# Mode expansion for a closed string?

1. Apr 12, 2006

### coalquay404

Hi. Suppose that I want to look at the BDH action for a bosonic string:

$$S_{BDH} = -\frac{T}{2}\int d^2\sigma \sqrt{\gamma}\gamma^{ij}\partial_iX^a\partial_jX^b$$

The string is in flat spacetime and $$\gamma_{ij}$$ is an independent world-sheet metric. I know that going to conformal gauge allows me to write the equation of motion for this string as

$$\left(\frac{\partial}{\partial\tau} - \frac{\partial}{\partial\sigma}\right)X^a(\tau,\sigma)$$.

This is just the massless wave equation and needs to be supplemented with boundary conditions. However, the case for an open string with Neumann boundary conditions is straightforward. I can express the solution as a combination of left and right-moving solutions, apply the boundary conditions, and obtain a mode expansion for the solution.

What I'm having trouble with is the mode expansion for the closed string. I've had a browse through chapter 12 of Zweibach and, while he mentions the mode expansion for the closed string, he doesn't go through the derivation. Can anyone give me a couple of pointers as to how I can produce the mode expansion for the closed string?

Thanks.

2. Apr 14, 2006

### Timbuqtu

I suppose you ment:

$$\left(\frac{\partial^2}{\partial\tau^2} - \frac{\partial^2}{\partial\sigma^2}\right)X^a(\tau,\sigma) = 0$$

The closed string case is actually easier than the open one. The only boundary condition you have is the periodicity at $$\sigma=0,2 \pi$$. So the general solution to the wave equation is just a combination of a left moving periodic function (i.e. of $$\tau+\sigma$$) and a right moving periodic function (i.e. of $$\tau-\sigma$$). So contrary to the open string case left and right moving modes are independent of eachother.

3. Apr 20, 2006

### coalquay404

Thanks for the reply. I managed to figure it out after a little bit of thought and you're right, it is easier than the open string case.

I do have another quick question if anyone's willing to have a go. I've been reading Green, Schwarz and Witten and their take on canonical quantization of the bosonic string. Ordinarily, I'd define the Poisson brackets of two functions $$F$$ and $$G$$ in a discrete classical theory as

$$\{F,G\} = \sum_{k}\left(\frac{\partial F}{\partial q^k}\frac{\partial G}{\partial p_k} - \frac{\partial F}{\partial p_k}\frac{\partial G}{\partial q^k}\right)$$

Now, at the moment I'm looking at the canonical quantization of the closed bosonic string. I take the BDH action (or Polyakov action, if you prefer) and put it in conformal gauge. I can calculate the canonical momentum density to be

$$\pi^\mu \equiv \frac{\partial \mathcal{L}}{\partial(\partial_\tau X_\mu)} = \frac{1}{2\pi\alpha'}\partial_\tau X^\mu$$

At this stage, Green et al define (it's equation 2.1.52 in the book) the equal time Poisson brackets of $$X^\mu$$ and $$\pi^\mu$$ as

$$\{\partial_\tau X^\mu(\sigma),X^\nu(\sigma')\} = 2\pi\alpha'\eta^{\mu\nu}\delta(\sigma-\sigma')$$

The problem I have with this is that it's actually the opposite of the definition of Poisson brackets that I gave at the start. In particular, this definition of the Poisson brackets would seem to correspond to

$$\{F,G\} = \sum_{k}\left(\frac{\partial F}{\partial p_k}\frac{\partial G}{\partial q^k} - \frac{\partial F}{\partial q^k}\frac{\partial G}{\partial p_k} \right)$$

I can calculate all of the Poisson brackets for the closed and open strings, including those for the oscillator modes. The thing is that I'm kind of concerned about the minus sign error that I'm getting with my definition of Poisson brackets compared with that used by GSW. Is there a particular reason why they've chosen to define the Poisson brackets in this way, or am I making a terribly simple mistake somewhere?

4. Apr 21, 2006

### Timbuqtu

I think this minus sign is just their convention. But indeed there seems to be a small mistake in the book when they start quantizing the bosonic string.

They first define

$$\left[ \dot{X}^{\mu}(\sigma),X^{\nu}(\sigma')\right]_{P.B.} = T^{-1} \delta (\sigma - \sigma') \eta^{\mu\nu} (2.1.52)$$

i.e.

$$\left[ P^{\mu}(\sigma),X^{\nu}(\sigma')\right]_{P.B.} = \delta (\sigma - \sigma') \eta^{\mu\nu}$$

Then they say they replace the poisson brackets by commutators via the substitution

$$\left[\ldots\right]_{P.B.} \rightarrow - i \left[\ldots\right] (2.2.4)$$

concluding that

$$\left[ P^{\mu}(\sigma),X^{\nu}(\sigma')\right] = -i\delta (\sigma - \sigma') \eta^{\mu\nu} (2.2.5)$$

but this minus sign does not seem to be consistent.

5. Apr 21, 2006

### coalquay404

Thanks for that, I thought that it didn't make sense alright. I've seen several mistakes like this in both GSW and Clifford Johnson's book. They're really frustrating.

