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Let us denote $L$ the Lagrangian, and $\mathcal{L}$ the Lagrangian density, and the action $S$.

It is common to find the action based on the Lagrangian. Here, however, I am interested in the reverse problem.

Going from $S$ to $L$ my intuition is as follows: one derives $S$ with respect to $t$

$$ S=\int L dt \implies \frac{dS(t)}{dt}=L $$

However, this does not appear correct to me because the relation is actually $S=\int_a^b Ldt$ (a definite integral).

Can someone chip in as to if this is valid for definite integrals?

It is even less clear to me how one goes from $S$ to $\mathcal{L}$. I know how to get $S$ from $\mathcal{L}$:

$$ S=\int_M \mathcal{L}\sqrt{-g}d^nx $$

where $M$ is an $n$-dimensional manifold.

But, what operation takes me from $S$ to $\mathcal{L}$?

Qmechanic
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Anon21
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  • Have you ever seen an action where a Lagrangian density wasn’t just sitting there inside the integral? The “operation” is “erase the integral sign and the $\sqrt{-g}d^nx$”. How can you have an explicit functional without an explicit function being integrated? – G. Smith Aug 22 '19 at 00:17

2 Answers2

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  1. The action is a functional on paths in configuration space, i.e. $$ S[q(t)] = \int^b_a L(q(t), \dot{q}(t), t)\mathrm{d}t,$$ see also the last paragraphs of this answer of mine. Your expression doesn't make any sense because the action is not a function of $t$ and hence you cannot differentiate it with respect to time.

  2. What you can take is its variation, and the condition that the variation of the action vanishes is precisely the principle of extremal action and yields the Euler-Lagrange equations, see e.g. Wikipedia.

  3. You cannot derive uniquely a Lagrangian from an action because there are infinitely many Lagrangians with the same actions - adding a total derivative with respect to time to the integrand does not change the equations of motion, i.e. $ \int L\mathrm{d} t$ and $\int (L + \dot{f})\mathrm{d}t$ for any function of time $f(t)$ yields the same equations of motions, and if $f(a) = f(b)$ this doesn't even change the absolute value of the action (which doesn't interest us anyway).

  4. There is really no plausible way in which you could end up in a situation where you know the action of a system but not its Lagrangian (if one exists), so it is unclear what actual purpose the answer to your question would have even if it existed. Physical systems are usually defined by a Lagrangian or Hamiltonian, not by an action - the action is derived from that data, not given as an axiom. That is, this reverse problem simply does not occur "in the wild".

ACuriousMind
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  • I just need to know a derivation operator that produces the term $\sqrt{-g}d^4x$ when applied to a function such as $S$. – Anon21 Aug 21 '19 at 21:22
  • re 4: the question is not really that naive. In QFT it is a common question whether there is some Lagrangian that reproduces a certain action, e.g. to reabsorb a potential divergence. Of course there are additional assumptions, such as the Lagrangian being local and consistent with the symmetries, etc. – AccidentalFourierTransform Aug 21 '19 at 21:57
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FWIW, if we know the action $$ S_V[\phi]~=~\int_V \! d^nx~{\cal L}(\phi(x),\partial\phi(x),x) \tag{1}$$ for an arbitrary spacetime integration region $V$, it is in principle possible to recover the Lagrangian density ${\cal L}(\phi(x),\partial\phi(x),x)$ by localization, i.e. shrinking the integration region $|V|\to 0$.

E.g. in point mechanics the formula $$\frac{d}{dt_f}\int_{t_i}^{t_f} \! dt ~L(q(t),\dot{q}(t),t)~=~ L(q(t_f),\dot{q}(t_f),t_f)\tag{2}$$ explicitly realizes this recovery.

Qmechanic
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