# Metrics and Spinors

this might be better posed in mathematics but I'll ask here anyway.

So the Lagrangian for the spinor field can be viewed as follows. Let $(M,g,\nabla)$ denote a locally Minkowskian spacetime, Where $\varphi:U\rightarrow \varphi(U)\subset\mathbb{R}^{n}$ denotes an chart on an open set containing some point $p$ of interest.

Let $s=(\varphi,\psi\circ\varphi): U\rightarrow \varphi(U)\times\mathbb{C}^d$ define a $d$ component spinor $\psi$ locally defined on some coordinate patch containing $p$ (here $\psi$ is a section of a locally trivialised spinor bundle; of course one could generalise to a more abstract fibre).

Then I can define a Dirac Lagrangian $\mathcal{L}: \varphi(U)\times\mathbb{C}^d\rightarrow \mathbb{R}$ as

$$\mathcal{L}_D = \bar{\psi}(\varphi(p))(i\gamma^\mu \partial_\mu - m) \psi(\varphi(p))$$

Now in effect, the only information from the manifold needed to feed into this Lagrangian is a single point - I do not need any other forms of data to retrieve this number (except $m$ which is a "constant").

Compare this to the Lagrangian of the metric: A metric requires three pieces of data to compute at a point:

$$g(\cdot,\cdot)(\cdot) : \Gamma(TM\otimes TM)\times U \rightarrow \mathbb{R}$$

Now when I define the Einstein Hilbert Lagrangian, I must consider terms of form $\sim\partial ^2g, (\partial g)^2$.

The point being, that I can consider a tensor field $g(X,Y):U \rightarrow\mathbb{R}$, but actually looking at the metric without specifying these vector fields is problematic as I need more than just a point to work with its lagrangian: $$\mathcal{L}_g \sim \partial^2 g + (\partial g)^2$$

So to well pose this Lagrangian, I ultimately need to consider two vector fields as data, as well as the point.

Does this mean we should really consider the Lagrangians for the scalars/spinors and metrics as in different classes of Lagrangian, based on the amount of data required in each case?

• Actually, the Dirac Lagrangian written here is not diffeomorphism invariant. In order to get this the spin connection is required. So it will not just depend on points of the manifold. – Frederic Thomas Aug 22 '18 at 8:13
• Sticking in a factor $\sqrt{-g}$ then? Or send $\partial \rightarrow D$ eg Fock Ivanenko derivative $D = \partial + \omega \gamma\gamma$ – MKF Aug 22 '18 at 9:03
• But the point is the actual fields themselves don't depend on the same number of parameters - the metric (graviton) depends on two vectors, as opposed to the fermion $\psi(x), x=\phi(p)$ – MKF Aug 22 '18 at 10:09
• As you apparently refer to the Lagrangian of the free gravitational field the curvature scalar being an expression like $\partial^2 g + (\partial g)^2$, it's the double contracted Riemann tensor, it is a scalar which does not act anymore on tangential vectors. For the contraction one might need some tangential vectors (on which is summed over), which can be chosen freely, i.e. $\partial_i$ or more generally without changing $R$. Even more exactly $L_{EH}\sim R\sqrt{-g}$ is a scalar density which transforms like $L'_{EH}=| det \frac{\partial x}{\partial x'}|L_{EH}$. – Frederic Thomas Aug 22 '18 at 10:45
• A Lagrangian is mathematically always a scalar density, so it does not need tangential vectors and it always transforms like $L' = |\frac{\partial x}{\partial x'}| L$ to make the action $S= \int L d^{n}x$ invariant. But already L is completely tangential vector independent. This is true for every $L$. If you like you can consider the integrant as 4-form: $L d^{n}x$, in order to evaluate the integral $S$ you need to evaluate the 4-form on tangential vectors which are actually given by the volume element. This is also valid for all $L d^{n}x$, whether Diracian or Hilbert-Einstein or other. – Frederic Thomas Aug 22 '18 at 11:09