# 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, $$(E_p)_i + (E_k)_i = (E_p)_f + (E_k)_f$$.

## Example

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, $$\tau \frac{\pi ab}{\ell/2m} = 2\pi a^{3/2}\sqrt{\frac{m}{k}}$$.

Created: 2024-05-30 Thu 21:20

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