Classical Mechanics HW3 - Taishan Seminar

Derivations

6. Because we have, $$
\varTheta =\pi -2\int_0^1{\frac{u_ms\,\,dy}{\sqrt{1-\frac{V(u)}{E}-s^2u^2}}},
$$ using $l=s\sqrt{2mE}$, we can simplify this as, $$
\varTheta =\pi -2\int_0^1{\frac{dy}{\sqrt{\frac{2m}{u_m^2l^2}\left( E-V(u) \right) -y^2}}}.
$$ Notice that $
\frac{2m}{u_ml^2}=\frac{1}{E-V(u_m)},
$ then, $$
\varTheta =\pi -2\int_0^1{\frac{dy}{\sqrt{\frac{E-V\left( u \right)}{E-V\left( u_m \right)}-y^2}}}=\pi -2\int_0^1{\frac{dy}{\sqrt{1-y^2+\frac{V\left( u_m \right) -V\left( u \right)}{E-V\left( u_m \right)}}}}
$$ \begin{align*}
\Longrightarrow \varTheta \,\approx\, &\pi -2\int_0^1{\left( \frac{1}{\sqrt{1-y^2}}-\frac{V\left( u_m \right) -V\left( u \right)}{2\left[ E-V\left( u_m \right) \right] \left( 1-y^2 \right) ^{3/2}} \right) dy}\\=&\frac{1}{E}\int_0^1{\frac{V\left( u_m \right) -V\left( u \right)}{\left( 1-y^2 \right) ^{3/2}}dy}.\end{align*} In this approximation, $E\approx\frac{l^2u_m^2}{2m}=Es^2u_m^2$, therefore $s=\frac{1}{u_m}.$ When $V(u)=Cu^n$, we have $$
\varTheta =\frac{C}{E}{u_m}^n\int_0^1{\frac{1-y^n}{\left( 1-y^2 \right) ^{3/2}}dy}\propto s^{-n}.
$$ Therefore, $$s\propto\varTheta^{-1/n}.$$ Then we have the differential cross section, $$\sigma=\frac{s}{\sin{\varTheta}}\left | \frac{ds}{d\varTheta} \right |\approx\frac{s}{\varTheta}\left | \frac{ds}{d\varTheta} \right |\propto\varTheta^{-2(1/n+1)}.$$

Exercises

12. For each atom, it experiences a force, so that, $$\vec{F_{ij}}\cdot\vec{r_{ij}}=-\nabla U_{ij}\cdot \vec{r_{ij}}=\frac{mk}{r_{ij}^m},$$ in which $U_{ij}=\frac{k}{r_{ij}^m}.$ Therefore we can conclude that, $$<\sum_i\vec{F_{i}}\cdot\vec{r_{i}}>=\frac{N}{2}\sum_j\vec{F_{ij}}\cdot\vec{r_{ij}}=\frac{N}{2}\int_0^R \frac{mk}{r^m}\rho(r)4\pi r^2 dr=\int_0^R\frac{2\pi mkN^2}{V}\frac{e^{-1/Tr^m}}{r^{m-2}}dr,$$ in which $R=\sqrt[3]{3V/4\pi}.$ Thus the addition to the virial of Clausius is, $$\Delta<T>=-\frac{1}{2}<\sum_i\vec{F_{i}}\cdot\vec{r_{i}}>=-\frac{\pi mkN^2}{V}\int_0^R\frac{e^{-1/Tr^m}}{r^{m-2}}dr.$$ Therefore for the perfect gas, it satisfy, $$NkT=PV-\frac{2\pi mkN^2}{3V}\int_0^R\frac{e^{-1/Tr^m}}{r^{m-2}}dr.$$ Then we can derive the Boyle’s law, $$P_1V_1-\frac{2\pi mkN^2}{3V_1}\int_0^{\sqrt[3]{\frac{3V_1}{4\pi}}}\frac{e^{-1/Tr^m}}{r^{m-2}}dr=P_2V_2-\frac{2\pi mkN^2}{3V_2}\int_0^{\sqrt[3]{\frac{3V_2}{4\pi}}}\frac{e^{-1/Tr^m}}{r^{m-2}}dr$$

20.

(a) The force is $$
F=\frac{GMm}{r^2}+mCr.
$$ Therefore the angular velocity is $$
\omega =\sqrt{\frac{GM}{r_0^3}+C}.
$$ The period is $$
T=\frac{2\pi}{\omega}=\frac{2\pi}{\sqrt{\frac{GM}{r_0^3}+C}}.
$$

(b) In the polar coordinate, we have $$
\ddot{r}-\frac{L^2}{m^2 r^3}=-\frac{GM}{r^2}-Cr,
$$ in which $\dot{\theta}=\omega.$ A slight disturbulance means $r=r_0+\delta,\,\delta\ll r_0.$ $$
\Longrightarrow \ddot{\delta}+\frac{3L^2}{m^2{r_0}^4}\delta =\frac{2GM}{{r_0}^3}\delta -C\delta.
$$ Put $L=m\omega r^2$, we have,$$
\ddot{\delta}=-\left( \frac{GM}{{r_0}^3} +4C \right) \delta,
$$ Therefore, $$\omega_o=\sqrt{\frac{GM}{{r_0}^3}+4C}.
$$

(c) The precessing frequency is $$\omega_p=\omega_o-\omega=\sqrt{\frac{GM}{{r_0}^3}+4C}-\sqrt{\frac{GM}{{r_0}^3}+C},$$ it is opposite the direction of angular velocity.

29. We have $$
\vec{L}\times \vec{A}=L^2\vec{p}-mk\vec{L}\times \hat{r}.
$$ Therefore, $$
\vec{L}\times \vec{A}=L^2\left( p_x\hat{x}+p_y\hat{y} \right) -mkL\left( -\sin \theta \hat{x}+\cos \theta \hat{y} \right) =LA\hat{y}.
$$ Thus, \begin{align*}\begin{cases}L^2p_y-mkL\cos \theta =LA,\\L^2p_x+mkL\sin \theta =0.
\end{cases}\end{align*} $$
\Longrightarrow \left( p_y-\frac{A}{L} \right) ^2+{p_x}^2=\frac{m^2k^2}{L^2}.
$$ The hodograph is a circle of radius $mk/L$ with origin on the y axis displaced a distance $A/L$ from the center of force.

36.

(a) The effective potential is $$
V(x,y)=-\frac{GM_1}{\sqrt{x^2+y^2}}-\frac{GM_1}{\sqrt{\left( x-L \right) ^2+y^2}}-\frac{1}{2}\omega ^2\left( x^2+y^2 \right)
$$
Taking some parameter, we can sketch the function $V(x,0).$