---
title: "Beltrami identity"
sort_title: "Beltrami identity"
date: 2022-09-17
categories:
- Physics
- Mathematics
layout: "concept"
---

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](/know/concept/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](/know/concept/noethers-theorem/).



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    unpublished.