From the Lagrangian to the Hamiltonian

1. Why replace velocities by momenta?

A position alone does not specify how a system will evolve. The pendulum can pass the same angle in either direction and with many different speeds. Lagrangian mechanics completes the state with generalized velocities. Hamiltonian mechanics uses their conjugate momenta instead.

This is useful because momenta carry the conservation laws found in Unit 2, and because position and momentum can be treated as independent coordinates of a state space. The Hamiltonian then supplies two first-order evolution equations for each pair. It does not require a new force law.

The replacement must be possible mathematically. Starting from $p_i=\partial L/\partial\dot q_i$, we must solve for all $\dot q_i$ as functions of $q,p,t$. A sufficient local condition is an invertible velocity Hessian $\partial^2L/\partial\dot q_i\partial\dot q_j$. For the regular mechanical systems studied here, this is the positive-definite inertia matrix. Singular Lagrangians need a different constraint theory and are outside this syllabus.

2. Constructing the Hamiltonian

Definition. The Hamiltonian obtained from a regular Lagrangian is

with every $\dot q_i$ on the right replaced by its expression in $q,p,t$. Before that substitution, the right-hand side is the energy function written using velocities. After substitution it is a function on phase space.

This operation is a Legendre transformation. The derivative $\partial L/\partial\dot q_i$ supplies the new variable, and subtracting $L$ from $\sum p_i\dot q_i$ removes the old velocity as an independent argument. A completed Hamiltonian may contain prescribed parameters such as $\Omega$; it must not contain an unresolved dynamical $\dot q$.

For one free particle, $L=m\dot x^2/2$ gives $p=m\dot x$, so $\dot x=p/m$. Then

The two terms in the definition are both needed. Merely replacing $\dot x$ by $p/m$ in $L$ would happen to work for a free particle but gives the wrong sign of potential energy for a general system.

3. A useful matrix construction

For $L=\tfrac12\dot q^{\mathsf T}M\dot q+b^{\mathsf T}\dot q+c-V$, the superscript $\mathsf T$ means transpose, so the products are sums over coordinates. The momenta satisfy

Let $v=M^{-1}(p-b)$. Substitution in $p^{\mathsf T}v-L$ gives

The second equality follows because $p-b=Mv$ and $M$ is symmetric. This is a compact construction rule; the component calculations below remain sufficient without matrix notation.

4. Applications

Example 1 · The spring oscillator

Choose displacement $x$. Then $T=m\dot x^2/2$, $V=kx^2/2$, and $L=T-V$. Since $p=m\dot x$,

The potential changes sign in passing from $L$ to $H$. Here $H$ equals total mechanical energy, because the coordinate transformation is stationary and $T$ is quadratic homogeneous. For $m=2\,\mathrm{kg}$, $p=4\,\mathrm{kg\,m\,s^{-1}}$, $k=8\,\mathrm{N\,m^{-1}}$, and $x=0.5\,\mathrm m$, the kinetic and potential contributions are $4\,\mathrm J$ and $1\,\mathrm J$, so $H=5\,\mathrm J$.

Example 2 · Polar coordinates

For a particle in $V(r)$,

The momenta are $p_r=m\dot r$ and $p_\theta=mr^2\dot\theta$. Invert them: $\dot r=p_r/m$, $\dot\theta=p_\theta/(mr^2)$. The sum $p_r\dot r+p_\theta\dot\theta$ is $p_r^2/m+p_\theta^2/(mr^2)$, while $T$ is half that sum. Hence

Do not substitute $p_\theta=m\dot\theta$: an angular conjugate momentum has different dimensions. When differentiating $H$ with respect to $r$, hold $p_\theta$ fixed. Holding $\dot\theta$ fixed would mean differentiating a different function on a different state space.

Example 3 · Coupled velocities

Let $L=\tfrac12m(\dot q_1^2+2\alpha\dot q_1\dot q_2+\dot q_2^2)-V(q_1,q_2)$ with $|\alpha|<1$. From Chapter 10,

Their momentum-weighted sum is $(p_1^2-2\alpha p_1p_2+p_2^2)/[m(1-\alpha^2)]$. Since $\sum_i p_i\dot q_i=2T$, subtracting $L=T-V$ gives

The cross term changes sign in the inverse quadratic form. For $\alpha=0$ the particles' kinetic terms separate; as $|\alpha|\to1$ the inverse fails, matching the loss of regularity already identified.

Example 4 · A translating coordinate origin

A free particle on a line has laboratory position $x=q+Ut$, with constant prescribed $U$. Its physical kinetic energy is $T=m(\dot q+U)^2/2$, and $V=0$. Thus $L=T$ and

The Hamiltonian is

Laboratory energy is $E_{\rm lab}=p^2/(2m)$, so $H\ne E_{\rm lab}$ even though both are constant for this free motion. The term $-Up$ is a consequence of the explicitly time-dependent coordinate transformation. Relative kinetic energy $m\dot q^2/2$ equals $H+mU^2/2$; the difference is a constant, but neither equals laboratory energy for arbitrary $p$.

5. Momentum is coordinate-dependent

The Cartesian momentum of a particle is the vector $m\mathbf v$ in this ordinary mechanical setting. The generalized momentum $p_i$ is its projection weighted by the positional response to $q_i$, possibly including terms from a more general velocity-dependent Lagrangian. It is not generally a Cartesian vector component.

For example, $p_\theta$ in polar coordinates equals angular momentum, whereas $p_r$ is radial linear momentum. Treating them as interchangeable “momenta” without units obscures the mechanics. Their shared role is that each is conjugate to a coordinate and that their product $p_i\dot q_i$ has units of power times time, namely energy.

6. Construction checks

A correct construction defines the coordinates, writes $T,V,L$, differentiates to obtain every momentum, inverts the complete set, and substitutes into $H$. Check that all dynamical velocities have disappeared and that dimensions remain those of energy. The meaning of the resulting function, including when it is conserved and when it equals mechanical energy, will be examined after introducing its equations of motion.

Exercises

Hints and solutions are collected separately.

1. 12.1 What must be checked before replacing velocities by momenta?

2. 12.2 Construct $H$ for $L=m\dot x^2/2-mgx$.

3. 12.3 For $L=I\dot\theta^2/2-U(\theta)$ with constant $I>0$, construct $H$.

4. 12.4 Why must a completed Hamiltonian contain no unresolved generalized velocities?

5. 12.5 Construct $H$ for $L=\tfrac12a(q)\dot q^2-V(q)$, $a(q)>0$.

6. 12.6 For $L=m\dot q^2/2+B\dot q-V(q)$ with constant $B$, find $p$ and $H$.

7. 12.7 Invert $p_1=m(2\dot q_1+\dot q_2)$, $p_2=m(\dot q_1+3\dot q_2)$ and construct kinetic $H$.

8. 12.8 For $x=q+Ut$, prove $H=p^2/(2m)-Up$ for a free particle.

9. 12.9 For $H=p_r^2/(2m)+p_\theta^2/(2mr^2)+V(r)$, calculate $H_r$ at fixed momenta.

10. 12.10 Why is the Legendre construction not merely changing the sign of $V$?

11. 12.11 If $m=3$, $p=6$, $k=2$, and $x=2$ in SI units, find the spring Hamiltonian.

12. 12.12 Explain why polar momentum coordinates fail at the origin although the free-particle problem is regular.