Hamilton's principle/Mathematics/The Beltrami identity

Plain-language version: Free ends

When the integrand does not contain the independent variable, the Euler-Lagrange equation admits a first integral. It reduces a second-order equation to a first-order one and is the single most useful labour-saving result in the applications track. It is called the Beltrami identity, and in mechanics it is conservation of energy.

Statement

A quantity conserved along the whole path.

Theorem. Let $L=L(y,y^{\prime })$ contain no explicit dependence on $x$, so $\partial L/\partial x=0$. Then along any extremal,

$L-y^{\prime }\frac{\partial L}{\partial y^{\prime }}=C\mathrm{(}\mathrm{1}\mathrm{)}$

for some constant $C$.

Derivation

The proof does not integrate the Euler-Lagrange equation. It differentiates the left side of (1) and shows the result vanishes.

Take the total derivative of $L$ along a path, treating $L$ as a function of its three slots and using the chain rule:

$\frac{dL}{dx}=\frac{\partial L}{\partial x}+\frac{\partial L}{\partial y}{y}^{\prime }+\frac{\partial L}{\partial y^{\prime }}{y}^{″}\mathrm{(}\mathrm{2}\mathrm{)}$

Now differentiate the product term:

$\frac{d}{dx}\left(y^{\prime }\frac{\partial L}{\partial y^{\prime }}\right)=y^{″}\frac{\partial L}{\partial y^{\prime }}+y^{\prime }\frac{d}{dx}\left(\frac{\partial L}{\partial y^{\prime }}\right)\mathrm{(}\mathrm{3}\mathrm{)}$

Subtract (3) from (2). The $y^{″}$ terms cancel exactly:

$\frac{d}{dx}(L-y^{\prime }\frac{\partial L}{\partial y^{\prime }})=\frac{\partial L}{\partial x}+y^{\prime }[\frac{\partial L}{\partial y}-\frac{d}{dx}\left(\frac{\partial L}{\partial y^{\prime }}\right)]\mathrm{(}\mathrm{4}\mathrm{)}$

Equation (4) is an identity. It holds for any $L$ and any path, extremal or not, and it is worth recording in that generality.

Two hypotheses now kill the two terms on the right. Along an extremal the bracket vanishes, by equation (2) of the Euler-Lagrange equation. And $\partial L/\partial x=0$ by assumption. So the left side is zero and the bracketed quantity is constant. $◼$

The cancellation of $y^{″}$ between (2) and (3) is the mechanism, and it is the reason the combination $L-y^{\prime }{L}_{y^{\prime }}$ and not some other combination is the conserved one.

The two hypotheses are separable

Equation (4) shows the structure clearly:

condition dropped	what survives
neither dropped	$L-y^{\prime }{L}_{y^{\prime }}$ is constant
$L$ depends on $x$	$\frac{d}{dx}(L-y^{\prime }{L}_{y^{\prime }})=\partial L/\partial x$, still useful
path is not an extremal	no conclusion

The middle row is the honest general statement, and it is how the quantity is handled in time-dependent mechanics.

Relation to the Hamiltonian

The conserved quantity in (1) is, up to sign, the Hamiltonian. Writing $p=L_{y^{\prime }}$ for the conjugate momentum,

$H=y^{\prime }p-L=-C\mathrm{(}\mathrm{5}\mathrm{)}$

This is not a coincidence and it is not merely notation. The map from $(y,y^{\prime },L)$ to $(y,p,H)$ is the Legendre transform, developed on the Legendre transform, and Beltrami's identity is the statement that $H$ is conserved when $L$ carries no explicit $x$. When $x$ is time, that is conservation of energy, and it is one entry in the symmetry-to-conservation table on Noether's first theorem.

The pairing is worth setting out, since the machinery track has now produced both halves:

$L$ independent of	first integral	mechanical reading
$y$	$L_{y^{\prime }}=\text{const}$	momentum conserved
$x$	$L-y^{\prime }{L}_{y^{\prime }}=\text{const}$	energy conserved

Worked example: the catenary

A uniform chain of fixed length hangs between two points. It minimises potential energy, which up to constants is

$J[y]={\int }_a^{b}y\sqrt{1+y{\text{'}}^2}dx\mathrm{(}\mathrm{6}\mathrm{)}$

$L=y\sqrt{1+y{\text{'}}^2}$ is free of $x$, so (1) applies:

$y\sqrt{1+y{\text{'}}^2}-y^{\prime }\cdot \frac{y{y}^{\prime }}{\sqrt{1+y{\text{'}}^2}}=C\mathrm{(}\mathrm{7}\mathrm{)}$

Put the left side over a common denominator. The numerator is $y(1+y{\text{'}}^2)-yy{\text{'}}^2=y$, so (7) collapses to

$\frac{y}{\sqrt{1+y{\text{'}}^2}}=C\mathrm{(}\mathrm{8}\mathrm{)}$

A second-order problem has become a first-order separable one in two lines. Rearranging, $y^{\prime }=\sqrt{y^2/C^2-1}$, which separates and integrates to

$y=C\cosh \left(\frac{x-x_0}{C}\right)\mathrm{(}\mathrm{9}\mathrm{)}$

The hanging chain is a catenary. The constants $C$ and $x_0$ are fixed by the two endpoint conditions.

Two caveats. First, the length constraint has been suppressed here; treating it properly requires a multiplier, and the full problem is worked on Isoperimetric problems and the catenary. Second, (9) is an extremal. That it is the minimum is a separate claim, addressed on the second variation.

Naming

The identity is named for Eugenio Beltrami, though the result is older and the attribution is conventional rather than historical. This volume uses the name because it is standard, and records here that it is a label of convenience.

Next

Several dependent variables extends the machinery from one unknown function to many.