# Beltrami identity

Let $q(t)$ be a function $\mathbb{R}\to\mathbb{R}$, $\dot{q}=\frac{d}{d{t}}{q}$, and $L=L(q,\dot{q},t)$. Begin with the time-relative Euler-Lagrange condition

 $\frac{\partial}{\partial{q}}L-\frac{d}{d{t}}\left(\frac{\partial}{\partial{% \dot{q}}}L\right)=0.$ (1)

If $\frac{\partial}{\partial{t}}L=0$, then the Euler-Lagrange condition reduces to

 $L-\dot{q}{\frac{\partial}{\partial{\dot{q}}}L}=C,$ (2)

which is the 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 $\frac{\partial}{\partial{t}}L=0$.

In space-relative terms, with $q^{\prime}:=\frac{d}{d{x}}q$, we have

 $\frac{\partial}{\partial{q}}L-\frac{d}{d{x}}{\frac{\partial}{\partial{q^{% \prime}}}L}=0.$ (3)

If $\frac{\partial}{\partial{x}}L=0$, then the Euler-Lagrange condition reduces to

 $L-q^{\prime}{\frac{\partial}{\partial{q^{\prime}}}L}=C.$ (4)

To derive the Beltrami identity, note that

 $\frac{d}{d{t}}\left(\dot{q}\frac{\partial}{\partial{\dot{q}}}L\right)=\ddot{q}% \frac{\partial}{\partial{\dot{q}}}L+\dot{q}\frac{d}{d{t}}\left(\frac{\partial}% {\partial{\dot{q}}}L\right)$ (5)

Multiplying (1) by $\dot{q}$, we have

 $\dot{q}\frac{\partial}{\partial{q}}L-\dot{q}\frac{d}{d{t}}\left(\frac{\partial% }{\partial{\dot{q}}}L\right)=0.$ (6)

Now, rearranging (5) and substituting in for the rightmost term of (6), we obtain

 $\dot{q}\frac{\partial}{\partial{q}}L+\ddot{q}\frac{\partial}{\partial{\dot{q}}% }L-\frac{d}{d{t}}\left(\dot{q}\frac{\partial}{\partial{\dot{q}}}L\right)=0.$ (7)

Now consider the total derivative

 $\frac{d}{d{t}}L(q,\dot{q},t)=\dot{q}\frac{\partial}{\partial{q}}L+\ddot{q}% \frac{\partial}{\partial{\dot{q}}}L+\frac{\partial}{\partial{t}}L.$ (8)

If $\frac{\partial}{\partial{t}}L=0$, then we can substitute in the left-hand side of (8) for the leading portion of (7) to get

 $\frac{d}{d{t}}L-\frac{d}{d{t}}\left(\dot{q}\frac{\partial}{\partial{\dot{q}}}L% \right)=0.$ (9)

Integrating with respect to $t$, we arrive at

 $L-\dot{q}{\frac{\partial}{\partial{\dot{q}}}L}=C,$ (10)

which is the Beltrami identity.

Title Beltrami identity BeltramiIdentity 2013-03-22 12:21:08 2013-03-22 12:21:08 PrimeFan (13766) PrimeFan (13766) 8 PrimeFan (13766) Definition msc 47A60 CalculusOfVariations EulerLagrangeDifferentialEquation