Atanas K. Stefanov

A guided derivation of the Noether theorem

In my UCL Galaxy Dynamics course (PHAS0065), I struggled with the derivation of the Noether theorem for quite a while. What I present here is just a step-by-step solution of the treatment given in Dr Ralph Schönrich's lecture notes of PHAS0065. This treatment is in the German theoretical physics tradition, and is particularly contained in "Mechanik: Lehrbuch zur Theoretischen Physik" by Torsten Fließbach. All errors in this solution are my own.

Consider a dynamical system with generalised space coordinates q‾∈Ri\underline{q}\in\mathbb{R}^i, a time coordinate tt and an associated Lagrangian LL. Let there be two perturbing maps of the form

∣q‾′:=q‾+λψ‾(q‾,q˙‾,t)t′:=t+λϕ(q‾,q˙‾,t)\left|\begin{array}{l} \underline{q}':= \underline{q}+\lambda\underline{\psi}(\underline{q},\underline{\dot{q}},t)\\[1ex] t':= t+\lambda\phi(\underline{q},\underline{\dot{q}},t) \end{array}\right.

with λ∈R\lambda \in \mathbb{R}. We require symmetry of action:

∫t1t2L(q‾,q˙‾,t)dt=∫t1t2L(q‾′,q˙‾′,t′)dt′, \int^{t_2}_{t_1} L \left( \underline{q},\underline{\dot{q}},t \right)\mathrm{d}t = \int^{t_2}_{t_1} L \left( \underline{q}',\underline{\dot{q}}',t' \right)\mathrm{d}t',

where we define q˙‾′:=dq‾′/dt′\underline{\dot{q}}':=\mathrm{d}\underline{q}'/\mathrm{d}t' for the rest of the derivation. Taylor-expanding the right-hand side gives

∫t1t2L(q‾,q˙‾,t)dt=∫t1t2L(q‾,q˙‾,t)dt+λ∫t1t2ddλ[L(q‾′,q˙‾′,t′)×dt′dt]λ=0dt+O(λ2), \cancel{\int^{t_2}_{t_1} L \left( \underline{q},\underline{\dot{q}},t \right)\mathrm{d}t} = \cancel{\int^{t_2}_{t_1} L \left( \underline{q},\underline{\dot{q}},t \right)\mathrm{d}t} + \lambda \int^{t_2}_{t_1} \frac{\mathrm{d}}{\mathrm{d}\lambda} \left[ L \left( \underline{q}',\underline{\dot{q}}',t' \right)\times\frac{\mathrm{d}t'}{\mathrm{d}t} \right]_{\lambda=0}\mathrm{d}t + \mathcal{O}(\lambda^2),

which holds for all λ\lambda if and only if

0={ddλ[L(q‾′,q˙‾′,t′)×dt′dt]}λ=0. 0 = \left\{ \frac{\mathrm{d}}{\mathrm{d}\lambda} \left[ L \left( \underline{q}',\underline{\dot{q}}',t' \right)\times\frac{\mathrm{d}t'}{\mathrm{d}t} \right] \right\}_{\lambda=0}.

From the definition of t′t',

dt′dt=1+λdϕdt, \frac{\mathrm{d}t'}{\mathrm{d}t} = 1+\lambda\frac{\mathrm{d}\phi}{\mathrm{d}t},

which further reduces the action equation to

0={ddλ[L(1+λdϕdt)]}λ=0={dLdλ(1+λdϕdt)+Ldϕdt}λ=0. \begin{aligned} 0 &= \left\{ \frac{\mathrm{d}}{\mathrm{d}\lambda} \left[ L\left( 1+\lambda\frac{\mathrm{d}\phi}{\mathrm{d}t} \right) \right] \right\}_{\lambda=0}\\[3ex] &= \left\{ \frac{\mathrm{d}L}{\mathrm{d}\lambda}\left( 1+\lambda\frac{\mathrm{d}\phi}{\mathrm{d}t} \right) + L \frac{\mathrm{d}\phi}{\mathrm{d}t} \right\}_{\lambda=0}. \end{aligned}

We then expand dL/dλ\mathrm{d}L/\mathrm{d}\lambda by the chain rule, keeping in mind that the Lagrangian is a function of the perturbed coordinates:

