thesis_orthog_gauge.lyx

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\begin_body

\begin_layout Title
Tel Aviv University
\end_layout

\begin_layout Author
Faculty of Natural Sciences
\end_layout

\begin_layout Author
School of Graduate Studies
\end_layout

\begin_layout Title
Charged Holographic Strings
\end_layout

\begin_layout Author
A Thesis submitted toward the degree of Master of Science in Physics
\end_layout

\begin_layout Author
by Lior Blech
\end_layout

\begin_layout Section*
Acknowledgements
\end_layout

\begin_layout Section*
Abstract
\end_layout

\begin_layout Standard
\begin_inset CommandInset toc
LatexCommand tableofcontents

\end_inset


\end_layout

\begin_layout Standard
\begin_inset FloatList figure

\end_inset


\end_layout

\begin_layout Standard
\begin_inset FloatList table

\end_inset


\end_layout

\begin_layout Part
Introduction
\end_layout

\begin_layout Section
Regge Trajectories and the Spinnig string
\end_layout

\begin_layout Standard
\begin_inset Formula $S=-T\iint_{(0,-l_{2})}^{(T,l_{1})}d^{2}\sigma\sqrt{-\det h}-\int d\tau\sqrt{u^{2}}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $h=\left(\begin{array}{cc}
\cfrac{dx^{\mu}}{d\tau}\cfrac{dx_{\mu}}{d\tau} & \cfrac{dx^{\mu}}{d\tau}\cfrac{dx_{\mu}}{d\sigma}\\
\cfrac{dx^{\mu}}{d\tau}\cfrac{dx_{\mu}}{d\sigma} & \cfrac{dx^{\mu}}{d\sigma}\cfrac{dx_{\mu}}{d\sigma}
\end{array}\right)$
\end_inset


\end_layout

\begin_layout Standard
We can write for short 
\begin_inset Formula $V^{\mu}=\cfrac{dx^{\mu}}{d\sigma},u^{\mu}\cfrac{dx^{\mu}}{d\tau}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $h=\left(\begin{array}{cc}
u^{2} & u\cdot V\\
u\cdot V & V^{2}
\end{array}\right)$
\end_inset


\end_layout

\begin_layout Standard
We will work in the orthogonal gauge in which 
\begin_inset Formula $u\cdot V=0$
\end_inset

, 
\begin_inset Formula $u^{2}=-V^{2}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula $h=\left(\begin{array}{cc}
u^{2} & 0\\
0 & -u^{2}
\end{array}\right)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $S=-T\iint_{(0,-l_{2})}^{(T,l_{1})}d^{2}\sigma u^{2}=-\frac{1}{2}T\iint_{(0,-l_{2})}^{(T,l_{1})}d^{2}\sigma\left(u^{2}-V^{2}\right)$
\end_inset


\end_layout

\begin_layout Standard
The equations of motion are
\end_layout

\begin_layout Standard
\begin_inset Formula $\frac{\partial\mathcal{L}}{\partial x^{\mu}}-\partial_{\alpha}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\alpha}x^{\mu})}\right)=0$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\partial_{0}\left(\frac{\partial\mathcal{L}}{\partial u^{\mu}}\right)+\partial_{1}\left(\frac{\partial\mathcal{L}}{\partial V^{\mu}}\right)=0$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $-\ddot{x^{\mu}}+x^{\mu}''=0$
\end_inset


\end_layout

\begin_layout Standard
Whose solution is (basically a two dimensional wave equation)
\end_layout

\begin_layout Standard
\begin_inset Formula $x^{\mu}=\sum_{n}e^{i\omega_{n}\left(\tau-\sigma\right)}\frac{\alpha_{n}^{\mu}}{\omega_{n}}+e^{i\omega_{n}\left(\tau+\sigma\right)}\frac{\beta_{n}^{\mu}}{\omega_{n}}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\ddot{x^{\mu}}=-\sum_{n}\omega_{n}\left(e^{i\omega_{n}\left(\tau-\sigma\right)}\alpha_{n}^{\mu}+e^{i\omega_{n}\left(\tau+\sigma\right)}\beta_{n}^{\mu}\right)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $x^{\mu}''=-\sum_{n}\omega_{n}\left(e^{i\omega_{n}\left(\tau-\sigma\right)}\alpha_{n}^{\mu}+e^{i\omega_{n}\left(\tau+\sigma\right)}\beta_{n}^{\mu}\right)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\dot{x^{\mu}}=i\sum_{n}e^{i\omega_{n}\left(\tau-\sigma\right)}\alpha_{n}^{\mu}+e^{i\omega_{n}\left(\tau+\sigma\right)}\beta_{n}^{\mu}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $x^{\mu}'=i\sum_{n}-e^{i\omega_{n}\left(\tau-\sigma\right)}\alpha_{n}^{\mu}+e^{i\omega_{n}\left(\tau+\sigma\right)}\beta_{n}^{\mu}$
\end_inset


\end_layout

\begin_layout Standard

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\begin_inset Formula $\dot{x^{\mu}}\cdot x_{\mu}'=\dot{x^{\mu}}=-\left(\sum_{n}e^{i\omega_{n}\left(\tau-\sigma\right)}\alpha_{n}^{\mu}+e^{i\omega_{n}\left(\tau+\sigma\right)}\beta_{n}^{\mu}\right)\left(\sum_{m}-e^{i\omega_{m}\left(\tau-\sigma\right)}\alpha_{m\mu}+e^{i\omega_{m}\left(\tau+\sigma\right)}\beta_{m\mu}\right)$
\end_inset


\end_layout

\begin_layout Standard

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\begin_inset Formula $=-\left(\sum_{n}\sum_{m}\left(-e^{i\left(\omega_{n}+\omega_{m}\right)\left(\tau-\sigma\right)}\alpha_{n}^{\mu}\alpha_{m\mu}+e^{i\left(\omega_{m}+\omega_{n}\right)\left(\tau+\sigma\right)}\beta_{n}^{\mu}\beta_{m\mu}\right)+\sum_{n}\sum_{m}\left(e^{i\omega_{n}\left(\tau+\sigma\right)}e^{i\omega_{n}\left(\tau-\sigma\right)}\alpha_{n}^{\mu}\beta_{m\mu}-e^{i\omega_{m}\left(\tau-\sigma\right)}e^{i\omega_{n}\left(\tau+\sigma\right)}\beta_{n}^{\mu}\alpha_{m\mu}\right)\right)$
\end_inset


\end_layout

\begin_layout Standard
Generaly from some boundary conditions we would get 
\begin_inset Formula $\alpha\propto\beta$
\end_inset

 so
\end_layout

\begin_layout Standard

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\begin_inset Formula $\propto-\sum_{n}\sum_{m}\left(e^{i\left(\omega_{m}+\omega_{n}\right)\left(\tau+\sigma\right)}-e^{i\left(\omega_{n}+\omega_{m}\right)\left(\tau-\sigma\right)}\right)\alpha_{n}^{\mu}\alpha_{m\mu}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $S=\int\mathcal{L}d^{2}\sigma+\int Ld\tau$
\end_inset


\end_layout

\begin_layout Standard
variate the solution by an infinitesimal function 
\begin_inset Formula $\delta x^{\mu}(\sigma,\tau)$
\end_inset

 such that 
\begin_inset Formula $\delta x^{\mu}(0)=\delta x^{\mu}(T)=0$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\delta S=\int\left(\frac{\partial\mathcal{L}}{\partial x^{\mu}}\delta x^{\mu}+\frac{\partial\mathcal{L}}{\partial(\partial_{\alpha}x^{\mu})}\partial_{\alpha}\delta x^{\mu}\right)d^{2}\sigma=\int\left(\frac{\partial\mathcal{L}}{\partial x^{\mu}}-\partial_{\alpha}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\alpha}x^{\mu})}\right)\right)\delta x^{\mu}d^{2}\sigma+\int\frac{\partial\mathcal{L}}{\partial(\partial_{\sigma}x^{\mu})}\delta x^{\mu}|_{\sigma=-l_{2}}^{\sigma=l_{1}}d\tau$
\end_inset


\end_layout

\begin_layout Standard
So requiring no variation to first order in the action:
\end_layout

\begin_layout Standard
\begin_inset Formula $\frac{\partial\mathcal{L}}{\partial x^{\mu}}-\partial_{\alpha}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\alpha}x^{\mu})}\right)=0$
\end_inset


\end_layout

\begin_layout Standard
and 
\begin_inset Formula $\frac{\partial\mathcal{L}}{\partial(\partial_{\sigma}x^{\mu})}=-T\cfrac{\left(u\cdot V\right)u_{\mu}-u^{2}V_{\mu}}{\sqrt{\left(u\cdot V\right)^{2}-u^{2}V^{2}}}=0$
\end_inset

 or 
\begin_inset Formula $\delta x^{\mu}=0$
\end_inset

 at the end points.
\end_layout

\begin_layout Standard

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\begin_inset Formula $\frac{\partial\mathcal{L}}{\partial(\partial_{\sigma}x^{\mu})}|_{\sigma=-l_{2}}^{\sigma=l_{1}}=\frac{\partial\mathcal{L}}{\partial V^{\mu}}|_{\sigma=-l_{2}}^{\sigma=l_{1}}=0$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $u=\left(1,\vec{v}\right)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $V=\left(0,\cfrac{\vec{x}}{r}\right)$
\end_inset


\end_layout

\begin_layout Standard
One possible solution of the equations of motion is that of an open rotating
 string.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\begin{cases}
t=\tau\\
x=\sigma\cdot sin\left(\omega\tau\right) & y=\sigma\cdot cos\left(\omega\tau\right)\\
z=const
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
u=\left(1,\omega\sigma\cdot cos\left(\omega\tau\right),-\omega\sigma\cdot sin\left(\omega\tau\right),0\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V=\left(0,sin\left(\omega\tau\right),cos\left(\omega\tau\right),0\right)
\]

\end_inset


\end_layout

\begin_layout Standard
The free endpoint boundary condition translates to
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\begin{cases}
\mu=0 & T\cfrac{\vec{v}\cdot\vec{V}}{\sqrt{1-\omega^{2}\sigma^{2}}}=0\\
\mu=i & T\sqrt{1-\omega^{2}\sigma^{2}}V^{i}=0
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
The timelike part is satisfied for any rotating string.
 The spacelike parts are satisfied when 
\begin_inset Formula $\omega l=1$
\end_inset

