# How to apply Noether's theorem

Say I have a point transformation:

$$x' ~=~ (1 +\epsilon)x,$$

$$t' ~=~ (1 +\epsilon)^2t,$$

and Lagrangian

$$L ~=~ \frac{1}{2}m\dot{x}^2 - \frac{\alpha}{x^2}.$$

How do I go out about showing that this transformation is a symmetry of the Lagrangian?

When I plug in $x'$ and $t'$ I do not get the same Lagrangian at all, it is off by factors of $(1+\epsilon)^2$ in both terms. I am not sure how I to calculate this. What sort of "symmetry" is this?

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1) In this answer we provide more details for David Bar Moshe's correct answer. The action reads

$$\tag{1}S[q; t_i,t_f]~:=~\int_{t_i}^{t_f} \!dt~ L, \quad L~:=~T-V ,\quad T~:=~\frac{m}{2}\dot{q}^2,\quad V~:=~\frac{\alpha}{q^2}.$$

It is not hard to check that the action has an exact symmetry

$$\tag{2} S[q; t_i,t_f] \quad \longrightarrow \quad S[q^{\prime}; t^{\prime}_i,t^{\prime}_f] ~=~S[q; t_i,t_f]$$

under the following scaling

$$\tag{3}t \quad \longrightarrow \quad t^{\prime}~=~\lambda^2 t, \qquad q(t) \quad \longrightarrow \quad q^{\prime}(t^{\prime})~=~\lambda q(t),$$

with a non-negative parameter $\lambda\neq 0$, if we also scale the initial and final time integration limits in the same way as the time parameter $t$:

$$\tag{4}t_i\quad \longrightarrow \quad t^{\prime}_i~=~\lambda^2 t_i, \qquad t_f\quad \longrightarrow \quad t^{\prime}_f~=~\lambda^2 t_f.$$

Interestingly, the transformation (3) is not a symmetry of the Lagrangian

$$\tag{5}L(t) \quad \longrightarrow \quad L^{\prime}(t^{\prime})~=~\frac{L(t)}{\lambda^2},$$ as OP already observes in the question(v1). This is a good opportunity to recall that Noether's theorem is about (quasi)-symmetries of the action rather than the Lagrangian.

2) Next let us consider the corresponding infinitesimal transformation. Assume that $\lambda=1+\epsilon$, where $\epsilon$ is infinitesimal, i.e. neglect higher-order terms in $\epsilon$. The so-called horizontal infinitesimal variation is

$$\tag{6}\delta t ~:=~t^{\prime}-t ~=~ \epsilon \cdot 2t.$$

The infinitesimal variation of the dynamical variable $q$ is

$$\tag{7} \delta q(t)~:= ~q^{\prime}(t^{\prime})-q(t)~=~\epsilon \cdot q(t),$$

so the vertical infinitesimal variation is

$$\tag{8} \delta_0 q(t)~:= ~q^{\prime}(t)-q(t)~=~\epsilon \cdot(q(t)-2 t\dot{q}(t)).$$ In other words, the transformation (3) has horizontal generator $2t$ and vertical generator $q-2 t\dot{q}$.

The bare Noether current (=charge) $Q$ is defined as the momentum times the vertical generator plus the Lagrangian times the horizontal generator:

$$\tag{9} Q~:=~ \frac{\partial L}{\partial \dot{q}}\cdot(q-2 t\dot{q})+ L\cdot 2t ~=~mq\dot{q} -2t(T+V).$$

[In general, if the infinitesimal transformation $\delta S$ of the action is only invariant up to boundary terms, it is called a quasi-symmetry, and the full Noether current would then receive boundary contributions. However in our case, the symmetry (3,4) is actually exact (2), i.e. without any boundary terms, so the full Noether current is just the bare Noether current (9).]

It is easy to check that the Noether charge (9) is conserved on-shell

$$\tag{10}\frac{dQ}{dt} ~=~-(q-2 t\dot{q})\frac{\delta S}{\delta q}~\approx~0,$$

where the $\approx$ sign means equality modulo equation of motion

$$\tag{11} 0~\approx~ \frac{\delta S}{\delta q} ~=~\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} ~=~\frac{2\alpha}{q^3}-m\ddot{q},$$

i.e. on-shell.

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I prefer this solution as certain items are dealt with more explicitly so I think this solution is more useful. Thank you. –  Timtam Oct 22 '12 at 22:43
The symmetry is required to leave the full action $A= \int L dt$ invariant. As can be seen the action is invariant because you get the missing $(1+\epsilon)^2$ factor from the measure $dt$. Now, you can apply the Noether theorem to find the conserved charge which is in this case:
$Q = 2 E t - Px$
with $E = T+V = \frac{1}{2}m\dot{x}^2 + \frac{\alpha}{x^2}$, and $P = m \dot{x}$.
It is not hard to verify that $\frac{dQ}{dt} = 0$ when the equations of motion are satisfied.