Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 306 / 1

Work away from $x=0$, where the inverse-square potential is singular. Variation of the phase-space action givesThe last equation is the constraint imposed by the Lagrange multiplier $e$. The canonical momentum conjugate to $t$ is $-E$, not $E$. Consequently the nonzero canonical Poisson brackets areIn particular,With the convention $\delta z=\alpha \{z,Q\}$, the Noether charge generating $\delta t=\alpha$ is $Q=-E$. Equivalently, the conserved physical energy is $E$; a convention that calls the energy the time-translation charge absorbs this minus sign into its parameter.

For the dilation charge,on the constraint surface. Its infinitesimal canonical transformations areFor constant $\alpha$, extending this by ${\delta }_{D}e=-\alpha e$ preserves the action: the kinetic terms are invariant and $E-H$ scales oppositely to $e$. Thus the dilation acts on time and position with their nonrelativistic relative scaling. Its Poisson bracket with the energy is

Put $C=E-H$. Direct variation of the Hamiltonian givessoThe supplied transformations are generated by the special conformal charge of inverse-square mechanicsChooseThen $\delta (eC)=0$ even for a time-dependent parameter. Expanding the kinetic-term variation, keeping the terms in $\dot{\beta }$, givesHenceFor a constant parameter only the boundary term remains, identifying $K$ as the Noether charge. Independently, the equations of motion giveUsing the same canonical Poisson bracket convention,Together, $\{D,E\}=-E$, $\{D,K\}=K$ and $\{E,K\}=2D$ form the sl2R Lie algebra of conformal mechanics. These are time translation, dilation and special conformal transformation. A useful normalization check is the Casimir elementon the constraint surface.

Define the open-string endpoint momentum flux along the open string by $J_m=e{T}^2{X}_m^{\prime }+u{P}_m$. Variation of the Nambu-Goto phase-space action givesThe two last equations are the Virasoro constraints. Integration of the momentum equation givesThus the necessary and sufficient condition for total momentum conservation is equality of the two endpoint fluxes, component by component. Vanishing of both fluxes is a sufficient local boundary condition.

The spatial boundary term in the action variation is $-\int dt[J_m\delta X^m{]}_0^{\pi }$. For free ends the endpoint variations are arbitrary, soIn a gauge with $u=0$ and nonzero $e$ this reduces to the usual Neumann boundary condition $X^{\prime m}=0$. The conclusion is conservation of total momentum, not that total momentum must vanish. The PDF contains the derivative in this conclusion; the TeX transcription drops it.

For endpoints confined to $x^1=0$, the Dirichlet boundary condition fixes $\delta X^1=0$. It therefore imposes no requirement that $J_1$ vanish. Preserving the fixed position requires ${\dot{X}}^1=e{P}^1+u{X}^{\prime 1}=0$ at each endpoint, but that is a different condition. For example, in $u=0$ gauge the endpoint momentum $P^1$ is zero while $e{T}^2{X}^{\prime 1}$ can be nonzero. Consequently ${\mathcal{P}}_1$ is generally not conserved: the normal endpoint forces transfer momentum to the hyperplane. A dynamical D-brane recoils and carries the missing momentum. If the hyperplane is idealized as fixed, its external support absorbs the momentum. The combined string-and-support system conserves momentum; freely varying directions tangent to the hyperplane retain their vanishing-flux condition.