.
 Later when dealing with massive endpoints this condition will be relaxed.
\end_layout

\begin_layout Standard
We can calculate the energy and angular momentum of this configuration in
 poincare coordinates.
\end_layout

\begin_layout Standard
Because we use the static gauge, the energy can either be calculated as
 the hamiltonian in the static gauge, or gauge independently as the Noether
 current for the transformation:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\delta t=\epsilon
\]

\end_inset


\end_layout

\begin_layout Standard
Using the hamiltonian approach we get:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
S=-T\iint_{(0,-l_{2})}^{(T,l_{1})}d^{2}\sigma\sqrt{\left(u\cdot V\right)^{2}-u^{2}V^{2}}=-T\iint_{(0,-l_{2})}^{(T,l_{1})}dtd\sigma\sqrt{\left(\vec{v}\cdot\vec{V}\right)^{2}+\left(1-\vec{v}^{2}\right)\vec{V}^{2}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
E^{gf}=\int_{-l_{2}}^{l1}d\sigma\left(\cfrac{\partial\mathcal{L}}{\partial v^{i}}v^{i}-\mathcal{L}\right)=T\int_{-l_{2}}^{l1}d\sigma\cfrac{\vec{V}^{2}}{\sqrt{\left(\vec{v}\cdot\vec{V}\right)^{2}+\left(1-\vec{v}^{2}\right)\vec{V}^{2}}}
\]

\end_inset


\end_layout

\begin_layout Standard
Inserting the spinning string solution
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
u=\left(1,\omega\sigma\cdot cos\left(\omega\tau\right),-\omega\sigma\cdot sin\left(\omega\tau\right),0\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
V=\left(0,sin\left(\omega\tau\right),cos\left(\omega\tau\right),0\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
E=T\int_{-l_{2}}^{l1}d\sigma\cfrac{1}{\sqrt{1-\omega^{2}\sigma^{2}}}=T\int_{-l_{2}}^{l1}d\sigma\cfrac{1}{\sqrt{1-\omega^{2}\sigma^{2}}}=\cfrac{T}{\omega}sin^{-1}\left(\omega l\right)=\cfrac{2T}{\omega}sin^{-1}\left(1\right)=\cfrac{\pi T}{\omega}
\]

\end_inset


\end_layout

\begin_layout Standard
The angular momentum is calculated as the Noether current for the rotation
 transformation:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\delta x=-y\epsilon
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\delta y=x\epsilon
\]

\end_inset


\end_layout

\begin_layout Standard
More convinently we can go to polar coordinates and formulate the rotation
 as
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\delta\theta=\epsilon
\]

\end_inset


\end_layout

\begin_layout Standard
That is
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
S=-T\iint_{(0,-l_{2})}^{(T,l_{1})}dtdr\sqrt{1-\dot{\theta}^{2}r^{2}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
J=\int_{-l_{2}}^{l1}d\sigma\cfrac{\partial\mathcal{L}}{\partial\dot{\theta}}\dot{\theta}=T\int_{-l_{2}}^{l_{1}}dr\cfrac{\dot{\theta}^{2}r^{2}}{\sqrt{1-\dot{\theta}^{2}r^{2}}}=\cfrac{T}{\omega}\left(sin^{-1}\left(\omega r\right)-\omega r\sqrt{1-\omega^{2}r^{2}}\right)=\cfrac{\pi T}{2\omega}
\]

\end_inset


\end_layout

\begin_layout Standard
We can combine the two relations and obtain
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
J=\frac{1}{2}\cfrac{T}{\omega}\left(sin^{-1}\left(\omega r\right)-\omega r\sqrt{1-\omega^{2}r^{2}}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
E=\cfrac{T}{\omega}sin^{-1}\left(\omega r\right)
\]

\end_inset


\end_layout

\begin_layout Section
The AdS/CFT Duality
\end_layout

\begin_layout Section
Confining backgrounds
\end_layout

\begin_layout Section
Spinning Massive Strings
\end_layout

\begin_layout Part
Charged Holographic Strings
\end_layout

\begin_layout Standard
In this section, we calculate the energy and angular momentum for a massive
 spinning string with charges added to the endpoints.
 This also necessiatates the calculation of the electric and magnetic fields
 at the endpoints.
\end_layout

\begin_layout Section
The action
\end_layout

\begin_layout Standard
We add charges to the end points by considering the action from the previous
 section and adding two electromagnetic interaction lagrangians at the endpoints.
 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
S=-T\iint_{(0,-l_{2})}^{(T,l_{1})}d^{2}\sigma\sqrt{-\det h}-m\int_{0}^{T}d\tau\sqrt{u^{\mu}u_{\mu}}|_{\sigma=-l_{2}}^{\sigma=l_{1}}-q\int_{0}^{T}u^{\mu}A_{\mu}d\tau|_{\sigma=-l_{2}}^{\sigma=l_{1}}
\]

\end_inset


\end_layout

\begin_layout Standard
The field lagrangian density for the electromagnetic fields is not included
 to avoid divergent self energy terms .Instead, the electromagnetic potentials
 and fields are determined directly by the Lienard-Wichert formulas.
 The worldsheet metric is as usual:
\end_layout

\begin_layout Standard
\begin_inset Formula $h=\left(\begin{array}{cc}
\cfrac{dx^{\mu}}{d\tau}\cfrac{dx_{\mu}}{d\tau} & \cfrac{dx^{\mu}}{d\tau}\cfrac{dx_{\mu}}{d\sigma}\\
\cfrac{dx^{\mu}}{d\tau}\cfrac{dx_{\mu}}{d\sigma} & \cfrac{dx^{\mu}}{d\sigma}\cfrac{dx_{\mu}}{d\sigma}
\end{array}\right)$
\end_inset


\end_layout

\begin_layout Standard
We can write for short 
\begin_inset Formula $V^{\mu}=\cfrac{dx^{\mu}}{d\sigma},u^{\mu}\cfrac{dx^{\mu}}{d\tau}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $h=\left(\begin{array}{cc}
u^{2} & u\cdot V\\
u\cdot V & V^{2}
\end{array}\right)$
\end_inset


\end_layout

\begin_layout Standard
The action can be writen then as:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
S=-T\iint_{(0,-l_{2})}^{(T,l_{1})}d^{2}\sigma\sqrt{\left(u\cdot V\right)^{2}-u^{2}V^{2}}-m\int_{0}^{T}d\tau\sqrt{u^{\mu}u_{\mu}}|_{\sigma=-l_{2}}^{\sigma=l_{1}}-q\int_{0}^{T}u^{\mu}A_{\mu}d\tau|_{\sigma=-l_{2}}^{\sigma=l_{1}}
\]

\end_inset


\end_layout

\begin_layout Standard
In the gauge 
\begin_inset Formula $\begin{cases}
t=\tau\\
r=\sigma
\end{cases}$
\end_inset

 :
\end_layout

\begin_layout Standard
The 4 velocity is
\end_layout

\begin_layout Standard
\begin_inset Formula $u^{\mu}=\left(\cfrac{dt}{d\tau},\cfrac{dx}{d\tau},\cfrac{dy}{d\tau},\cfrac{dz}{d\tau}\right)=\left(1,\sigma\cfrac{dcos\left(\theta\right)}{d\tau},\sigma\cfrac{dsin\left(\theta\right)}{d\tau},\cfrac{dz}{d\tau}\right)=\left(1,-\sigma\dot{\theta}sin\left(\theta\right),\sigma\dot{\theta}cos\left(\theta\right),\cfrac{dz}{d\tau}\right)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $V^{\mu}=\left(\cfrac{dt}{d\sigma},\cfrac{dx}{d\sigma},\cfrac{dy}{d\sigma},\cfrac{dz}{d\sigma}\right)=\left(0,cos\left(\theta\right),sin\left(\theta\right),\cfrac{dz}{d\sigma}\right)$
\end_inset


\end_layout

\begin_layout Standard
The basis vectors for polar coordinates are
\end_layout

\begin_layout Standard
\begin_inset Formula $\hat{r}=\cfrac{1}{r}\left(x\hat{x}+y\hat{y}\right)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\hat{\theta}=\cfrac{1}{r}\left(x\hat{y}-y\hat{x}\right)$
\end_inset


\end_layout

\begin_layout Standard
Both of which are normalized.
\end_layout

\begin_layout Standard
The components for 
\begin_inset Formula $u^{\mu}$
\end_inset

 and 
\begin_inset Formula $A^{\mu}$
\end_inset

 in polar coordinates are thus:
\end_layout

\begin_layout Standard
\begin_inset Formula $u^{0}=1$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $u^{z}=\frac{dz}{d\sigma}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $u^{r}=\vec{u}\cdot\hat{r}=\cfrac{1}{\sigma}\left(-\sigma^{2}\dot{\theta}sin\left(\theta\right)cos\left(\theta\right)+\sigma^{2}\dot{\theta}sin\left(\theta\right)cos\left(\theta\right)\right)=0$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $u^{\theta}=\vec{u}\cdot\hat{\theta}=\left(\hat{t}-\sigma\dot{\theta}sin\left(\theta\right)\hat{x}+\sigma\dot{\theta}cos\left(\theta\right)\hat{y}\right)\cdot\cfrac{1}{r}\left(x\hat{y}-y\hat{x}\right)=\sigma\dot{\theta}\left(ysin\left(\theta\right)+cos\left(\theta\right)x\right)\cdot\cfrac{1}{\sigma}=\sigma\dot{\theta}\cfrac{1}{\sigma}\left(\sigma sin^{2}\left(\theta\right)+\sigma cos^{2}\left(\theta\right)\right)=\dot{\theta}\sigma$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\begin{cases}
A^{r}=\vec{A}\cdot\hat{r}=\cfrac{xA^{x}+yA^{y}}{\sigma}=cos\left(\theta\right)A^{x}+sin\left(\theta\right)A^{y}\\
A^{\theta}=\vec{A}\cdot\hat{\theta}=\cfrac{-yA^{x}+xA^{y}}{\sigma}=-sin\left(\theta\right)A^{x}+cos\left(\theta\right)A^{y}
\end{cases}$
\end_inset

 
\begin_inset Formula $\begin{cases}
A^{x}=cos\left(\theta\right)A^{r}-sin\left(\theta\right)A^{\theta}\\
A^{y}=sin\left(\theta\right)A^{r}+cos\left(\theta\right)A^{\theta}
\end{cases}$
\end_inset


