Intro to elementary particles, electron - nucleus collision

[Coffee_]

1.Problem: An electron with energy $E$ which is much higher than its restmass collides with a much much heavier particle "A" of mass $m$ which is at rest. Find the maximal transfer of four-momentum. (Elastic collision)2. Conservation of four momentum3. Everything in natural units. So I go look at the particle A in its restframe and go look at this particle before and after the collision. The square of the four-momentum transfer is then:

$q^{2}=(p_{A1}-p_{A2})^{2}$

Working this out gives me:

$q^{2}=2m_{A}^{2}-2E_{A2}m_{A}$

When $E_{A2}$ is minimal, namely just $m_{A}$, $q^{2}=0$. This means that for any real collision the square of the momentum transfer will be negative. I assume the question is asking when this value is the most negative? In that case, it's obvious, when $E_{A2}$ is maximal.

Now I tried to find some expression for a maximal $E_{A2}$ but not really succeeded yet.

 

[Orodruin]

The much much in front of the mass of the heavy particle seems to suggest you can approximate its mass as infinite. In this approximation, just as in classical mechanics, you can always fix the 3-momentum conservation by giving the heavy particle the correct momentum. What would this tell you about the post-collision 4-momentum of the light particle?

 

[Coffee_]

The much much in front of the mass of the heavy particle seems to suggest you can approximate its mass as infinite. In this approximation, just as in classical mechanics, you can always fix the 3-momentum conservation by giving the heavy particle the correct momentum. What would this tell you about the post-collision 4-momentum of the light particle?

Hmm, this is how I worded the question from memory when seeing it in class. It might be just the way I wrote it is suggestive. I remember that this particle A was something of the order of a heavier nucleus mass but couldn't remember for sure so I wrote much much heavier. It is also given that the energy of the electron is much higher than the electron mass. It is possible that that electron energy is in the order of the rest mass of particle A? Anyway thanks for the answer, any way to do this without approximating mass A as infinite?

 

[Orodruin]

Of course, you can always compute the momentum transfer as a function of the electron scattering angle and maximize it.

 

[Orodruin]

Also, the fact that the incoming particle mass is very low suggests you could also neglect it in comparison with the energy and target mass. You should end up with something similar to Compton scattering.

 

[Coffee_]

Also, the fact that the incoming particle mass is very low suggests you could also neglect it in comparison with the energy and target mass. You should end up with something similar to Compton scattering.

Was I correct to interpret ''Find the maximal four momentum transfer'' the way I described it? Basically $q^{2}$ being as negative as possible?

 

[Coffee_]

Also, the fact that the incoming particle mass is very low suggests you could also neglect it in comparison with the energy and target mass. You should end up with something similar to Compton scattering.

It's not working out, just tried for like 40 minutes with any combinations of the mandelstam variables. The closes I came was relating the energy of the electron after the collision and the angle of the electron after the collision in a quite ugly algebraic expression. Also from conservation of energy I know that for my case this energy of the electron after the collision should be minimal. I'd be inclined to play around with this expression to see if I can find this angle for the minimal value of the energy of the electron but I'm a bit intimidated since it contains energy under roots and outside roots and so on. Is there a faster way?

 

[Orodruin]

Neglect the electron mass.

 

[Coffee_]

Neglect the electron mass.

I did and I seem to find that no matter what it has to be a frontal collision so that the electron bounces back from where it came. Weird result.

It's this expression that I get:

$2E_{2}(E_{1}(1-cos(\theta))+2M)=2ME_{1}+M^{2}$

Subscript 1 points to the electron before the collision, 2 to the electron after the collision. M is the mass of the nucleus and theta is the angle between the electron after the collision and the x-axis.

My work: http://i.imgur.com/xhtEBoe.jpg

Two questions:1) The only thing I worry about is the plugging in of cos(theta)=-1. While this equation seem to allow it, maybe there are other restrictions on theta I haven't thought about?

2) Was my interpretation of $q^{2}$ being as negative as possible in the original question correct?

 

[Orodruin]

I did and I seem to find that no matter what it has to be a frontal collision so that the electron bounces back from where it came. Weird result.

It's this expression that I get:

$2E_{2}(E_{1}(1-cos(\theta))+2M)=2ME_{1}+M^{2}$

Subscript 1 points to the electron before the collision, 2 to the electron after the collision. M is the mass of the nucleus and theta is the angle between the electron after the collision and the x-axis.

My work: http://i.imgur.com/xhtEBoe.jpg

Two questions:1) The only thing I worry about is the plugging in of cos(theta)=-1. While this equation seem to allow it, maybe there are other restrictions on theta I haven't thought about?

2) Was my interpretation of $q^{2}$ being as negative as possible in the original question correct?

2) it is how I would interpret it.

1) the easiest way is to express E2 as a function of the angle and then compute q^2 from there. Doing this is exactly analogous to finding the Compton formula.

 

[Coffee_]

Thanks for answering, is my conclusion in 1) about if it would be frontal collision where the electron bounces back then the E2 would be minimal, correct or can't you tell from the equation alone?

 

[Orodruin]

If you solved for E2 and inserted it into the equation for momentum transfer, you would be able to see it directly.