Show that if there are $M$ independent constraints $\phi_m(x_\mu,p_\mu)$ there are $M$ of the $\ddot{x}_i$'s that the Euler-Lagrange equations cannot be solved for.
Attempt of solution:
Assume that we are in a $d$-dimensional spacetime (i.e. $\mu = 0,...,d-1$) and $m=1,...,M$.
Define $M_{\mu\nu}:=\frac{\partial^2\mathcal{L}}{\partial \dot{x}^\nu \partial \dot{x}^mu}=\frac{\partial}{\partial \dot{x}^\nu}p_\mu$. Then the Euler Lagrange equations are: \begin{equation} M_{\mu\nu}\ddot{x}^\nu+\frac{\partial^2\mathcal{L}}{\partial x^\nu\partial\dot{x}^\mu} \dot{x}^\nu-\frac{\partial \mathcal{L}}{\partial x^\mu}=0 \end{equation} and if $M_{\mu \nu}$ is invertible (i.e. has full rank) I can solve for all the $\ddot{x}_i$'s.
However, having constraints is equivalent to ask that \begin{equation} 0=\phi_m(x,p(x,\dot{x})) \end{equation} for all $x$ and $\dot{x}$. Hence, taking an infinitesimal transformation of the $x$ we need that $\frac{\partial \phi_m}{\partial \dot{x}^\sigma} = 0$, but this is equivalent to \begin{equation} \frac{\partial \phi_m}{\partial p_\mu}\frac{\partial p_\mu}{\partial \dot{x}^\sigma} = \frac{\partial \phi_m}{\partial p_\mu}M_{\mu \sigma} = 0. \end{equation} Now wlog,
- $\frac{\partial \phi_m}{\partial p_\mu} = 0$ for m=1,...,k. This implies that $\phi_m = \phi_m(x)$ and therefore I can express $x^j=f^j(x^\alpha)$, with $j=0,...,k-1$ and $\alpha=k,...,d-1$.
- $\frac{\partial \phi_m}{\partial p_\mu} \neq 0$ for $m=k+1,...,M$.
Now I'm stucked. I need to show the the rank of $M_{\mu\nu} = d-m$ in order to solve the problem, but I don't understand how.