\end_layout

\begin_layout Standard
Computing the product-sum 
\begin_inset Formula $u^{\mu}A_{\mu}$
\end_inset

 starting from minkowski metric:
\end_layout

\begin_layout Standard
\begin_inset Formula $u^{\mu}A_{\mu}=u^{t}A^{t}-u^{x}A^{x}-u^{y}A^{y}-u^{z}A^{z}=A^{t}+\sigma\dot{\theta}sin\left(\theta\right)\left(cos\left(\theta\right)A^{r}-sin\left(\theta\right)A^{\theta}\right)-\sigma\dot{\theta}cos\left(\theta\right)\left(sin\left(\theta\right)A^{r}+cos\left(\theta\right)A^{\theta}\right)-\frac{dz}{d\sigma}A^{z}=A^{t}-\sigma\dot{\theta}A^{\theta}-\frac{dz}{d\sigma}A^{z}$
\end_inset


\end_layout

\begin_layout Standard
The action becomes
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
S=-T\iint_{(0,-l_{2})}^{(t,l_{1})}d\sigma dt\sqrt{1+r^{2}\theta'^{2}-r^{2}\dot{\theta}^{2}+z'^{2}-\dot{z}^{2}-r^{2}\left(\dot{\theta}z'-\theta'\dot{z}\right)^{2}}-m\int dt\sqrt{1-r^{2}\dot{\theta}^{2}-\dot{z}^{2}}|_{\sigma=-l_{2}}^{\sigma=l_{1}}-q\int_{0}^{t}\left(A^{t}-\sigma\dot{\theta}A^{\theta}-z'A^{z}\right)d\tau|_{\sigma=-l_{2}}^{\sigma=l_{1}}
\]

\end_inset


\end_layout

\begin_layout Standard
quadratic approximation*****
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
S=-TL\int_{0}^{t}dt-T\iint_{(0,-l_{2})}^{(T,l_{1})}d\sigma dt\frac{1}{2}\left[r^{2}\theta'^{2}-r^{2}\dot{\theta}^{2}+z'^{2}-\dot{z}^{2}\right]-m\delta t+\frac{1}{2}m\int dt\left[r^{2}\dot{\theta}^{2}+\dot{z}^{2}\right]|_{\sigma=-l_{2}}^{\sigma=l_{1}}-q\int_{0}^{T}\left(A^{t}-\sigma\dot{\theta}A^{\theta}-z'A^{z}\right)d\tau|_{\sigma=-l_{2}}^{\sigma=l_{1}}
\]

\end_inset


\end_layout

\begin_layout Standard
equations of motion are then
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
S=-T\iint_{(0,-l_{2})}^{(T,l_{1})}d^{2}\sigma\sqrt{\left(u\cdot V\right)^{2}-u^{2}V^{2}}-m\int_{0}^{T}d\tau\sqrt{u^{\mu}u_{\mu}}|_{\sigma=-l_{2}}^{\sigma=l_{1}}-q\int_{0}^{T}u^{\mu}A_{\mu}d\tau|_{\sigma=-l_{2}}^{\sigma=l_{1}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $x''-\ddot{x}=0$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $-2\theta'sin\theta-r\theta''sin\theta-r\left(\theta'\right)^{2}cos\theta+r\left(\dot{\theta}\right)^{2}cos\theta+r\ddot{\theta}sin\theta=0$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $r\theta''cos\theta-r\left(\theta'\right)^{2}sin\theta-r\ddot{\theta}cos\theta+r\left(\dot{\theta}\right)^{2}sin\theta=0$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\begin{cases}
\frac{\partial\mathcal{L}}{\partial\theta}-\partial_{\alpha}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\alpha}\theta)}\right)=0 & \Rightarrow\partial_{r}\left(\frac{\partial\mathcal{L}}{\partial\theta'}\right)+\partial_{t}\left(\frac{\partial\mathcal{L}}{\partial\dot{\theta}}\right)=\partial_{r}\left(r^{2}\theta'\right)-r^{2}\ddot{\theta}=0\\
\frac{\partial\mathcal{L}}{\partial z}-\partial_{\alpha}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\alpha}z)}\right)=0 & \Rightarrow\partial_{r}\left(\frac{\partial\mathcal{L}}{\partial z'}\right)+\partial_{t}\left(\frac{\partial\mathcal{L}}{\partial\dot{z}}\right)=z''-\ddot{z}=0
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
I split the functions describing the string into two halves describing the
 
\begin_inset Quotes eld
\end_inset

left
\begin_inset Quotes erd
\end_inset

 and 
\begin_inset Quotes eld
\end_inset

right
\begin_inset Quotes erd
\end_inset

 halves.
 This is needed because 
\begin_inset Formula $r>0$
\end_inset

 while the string need not be completely symmetric around the center (and
 may also have two different 
\begin_inset Quotes eld
\end_inset

lengths
\begin_inset Quotes erd
\end_inset

 from the center to the edges).
\end_layout

\begin_layout Standard
In doing so I add continuity conditions around the center:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\begin{cases}
z^{L}\left(0,t\right)=z^{R}\left(0,t\right) & \partial_{t,r}z^{L}\left(0,t\right)=\partial_{t,r}z^{R}\left(0,t\right)\\
\theta^{L}\left(0,t\right)=\theta^{R}\left(0,t\right)=0 & \partial_{t,r}\theta^{L}\left(0,t\right)=\partial_{t,r}\theta^{R}\left(0,t\right)
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
z^{L}\left(r,t\right)=\sum e^{i\omega_{n}t}\frac{z_{n}^{L}\left(r\right)}{\omega_{n}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
z^{R}\left(r,t\right)=\sum e^{i\omega_{n}t}\frac{z_{n}^{R}\left(r\right)}{\omega_{n}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\theta^{L}\left(r,t\right)=\sum e^{i\omega_{n}t}\frac{\theta_{n}^{L}\left(r\right)}{\omega_{n}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\theta^{R}\left(r,t\right)=\sum e^{i\omega_{n}t}\frac{\theta_{n}^{R}\left(r\right)}{\omega_{n}}
\]

\end_inset


\end_layout

\begin_layout Standard
The boundary equations are derived to be:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial L}{\partial x^{\mu}}-\partial_{\tau}\left(\frac{\partial L}{\partial(\partial_{\tau}x^{\mu})}\right)+\frac{\partial\mathcal{L}}{\partial(\partial_{\sigma}x^{\mu})}|_{\sigma=-l_{2}}^{\sigma=l_{1}}=0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\begin{cases}
\frac{\partial L}{\partial\theta}-\partial_{t}\left(\frac{\partial L}{\partial\dot{\theta}}\right)\pm\frac{\partial\mathcal{L}}{\partial\theta'}|_{\sigma=-l_{R}}^{\sigma=l_{L}}=0 & \Rightarrow-mr^{2}\ddot{\theta}\pm Tr^{2}\theta'=0\\
\frac{\partial L}{\partial z}-\partial_{\tau}\left(\frac{\partial L}{\partial\dot{z}}\right)\pm\frac{\partial\mathcal{L}}{\partial z'}|_{\sigma=-l_{R}}^{\sigma=l_{L}}=0 & \Rightarrow-m\ddot{z}\pm Tz'=0
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
And after the split into two halves they are:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\begin{cases}
m_{R}r^{2}\ddot{\theta}^{L}+Tr^{2}\theta^{L}'=0 & m_{R}r^{2}\ddot{\theta}^{R}-Tr^{2}\theta^{R}'=0\\
m_{L}\ddot{z}^{L}+Tz'^{L}=0 & m_{R}\ddot{z}^{R}-Tz'^{R}=0
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
z^{L}\left(r,t\right)=\sum e^{i\omega_{n}t}\frac{1}{\omega_{n}}\left(A_{n}^{L}cos\left(\omega_{n}r\right)+B_{n}^{L}sin\left(\omega_{n}r\right)\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
z^{R}\left(r,t\right)=\sum e^{i\omega_{n}t}\frac{1}{\omega_{n}}\left(A_{n}^{R}cos\left(\omega_{n}r\right)+B_{n}^{R}sin\left(\omega_{n}r\right)\right)
\]

\end_inset


\end_layout

\begin_layout Standard
From the continuity conditions we get
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\begin{cases}
A_{n}^{L}=A_{n}^{R}\\
B_{n}^{L}=B_{n}^{R}
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
So we really only have one function 
\begin_inset Formula 
\[
z^{L}\left(r,t\right)=\sum e^{i\omega_{n}t}\frac{1}{\omega_{n}}\left(A_{n}cos\left(\omega_{n}r\right)+B_{n}sin\left(\omega_{n}r\right)\right)
\]

\end_inset


\end_layout

\begin_layout Standard
with 
\begin_inset Formula $-l_{R}<r<l_{L}$
\end_inset

.
\end_layout

\begin_layout Standard
Putting this into the corresponding endpoint equation we get 
\end_layout

\begin_layout Standard

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\color none
\begin_inset Formula 
\[
\begin{cases}
m_{L}\ddot{z}^{L}+Tz'^{L}=0 & \Rightarrow B_{n}\left(cos\left(\omega_{n}l_{L}\right)-\frac{m_{L}\omega_{n}}{T}sin\left(\omega_{n}l_{L}\right)\right)=A_{n}\left(\frac{m_{L}\omega_{n}}{T}cos\left(\omega_{n}l_{L}\right)+sin\left(\omega_{n}l_{L}\right)\right)\\
m_{R}\ddot{z}^{R}-Tz'^{R}=0 & \Rightarrow B_{n}\left(cos\left(\omega_{n}l_{R}\right)-\frac{m_{R}\omega_{n}}{T}sin\left(\omega_{n}l_{R}\right)\right)=A_{n}\left(-sin\left(\omega_{n}l_{R}\right)-\frac{m_{R}\omega_{n}}{T}cos\left(\omega_{n}l_{R}\right)\right)
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard

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\noun off
\color none
\begin_inset Formula 
\[
\begin{cases}
 & A_{n}\left(\frac{m_{L}\omega_{n}}{T}cos\left(\omega_{n}l_{L}\right)+sin\left(\omega_{n}l_{L}\right)\right)+B_{n}\left(\frac{m_{L}\omega_{n}}{T}sin\left(\omega_{n}l_{L}\right)-cos\left(\omega_{n}l_{L}\right)\right)=0\\
 & A_{n}\left(-sin\left(\omega_{n}l_{R}\right)-\frac{m_{R}\omega_{n}}{T}cos\left(\omega_{n}l_{R}\right)\right)+B_{n}\left(\frac{m_{R}\omega_{n}}{T}sin\left(\omega_{n}l_{R}\right)-cos\left(\omega_{n}l_{R}\right)\right)=0
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
We can represent the previous equations by the linear system
\end_layout

