L06E06.tex

\documentclass[solutions.tex]{subfiles}

\xtitle

\begin{document}
\maketitle
\begin{exercise}
Explain how we derived this.
\end{exercise}

Let us recall that "this" refers to the following expression for the
kinetic energy:

\[
	T = m (\dot{x_+}^2+\dot{x_-}^2)
\]

Starting from the following Lagrangian, involving two particles
moving on a line with respective position and velocity $x_i$, $\dot{x_i}$:

\begin{align}\label{l06e06:lagrangian}
	L = \frac{m}{2}(\dot{x_1}^2+\dot{x_2}^2)-V(x_1-x_2)
\end{align}

After having performed the following change of coordinates:

\begin{align}
	x_+ &= \dfrac{x_1+x_2}{2} &
	x_- &= \dfrac{x_1-x_2}{2} \label{l06e06:coord-update}
\end{align}

From the Lagrangian, \eqref{l06e06:lagrangian} we have the kinetic energy:

\begin{align}\label{l06e06:k}
	T = \frac{m}{2}(\dot{x_1}^2+\dot{x_2}^2)
\end{align}

By first both summing and subtracting the two equations of
\eqref{l06e06:coord-update}, and then by linearity of the derivation,
we get:
\begin{align}
	x_+ + x_- &= x_1 &
	x_+ - x_- &= x_2 \nonumber \\
	\dot{x_+} + \dot{x_-} &= \dot{x_1} &
	\dot{x_+} - \dot{x_-} &= \dot{x_2} \label{l06e06:coord-update-bis}
\end{align}

It's now simply a matter of injecting \eqref{l06e06:coord-update-bis}
into \eqref{l06e06:k}:

\begin{align*}
	T &= \dfrac{m}{2}(\dot{x_1}^2+\dot{x_2}^2) \\
	~ &= \dfrac{m}{2}((\dot{x_+} + \dot{x_-})^2+(\dot{x_+} - \dot{x_-})^2) \\
	~ &= \dfrac{m}{2}(
		2\dot{x_+}^2 +
		2\dot{x_-}^2 +
		2\dot{x_+}\dot{x_-} - 2\dot{x_+}\dot{x_-}
	) \\
	~ &= m(\dot{x_+}^2 +\dot{x_-}^2) \qed
\end{align*}

\end{document}