Effect of magnetic field on pair production

[aurao2003]

Homework Statement

Hi
Its just me again! I kindly need help on this question. Its as follows:

The tracks of a positron and an electron created by pair production in a magnetic field curve in opposite directions.
a) Why do they curve in opposite directions
b)Both particles spiral inwards. What can you deduce from this observation about their kinetic energy.

Please advise.

Regards
Ben

Homework Equations

Momentum= mass x velocity
Kinetic energy= 0.5 x mass x (velocity)^2

The Attempt at a Solution

a)I surmised the following. The electron and positron are obviously particles/antiparticle. Therefore, they have opposite charges. A magnetic field can act as a south and north pole. Based on this, charges can be alternately attracted and repelled. This is as far as I got. However, I don't know whether I am on the right line of reasoning. I am wondering whether I should consider the effect of charges in a magnetic field.

b)Momentum will be conserved during the porcess of particle production. Since momentum is related mass and velocity, they will have the same but opposite amount of momentum. Kinetic energy is related to velocity. If they both spiral inwards, they seem to have the same amount of kinetic energy. Therefore, kinetic energy seems to the same(LOOKING LIKE AN ELASTIC COLLISION).

 

[collinsmark]

a)I surmised the following. The electron and positron are obviously particles/antiparticle. Therefore, they have opposite charges. A magnetic field can act as a south and north pole. Based on this, charges can be alternately attracted and repelled. This is as far as I got. However, I don't know whether I am on the right line of reasoning. I am wondering whether I should consider the effect of charges in a magnetic field.

I'm not sure about your wording, but you're right about the opposite charges. That's the important part here.

For a charged particle moving in a magnetic field, the magnetic force on that particle is:

$$\vec F = q \vec v \times \vec B$$

q can be positive or negative.

b)Momentum will be conserved during the porcess of particle production. Since momentum is related mass and velocity, they will have the same but opposite amount of momentum. Kinetic energy is related to velocity. If they both spiral inwards, they seem to have the same amount of kinetic energy. Therefore, kinetic energy seems to the same(LOOKING LIKE AN ELASTIC COLLISION).

You might want to redo b). I think the question is asking you about how the kinetic energy of each particle changes with time. The question states that the particles spiral inward. What does this "spiraling inward" say about the how particles' kinetic energies change with time, as opposed to some other situation where the particles travel in perfect circles or spiral outward?

(Hint: in order to produce tracks in a bubble chamber or cloud chamber, the particles have to interact with surroundings in some way or another. In other words, there may be some friction like forces involved. )

 

[aurao2003]

Thanks for your speedy reply. Regarding question b, would it be safe to say that the'spiralling inward' would be due to collision with particles of the same charge? If the kinetic energy is maintained the particles might travel in a straight line or a circle. The' spiralling inwards' almost suggests the particles undergoing a displacement to their position of origin. So, there seems to be a loss of KINETIC ENERGY.

I am looking at a situation where the particles are being produced with different Kinetic energies. Sorry if I am rambling!

 

[collinsmark]

Thanks for your speedy reply. Regarding question b, would it be safe to say that the'spiralling inward' would be due to collision with particles of the same charge? If the kinetic energy is maintained the particles might travel in a straight line or a circle. The' spiralling inwards' almost suggests the particles undergoing a displacement to their position of origin. So, there seems to be a loss of KINETIC ENERGY.

Essentially, yes. Well, almost.

On the submicroscopic level, there are collisions with mostly uncharged particles -- namely the neutral atoms of gas that eventually end up forming the tracks. Don't overthink this though. On a macroscopic level, you can think of this as being kind of like friction.

Keep this in mind. If a charged particle moves though a constant magnetic field totally unimpeded, without any other forces involved, it will move in a circle. The magnetic force is perpendicular to the particle's velocity. Thus in this situation, the particle will neither speed up or slow down.

Keeping that in mind, getting back to the original problem, what does the spiraling inward imply about the particle's kinetic energy as time moves forward? I think that is what the problem is asking.

I am looking at a situation where the particles are being produced with different Kinetic energies. Sorry if I am rambling!

If the two particles always have the same kinetic energy, and since they have the same mass, their respective spirals would be the same size. That's fine and good, but I don't think the question is asking for you to comment on that. Rather I think the question is asking you about what spiraling inwards implies, as opposed to traveling in a perfect circle or spiraling outward.

 

[aurao2003]

Essentially, yes. Well, almost.

On the submicroscopic level, there are collisions with mostly uncharged particles -- namely the neutral atoms of gas that eventually end up forming the tracks. Don't overthink this though. On a macroscopic level, you can think of this as being kind of like friction.

Keep this in mind. If a charged particle moves though a constant magnetic field totally unimpeded, without any other forces involved, it will move in a circle. The magnetic force is perpendicular to the particle's velocity. Thus in this situation, the particle will neither speed up or slow down.

Keeping that in mind, getting back to the original problem, what does the spiraling inward imply about the particle's kinetic energy as time moves forward? I think that is what the problem is asking.

If the two particles always have the same kinetic energy, and since they have the same mass, their respective spirals would be the same size. That's fine and good, but I don't think the question is asking for you to comment on that. Rather I think the question is asking you about what spiraling inwards implies, as opposed to traveling in a perfect circle or spiraling outward.

So, would the spiraling occur due to opposition from other particles? Based on your analysis, there would more friction opposing its motion thereby causing continuous spiralling inwards.

 

[collinsmark]

So, would the spiraling occur due to opposition from other particles? Based on your analysis, there would more friction opposing its motion thereby causing continuous spiralling inwards.

Essentially, yes. Below is a method you can use for a more detailed analysis.

