1  ^0.40 For a real particle of mass \(m\) in free motion, its kinetic energy \(K\) can be expressed in terms of the magnitude of its momentum \(p\) by \(K=p^{2} / 2 m\). Express its velocity \(\vec{v}\) in terms of its momentum \(\vec{p}\) using the work-energy theorem.

The work-energy theorem gives us \[ \mathrm{d} K=\vec{F} \cdot \mathrm{d} \vec{r}=\frac{\mathrm{d} \vec{p}}{\mathrm{ d} t} \cdot \mathrm{ d} \vec{r}=\vec{v} \cdot \mathrm{d} \vec{p} \tag{1} \] Differentiating the given energy-momentum relation for real particles on both sides with respect to \(\vec{p}\) also gives us \[ \mathrm{d} K=\frac{\vec{p}}{m} \cdot \mathrm{ d} \vec{p} \tag{2} \] in which we note that \(\mathrm{d}\left(p^{2}\right)=\mathrm{d}(\vec{p} \cdot \vec{p})=\vec{p} \cdot \mathrm{d} \vec{p}+\mathrm{d} \vec{p} \cdot \vec{p}=2(\vec{p} \cdot \mathrm{~d} \vec{p})\). Setting the RHS's of (1) and (2) equal, we have \[ \left(\frac{\vec{p}}{m}-\vec{v}\right) \cdot \mathrm{d} \vec{p}=0 \] and combined with the fact that the above is true for all \(\mathrm{d} \vec{p}\), we confirm that \[ \vec{v}=\frac{\vec{p}}{m}\tag{3} \] and that we haven't gone insane (yet).

2  ^0.50 Using a similar method, express the velocity \(\vec{v}\) of a quasiparticle in terms of its momentum \(\vec{p}\).

Differentiating the given energy-momentum relation for quasiparticles on both sides with respect to \(\vec{p}\) gives us \[ \mathrm{d} K=\frac{\vec{p}}{m} \cdot \mathrm{ d} \vec{p}+\alpha \hat{\vec{p}} \cdot \mathrm{d} \vec{p}\tag{4} \] Again setting the RHS's of (1) and (4) equal, we have \[ \left(\frac{\vec{p}}{m}+\alpha \hat{\vec{p}}-\vec{v}\right) \cdot \mathrm{d} \vec{p}=0 \] and by the same argument \[ \vec{v}=\frac{\vec{p}}{m}+\alpha \hat{\vec{p}}\tag{5} \]

3  ^0.40 Express \(v=|\vec{v}|\) in terms of \(K\).

Squaring (5) on both sides, we obtain \[ v^{2}=\frac{p^{2}}{m^{2}}+2 \alpha \frac{p}{m}+\alpha^{2}=\left(\frac{p}{m}+\alpha\right)^{2} \] so \[ v=\frac{p}{m}+\alpha\tag{6} \] Making \(p\) the subject of the energy-momentum relation for quasiparticles, we have \[ p=-m \alpha+\sqrt{m^{2} \alpha^{2}+2 m K}\tag{7} \] so substituting (7) into (6), we obtain \[ v=\sqrt{\alpha^{2}+\frac{2 K}{m}}\tag{8} \]

4  ^1.10 We now place the two-dimensional interface within a uniform magnetic field of magnitude \(B\) and pointing in the \(+z\) direction. For a quasiparticle of kinetic energy \(K\), which will undergo uniform circular motion, find the radius of its trajectory, the period of its motion, and the magnitude of its angular momentum.

In a uniform magnetic field, the quasiparticle undergoes uniform circular motion with speed \(v=\omega R\) and rate of change of momentum \(|\mathrm{d} \vec{p} / \mathrm{d} t|=\omega p\), so \[ \left|\frac{\mathrm{d} \vec{p}}{\mathrm{ d} t}\right|=\frac{v}{R} p\tag{9} \] The quasiparticle experiences a Lorentz force related to the above by \[ p \frac{v}{R}=q v B\tag{10} \] so substituting in (7), we obtain the radius of the quasiparticle's trajectory \[ R=\frac{p}{q B}=\frac{-m \alpha+\sqrt{m^{2} \alpha^{2}+2 m K}}{q B}\tag{11} \]