0={[∑i∂L∂qi′dqi′dλ+∑i∂L∂q˙i′dq˙i′dλ+∂L∂t′dt′dλ+∂L∂λ](1+λdϕdt)+Ldϕdt}λ=0={[∑i∂L∂qi′ψi+∑i(∂L∂q˙i′×ddλdqi′dt′)+∂L∂t′ϕ](1+λdϕdt)+Ldϕdt}λ=0, \begin{aligned} 0 &= \left\{ \left[ \sum_i \frac{\partial L}{\partial q'_i} \frac{\mathrm{d} q'_i}{\mathrm{d} \lambda} + \sum_i \frac{\partial L}{\partial \dot{q}'_i} \frac{\mathrm{d} \dot{q}'_i}{\mathrm{d} \lambda} + \frac{\partial L}{\partial t'} \frac{\mathrm{d} t'}{\mathrm{d} \lambda} + \cancel{ \frac{\partial L}{\partial \lambda} } \right] \left( 1+\lambda\frac{\mathrm{d}\phi}{\mathrm{d}t} \right) + L \frac{\mathrm{d}\phi}{\mathrm{d}t} \right\}_{\lambda=0}\\[3ex] &= \left\{ \left[ \sum_i \frac{\partial L}{\partial q'_i} \psi_i + \sum_i \left( \frac{\partial L}{\partial \dot{q}'_i} \times \frac{\mathrm{d}}{\mathrm{d} \lambda} \frac{\mathrm{d}q'_i}{\mathrm{d} t'} \right) + \frac{\partial L}{\partial t'} \phi \right] \left( 1+\lambda\frac{\mathrm{d}\phi}{\mathrm{d}t} \right) + L \frac{\mathrm{d}\phi}{\mathrm{d}t} \right\}_{\lambda=0}, \end{aligned}

where in the first line ∂L/∂λ\partial L/\partial \lambda vanishes, since LL is unperturbed; and in the second line we used the definitions of qi′q'_i, q˙i′\dot{q}'_i and t′t'. We now make the following approximations: While the fourth-line approximation is the standard binomial approximation, the second-line one (i.e. the act of “flipping” the derivative) is dubious, and I leave it with no formal justification.

dqi′dt′=dqi′dtdtdt′≈dqi′dt(dt′dt)−1=(dqidt+λdψidt)(1+λdϕdt)−1≈(dqidt+λdψidt)(1−λdϕdt). \begin{aligned} \frac{\mathrm{d}q_i'}{\mathrm{d}t'} &= \frac{\mathrm{d}q_i'}{\mathrm{d}t} \frac{\mathrm{d}t}{\mathrm{d}t'}\\[2ex] &\approx \frac{\mathrm{d}q_i'}{\mathrm{d}t} \left( \frac{\mathrm{d}t'}{\mathrm{d}t} \right)^{-1}\\[2ex] &= \left( \frac{\mathrm{d}q_i}{\mathrm{d}t} + \lambda\frac{\mathrm{d}\psi_i}{\mathrm{d}t} \right) \left( 1 + \lambda\frac{\mathrm{d}\phi}{\mathrm{d}t} \right)^{-1}\\[2ex] &\approx \left( \frac{\mathrm{d}q_i}{\mathrm{d}t} + \lambda\frac{\mathrm{d}\psi_i}{\mathrm{d}t} \right) \left( 1 - \lambda\frac{\mathrm{d}\phi}{\mathrm{d}t} \right). \end{aligned}

Its derivative with respect to λ\lambda reads

ddλdqi′dt′=ddλ[(dqidt+λdψidt)(1−λdϕdt)]=ddλ[dqidt+λdψidt−λdqidtdϕdt+O(λ2)]=dψidt−dqidtdϕdt. \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}\lambda}\frac{\mathrm{d}q_i'}{\mathrm{d}t'} &= \frac{\mathrm{d}}{\mathrm{d}\lambda} \left[ \left( \frac{\mathrm{d}q_i}{\mathrm{d}t} + \lambda\frac{\mathrm{d}\psi_i}{\mathrm{d}t} \right) \left( 1 - \lambda\frac{\mathrm{d}\phi}{\mathrm{d}t} \right) \right]\\[3ex] &= \frac{\mathrm{d}}{\mathrm{d}\lambda} \left[ \frac{\mathrm{d}q_i}{\mathrm{d}t} + \lambda\frac{\mathrm{d}\psi_i}{\mathrm{d}t} - \lambda\frac{\mathrm{d}q_i}{\mathrm{d}t}\frac{\mathrm{d}\phi}{\mathrm{d}t} + \mathcal{O}(\lambda^2) \right]\\[3ex] &= \frac{\mathrm{d}\psi_i}{\mathrm{d}t} - \frac{\mathrm{d}q_i}{\mathrm{d}t}\frac{\mathrm{d}\phi}{\mathrm{d}t}. \end{aligned}

