String motion

  • Thread starter moriheru
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  • #1
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My question concerns motion of relativistic Strings, as the reader with a great a mount of deductive skills can deduce :).
img004.jpg

The question is: How can I derive the P's. More acuratly speaking why is the numerator how it is.
I am refering to Zwiebach chapter 6.
Thanks for any clarifications and sorry about the zoom.
 

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Answers and Replies

  • #2
Vanadium 50
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Please type your question in, using LaTex if you need to express a formula. That's better than asking us to try and figure out what is in a tiny blurry picture.
 
  • #3
ChrisVer
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Take the Lagrangian you have, and find its derivative...
The Nambu Goto lagrangian he has derived is : [itex] L = - \frac{T_0}{c} \sqrt{ - det \gamma } = - \frac{T_0}{c} \sqrt{ ( \dot{X} X' )^2 - \dot{X}^2 X'^2 } [/itex]
Forgeting the prefactor:
[itex] \frac{\partial L}{\partial \dot{X}^\rho} \propto \frac{2 ( \dot{X} X' ) X'_\rho - 2 X'^2 \dot{X}_\rho}{2 \sqrt{ ( \dot{X} X' )^2 - \dot{X}^2 X'^2 } }[/itex]
Similarily for the other...
If you have some problems with these derivatives of inner products:
[itex] \frac{\partial }{\partial \dot{X}^\rho} ( \dot{X}^\mu X'_\mu )^2 =\frac{\partial }{\partial \dot{X}^\rho} \dot{X}^\mu X'_\mu \dot{X}^\nu X'_\nu = \delta_{\rho}^{\mu} X'_\mu \dot{X}^\nu X'_\nu + \dot{X}^\mu X'_\mu \delta_{\rho}^{\nu} X'_\nu = X'_\rho \dot{X}^\nu X'_\nu + \dot{X}^\mu X'_\mu X'_\rho =2 (\dot{X}^\nu X'_\nu ) X'_\rho [/itex]

The only thing you use is the orthonormality relation [itex] \frac{\partial x^a}{\partial x^b}= \delta_b^a[/itex]
 
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  • #4
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Take the Lagrangian you have, and find its derivative...
The Nambu Goto lagrangian he has derived is : [itex] L = - \frac{T_0}{c} \sqrt{ - det \gamma } = - \frac{T_0}{c} \sqrt{ ( \dot{X} X' )^2 - \dot{X}^2 X'^2 } [/itex]
Forgeting the prefactor:
[itex] \frac{\partial L}{\partial \dot{X}^\rho} \propto \frac{2 ( \dot{X} X' ) X'_\rho - 2 X'^2 \dot{X}_\rho}{2 \sqrt{ ( \dot{X} X' )^2 - \dot{X}^2 X'^2 } }[/itex]
Similarily for the other...
If you have some problems with these derivatives of inner products:
[itex] \frac{\partial }{\partial \dot{X}^\rho} ( \dot{X}^\mu X'_\mu )^2 =\frac{\partial }{\partial \dot{X}^\rho} \dot{X}^\mu X'_\mu \dot{X}^\nu X'_\nu = \delta_{\rho}^{\mu} X'_\mu \dot{X}^\nu X'_\nu + \dot{X}^\mu X'_\mu \delta_{\rho}^{\nu} X'_\nu = X'_\rho \dot{X}^\nu X'_\nu + \dot{X}^\mu X'_\mu X'_\rho =2 (\dot{X}^\nu X'_\nu ) X'_\rho [/itex]

The only thing you use is the orthonormality relation [itex] \frac{\partial x^a}{\partial x^b}= \delta_b^a[/itex]

Thanks! A like is yours.
 
  • #5
272
17
Take the Lagrangian you have, and find its derivative...
The Nambu Goto lagrangian he has derived is : [itex] L = - \frac{T_0}{c} \sqrt{ - det \gamma } = - \frac{T_0}{c} \sqrt{ ( \dot{X} X' )^2 - \dot{X}^2 X'^2 } [/itex]
Forgeting the prefactor:
[itex] \frac{\partial L}{\partial \dot{X}^\rho} \propto \frac{2 ( \dot{X} X' ) X'_\rho - 2 X'^2 \dot{X}_\rho}{2 \sqrt{ ( \dot{X} X' )^2 - \dot{X}^2 X'^2 } }[/itex]
Similarily for the other...
If you have some problems with these derivatives of inner products:
[itex] \frac{\partial }{\partial \dot{X}^\rho} ( \dot{X}^\mu X'_\mu )^2 =\frac{\partial }{\partial \dot{X}^\rho} \dot{X}^\mu X'_\mu \dot{X}^\nu X'_\nu = \delta_{\rho}^{\mu} X'_\mu \dot{X}^\nu X'_\nu + \dot{X}^\mu X'_\mu \delta_{\rho}^{\nu} X'_\nu = X'_\rho \dot{X}^\nu X'_\nu + \dot{X}^\mu X'_\mu X'_\rho =2 (\dot{X}^\nu X'_\nu ) X'_\rho [/itex]

The only thing you use is the orthonormality relation [itex] \frac{\partial x^a}{\partial x^b}= \delta_b^a[/itex]

But why would one have a inner product?
 
  • #6
ChrisVer
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What do you mean by why you have inner products? they are there when you wrote:

[itex] \gamma = \begin{pmatrix} \dot{X} \cdot \dot{X} & \dot{X} \cdot X' \\ X' \cdot \dot{X} & X' \cdot X' \end{pmatrix}[/itex]
The [itex]\cdot[/itex] denotes inner product, and in this case the Lorentzian product [itex] x \cdot y = \eta_{\mu \nu} x^\mu y^\nu [/itex]...
Now why is that the case? because you have already derived the induced metric:
[itex] \gamma_{ab} = \eta_{\mu \nu} \partial_a X^\mu \partial_b X^\nu \equiv \partial_a X \cdot \partial_b X[/itex]
 
  • #7
ChrisVer
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Or if you didn't ask exactly that, then remember:
[itex] \frac{d}{dx} f(g(x)) = \frac{d}{dg}f(g) \frac{dg(x)}{dx} [/itex]

So the [itex]\sqrt{...}[/itex] will give the [itex] \frac{1}{2\sqrt{...}}[/itex]
and you still have to take the derivatives of [itex](...)[/itex] wrt your initial derivative.
 
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  • #8
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Thanks fine from here!
 

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