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?
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3$\begingroup$ 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 ${\cal L} = \text{Tr} [ \partial_\mu \phi \partial^\mu \phi ]$. $\endgroup$– PraharCommented Jul 18 at 8:22
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$\begingroup$ 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 $\mathcal{L} = det(\partial_{\mu}\phi\partial^{\mu}\phi)$. $\endgroup$– physics_2015Commented Jul 18 at 17:20
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1$\begingroup$ 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 $\endgroup$– PraharCommented Jul 18 at 18:49
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$\begingroup$ 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. $\endgroup$– LPZCommented Jul 19 at 14:25
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