Newtonian Mechanics

Inertial Reference Frame

Definition of a inertial reference frame: a reference frame that undergoes no accelertaion with respect to a fixed point, or a point of constant velocity.

Thought, suppose we have an accelerating reference frame with respect to a star. Can we use regular Newtonian mechanics to describe motion if we are far out enough and with sufficiently slow acceleration?

Newton’s Laws

  1. An object is stationary or moves at a constant velocity unless an external force is applied to the object. \(\vec{r}\): Position vector. \(\vec{v}\): Velocity vector. \(\vec{a}\): Acceleration vector.
  2. \(\frac{d}{dt}[\vec{p}] = \vec{F}_e\), \(\frac{d}{dt}(m\vec{v}) = m\frac{d}{dt}(\vec{v}) = \vec{F}_e \Rightarrow m\vec{a} = \vec{F}_e\).
  3. \(\vec{L} = \vec{r}\times\vec{p}\). \(\frac{d}{dt}(\vec{A}\times\vec{B}) = \vec{A}\times\frac{d}{dt}\vec{B} + \frac{d}{dt}\vec{A}\times\vec{B}\). \(\frac{d}{dt}\vec{L} = \vec{r}\times\vec{F}_e = \vec{\tau}_e\). Hence, the angular momentum remains invariant in the motion unless there is a force acting on it.

Work-Energy Theorem

\(dW = \vec{F}_e\cdot d\vec{r}\).

\(\Delta W = \int\vec{F}_e\cdot d\vec{r} = \int \frac{d\vec{p}}{dt}\cdot\vec{v}dt = m\int \vec{v}\cdot d\vec{v} = \frac{m}{2}\int d(\vec{v}\cdot\vec{v}) = \left.\frac{1}{2}mv^2\right|_i^f = E_{k,2} - E_{k,1}\) (for constant mass).

For conservative forces \(\vec{F}_e=-\nabla f\), we can take any path from the initial point to the final point. Then, \(\Delta W = \int_i^f \vec{F}_e\cdot d\vec{r} = -\int_i^f \nabla f\cdot d\vec{r} = -\int_{i}^{f} df = \int_{f}^{i} df = f(i) - f(f)\). Note, we call \(f\) is the potential energy of the conservative force and \(f(i) = E_{p,i}\).

Additionally, if \(\vec{F}_e = -\nabla f\) then \(\nabla \times \vec{F}_e = 0\). Hence, conservative forces are irrotational.

Further, \((E_p)_i + (E_k)_i = (E_p)_f + (E_k)_f\).


Suppose we are looking at a particle under the influence of a force \(\vec{f}\). Suppose this force is a central force, \(\vec{F} = \hat{r}f(r)\).

Starting with the angular momentum, \(\frac{d}{dt}\vec{L} = \vec{r}\times \hat{r}f(r) = rf(r) \hat{r}\times\hat{r} = 0\).

Note, central forces are irrotational since there are no spherical coordinates in the function. Hence, Energy is Conserved.

Since, \(\vec{L} = \vec{r}\times\vec{p}\) then if we align \(z\) with \(\vec{L}\) then \(\vec{r},\vec{v}\) are confined to the xy-plane. Note, \(\vec{L} = (xp_y-yp_x)\hat{z}\). In cylindrical coordinates: \(x=r\cos\varphi,y=r\sin\varphi\) and \(\dot{x}=\dot{r}\cos\varphi - r\sin\varphi\dot{\varphi}\). \(\dot{y}=\dot{r}\sin\varphi + r\cos\varphi\dot{\varphi}\). Then, \(\vec{L} = (r\cos\varphi m(\dot{r}\sin\varphi+r\cos\varphi\dot{\varphi}) - r\sin\varphi m(\dot{r}\cos\varphi - r\sin\varphi\dot{\varphi})) = mr^2\dot{\varphi} = \mathbb{C}\) is a constant.

The amount of area (triangular) swept out in \(dt\) time: \(dA = \frac{1}{2} r^2d\varphi\). So, \(\frac{dA}{dt} = \frac{1}{2}r^2\dot{\varphi} = \frac{L_z}{m} \equiv \frac{\ell}{2m}\) is a constant.

In spherical coordinates: \(d\vec{r} = \hat{r}dr + \hat{\theta}rd\theta + \hat{\varphi}r\sin\theta d\varphi\). Note: \(\nabla f\cdot d\vec{r} = df\).

Hence, in cylindrical coordinates: \(d\vec{r} = \hat{r}dr + \hat{\theta}rd\theta + \hat{z}dz\). Then, \(\nabla = \hat{r}\frac{\partial}{\partial r} + \hat{\theta}\frac{1}{r}\frac{\partial}{\partial \theta} + \hat{z}\frac{\partial}{\partial z}\). Note: \(\nabla f\cdot d\vec{r} = df\).

