Beltrami identity

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

$$\frac{\partial }{\partial q}L-\frac{d}{dt}\left(\frac{\partial }{\partial \dot{q}}L\right)=0.$$(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 $q^{\prime }:=\frac{d}{dx}q$, we have

$$\frac{\partial }{\partial q}L-\frac{d}{dx}\frac{\partial }{\partial q^{\prime }}L=0.$$(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

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

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

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

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

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

Now consider the total derivative

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

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