Point particles as the limit of a short string

Question

There's a common saying in the domain of the study of classical relativistic strings, that in the limit of a very short string, the action reduces to that of a point particle (there is for instance a sketch of a proof in Barbashov). I've been trying to show it.

Consider the Nambu-Goto string action in Minkowski space:

\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}

We want this in the limit that $\sigma_2 \to \sigma_1$, let's say for instance replacing $[0, 2\pi]$ by $[0, \lambda 2\pi]$. Obviously this does nothing but give us $S = 0$, so first we have to change the action very slightly. Take our tension $T$. As it is a tension, we can express it in terms of a linear mass density. We'll choose to express it as

$$T = \frac{mc}{\int_{0}^{2\pi}d\sigma} = \frac{mc}{l}$$

For any fixed $\lambda > 0$, this doesn't change our dynamics. So we can say get it parametrized by $\lambda$:

$$T_\lambda = \frac{mc}{\int_{0}^{\lambda 2\pi}d\sigma} = \frac{mc}{l_\lambda}$$

with $l_\lambda = \lambda 2\pi$.

Let's consider our parametrized action now:

\begin{equation} S = -\frac{mc}{l_\lambda} \int_{t_a}^{t_b} \int_{0}^{\lambda 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}

By manipulating it somewhat, we get

\begin{equation} S = -\frac{mc}{l_\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}

Using the mean value theorem, there exists a $\sigma^* \in [0, \lambda 2\pi]$ such that

\begin{equation} S_\lambda = -\frac{mc}{l_\lambda} \int_{t_a}^{t_b} 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}

From the mouth of Goto himself, the second integral is nothing but the length of our string at a time $\tau$. We can then rewrite the action as

\begin{equation} S_\lambda = -mc \int_{t_a}^{t_b} d\tau \sqrt{(\dot{X}(\tau,\sigma^*))^2} \frac{L_\lambda(\tau)}{l_\lambda} \end{equation}

From the fundamental theorem of calculus, this is just

\begin{eqnarray} \frac{\int_0^{\lambda} f(\sigma) d\sigma}{\lambda} &=& \frac{F(\lambda) - F(0)}{\lambda} \end{eqnarray}

for which the limit should just be $f(0)$, in other words:

$$\lim_{\lambda \to 0} \frac{L_\lambda(\tau)}{l_\lambda} = \sqrt{\frac{(\dot{X}^\mu(\tau,0) X'_\mu(\tau,0))^2}{(\dot{X}(\tau,0))^2} - (X'(\tau,0))^2 }$$

If this quantity is a constant $C$, lucky us we get

\begin{equation} \lim_{\lambda \to 0} S_\lambda = -Cmc \int_{t_a}^{t_b} d\tau \sqrt{(\dot{X}(\tau,0))^2} \end{equation}

Which would indeed be the correct action. But unfortunately I'm not sure how to show this. For a start, things get tricky assuming Neumann boundary conditions. Is there a way to show that this quantity indeed converges to a proper finite, non-zero limit?

Answer 1

Perhaps the simplest derivation that the string stiffens $X^{\prime}\to 0$ in the point particle limit with infinite string tension $T_0\to \infty$ comes from the corresponding Hamiltonian formulations:

1. On one hand, the Nambu-Goto Hamiltonian Lagrangian is $$ L_H~:=~\int_0^{\ell}\! d\sigma~{\cal L}_H, \qquad {\cal L}_H ~:=~ P\cdot \dot{X}-{\cal H}, \qquad {\cal H}~=~\lambda^{\alpha}\chi_{\alpha}, \qquad \alpha~\in~\{0,1\},\tag{1}$$ with two first class constraints $$\chi_0~:=~P\cdot X^{\prime}~\approx~0, \qquad \chi_1~:=~\frac{P^2}{2T_0}+\frac{T_0}{2}(X^{\prime})^2~\approx~0,\tag{2}$$ cf. e.g. this Phys.SE post.

2. On the other hand, the point particle Hamiltonian Lagrangian is $$ L_H~:=~p\cdot \dot{x} -H \qquad H~=~ \frac{e}{2}(p^2+m^2) ,\tag{3}$$ cf. e.g. this Phys.SE post.