Rigidly rotating relativistic string

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SUMMARY

The discussion focuses on calculating the angular momentum of a rigidly rotating relativistic string as described in Zwiebach's 'A First Course in String Theory'. The string is parameterized with time τ=t and spatial coordinate σ=s, where s ranges from -l/2 to l/2. The solution for the string's position is given by vec{X}(t,s) = s(cos(ωt), sin(ωt)). The user encounters difficulties in determining the correct expression for momentum density vec{P}, leading to discrepancies in angular momentum calculations.

PREREQUISITES
  • Understanding of string theory concepts, particularly rigidly rotating strings.
  • Familiarity with the equations of motion for relativistic strings.
  • Knowledge of angular momentum calculations in relativistic contexts.
  • Proficiency in tensor calculus and the use of Lagrangians in physics.
NEXT STEPS
  • Research the derivation of angular momentum for relativistic strings in string theory.
  • Study the relationship between energy and tension in string theory, specifically l = (2/π)(E/T_0).
  • Learn about the implications of free endpoint conditions on string dynamics.
  • Investigate the role of the Lorentz factor γ(v) in relativistic momentum density calculations.
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This discussion is beneficial for theoretical physicists, graduate students in string theory, and researchers focusing on relativistic dynamics and angular momentum in high-energy physics.

da_willem
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I am reading Zwiebach's 'a first course in string theory' and I have a question. As an example of a solution to the equations of motion he uses a rigidly rotating string with length l in the x-y plane with it's center at the origin. I would like to do some calculations using as parameters

\tau=t and \sigma =s

with s the distance from the origin, and t just time. s runs from -l/2 to l/2. This should be possible right, as I'm free to choose which way to paramterize the string.

Now the solution is (given that the string lies along the x-axis at t=0) ofcourse:

\vec{X} (t,s) = s(cos(\omega t), sin(\omega t))

As the endpoints are not tied down to anything I have to use free enpoint conditions which means the endpoints move at the speed of light. This yields for the enpoints c= \omega l/2. Futhermore you can calculate that l is related to the energy of the string E by:

\frac{2}{\pi} \frac{E}{T_0}

With T0 the tension of the string. Now the solution is complete as both l and omega are found.

Now I would like to calculate the angular momentum, so I can find \alpha &#039;[/tex] the constant of proprtianality between the angular momentum and the energy squared (in terms of hbar). I tried:<br /> <br /> J= \int _{-l/2} ^{l/2} X_1 P_2 - X_2 P_1 ds should be \frac{E^2}{2 \pi T_0 x}<br /> <br /> but now I don&#039;t know what expression to use for P the momentum density. I tried:<br /> <br /> \vec{P} =\frac{T_0}{c^2} \frac{\partial \vec{X}}{\partial t}<br /> <br /> but this yields a result a factor 4/3pi off. So I guess there is something wrong with my expression for P. T_0/c^2 is the rest mass density of a relativistic string. But my string is moving, so I tried<br /> <br /> \vec{P}=\gamma (v) \frac{T_0}{c^2} \frac{\partial \vec{X}}{\partial t}<br /> <br /> with v= \omega s, but this diverges bacause gamma explodes at the string endpoints. What&#039;s wrong? How to calculate the angular momentum using \sigma =s?
 
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would be nice to have an answer to this question...I would be very interested to see how alpha prime is derived.
 

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