# Hamiltonian systems without a corresponding Lagrangian system

I was playing around with a Hamiltonian model for the propagation of photons:

$$H = c \sqrt{p \cdot p} + V(q) \tag{1}$$

which gives a meaningful set of equations of motion,

$$\dot{q}_i = c \frac{p_i}{\sqrt{p \cdot p}} \quad \quad \dot{p}_i = - \frac{\partial V(q)}{\partial q_i}. \tag{2}$$

Notice that $$\dot{q} \cdot \dot{q} = c^2\tag{3}$$ always which is why I considered this as modeling the propagation of a massless particle.

However, this Hamiltonian has the following strange feature. If we perform a Legendre transformation to find an associated Lagrangian, $$\mathcal{L} = p \cdot \dot{q} - H = c \frac{p \cdot p}{\sqrt{p \cdot p}} - \left( c \sqrt{p \cdot p} + V(q) \right) = - V(q) \tag{4}$$ which is not dynamical since $$\frac{\partial \mathcal{L}}{\partial \dot{q}} = 0$$.

A similar problem arises if I consider a dual Lagrangian system in $$q, \dot{q}$$ variables and attempt to find a Hamiltonian by Legendre transformation: $$\mathcal{L} = c \sqrt{\dot{q} \cdot \dot{q}} - V(q) \tag{5}$$ then we get well-defined Euler-Lagrange equations: $$\frac{\mathrm{d}}{\mathrm{d} t} \left( c \frac{\dot{q}_i}{\sqrt{\dot{q}_i \cdot \dot{q}_i}} \right) = - \frac{\partial V(q)}{\partial q_i}\tag{6}$$ which becomes: $$(\dot{q} \cdot \dot{q}) \ddot{q}_i - (\dot{q} \cdot \ddot{q}) \dot{q}_i + (\dot{q} \cdot \dot{q})^{3/2} \frac{\partial V(q)}{\partial q_i} = 0.\tag{7}$$ However, if we try to find an associated Hamiltonian, $$H = p \cdot \dot{q} - \mathcal{L} = c \frac{\dot{q} \cdot \dot{q}}{\sqrt{\dot{q} \cdot \dot{q}}} - \left( c \sqrt{\dot{q} \cdot \dot{q}} - V(q) \right) = V(q) \tag{8}$$ which is again non-dynamical.

What is going on here? Is there an interesting reason that these systems should not admit Lagrangian/Hamiltonian descriptions? In general, when should I expect the Legendre transformation to give me a well-behaved system that reproduces the physics I started with?

• It may be related to this, since the momentum dependence is linear.
– J.G.
Jun 19, 2021 at 18:43
• Jun 19, 2021 at 19:12

1. The Lagrangian can be constructed directly by performing a Dirac-Bergmann constraint analysis of OP's Hamiltonian (1). In eq. (3) OP has already correctly identified the primary constraint$$^1$$ $$\dot{x}^2~:=~g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}~\approx ~0, \qquad \dot{x}^{\mu}~:=~\frac{dx^{\mu}}{d\tau}, \tag{A}$$ where $$\tau$$ is the world-line (WL) parameter (which does not have to be the proper time).

2. The Lagrangian becomes the massless limit of$$^2$$ $$L~=~\lambda \dot{x}^2-\frac{m^2}{4\lambda} - V,\tag{B}$$ where $$\lambda(\tau)$$ is a Lagrange multiplier, cf. e.g. this Phys.SE post.

3. The momentum for the Lagrangian is $$p_{\mu}~:=~\frac{\partial L}{\partial \dot{x}^{\mu}}~=~2\lambda g_{\mu\nu}(x)~\dot{x}^{\nu}, \tag{C}$$ so that the corresponding Hamiltonian is $$H~=~\frac{p^2+m^2}{4\lambda} + V. \tag{D}$$

4. Therefore the Hamiltonian Lagrangian becomes $$L_H~:=~ p_{\mu} \dot{x}^{\mu} - H. \tag{E}$$

5. Let us now go to the static gauge $$x^0=\tau$$. If we integrate out $$p^0$$ and $$\lambda$$, we get$$^3$$ \begin{align} \left. L_H\right|_{x^0=\tau} \quad\stackrel{p^0}{\longrightarrow}&\quad {\bf p}\cdot \dot{\bf x}- \underbrace{\left(\lambda + \frac{{\bf p}^2+m^2}{4\lambda} + V\right)}_{\text{Hamiltonian}}\cr\cr \quad\stackrel{\lambda}{\longrightarrow}&\quad {\bf p}\cdot \dot{\bf x} - \underbrace{\left(\sqrt{{\bf p}^2+m^2}+V\right)}_{\text{Hamiltonian}} .\end{align}\tag{F}

6. If we put the mass $$m\to 0$$ then the square-root Hamiltonian (F) becomes precisely OP's Hamiltonian (1). This confirms our claim that the massless limit of eq. (B) is OP's sought-for Lagrangian.

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$$^1$$ Let us work in units where the speed of light $$c=1$$ and with Minkowski sign convention $$(-,+,+,+)$$.

$$^2$$ The massterm in eq. (B) is included for generality and is not essential. The only slightly strange thing is that we restrict the $$\lambda$$ target-space from $$\mathbb{R}$$ to $$\mathbb{R}_+$$. This latter point is also discussed in my Phys.SE answer here.

$$^3$$ A similar argument was given in eq. (3) of my Phys.SE answer here, where the Lagrange multiplier $$\lambda=\frac{1}{2e}$$ is replaced by an einbein field $$e$$.