# Calculating the (on-shell) action of a free particle

I am having difficulty with the first problem from Feynman and Hibbs' book.

For a free particle $L = (m/2)\dot{x}^2$. Show that the (on-shell) action $S_{cl}$ corresponding to the classical motion of a free particle is $$S_{cl} ~=~ \frac{m}{2}\frac{(x_b - x_a)^2}{t_b - t_a}$$ where we have that $x(t_a) = x_a$ and $x(t_b) = x_b$.

I understand that the action is

$$S ~=~ \int_{t_a}^{t_b} \frac{m}{2}\dot{x}^2 \,dt.$$

But I do not know how to solve the integral $\int \dot{x}^2\,dt$. Any help is appreciated.

Following Noldig's comment, integrating by parts we have that:

\begin{eqnarray} \int_{t_a}^{t_b} \dot{x} \dot{x} \,dt & = & \left. \dot{x} x\right|_{t_a}^{t_b} - \int x \ddot{x} \,dt \\ & = & \dot{x}x(t_b)-\dot{x}x(t_a) \end{eqnarray}

As the velocity is constant it is given by $\dot{x} = (x_b-x_a)/(t_b-t_a)$ and the result follows.

## 3 Answers

Use integration by parts and the fact, that for a free particle $\frac{d^{2}x}{dt^{2}}=0$. In addition you know, that the velocity is constant, therfore you can solve the first part too.

Also, you can use a sort of change of variables. $\int_a^b \dot{x}^2 dt = \int_a^b v^2 dt = v^2 \int_a^b dt = v^2({t_b}-{t_a})$ where the last part is using the fact v is constant for a line to pull it out of the integral. The result follows.

What can be frustrating is to start from the idea that we are calculating the general action for any path connecting $x_a$ and $x_b$. Then is when you meet the uncomfortable expression $S = \frac{m}{2}\int_{t_a}^{t_b} \dot{x}^2 dt =\frac{m}{2}\{[x \dot{x}]_{t_a}^{t_b}-\int_{t_a}^{t_b}x\ddot{x} dt\}$

However, we can still write such general action as $S = \frac{m}{2}\int_{t_a}^{t_b} \dot{x}^2 dt = \frac{m}{2}\int_{t_a}^{t_b} [\frac{dx}{dt}|_{x=x_a}+\frac{d^2x}{dt^2}|_{x=x_a} {\small(t-t_a)}+\frac{1}{2!}\frac{d^3x}{dt^3}|_{x=x_a}{\small(t-t_a)^2}+o(|{\small t}|^3)]^2 {\small dt}$

For the classical trajectory which minimizes $S$, we obtain

$S_{\scriptsize\mbox{min}}=S_{\scriptsize\mbox{cl}}=\frac{m}{2}\int_{t_a}^{t_b} [\frac{dx}{dt}|_{x=x_a}]^2 dt$

from which it is straightforward to obtain the given result. So this is all to stress the difference between $S$ and $S_{{\scriptsize\mbox{cl}}}$