11.6: Hamiltonian

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Since the center-of-mass Lagrangian is not an explicit function of time, then

\[\frac{dH_{cm}}{dt}=\mathcal{-}\frac{\partial L_{cm}}{\partial t}=0\]

Thus the center-of mass Hamiltonian \(H_{cm}\) is a constant of motion. However, since the transformation to center of mass can be time dependent, then \(H_{cm}\neq E,\) that is, it does not include the total energy because the kinetic energy of the center-of-mass motion has been omitted from \(H_{cm}\). Also, since no transformation is involved, then

\[H_{cm}=T_{cm}+U=E_{cm}\]

That is, the center-of-mass Hamiltonian \(H_{cm}\) equals the center-of-mass total energy. The center-of-mass Hamiltonian then can be written using the effective potential \((11.4.6)\) in the form \[H_{cm}= \frac{p_{r}^{2}}{2\mu }+\frac{p_{\theta }^{2}}{2\mu r^{2}}+U(r)=\frac{ p_{r}^{2}}{2\mu }+\frac{l^{2}}{2\mu r^{2}}+U(r)=\frac{p_{r}^{2}}{2\mu } +U_{eff}(r)=E_{cm} \label{11.42}\]

It is convenient to express the center-of-mass Hamiltonian \(H_{cm}\) in terms of the energy equation for the orbit in a central field using the transformed variable \(u=\frac{1}{r}\). Substituting equations \((11.4.6)\) and \((11.5.3)\) into the Hamiltonian Equation \ref{11.42} gives the energy equation of the orbit \[\frac{l^{2}}{2\mu }\left[ \left( \frac{du}{d\psi }\right) ^{2}+u^{2}\right] +U\left( u^{-1}\right) =E_{cm}\]

Energy conservation allows the Hamiltonian to be used to solve problems directly. That is, since

\[H_{cm}=\frac{\mu \dot{r}^{2}}{2}+\frac{l^{2}}{2\mu r^{2}}+U(r)=E_{cm}\]

then

\[\dot{r}=\frac{dr}{dt}=\pm \sqrt{\frac{2}{\mu }\left( E_{cm}-U-\frac{l^{2}}{ 2\mu r^{2}}\right) }\label{11.45}\]

The time dependence can be obtained by integration

\[t=\int \frac{\pm dr}{\sqrt{\frac{2}{\mu }\left( E_{cm}-U-\frac{l^{2}}{2\mu r^{2}}\right) }}+\text{ constant}\label{11.46}\]

An inversion of this gives the solution in the standard form \(r=r\left( t\right) .\) However, it is more interesting to find the relation between \(r\) and \(\theta .\) From relation \ref{11.46} for \(\frac{dr}{dt}\) then

\[dt=\frac{\pm dr}{\sqrt{\frac{2}{\mu }\left( E_{cm}-U-\frac{l^{2}}{2\mu r^{2}} \right) }}\]

while equation \((11.4.2)\) gives

\[d\psi =\frac{ldt}{\mu r^{2}}=\frac{\pm ldr}{r^{2}\sqrt{2\mu \left( E_{cm}-U- \frac{l^{2}}{2\mu r^{2}}\right) }}\]

Therefore

\[\psi =\int \frac{\pm ldr}{r^{2}\sqrt{2\mu \left( E_{cm}-U-\frac{l^{2}}{2\mu r^{2}}\right) }}+\text{ constant}\label{11.49} \]

which can be used to calculate the angular coordinate. This gives the relation between the radial and angular coordinates which specifies the trajectory.

Although equations \ref{11.45} and \ref{11.49} formally give the solution, the actual solution can be derived analytically only for certain specific forms of the force law and these solutions differ for attractive versus repulsive interactions.