# The string Poisson bracket

+ 5 like - 0 dislike
1409 views

Where does the factor $\frac{1}{T}$ ($T$ is the string tension) in this Poisson bracket come from?

$$\{X^{\mu}(\tau,\sigma),\dot{X}^{\nu}(\tau,\sigma')\} ~=~ \frac{1}{T}\delta(\sigma-\sigma')\eta_{\mu\nu}.$$

I think I can see from remembering the definition of a Poisson bracket (for example in canonical coordinates) why in terms of momentum we have

$$\{P^{\mu}(\tau,\sigma),X^{\nu}(\tau,\sigma')\} ~=~ \delta(\sigma-\sigma')\eta_{\mu\nu}$$

but I don't see why this factor in the first equation has to be there.

In addition to deriving it by calculation, is there an intuitive physical way how one can see why the factor of inverse tension has to be there, similar to explaining the appearance of the tension in front of the integral in the action by the fact that it costs energy to stretch the world-sheet?

I somehow feel very stupid that I dont see this :-/

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

This post imported from StackExchange Physics at 2014-03-12 15:21 (UCT), posted by SE-user Michael Brown
@MichaelBrown Maybe ... but I am not too fond of purely dimensional arguments and nothing else ... ;-)

This post imported from StackExchange Physics at 2014-03-12 15:21 (UCT), posted by SE-user Dilaton
Agreed. I just didn't know what else it could be. Defering to others on this one. :)

This post imported from StackExchange Physics at 2014-03-12 15:21 (UCT), posted by SE-user Michael Brown
@Dilation, does it help if you rewrite $\frac{1}{T}\delta(\sigma-\sigma')=\delta(T(\sigma-\sigma'))=$?

This post imported from StackExchange Physics at 2014-03-12 15:21 (UCT), posted by SE-user santa claus
@joshphysics ah ok, I see. I will do, if I'll get my WLAN running again at home :-/

This post imported from StackExchange Physics at 2014-03-12 15:21 (UCT), posted by SE-user Dilaton
I put the string-theory tag on it as a start.

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

+ 4 like - 0 dislike

It comes from the normalization of the Polyakov action, \begin{align*} S=\frac{T}{2}\int d^2\sigma\, (\dot{X}^\mu\dot{X}_\mu-{X'}^\mu{X'}_\mu). \end{align*} The canonical momentum is \begin{align*} \frac{\partial \mathcal{L}}{\partial \dot{X}^\mu}=T\dot{X}_\mu, \end{align*} and this gives the equal time commutator (or Poisson bracket) that you wrote down, \begin{align*} [X^\mu(\tau,\sigma),\dot{X}^\nu(\tau,\sigma')]=\frac{i}{T}\eta^{\mu\nu}\delta(\sigma-\sigma'). \end{align*} The way I think about this is that as the tension of the string goes to zero, the string becomes less and less classical (alternatively, $T$ plays the role of $1/\hbar$ on the worldsheet).

This post imported from StackExchange Physics at 2014-03-12 15:21 (UCT), posted by SE-user Matthew
answered Mar 15, 2013 by (320 points)
+ 3 like - 0 dislike

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
answered Mar 15, 2013 by (3,120 points)
Thanks Qmechanic, I like these additional explanations.

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

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:$\varnothing\hbar$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.