This question is related (but not similar) to this old one of mine:

How to derive the two Friedmann-Lemaître equations from a Lagrangian?

Consider the Lagrangian of an isotropic-homogeneous spacetime (Robertson-Walker metric), containing a simple scalar field and a cosmological constant (this expression comes from the standard Hilbert-Einstein and scalar field action, in an isotropic-homogeneous spacetime. The space coordinates were integrated out and an hypersurface term was discarded) : \begin{equation}\tag{1} \mathcal{L} = -\: \frac{1}{8 \pi G} (3 \, a \, \dot{a}^2 - 3 k \, a + \Lambda \, a^3) + \frac{1}{2} \; \dot{\phi}^2 a^3 - \mathcal{V}(\phi) \, a^3. \end{equation} The function $\mathcal{V}(\phi)$ is the potential energy density of the scalar field. The action is simply $$\tag{2} S = \int_{t_1}^{t_2} \mathcal{L} \, dt. $$ Just for reference, the hypersurface term that was neglected is the following: $$\tag{1b} \mathcal{L}_{\text{surf}} =\frac{1}{8 \pi G} \, \frac{d}{dt} ( 3 \, a^2 \, \dot{a}). $$ It's easy to find the Hamiltonian: \begin{equation}\tag{3} \mathcal{H} \equiv \dot{a} \, \frac{\partial \mathcal{L}}{\partial \, \dot{a}} + \dot{\phi} \, \frac{\partial \mathcal{L}}{\partial \, \dot{\phi}} - \mathcal{L} = -\: \frac{3}{8 \pi G} \Big( \frac{\dot{a}^2}{a^2} + \frac{k}{a^2} - \frac{\Lambda}{3} \Big) \, a^3 + \Big( \frac{1}{2} \, \dot{\phi}^2 + \mathcal{V}(\phi) \Big) \, a^3. \end{equation} That Hamiltonian can be proved to be 0, by a transformation of the time variable $dt \Rightarrow N \, dt$, where $N$ is an arbitrary function of $t$ (the "lapse" function). This changes the lagrangian (1): \begin{equation}\tag{4} \tilde{\mathcal{L}} = -\: \frac{1}{8 \pi G} (3 \, a \, \dot{a}^2/N - 3 k \, a N + \Lambda \, a^3 N) + \frac{1}{2} \; \dot{\phi}^2 a^3 / N - \mathcal{V}(\phi) \, a^3 N. \end{equation} Since $N$ is arbitrary, we then could consider it as a new variable that can be varied. The Euler-Lagrange equation applied to the lapse function implies that $\mathcal{H} = 0$.

What confuses me is that this reasoning could also be applied to any other Lagrangian, starting from its action and applying an arbitrary time transformation that doesn't change the end points: $$\tag{5} S = \int_{t_1}^{t_2} L \Big(q, \frac{dq}{dt} \Big) \, dt = \int_{t_1}^{t_2} L \Big(q, \frac{1}{N} \, \frac{dq}{d\tilde{t}} \Big) \, N \, d\tilde{t}, $$ so that $$\tag{6} \tilde{L} = N \, L \Big(q, \frac{1}{N} \, \frac{dq}{d\tilde{t}} \Big). $$ Applying the Euler-Lagrange equation to $N$ (which is still arbitrary) then gives an absurdity: $H = 0$ for any Lagrangian! Of course, this cannot be true!

So two questions:

  1. Where did I made a stupid mistake in this reasoning? It certainly should be obvious, but I don't see it!
  2. I often read that $H = 0$ comes from the parameterization invariance of the action. But then, like most classical Lagrangians, (1) and (2) don't seem to be independent of the time parameterization (even if I bring back the surface term that was discarded at the beginning). So how can we show that (1)-(2) are actually independent of the time parameterization and show the relation to the Hamiltonian being zero?
  • 1
    $\begingroup$ A comment: the cosmology stuff is not very relevant to the actual question, so it might be clearer to edit it down a bit. $\endgroup$ – Javier Jun 21 at 13:58
  1. First of all, reparametrizing the Lagrangian 1-form$^1$ $$\mathbb{L}_1~=~L_1\left(q,\frac{dq}{d\tilde{t}}\right)\mathrm{d}\tilde{t} ~=~L_1\left(q,\frac{1}{N}\frac{dq}{dt}\right)N\mathrm{d}t \tag{A}$$ using a fixed function $N$ $$ \frac{d\tilde{t}}{dt}~=~N(t)\tag{B} $$ does not change the model.

  2. However, if we promote the lapse function $N$ as a new variable to be varied in the stationary action principle with new Lagrangian $$L(q,\dot{q},N)~=~NL_1\left(q,\frac{\dot{q}}{N}\right),\tag{C} $$ then we do change the model, cf. Javier's answer. For starters we get an algebraic EOM for $N$. Also the new Lagrangian (C) has a local worldline (WL) reparametrization gauge symmetry.

