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1. The Lagrangian can be constructed directly by performing a Dirac-Bergmann constraint analysis of OP's Hamiltonian (1). 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 $$\begin{align} \left. L_H\right|_{x^0=\tau} \quad\stackrel{p^0}{\longrightarrow}&\quad {\bf p}\cdot \dot{\bf x}- \underbrace{\left(\lambda + \frac{{\bf p}^2+m^2}{4\lambda} + V\right)}_{\text{Hamiltonian}}\cr\cr \quad\stackrel{\lambda}{\longrightarrow}&\quad {\bf p}\cdot \dot{\bf x} - \underbrace{\left(\sqrt{{\bf p}^2+m^2}+V\right)}_{\text{Hamiltonian}} .\end{align}\tag{F} $$ A similar argument was given in eq. (3) of my Phys.SE answer here.

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

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.

--

$^1$ 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 $e$.

Let us work in units where the speed of light $c=1$.

1. OP's Hamiltonian is then just the massless version of the standard Hamiltonian $$\sqrt{{\bf p}^2+m^2} +V$$ for a relativistic point particle. It is known to be the Legendre transformation of the standard Lagrangian $$L~=~\frac{\dot{x}^2}{2e}-\frac{e m^2}{2}-V$$ for a relativistic point particle, cf. eq. (3) in my Phys.SE answer here. In particular, the Lagrangian does exist.

2. The Lagrangian can also be constructed directly by performing a Dirac-Bergmann constraint analysis of OP's Hamiltonian. OP has already correctly identified the primary constraint.

1. The Lagrangian can be constructed directly by performing a Dirac-Bergmann constraint analysis of OP's Hamiltonian (1). 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 $$\begin{align} \left. L_H\right|_{x^0=\tau} \quad\stackrel{p^0}{\longrightarrow}&\quad {\bf p}\cdot \dot{\bf x}- \underbrace{\left(\lambda + \frac{{\bf p}^2+m^2}{4\lambda} + V\right)}_{\text{Hamiltonian}}\cr\cr \quad\stackrel{\lambda}{\longrightarrow}&\quad {\bf p}\cdot \dot{\bf x} - \underbrace{\left(\sqrt{{\bf p}^2+m^2}+V\right)}_{\text{Hamiltonian}} .\end{align}\tag{F} $$ A similar argument was given in eq. (3) of my Phys.SE answer here.

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

--

$^1$ 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.

Let us work in units where the speed of light $c=1$.

1. OP's Hamiltonian is then just the massless version of the standard Hamiltonian $$\sqrt{{\bf p}^2+m^2} +V$$ for a relativistic point particle. It is known to be the Legendre transformation of the standard Lagrangian $$L~=~\frac{\dot{x}^2}{2e}-\frac{e m^2}{2}-V$$ for a relativistic point particle, cf. eq. (3) in my Phys.SE answer here. In particular, the Lagrangian does exist.

2. The Lagrangian can also be constructed directly by performing a Dirac-Bergmann constraint analysis of OP's Hamiltonian.

Let us work in units where the speed of light $c=1$.

1. OP's Hamiltonian is then just the massless version of the standard Hamiltonian $$\sqrt{{\bf p}^2+m^2} +V$$ for a relativistic point particle. It is known to be the Legendre transformation of the standard Lagrangian $$L~=~\frac{\dot{x}^2}{2e}-\frac{e m^2}{2}-V$$ for a relativistic point particle, cf. eq. (3) in my Phys.SE answer here. In particular, the Lagrangian does exist.

2. The Lagrangian can also be constructed directly by performing a Dirac-Bergmann constraint analysis of OP's Hamiltonian. OP has already correctly identified the primary constraint.