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The background:

If we have a spacetime path $x^\mu(t)$ parameterized by arbitrary parameter $t$, the proper time along the path between $t_1$ and $t_2$ is $$ \int_{t_1}^{t_2} (g_{\mu \nu} \dot x^\mu \dot x^\nu)^{1/2}dt. $$ This action has a reparameterization symmetry. If we take the Lagrangian to be $$L = (g_{\mu \nu} \dot x^\mu \dot x^\nu)^{1/2}$$ then the Euler Lagrange equation ends up being (after multiplying by the inverse metric) $$ \ddot x^\mu = - \Gamma^\mu_{\; \beta \gamma} \dot x^\beta \dot x^\gamma + \dot x^\mu \frac{d}{dt} \ln(L). $$ If we use an affine parameter $t$ such that $L$ is constant along the path, then this is just the regular geodesic equation.

We can see that the reparameterization (gauge) symmetry is a huge pain. For example, it seems to render the Hamiltonian to be $0$.

\begin{align*} H &= \Big( \frac{\partial L}{\partial\dot x^\mu} \Big)\dot x^\mu - L \\ &= 2\frac{1}{2}\frac{g_{\mu \nu} \dot x^\nu}{(g_{\alpha \beta} \dot x^\alpha \dot x^\beta )^{1/2}} \dot x^\mu - (g_{\mu \nu} \dot x^\mu \dot x^\nu)^{1/2} \\ & = 0. \end{align*} This seems to be related to the reparameterization symmetry because $H$ should generate time translations. However, if time evolution is not deterministic, then there's nothing for $H$ to reasonably be.

We would therefore like to find a less "pathological" Lagrangian and Hamiltonian with the same equations of motion, but in an automatically affinely parameterized form. The answer is to take $$ L = \frac{1}{2} g_{\mu \nu} \dot x^\mu \dot x^\nu \implies H = \frac{1}{2} g^{\mu \nu} p_{\mu} p_{\nu}. $$ Note that $L_{\rm new} = \tfrac{1}{2}L_{\rm old}^2$. The Euler Lagrange equations are indeed the affinely parameterized geodesic equations. The same goes for Hamilton's equations $$ \dot x^\mu = \frac{\partial H}{\partial p_{\mu}} , \hspace{1 cm} \dot p_{\mu} = - \frac{\partial H}{\partial x^{\mu}} \implies \ddot x^\mu = - \Gamma^\mu_{\; \beta \gamma} \dot x^\beta \dot x^\gamma. $$

The question:

It seems like a complete coincidence that squaring the Lagrangian gives us sort of "gauge fixed" version of our original Lagrangian. Is there a principled, philosophical approach we can take to move from $L_{\rm old}$ to $L_{\rm new}$, rather than just seeing that the equations of motion give us what we want? Is this a special case of a more general procedure?

Birrabenzina
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user1379857
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  • Are you aware of the same difference appearing between the Nambu-Goto action and the Polyakov action? – octonion Oct 23 '19 at 19:38
  • Yes, which is part of the reason I am interested in the answer. However, even that procedure, as I have seen it described, is a little "ad hoc." – user1379857 Oct 23 '19 at 19:39
  • Well, minimizing the proper time is the same as the proper time squared. But the proper time squared action is just the ordinary Lagrangian for a free particle (the kinetic term) generalized to a curved background. You can think of this as a non-linear sigma model too, if you like – octonion Oct 23 '19 at 19:59
  • Well, the integral of the (proper times) squared is different from (the integral of the proper time)^2 – user1379857 Oct 23 '19 at 20:02
  • Yes the quantum theory built on the two is different, but the saddle point is the same – octonion Oct 23 '19 at 20:02

1 Answers1

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  1. Physicists conventionally normalize the square root Lagrangian a bit differently, namely as$^1$ $$L_0~:=~ -m\sqrt{-\dot{x}^2}, \qquad \dot{x}^2~:=~g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}~<~0,\tag{1}$$ where $m>0$ is the mass. (OP assumes that $m=1$.)

  2. It is important to note that the Legendre transformation of the square root Lagrangian (1) is singular: the momentum has to satisfy the mass-shell constraint $$ p^2+m^2~\approx~0 , \qquad p^2~:=~g^{\mu\nu}(x)~ p_{\mu}p_{\nu}.\tag{2} $$ Therefore, even though OP is correct that the original Hamiltonian is zero because of worldline (WL) reparametrization invariance, the full Hamiltonian $$ H~=~ \frac{e}{2}(p^2+m^2)\tag{3} $$ becomes a Lagrange multiplier $e$ times the mass-shell constraint (2).

  3. The inverse Legendre transformation of the Hamiltonian (3) leads to the Lagrangian $$L~=~\frac{\dot{x}^2}{2e}-\frac{e m^2}{2}.\tag{4}$$ If we integrate out the $e$ field, the Lagrangian (4) becomes the square root Lagrangian (1), cf. e.g. this related Phys.SE post.

  4. The main point is now that OP's quadratic Hamiltonian and Lagrangian are the $e=1$ gauge of the Hamiltonian (3) and the Lagrangian (4), respectively, up to irrelevant constant terms. In that sense they follow systematically from the Dirac-Bergmann algorithm for constrained systems.

  5. For further details, see e.g. this related Phys.SE post.

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$^1$ We use the sign convention $(-,+,+,+)$, and put the speed-of-light $c=1$.

Qmechanic
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  • But if we wish to fix the constraint $(p^2 + m^2) = 0$, wouldn't adding any function of that constraint to the Lagrangian, $ef(p^2 + m^2)$, do the trick? Why not use $(p^2 + m^2)^{1/2}$ for instance? Isn't there something special about the squared Langrangian? – user1379857 Oct 23 '19 at 21:18
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    Note that the mass-shell constraint (2) is already automatically satisfied in the Lagrangian formalism that is based on the the square root Lagrangian (1). – Qmechanic Oct 23 '19 at 22:38