Exercise 6

The new coordinates \(x_+\) and \(x_-\) are defined as follows: \[ x_+ = \frac{x_1 + x_2}{2} \Rightarrow x_1 = 2 x_+ - x_2\] \[ x_- = \frac{x_1 - x_2}{2} \Rightarrow x_2 = -2x_- + x_1\]

The equations of transformation are: \[ x_1 = x_+ + x_- \] \[ x_2 = x_+ - x_- \] \[ \dot{x}_1 = \dot{x}_+ + \dot{x}_- \] \[ \dot{x}_2 = \dot{x}_+ - \dot{x}_- \]

\[ \dot{x}_1^2 = (\dot{x}_+ + \dot{x}_-)^2 = \dot{x}_+^2 + 2 \dot{x}_+\dot{x}_- + \dot{x}_-^2\] \[ \dot{x}_2^2 = (\dot{x}_+ - \dot{x}_-)^2 = \dot{x}_+^2 - 2 \dot{x}_+\dot{x}_- + \dot{x}_-^2\]

\[ \dot{x}_1^2 + \dot{x}_2^2 = 2\dot{x}_+^2 + 2 \dot{x}_-^2 \Rightarrow \frac{1}{2} (\dot{x}_1^2 + \dot{x}_2^2) = \dot{x}_+^2 + \dot{x}_-^2 \]

Hence, the kinetic energy is \[ T = m (\dot{x}_+^2 + \dot{x}_-^2) \]