The string Poisson bracket

Question

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?

Comments

You might want to also mention what the context of this question is.

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

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@joshphysics huh? I do not know what you exactly mean. Can you give an answer? I suspect that it is a darn embarassingly simple thing I just dont see ...

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

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Oh I don't have an answer, I just mean that someone seeing this question may not recognize that it's being asked in the context of string world sheets, so it might help to mention that.

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

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@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

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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

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@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

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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

Answer 1

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

Answer 2

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

Comments

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