\begin_layout Standard
\begin_inset Formula $C_{n}\vec{X_{n}}=0$
\end_inset


\end_layout

\begin_layout Standard
Where
\end_layout

\begin_layout Standard
\begin_inset Formula $\left(\begin{array}{c}
A_{n}\\
B_{n}
\end{array}\right)$
\end_inset

 and 
\begin_inset Formula $C_{n}=\left(\begin{array}{cc}
\frac{m_{L}\omega_{n}}{T}cos\left(\omega_{n}l_{L}\right)+sin\left(\omega_{n}l_{L}\right) & \frac{m_{L}\omega_{n}}{T}sin\left(\omega_{n}l_{L}\right)-cos\left(\omega_{n}l_{L}\right)\\
-sin\left(\omega_{n}l_{R}\right)-\frac{m_{R}\omega_{n}}{T}cos\left(\omega_{n}l_{R}\right) & \frac{m_{R}\omega_{n}}{T}sin\left(\omega_{n}l_{R}\right)-cos\left(\omega_{n}l_{R}\right)
\end{array}\right)$
\end_inset


\end_layout

\begin_layout Standard
And because we want a nontrivial solution for 
\begin_inset Formula $X_{n}$
\end_inset

 we require 
\begin_inset Formula $det\left(C_{n}\right)=0$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
det\left(C_{n}\right)=\frac{sin\left(\omega_{n}\left(l_{L}+l_{R}\right)\right)\left(m_{L}m_{R}\omega_{n}^{2}-T^{2}\right)-T\omega_{n}\left(m_{L}+m_{R}\right)cos\left(\omega_{n}\left(l_{L}-l_{R}\right)\right)}{T^{2}}=0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
tan\left(\omega_{n}\left(l_{L}+l_{R}\right)\right)=\frac{T\omega_{n}\left(m_{L}+m_{R}\right)}{m_{L}m_{R}\omega_{n}^{2}-T^{2}}
\]

\end_inset


\end_layout

\begin_layout Standard
From which we know
\end_layout

\begin_layout Standard

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\color none
\begin_inset Formula $B_{n}=A_{n}\frac{\frac{m_{L}\omega_{n}}{T}cos\left(\omega_{n}l_{L}\right)+sin\left(\omega_{n}l_{L}\right)}{cos\left(\omega_{n}l_{L}\right)-\frac{m_{L}\omega_{n}}{T}sin\left(\omega_{n}l_{L}\right)}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $z^{L}\left(r,t\right)=\sum e^{i\omega_{n}t}\frac{1}{\omega_{n}}A_{n}\left(cos\left(\omega_{n}r\right)+\frac{\frac{m_{L}\omega_{n}}{T}cos\left(\omega_{n}l_{L}\right)+sin\left(\omega_{n}l_{L}\right)}{cos\left(\omega_{n}l_{L}\right)-\frac{m_{L}\omega_{n}}{T}sin\left(\omega_{n}l_{L}\right)}sin\left(\omega_{n}r\right)\right)$
\end_inset


\end_layout

\begin_layout Standard
The solutions for the angle are:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\theta^{L}\left(r,t\right)=\sum e^{i\omega_{n}t}\frac{2C_{n}^{L}\omega_{n}e^{-r\omega_{n}}+D_{n}^{L}e^{r\omega_{n}}}{2r\omega_{n}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\theta^{R}\left(r,t\right)=\sum e^{i\omega_{n}t}\frac{2C_{n}^{R}\omega_{n}e^{-r\omega_{n}}+D_{n}^{R}e^{r\omega_{n}}}{2r\omega_{n}}
\]

\end_inset


\end_layout

\begin_layout Standard
From the continuity conditions we again get that the two sides are identified.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\begin{cases}
lim_{\rightarrow0}r\theta^{L}\left(0,t\right)=lim_{\rightarrow0}r\theta^{R}\left(0,t\right)=0 & lim_{\rightarrow0}\partial_{r}r\theta^{L}\left(0,t\right)=lim_{\rightarrow0}\partial_{r}r\theta^{R}\left(0,t\right)\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\begin{cases}
2\omega_{n}C_{n}^{L}+D_{n}^{L}=2\omega_{n}C_{n}^{R}+D_{n}^{R} & \Rightarrow D_{n}^{L}=D_{n}^{R}\\
-2\omega_{n}C_{n}^{L}+D_{n}^{L}=-2\omega_{n}C_{n}^{R}+D_{n}^{R} & \Rightarrow C_{n}^{L}=C_{n}^{R}
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\theta\left(r,t\right)=\sum e^{i\omega_{n}t}\frac{2C_{n}\omega_{n}e^{-r\omega_{n}}+D_{n}e^{r\omega_{n}}}{2\omega_{n}r}
\]

\end_inset


\end_layout

\begin_layout Standard
From the endpoint equations we get
\end_layout

\begin_layout Standard
\begin_inset Formula $\omega_{n}\left(m_{R}l_{R}+m_{L}l_{L}+\frac{T}{\omega_{n}^{2}}+\frac{1}{T}\left(m_{L}m_{R}\omega_{n}^{2}+T^{2}\right)l_{L}l_{R}\right)tanh\left(\omega_{n}\left(l_{L}+l_{R}\right)\right)=T\left(l_{R}+l_{L}\right)+l_{L}l_{R}\omega_{n}^{2}\left(m_{L}+m_{R}\right)$
\end_inset


\end_layout

\begin_layout Section
Electromagnetic Fields
\end_layout

\begin_layout Standard
The Lienard-Wichert formulas for the fields of a point charge are
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\phi\left(\vec{x},t\right)=\left[\cfrac{e}{\left(1-\vec{\beta}\cdot\vec{n}\right)R}\right]_{ret}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\vec{A}\left(\vec{x},t\right)=\left[\cfrac{e\vec{\beta}}{\left(1-\vec{\beta}\cdot\vec{n}\right)R}\right]_{ret}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\vec{E}\left(\vec{x},t\right)=\left[e\cfrac{\vec{n}-\vec{\beta}}{\gamma^{2}\left(1-\vec{\beta}\cdot\vec{n}\right)^{3}R^{2}}+\cfrac{e}{c}\cfrac{\vec{n}\times\left\{ \left(\vec{n}-\vec{\beta}\right)\times\dot{\vec{\beta}}\right\} }{\left(1-\vec{\beta}\cdot\vec{n}\right)^{3}R}\right]_{ret}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\vec{B}\left(\vec{x},t\right)=\left[\vec{n}\times\vec{E}\right]_{ret}
\]

\end_inset


\end_layout

\begin_layout Standard
Where
\end_layout

\begin_layout Standard
\begin_inset Formula $\vec{\beta}$
\end_inset

 is the velocity of the point charge source, divided by 
\begin_inset Formula $c$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula $\gamma=\left(1-\beta^{2}\right)^{-1/2}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\vec{n}$
\end_inset

 is the unit vector from the source to the point where we evaluate the fields.
\end_layout

\begin_layout Standard
\begin_inset Formula $R$
\end_inset

 is the distance from the source to the point where we evaluate the fields.
\end_layout

\begin_layout Standard
\begin_inset Formula $e$
\end_inset

 is the charge of the source.
\end_layout

\begin_layout Standard
The formulas are evaluated with the retardation condition (also called the
 light cone condition) 
\begin_inset Formula $t-t_{ret}=R_{ret}$
\end_inset

.
 
\begin_inset Formula $R_{ret}$
\end_inset

 denotes the distance from the source 
\bar under
at retarded time
\bar default
 to the point where we evaluate the fields.
\end_layout

\begin_layout Standard
The calculation of the various fields is then reduced to a geometric calculation
 where everything will depend on the length of the string and the rate of
 rotation.
\end_layout

\begin_layout Standard
Diagram (**) displays the geometry of the spinning string, with retarded
 positions.
\end_layout

\begin_layout Standard
The origin of the coordinate system is at the center of the circle.
\end_layout

\begin_layout Standard
P' is the location of particle P in retarted time (t<x0)
\end_layout

\begin_layout Standard
Q is the observation point (other particle)
\end_layout

\begin_layout Standard
From this geometry (the cosine law) and the retardation condition 
\begin_inset Formula $\left|P'Q\right|=\tau=\cfrac{\Delta\phi}{\omega}$
\end_inset

 we get
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\Delta\phi^{2}=\beta_{1}^{2}+\beta_{2}^{2}+2\beta_{1}\beta_{2}cos\left(\Delta\phi\right)
\]

\end_inset


\end_layout

\begin_layout Standard
This condition relates the 
\begin_inset Quotes eld
\end_inset

retardation angle
\begin_inset Quotes erd
\end_inset

 
\begin_inset Formula $\Delta\phi$
\end_inset

 and the velocities of the string end points.
 This equation is solved numerically, giving implicitly the function 
\begin_inset Formula $\Delta\phi=f\left(\beta_{1}\right)$
\end_inset

, or it can be solved approximately by guessing (for the case 
\begin_inset Formula $\beta_{1}\rightarrow0$
\end_inset

) 
\begin_inset Formula $\Delta\phi=a_{0}\beta_{1}+a_{1}\beta_{1}^{2}+...$
\end_inset

 and solving for the coefficients iterativley.
\end_layout

\begin_layout Standard
The electromagnetic fields and potentials are calculated to be:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
A^{0}\left(l_{1}\right)=\frac{q\omega}{\beta_{1}\beta_{2}\sin(\Delta\theta)+\Delta\theta}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
A^{\theta}\left(l_{1}\right)=-\frac{\beta_{2}q\omega\cos(\Delta\theta)}{\beta_{1}\beta_{2}\sin(\Delta\theta)+\Delta\theta}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\vec{E}=\left\{ \begin{array}{c}
-\frac{q\omega^{2}\left(\beta_{1}\left(\beta_{2}^{2}\cos(2\Delta\theta)+\beta_{2}^{2}-2\right)+2\beta_{2}\cos(\Delta\theta)\left(\beta_{2}^{2}-1\right)-2\Delta\theta\beta_{2}\sin(\Delta\theta)\right)}{2\left(\Delta\theta+\beta_{1}\beta_{2}\sin(\Delta\theta)\right)^{3}}\\
\frac{q\beta_{2}\omega^{2}\left(\sin(\Delta\theta)\left(\beta_{1}^{2}+\beta_{2}^{2}+\beta_{1}\beta_{2}\cos(\Delta\theta)-1\right)+\Delta\theta\left(\cos(\Delta\theta)+\beta_{1}\beta_{2}\right)\right)}{\left(\Delta\theta+\beta_{1}\beta_{2}\sin(\Delta\theta)\right)^{3}}\\
0
\end{array}\right\} 
\]

