1
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Excuse me, maybe my question stayed a bit confusing. Let me clarify. The Taylor expansion of the T and Q functions around ε=0 should be \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\left( t,q^{\mu},\varepsilon\right) & =Q\left( t,q^{\mu},0\right) +\left( \dfrac{\partial Q}{\partial\varepsilon}\right) _{\varepsilon =0}+O\left( \varepsilon^{2}\right), \end{align} thats ok! $\endgroup$ – lucenalex Apr 29 at 11:05
  • $\begingroup$ But, according to the author, on the left side, we have $t^\prime=T\big(t,q^\mu,\varepsilon\big)$ and ${q^\prime}^\mu=Q^\mu\big(t,q^\nu,\varepsilon\big)$. On the other hand, on the right side, the zero order term is $t = T\big(t,q^\mu,0\big)$ and $q^\mu = Q^\mu=\big(t,q^\mu,0\big)$, while the first order term $\tau\big(t,q^\mu\big)=\bigg(\frac{\partial T}{\partial\varepsilon}\bigg)_{\varepsilon=0}$ and $\zeta^\mu\big(t,q^\nu\big)=\bigg(\frac{\partial Q^\mu}{\partial\varepsilon}\bigg)_{\varepsilon=0}$. $\endgroup$ – lucenalex Apr 29 at 11:15
  • $\begingroup$ My question is centered on the zero-order term because I think that instead of being $t = T\big(t,q^\mu,0\big)$ and $q^\mu = Q^\mu\big(t,q^\mu,0\big)$ it should be $\mathcal{T}\big(t,q^\mu\big) = T\big(t,q^\mu,0\big)$ and $\mathcal{Q}^\mu\big(t,q^\nu\big) = T\big(t,q^\nu,0\big)$ since we are searching for the most general form possible. So, what should be the argument for making a more restrictive choice such as $t = T\big(t,q^\mu,0\big)$ and $q^\mu = Q^\mu\big(t,q^\mu,0\big)$? $\endgroup$ – lucenalex Apr 29 at 11:23
  • $\begingroup$ I updated the question. $\endgroup$ – lucenalex Apr 29 at 11:58
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Yes, this is the general idea. By hyopthesis we must recover the original value of the coordinates for some value of $\varepsilon$, which we may as well take to be $0$. $\endgroup$ – Javier Apr 29 at 20:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.