# What's the Lagrangian density means?

I remember from my classical mechanics class the Lagrangian function, but I don´t understand what's the meanning of the lagrangian density in General Relativity.

What's the difference between one and other? Why in some books is marked whit sqrt(-g) and others without it? Example: $L=sqrt(-g)R$ or $L=R$ for the Einstein Hilbert Action. What's the correct?

• – Qmechanic Aug 28 '18 at 16:20
• Hint: If the action integral $S=\int \!d^nx ~{\cal L}$ should be a scalar quantity, how should its integrand ${\cal L}$ transform? – Qmechanic Aug 28 '18 at 16:23

## 1 Answer

A Lagrangian Density is used in field theories. It is easiest to see the difference when you look at the integral for the action. Given a Lagrangian, $L$, the action integral is: $$S=\int L dt.$$ Given a Lagrangian Density, $\mathcal{L}$, the action integral is: $$S=\int\int\mathcal{L}d^3xdt.$$ Notice the extra integral over spatial coordinates. Essentially, the Lagrangian is the integral of the Lagrangian density over all of space. And in this way there is a perfect analogy between Mass and Density and the Lagrangian and Lagrangian density. The Lagrangian density measures the "distribution of the Lagrangian over space".

For General Relativity, there is no obvious way (lacking some symmetry) in which one should split space-time up into space and time, and so it becomes natural to use the Lagrangian density and integrate over all of spacetime rather than trying to figure out how one should split that integral into $d^3x$ and $dt$. In addition, in General Relativity, we no longer are working with flat spacetimes and linear (Cartesian) coordinates. Because of this, the volume element of space-time will no longer be the simple $d^4x$.Therefore, the Einstein-Hilbert action within some neighborhood $O$ in the manifold which we optimize to get the Einstein Field equations is (in units where $G=c=1$): $$S=\frac{1}{16\pi}\int_{O} R\epsilon$$ Where $\epsilon$ is the natural volume element on the manifold. Given a metric with determinant $g$, and a set of coordinates $x^1,...,x^4$ corresponding to a mapping $\psi:M\rightarrow\mathbb{R}^4$ the natural volume element $\epsilon$ can be expressed as $\epsilon=\sqrt{-g}dx^1\wedge ...\wedge dx^4$ and so the integral can be expressed as: $$S=\frac{1}{16\pi}\int_{\psi[O]} R\sqrt{-g}d^4x.$$

I've restricted attention here to one neighborhood, $O$. In order to generalize the integral over the entire manifold $M$ one would have to introduce partitions of unity which I didn't want to go over here - but that's basically just a detail to ensure one doesn't double count the integral in overlapping regions of the charts.

As you can see, for the Einstein Hilbert action, the Lagrangian density is $R$ (or $R/16\pi$ if you want to include the proportionality constant). The factor of $\sqrt{-g}$ arises as part of the volume element when we choose a coordinate system over which to integrate. It arises because we are not integrating over a flat manifold using Cartesian coordinates.