Then, in our irrotational force, \(E_{total} = T + V = \mathbb{C}\) is a constant. So, \(\frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\dot{y}^2 + V(r) = \frac{1}{2}m\dot{r}^2 + \frac{1}{2}mr^2\dot{\varphi}^2 + V(r) = \frac{\vec{p}_r^2}{2m} + \frac{\vec{L}^2}{2mr^2} + V(r) = \frac{p_r^2}{2m} + V_{eff}(r)\), since the angular momentum is constant. Hence, \(E_{total}(r,\dot{r})\). So, to get \(r\) as a function of time, lets rearrange our differential equation: \(E_{total} = \frac{1}{2}m\dot{r}^2 + V_{eff}(r)\) becomes \(\dot{r} = \sqrt{\frac{2}{m}\left(E-V-\frac{1}{2}\frac{\ell^2}{mr^2}\right)}\). So, \(\frac{dr}{\sqrt{\frac{2}{m}\left(E-V-\frac{1}{2}\frac{\ell^2}{mr^2}\right)}} = dt\). Then, to get to \(r(t)\) to \(r(\varphi)\) we use the fact that \(mr^2\dot{\varphi} = \ell\). Hence, \(dt = \frac{mr^2}{\ell}d\varphi\). Therefore, \(\frac{dr}{\sqrt{\frac{2}{m}\left(E-V-\frac{1}{2}\frac{\ell^2}{mr^2}\right)}} = \frac{mr^2}{\ell}d\varphi\). Hence, \(\frac{\ell dr}{\sqrt{2m}\sqrt{(E-V)r^2 - \frac{\ell^2}{2m}}} = d\varphi\). Let, \(u=\frac{1}{r}\) so \(dr = -\frac{1}{u^2}du\). Then, \(\varphi = \varphi_0 - \int_{u_0}^u \frac{du}{\sqrt{\frac{2mE}{\ell^2} - \frac{2mV}{\ell^2} - u^2}}\).

Most important potentials are power law functions of \(r\): \(V(r) = ar^{n+1}\). For \(n=0,-2\) the integral is solvable in terms of trig functions.

Suppose \(n=-2\). Then, \(\varphi = \varphi_0 - \int_{u_0}^u\frac{du}{\sqrt{\frac{2mE}{\ell^2} - \frac{2mau}{\ell^2} - u^2}}\) \(\varphi = \varphi_0 - \int_{u_0}^u\frac{du}{\sqrt{\frac{2mE}{\ell^2} + \frac{2mku}{\ell^2} - u^2}} \Rightarrow \frac{1}{r} = \frac{mk}{\ell^2}\left(1 + \sqrt{1+\frac{2E\ell^2}{mk^2}}\cos(\varphi-\varphi_0)\right)\). In the normal form, \(\frac{1}{r} = \frac{\ell^2}{mk^2}\left(1 + e\cos(\varphi-\varphi_0)\right)^{-1}\) where \(e=\sqrt{1 + \frac{2E\ell^2}{mk^2}}\) is the eccentricity.

So, for \(e=0\) we get a circle, \(e\in(0,1)\) an ellipse, \(e\in(1,\infty)\) a parabola, \(e>1\) a hyperbola$.

Kepler’s Third Law

With \(V(r) = -\frac{k}{r}\) with \(k>0\).

Then, for \(e\in(0,1)\) we have an ellipse with focal points at a length \(f\) away from the origin, \(a\) major axis, and \(b\) minor axis.

Then, \(r = \frac{\ell^2/mk}{1+e\cos\varphi}\).

So, \(E = -\frac{\ell^2}{2mr^2} + \frac{k}{r} = 0\) so \(r^2 + \frac{k}{E}r - \frac{\ell^2}{2mE} = 0\). So \(r_1+r_2=2a = -\frac{k}{E}\) then. Therefore, \(a = -\frac{k}{2E}\). Since, \(f=ae\) and \(b=\sqrt{a^2 - f^2}\) then we have both \(f\) and \(b\) determined. So, \(b = a\sqrt{1-e^2} = \sqrt{\frac{a\ell^2}{mk}}\).

The area of the ellipse is \(A = \pi ab\). Then, \(A\frac{dt}{dA} = \pi ab \frac{2m}{\ell}\). So, \(\tau \frac{\pi ab}{\ell/2m} = 2\pi a^{3/2}\sqrt{\frac{m}{k}}\).

Author: Christian Cunningham

Created: 2024-05-30 Thu 21:20