We repeat the action-symmetry equation and continue to reduce it with this new result:

0={[∑i∂L∂qi′ψi+∑i(∂L∂q˙i′×ddλdqi′dt′)+∂L∂t′ϕ](1+λdϕdt)+Ldϕdt}λ=0={[∑i∂L∂qi′ψi+∑i(∂L∂q˙i′×(dψidt−dqidtdϕdt))+∂L∂t′ϕ](1+λdϕdt)+Ldϕdt}λ=0. \begin{aligned} 0 &= \left\{ \left[ \sum_i \frac{\partial L}{\partial q'_i} \psi_i + \sum_i \left( \frac{\partial L}{\partial \dot{q}'_i} \times \frac{\mathrm{d}}{\mathrm{d} \lambda} \frac{\mathrm{d}q'_i}{\mathrm{d} t'} \right) + \frac{\partial L}{\partial t'} \phi \right] \left( 1+\lambda\frac{\mathrm{d}\phi}{\mathrm{d}t} \right) + L \frac{\mathrm{d}\phi}{\mathrm{d}t} \right\}_{\lambda=0}\\[3ex] &= \left\{ \left[ \sum_i \frac{\partial L}{\partial q'_i} \psi_i + \sum_i \left( \frac{\partial L}{\partial \dot{q}'_i} \times \left( \frac{\mathrm{d}\psi_i}{\mathrm{d}t} - \frac{\mathrm{d}q_i}{\mathrm{d}t} \frac{\mathrm{d}\phi}{\mathrm{d}t} \right) \right) + \frac{\partial L}{\partial t'} \phi \right] \left( 1+\lambda\frac{\mathrm{d}\phi}{\mathrm{d}t} \right) + L \frac{\mathrm{d}\phi}{\mathrm{d}t} \right\}_{\lambda=0}. \end{aligned}

This expression is finally ready to be evaluated at λ=0\lambda=0. Note that not only λ\lambda-terms vanish, but partial derivatives change to their “unperturbed” equivalents, This is perhaps another mathematical sin, though I would argue it is of not such severity. i.e.

lim⁡λ→0∂∂qi′=∂∂qilim⁡λ→0∂∂q˙i′=∂∂q˙ilim⁡λ→0∂∂t′=∂∂t. \begin{aligned} \lim_{\lambda\to 0} \frac{\partial}{\partial q_i'} &= \frac{\partial}{\partial q_i} \\[2ex] \lim_{\lambda\to 0} \frac{\partial}{\partial \dot{q}_i'} &= \frac{\partial}{\partial \dot{q}_i} \\[2ex] \lim_{\lambda\to 0} \frac{\partial}{\partial t'} &= \frac{\partial}{\partial t}. \end{aligned}

Thus the evaluation at λ=0\lambda=0 yields

0=∑i∂L∂qiψi+∑i(∂L∂q˙i×(dψidt−dqidtdϕdt))+∂L∂tϕ+Ldϕdt=∑i[∂L∂qiψi+∂L∂q˙i×(dψidt−dqidtdϕdt)]+∂L∂tϕ+Ldϕdt=∑i[∂L∂qiψi+∂L∂q˙idψidt−∂L∂q˙idqidtdϕdt]+∂L∂tϕ+Ldϕdt. \begin{aligned} 0 &= \sum_i \frac{\partial L}{\partial q_i} \psi_i + \sum_i \left( \frac{\partial L}{\partial \dot{q}_i} \times \left( \frac{\mathrm{d}\psi_i}{\mathrm{d}t} - \frac{\mathrm{d}q_i}{\mathrm{d}t} \frac{\mathrm{d}\phi}{\mathrm{d}t} \right) \right) + \frac{\partial L}{\partial t} \phi + L \frac{\mathrm{d}\phi}{\mathrm{d}t}\\[3ex] &= \sum_i \left[ \frac{\partial L}{\partial q_i} \psi_i + \frac{\partial L}{\partial \dot{q}_i} \times \left( \frac{\mathrm{d}\psi_i}{\mathrm{d}t} - \frac{\mathrm{d}q_i}{\mathrm{d}t} \frac{\mathrm{d}\phi}{\mathrm{d}t} \right) \right] + \frac{\partial L}{\partial t} \phi + L \frac{\mathrm{d}\phi}{\mathrm{d}t}\\[3ex] &= \sum_i \left[ \frac{\partial L}{\partial q_i} \psi_i + \frac{\partial L}{\partial \dot{q}_i} \frac{\mathrm{d}\psi_i}{\mathrm{d}t} - \frac{\partial L}{\partial \dot{q}_i} \frac{\mathrm{d}q_i}{\mathrm{d}t} \frac{\mathrm{d}\phi}{\mathrm{d}t} \right] + \frac{\partial L}{\partial t} \phi + L \frac{\mathrm{d}\phi}{\mathrm{d}t}. \end{aligned}

