Non-relativistic and relativistic field theory Lagrangian confusion

Question

In non-relativistic classical mechanics if I was to describe the waves within a three dimensional medium, I would write down a Lagrangian $L$ in terms of the Lagrangian density $\mathcal{L}$ as

$$L = \int\mathcal{L}dxdydz $$

and the action as

$$ S = \int L dt = \int \mathcal{L}dxdydzdt. $$

When transitioning to relativistic field theory, we are now considering space-time and therefore an extra dimension. If time is to be treated on an equal footing as space, then why do the integrals not increase by one dimension with proper time as the new path parameter i.e.

$$ L = \int \mathcal{L}cdtdxdxdz $$

and the action as

$$ S = \int L d\tau = \int \mathcal{L}cdtdxdydzd\tau. $$

$L$ and $S$ do not change when transitioning to relativistic field theory. So essentially my question is: why is the Lagrange density in relativistic field theory not defined as the Lagrangian per unit volume of the space, when in non-relativistic field theory it is treated as Lagrangian per unit volume of that space?

Answer 1

It's really the same in both cases. We want both the action and the Lagrangian density to be Lorentz invariant, so we define $$S = \int d^4x \, \mathcal{L}.$$ This is exactly the same as the situation in nonrelativistic field theory. We no longer care much about the Lagrangian, $$L = \int d^3x \, \mathcal{L}$$ as much because, as you noted, it's not Lorentz invariant as it picks out a direction of time. However, it's still useful, e.g. it still Legendre transforms to the Hamiltonian.

It is not useful to define $L = \int d^4x \, \mathcal{L}$, because the expression $$S = \int d\tau \, d^4x \, \mathcal{L}$$ is nonsensical. Not only is it not dimensionally correct, but the $d\tau$ doesn't have any meaning -- who is measuring this proper time? (It makes sense for a particle, since you can put a clock along its worldline, but not for a field.)