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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.

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up vote 4 down vote accepted

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.

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Thanks Noldig. $~$ – anthus Feb 3 '13 at 13:32

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.

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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 $S_{{\scriptsize\mbox{cl}}}$

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