Newtonian Mechanics

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?

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

$dW={\overrightarrow{F}}_e\cdot d\overrightarrow{r}$.

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

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

Additionally, if ${\overrightarrow{F}}_e=-\mathrm{\nabla }f$ then $\mathrm{\nabla }\times {\overrightarrow{F}}_e=0$. 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 $\overrightarrow{f}$. Suppose this force is a central force, $\overrightarrow{F}=\hat{r}f(r)$.

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

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

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

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

In spherical coordinates: $d\overrightarrow{r}=\hat{r}dr+\hat{\theta }rd\theta +\hat{\phi }r\sin \theta d\phi$. Note: $\mathrm{\nabla }f\cdot d\overrightarrow{r}=df$.

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

Then, in our irrotational force, $E_{total}=T+V=\mathbb{C}$ is a constant. So, $\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}m{\dot{y}}^2+V(r)=\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{r}^2{\dot{\phi }}^2+V(r)=\frac{{\overrightarrow{p}}_r^2}{2m}+\frac{{\overrightarrow{L}}^2}{2m{r}^2}+V(r)=\frac{p_r^2}{2m}+V_{eff}(r)$, 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 $\dot{r}=\sqrt{\frac{2}{m}(E-V-\frac{1}{2}\frac{{\ell }^2}{m{r}^2})}$. So, $\frac{dr}{\sqrt{\frac{2}{m}(E-V-\frac{1}{2}\frac{{\ell }^2}{m{r}^2})}}=dt$. Then, to get to $r(t)$ to $r(\phi )$ we use the fact that $m{r}^2\dot{\phi }=\ell$. Hence, $dt=\frac{m{r}^2}{\ell }d\phi$. Therefore, $\frac{dr}{\sqrt{\frac{2}{m}(E-V-\frac{1}{2}\frac{{\ell }^2}{m{r}^2})}}=\frac{m{r}^2}{\ell }d\phi$. Hence, $\frac{\ell dr}{\sqrt{2m}\sqrt{(E-V)r^2-\frac{{\ell }^2}{2m}}}=d\phi$. Let, $u=\frac{1}{r}$ so $dr=-\frac{1}{u^2}du$. Then, $\phi ={\phi }_0-{\int }_{u_0}^u\frac{du}{\sqrt{\frac{2mE}{{\ell }^2}-\frac{2mV}{{\ell }^2}-u^2}}$.

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

Suppose $n=-2$. Then, $\phi ={\phi }_0-{\int }_{u_0}^u\frac{du}{\sqrt{\frac{2mE}{{\ell }^2}-\frac{2mau}{{\ell }^2}-u^2}}$ $\phi ={\phi }_0-{\int }_{u_0}^u\frac{du}{\sqrt{\frac{2mE}{{\ell }^2}+\frac{2mku}{{\ell }^2}-u^2}}\Rightarrow \frac{1}{r}=\frac{mk}{{\ell }^2}(1+\sqrt{1+\frac{2E{\ell }^2}{m{k}^2}}\cos (\phi -{\phi }_0))$. In the normal form, $\frac{1}{r}=\frac{{\ell }^2}{m{k}^2}{(1+e\cos (\phi -{\phi }_0))}^{-1}$ where $e=\sqrt{1+\frac{2E{\ell }^2}{m{k}^2}}$ is the eccentricity.

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