Worldline formalism
Point particle as a limit of string action :
\begin{equation} S = -T \int_{\Sigma} d\mu[X^*g] \end{equation}For now, let's consider the Minkowski case, with two coordinates $\sigma, \tau$
\begin{equation} S = -T \int_{t_a}^{t_b} \int_{0}^{2\pi} d\tau d\sigma \sqrt{(\dot{X}^\mu(\tau,\sigma) X'_\mu(\tau,\sigma))^2 - (\dot{X}(\tau,\sigma))^2 (X'(\tau,\sigma))^2 } \end{equation}The obvious method by which to perform a deformation retract of our string to a point is to change the integration boundaries from $2\pi$ to $\lambda 2\pi$, for $\lambda \in [0,1]$. Naively this would just make the action zero. To remedy this somewhat, let's consider the length of our string based purely on parameters
\begin{equation} l_\lambda = \int_0^{\lambda 2\pi} d\sigma = \lambda 2\pi \end{equation}For any fixed $\lambda > 0$, this will not change the dynamics of our theory. Now, we know that the constant $T$ can be considered as the tension of our string. From classical formulas, let's consider the string tension as
\begin{equation} T_\lambda = \frac{mc}{l_\lambda} \end{equation}such that the string tension is proportional to its linear density. Then let's recast our action a bit
\begin{equation} S = -T_\lambda \int_{t_a}^{t_b} \int_{0}^{\lambda 2\pi} d\tau d\sigma \sqrt{ (\dot{X}(\tau,\sigma))^2 } \sqrt{\frac{(\dot{X}^\mu(\tau,\sigma) X'_\mu(\tau,\sigma))^2}{(\dot{X}(\tau,\sigma))^2} - (X'(\tau,\sigma))^2 } \end{equation}By the mean value theorem, there exists some $\sigma^* \in [0, \lambda 2 \pi]$ such that our action can be expressed as
\begin{equation} S = -T_\lambda \int_{t_a}^{t_b} \int_{0}^{\lambda 2\pi} d\tau \sqrt{ (\dot{X}(\tau,\sigma^*))^2 } \int_{0}^{\lambda 2\pi} d\sigma \sqrt{\frac{(\dot{X}^\mu(\tau,\sigma) X'_\mu(\tau,\sigma))^2}{(\dot{X}(\tau,\sigma))^2} - (X'(\tau,\sigma))^2 } \end{equation}The term in the $\sigma$ integral can be recognized as the length of the string as a time $\tau$, let's call it $L_\lambda(\tau)$. We can now express our action as
\begin{equation} S = -mc \int_{t_a}^{t_b} \int_{0}^{\lambda 2\pi} d\tau \sqrt{ (\dot{X}(\tau,\sigma^*))^2 } \frac{L_\lambda(\tau)}{l_\lambda} \end{equation}Considering the fundamental theorem of calculus, we can express the limit as $\lambda \to 0$ by
\begin{equation} \lim_{\lambda \to 0} \frac{L_\lambda(\tau)}{l_\lambda} = \sqrt{\frac{(\dot{X}^\mu(\tau, 0) X'_\mu(\tau, 0))^2}{(\dot{X}(\tau, 0))^2} - (X'(\tau, 0))^2 } \end{equation}Vertex in the zero length limit
Something more interesting is the limit of a non-trivial
Last updated : 2019-12-02 10:13:10