From the action, $$\int L\,\mathrm dt=\int R \sqrt{|g|}\,\mathrm d^4x,$$ why is the Lagrangian density for the gravitational field replaced by the Ricci scalar, which yield field equations in vacuum $$ R_{\mu\nu}-R\frac{g_{\mu\nu}}{2}=0.$$ Is it that the Lagrangian density for vacuum is just the Ricci scalar?

  • $\begingroup$ "Is it that the lagrangian density for vaccum is just the Ricci scaler??" Yes. $\endgroup$ – Bence Racskó Feb 14 '19 at 14:29
  • $\begingroup$ Ok but why is for the gravitational field the lagrangian density is Ricci scaler??I read somewhere that lagrangian density is scaler and only scaler term in gravitational field is the Ricci scaler?? Is it so $\endgroup$ – Apashanka Das Feb 14 '19 at 14:36

The Ricci scalar is the simplest curvature scalar, and the only one that is linear in the curvature.

Choosing it as the Lagrangian density gives a theory which reduces to Newtonian gravity in the weak-field limit.

It is common to also include a term that is zeroth-order in the curvature rather than first-order. This is the “cosmological constant” term and can explain the observed acceleration of the expansion of the universe.

More speculative theories include higher-order scalars. For example, there are three scalars that are quadratic in the curvature: the Kretschmann scalar $R_{abcd}R^{abcd}$, the Chern-Pontryagin scalar $^*R_{abcd}R^{abcd}$, and the Euler scalar ${^*R^*}_{abcd}R^{abcd}$.


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