Theoretical Physics II – Exam Essentials

Inclined Plane (Lagrange 1st Kind) - Constraint (plane moves right with $s(t)=\tfrac{1}{2}at^2$):

$\phi(x,y,t)=y-\tan\alpha\,[x-s(t)]=0$

• Cartesian Lagrangian with multiplier $\lambda$:

  $L=\tfrac{1}{2}m(\dot x^2+\dot y^2)-mg y+\lambda\,\phi$

  → Equations

  $m\ddot x = -\lambda\tan\alpha,\quad m\ddot y = -mg+\lambda$

  $\phi=0$ - Eliminate $\lambda$ → single equation along the slope (coordinate $q$):

  $\ddot q = -\sin\alpha\,g- a\cos\alpha$

  (down-slope positive). - Solution (with constants fixed by initial data):

  $q(t)=q<em>0+\dot q</em>0 t-\tfrac{1}{2}(g\sin\alpha + a\cos\alpha)t^2$

• Constraint force magnitude:

  $\Vert\mathbf F_c\Vert=\Vert\lambda\Vert=m\,(g\cos\alpha-a\sin\alpha)$

Atwood Machine (Lagrange 2nd Kind) - Masses $m<em>1,m</em>2$; massless pulley, rope length $L=l+\pi R$.

• One degree of freedom: choose downward displacement of $m<em>1$ ⇒ $q=l</em>1$ (then $l_2=L-\pi R-q$).

• Kinetic & potential

  $T=\tfrac{1}{2}(m<em>1+m</em>2)\dot q^2 ,\qquad V = m<em>1 g q - m</em>2 g (L-\pi R-q)$

  $L=T-V$

• Lagrange eq.:

  $\frac{d}{dt}\big[(m<em>1+m</em>2)\dot q\big]=-(m<em>1-m</em>2)g$

  ⇒ $\ddot q=\frac{m<em>2-m</em>1}{m<em>1+m</em>2}g$

• General solution:

  $q(t)=q<em>0+\dot q</em>0 t+\tfrac{1}{2}\ddot q\,t^2$

Free Fall & Action Integral - Trial paths $z<em>1=at,\;z</em>2=bt^2,\;z_3=ct^3$ with

$z(0)=0,\;z(T)=-\tfrac{g}{2}T^2$ give

• Action (one-dim.

  $L=\tfrac{1}{2}m\dot z^2-mg z$):

  $S<em>1=\tfrac{3}{8}m g^2 T^3,\;\; S</em>2=\tfrac{1}{3}m g^2 T^3,\;\; S_3=\tfrac{5}{16}m g^2 T^3$

• Minimum action ⇒ parabolic path $z<em>2(bt^2)$ is physical; $z</em>1,z_3$ ruled out by Hamilton’s principle.

Rotating Water Surface (Variational Method) - Potentials per mass element $dm$ (in co-rotating frame):

$dU=\big( -\tfrac{1}{2}\omega^2\rho^2 + g z \big) dm$

• Total potential functional (axial symmetry):

  $U[f]=\int<em>0^{2\pi}!d\varphi\int</em>0^R!d\rho\,\rho\int_0^{f(\rho)}!dz\,\mu\Big(-\tfrac{1}{2}\omega^2\rho^2+g z\Big)$

• Volume functional (constraint):

  $V[f]=\int<em>0^{2\pi}!d\varphi\int</em>0^R!d\rho\,\rho\,f(\rho)=\text{const}=\tfrac{V}{\mu}$

• Minimise $U[f]+\lambda V[f]$ → Euler-Lagrange:

  $g f'(\rho)-\omega^2\rho + \lambda =0$

  ⇒ $f(\rho)=\tfrac{\omega^2}{2g}\rho^2 + C$ (paraboloid; constant fixed by volume).

Planar Pendulum (Hamilton Formalism)

• One degree of freedom: angle $\theta$; conjugate momentum $p = m l^2 \dot\theta$.

• Hamiltonian:

  $H(\theta,p)=\tfrac{p^2}{2 m l^2}+m g l (1-\cos\theta)$

• Hamilton equations:

  $\dot\theta = \frac{\partial H}{\partial p}=\frac{p}{m l^2},\quad \dot p = -\frac{\partial H}{\partial \theta}= -m g l \sin\theta$

• Combine → nonlinear pendulum equation:

  $\ddot\theta+\frac{g}{l}\sin\theta=0$