Invoking the Euler-Lagrange equation in the first term under the sum:

0=∑i[ddt(∂L∂q˙i)ψi+∂L∂q˙idψidt−∂L∂q˙idqidtdϕdt]+∂L∂tϕ+Ldϕdt, 0 = \sum_i \left[ \frac{\mathrm{d}}{\mathrm{d}t}\left( \frac{\partial L}{\partial \dot{q}_i} \right) \psi_i + \frac{\partial L}{\partial \dot{q}_i} \frac{\mathrm{d}\psi_i}{\mathrm{d}t} - \frac{\partial L}{\partial \dot{q}_i} \frac{\mathrm{d}q_i}{\mathrm{d}t} \frac{\mathrm{d}\phi}{\mathrm{d}t} \right] + \frac{\partial L}{\partial t} \phi + L \frac{\mathrm{d}\phi}{\mathrm{d}t},

but now the first two terms under the sum can be combined through the product rule:

0=∑i[ddt(∂L∂q˙iψi)−∂L∂q˙idqidtdϕdt]+∂L∂tϕ+Ldϕdt=ddt[∑i∂L∂q˙iψi]+(−∑i∂L∂q˙idqidt+L)dϕdt+∂L∂tϕ. \begin{aligned} 0 &= \sum_i \left[ \frac{\mathrm{d}}{\mathrm{d}t}\left( \frac{\partial L}{\partial \dot{q}_i} \psi_i \right) - \frac{\partial L}{\partial \dot{q}_i} \frac{\mathrm{d}q_i}{\mathrm{d}t} \frac{\mathrm{d}\phi}{\mathrm{d}t} \right] + \frac{\partial L}{\partial t} \phi + L \frac{\mathrm{d}\phi}{\mathrm{d}t}\\[3ex] &= \frac{\mathrm{d}}{\mathrm{d}t} \left[ \sum_i \frac{\partial L}{\partial \dot{q}_i} \psi_i \right] + \left( -\sum_i\frac{\partial L}{\partial \dot{q}_i} \frac{\mathrm{d}q_i}{\mathrm{d}t} + L \right) \frac{\mathrm{d}\phi}{\mathrm{d}t} + \frac{\partial L}{\partial t} \phi. \end{aligned}

Finally, we note that the Hamiltonian of the system is defined H:=∑i(∂L/q˙i)q˙i−LH:=\sum_i (\partial L/\dot{q}_i)\dot{q}_i-L, and that dH/dt=−∂L/∂t\mathrm{d}H/\mathrm{d}t=-\partial L/\partial t. This leads to

0=ddt[∑i∂L∂q˙iψi]−Hdϕdt−dHdtϕ=ddt[∑i∂L∂q˙iψi]−ddt[Hϕ], \begin{aligned} 0 &= \frac{\mathrm{d}}{\mathrm{d}t} \left[ \sum_i \frac{\partial L}{\partial \dot{q}_i} \psi_i \right] - H\frac{\mathrm{d}\phi}{\mathrm{d}t} - \frac{\mathrm{d} H}{\mathrm{d} t} \phi \\[3ex] &= \frac{\mathrm{d}}{\mathrm{d}t} \left[ \sum_i \frac{\partial L}{\partial \dot{q}_i} \psi_i \right] - \frac{\mathrm{d}}{\mathrm{d}t}\left[ H\phi \right], \end{aligned}

implying that the quantity

Q:=∑i∂L∂q˙iψi−Hϕ≡∑ipiψi−Hϕ\begin{aligned} Q&:= \sum_i \frac{\partial L}{\partial \dot{q}_i} \psi_i - H\phi\\[3ex] &\equiv \sum_i p_i \psi_i - H\phi \end{aligned}

is conserved. This is the Noether theorem.