I'm trying to understand something with the lagrangian and the hamiltonian formalisms in relativity theory, and why the following result cannot be the same in classical (non-relativistic) mechanics. There's something missing or badly defined in my reasoning, and I don't see what yet.

Consider a system of "particles", of generalized coordinates $q^i(t)$ in some reference frame. The *action* of the system is defined as the following integral :

\begin{equation}\tag{1}

S = \int_{t_1}^{t_2} L(q^i, \dot{q}^i) \, dt.

\end{equation}

The *hamiltonian* of the system is defined as this (summation is implied on the repeated indices) :

\begin{equation}\tag{2}

H = \dot{q}^i \, \frac{\partial L}{\partial \dot{q}^i} – L.

\end{equation}

Now consider a change of parametrization in integral (1) ; $dt = \theta \, ds$, where $s$ is a new integration variable and $\theta(s)$ is an arbitrary function that could be considered as a new dynamical variable (there's something missing in my interpretation, and I need to find what is "wrong" here). The action is now this (the prime is the derivative with respect to $s$. Notice the change of limits, on this integral) :

\begin{equation}\tag{3}

S = \int_{s_1}^{s_2} L(q^i, \frac{q^{i \, \prime}}{\theta}) \; \theta \; ds.

\end{equation}

So the new lagrangian is

\begin{equation}\tag{4}

\tilde{L} = L(q^i, \frac{q^{i \, \prime}}{\theta}) \; \theta.

\end{equation}

Now, if $\theta$ is considered as a dynamical variable, we can apply the Euler-Lagrange to this new lagrangian:

\begin{equation}\tag{5}

\frac{d}{d s} \Big( \frac{\partial \tilde{L}}{\partial \, \theta^{\, \prime}} \Big) – \frac{\partial \tilde{L}}{\partial \, \theta} = 0.

\end{equation}

Since there is no $\theta^{\, \prime}$ in $\tilde{L}$, the first part is 0. After some simple algebra, the second part implies that the hamiltonian (2) should be 0 !

\begin{equation}\tag{6}

H \equiv 0.

\end{equation}

So my questions are these:

Is there something wrong in the previous reasoning? Or what

am I missing? Where is relativity in this?implicit assumptionsIf the reasoning is valid, why can't we apply the result to any classical lagrangian, which would give non-sense?

Of course, I know that the hamiltonian of free relativistic particles isn't 0! But I also know that $H = 0$ is a well known property of systems which have an action with *parametrization independence*. I'm missing some parts related to this constraint and I don't see what yet. I need help to disentangle this subject.

## Best Answer

onlyif the $q,p$ transform as scalars under the time-reparametrization $t\mapsto \tau(t)$ (since then $\dot{q}^i p_i \mapsto \frac{\mathrm{d}\tau}{\mathrm{d}t} q'^i p_i$ (the prime $'$ denotes the derivative w.r.t. $\tau$)) and if the constraint terms $u^\alpha\chi_\alpha$ also transform as scalar densities $u^\alpha \chi_\alpha \mapsto \frac{\mathrm{d}\tau}{\mathrm{d}t} u^\alpha \chi_\alpha$. Under these assumptions, $H=0$ follows because if $q,p$ are scalars then so is $H(q,p)$, and $H(q,p)$ therefore cannot transform as a scalar density to preserve invariance of the action, and must be zero.Note that the derivative in $\mathrm{d}t = \frac{\mathrm{d}t}{\mathrm{d}\tau}\mathrm{d}\tau$ (which is called $\theta$ in the question) is

not a dynamical variablefrom the viewpoint of the original system. Your Euler-Lagrange equation does not exist because it is not a function/coordinate on phase space.Note also that, if $s_1 \neq t_1$ and $s_2\neq t_2$, the transformation $t\mapsto s$ you perform is not what is generally called a

time-reparametrizationin the context where we talk about reparametrization invariance and vanishing Hamiltonians. It is usually assumed that the start and end of the parameter stay fixed.Your new system $\bar{L}$ is not equivalent to the original one: You can, however, define a new Lagrangian system $\bar L(q,q',\theta,{\theta}') := L(q,q'/\theta)\theta$, with an additional variable $\theta$, if you so desire. Let's examine that new system a bit:

