# Lagrangian of a multi-dimensional scalar field

We know that the Lagrangian has to be a scalar. Would it be possible if this scalar is multi-dimensional (for example $$m\times m$$)? Let's say a field $$\phi$$ is represented with an $$m\times m$$ matrix where the elements are constants and the Lagrangian for this scalar field is: $$\mathcal{L} = \partial_{\mu}\phi\partial^{\mu}\phi$$ In this case the Lagrangian would have dimension $$m\times m$$. Would that be acceptable? If so, how would one derive the Euler-Lagrange equations? Does one consider the variation of each element of the $$m\times m$$ representation separately?

• It would not be acceptable as the Lagrangian has to be a number, not a matrix. What you can do is take a trace and define ${\cal L} = \text{Tr} [ \partial_\mu \phi \partial^\mu \phi ]$. Commented Jul 18 at 8:22
• Hi @Prahar, the trace would contain only certain elements of $\phi$. Would taking the determinant of the matrix be acceptable as well? in other words, having $\mathcal{L} = det(\partial_{\mu}\phi\partial^{\mu}\phi)$. Commented Jul 18 at 17:20
• You could add such a term, yes. However, that's not a quadratic action, so it is not a kinetic term. The trace, on the other hand, is a linear function Commented Jul 18 at 18:49
• A classic example would be the $Q$ tensor for liquid crystals. Without spatial correlations you get Landau- de Gennes theory but just as for Landau-Ginzburg for superconductivity, you can promote it to a field theory.
– LPZ
Commented Jul 19 at 14:25