The Hamiltonian

The Hamiltonian

Now that we know the Lagrangian \(L\), we can directly calculate the Hamiltonian \(H\) by means of the Legendre transform [1].

\[p = \frac{\partial L}{\partial \dot{q}},\qquad H = p\cdot\dot{q} - L\]

We’ll step through applying these equations, but the Hamiltonian will eventually end up in terms of canonical variables \(q = (x,y,z)\) and \(p = (p_x,p_y,p_z)\) taking the following form.

\[H = \frac{1}{2}\left((p_x + y)^2 + (p_y - x)^2 + p_z^2\right) + \underbrace{\left(-\frac{1}{2}(x^2 + y^2) - \frac{1-\mu}{r_1(x,y,z)} - \frac{\mu}{r_2(x,y,z)} - \frac{1}{2}\mu(1-\mu)\right)}_{\bar{U}(x,y,z)}\]

Legendre Transform

Knowing the Lagrangian allows us to find the canonical coordinates, where the momenta are found from \(p = L_\dot{q}\). Performing the calculations provides the following relationships between the momenta \((p_x, p_y, p_z)\) and rotating coordinates \((x,y,z,\dot{x},\dot{y},\dot{z})\).

\[\begin{split}p_x &= \dot{x} - y \\ p_y &= \dot{y} + x \\ p_z &= \dot{z}\end{split}\]

Proof

Recall the Lagrangian \(L = T-U\),

\[L = \frac{1}{2}\Big((\dot{x} - y)^2 + (\dot{y} + x)^2 + \dot{z}^2\Big) + \frac{1-\mu}{r_1(x,y,z)} + \frac{\mu}{r_2(x,y,z)} + \frac{1}{2}\mu(1-\mu).\]

We need to calculate \(L_\dot{q}\) for each component of \(\dot{q}\). Note that the velocities \(\dot{q} = (\dot{x}, \dot{y}, \dot{z})\) only appear in the kinetic energy, so we know \(L_\dot{q} = T_\dot{q}\), where

\[T = \frac{1}{2}\Big((\dot{x} - y)^2 + (\dot{y} + x)^2 + \dot{z}^2\Big)\]

Carrying out the calculations brings us to the answer.

\[\begin{split}p_x &= \frac{\partial T}{\partial \dot{x}} = \dot{x} - y \\ p_y &= \frac{\partial T}{\partial \dot{y}} = \dot{y} - x \\ p_z &= \frac{\partial T}{\partial \dot{z}} = \dot{z}\end{split}\]

The theory of Hamiltonian systems assumes that these equations are invertible in favor of the velocities. In our present case, this task is trivial as we can simply solve for \((\dot{x},\dot{y},\dot{z})\) by inspection.

\[\begin{split}\dot{x} &= p_x + y \\ \dot{y} &= p_y - x \\ \dot{z} &= p_z\end{split}\]

Now we’re in a position to finish the Legendre transformation to find the Hamiltonian. Recalling that the Lagrangian was found to be

\[L(x,y,z,\dot{x},\dot{y},\dot{z}) = \frac{1}{2}\Big((\dot{x} - y)^2 + (\dot{y} + x)^2 + \dot{z}^2\Big) + \underbrace{\frac{1-\mu}{r_1(x,y,z)} + \frac{\mu}{r_2(x,y,z)} + \frac{1}{2}\mu(1-\mu)}_{-U(x,y,z)},\]

simply plug into the equation with \(\dot{q} = \dot{q}(q,p)\), where \(q = (x,y,z)\) and \(p = (p_x, p_y, p_z)\), as follows.

\[\begin{split}H &= \begin{pmatrix}p_x \\ p_y \\ p_z\end{pmatrix} \cdot \begin{pmatrix}p_x + y \\ p_y - x \\ p_z\end{pmatrix} - L(x,y,z,p_x+y,p_y-x,p_x) \\ &= p_x (p_x + y) + p_y (p_y - x) + p_z^2 - \left(\frac{1}{2}(p_x^2 + p_y^2 + p_z^2) - U(x,y,z)\right) \\ &= p_x (p_x + y) + p_y (p_y - x) + p_z^2 - \frac{1}{2}(p_x^2 + p_y^2 + p_z^2) + U(x,y,z) \\ &= \frac{1}{2}(p_x^2 + p_y^2 + p_z^2) + p_x y - p_y x + U(x,y,z) \\ &= \frac{1}{2}\Big((p_x + y)^2 + (p_y - x)^2 + p_z^2\Big) -\frac{1}{2}(x^2 + y^2) + U(x,y,z)\end{split}\]

Finally, we have the Hamiltonian.

\[H = \frac{1}{2}\left((p_x + y)^2 + (p_y - x)^2 + p_z^2\right) - \frac{1}{2}(x^2 + y^2) - \frac{1-\mu}{r_1(x,y,z)} - \frac{\mu}{r_2(x,y,z)} - \frac{1}{2}\mu(1-\mu)\]

In the midst of simplifying the Hamiltonian to preserve some nice structure, a new term appeared which depends only on position in the rotating frame. We group this term in with the gravitational potential \(U\) to form the pseudopotential \(\bar{U}\),

\[\bar{U}(x,y,z) = -\frac{1}{2}(x^2 + y^2) + \underbrace{\left(-\frac{1-\mu}{r_1(x,y,z)} - \frac{\mu}{r_2(x,y,z)} - \frac{1}{2}\mu(1-\mu)\right)}_{U(x,y,z)},\]

such that the Hamiltonian can be written in the form

\[H = \frac{1}{2}\Big((p_x + y)^2 + (p_y - x)^2 + p_z^2\Big) + \bar{U}(x,y,z).\]

Importantly, notice that the Hamiltonian is explicitly independent of time. This will be an important feature in CRTBP motion that we will explore in detail next.

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References

1. Koon, W.S., Lo, M.W., Marsden, J.E., Ross, S.D. (2022) Dynamical Systems, the Three-Body Problem and Space Mission Design. Marsden Books, ISBN 978-0-615-24095-4.