\end_inset


\end_layout

\begin_layout Standard

\series bold
\begin_inset Formula 
\[
\vec{B}=\left\{ \begin{array}{c}
0\\
0\\
\frac{q\beta_{2}\omega^{2}\left(\Delta\theta\left(\beta_{1}^{2}\beta_{2}+\beta_{1}\cos(\Delta\theta)\left(\beta_{2}^{2}+1\right)+\beta_{2}\right)+\beta_{1}\sin(\Delta\theta)\left(\beta_{1}^{2}+2\beta_{1}\beta_{2}\cos(\Delta\theta)+\beta_{2}^{2}\right)\right)}{\Delta\theta\left(\Delta\theta+\beta_{1}\beta_{2}\sin(\Delta\theta)\right)^{3}}
\end{array}\right\} 
\]

\end_inset


\end_layout

\begin_layout Section
Equations of Motion
\end_layout

\begin_layout Subsection
combining fields and point particles
\end_layout

\begin_layout Standard
In combining point particle lagrangians and field lagrangian densities,
 we obtain slightly non standard euler lagrange equations.
 The action is the sum:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
S=\int\mathcal{L}d^{2}\sigma+\int Ld\tau
\]

\end_inset


\end_layout

\begin_layout Standard
where 
\begin_inset Formula $\mathcal{L}$
\end_inset

 is specifically the lagrangian density and 
\begin_inset Formula $L$
\end_inset

 is a lagrangian.
\end_layout

\begin_layout Standard
We obtain the equations of motion by varying the solution by an infinitesimal
 function 
\begin_inset Formula $\delta x^{\mu}(\sigma,\tau)$
\end_inset

 such that 
\begin_inset Formula $\delta x^{\mu}(0)=\delta x^{\mu}(T)=0$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\delta S=\int\left(\frac{\partial\mathcal{L}}{\partial x^{\mu}}\delta x^{\mu}+\frac{\partial\mathcal{L}}{\partial(\partial_{\alpha}x^{\mu})}\partial_{\alpha}\delta x^{\mu}\right)d^{2}\sigma+\int\left(\frac{\partial L}{\partial x^{\mu}}\delta x^{\mu}+\frac{\partial L}{\partial(\partial_{\tau}x^{\mu})}\partial_{\tau}\delta x^{\mu}\right)d\tau=$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $=\int\left(\frac{\partial\mathcal{L}}{\partial x^{\mu}}-\partial_{\alpha}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\alpha}x^{\mu})}\right)\right)\delta x^{\mu}d^{2}\sigma+\int\left(\frac{\partial L}{\partial x^{\mu}}-\partial_{\tau}\left(\frac{\partial L}{\partial(\partial_{\tau}x^{\mu})}\right)+\frac{\partial\mathcal{L}}{\partial(\partial_{\sigma}x^{\mu})}\right)|_{\sigma=-l_{2}}^{\sigma=l_{1}}\delta x^{\mu}d\tau$
\end_inset


\end_layout

\begin_layout Standard
So requiring no variation to first order in the action gives us the euler
 lagrange equations of motion for the field:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial\mathcal{L}}{\partial x^{\mu}}-\partial_{\alpha}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\alpha}x^{\mu})}\right)=0
\]

\end_inset

 and at each end point the Neumann boundary condition 
\begin_inset Formula 
\[
\frac{\partial L}{\partial x^{\mu}}-\partial_{\tau}\left(\frac{\partial L}{\partial(\partial_{\tau}x^{\mu})}\right)+\frac{\partial\mathcal{L}}{\partial(\partial_{\sigma}x^{\mu})}|_{\sigma=-l_{2}}^{\sigma=l_{1}}=0
\]

\end_inset


\end_layout

\begin_layout Standard
or at 
\begin_inset Formula $\mu\neq0$
\end_inset

 the Dirichlet boundary condition:
\end_layout

\begin_layout Standard

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\begin_inset Formula 
\[
\delta x^{\mu}=0
\]

\end_inset

We will use the first (Neumann) boundary condition, as the string endpoints
 are free to move.
\end_layout

\begin_layout Subsection
Masses equation
\end_layout

\begin_layout Standard
We turn to the boundary equations and write it explicitly and get a forces
 equation for the end points.
\end_layout

\begin_layout Standard
The derivatives:
\end_layout

\begin_layout Standard
\begin_inset Formula $\frac{\partial L}{\partial x^{\mu}}=-q_{1}u^{\nu}\cfrac{\partial A_{\nu}\left(l_{1}\right)}{\partial x^{\mu}}-q_{2}u^{\nu}\cfrac{\partial A_{\nu}\left(l_{2}\right)}{\partial x^{\mu}}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\partial_{\tau}\frac{\partial L}{\partial(\partial_{\tau}x^{\mu})}=-m_{1}\partial_{\tau}\left(\cfrac{u_{\mu}}{\sqrt{u^{\nu}u_{\nu}}}\right)-m_{2}\partial_{\tau}\left(\cfrac{u_{\mu}}{\sqrt{u^{\nu}u_{\nu}}}\right)-q_{1}u^{\nu}\cfrac{\partial A_{\mu}\left(l_{1}\right)}{\partial x^{\nu}}-q_{2}u^{\nu}\cfrac{\partial A_{\mu}\left(l_{2}\right)}{\partial x^{\nu}}$
\end_inset


\end_layout

\begin_layout Standard

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\color none
\begin_inset Formula $\frac{\partial\mathcal{L}}{\partial(\partial_{\sigma}x^{\mu})}|^{\sigma=l_{1}}=\frac{\partial\mathcal{L}}{\partial V^{\mu}}|^{\sigma=l_{1}}=-T\cfrac{\left(u\cdot V\right)u_{\mu}-u^{2}V_{\mu}}{\sqrt{\left(u\cdot V\right)^{2}-u^{2}V^{2}}}|^{\sigma=l_{1}}$
\end_inset


\end_layout

\begin_layout Standard
The equation now becomes
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
-q_{1}u^{\nu}\cfrac{\partial A_{\nu}\left(l_{1}\right)}{\partial x^{\mu}}-q_{2}u^{\nu}\cfrac{\partial A_{\nu}\left(l_{2}\right)}{\partial x^{\mu}}-\left[-m_{1}\partial_{\tau}\left(\cfrac{u_{\mu}}{\sqrt{u^{\nu}u_{\nu}}}\right)-m_{2}\partial_{\tau}\left(\cfrac{u_{\mu}}{\sqrt{u^{\nu}u_{\nu}}}\right)-q_{1}u^{\nu}\cfrac{\partial A_{\mu}\left(l_{1}\right)}{\partial x^{\nu}}-q_{2}u^{\nu}\cfrac{\partial A_{\mu}\left(l_{2}\right)}{\partial x^{\nu}}\right]+
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
-T\cfrac{\left(u\cdot V\right)u_{\mu}-u^{2}V_{\mu}}{\sqrt{\left(u\cdot V\right)^{2}-u^{2}V^{2}}}=0
\]

\end_inset


\end_layout

\begin_layout Standard
Which can be written in terms of the electromagnetic fields as:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
m_{1}\partial_{\tau}\left(\cfrac{u_{\mu}}{\sqrt{u^{\nu}u_{\nu}}}\right)+m_{2}\partial_{\tau}\left(\cfrac{u_{\mu}}{\sqrt{u^{\nu}u_{\nu}}}\right)+q_{1}u^{\nu}F_{\nu\mu}\left(l_{1}\right)-q_{2}u^{\nu}F_{\mu\nu}\left(l_{2}\right)-T\cfrac{\left(u\cdot V\right)u_{\mu}-u^{2}V_{\mu}}{\sqrt{\left(u\cdot V\right)^{2}-u^{2}V^{2}}}|^{\sigma=l_{1}}=0
\]

\end_inset


\end_layout

\begin_layout Standard
Inserting the spinning string solution and Specializing to 
\begin_inset Formula $\mu=1$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $x=\left(\tau,\sigma cos\left(\omega\tau\right),\sigma sin\left(\omega\tau\right)\right)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $u=\left(1,-\omega\sigma sin\left(\omega\tau\right),\omega\sigma cos\left(\omega\tau\right)\right)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $V=\left(0,cos\left(\omega\tau\right),sin\left(\omega\tau\right)\right)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $r=\left(\tau,\sigma cos\left(\omega\tau+\pi\right),\sigma sin\left(\omega\tau+\pi\right)\right)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $u_{r}=\left(1,-\omega\sigma sin\left(\omega\tau+\pi\right),\omega\sigma cos\left(\omega\tau+\pi\right)\right)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $V_{r}=\left(0,cos\left(\omega\tau+\pi\right),sin\left(\omega\tau+\pi\right)\right)$
\end_inset


\end_layout

\begin_layout Standard
Where we denoted retarded quantities with index 
\begin_inset Formula $r$
\end_inset

.
\end_layout

\begin_layout Standard
We now get
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
-m_{1}\cfrac{-\omega^{2}\sigma cos\left(\omega\tau\right)}{\sqrt{1-\omega^{2}\sigma^{2}}}+q_{1}\omega\sigma cos\left(\omega\tau\right)B_{3}\left(\sigma\right)+q_{1}E_{1}\left(\sigma\right)-T\sqrt{1-\omega^{2}\sigma^{2}}cos\left(\omega\tau\right)=0
\]

\end_inset


\end_layout

\begin_layout Standard
Which at an endpoint is:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
m_{1}\cfrac{\omega^{2}l_{1}cos\left(\omega\tau\right)}{\sqrt{1-\omega^{2}l_{1}^{2}}}+q_{1}\omega l_{1}cos\left(\omega\tau\right)B_{3}\left(l_{1}\right)+q_{1}E_{1}\left(l_{1}\right)-T\sqrt{1-\omega^{2}l_{1}^{2}}cos\left(\omega\tau\right)=0
\]

