Beam of particles in momentum space

Habeebe
Messages
37
Reaction score
1
I'm mostly concerned with whether or not I understand this problem intuitively in order to answer the final part of this problem.

Homework Statement


Discuss the implications of Liouville's theorem on the focusing of beams of charged particles by considering the following case. An electron beam of circular cross section (radius R0) is directed along the z-axis. The density of electrons across the beam is constant, but the momentum components transverse to the beam are disctributed uniformly over a circle of radius p0 in momentum space. If some focusing system reduces the beam radius from R0 to R1 find the resulting distribution of the transverse momentum components. What is the physical meaning of this result? (Consider the angular divergence of the beam.)


Homework Equations


Aellipse=πr1r2

The Attempt at a Solution



I answered that the circle in momentum space would become an ellipse of equal area, thereby satisfying the equation R_0^2=R_1R_p where R_p is the radius of the ellipse along the momentum axis. The next part is what I'm feeling sketchy on: the physical meaning. It seems like the focusing causes an increased tendency of the beam to want to converge/diverge, that is, the divergence of the beam is increased proportional to the change of radius from R0 to R1.

Does this sound right?
 
Physics news on Phys.org
Yes, that is correct. The focusing of the beam increases its angular divergence, which means that the particles will have a greater tendency to spread out over a wider area in space. This is because the transverse momentum components are distributed uniformly over an ellipse rather than a circle.
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top