With the choice of equation the momentum density in the $+$ direction becomes
$$
P^{+}=\frac{1}{2 \pi \alpha^{\prime}} c,
$$
which is independent of $\sigma$ and $\tau$. The total momentum $p^{\mu}$, which is of course conserved, is given by
$$
p^{\mu}=\int_{a}^{\pi} d \sigma P^{\mu}(\sigma),
$$
where $a=0$ for the open string and $\sigma=-\pi$ for the closed string. Using equation (4.1.23) we find
$$
p^{+}=\int_{a}^{\pi} d \sigma P^{+}=\frac{(\pi-a)}{2 \pi \alpha^{\prime}} c .
$$
Consequently, for the open string
$$
c=2 \alpha^{\prime} p^{+} \text {implying } P^{+}=\frac{p^{+}}{\pi},
$$
while for the closed string
$$
c=\alpha^{\prime} p^{+}, \text {implying } P^{+}=\frac{p^{+}}{2 \pi} .
$$
7CCMMS34 COURSE NOTES ：
$$
\delta x^{\mu}=-\omega_{\nu}^{\mu} x^{\nu}+\eta^{\alpha} \partial_{\alpha} x^{\mu}
$$
and in particular
$$
\delta x^{+}=-\omega_{v}^{+} x^{v}+\eta^{0} 2 \alpha^{\prime} p^{+} .
$$
On the other hand,
$$
\delta\left(2 \alpha^{\prime} p^{+} \tau\right)=-2 \alpha^{\prime} p^{v} \tau \omega_{v^{*}}^{+}
$$
Equating these two results we find that
$$
\eta^{0}=\frac{\omega_{v}^{+}}{p^{+}}\left(x^{v}-2 \alpha^{\prime} p^{v} \tau\right)=\frac{\omega_{k}^{+}}{p^{+}}\left(x^{k}-2 \alpha^{\prime} p^{k} \tau\right) .
$$