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String Theory

  • #1
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http://www.damtp.cam.ac.uk/user/tong/string/string.pdf

In these notes, I am trying to derive the equation of motion (1.21) corresponding to the Nambu-Goto action:

[itex]\partial_\alpha ( \sqrt{ - \text{det} \gamma} \gamma^{\alpha \beta} \partial_\beta X^\mu ) =0[/itex]

I have found and proved all the stuff he says we need in order to do it in the above paragraph but just cannot get it to work out.

Thanks to anybody who can help!
 

Answers and Replies

  • #2
fzero
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I'm not sure if you're having an algebraic issue or a conceptual issue. (1.21) follows from the formula in the text above via the chain rule. The equation of motion is

[tex]0=\partial_\epsilon \left( \frac{\delta\sqrt{-\gamma}}{\delta(\partial_\epsilon X^\mu)} \right) = \partial_\epsilon \left( \frac{\delta\sqrt{-\gamma}}{\delta\gamma_{\alpha\beta} } \frac{\delta\gamma_{\alpha\beta}}{\delta(\partial_\epsilon X^\mu)} \right) .[/tex]

(1.21) follows from putting everything together.
 
  • #3
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I'm not sure if you're having an algebraic issue or a conceptual issue. (1.21) follows from the formula in the text above via the chain rule. The equation of motion is

[tex]0=\partial_\epsilon \left( \frac{\delta\sqrt{-\gamma}}{\delta(\partial_\epsilon X^\mu)} \right) = \partial_\epsilon \left( \frac{\delta\sqrt{-\gamma}}{\delta\gamma_{\alpha\beta} } \frac{\delta\gamma_{\alpha\beta}}{\delta(\partial_\epsilon X^\mu)} \right) .[/tex]

(1.21) follows from putting everything together.
Algebraic. Can you explain this in more detail please. Why are there deltas involved?

Thanks.
 
  • #4
fzero
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The equation of motion is obtained by using the variational principle

[tex]
0=\partial_\epsilon \left( \frac{\delta S}{\delta(\partial_\epsilon X^\mu)} \right) - \frac{\delta S}{\delta( X^\mu)}.
[/tex]

The 2nd term vanishes, while the equation I wrote in post 2 follows from the first term.
 
  • #5
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The equation of motion is obtained by using the variational principle

[tex]
0=\partial_\epsilon \left( \frac{\delta S}{\delta(\partial_\epsilon X^\mu)} \right) - \frac{\delta S}{\delta( X^\mu)}.
[/tex]

The 2nd term vanishes, while the equation I wrote in post 2 follows from the first term.
Ok. But the answer doesn't ahve any deltas in it so why are we using the variational EL equations? Why not use

[itex]\partial_\mu \left( \frac{\partial L}{\partial ( \partial_\mu X^\nu )} \right) = \frac{\partial L}{\partial X^\nu}[/itex]
 
  • #6
fzero
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Using [tex]\delta[/tex] here is partially a matter of notation and partially an attempt to distinguish that the functional derivative is not mathematically the same as a normal derivative. I don't believe that this issue would prevent you from obtaining the result, so you might want to explain in more detail where you got stuck in your derivation.
 

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