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Just a silly question, but shall we include $\sqrt{-g}$ inside the Lagrangian density in GR? In other words, is it

$$S = \int{\mathcal{L}d^4x} \Longrightarrow \mathcal{L} = \sqrt{-g}\left(\frac{1}{2\kappa}R + \frac{\Lambda}{\kappa} + \mathcal{L}_m\right)~?\tag{1}$$

Or the following

$$S = \int{\sqrt{-g}\mathcal{L}d^4x} \Longrightarrow \mathcal{L} = \frac{1}{2\kappa}R + \frac{\Lambda}{\kappa} + \mathcal{L}_m~?\tag{2}$$

Most of the actions I've seen out there (like this one) study classical particles moving in curved spacetime rather than the Einstein-Hilbert Lagrangian density itself.

Qmechanic
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Antoniou
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1 Answers1

6

The logic is as follows:

  1. The Lagrangian density ${\cal L}$ and $\sqrt{|g|}$ are examples of a density $\rho$; it transforms with the inverse Jacobian $J=\det\frac{\partial x^{\prime}}{\partial x}$ under general coordinate transformations $x\to x^{\prime}$.

  2. Conversely, the measure $d^dx$ transforms with the Jacobian $J$ under general coordinate transformations $x\to x^{\prime}$.

  3. The product $\rho d^dx$ and the quotient $\frac{\cal L}{\sqrt{|g|}}$ are examples of a scalar; it is invariant under general coordinate transformations $x\to x^{\prime}$.

  4. In particular the action $S=\int {\cal L} d^dx$ is an integrated scalar and a geometric object; it is invariant under general coordinate transformations $x\to x^{\prime}$.

  5. In contrast the quantity ${\cal L}$ in OP's eq. (2) is a scalar; not a density.

Qmechanic
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