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The change in a Lagrangian with no explicit time dependence $L(\mathbf{q},\mathbf{\dot q})$ can be written using the chain rule:

$$δL = \frac{\partial L}{\partial \mathbf{q}}\cdot δ\mathbf{q} + \frac{\partial L}{\partial \mathbf{\dot q}}\cdot δ\mathbf{\dot q} \ \ \ \ \ \ \ \ (\mathbf{*})$$

Where a derivative with respect to a vector $\mathbf{q}$ denotes a vector with components $\frac{\partial L}{\partial q_1}$, $\frac{\partial L}{\partial q_2}$, $\frac{\partial L}{\partial q_3}$, etc.

Consider the Lagrangian: $$L=\dot q_1^2 + q_1^2(\dot q_2^2 + \dot q_3^2) -q_2^2-q_3^2$$

and the transformation

$q_1 \rightarrow q_1$,

$q_2 \rightarrow q_2 + q_3 δλ$,

$q_3 \rightarrow q_3 - q_2 δλ$,

(in other terms, $δq_1 = 0$, $δq_2 = q_3δλ$, $δq_3 = -q_2δλ$

and also $δ\dot q_1 = 0$, $δ\dot q_2 = \dot q_3δλ$, $δ\dot q_3 = -\dot q_2δλ$)

Under this transformation the lagrangian is unchanged according to $(\mathbf{*})$

$$δL = q_1^2(2\dot q_2 \dot q_3 δλ - 2\dot q_3 \dot q_2 δλ) - (2q_2q_3δλ-2q_3q_2δλ) = q_1^2(0)+(0) = 0 $$

However direct substitution of the transformation into the Lagrangian yields:

$$L'=\dot q_1^2 + q_1^2((\dot q_2 + \dot q_3 δλ)^2 + (\dot q_3 - \dot q_2 δλ)^2)-(q_2+q_3δλ)^2 - (q_3-q_2δλ)^2$$ $$=\dot q_1^2 + q_1^2(\dot q_2^2 + \dot q_3^2 + \dot q_3^2δλ^2 + \dot q_2^2 δλ^2)-q_2^2-q_3^2 -q_3^2δλ^2-q_2^2δλ^2$$

And so the terms in $δλ^2$ are still lingering, and $L' \neq L$. What's gone wrong?

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2 Answers 2

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TL;DR: The infinitesimal symmetry transformation needs to be properly integrated into a corresponding finite symmetry transformation in order to preserve the invariance of $L$ to all order in the parameter $\lambda$.

Hint: (The variable $q_1$ is a passive spectator that we ignore in what follows.) If we introduce a complex notation $$q~:=~q_2+iq_3~\in~\mathbb{C},$$ then OP's infinitesimal transformation becomes $$q^{\prime}-q~=~\delta q ~=~ -iq\delta\lambda.$$ We can write this as an initial value problem (IVP) for a first-order ODE $$ \frac{dq(\lambda)}{d\lambda}~=~-iq(\lambda), \qquad q(\lambda\!=\!0)~=~q,$$ which has a unique solution. Clearly the corresponding finite transformation solution is a rotation $$ q^{\prime}~=~e^{-i\lambda}q, \qquad \lambda~\in~\mathbb{R} . $$ It remains to check that OP's Lagrangian $L$ is invariant to all orders in $\lambda$.

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  • $\begingroup$ I think "Clearly the corresponding finite transformation is a rotation" might be an overstatement (although experience dictates this identification:-), since there is an infinite set of transformations which could be approximated linearly around unity in the given way. So there is a bit more we need to know about the transformation to make this identification unique. $\endgroup$ Commented May 27, 2022 at 13:16
  • $\begingroup$ Hi @AlmostClueless: Fair point. I updated the answer. $\endgroup$
    – Qmechanic
    Commented May 27, 2022 at 13:50
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In one case you've made an infinitesimal transformation, in the other a finite one. They're only equal to first order. Or to put it another way, your first equation $(*)$ is not exact: to make it exact you need to include higher order terms. In particular, if you work out the second order variation, it should match what you found by direct substitution, since the difference is quadratic in $\delta \lambda$.

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