Why the zero-order term in a variational transformation of coordinates should be identically the same as the old coordinates?

Question

In the Ref.[1, page 61] the author proposes that transformations between two coordinate systems can be described by a continuous parameter $\varepsilon$ such that when $\varepsilon=0$ the original coordinates are recovered.

The mapping between these two systems imply the existence of functions $T$ and $Q$ such that \begin{align} t\rightarrow t^\prime &=T\big(t,q^\mu,\varepsilon\big),\tag{4.1.1}\label{eq1}\\ q^\mu\rightarrow {q^\prime}^\mu &=Q^\mu\big(t,q^\nu,\varepsilon\big).\tag{4.1.2}\label{eq2} \end{align}

The author states that if $\varepsilon$ is sufficiently small, then we can to expand the functions $T$ and $Q$ in Taylor series about $\varepsilon=0$, such that: \begin{align} t^\prime &=t+\varepsilon\bigg(\frac{\partial T}{\partial\varepsilon}\bigg)_{\varepsilon=0}+O\big(\varepsilon^2\big),\tag{4.1.7}\label{eq7}\\ {q^\prime}^\mu &=q^\mu+\varepsilon\bigg(\frac{\partial Q^\mu}{\partial\varepsilon}\bigg)_{\varepsilon=0}+O\big(\varepsilon^2\big),\tag{4.1.8}\label{eq8} \end{align} where the author the author identifies $t^\prime=t$ and ${q^\prime}^\mu=q^\mu$ when $\varepsilon=0$.

Also, according to the author, the coefficients of $\varepsilon$ to the first power are called the ''generators'' of the transformation and they can be denoted by \begin{align} \tau &\equiv\bigg(\frac{\partial T}{\partial\varepsilon}\bigg)_{\varepsilon=0}=\tau\big(t,q^\mu\big),\tag{4.1.9}\label{eq9}\\ \zeta^\mu &\equiv\bigg(\frac{\partial Q^\mu}{\partial\varepsilon}\bigg)_{\varepsilon=0}=\zeta^\mu\big(t,q^\mu\big).\tag{4.1.10}\label{eq10} \end{align}

Now, let me introduce how I'm seeing the problem. The Taylor series expansion around $\varepsilon = 0$ are written as \begin{align} T\left( t,q^{\mu},\varepsilon\right) & =T\left( t,q^{\mu},0\right) +\left( \dfrac{\partial T}{\partial\varepsilon}\right) _{\varepsilon =0}+O\left( \varepsilon^{2}\right), \\ Q^\mu\left( t,q^{\mu},\varepsilon\right) & =Q^\mu\left( t,q^{\mu},0\right) +\left( \dfrac{\partial Q^\mu}{\partial\varepsilon}\right) _{\varepsilon =0}+O\left( \varepsilon^{2}\right). \end{align} But, according to the author, on the left side, we have $t^\prime=T\big(t,q^\mu,\varepsilon\big)$ and ${q^\prime}^\mu=Q^\mu(t,qν,\varepsilon\big)$. On the other hand, on the right side, the zero-order term is $t=T\big(t,q^\mu,0\big)$ and $q^\mu=Q^\mu(t,q^\mu,0\big)$, while the first order term $\tau\big(t,q^\mu\big)=\big(\partial T/\partial\varepsilon\big)_{\varepsilon=0}$ and $\zeta^\mu\big(t,q^\nu\big)=\big(\partial Q^\mu/\partial\varepsilon\big)_{\varepsilon=0}$.

My question is centered on the zero-order term because I think that instead of being $T\big(t,q^\mu,0\big)=t$ and $Q^\mu\big(t,q^\nu,0\big)=q^\mu$ it should be $T\big(t,q^\mu,0\big)=\mathcal{T}\big(t,q^\mu\big)$ and $Q^\mu\big(t,q^\nu,0\big)=\mathcal{Q}^\mu\big(t,q^\nu\big)$ since we are searching for the most general form possible.

So, what should be the argument for making a more restrictive choice such as $T\big(t,q^\mu,0\big)=t$ and $Q^\mu\big(t,q^\nu,0\big)=q^\mu$?

In my opinion, Eqs. \eqref{eq7} and \eqref{eq8} should be written as \begin{align} t^\prime &=\mathcal{T}\big(t,q^\mu\big)+\varepsilon\tau\big(t,q^\mu\big)+O\big(\varepsilon^2\big),\tag{B1}\label{eqB1}\\ {q^\prime}^\mu &=\mathcal{Q}^\mu\big(t,q^\mu\big)+\varepsilon\zeta^\mu\big(t,q^\mu\big)+O\big(\varepsilon^2\big).\tag{B2}\label{eqB2} \end{align}

$^{[1]}$ Dwight E. Neuenschwander, Emmy Noether's Wonderful Theorem

Answer 1

If you replace eqs. (4.1.1) and (4.1.2) by your (A) equations you actually kill the extra parameter introduced $\varepsilon$, rendering the rest of the chapter pointless.

The "generator" term comes from group theory, and in page 64 you find a very brief note: "[...] Because a set of such transformations has an identity element and each transformation has an inverse, the transformations form a group [...]".

And $t'=t$ holds in the limiting case $\varepsilon \rightarrow 0$, from where you can deduce the form of your function $\mathcal{T}$ not to include $q^\mu$, and $\mathcal{Q}$ not to include $t$.

Look at example in equation (4.1.3) with $\varepsilon$ being a rotation angle about the $z$ axis, how would a series expansion for $t'$ look like in that case?

Answer 2

After some reflection, I think I came to an understanding of what should be the answer to the problem. Transformations \eqref{eqB1} and \eqref{eqB2} should be viewed not only as transformations between coordinates but also as perturbations around the original coordinates such that at the boundary where $\varepsilon\rightarrow 0$ these must be retrieved to recover. In this sense, in order for such a criterion to be satisfied, we must choose (or configure) the functions $\mathcal{T}$ and $\mathcal{Q}^\mu$ as $\mathcal{T}\big(t,q^\mu\big) = t$ and $\mathcal{Q}^\mu\big(t,q^\mu\big) = q^\mu$.

Basically, I think this is the same answer that @fredwhileshavin wanted to give in the previous post, but in a veiled way.

Well, I think I'm on the right track for a good answer, but I leave it to the community for judgment.