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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?

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  • $\begingroup$ Link to lecture? $\endgroup$ – Qmechanic May 9 '19 at 7:48
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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.

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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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