Noether charges for Poincaré Symmetry

Poincaré Invariance is a global symmetry in the D-dimensionalMinkowski space-time ${\mathbb{R}}^{1,D-1}$. Hence, by Noether’s Theorem, there are associated conserved charges and currents, which are spacetime momentum and angular momentum.

Spacetime Momentum

For an infinitesimal symmetry transformation acting on a field $\varphi (\sigma )$,

$\varphi (\sigma )\mapsto \varphi (\sigma )+ϵ\delta \varphi (\sigma )$,

the Noether current is given by $J^a=\delta \varphi \frac{\delta S}{\delta ({\partial }_a\varphi )}$.

This current is conserved, ${\partial }_a{J}^a=0$ as a consequence of the equations of motion and this leads to a conserved charge

$Q={\int }_0^{\pi }d\sigma J^0$.

This satisfies ${\partial }_{\tau }Q=0$, with appropriate boundary conditions.

We can now apply this to infinitesimal translations $X^{\mu }\mapsto X^{\mu }+a^{\mu }$ with $\delta X^{\mu }={ϵ}^{\mu }$, a constant. Then, we obtain D conserved currents

$P_a^{\mu }=\frac{\delta S}{\delta ({\partial }^a{X}^{\mu })}=-T{\partial }_a{X}^{\mu }$

where $S=-\frac{T}{2}\int d^2\sigma {\partial }^a{X}^{\mu }{\partial }_a{X}_{\mu }$.

Now, ${\partial }^a{P}_a^{\mu }=-T{\partial }^a{\partial }_a{X}^{\mu }=0$ since the equations of motion ${\partial }^a{\partial }_a{X}^{\mu }=0$.

$P_a^{\mu }$ describes the spacetime momentum density, the D-dimensional momentum carried by the world-sheet. Then, the total momentum carried by a string is

$P^{\mu }:={\int }_0^{\pi }d\sigma (P^{\mu }{)}^0=T{\int }_0^{\pi }d\sigma {\partial }_{\tau }{X}^{\mu }$

We claim that $P^{\mu }=p^{\mu }$ (check Boundary Conditions), which justifies our identification of $p^{m}u$ as the total spacetime momentum carried by the string.

Spacetime Angular Momentum

Similarly, if we consider infinitesimal Lorentz transformations $\delta X^{\mu }={ϵ}_{\nu }^{\mu }{X}^{\nu }$ with ${ϵ}_{\mu \nu }=-{ϵ}_{\nu \mu }$, we obtain conserved currents

$J_a^{\mu \nu }=T(X^{\mu }{\partial }_a{X}^{\nu }-X^{\nu }{\partial }_a{X}^{\mu })$,

which describe the D-dimensional angular momentum density. Then, the total angular momentum is

$J^{\mu \nu }=T{\int }_0^{\pi }d\sigma (X^{\mu }{\partial }_{\tau }{X}^{\nu }-X^{\nu }{\partial }_{\tau }{X}^{\mu })$.

For Lorentz rotations, $J^{\mu \nu }=l^{\mu \nu }+E^{\mu \nu }$ with $l^{\mu \nu }=x^{\mu }{p}^{\nu }-x^{\nu }{p}^{\mu }$ and $E^{\mu \nu }=-i{\sum }_{n=1}^{\infty }\frac{1}{n}({\alpha }_{-n}^{\mu }{\alpha }_n^{\nu }-{\alpha }_{-n}^{\nu }{\alpha }_n^{\mu })$.

We can also show that $P_{\sigma }^{\mu }=0$ at $\sigma =0,\pi$, meaning that no momentum flows out of the string ends, by taking ${\partial }_{\mu }{X}^{\mu }{|}_{\sigma =0,\pi }$.