I know that an action must be locally-Lorentz invariant based on physical reasons, but is there any requirement for it to be a coordinate pseudo-scalar (up to surface terms)? In particular, would an action of the following form be permissible, $$ S=\int \sqrt{-g} L \, d^4x \ , $$ where $L$ is not a coordinate scalar (i.e. has no free indices but explicitly depends on coordinates) but is still a local Lorentz scalar. An example of such a Lagrangian that isn't a coordinate scalar would be using a contraction of the Christoffel symbols $$ L= g^{\mu \nu}\Gamma_{\mu \lambda}^{\rho} \Gamma^{\lambda}_{\nu \rho} \ . $$

Will this still give a well-defined variation principle $\delta S = 0$ despite $S$ being non-covariant and depending on the choice of coordinates?

A similar question but regarding Lorentz invariance is asked here Must the action be a Lorentz scalar? but I'm unsure if the argument also applies here too (with diffeomorphism invariance instead of Lorentz invariance).

Also note I do not care if the equations of motion resulting from $\delta S = 0$ are not covariant.

-- On further thought, rather than Lagrangian wrote above one could just consider something simple like $$ L = \partial_{\mu}\partial_{\nu}g^{\mu \nu} $$ which is probably easier to work with. One potential problem I see is that the action could vanish or diverge to plus or minus infinity in some coordinate systems - is this a problem for the variation principle $\delta S= 0$?

  • $\begingroup$ I'm not sure what exactly you mean by "coordinate scalar" - can you give an example of such a quantity that is "not a coordinate scalar but still a Lorentz scalar"? $\endgroup$
    – ACuriousMind
    Commented Mar 14, 2021 at 16:09
  • $\begingroup$ @ACuriousMind Sure, that's a good clarification and I'll edit it into the question. $\endgroup$
    – user290954
    Commented Mar 14, 2021 at 16:12
  • $\begingroup$ It probably depends on the theory you're talking about, but generally the action will be something that you don't want to depend on the coordinate system you're using. Even in more general non-Lorentz coordinate systems (i.e. diffeomorphism invariant). $\endgroup$
    – Charlie
    Commented Mar 14, 2021 at 16:18
  • $\begingroup$ Realised I should have said local Lorentz invariance (talking about the tangent structure, not spacetime transformations). @Charlie yes this was my suspicion, but I didn't know if the variational principles themselves would be well-posed or not. $\endgroup$
    – user290954
    Commented Mar 14, 2021 at 16:21

1 Answer 1


When we write an integral like $$ \int f\sqrt{g}\mathrm{d}^4x$$ then $f : M \to \mathbb{R}$ is a scalar function on our spacetime $M$. Coordinate-dependent expressions do not define such functions, when you write something like $$ L = g^{\mu \nu}\Gamma_{\mu \lambda}^{\rho} \Gamma^{\lambda}_{\nu \rho}, $$ then that actually doesn't define anything - the expression on the r.h.s. does not have a single value, but infinitely many, depending on what coordinate system you choose.

Of course you can define a function if you declare that $L$ always takes the value that the r.h.s. has in one fixed coordinate system, but since we usually do not want to fix a coordinate system this is rather pointless - there is no naturally "preferred" coordinate system on manifold so it is hard to see how any quantity of interest could ever be of this form.


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