles. I recommend looking into Magnetohydrodynamics if you're interested in calculating the size and shape of a magnetized plasma. First make sure you understand the assumptions that go into MHD. Then look into MHD equilibrium and force balance.

Another good excersise for you might be to calculate a characteristic gyroradius for particles in the sun. Then compare that to the size of the sun. I think the solar magnetic field is around 10 Gauss. I'm not sure what the density and temperature. But thats what google and wikipedia are for.

15. Oct 14, 2014

### rogerk8

For now I just want to say that I totally understand nothing.

But I believe I sometimes understand something.

Best regards, Roger

16. Oct 16, 2014

### rogerk8

According to Wikipedia the temperature of the corona is some 5MK (and the surface temperature some 5kK).

Using this, protons and 1mT (10 Gauss) gives a Larmor radius of some 3m.

Which somehow was supposed to be compared to the diameter of the Sun, i.e some 1 400 000km

I have now read the major part of the nicely provided link.

I have however collision theory to study some more (and of course MHD).

Repeating the collision frequency formula (from the above link) for two particles (A & B):

$$Z_{AB} = N_{A}N_{B}\left(r_{A} + r_{B}\right)^2\sqrt{ \frac{8\pi K_{B}T}{\mu_{AB}}}$$

And simplifying it for our case:

$$Z_{AB} \propto N_{A}N_{B}(r_{A} + r_{B})^2\sqrt{kT}$$

where the parallelling of the masses $$\mu_{AB}$$ has been omitted due to considering only the change of the number of same particles and their temperature.

Before the plama ignites we have only electrons and protons (i.e Deuterium, right?).

And while the electron radius is much less than the proton we actually have

$$Z_{ie} \propto N_{i}N_{e}r_{i}^2\sqrt{kT}$$

as the collision frequency.

Best regards, Roger
PS
I found a funny sentence in the provided link. Down at the bottom under "Density" it actually says "if the density is increased, the number of molecules per volume is also increased" :D

The "system of hydrogen in a jar" would have been better to explain...

17. Oct 17, 2014

### rogerk8

Thinking some more about this it is clear that using the actual number of particles/things always is within a certain volume/space.

So the use of actual numbers of things is obsolete. It is however more easilly uderstood but density is what counts.

Also, it is interesting to take the parallell particle mass

$$\mu_{AB}$$

into account (even though it does not change) while it in our case simplifies to the weight of the electron.

So I now wish to rewrite the above equation like

$$Z_{ie} \propto n_{i}n_{e}r_{i}^2\sqrt{\frac{kT}{m_e}}$$

where I have changed the particle numbers to their densities as well as added the resulting mass of the electron.

It is easy to understand that the "Collision Frequency" is reversely "proportional" to the lightest particles because they can move around more easily.

Best regards, Roger

Last edited: Oct 17, 2014
18. Oct 20, 2014

### rogerk8

Consider a plasma of preliminary only protons and electrons.

Consider also that the collision frequency may be the recombination rate for a not-so-warm plasma.

Consider also that we are only interested in what happens to recombination within the same "soup" (jiggeling with density and temperature only).

We may then write the simplified equation for collision frequency as equal to the rate of recombination:

$$Z_{ie} \propto n_{i}n_{e}r_{i}^2\sqrt{kT}$$

Am I right or wrong?

Best regards, Roger

19. Oct 23, 2014

### rogerk8

$$U_2=\sqrt{U_1^2(1-4\frac{m}{M})-4\frac{m}{M}U_1v_1}<U_1$$

or even (where U is the ion speed and v is the electron speed)

$$U_2=\sqrt{U_1(U_1-4\frac{m}{M}v_1)}<U_1$$

Because m<<M.

This solution is however only true for a 180 degrees "sling shot".

But it is interesting to note that not all of the "incoming" electrons are absorbed by the ions.

Some actually just change trajectory.

And the reduction of ion speed actually decreases their temperature.

At least, this is my thought.

Making the recombination rate calculation above kind of academic.

Anyway, considering a proton/electron plasma and the jiggeling of densities and temperature only, the preliminary recombination rate equation should read:

$$Z_{ie} \propto n_{i}n_{e}\sqrt{kT}$$

And I rest my case :)

Best regards, Roger