Again, consider an electron moving at some speed in a uniform magnetic field, without any other particles or forces around. The electron would travel in a perfect circle. You don't need a positron or any other particles around for this. It's simply the interaction between the moving charged particle and the magnetic field.

You can calculate the radius of of this circle by noting that the magnetic force equals the centripetal force. So, considering the situation that the particle's motion is also perpendicular to the magnetic field (i.e the magnetic field, the particle's motion, and the resulting force are all perpendicular to each other), we have

$$F = q v B = m\frac{v^2}{r}$$

Solve for v in terms of q, m, r, and B. Then treat q, m and B as constants (which is okay since the particle's charge & mass and the magnetic field are not changing significantly).

So, getting back to our original problem where there are lots of other atoms and frictional forces around, if r becomes smaller as time goes on, what does that tell you about v as time moves forward?

Kinetic energy is

$$KE = \frac{1}{2}mv^2$$

How does a change in v relate to KE?

(The above applies to particles moving significantly slower than the speed of light, such that v << c. However, even when v approaches c the subjective conclusions are still the same regarding the "spiraling inward," even the though the detailed analysis is more complicated.)

[Edit: and by the way, none of the discussion on this post even requires a positron.]

 

[aurao2003]

Hi
Sorry for my horrible equationss but I am new to this forum and not conversant with Latex. I promise to get better! Now! Back to business.

I obtained a directly proportional relationship between
V=(qb/m)r

Therefore v will get smaller as r decreases.

A decrease in velocity results in decreased KE (Since mass stays constant).This seems to suggest an inelastic collision. Will the particle come to rest at some point?

 

[collinsmark]

I obtained a directly proportional relationship between
V=(qb/m)r

Therefore v will get smaller as r decreases.

A decrease in velocity results in decreased KE

That's it! If, in the presence of a uniform magnetic field, the particle's tracks spiral inward (as opposed to some other type of circle or spiral), it implies the particle is losing kinetic energy.

(Since mass stays constant).This seems to suggest an inelastic collision.

Well, be careful about reading too much into it. It's legitimate to say that the particle's kinetic energy is decreasing. Beyond that it gets quite tricky.

But since we're on the topic, I'll continue. On the sub-microscopic scale, there are a whole bunch of collisions going on. Don't think of it as just one collision -- there are many things going on including individual elastic collisions; and the possible excitation -- maybe even ionization -- of other atoms in the surrounding gas. However, the mechanisms are not important for this particular problem. What is important is that the particle's kinetic energy (remember it has a lot of energy because it is moving very fast) is transferred to the surroundings in some way or another, in a thermodynamically irreversible process. The atoms in the cloud increase in energy either by gaining atomic momentum, or by excitation, and even by ionization. This increase in energy comes from somewhere, and it comes from the original particle's kinetic energy. The surrounding gas increases in energy, and the particle's kinetic energy is reduced.

So some of the atoms in the gas were ionized. These ions form condensation seeds causing a mist to form. These trails of mist are the observed tracks.

In short, it takes energy to create the tracks. That energy has to come from somewhere, and it comes from the kinetic energy of the particle.

Will the particle come to rest at some point?

Well, let's not get too crazy :tongue2: -- The cloud isn't at absolute zero and there is still all the quantum weirdness going on. But suffice it to say that the particle will eventually slow down enough such that it no longer creates a track.

 

[aurao2003]

Okay sir! I appreciate your help and insight.

Cheers.

 

[collinsmark]

One last thought on the subject. This information isn't necessary to solve this particular problem. But I think it's useful to know and consider as you continue your quantum physics course. I made some useful approximations in my previous explanations, but I don't want to mislead you.

On the sub-microscopic level, there is really no such thing as friction. And when all factors are considered, there are not such things as an "inelastic collisions" either (at least not in the classical sense). All individual interactions between the particle and the atoms (and individual interactions between photons and other particles) are completely reversible. All individual collision are essentially elastic, and everything can essentially happen backwards as easily as it did forwards.

It's only when you consider groups of particles (including atoms) that thermodynamics come into play. If you consider a vast number of non-radioactive atoms and only one free electron, there are a vast number of possible states where kinetic energy of the free electron is roughly closer to the kinetic energy of a typical atom in the mix. For comparison, there is a tiny, tiny number of states where the one free electron has huge amount of energy compared to the average kinetic energy of the surrounding atoms (such that the total energy of all atoms and particles is the same in both cases).

Given the above, it is far more probable that the fast moving electron will transfer its kinetic energy to the surrounding atoms rather than the surrounding atoms transferring their energy to the particle making it move even faster. On the sub-microscopic level, this is all a matter of probability.

Sure, it is within the realm of quantum mechanics for the tracks to decay in just such a way that the energy is transferred back to the original particle making it move faster, perhaps back to the speed it was originally. But that is so inconceivably, astronomically improbably that its not even considered for all practical purposes. It's far, far more likely that when the tracks decay, the energy will be more evenly distributed to other atoms/particles in the cloud chamber. This is what is meant by "thermodynamically irreversible process" on the sub-microscopic scale. It's not that the total reversal is impossible, but rather outlandishly improbable.

[Edit: It is expected that the energy of the fast moving particle will fluctuate such that sometimes its energy actually increases on some collisions (sometimes). But the probability of this happening over many, many collisions such that it retains its original, huge kinetic energy, or gains it all back after losing it, is incredibly unlikely.]

Moving up to the macroscopic scale, this same property manifests itself in what we approximate as friction, inelastic collisions, and non-conservative forces. That's what I meant when I said "friction like" forces. I was making a macroscopic approximation of the microscopic world.