Past exam of the theoretical physics I course of the University of Cambridge / 2022 / 4

Apply the linear transformation of the scalar fields to either time- or space-derivative product. Its coefficients become $\alpha \gamma$ multiplying the square of the first derivative, $\beta \delta$ multiplying the square of the second, and $\alpha \delta +\beta \gamma$ multiplying their cross product. Invariance for arbitrary fields therefore requiresThe complete real solution isThere are four connected components, distinguished by the diagonal/antidiagonal choice and the sign of $r$. Only the positive diagonal family is in the identity component; write $r=e^s$ to obtain ${\varphi }_1\mapsto e^s{\varphi }_1$, ${\varphi }_2\mapsto e^{-s}{\varphi }_2$. This is the reciprocal rescaling symmetry of two cross-coupled scalar fields. The antidiagonal transformations exchange the fields as well as rescale them.

The infinitesimal connected internal symmetry of a classical field theory is $\delta {\varphi }_1=ϵ{\varphi }_1$, $\delta {\varphi }_2=-ϵ{\varphi }_2$. For this exact invariance of the Lagrangian density, the Noether current is $J^{\mu }={\sum }_j{\mathcal{L}}_{{\partial }_{\mu }{\varphi }_j}\delta {\varphi }_j/ϵ$. In coordinates $(t,x)$ it isVarying the two fields gives the two wave equations ${\varphi }_{1,tt}-c^2{\varphi }_{1,xx}=0$ and ${\varphi }_{2,tt}-c^2{\varphi }_{2,xx}=0$. Direct differentiation cancels the first-derivative cross terms:The Noether charge $Q=\int J^{0}dx$ is constant when the spatial boundary flux vanishes. The disconnected transformations from part (a) do not have independent infinitesimal generators of this type.

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.

The velocity coupling changes the momenta and the undifferentiated field terms. Variation with respect to ${\varphi }_2$ gives the equation for ${\varphi }_1$:so the two Euler-Lagrange field equations areThey describe damped and antidamped fields from an antisymmetric velocity coupling. With the convention ${\varphi }_j=A_j{e}^{i(k_{j}x-{\omega }_{j}t)}$, their characteristic equations areThus the dispersion relations and modes areFor each spatial Fourier mode, the general solution is the linear combination of its two temporal branches; spatial superposition gives the general field. Take real parts or suitable complex conjugate pairs to obtain the required real fields. At the repeated-root threshold, the independent second temporal solution contains an extra factor $t$, as detailed next.

Above the threshold, $c^2{k}_j^2>1/(4{\tau }^2)$, the square root in part (d) is real. Both fields oscillate, but the first has envelope $\boxed{e^{-t/(2\tau )}}$ and the second $\boxed{e^{+t/(2\tau )}}$. Thus the first is damped and the second amplified.

Below the threshold, set ${\sigma }_j=\sqrt{1/(4{\tau }^2)-c^2{k}_j^2}$. The temporal factors areThey are nonoscillatory. For nonzero $k_j$, $0<{\sigma }_j<1/(2\tau )$, so both first-field branches decay and both second-field branches grow. At $k_j=0$, one branch is constant; the other is $e^{-t/\tau }$ for ${\varphi }_1$ or $e^{t/\tau }$ for ${\varphi }_2$.

At equality, the repeated temporal root gives $\boxed{(A+Bt)e^{-t/(2\tau )}}$ for the first field and $\boxed{(C+Dt)e^{t/(2\tau )}}$ for the second. The full two-field system is generically unstable on both sides because of the amplified sector. This is compatible with its autonomous Lagrangian: the total conserved energy is an indefinite quadratic form, rather than a positive energy forcing every component to remain bounded.