6. Apr 21, 2006

### josh1

If we write

$$\left[\ldots\right]_{P.B.}=RHS$$,

then

$$\left[\ldots\right]_{P.B.} \rightarrow - i \left[\ldots\right]$$

means

$$RHS \rightarrow - iRHS$$

Unless you interpret things this way, it looks like there are lots of sign errors.

7. Apr 21, 2006

### coalquay404

The way I view it is as follows. We have a Poisson bracket equal to something:

$$\{\ldots\}_{\textrm{P.B.}} = \textrm{something}$$

Then following the prescription

$$\{\ldots\}_{\textrm{P.B.}} \to -i[\ldots]$$

we must have

$$-i[\ldots]=\{\ldots\}_{\textrm{P.B.}}=\textrm{something}$$

Therefore, I read this as saying

$$[\ldots] = i(\textrm{something})$$

8. Apr 21, 2006

### josh1

Hi coalquay404,

I read it the same way you and I'm sure any sane person does. It's just that if you read it the alternative way I described, the signs are all consistent. So you have a choice: Read it the natural way and the signs are wrong; Read it the retarded way I described and the signs are right. Somebody should email witten and ask him.

Last edited by a moderator: Apr 21, 2006
9. Apr 21, 2006

### coalquay404

Funnily enough, a couple of days ago Michael Green swore blind to me that there were *no* mistakes in the book at all, claiming that they were all ironed out after the second printing. This is obviously incorrect (check the first two paragraphs in chapter 2 for an absolute hummer of a contradiction).

10. Apr 22, 2006

### josh1

11. Apr 22, 2006

### coalquay404

Well, consider the following quote from the first paragraph in section 2.1:

Thus, let us consider the motion of a point particle of mass $$m$$ in a background gravitational field, i.e., in a curved Riemannian geometry described by a metric tensor $$g_{\mu\nu}(x)$$. The metric is assumed to have $$D-1$$ positive eigenvalues and one negative eigenvalue corresponding to the Minkowski signature of $$D$$-dimensional space-time.

The point about this is that they claim that the background field is described by a Riemannian manifold, i.e., something which is locally homeomorphic to $$\mathbb{R}^{D}$$. This isn't correct. If it's a background space-time it's got to be locally Minkowskian, not locally Euclidean. That's why it doesn't make sense to claim that the background is Riemannian and yet it has a metric with a negative eigenvalue.

12. Apr 22, 2006

### josh1

A manifold is Riemannian if it's tangent space has a smoothly varying inner product g no matter what g's signature is.

13. Apr 22, 2006

Staff Emeritus

A manifold which behaves like a Riemannian one but has a Minkowskian signature is now called Lorentzian[/i}, a bad name IMHO becuse Lorentz had nothing to do with it. It should have been called Einsteinian, as he was the one who really invented this geometry. The name Riemannian is now reserved for manifolds with Euclidean signature.

Anyway, that's the "error", just a difference of terminology.

14. Apr 22, 2006

### marcus

Wiki "pseudo-Riemannian"
http://en.wikipedia.org/wiki/Pseudo-Riemannian

"In differential geometry, a pseudo-Riemannian manifold ... is a generalization of a Riemannian manifold. The key difference between the two is that on a pseudo-Riemannian manifold the metric tensor need not be positive-definite. Instead a weaker condition of nondegeneracy is imposed.
Arguably, the most important type of pseudo-Riemannian manifold is a Lorentzian manifold. Lorentzian manifolds occur in the general theory of relativity as models of curved 4-dimensional spacetime. Just as Riemannian manifolds may be thought of a being locally modeled on Euclidean space, Lorentzian manifolds are locally modeled on Minkowski space."

15. Apr 22, 2006

Staff Emeritus
Yes, thank you Marcus. That gives us the full picture. My point was that many people from the older tradition continue to use the "abus de langage" of calling all of them Riemannian without regard for signature.

16. Apr 22, 2006

### josh1

Yes, describing spacetime in terms of "Riemannian manifolds with lorentzian metrics" is common among physicists. I guess we shouldn't feel too bad about viewing Riemannian manifolds in this way since Witten who is one of the worlds greatest mathematicians apparently has no problem with it, at least as far as his physics book is concerned. Hawking in his famous book "The large scale structure of spacetime" also fails to make the distinction. With this understanding, I can't say that GSW erred.

17. Apr 26, 2006

### coalquay404

Hi, sorry for the delay in replying. I appreciate now that it may have been a difference in terminology that caused me to think that it was actually a mistake. What's even more confusing, now that I look back on it, is that I talked to Michael Green over the weekend about this point and he admitted that it should have read "locally Lorentzian" where it says "Riemannian." He claims that John Schwarz wrote that chapter and, even more interestingly, that he has read hardly any of volume 1.

Anyhow, thanks for the help.

18. Apr 27, 2006

### George Jones

Staff Emeritus
Another possible point of confusion: mathematicians and physicists tend to mean different things by the term "locally". For a mathematician, a space is locally #$!% if every element of the space is contained in a neighbourhood that is #$!%. A common example takes #\$!% = compact.

Often, e.g., in this thread, physicists mean something quite different by "locally".

Regards,
George

19. Apr 28, 2006