\end_inset


\end_layout

\begin_layout Standard
We want the equation in the radial direction, so we also specialize to 
\begin_inset Formula $\tau=0$
\end_inset

, and In terms of 
\begin_inset Formula $\beta_{i}=\omega l_{i}$
\end_inset

 and 
\begin_inset Formula $\gamma_{i}=\left(1-\beta_{i}^{2}\right)^{-1/2}$
\end_inset

 this is:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
m_{1}\cfrac{\omega\beta_{1}}{\sqrt{1-\beta_{1}^{2}}}+q_{1}\beta_{1}B_{3}\left(l_{1}\right)+q_{1}E_{1}\left(l_{1}\right)-T\sqrt{1-\beta_{1}^{2}}=0
\]

\end_inset


\end_layout

\begin_layout Standard
We can see a string tension term, a centrifugal term and an electromagnetic
 term.
\end_layout

\begin_layout Standard
This equation constitutes a constraint implicitly giving 
\begin_inset Formula $\omega=\omega\left(\beta_{1},\beta_{2},m_{1},m_{2},q_{1},q_{2},T\right)$
\end_inset

.
\end_layout

\begin_layout Standard
Using the fact that 
\begin_inset Formula $T,q^{2},\omega$
\end_inset

 and 
\begin_inset Formula $\Delta\phi$
\end_inset

 are the same for both end points, the force equations for the two end points
 also provide a relation between 
\begin_inset Formula $\left\{ m_{1},m_{2},\beta_{1},\beta_{2}\right\} $
\end_inset

.
 In the case of 
\begin_inset Formula $q^{2}=0$
\end_inset

 it is just:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\gamma_{1}^{2}\beta_{1}m_{1}\omega=T=\gamma_{2}^{2}\beta_{2}m_{2}\omega
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\gamma_{1}^{2}\beta_{1}m_{1}=\gamma_{2}^{2}\beta_{2}m_{2}
\]

\end_inset


\end_layout

\begin_layout Standard
Which can be argued to be approximately true (and much simpler) as long
 as 
\begin_inset Formula $F_{tension}\approx F_{centrifugal}$
\end_inset

.
 This turns out to be true for hadrons (?) as long as the state has nonzero
 angular momentum.
 In the case of zero angular momentum we get the trivial solution 
\begin_inset Formula $\beta_{1}=\beta_{2}=0$
\end_inset

.
\end_layout

\begin_layout Subsection
String equations
\end_layout

\begin_layout Standard
We wish to verify that the spinning string ansatz is truly a solution to
 the equations of motion.
\end_layout

\begin_layout Standard
The derivatives are:
\end_layout

\begin_layout Standard

\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\uuline off
\uwave off
\noun off
\color none
\begin_inset Formula 
\[
\frac{\partial\mathcal{L}}{\partial(\partial_{\tau}x^{\mu})}=\frac{\partial\mathcal{L}}{\partial u^{\mu}}=-T\cfrac{V_{\mu}\left(u\cdot V\right)-u_{\mu}V^{2}}{\sqrt{\left(u\cdot V\right)^{2}-u^{2}V^{2}}}
\]

\end_inset


\end_layout

\begin_layout Standard

\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\uuline off
\uwave off
\noun off
\color none
\begin_inset Formula 
\[
\frac{\partial\mathcal{L}}{\partial(\partial_{\sigma}x^{\mu})}=\frac{\partial\mathcal{L}}{\partial V^{\mu}}=-T\cfrac{u_{\mu}\left(u\cdot V\right)-u^{2}V_{\mu}}{\sqrt{\left(u\cdot V\right)^{2}-u^{2}V^{2}}}
\]

\end_inset


\end_layout

\begin_layout Standard
Which give the equationf of motion:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{\partial\mathcal{L}}{\partial x^{\mu}}-\partial_{\alpha}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\alpha}x^{\mu})}\right)=-\partial_{\tau}\left(\frac{\partial\mathcal{L}}{\partial u^{\mu}}\right)-\partial_{\sigma}\left(\frac{\partial\mathcal{L}}{\partial V^{\mu}}\right)=0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
-\partial_{\tau}\left(-T\cfrac{V_{\mu}\left(u\cdot V\right)-u_{\mu}V^{2}}{\sqrt{\left(u\cdot V\right)^{2}-u^{2}V^{2}}}\right)-\partial_{\sigma}\left(-T\cfrac{u_{\mu}\left(u\cdot V\right)-u^{2}V_{\mu}}{\sqrt{\left(u\cdot V\right)^{2}-u^{2}V^{2}}}\right)=0
\]

\end_inset


\end_layout

\begin_layout Standard
Putting in the spinning string solution we get
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\cfrac{-\partial_{\tau}u_{\mu}}{\sqrt{1-\omega^{2}\sigma^{2}}}+\cfrac{-\omega^{2}\sigma}{\sqrt{1-\omega^{2}\sigma^{2}}}V_{\mu}+\sqrt{1-\omega^{2}\sigma^{2}}\partial_{\sigma}V_{\mu}=0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
-\partial_{\tau}u_{\mu}-\omega^{2}\sigma V_{\mu}+\left(1-\omega^{2}\sigma^{2}\right)\partial_{\sigma}V_{\mu}=0
\]

\end_inset


\end_layout

\begin_layout Standard
Specializing to the time component it is trivially true 
\begin_inset Formula $\left(V_{0}=\partial_{\tau}u_{0}=0\right)$
\end_inset

.
\end_layout

\begin_layout Standard
Specializing to the spatial component 
\begin_inset Formula $\mu=1$
\end_inset

 we get:
\end_layout

\begin_layout Standard
\begin_inset Formula $-\omega^{2}\sigma cos\left(\omega\tau\right)+\omega^{2}\sigma cos\left(\omega\tau\right)-\left(1-\omega^{2}\sigma^{2}\right)\partial_{\sigma}cos\left(\omega\tau\right)=0$
\end_inset


\end_layout

\begin_layout Section
Energy
\end_layout

\begin_layout Standard
Calculating the energy can be done in two ways.
 Either as the generator of poincare time translations, or as the hamiltonian
 of the gauge fixed system.
 We show the second choice here:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
E=\int d\sigma\left(\cfrac{\partial\mathcal{L}}{\partial\dot{\theta}}\dot{\theta}-\mathcal{L}\right)+\cfrac{\partial L}{\partial\dot{\theta}}\dot{\theta}-L
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
E=T\int_{-l_{2}}^{l_{1}}d\sigma\cfrac{1+\sigma^{2}\theta'^{2}}{\sqrt{1+\sigma^{2}\left(\theta'^{2}-\dot{\theta}^{2}\right)}}+m\cfrac{1}{\sqrt{1-\sigma^{2}\dot{\theta}^{2}}}|_{\sigma=-l_{2}}^{\sigma=l_{1}}+q_{1}A^{t}\left(l_{1}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
Notice in the electromagnetic part we took just one contribution out of
 the two end points.
 This is because it already contains the coloumb interaction energy, and
 taking both 
\begin_inset Quotes eld
\end_inset

contibutions
\begin_inset Quotes erd
\end_inset

 would be wrong.
\end_layout

\begin_layout Standard
Inserting the spinning string solution and potential we get
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
E=T\int_{-l_{2}}^{l_{1}}d\sigma\cfrac{1}{\sqrt{1-\omega^{2}\sigma^{2}}}+m\cfrac{1}{\sqrt{1-\sigma^{2}\omega^{2}}}|_{-l_{2}}^{l_{1}}+\frac{q_{1}q_{2}\omega}{\beta_{1}\beta_{2}\sin(\Delta\theta)+\Delta\theta}=
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
E=T\frac{sin^{-1}(\beta_{1})}{\omega}+T\frac{sin^{-1}\left(\beta_{2}\right)}{\omega}+\cfrac{m_{1}}{\sqrt{1-\beta_{1}^{2}}}+\cfrac{m_{2}}{\sqrt{1-\beta_{2}^{2}}}+\frac{q_{1}q_{2}\omega}{\beta_{1}\beta_{2}\sin(\Delta\theta)+\Delta\theta}
\]

\end_inset


\end_layout

\begin_layout Section
Angular momentum
\end_layout

\begin_layout Standard
The angular momentum is obtained as the generator of rotations.
 For the rotations in the rotation plane of the string:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
J=\int d\sigma\cfrac{\partial\mathcal{L}}{\partial\dot{\theta}}+\cfrac{\partial L}{\partial\dot{\theta}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
J=T\int_{-l_{2}}^{l_{1}}d\sigma\cfrac{\sigma^{2}\dot{\theta}}{\sqrt{1+\sigma^{2}\left(\theta'^{2}-\dot{\theta}^{2}\right)}}+m_{1}\cfrac{l_{1}^{2}\dot{\theta}}{\sqrt{1-l_{1}^{2}\dot{\theta}^{2}}}+m_{1}\cfrac{l_{2}^{2}\dot{\theta}}{\sqrt{1-l_{2}^{2}\dot{\theta}^{2}}}+q_{1}l_{1}A^{\theta}\left(l_{1}\right)+q_{2}l_{2}A^{\theta}\left(l_{2}\right)
\]

\end_inset


\end_layout

\begin_layout Standard
Inserting the spinning string solution and potential we get
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
J=T\int_{-l_{2}}^{l_{1}}d\sigma\cfrac{\sigma^{2}\omega}{\sqrt{1-\sigma^{2}\omega^{2}}}+m\cfrac{l_{1}^{2}\omega}{\sqrt{1-l_{1}^{2}\omega^{2}}}+m\cfrac{l_{2}^{2}\omega}{\sqrt{1-l_{2}^{2}\omega^{2}}}-q_{1}q_{2}\frac{\beta_{1}\beta_{2}\cos(\Delta\theta)}{\beta_{1}\beta_{2}\sin(\Delta\theta)+\Delta\theta}-q_{1}q_{2}\frac{\beta_{1}\beta_{2}\cos(\Delta\theta)}{\beta_{1}\beta_{2}\sin(\Delta\theta)+\Delta\theta}=
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
J=T\cfrac{-\cfrac{\beta_{1}}{\gamma_{1}}+sin^{-1}\left(\beta_{1}\right)-\cfrac{\beta_{2}}{\gamma_{2}}+sin^{-1}\left(\beta_{2}\right)}{2\omega^{2}}+m\cfrac{\gamma_{1}\beta_{1}^{2}}{\omega}+m\cfrac{\gamma_{2}\beta_{2}^{2}}{\omega}-q_{1}q_{2}\frac{2\beta_{1}\beta_{2}\cos(\Delta\theta)}{\beta_{1}\beta_{2}\sin(\Delta\theta)+\Delta\theta}
\]

