 Solution

Use coordinates $x^0=t$, $x^1=x$ and first compute the mixed canonical stress-energy tensor supplied by spacetime translations:Define the energy density $\mathcal{E}={\varphi }_{1,t}{\varphi }_{2,t}+c^2{\varphi }_{1,x}{\varphi }_{2,x}$ and $\mathcal{P}={\varphi }_{1,t}{\varphi }_{2,x}+{\varphi }_{2,t}{\varphi }_{1,x}$. Direct substitution givesThe two wave equations implyFor example, differentiating $\mathcal{E}$ and replacing each ${\varphi }_{j,tt}$ by $c^2{\varphi }_{j,xx}$ gives exactly $c^2{\partial }_x\mathcal{P}$ by the product rule. These are respectively the two component equations $\boxed{{\partial }_{\mu }{T}^{\mu }{}_{\nu }=0}$, establishing stress-energy conservation explicitly.

If the given metric $g^{\mu \nu }=\mathrm{diag}(1,-1)$ is used to raise the second index in these unrescaled coordinates, the result is $T^{\mu \nu }=\left(\begin{array}{cc}\mathcal{E}& -\mathcal{P}\\ -c^2\mathcal{P}& \mathcal{E}\end{array}\right)$, still conserved. To display the symmetric relativistic form with light speed $c$, choose $X^0=ct$, $X^1=x$ instead. With ${\partial }_{X^0}=c^{-1}{\partial }_t$, the corresponding tensor isThis states the coordinate convention explicitly rather than setting $c=1$ in some components but not others.