2024-02-16 Central Forces

Fields and forces

• Fields are a useful concept we use to describe action at a distance
• Examples include electricity, magnetism and gravitation.
• There is a time dependent component to field that requires a certain amount of time for the field to “update” at a point

Gravitational Potential

• Gravitational potential $\ne$ Gravitational potential energy

$$\vec{g}=-\vec{\nabla }\varphi$$ $$-\frac{d\varphi }{dr}=-\frac{GM}{r^2}⟹\varphi =\frac{-GM}{r}$$

• Force and Gravitational potential energy on one side, Fields and potentials on the other side.
• This dichotomy divides between physical quantities and mathematical concepts.

What is phi? Phi is gravitational potential.

$$\vec{F}=-\vec{\nabla }u$$ $$\vec{g}=-\vec{\nabla }\varphi$$ $$\therefore u=m\varphi$$

For an extended object, the force of gravity is defined as below

$${\vec{F}}_g=-Gm\int \frac{\rho (\vec{r})}{r^2}{\hat{e}}_{r}dV$$

How to get little g?

$$\vec{g}=-G\underset{V}{\int }\frac{\rho (\vec{r})}{r^2}{\hat{e}}_{r}dV$$

$$\varphi =-\frac{G{m}_1}{r_1}-\frac{G{m}_2}{r_2}$$ $$-Gm\left(\frac{1}{r_1}+\frac{1}{r_2}\right)$$

Want to generalize to all possible locations of $p$.

$$r_1=\sqrt{(x-a{)}^2+y^2}$$ $$r^2=\sqrt{(x+a{)}^2+y^2}$$ $$\varphi =-Gm\left[\frac{1}{\sqrt{V(x-a{)}^2}}+\frac{1}{\sqrt{(x+a{)}^2+y^2}}\right]$$ $$\vec{g}=-\vec{\nabla }=-\frac{\partial \varphi }{\partial x}{\hat{e}}_x-\frac{\partial \varphi }{\partial y}{\hat{e}}_y$$ $$g_x=-\frac{\partial \varphi }{\partial x}=\frac{d}{dx}(-Gm[((x-a{)}^2+y^2{)}^{-1/2}+((x+a{)}^2+y^2{)}^{-1/2}])$$ $$\frac{d\varphi }{dx}=-Gm\left[-\frac{1}{2}((x-a{)}^2+y^2{)}^{-3/2}\cdot (2(x-a))-((x+a{)}^2+y^2{)}^{-3/2}\cdot (\cancel{2}(x+a))\right]$$ $$-\frac{\partial \varphi }{\partial x}=-Gm\left[\frac{x-a}{((x-a{)}^2+y^2{)}^{3/2}}+\frac{x+a}{((x+a{)}^2+y^2{)}^{3/2}}\right]=g_y$$

Example 1

Gravitational field produced by a thin layer of material spread over an infinite plane

Where $\sigma$ is mass/unit area

We have divided the plane into a series of rings of radius R and thickness dR

Area ring $2\pi RdR$ Mass ring

$$dm=\sigma 2\pi RdR$$

Potential @ point P

$$d\varphi =-\frac{Gdm}{r}=-\frac{G\sigma 2\pi RdR}{(z^2+r^2{)}^{1/2}}$$ $$\varphi =\int d\varphi =-G{\sigma }_{2}p\underset{0}{\overset{\infty }{\int }}\frac{R}{(z^2+R^2{)}^{1/2}}dR$$

Do u-sub

$$u=z^2+R^2$$ $$du=2RdR$$ $$\frac{1}{2}\int \frac{du}{u^{1/2}}=\cancel{\frac{1}{2}2}{u}^{1/2}=u^{1/2}$$ $$\varphi =-G\sigma 2\pi (z^2+R^2{)}^{1/2}{|}_0^{\infty }$$ $$=-G{\sigma }_2\pi \infty +G\sigma 2\pi z$$ $$\varphi =G\sigma 2\pi (z-\infty )$$

However, we don’t care about the potential. We are only interested in the potential difference

$$\vec{g}=-\vec{\nabla }\varphi$$

relative to the plane

$$\varphi =G\sigma 2\pi (z-\infty )+G\sigma 2\pi \infty$$

This is the process of re-normalizing

$$\varphi =G\sigma 2\pi z$$ $$\vec{g}=-\vec{\nabla }\varphi$$ $$\vec{g}=-2G\sigma \pi$$ $$=-2\pi G\sigma$$

above and below the plane

Example 2

2- body central force problem (using Lagrangian Mechanics)

$$L=T-U=\frac{1}{2}{m}_1{\vec{r}}_1^2+\frac{1}{2}{m}_2{\vec{r}}_2^2-u({\vec{r}}_2-{\vec{r}}_1)$$

What is a central force, really?

Generalized coordinates $\vec{R}$ and $\vec{r}$

$$\vec{R}=\frac{m_1{\vec{r}}_1+m_2{\vec{r}}_2}{m_1+m_2}\text{ (C.O.M)}$$ $$\vec{r}={\vec{r}}_2+{\vec{r}}_1$$ $$\vec{r}=\vec{R}-\frac{m_2}{M}\vec{r},{\vec{r}}_2=\vec{R}+\frac{m_1}{M}\vec{r}$$

with $M=m_1+m_2$

$${\vec{\dot{r}}}_1=\vec{\dot{R}}-\frac{m_2}{m}\vec{\dot{r}},{\vec{\dot{r}}}_2=\vec{\dot{R}}+\frac{m_1}{m}\vec{\dot{r}}$$ $${\vec{\dot{r}}}_1^2={\vec{\dot{R}}}^2+\frac{m_2^2}{m_2}{\vec{\dot{r}}}^2-2\frac{m_2}{m}\vec{\dot{R}}$$

etc etc for ${\dot{\vec{r}}}_2$

$$L=\frac{m_1+m_2}{2}{\dot{\vec{R}}}^2+\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}{\dot{\vec{r}}}^2-u(|\vec{r}|)$$

Inspect second element

$$\frac{1}{2}{m}_1\frac{m_2^2}{m_2}{\dot{\vec{r}}}^2+\frac{\frac{1}{2}{m}_2{m}_1^2}{m_2}{\dot{\vec{r}}}^2$$ $$=\frac{1}{2}\frac{{\dot{\vec{r}}}^2}{m^2}(m_1{m}_2^2+m_1^2{m}_2)$$ $$=\frac{1}{2}\frac{m_1{m}_2}{M^2}{\dot{\vec{r}}}^2(m_2+m_1)=\frac{1}{2}\frac{m_1{m}_2}{M}{\dot{\vec{r}}}^2$$ $$\frac{m_1{m}_2}{M}=\mu \text{ Reduced mass}$$

Equations of motion for $\vec{R}$

$$\frac{\partial L}{\partial \dot{R}}=0$$

This means that R is a cyclic coordinate. Aka, the Lagrangian is invariant under translation of R.

$$\vec{P}=M\dot{\vec{R}}$$ $$\frac{d}{dt}\frac{\partial L}{\partial \vec{R}}=0$$

Motion of COM moves at constant velocity

This permits the reduction of the two body system (central force problem) to a one body problem by dropping the $\frac{1}{2}(m_1+m_2){\dot{\vec{R}}}^2$ term.