Categories: Mathematics, Physics.

# Beltrami identity

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

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

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

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

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

\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:

\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 $\idv{L}{x}$, we arrive at an expression for the “soft” derivative $\ipdv{L}{x}$ describing only the explicit dependence of $L$ on $x$:

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

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

\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 $x$, which could be useful to know. Furthermore, for some Lagrangians $L$, the Beltrami identity is easier to solve for $f$ than the full Euler-Lagrange equation. The condition $\ipdv{L}{x} = 0$ is often justified: for example, if $x$ is time, it simply means that the potential is time-independent.

When we add more dimensions, e.g. for $L(f, f_x, f_y, x, y)$, 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.

## References

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 O. Bang, Nonlinear mathematical physics: lecture notes, 2020, unpublished.