The period of its motion \[ T=\frac{2 \pi R}{v}=\frac{2 \pi\left(-m \alpha+\sqrt{m^{2} \alpha^{2}+2 m K}\right)}{q B \sqrt{\alpha^{2}+\frac{2 K}{m}}}\tag{12} \]

Since the magnitude of its angular momentum is given by \[ L=p R\tag{13} \] we may substitute (7) and (11) into (13) to obtain \[ L=\frac{2 m^{2} \alpha^{2}+2 m K-2 m \alpha \sqrt{m^{2} \alpha^{2}+2 m K}}{q B}\tag{14} \]

5  ^1.10 We replace the magnetic field with a uniform electric field of magnitude \(E\) and pointing in the \(+x\) direction. Note that the component of the quasiparticle's acceleration perpendicular to the electric field may be nonzero. Find the components of the quasiparticle's acceleration \(a_{x}\) and \(a_{y}\) when it moves with speed \(v\) and its velocity makes an angle \(\theta\) with the electric field.

(5) gives us \[ \vec{p}=m \vec{v}-m \alpha \hat{\vec{v}} \] keeping in mind that \(\hat{\vec{p}}=\vec{p} / p\) and similar. Newton's second law yields \[ \frac{\mathrm{d} \vec{p}}{\mathrm{ d} t}=q \vec{E} \] Alternative 1. The acceleration of the quasiparticle is given by \[ \frac{\mathrm{d} v}{\mathrm{ d} t}=\alpha \frac{\mathrm{d}}{\mathrm{ d} t}\left(\frac{\vec{v}}{v}\right)+\frac{q}{m} \vec{E} \] or, written out by components, \[ \begin{aligned} & a_{x}=\frac{\alpha}{v} \frac{\mathrm{ d} v_{x}}{\mathrm{ d} t}-\frac{\alpha v_{x}}{v^{2}} \frac{\mathrm{ d} v}{\mathrm{ d} t}+\frac{q E}{m} \\ & a_{y}=\frac{\alpha}{v} \frac{\mathrm{ d} v_{y}}{\mathrm{ d} t}-\frac{\alpha v_{y}}{v^{2}} \frac{\mathrm{ d} v}{\mathrm{ d} t} \end{aligned} \] Using the kinematic relation \[ \frac{\mathrm{d} v}{\mathrm{ d} t}=\frac{v_{x}}{v} a_{x}+\frac{v_{y}}{v} a_{y} \] we may rewrite (17) as \[ \left\{\begin{array}{l} \left(1-\frac{\alpha}{v}+\frac{\alpha v_{x}^{2}}{v^{3}}\right) a_{x}+\frac{\alpha v_{x} v_{y}}{v^{3}} a_{y}=\frac{q E}{m} \\ \left(1-\frac{\alpha}{v}+\frac{\alpha v_{y}^{2}}{v^{3}}\right) a_{y}+\frac{\alpha v_{x} v_{y}}{v^{3}} a_{x}=0 \end{array}\right. \] which, in turn, may be rewritten without \(v_{x}\) and \(v_{y}\) as \[ \left\{\begin{array}{l} \left(1-\frac{\alpha \sin ^{2} \theta}{v}\right) a_{x}+\frac{\alpha \sin \theta \cos \theta}{v} a_{y}=\frac{q E}{m} \\ \left(1-\frac{\alpha \cos ^{2} \theta}{v}\right) a_{y}+\frac{\alpha \sin \theta \cos \theta}{v} a_{x}=0 \end{array}\right. \] This system of equations may be solved to obtain our desired result, \[ \left\{\begin{array}{l} a_{x}=\frac{q E}{m}+\frac{q E}{m} \frac{\alpha \sin ^{2} \theta}{v-\alpha} \\ a_{y}=-\frac{q E}{m} \frac{\alpha \cos \theta \sin \theta}{v-\alpha} \end{array}\right. \] Alternative 2. In Cartesian coordinates, \[ \left\{\begin{array}{l} \frac{\mathrm{d} p}{\mathrm{ d} t} \cos \theta-p \sin \theta \frac{\mathrm{ d} \theta}{\mathrm{ d} t}=q E \\ \frac{\mathrm{ d} p}{\mathrm{ d} t} \sin \theta+p \cos \theta \frac{\mathrm{ d} \theta}{\mathrm{ d} t}=0 \end{array}\right. \] from which we solve for the time derivatives and obtain \[ \left\{\begin{array}{l} \frac{\mathrm{d} p}{\mathrm{ d} t}=q E \cos \theta \\ \frac{\mathrm{ d} \theta}{\mathrm{ d} t}=-\frac{q E}{p} \sin \theta \end{array}\right. \] Splitting (5) into components and differentiating with respect to time yields \[ \left\{\begin{array}{l} a_{x}=\frac{\mathrm{d} v_{x}}{\mathrm{ d} t}=\frac{\mathrm{d}}{\mathrm{ d} t}\left(\left(\frac{p}{m}+\alpha\right) \cos \theta\right)=\frac{1}{m} \cos \theta \frac{\mathrm{ d} p}{\mathrm{ d} t}-\left(\frac{p}{m}+\alpha\right) \sin \theta \frac{\mathrm{d} \theta}{\mathrm{ d} t} \\ a_{y}=\frac{\mathrm{d} v_{y}}{\mathrm{ d} t}=\frac{\mathrm{d}}{\mathrm{ d} t}\left(\left(\frac{p}{m}+\alpha\right) \sin \theta\right)=\frac{1}{m} \sin \theta \frac{\mathrm{ d} p}{\mathrm{ d} t}+\left(\frac{p}{m}+\alpha\right) \cos \theta \frac{\mathrm{d} \theta}{\mathrm{ d} t} \end{array}\right. \] Substituting (18) into (19) and using (5) yields our final result, \[ \left\{\begin{array}{l} a_{x}=\frac{q E}{m}+\frac{q E}{m} \frac{\alpha \sin ^{2} \theta}{v-\alpha} \\ a_{y}=-\frac{q E}{m} \frac{\alpha \cos \theta \sin \theta}{v-\alpha} \end{array}\right. \] Alternative 3. (15) can be decomposed into its components and written as \[ \left\{\begin{array}{l} p_{x}=m(v-\alpha) \cos \theta \\ p_{y}=m(v-\alpha) \sin \theta \end{array}\right. \] From (16) and (17) we obtain \[ \left\{\begin{array}{c} -m(v-\alpha) \sin \theta \frac{\mathrm{d} \theta}{\mathrm{ d} t}+m \cos \theta \frac{\mathrm{ d} v}{\mathrm{ d} t}=q E \\ m(v-\alpha) \cos \theta \frac{\mathrm{d} \theta}{\mathrm{ d} t}+m \sin \theta \frac{\mathrm{ d} v}{\mathrm{ d} t}=0 \end{array}\right. \] from which we solve for the time derivatives and obtain \[ \left\{\begin{array}{l} \frac{\mathrm{d} v}{\mathrm{ d} t}=\frac{q E}{m} \cos \theta \\ \frac{\mathrm{ d} \theta}{\mathrm{ d} t}=-\frac{q E}{m} \frac{\sin \theta}{v-\alpha} \end{array}\right. \] Substituting (19) into the system \[ \left\{\begin{array}{l} a_{x}=\frac{\mathrm{d} v_{x}}{\mathrm{ d} t}=\frac{\mathrm{d} v}{\mathrm{ d} t} \cos \theta-\sin \theta \frac{\mathrm{d} \theta}{\mathrm{ d} t} \\ a_{y}=\frac{\mathrm{d} v_{y}}{\mathrm{ d} t}=\frac{\mathrm{d} v}{\mathrm{ d} t} \sin \theta+\cos \theta \frac{\mathrm{d} \theta}{\mathrm{ d} t} \end{array}\right. \] we obtain our desired result, \[ \left\{\begin{array}{l} a_{x}=\frac{q E}{m}+\frac{q E}{m} \frac{\alpha \sin ^{2} \theta}{v-\alpha} \\ a_{y}=-\frac{q E}{m} \frac{\alpha \cos \theta \sin \theta}{v-\alpha} \end{array}\right. \]