3

Given the action $$ A = \int_{M} d^{4}x \ \mathcal{L}(\phi, \nabla \phi) $$ where $\mathcal{L}$ is a lagrangian density, or if you prefer $\mathcal{L} = \sqrt{-g} \mathcal{\tilde{L}}$ and $\mathcal{\tilde{L}}$ a function of spacetime.

From Noether second theorem we know that diffeomorphism invariance produces the following identity:

$$ \nabla_{\mu} T^{\mu \nu } = \nabla_{\mu} \bigg( \frac{2}{\sqrt{|g|}}\frac{\delta \mathcal L}{\delta g_{\mu\nu}} \bigg) = 0$$

Where we have an infinite dimensional Lie Algebra $ \big( \ \Gamma (M) \ , \ [ , ] \ \big)$ where $\Gamma ( M ) $ are the smooth sections of the Tangent bundle $TM$. So, smooth vector field generates the lie group of diffeomorphism, hence diffeomorphism invariance is arena for the Second Noether Theorem.

In principle I know that if we have a family $\xi_{(i)}^{\mu}$ of killing vector fields($\mathscr{L}_{\xi_{(i)}^{\mu}} g_{ab} = 0$), we can construct from $T^{\mu \nu}$ a conserved current and a corresponging conserved quantity integrating the current over a Cauchy Hypersurface.

This I know. But my question is: If in principle we start from a diffeomorphism invariant action, what stops me from defining arbitrary $\xi_{(i)}^{\mu}$ such that with the usual Lie Bracket $[ , ]$ form a finite dimensional Lie subalgebra of $ \big( \ \Gamma (M) \ , \ [ , ] \ \big)$. The action would still be invariant w.r.t. the action of the Lie Group generated by this Lie Subalgebra, hence I can apply Noether First Theorem obtaining free a conserved current and an integral conserved charge. All this without the assumption that $\mathscr{L}_{\xi_{(i)}^{\mu}} g_{ab} = 0$.

I mean Invariance from an infinite dimensional Lie Algebra implies invariance from a finite dimensional Lie Subalgebra (es. $SO(3)$). So why we need a priori killing vector field in this picture? Where Killing vector field enters the picture if my construction doesn't need them and is arbitrary on the choise of the Lie Subalgebra generators?

Qmechanic
  • 201,751
Coltrane8
  • 107

1 Answers1

1

  1. OP is considering the matter action $S_m\equiv A$ and its corresponding symmetric SEM tensor $T^{\mu\nu}$ of matter.

  2. Notice that the gravitational Einstein-Hilbert (EH) action $S_{EH}$ is not included. For a good reason: The SEM pseudo-tensor $t^{\mu\nu}$ to the gravitational field is marred with problems.

  3. If one reads the fineprint of Noether's first theorem, one will notice that its conservation laws are only satisfied modulo terms proportional to $\delta S_m / \delta g_{\mu\nu}$ and $\delta S_m /\delta\phi$. The latter vanishes on-shell since it is the EOM for $\phi$, but the former does not since the gravitational sector is incomplete. Hence its conservation laws are violated.

  4. See also this related Phys.SE post.

Qmechanic
  • 201,751
  • I read your post 4. and I have great esteem for you, so I appreciate your answer, nevertheless I wanted to know more. From point 5. of your linked post we have $ d_{\mu}\Theta^{\mu}{}{\lambda}~\stackrel{m}{\approx}~-\frac{\delta S_m}{\delta g{\mu\nu}}g_{\mu\nu,\lambda} = T^{\mu\nu}\frac{\sqrt{|g|}}{2}g_{\mu\nu,\lambda} $. If I understand it correctly $d_{\mu}\Theta^{\mu}{}{\lambda} =\partial{\mu}\Theta^{\mu}{}_{\lambda}$, hence we have a trivial identity. (If we add the gravitational action what would have changed?). – Coltrane8 Nov 26 '20 at 15:30
  • Second, in your post 4. I don't see how esplicitly the lie sub algebra generators plays a role and where we arrive to the Killing condition by Noether first. I mean explicitly I didn't find a reference to see the details of this doubt, and a priori it seems to me that the two procedure should be the same, just specializing the second to a finite dim. lie subalgebra – Coltrane8 Nov 26 '20 at 15:32