Series Quick-Reference Overview [Classical Mechanics I Studied #12]

At a glance! (Don’t look with both eyes, okay? >0<)

Handwritten notes on the 2D harmonic oscillator and the shape of its orbit

Handwritten notes on 3D isotropic and anisotropic harmonic oscillators and Lissajous figures

Handwritten notes on non-inertial reference frames and the Galilean transformation

Handwritten notes on the rotating coordinate system and the time-derivative operator

Handwritten notes on acceleration in a rotating frame with the transverse, Coriolis and centrifugal terms

Handwritten notes on Kepler’s second law, the equal-area law

Handwritten notes on Kepler’s first law and the equations of motion in polar coordinates

Handwritten notes on the substitution r = 1/u that leads to the orbit equation

Handwritten notes on the inverse-square orbit equation and its solution

Handwritten notes on the ellipse in polar coordinates and the eccentricity

Handwritten notes on Kepler’s third law, the harmonic law

Handwritten notes on deriving the orbit equation from the energy relation

Handwritten notes on the effective potential

Handwritten notes on the stability of a circular orbit

Handwritten notes on Rutherford alpha-particle scattering and its orbit equation

Handwritten notes on the scattering angle and the Rutherford cross-section

Heads-up. The figures here are the original notes, slips and all. A few things to watch for:

In the anisotropic 3D oscillator, $z$ has its own phase: $z = C\cos(\omega_3 t + \gamma)$; the picture reuses $\beta$ from $y$.
In the rotating-frame equation, the three braced terms stand on the side of $m\vec a$; the transverse, Coriolis and centrifugal forces felt in the rotating frame are their negatives: $-m\dot{\vec\omega}\times\vec{r}'$, $-2m\vec\omega\times\vec{v}'$ and $-m\vec\omega\times(\vec\omega\times\vec{r}')$.
In the equal-area law, the external force is not zero: it is central, so the torque $\vec r\times\vec F$ is zero and $\vec L$ is constant.
In the orbit-equation derivation, the first equation needs its $m$: $m(\ddot r - r\dot\theta^2) = f(r)$.
In the inverse-square solution, $r = \frac{1}{A\cos(\theta-\theta_0) + \frac{k}{ml^2}}$, with a plus sign; the boxed formula under it is right.
In the two energy-method derivations (the orbit and the Rutherford scattering), the integral table has a sign slip: for $a \lt 0$, $\int\frac{du}{\sqrt{au^2+bu+c}} = \frac{1}{\sqrt{-a}}\cos^{-1}\frac{b+2au}{\sqrt{b^2-4ac}}$ plus a constant, and with $a = -1$ the numerator is $b - 2u$. The orbit equations that follow are right.
In the orbit-stability notes, $\dot\theta = \frac{l}{r^2}$, so $r\dot\theta^2 = \frac{l^2}{r^3}$; the circular-orbit condition gives only $f(a) = -\frac{ml^2}{a^3}$, not $f'(a) = \frac{3ml^2}{a^4}$; and the last equation is for the displacement $x = r - a$: $m\ddot x - \left[f'(a) + \frac{3}{a}f(a)\right]x = 0$. The stability condition $f'(a) + \frac{3}{a}f(a) \lt 0$ is right.
In the Rutherford derivation, half of $\sqrt{\frac{4Q^2q^2}{m^2l^4} + \frac{8E}{ml^2}}$ is $\sqrt{\frac{Q^2q^2}{m^2l^4} + \frac{2E}{ml^2}}$; the picture writes $4E$ there, and the next line has $2E$.
In the cross-section notes, differentiating $\cot\frac{\theta_s}{2} = \frac{2E}{Qq}b$ gives $\frac{1}{2\sin^2(\theta_s/2)} = -\frac{2E}{Qq}\frac{db}{d\theta_s}$, and $\sigma(\theta_s) = \frac{b}{\sin\theta_s}\left|\frac{db}{d\theta_s}\right|$; the final $\frac{Q^2q^2}{16E^2}\frac{1}{\sin^4(\theta_s/2)}$ is right.

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