  3. Example: $$L_1(q,\dot{q})~=~\frac{m}{2}\dot{q}^2-V(q).\tag{D}$$ The energy function $$ h_1(q,\dot{q}) ~:=~\left( \dot{q}\frac{\partial}{\partial\dot{q}}-1\right) L_1 ~\stackrel{(D)}{=}~ \frac{m}{2}\dot{q}^2+V(q) \tag{E}$$ does in general not vanish. Then the new Lagrangian becomes $$L(q,\dot{q},N)~\stackrel{(C)+(D)}{=}~\frac{m}{2N}\dot{q}^2 - N V(q).\tag{F} $$ The algebraic EOM for $N$ is$^2$ $$ N^2~\approx~ \frac{m\dot{q}^2}{-2V}.\tag{G} $$ The first lesson is that we must demand that the potential $V<0$ is negative in order to have solutions to eq. (G). Secondly, we declare that the lapse function $N>0$ is positive in order to avoid an unphysical negative square root branch. If we integrate out $N$, the Lagrangian (F) takes a square root form: $$ L_0(q,\dot{q})~\stackrel{(F)+(G)}{=}~\sqrt{-2Vm\dot{q}^2},\tag{H}$$ which still has WL reparametrization gauge symmetry. The corresponding energy function $$ h_0(q,\dot{q}) ~:=~\left( \dot{q}\frac{\partial}{\partial\dot{q}}-1\right) L_0 ~\stackrel{(H)}{=}~0, \tag{I}$$ vanishes identically, cf. e.g. this Phys.SE post. This should be contrasted with the energy function for the new Lagrangian (F), $$ \begin{align} h(q,\dot{q},N) ~:=~&\left( \dot{q}\frac{\partial}{\partial\dot{q}}+\dot{N}\frac{\partial}{\partial\dot{N}}-1\right) L\cr ~\stackrel{(F)}{=}~& \frac{m}{2N}\dot{q}^2 + N V(q) ~\stackrel{(G)}{\approx}~ 0, \end{align} \tag{J}$$ which only vanishes on-shell.


$^1$ For later convenience we have exchanged the notation of $t\leftrightarrow \tilde{t}$ as compared to OP, but that's immaterial.

$^2$ The $\approx$ symbol means equality modulo EOM.

  • $\begingroup$ Excellent answer! Thanks! $\endgroup$ – Cham Jun 23 at 13:57

There's no mistake, everything is correct. The thing is that by considering $N$ to be a dynamical variable, you've changed the Lagrangian, by introducing gauge invariance - your action is now time reparametrization invariant. And the Hamiltonian of a reparametization invariant action is zero, by the simple argument you've given.

This makes sense since the Hamiltonian is linked to the time variable, so to speak: it generates time translations. And since you've effectively gotten rid of the physical time by introducing the lapse function, the Hamiltonian is now zero. It's not absurd, and the equations of motion still work.

  • $\begingroup$ Then, there's something I don't understand clearly. In the case of a simple classical action and Lagrangian ((5) and (6) above), can you show explicitly that the equations of motion still work? If I take an harmonic oscillator for example, $H = 0$ imposes a single trivial solution (i.e. no motion at all!). This doesn't make any sense to me. Introducing $N$ into the Lagrangian makes the time variable a dynamical field. This shouldn't be allowed in a classical theory. $\endgroup$ – Cham Jun 21 at 14:15
  • $\begingroup$ I now think that my last comment is a clue to my misunderstanding... (5) should not be permitted in a classical setting, since time is an "exterior" variable, not a dynamical field.. $\endgroup$ – Cham Jun 21 at 14:19
  • $\begingroup$ By introducing the new variable, you've changed the Hamiltonian, by subtracting it from itself! It's the new Hamiltonian that is identically zero, not the old one. $\endgroup$ – Javier Jun 21 at 15:34
  • $\begingroup$ Am I right in saying that the transformation $dt \Rightarrow N dt$ is "illegal", or forbidden, in a classical (i.e non-relativistic) theory, since time is considered as an absolute and immutable parameter? In other words, (5)-(6) aren't permitted in a classical theory, right? $\endgroup$ – Cham Jun 22 at 0:56
  • $\begingroup$ It's not illegal, but it changes the theory by adding a dynamical variable. Lagrangian mechanics is just math, it doesn't care about what your variable $t$ represents. And the same thing will happen in a relativistic theory, the vanishing of the Hamiltonian is because of reparametrization invariance. $\endgroup$ – Javier Jun 22 at 1:58

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