What is the solution to Zwiebach eqn (17.44) on page 395?

  • Thread starter Jimmy Snyder
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In summary, the given equation (17.44) on page 395 can be easily derived if \frac{\ell}{R} is small, but since \ell can be any integer, there are no conditions necessary for it to be true to all orders. Using the suggestion provided, the equation can be simplified to p + \frac{\ell}{R}.
  • #1
Jimmy Snyder
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[SOLVED] Zwiebach eqn (17.44) page 395

Homework Statement


Equation (17.44) is:
[tex]e^{-i\ell x_0/R}pe^{i\ell x_0/R} = p + \frac{\ell}{R}[/tex]



Homework Equations


Equation (17.43)
[tex]\ell \in \mathbb{Z}[/tex]



The Attempt at a Solution


Equation (17.44) is easy to derive if [itex]\frac{\ell}{R}[/itex] is small. but since [itex]\ell[/itex] can be any integer, there doesn't seem to be any way to insure that. Did the author simply forget to mention some conditions necessary for it to be true?
 
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  • #2
jimmysnyder said:

Homework Statement


Equation (17.44) is:
[tex]e^{-i\ell x_0/R}pe^{i\ell x_0/R} = p + \frac{\ell}{R}[/tex]



Homework Equations


Equation (17.43)
[tex]\ell \in \mathbb{Z}[/tex]



The Attempt at a Solution


Equation (17.44) is easy to derive if [itex]\frac{\ell}{R}[/itex] is small. but since [itex]\ell[/itex] can be any integer, there doesn't seem to be any way to insure that. Did the author simply forget to mention some conditions necessary for it to be true?

It's true to all orders. Just use [itex] [ x_0^n, p] = i n x_0^{n-1} [/itex].
 
  • #3
nrqed said:
It's true to all orders. Just use [itex] [ x_0^n, p] = i n x_0^{n-1} [/itex].
Thanks for taking a look at this nrqed. If [itex]\frac{\ell}{R}[/itex] were small, then I would expand the exponentials to linear terms. But it isn't small, so it would seem that I need to keep all terms. But if I consider up to quadratic terms I get:
[tex]p + \frac{\ell}{R} + \frac{\ell^2}{2!R^2}(x_0^2p + x_0px_0 - px_0^2) + \mathcal{O}({\frac{\ell^3}{R^3}})[/itex]
Using your suggestion this becomes:
[tex]p + \frac{\ell}{R} + \frac{\ell^2}{2!R^2}(2ix_0 + x_0px_0) + \mathcal{O}(\frac{\ell^3}{R^3})[/itex]
This doesn't seem to improve things. Is there a simpler approach that I am missing?
 
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  • #4
jimmysnyder said:
Thanks for taking a look at this nrqed. If [itex]\frac{\ell}{R}[/itex] were small, then I would expand the exponentials to linear terms. But it isn't small, so it would seem that I need to keep all terms. But if I consider up to quadratic terms I get:
[tex]p + \frac{\ell}{R} + \frac{\ell^2}{2!R^2}(x_0^2p + x_0Px_0 - px_0^2) + \mathcal{O}({\frac{\ell^3}{R^3}})[/itex]
Using your suggestion this becomes:
[tex]p + \frac{\ell}{R} + \frac{\ell^2}{2!R^2}(2ix_0 + x_0Px_0) + \mathcal{O}(\frac{\ell^3}{R^3})[/itex]
This doesn't seem to improve things. Is there a simpler approach that I am missing?

Hi Jimmy,

I am in a hurry because I am teaching a class in 15 minutes and have stuff to do. I will get back to you early this afternoon.
 
  • #5
nrqed said:
I will get back to you early this afternoon.
Take your time. When you get back to it, you will find that I erroniously put upper case P's in my equations. They should be lower case. I fixed it in my post.

Edit - With your help, I have figured this out. Thanks nrqed.
 
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  • #6
jimmysnyder said:
Take your time. When you get back to it, you will find that I erroniously put upper case P's in my equations. They should be lower case. I fixed it in my post.

Edit - With your help, I have figured this out. Thanks nrqed.

I just got back from my classes. I am glad it worked out! As you see, it works to all orders.

Glad I could help!


Patrick
 
  • #7
Just for the record, here is the solution:
[tex]e^{-ilx_0/R}pe^{ilx_0/R}[/tex]

[tex]= (\Sigma\frac{1}{n!}(\frac{-il}{R})^nx_0^np)e^{ilx_0/R}[/tex]

Now using nrqed's suggestion:

[tex]= (\Sigma\frac{1}{n!}(\frac{-il}{R})^npx_0^n + \Sigma_0\frac{1}{n!}(\frac{-il}{R})^ninx_0^{n-1})e^{ilx_0/R}[/tex]

[tex]= (p\Sigma\frac{1}{n!}(\frac{-il}{R})^nx_0^n + \Sigma_1\frac{1}{n!}(\frac{-il}{R})^ninx_0^{n-1})e^{ilx_0/R}[/tex]

[tex]= (pe^{-ilx_0/R} + \frac{\ell}{R}\Sigma_1\frac{1}{(n-1)!}(\frac{-il}{R})^{n-1}x_0^{n-1})e^{ilx_0/R}[/tex]

[tex]= (pe^{-ilx_0/R} + \frac{\ell}{R}\Sigma_0\frac{1}{n!}(\frac{-il}{R})^n}x_0^n)e^{ilx_0/R}[/tex]

[tex]= (pe^{-ilx_0/R} + \frac{\ell}{R}e^{-ilx_0/R})e^{ilx_0/R}[/tex]

[tex]= p + \frac{\ell}{R}[/tex]
 
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1. What is the Zwiebach equation (17.44) on page 395?

The Zwiebach equation (17.44) on page 395 is a mathematical formula that describes the dynamics of open strings in string theory. It is used to calculate the scattering amplitudes of these strings.

2. How is the Zwiebach equation derived?

The Zwiebach equation is derived from the Polyakov action, which is the action that describes the dynamics of strings. By applying the equations of motion and using certain mathematical techniques, the equation can be derived.

3. What is the significance of the Zwiebach equation in string theory?

The Zwiebach equation is a fundamental equation in string theory, as it allows for the calculation of scattering amplitudes for open strings. This is important because it helps to understand the behavior of strings and their interactions.

4. Can the Zwiebach equation be applied to closed strings?

No, the Zwiebach equation is specifically designed for open strings in string theory. There are other equations and methods that are used to study closed strings.

5. Are there any real-world applications of the Zwiebach equation?

The Zwiebach equation is primarily used in theoretical physics to study the behavior of strings. However, some researchers have also applied it to other areas such as condensed matter physics and quantum gravity.

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