Effective 1-d potential

Results so far...

Two bodies, interacting through a central force (think gravity), with no external forces.

Center of mass moves with a constant velocity. So, working in the reference frame where the CM is at rest we have shown:

• The Lagrangian can be written in terms of the reduced mass, $\mu=m_1m_2/(m_1+m_2)$ and relative coordinate, $\myv r$, as: $${\cal L} = T-U=\frac 12\mu\dot{\myv r}^2-U(r).$$
• Total angular momentum is conserved, and is found, mathematically, to be equal to: $$\myv L = \myv r \times \mu\dot{\myv r}.$$
• Summarizing these last two in cartoon form:
  
  

  [I grabbed this drawing off the Internet, but there is an error in this cartoon. Considering that $\myv r=\myv r_1-\myv r_2$, can you see the problem??

Now our program will be:

1. Write the Lagrangian in terms of polar coordinates $(r,\phi)$.
2. Write the E-L equations for the coordinates $\phi$ and $r$.
3. Tease out the consequences: We'll find a simple form for $\ddot r$.

Lagrangian in polar coordinates

Our Lagrangian in vector form is: $${\cal L}=\frac12 \mu\dot{\myv r}^2-U(r),$$ where $\dot{\myv r}^2=\dot{\myv r} \cdot \dot{\myv r} = |\dot{\myv r}|^2$.

As we've seen before, writing $\myv r$ in terms of polar coordinates $(r,\phi)$ results in: $$\dot{\myv r}=\frac{d}{dt}\myv r = r\dot \phi\, \uv \phi+\dot r\, \uv r$$ Dotting this vector into itself, whilst recognizing that $\uv \phi$ and $\uv r$ are mutually perpendicular: $$\dot{\myv r}\cdot\dot{\myv r}=(r\dot \phi)^2+\dot r^2={\color{blue}r^2\dot\phi^2+\dot r^2},\label{vdotv}$$ so the Lagrangian can be written as: $${\cal L} = \frac{1}{2}\mu({\color{blue}\dot{r}^2+r^2\dot{\phi}^2}) - U(r).\label{Lagrangian}$$

E-L equation for $\phi$

There's no explicit $\phi$ dependence in ${\cal L}$. The $\phi$ coordinate is ignorable, and the generalized $\phi$ momentum will be conserved, that is: $$\frac{\del {\cal L}}{\del \phi} = 0 =\frac{d}{dt}\frac{\del {\cal L}}{\del \dot{\phi}}.$$

So, $$c o n s t=\frac{\del {\cal L}}{\del \dot{\phi}} =\mu r^2 \dot{\phi}=|\myv L|\equiv\ell.$$

Solving for $\dot{\phi}$ in terms of the angular momentum: $$\dot{\phi} = \frac{ \ell}{\mu r^2}$$

The radial equation

The E-L equation for $r$ results in

$$\mu \ddot{r} = -\frac{dU}{dr}+ \mu r \dot{\phi}^2.$$

That last term, the centripetal, or 'fictitious' centrifugal acceleration, probably looks more familiar written as $F_{cf} = \mu r\dot{\phi}^2=\mu(r \dot{\phi})^2/r = \mu v_{\phi}^2/r$. But I'd rather write it in terms of $\ell$ using $\dot{\phi}=\frac{\ell}{\mu r^2}$ as: $$F_{cf} = \frac{\ell^2}{\mu r^3} = -\frac{d}{dr}\left[\frac{\ell^2}{2\mu r^2}\right] \equiv -\frac{d}{dr}\left[U_{cf}(r)\right]$$

The quantity in $[...]$ defines what we'll call the "centrifugal potential energy" $U_{cf}$. The radial equation is now: $$\mu \ddot{r} = -\frac{d}{dr}[U(r)+U_{cf}] \equiv -\frac{dU_{eff}}{dr}.\label{radial_equation}$$

The "effective potential" is: $$U_{eff} = U(r)+\frac{\ell^2}{2\mu r^2}$$

Many features of how planets orbit stars can be shown as a result of considering this one-dimensional problem, now that we have an "effective" 1-d potential. We can bring to bear the mental picture of a ball rolling around on the 1-d potential curve that we've calculated...

Radial "motion" under the influence of gravity

Let $m_1$ be the mass of the sun and $m_2$ be the mass of a some hunk of rock near the sun. The effective radial potential energy is $$U_{eff}(r) = -G\frac{m_1m_2}{r} +\frac{\ell^2}{2\mu r^2}$$ We shall shortly find out that $U_\text{eff}(r)$ actually accounts for all the energy of the system, both kinetic energy + the potential energy, $U(r)$. Because of conservation of momentum, it can all be expressed in terms of $r$ alone. Thus, we can use our intuition about 1-d energy landscapes to understand some of the dynamics of the system just by examining the graph of $U\text{eff}(r)$:

Using that "marble rolling around in a 1-d potential" mental picture...

Here's the potential energy for a $1/r^4$ force.

Any "comet-like" orbits? Any stable orbits here?

But...Am I even warranted in talking in terms of a "total energy" as if it were conserved, when we're talking about these 'effective' potential energies?

Conservation of energy

Taking the radial equation ($\ref{radial_equation}$) above, and multiplying both sides by $\dot{r}$: $$\begineq \mu\ddot{r}\dot{r} =& -\frac{d}{dr}U_{eff}(r)\frac{dr}{dt}\\ \frac{d}{dt}\frac{1}{2}\mu \dot{r}^2 =& \frac{d}{dt}(-U_{eff}(r))\\ \frac{d}{dt}\left[\frac{1}{2}\mu\dot{r}^2 + U_{eff}(r)\right] =& 0.\endeq$$

Integrating this last expression (and referring back to Eq $\ref{vdotv}$ and Eq $\ref{Lagrangian}$ for the kinetic energy) means that $$\begineq C o n s t =&\frac{1}{2}\mu \dot{r}^2 + U_{eff}(r) \\ =&\frac{1}{2}\mu \dot{r}^2 + \frac{\ell^2}{2\mu r^2} +U(r) \\ =& {\color{blue}\frac{1}{2}\mu \dot{r}^2 + \frac{1}{2}\mu r^2 \dot{\phi}^2} +U(r) \\ =& {\color{blue}T} + U = E.\endeq$$

So, indeed, it appears that $U_{eff}$ comprises both the potential energy $U(r)$ as well as the tangential kinetic energy. So, adding this up with the radial kinetic energy gives just the total energy of the system which is conserved.

Homework

Chapter 8, problems: 1, 7, 13.

Image credits

Quantum Redpill