\end_inset


\end_layout

\begin_layout Section
Regge Trajectories
\end_layout

\begin_layout Standard
Plotting 
\begin_inset Formula $\left(J,E^{2}\right)$
\end_inset

 as a function of 
\begin_inset Formula $\beta_{1}$
\end_inset

 gives the predicted classical Regge trajectories.
 In order to account for the observed intercept in the Regge trajectories
 we can add it in using four methods:
\end_layout

\begin_layout Enumerate
Put an intercept by hand into the angular momentum: 
\begin_inset Formula $J\rightarrow J+a$
\end_inset


\end_layout

\begin_layout Enumerate
Put an intercept by hand into the energy squared: 
\begin_inset Formula $\alpha'E^{2}\rightarrow\alpha'E^{2}-a$
\end_inset


\end_layout

\begin_layout Enumerate
Introduce a constant repulsive force which induces an intercept.
\end_layout

\begin_layout Enumerate
Introduce a casimir energy 
\begin_inset Formula $E_{casimir}=-\frac{a-\frac{\pi}{2}S}{L}$
\end_inset

 which also induces a force and an intercept.
\end_layout

\begin_layout Standard
Focusing on the last option, the induced force is 
\begin_inset Formula $F_{casimir}=-\frac{d}{dL}E_{casimir}=-\frac{a-\frac{\pi}{2}S}{L^{2}}$
\end_inset

.
 This force is attractive for 
\begin_inset Formula $a>\frac{\pi}{2}S$
\end_inset

 and repulsive for 
\begin_inset Formula $a<\frac{\pi}{2}S$
\end_inset

.
 The new equations for the energy and forces are 
\begin_inset Formula $\left(L=\frac{\beta_{1}+\beta_{2}}{\omega}\right)$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
E=\cfrac{T}{\omega}\left(\sin^{-1}(\beta_{1})+\sin^{-1}(\beta_{2})\right)+\gamma_{1}m_{1}+\gamma_{2}m_{2}+\frac{q_{1}q_{2}\omega}{\beta_{1}\beta_{2}\sin(\Delta\theta)+\Delta\theta}-\frac{\left(a-\frac{\pi}{2}S\right)\omega}{\beta_{1}+\beta_{2}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
J=T\cfrac{-\cfrac{\beta_{1}}{\gamma_{1}}+sin^{-1}\left(\beta_{1}\right)-\cfrac{\beta_{2}}{\gamma_{2}}+sin^{-1}\left(\beta_{2}\right)}{2\omega^{2}}+m_{1}\cfrac{\gamma_{1}\beta_{1}^{2}}{\omega}+m_{2}\cfrac{\gamma_{2}\beta_{2}^{2}}{\omega}-2q_{1}q_{2}\beta_{1}\beta_{2}\frac{cos\left(\Delta\theta\right)}{\beta_{1}\beta_{2}sin\left(\Delta\theta\right)+\Delta\theta}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\begin{cases}
\gamma_{1}\beta_{1}m_{1}\omega+\frac{\beta_{1}\beta_{2}q^{2}\omega^{2}\left(\beta_{1}\left(\beta_{1}^{2}+\beta_{2}^{2}\right)\sin(\theta)+\beta_{2}\left(\beta_{1}^{2}\theta+\beta_{1}^{2}\sin(2\theta)+\theta\right)+\beta_{1}\left(\beta_{2}^{2}+1\right)\theta\cos(\theta)\right)}{\theta(\beta_{1}\beta_{2}\sin(\theta)+\theta)^{3}}+\\
-\frac{q^{2}\omega^{2}\left(\beta_{1}\beta_{2}^{2}\cos(2\theta)+\beta_{1}\beta_{2}^{2}-2\beta_{1}+2\left(\beta_{2}^{2}-1\right)\beta_{2}^{1}\cos(\theta)-2\beta_{2}\theta\sin(\theta)\right)}{2(\beta_{1}\beta_{2}\sin(\theta)+\theta)^{3}}-\frac{T}{\gamma_{1}}-\frac{\left(a-\frac{\pi}{2}S\right)\omega^{2}}{\left(\beta_{1}+\beta_{2}\right)^{2}}=0
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\Delta\phi^{2}=\beta_{1}^{2}+\beta_{2}^{2}+2\beta_{1}\beta_{2}cos\left(\Delta\phi\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\gamma_{1}^{2}\beta_{1}m_{1}=\gamma_{2}^{2}\beta_{2}m_{2}
\]

\end_inset


\end_layout

\begin_layout Standard
The force is radial and so does not contribute to angular momentum.
\end_layout

\begin_layout Standard
We would like to see the behaviour of the regge trajectory, in the two following
 limits:
\end_layout

\begin_layout Standard
\begin_inset Formula $\beta\rightarrow0$
\end_inset

: In this limit the quarks are slowly moving, implying most of the energy
 is related to the string and end points and the quarks are heavy.
\end_layout

\begin_layout Standard
\begin_inset Formula $\beta\rightarrow1$
\end_inset

: In this limit the quarks are relativistic, implying most of the energy
 is related to the motion and the quarks are light.
\end_layout

\begin_layout Standard
In both cases we expand the energy and angular momentum around the limit,
 obtaining series of either 
\begin_inset Formula $\beta$
\end_inset

 or 
\begin_inset Formula $\left(1-\beta\right)$
\end_inset

.
 From the series for the energy 
\begin_inset Formula $E=f\left(\beta\right)$
\end_inset

(or 
\begin_inset Formula $E=f\left(1-\beta\right)$
\end_inset

) we calculate the inverse series 
\begin_inset Formula $\beta=f^{-1}\left(E\right)$
\end_inset

 or 
\begin_inset Formula $1-\beta=f^{-1}\left(E\right)$
\end_inset

 and insert into the series for the angular momentum, and obtain the regge
 trajectory 
\begin_inset Formula $J\left(E\right)$
\end_inset

.
 
\end_layout

\begin_layout Standard
The choice of naming of the parameter 
\begin_inset Formula $S$
\end_inset

 comes from the following consideration.
 Taking the limit 
\begin_inset Formula $\beta\rightarrow0$
\end_inset

 we get the following energy, angular momentum and force equations (note
 that in that limit 
\begin_inset Formula $\beta_{1}=\beta_{2}=\frac{\omega L}{2}$
\end_inset

 where 
\begin_inset Formula $L$
\end_inset

 is the string length):
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
E=TL+m_{1}+m_{2}+\frac{q^{2}-\left(a-\frac{\pi}{2}S\right)}{L}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
J=0
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\frac{q^{2}-\left(a-\frac{\pi}{2}S\right)}{L^{2}}=T
\]

\end_inset


\end_layout

\begin_layout Standard
Putting the two together we get
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
E=2\sqrt{T}\sqrt{q^{2}-\left(a-\frac{\pi}{2}S\right)}+m_{1}+m_{2}
\]

\end_inset


\end_layout

\begin_layout Standard
Which is equivalent to:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\alpha'\left(E-m_{1}-m_{2}\right)^{2}-\frac{2}{\pi}q^{2}+\frac{2}{\pi}a=S
\]

\end_inset


\end_layout

\begin_layout Standard
By analogy to the case of mass alone the left hand side is identified with
 the angular momentum and we are led to find that 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
S\equiv Spin
\]

\end_inset


\end_layout

\begin_layout Subsection
Light quark limit
\end_layout

\begin_layout Standard
In this limit the quark velocity is large, 
\begin_inset Formula $\beta=1-\epsilon^{2}$
\end_inset

, 
\begin_inset Formula $\epsilon\rightarrow0$
\end_inset


\end_layout

\begin_layout Standard
In this case the regge trajectory is
\end_layout

\begin_layout Standard
The result when adding massive end points and no charge and casimir is:
\end_layout

\begin_layout Standard
\begin_inset Formula $J=1-\frac{4\sqrt{\pi}}{3}\left(\frac{1}{E}\right)^{3/2}\left(m_{1}^{3/2}+m_{2}^{3/2}\right)+\frac{1}{5}\pi^{3/2}\left(\frac{1}{E}\right)^{5/2}\left(m_{1}^{5/2}+m_{2}^{5/2}\right)+O\left(\left(\frac{1}{E}\right)^{3}\right)$
\end_inset


\end_layout

\begin_layout Standard
The result when adding casimir forces and charges is
\end_layout

\begin_layout Standard
\begin_inset Formula $J=\alpha'E^{2}+O\left(\frac{1}{\sqrt{\frac{1}{E}}}\right)$
\end_inset


\end_layout

\begin_layout Subsection
Heavy quark limit
\end_layout

\begin_layout Standard
In this limit the quark velocity is small, 
\begin_inset Formula $\beta\rightarrow0$
\end_inset


\end_layout

\begin_layout Standard
The equations are solved in the same manner and we get:
\end_layout

\begin_layout Standard
mass only
\end_layout

\begin_layout Standard
\begin_inset Formula $J=\frac{4}{3}\sqrt{\frac{2}{3}}\pi\alpha'\sqrt{\frac{m_{1}m_{2}}{m_{1}+m_{2}}}(E-m_{1}-m_{2})^{3/2}+\frac{7\sqrt{\frac{2}{3}}\pi\alpha'\left(m_{1}^{2}-m_{1}m_{2}+m_{2}^{2}\right)(E-m_{1}-m_{2})^{5/2}}{27\sqrt{m_{1}m_{2}(m_{1}+m_{2})^{3}}}+O\left((E-m_{1}-m_{2})^{7/2}\right)$
\end_inset


\end_layout

\begin_layout Standard
The result when adding casimir forces and charges is
\end_layout

\begin_layout Standard
\begin_inset Formula $-\frac{\sqrt{\frac{1}{6m\sqrt{T\left(q^{2}-a\right)}+aT+2q^{2}T}}\left(q^{2}\left(5\sqrt{T\left(q^{2}-a\right)}+6m\right)+a\left(\sqrt{T\left(q^{2}-a\right)}-6m\right)\right)\sqrt{2\sqrt{T\left(q^{2}-a\right)}+E-2m}}{\sqrt{6}\left(T\left(q^{2}-a\right)\right)^{1/4}}+O\left(\left(2\sqrt{T\left(q^{2}-a\right)}+E-2m\right)^{3/2}\right)$
\end_inset