The equations of motion for $q$ are $$ \frac{\mathrm{d}}{\mathrm{d}s}\left(\frac{\partial\bar{L}}{\partial q'}\right) - \frac{\partial\bar{L}}{\partial q} = \frac{\mathrm{d}}{\mathrm{d}s}\left(\frac{\partial(L\theta)}{\partial q'}\right) - \frac{\partial(L\theta)}{\partial q} = \frac{\mathrm{d}}{\mathrm{d}s}\left(\frac{\partial{L}}{\partial q'}\right)\theta + \frac{\partial L}{\partial q'}\theta' - \frac{\partial{L}}{\partial q}\theta = 0\tag{a}$$ and the equation of motion for $\theta$ is $$ \frac{\partial \bar{L}}{\partial \theta} = \frac{\partial L}{\partial \theta} \theta + L = 0.\tag{b}$$ Here, crucially, $L$ has to be read as the $L(q,q'/\theta)$ it appears as in the $\bar{L}$, so we have $$ \frac{\partial L}{\partial q'} = \frac{1}{\theta} \frac{\partial L}{\partial \dot{q}}\quad\text{and}\quad \frac{\partial L}{\partial \theta} = \frac{\partial (q'/\theta)}{\partial \theta}\frac{\partial L}{\partial \dot{q}} = -\frac{q'}{\theta^2}\frac{\partial L}{\partial \dot{q}} \tag{c}$$ in terms of the $\partial L /\partial \dot{q}$ that appears in the original E-L equations. Notably, eq. (b) indeed implies that $\dot{q}\frac{\partial L}{\partial \dot{q}} - L = 0$

but onlyunder the assumption that $q'/\theta = \dot{q}$. However, the moment we promoted $\theta$ to a dynamical variable, welost that relation- the object $\dot{q}$ does not exist in this new world anymore, the $\partial L/\partial \dot{q}$ is a function of $q,q'$ and $\theta$, not one of $q$ and $\dot{q}$. From the viewpoint fo the new system it should be read simply as $\frac{\partial L}{\partial(q'/\theta)}$.Let us now plug eq. (c) into eq. (a): $$ \frac{\mathrm{d}}{\mathrm{d}s}\left(\frac{1}{\theta} \frac{\partial L}{\partial \dot{q}}\right)\theta + \frac{1}{\theta} \frac{\partial L}{\partial \dot{q}}\theta' - \frac{\partial{L}}{\partial q}\theta = -\frac{\theta'}{\theta}\frac{\partial L}{\partial \dot{q}} + \frac{\mathrm{d}}{\mathrm{d}s}\left(\frac{\partial{L}}{\partial \dot{q}}\right) + \frac{\theta'}{\theta}\frac{\partial{L}}{\partial \dot{q}}- \frac{\partial{L}}{\partial q}\theta = 0$$ $$ \implies \frac{\mathrm{d}}{\mathrm{d}s}\left(\frac{\partial{L}}{\partial \dot{q}}\right) - \frac{\partial{L}}{\partial q}\theta = \frac{1}{\theta}\frac{\mathrm{d}}{\mathrm{d}s}\left(\frac{\partial{L}}{\partial \dot{q}}\right) - \frac{\partial{L}}{\partial q} = 0\tag{d}$$ If the two Lagrangian systems $L(q,\dot{q})$ and $\bar{L}(q,q',\theta,\theta')$ were equivalent, the equations of motion of $\bar{L}$ should reduce to the equations of motion of $L$ upon use of the equation of motion for $\theta$. But eq. (b) does not make that possible - inserting it into eq. (d) does not lend itself to simplications by which we would obtain the original equations of motion. The relation we

actuallyneed would be, once again, $q'/\theta = \dot{q}$.Here is the Lagrangian argument for how reparametrization invariance implies vanishing Hamiltonian: If the original action was time-reparameterization invariant, we have that, under an infintesimal reparametrization $\delta t = \theta(t)$ with induced changes $$\delta q = \dot{q}\theta \quad \delta \dot{q} = \dot{\delta q} = \ddot{q}\theta + \dot{q}\dot{\theta} \quad \theta(t_1) = \theta(t_2) = 0 $$ the change induced in the Lagrangian is zero, i.e. $$ \delta L = \frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial \dot{q}}\delta \dot{q} = 0.$$ But, due to a "trick" to derive the Noether current discussed, for example, in this question, we also know that $$ \delta S = \int j \frac{\partial \epsilon}{\partial t},$$ where $j$ is the Noether current associated to the symmetry with constant $\epsilon$, which is just time translation, and indeed a symmetry because we have implicitly assumed that the Lagrangian does not explicitly depend on time. Therefore, $j$ is the Hamiltonian, and reparametrization invariance obviously forces $j=0$.