Matthew Dodelson has already given the main reason in his answer: The string action $S$ (either the Nambu-Goto or the Polyakov action) is proportional to the string tension $T_0$. So the canonical momentum density

$$\tag{$\star$} {\cal P}^{\tau}_I ~:=~\frac{\partial {\cal L}}{\partial \dot{X}^{I}}~=~\frac{T_0}{c^2} \eta_{IJ}\dot{X}^{J}$$

becomes proportional to the string tension $T_0$ as well. (The latter equality in ($\star$) is only true with further assumptions. See e.g. B. Zwiebach, *A first course in string theory,* for details.)

Finally, let us restore the correct factors of speed of light $c$ in the Poisson brackets:

$$\tag{$\star\star$}\{X^{I}(\tau,\sigma),\dot{X}^{J}(\tau,\sigma')\}_{PB} ~=~ \frac{c^2}{T_0}\delta(\sigma-\sigma')\eta^{IJ}.$$

Both sides of ($\star\star$) have now dimension of inverse mass $M^{-1}$. This is a consequence of the following dimensional analysis:

$$[\tau]~=~ \text{dim. of time} ~=:~T,$$

$$[\sigma] ~=~ \text{dim. of length} ~=:~L~=~ [X],$$

$$[c]~=~ \text{dim. of speed} ~=~\frac{L}{T}, $$

$$[T_0] ~=~ \text{dim. of force} ~=~\frac{ML}{T^2},$$

$$[\text{Poisson bracket}] ~=~[\{\cdot, \cdot\}_{PB}] ~=~ \frac{1}{\text{dim. of angular momentum}}~=~\frac{T}{ML^2}. $$

This post imported from StackExchange Physics at 2014-03-12 15:21 (UCT), posted by SE-user Qmechanic