Lagrangian for the General Relativity

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SUMMARY

The discussion centers on the Lagrangian formulation for General Relativity, specifically the Einstein-Hilbert action represented as L=R, where R is the Ricci scalar. The full Lagrangian for Einstein's field equations is derived by combining the gravitational Lagrangian density L_G with the matter Lagrangian density L_M, expressed as L=L_G+L_M. The action for electromagnetic fields is detailed as S = ∫ ∗ℜ - (1/2) ∫ F ∧ ∗F, leading to the equation R_{μν} - (1/2)ℜ g_{μν} = (1/2)T_{μν}, where T_{μν} represents the energy-momentum tensor for electromagnetic fields.

PREREQUISITES
  • Understanding of the Einstein-Hilbert action in General Relativity
  • Familiarity with the Ricci scalar and its role in gravitational theories
  • Knowledge of Lagrangian density and variational principles
  • Basic concepts of electromagnetic field theory and the Hodge dual
NEXT STEPS
  • Study the derivation of the Einstein-Hilbert action in detail
  • Explore the implications of varying the action with respect to the metric tensor
  • Learn about the energy-momentum tensor T_{μν} in various field theories
  • Investigate the role of the Hodge dual in differential geometry and field theory
USEFUL FOR

This discussion is beneficial for theoretical physicists, researchers in gravitational physics, and students studying advanced topics in General Relativity and field theory.

MManuel Abad
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I've always found that the lagrangian for the Gravitational field is that from the Einstein-Hilbert action:

<i>L</i>=R (R is the Ricci scalar; I'm not including the factor of \sqrt{-g})

but when variational principles are applied, we get the vacuum field equations (obviously). I'd like someone to tell me which would be the FULL lagrangian (with matter coupled) for getting the Einstein's field equations.
 
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You simply take the gravitational Lagrangian density and add in the Lagrangian density of the matter fields: L=L_G+L_M.
 
For example, if you have electromagnetic fields, then the action is

S = \int \ast \mathcal{R} - \frac12 \int F \wedge \ast F

where \ast is the Hodge dual and F is the electromagnetic 2-form (rescaled up to some factors of 2\pi which I don't remember...I use the above normalization in my research). Using the normalization I've given here and varying with respect to the inverse metric, you obtain

R_{\mu\nu} - \frac12 \mathcal{R} g_{\mu\nu} = \frac12 T_{\mu\nu}

where

T_{\mu\nu} = F_{\mu\rho} F_\nu{}^\rho - \frac14 g_{\mu\nu} F_{\rho\sigma} F^{\rho\sigma}
 
Wow, thankyou both, that was very useful! :)
 

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