Categories: Mathematics, Physics.

Beltrami identity

Consider a general functional J[f]J[f] of the following form, with f(x)f(x) an unknown function:

J[f]=x0x1L(f,f,x)dx\begin{aligned} J[f] = \int_{x_0}^{x_1} L(f, f', x) \dd{x} \end{aligned}

Where LL is the Lagrangian. To find the ff that maximizes or minimizes J[f]J[f], the calculus of variations states that the Euler-Lagrange equation must be solved for ff:

0=Lfddx(Lf)\begin{aligned} 0 = \pdv{L}{f} - \dv{}{x} \Big( \pdv{L}{f'} \Big) \end{aligned}

We now want to know exactly how LL depends on the free variable xx. Of course, xx may appear explicitly in LL, but usually LL also has an implicit dependence on xx via f(x)f(x) and f(x)f'(x). To find a relation between this implicit and explicit dependence, we start by using the chain rule:

dLdx=Lfdfdx+Lfdfdx+Lx\begin{aligned} \dv{L}{x} = \pdv{L}{f} \dv{f}{x} + \pdv{L}{f'} \dv{f'}{x} + \pdv{L}{x} \end{aligned}

Substituting the Euler-Lagrange equation into the first term gives us:

dLdx=fddx(Lf)+dfdxLf+Lx=ddx(fLf)+Lx\begin{aligned} \dv{L}{x} &= f' \dv{}{x} \Big( \pdv{L}{f'} \Big) + \dv{f'}{x} \pdv{L}{f'} + \pdv{L}{x} \\ &= \dv{}{x} \bigg( f' \pdv{L}{f'} \bigg) + \pdv{L}{x} \end{aligned}

Although we started from the “hard” derivative dL/dx\idv{L}{x}, we arrive at an expression for the “soft” derivative L/x\ipdv{L}{x} describing only the explicit dependence of LL on xx:

Lx=ddx(fLfL)\begin{aligned} - \pdv{L}{x} = \dv{}{x} \bigg( f' \pdv{L}{f'} - L \bigg) \end{aligned}

What if LL does not explicitly depend on xx at all, i.e. L/x=0\ipdv{L}{x} = 0? In that case, the equation can be integrated to give the Beltrami identity, where CC is a constant:

fLfL=C\begin{aligned} \boxed{ f' \pdv{L}{f'} - L = C } \end{aligned}

This says that the left-hand side is a conserved quantity with respect to xx, which could be useful to know. Furthermore, for some Lagrangians LL, the Beltrami identity is easier to solve for ff than the full Euler-Lagrange equation. The condition L/x=0\ipdv{L}{x} = 0 is often justified: for example, if xx is time, it simply means that the potential is time-independent.

When we add more dimensions, e.g. for L(f,fx,fy,x,y)L(f, f_x, f_y, x, y), the above derivation no longer works due to the final integration step, so the name Beltrami identity is only used in 1D. Nevertheless, a generalization does exist that can handle more dimensions: Noether’s theorem.


  1. O. Bang, Nonlinear mathematical physics: lecture notes, 2020, unpublished.