In my classical mechanics lecture we derived the Noether-theorem for a coordinate transformation given by:

$$ q_i(t) \rightarrow q^{'}_i(t)=q_i(t) + \delta q_i(t) = q_i(t) + \lambda I_i(q,\dot q,t).$$

Then we calculated the action $S'$ and used a taylor expansion around $\lambda = 0$:

$$S' = \int_{t_1}^{t_2} dt\,L(q^{'}_i,\dot q^{'}_i,t) = \int_{t_1}^{t_2} dt\,[ L(q_i,\dot q_i,t) + \lambda \frac{d}{d\lambda}L(q^{'}_i,\dot q^{'}_i,t)\bigg \vert_{\lambda = 0}\,].$$

My question is why is there a $\lambda$ in the first order term of the expansion?

  • $\begingroup$ Link to lecture? $\endgroup$
    – Qmechanic
    Commented May 9, 2019 at 7:48

2 Answers 2


It seems that your $\lambda$ should be thought of as a small parameter. For example, you may write a first order Taylor expansion as

$$ f(x + \lambda) = f(x) + \lambda f'(x) + ... $$ You could also note that $$ f'(x) = \frac{d}{d \lambda}f(x + \lambda)\vert_{\lambda = 0} $$

The main thing is that $\lambda$ should be thought of as a tiny infinitesimal constant which parameterizes your symmetry transformation, i.e. $\delta q_i = \lambda I_i$. Therefore, $\lambda$ parameterizes $L'$. In more compact notation,

$$ L' = L(q_i + \delta q_i) = L(q_i + \lambda I_i) $$ This means that if we want the tiny change due to this transformation, we can just think of $\lambda$ itself as tiny, so

$$ dL = L(q_i + \lambda I_i) - L(q_i) = \lambda \frac{dL}{d \lambda}. $$

One final thing to mention is that there MUST be a $\lambda$ out front because the change to your action due to an infinitesimal transformation must be infinitesimal itself, and that change must be proportional to the infinitesimal parameter.


It seems to be a notational clash where the symbol $\lambda$ is used in 2 different meanings. It is similar to the Taylor expansion $f(a)=f(0)+ a \frac{df(x)}{dx}|_{x=0} +{\cal O}(a^2),$ with the poor decision to use the same symbol for $a$ and $x$...


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