Inspiration for this question: a this thread made me wonder if there is more to a story of a Lagrangian being equal to one of its first integral than "coincidence".


I know that the Schwarzschild metric is:

$$d s^{2}=c^{2}\left(1-\frac{2 \mu}{r}\right) d t^{2}-\left(1-\frac{2 \mu}{r}\right)^{-1} d r^{2}-r^{2} d \Omega^{2}$$

I know that if I divide by $d \lambda^2$, I obtain the Lagrangian:

$$ L=c^{2}\left(1-\frac{2 \mu}{r}\right) \dot{t}^{2}-\left(1-\frac{2 \mu}{r}\right)^{-1} \dot{r}^{2}-r^{2} \dot{\theta}^{2}-r^{2} \sin ^{2} \theta \dot{\phi}^{2} $$

(where we have also expanded $\Omega^{2}$ into $\theta$ and $\phi$ dependent parts but that's not tha main point).

Overdots denote differentiation with respect to affine parameter $\lambda$. Lets set $\theta=\pi/2$ for the remainder of this post.

The interesting bit

As we can see, the Lagrangian is not dependent on the affine parameter ($\lambda$) explicitly. So $\dot{t}\frac{\partial L}{\partial \dot{t}} - L=\operatorname{const}$ should hold.

We have:

$$\frac{\partial L}{\partial t}=2 c^{2}\left(1-\frac{2 H}{r}\right) \dot{t}$$

Substitute in to $\dot{t}\frac{\partial L}{\partial \dot{t}} - L=\operatorname{const}$:

$$\dot{t}\frac{\partial L}{\partial \dot{t}} - L = \dot{t} 2 c^{2}\left(1-\frac{2 H}{r}\right) \dot{t} - c^{2}\left(1-\frac{2 \mu}{r}\right) \dot{t}^{2}-\left(1-\frac{2 \mu}{r}\right)^{-1} \dot{r}^{2}-r^{2} \dot{\phi}^{2}=\operatorname{const}$$


$$c^{2}\left(1-\frac{2 H}{r}\right) \dot{t}^2 -\left(1-\frac{2 \mu}{r}\right)^{-1} \dot{r}^{2}-r^{2} \dot{\phi}^{2}=\operatorname{const}$$

Notice that the LHS just gave us back $L$!


Is this "coincidence", ie that $\dot{t}\frac{\partial L}{\partial \dot{t}} - L = L$ expected? or maybe there a property of Schwarzschild spacetime specifically which would make us expect this, or perhaphs is the Lagrangian derived from Schwarzschild solution is a member of a wider group (not necessarily in the mathematical sense) of Lagrangians for which this is true? It would be great if I would see why $\dot{t}\frac{\partial L}{\partial \dot{t}} - L= L$ is true while $\dot{r}\frac{\partial L}{\partial \dot{r}} - L\neq L$.


2 Answers 2


To answer your question, the way this Lagrangian problem is posed is special. As a counterexample to show what's going on, consider the Euclidean signature metric $ds^2=dx^2+x^2dy^2$. The Lagrangian in an arbitrary parametrization is not an integral of motion. To demonstrate, the arclength functional in this metric is


The EL equations read

$$\ddot{x}=x\dot{y}^2, ~~\ddot{y}=0$$

Using these we see that


which, given that $\dot{y}$ is constant, is a total derivative, but since $x(t)$ is decidedly non-constant it has a non trivial time dependence and therefore is not conserved.

Now I would like to expose a few points that get glossed over in the average general relativity course. Because of the reparametrization invariance of the general metric space length functional

$$L=\int\mathcal{L}dt=\int dt \sqrt{g_{\mu\nu}(dx^{\mu}/dt)(dx^{\nu}/dt)}$$

it is always possible to recast the problem as a minimization problem with constraints. In fact in Euclidean metrics where $g_{\mu\nu}(x)>0$, there always exists a parametrization (the arclength parametrization) such that $g_{\mu\nu}(dx^{\mu}/ds)(dx^{\nu}/ds)=1$ and then the geodesic problem can be rewritten as the minimization of the following functional

$$L'[\lambda(s), x^\mu(s),dx^{\mu}/ds]=\int ds-\int ds\mu(s)(g_{\mu\nu}(dx^{\mu}/ds)(dx^{\nu}/ds)-1)$$

which contains a Lagrange multiplier to enforce the constraint. Equivalently one can rephrase this in terms of a constrained minimization (which is the one you are using) $$\int ds, ~~\mathcal{L}^2=1\iff L''(x,\dot{x})=\int \mathcal L^2 ds, ~~\mathcal{L}^2=1$$

