Beltrami identityBeltramiIdentity2002-02-16 14:37:112006-08-10 19:43:18Definition\usepackage{amssymb}
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\newcommand{\braket}[2]{\langle #1 \ket{#2}}Let $q(t)$ be a function $\reals \to \reals$, $\dot{q} = \mderiv{t}{q}$, and $L = L(q, \dot{q}, t)$. Begin with the time-relative Euler-Lagrange condition
\begin{equation}\label{el}
\mpderiv{q} L - \mderiv{t}\left(\mpderiv{\dot{q}} L\right) = 0.
\end{equation}
If $\mpderiv{t}L = 0$, then the Euler-Lagrange condition reduces to
\begin{equation}
L - \dot{q}{\mpderiv{\dot{q}} L} = C,
\end{equation}
which is the \emph{Beltrami identity}. In the calculus of variations, the ability to use the Beltrami identity can vastly simplify problems, and as it happens, many physical problems have $\mpderiv{t}L = 0$.
In space-relative terms, with $q' \defined \mderiv{x}q$, we have
\begin{equation}
\mpderiv{q} L - \mderiv{x}{\mpderiv{q'} L} = 0.
\end{equation}
If $\mpderiv{x}L = 0$, then the Euler-Lagrange condition reduces to
\begin{equation}
L - q' {\mpderiv{q'} L} = C.
\end{equation}
To derive the Beltrami identity, note that
\begin{equation}\label{step2}
\mderiv{t}\left(\dot{q}\mpderiv{\dot{q}}L\right) = \ddot{q}\mpderiv{\dot{q}}L + \dot{q}\mderiv{t}\left(\mpderiv{\dot{q}}L\right)
\end{equation}
Multiplying (1) by $\dot{q}$, we have
\begin{equation}\label{step3}
\dot{q}\mpderiv{q} L - \dot{q}\mderiv{t}\left(\mpderiv{\dot{q}} L\right) = 0.
\end{equation}
Now, rearranging (5) and substituting in for the rightmost term of (6), we obtain
\begin{equation}\label{step4}
\dot{q}\mpderiv{q} L + \ddot{q}\mpderiv{\dot{q}}L - \mderiv{t}\left(\dot{q}\mpderiv{\dot{q}}L\right) = 0.
\end{equation}
Now consider the total derivative
\begin{equation}\label{step1}
\mderiv{t}L(q, \dot{q}, t) = \dot{q}\mpderiv{q}L + \ddot{q}\mpderiv{\dot{q}}L + \mpderiv{t}L.
\end{equation}
If $\mpderiv{t}L = 0$, then we can substitute in the left-hand side of (8) for the leading portion of (7) to get
\begin{equation}
\mderiv{t}L - \mderiv{t}\left(\dot{q}\mpderiv{\dot{q}}L\right) = 0.
\end{equation}
Integrating with respect to $t$, we arrive at
\begin{equation}
L - \dot{q}{\mpderiv{\dot{q}} L} = C,
\end{equation}
which is the Beltrami identity.