High School A bit confused about this variation

[etotheipi]

I came across a derivation of the Noether theorem with a step I don't understand; the transformation of the time and configuration space are written as$$\tau(t,\varepsilon) = t + \varepsilon \delta t, \quad Q^a(q,\varepsilon) = q^a + \varepsilon \delta q^a$$here $\varepsilon$ is an infinitesimal parameter, whilst it looks like $\delta t$ and $\delta q^a$ are in this notation the generators of the time evolution and generalised coordinates respectively. Then they write$$\varepsilon \delta \mathcal{L} = \mathcal{L}(Q(q,\varepsilon), \dot{Q}(q,\varepsilon), \tau(t,\varepsilon)) - \mathcal{L}(q,\dot{q}, t)$$ $$\delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial q^a} \delta q^a + \frac{\partial \mathcal{L}}{\partial \dot{q}^a} \delta \dot{q}^a + \frac{\partial \mathcal{L}}{\partial t} \delta t$$and from this we can go on to show that if $\delta \mathcal{L} = \dot{F}$ i.e. for some transformation that is a quasi-symmetry, then $J = \mathcal{H} \delta t - p_a \delta q^a + F$ is a constant of the motion.

I don't understand the third line, where does it come from? It's not the chain rule of multivariable calculus, because the generators $\delta q^a$ and $\delta t$ for instance are finite. Maybe I misunderstand how the author has defined the variation of a function in this case. I wondered if someone could help out? Thanks!

 

[Gaussian97]

It is indeed the chain rule of multivariable calculus, just use Taylor expansion arround $\varepsilon = 0$ in
$$\varepsilon \delta \mathcal{L} = \mathcal{L}(Q(q,\varepsilon), \dot{Q}(q,\varepsilon), \tau(t,\varepsilon)) - \mathcal{L}(q,\dot{q}, t)$$
And then let $\varepsilon \to 0$. From where follows
$$\delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial q^a} \delta q^a + \frac{\partial \mathcal{L}}{\partial \dot{q}^a} \delta \dot{q}^a + \frac{\partial \mathcal{L}}{\partial t} \delta t$$

 

[etotheipi]

Thanks, do you mean something like, let$$F(\varepsilon) = \mathcal{L}(q + \varepsilon \delta q, \dot{q} + \varepsilon \delta \dot{q}, t + \varepsilon \delta t)$$then$$
\begin{align*}
F(\varepsilon) &= \mathcal{L}(q, \dot{q},t) + \varepsilon \left[ \frac{d\mathcal{L}(q + \varepsilon \delta q, \dot{q} +\varepsilon \delta \dot{q}, t + \varepsilon \delta t)}{d\varepsilon} \right]_{\varepsilon = 0} + \mathcal{O}(\varepsilon^2) \\

&= \mathcal{L}(q, \dot{q},t) +

\varepsilon\left[

\frac{\partial \mathcal{L}}{\partial (q + \varepsilon \delta q)}\frac{d(q + \varepsilon \delta q)}{d\varepsilon} + \frac{\partial \mathcal{L}}{\partial (\dot{q} + \varepsilon \delta \dot{q})}\frac{d(\dot{q} + \varepsilon \delta \dot{q})}{d\varepsilon} + \frac{\partial \mathcal{L}}{\partial (t + \varepsilon \delta t)}\frac{d(t + \varepsilon \delta t)}{d\varepsilon} \right]_{\varepsilon = 0} + \mathcal{O}(\varepsilon^2) \\

&= \mathcal{L}(q, \dot{q},t) + \varepsilon \left[

\frac{\partial \mathcal{L}}{\partial (q + \varepsilon \delta q)}\delta q + \frac{\partial \mathcal{L}}{\partial (\dot{q} + \varepsilon \delta \dot{q})} \delta \dot{q} + \frac{\partial \mathcal{L}}{\partial (t + \varepsilon \delta t)} \delta t\right]_{\varepsilon = 0} + \mathcal{O}(\varepsilon^2)
\end{align*}$$And this implies that$$\begin{align*}

\delta \mathcal{L} &= \frac{1}{\varepsilon} \left[ \mathcal{L}(Q(q,\varepsilon), \dot{Q}(q,\varepsilon), \tau(t,\varepsilon)) - \mathcal{L}(q,\dot{q}, t) \right] \\

&= \frac{\partial \mathcal{L}}{\partial q}\delta q + \frac{\partial \mathcal{L}}{\partial \dot{q}} \delta \dot{q} + \frac{\partial \mathcal{L}}{\partial t} \delta t

\end{align*}$$as we let $\varepsilon \rightarrow 0$ so that the $\mathcal{O}(\varepsilon)$ terms drop out.

 

[Gaussian97]

Yeah, essentially note that
$$\left.\frac{\partial \mathscr{L}(q+\varepsilon \delta q)}{\partial (q + \varepsilon \delta q)}\right|_{\varepsilon=0} = \frac{\partial \mathscr{L}(q)}{\partial q}$$

 

[vanhees71]

Where is this derivation from? I'm a bit surprised that they don't consider the action,
$$S'=\int \mathrm{d} t' L(q',\dot{q}',t')=\int \mathrm{d} t \frac{\mathrm{d} (t+\epsilon \delta t)}{\mathrm{d} t} L(q',\dot{q}',t').$$
and why they don't carefully define
$$\delta \dot{q}=\frac{\mathrm{d} q'}{\mathrm{d} t'}-\frac{\mathrm{d} q}{\mathrm{d} t} = \frac{\mathrm{d} (q+\epsilon \delta q)}{\mathrm{d} t} \left (\frac{\mathrm{d}(t+\epsilon \delta t)}{\mathrm{d} t} \right)^{-1} -\frac{\mathrm{d} q}{\mathrm{d} t}=\epsilon \frac{\mathrm{d}}{\mathrm{d} t} \delta q-\epsilon\frac{\mathrm{d}q}{\mathrm{d} t} \frac{\mathrm{d} \delta t}{\mathrm{d} t} + \mathcal{O}(\epsilon^2).$$

 

[etotheipi]

Where is this derivation from?

This one is from Chapter 10 section 10.2.4 of Orodruin's book, in the section on the classical mechanics

 

[vanhees71]

On p. 623 he restricts himself to pure time translations, i.e., $\delta t=1=\text{const}$, which explains why he doesn't need to consider the complications of the more general case that $\delta t$ is a function of the $q$ and $t$ (which you need for special relativity and the Poincare symmetry but usually not for Newtonian physics and Galilei symmetry).

 

[etotheipi]

Thanks! I did find just now the section in your classical mechanics notes (https://itp.uni-frankfurt.de/~hees/publ/theo1-l3.pdf) on page 91 where the Noether theorem is also derived, so I'll read through that as well! It's in German but there is a high density of equations! ☺