Thus in this new formulation $L''$, the Lagrangian squared is the constraint itself (despite the fact that it is not necessarily conserved for all parametrizations), and sometimes in my experience physicists tend to confuscate the two. Nevertheless, this formulation of the problem is useful because the constraint is first order in the derivatives, and there is no annoying square root anymore-which is calculationally a blessing despite the two problems being equivalent.

All this is true in Euclidean spaces, where the metric is positive definite. However in Minkowski spaces, one encounters the problem that this functional can be negative. Nevertheless, we know that massive particles have $ds^2>0$, and massless particles must obey $ds^2=0$ along their trajectories (presumably tachyons have $ds^2<0$). Therefore we have to distinguish cases for the different kinds of particles. In the massive case, we can follow the steps of the Euclidean case and minimize the functional

$$L''=\int{\mathcal{L}^2}d\tau~, ~\mathcal{L}^2=c^2$$

For massless particles we need to devise a better way to simplify the problem since $\mathcal{L}^2=0$ and the way to define an affine parameter is by demanding that parallel transport keeps you along the lightlike geodesics.

Now all that being said, it shouldn't come as a surprise that with an affine parameter, the Lagrangian you have defined is an integral of motion. The last remark I would like to make is that when the Lagrangian doesn't depend on the affine parameter (which is as general as it gets in GR), the conserved quantity is the Hamiltonian defined as

$$\mathcal{H}=\sum_{\lambda}\dot{x}_{\lambda}\frac{\partial \mathcal{L}^2}{\partial \dot{x}_\lambda}-\mathcal {L}^2$$

and then you can show that given the form of $\mathcal{L}$, always, (exercise) $\mathcal{H}=\mathcal{L}$!

There is no new information on the conserved quantity there, just a validation that in this minimization problem we defined where the constraint is the Lagrangian itself, it is conserved, which is a pretty trivial fact in itself.

I hope this helps resolve the confusion!


Well, it is special that the Lagrangian equals a integral of motion. In this case, it happens because the Euler-Lagrangian equations derived give geodesics in that spacetime. The affine parameters used are of physical significance (for example, we often use the proper time for a massive particle. That conservation law will tell proper time experienced when a particle travels through that path is related to the speed of light), so that is significant. As for which formula will give that consevation...well, not so much. You could find infinitely many constants of motion, but only a few will be independent (four in the case of Schwarschild). So a certain combination of these constants will be the Lagrangian.

Edit: I originally claimed that the formula in the question would indeed give the Lagrangian, but that was a mistake. In fact, the Lagrangian will be a constant of the motion, but not one that is given by these formulas (which, of course, does not need to be the case for every integral of motion).

  • $\begingroup$ @zabop did this answer solve the problem? $\endgroup$
    – Othin
    Apr 3, 2021 at 22:42
  • $\begingroup$ Yeah, its great, just wanted to give time if others also want to add something. $\endgroup$
    – zabop
    Apr 4, 2021 at 11:18
  • 1
    $\begingroup$ Oh wait, it didn't, $2f(r,\theta,\phi)\dot{t}^2 - \left(f(r,\theta,\phi)\dot{t}^2 + g(r,\theta, \phi)\right) = f(r,\theta,\phi)\dot{t}^2 - g(r,\theta, \phi)$. Sign of $g$ is wrong. $\endgroup$
    – zabop
    Apr 4, 2021 at 11:35
  • $\begingroup$ @zabop yeah, you're right, I made a mistake and thought that constant of motion was the Lagrangian. In fact, you calculated $\dot{t}\frac{\partial L}{\partial \dot{t}} - L$ in the other question, so by now you've noticed it is not the Lagrangian. The Lagrangian is indeed a constant of the motion, but it i given by some linear combination of the other ones. $\endgroup$
    – Othin
    Apr 4, 2021 at 20:18

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