\end_layout

\begin_layout Section
Energy Differences
\end_layout

\begin_layout Standard

\series bold
\bar under
todo-introduction - how quark mass differences are measured in QCD
\end_layout

\begin_layout Standard
In the stationary string:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
E=2\sqrt{T}\sqrt{q^{2}-\left(a-\frac{\pi}{2}S\right)}+m_{1}+m_{2}
\]

\end_inset


\end_layout

\begin_layout Standard
And the orbital angular momentum is 
\begin_inset Formula $J=0$
\end_inset

.
\end_layout

\begin_layout Standard
The parameters 
\begin_inset Formula $a,T$
\end_inset

 are measured from the regge trajectory fits while the spin 
\begin_inset Formula $S$
\end_inset

 is related to the angular momentum and parity:
\end_layout

\begin_layout Standard
\begin_inset Formula $P_{meson}=\left(-1\right)^{L+1}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $P_{baryon}=\left(-1\right)^{L}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $J=L+S$
\end_inset


\end_layout

\begin_layout Standard
When the charge is negligible relative to the casimir contribution we can
 approximate:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
E\approx\sqrt{\frac{2}{\pi\alpha'}}\sqrt{\left(\frac{\pi}{2}S-a\right)}+\sqrt{\frac{1}{2\pi\alpha'}}\frac{q^{2}}{\sqrt{\left(\frac{\pi}{2}S-a\right)}}+m_{1}+m_{2}
\]

\end_inset


\end_layout

\begin_layout Standard
Taking the difference between masses of hadrons which correspond to differently
 charged versions of the same particle, for which the parameters 
\begin_inset Formula $a,S,\alpha'$
\end_inset

 are constants:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\Delta E\approx\sqrt{\frac{1}{2\pi\alpha'}}\frac{\Delta q^{2}}{\sqrt{\left(\frac{\pi}{2}S-a\right)}}+\Delta m_{quarks}
\]

\end_inset


\end_layout

\begin_layout Standard
From the mass difference of the hadrons we can infer mass differences between
 the quarks themselves.
 For example the proton and neutron are considered as a quark-diquark system
 of 
\begin_inset Formula $\left(ud-u\right)$
\end_inset

 and 
\begin_inset Formula $\left(ud-d\right)$
\end_inset

 respectively, and so the end point masses are 
\begin_inset Formula $m_{u}+m_{d}$
\end_inset

 and 
\begin_inset Formula $2m_{d}$
\end_inset

 respectively.
 Furthermore 
\begin_inset Formula $q^{2}$
\end_inset

 is calculated as 
\begin_inset Formula $q_{diquark}q_{quark}$
\end_inset

 so that 
\begin_inset Formula $q_{proton}^{2}=\left(\frac{2}{3}-\frac{1}{3}\right)\frac{2}{3}=\frac{2}{9}$
\end_inset

 , 
\begin_inset Formula $q_{neutron}^{2}=-\left(\frac{2}{3}-\frac{1}{3}\right)\frac{1}{3}=-\frac{1}{9}$
\end_inset

.
 If the hadron is a superposition of electromagnetic states (i.e.
 
\begin_inset Formula $\frac{1}{\sqrt{2}}\left(\left|u\bar{u}\right\rangle \pm\left|d\bar{d}\right\rangle \right)$
\end_inset

 ) then 
\begin_inset Formula $q^{2}$
\end_inset

 is calculated as the average 
\begin_inset Formula $q^{2}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
M_{neutron}-M_{proton}=2m_{d}-m_{d}-m_{u}+\sqrt{\frac{1}{2\pi\alpha'}}\frac{\left(-\frac{3}{9}\right)}{\sqrt{\left(\frac{\pi}{2}S-a\right)}}=m_{d}-m_{u}-\frac{3}{9}\sqrt{\frac{1}{2\pi\alpha'}}\frac{1}{\sqrt{\left(\frac{\pi}{2}S-a\right)}}
\]

\end_inset


\end_layout

\begin_layout Standard
Reversing the argument, the difference in the quark masses can be calculated
 from the hadron mass differences and the fit to the regge trajectory:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
m_{d}-m_{u}=M_{neutron}-M_{proton}+\frac{3}{9}\sqrt{\frac{1}{2\pi\alpha'}}\frac{1}{\sqrt{\left(\frac{\pi}{2}S-a\right)}}
\]

\end_inset


\end_layout

\begin_layout Standard
In the following section we present the calculation of the 
\begin_inset Formula $u/d$
\end_inset

 mass difference for several different hadrons using the parameters consistent
 with the regge fits, and analyse the results.
\end_layout

\begin_layout Part
Confronting with PDG Data
\end_layout

\begin_layout Section
Regge Trajectories
\end_layout

\begin_layout Standard
Using the methods outlined in the previous section, we fit the regge trajectorie
s of our model to PDG data for the hadrons.
 Specifically we assume 
\begin_inset Formula $m_{1}=m_{2}$
\end_inset

 , without loss of generality as can be seen from equation ** and the fits
 done in [dorin] which show this behaviour in general.
 Also we take 
\begin_inset Formula $\alpha_{e}=\cfrac{1}{137}$
\end_inset

.
\end_layout

\begin_layout Subsection
Plotting against J vs.
 Plotting against L
\end_layout

\begin_layout Standard
In previous work [dorin] the model with no electric charge was confronted
 with PDG data.
 There the intercept was using method 1 discussed in section **.
 When the fits were done, there was no way of distinguishing between fits
 to the total angular momentum 
\begin_inset Formula $J$
\end_inset

 of a hadron, and fits to the orbital angular momentum 
\begin_inset Formula $L$
\end_inset

, of the same hadron, as the intercept could always be adjusted to give
 either plot.
 However when putting in the intercept using a casimir energy prescription
 we obtain fundementaly different results for the two fits.
 We find that we cannot fit the model to the total angular momentum 
\begin_inset Formula $J$
\end_inset

 of the hadron.
 An example for the 
\begin_inset Formula $\rho$
\end_inset

 meson can be seen in figure **.
\end_layout

\begin_layout Standard
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
Fit to 
\begin_inset Formula $\left(J,E^{2}\right)$
\end_inset

 for the 
\begin_inset Formula $\rho$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption

\begin_layout Plain Layout
\begin_inset Graphics
	filename figures/Jtrajectory.png
	scale 50

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Plain Layout
Blue dots - model, red dots - data, green line - linear fit
\end_layout

\end_inset


\end_layout

\begin_layout Standard
In the next sections, all the fits will be done for 
\begin_inset Formula $\left(L,E^{2}\right)$
\end_inset

.
\end_layout

\begin_layout Subsection
Fit Plots
\end_layout

\begin_layout Standard
In the following, we will present the goodness of the fit in the parameter
 space, and determine values for the universal string tension and end point
 masses for various hadrons.
 The fits were done in Mathematica 10 (see code in the appendix) using numerical
 methods.
 The goodness of fit is measured in terms of the function
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\chi^{2}=\sum_{i\in trajectory}\cfrac{M_{i,model}^{2}-M_{i,pdg}^{2}}{M_{i,pdg}^{2}}
\]

\end_inset


\end_layout

\begin_layout Subsubsection
Meson Fits
\end_layout

\begin_layout Subsubsection
Baryon Fits
\end_layout

\begin_layout Subsection
Summary tables
\end_layout

\begin_layout Subsection
Comparison with purely massive fits
\end_layout

\begin_layout Subsubsection
Meson Fits
\end_layout

\begin_layout Standard
\begin_inset Tabular
<lyxtabular version="3" rows="5" columns="5">
<features tabularvalignment="middle">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\rho$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\omega$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\eta$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\pi$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\alpha'_{w/casimir}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0.86-0.95
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0.81-0.88
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0.735-0.767
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0.825-0.846
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\alpha'_{mass}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0.896
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0.909
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0.839
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0.844
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $m_{w/casimir}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
150-300
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0-196
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0-56
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0-203
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $m_{mass}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
65
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
170
\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\begin_layout Subsubsection
Baryon Fits
\end_layout

\begin_layout Standard
\begin_inset Tabular
<lyxtabular version="3" rows="5" columns="5">
<features tabularvalignment="middle">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $N$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\Lambda$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\Sigma$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\alpha'_{w/casimir}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0.94-1
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
1.06-1.1
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
1.11-1.48
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $\alpha'_{mass}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0.959
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0.955
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
1.502
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $m_{w/casimir}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0-150
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
390-410
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
500-569
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $m_{mass}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0-85
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
62.5
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
595
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\begin_layout Section
Mass Differences
\end_layout

\begin_layout Standard

\end_layout

\begin_layout Section
Decay Rates
\end_layout

\begin_layout Standard
\begin_inset Formula $\Gamma_{K^{*}\rightarrow K^{-}\pi^{+}}=L\times e^{-m_{q}^{2}/T}$
\end_inset


\end_layout

\begin_layout Part
Bibliography
\end_layout

\begin_layout Part
Appendix
\end_layout

\begin_layout Section
Conventions
\end_layout

\begin_layout Standard
The convention for the metric tensor is the same as in J.
 D.
 Jackson 3rd edition (equation 11.59, 11.81):
\end_layout

\begin_layout Standard
\begin_inset Formula $g_{\mu\nu}=sig(+,-,-,-)$
\end_inset


\end_layout

\begin_layout Standard
and for the electromagnetic field tensor (equations 11.136, 11.137, 11.138):
\end_layout

\begin_layout Standard
\begin_inset Formula $F^{\alpha\beta}=\left(\begin{array}{cccc}
0 & -E_{1} & -E_{2} & -E_{3}\\
E_{1} & 0 & -B_{3} & B_{2}\\
E_{2} & B_{3} & 0 & -B_{1}\\
E_{3} & -B_{2} & B_{1} & 0
\end{array}\right)$
\end_inset

 
\begin_inset Formula $F_{\alpha\beta}=\left(\begin{array}{cccc}
0 & E_{1} & E_{2} & E_{3}\\
-E_{1} & 0 & -B_{3} & B_{2}\\
-E_{2} & B_{3} & 0 & -B_{1}\\
-E_{3} & -B_{2} & B_{1} & 0
\end{array}\right)$
\end_inset


\end_layout

\begin_layout Section
Mathematica code
\end_layout

\end_body
\end_document