Noether's theorems/Applications/Variational integrators and discrete symmetry

This page works its case, and the result is exact rather than approximate, which is the reason the page exists.

The problem with simulating mechanics

Solve a mechanical system numerically the obvious way — step the equations of motion forward in small increments — and the conserved quantities drift. Energy creeps up or down, an orbit spirals slowly in or out, and refining the step size reduces the drift without removing it.

That is not a bug in the arithmetic. Nothing in the method knows the quantities are supposed to be conserved.

Discretise the action instead

The alternative is to discretise the object the theorems are about. Replace the integral by a sum over steps, with a function $L_d$ of consecutive positions standing in for the integral across one step:

$S_d=\sum _k{L}_d(q_k,q_{k+1})\mathrm{(}\mathrm{1}\mathrm{)}$

Require the sum to be stationary under variations of the interior points. Each $q_k$ appears in two terms, so differentiating with respect to it gives

$D_2{L}_d(q_{k-1},q_k)+D_1{L}_d(q_k,q_{k+1})=0\mathrm{(}\mathrm{2}\mathrm{)}$

where $D_1$ and $D_2$ denote differentiation with respect to the first and second argument. This is the equation the simulation steps forward.

The conserved quantity, exactly

Now suppose $L_d$ is invariant under a rigid shift of both its arguments:

$L_d(q_k+\epsilon ,q_{k+1}+\epsilon )=L_d(q_k,q_{k+1})\mathrm{(}\mathrm{3}\mathrm{)}$

Differentiate (3) with respect to $\epsilon$ at zero:

$D_1{L}_d+D_2{L}_d=0\mathrm{(}\mathrm{4}\mathrm{)}$

as an identity in both arguments. Define

$J_k=-D_1{L}_d(q_k,q_{k+1})\mathrm{(}\mathrm{5}\mathrm{)}$

By (4), $J_k=D_2{L}_d(q_k,q_{k+1})$, and therefore also $J_{k-1}=D_2{L}_d(q_{k-1},q_k)$. But (2) says $D_2{L}_d(q_{k-1},q_k)=-D_1{L}_d(q_k,q_{k+1})=J_k$. Hence

$J_{k-1}=J_k\mathrm{(}\mathrm{6}\mathrm{)}$

Exactly, at every step, for any step size. Not to first order, not to the accuracy of the scheme. The quantity is the same number for ever, and it is the discrete momentum.

What the derivation used

Only (3) and (2). No limit was taken, no step size was assumed small, and no property of the continuous theory was borrowed.

The theorem's structure did not need continuity. What it needed was an action and a symmetry of it, and both survive discretisation intact. That is why integrators built this way conserve momentum to machine precision while conventional ones do not.

The one that does not come free

Energy is different. Energy corresponds to translation in the independent variable, and discretising fixes the step, which breaks that symmetry.

So discrete-action methods conserve momentum exactly and energy only approximately — though they keep it bounded rather than letting it drift, for reasons beyond this page. The pattern is exactly what the theorems predict: the symmetries you keep are the quantities you conserve, and discretising is a choice about which symmetries to keep.

Provenance: (1) to (6) are standard in the numerical-analysis literature and are derived above rather than cited. No source was consulted. See Sources.