My questions pertain to the development of Euclidean 3-space orthogonal transformations presented in Einstein's The Meaning of Relativity. Also available at https://en.wikisource.org/wiki/The_Meaning_of_Relativity/Lecture_1

I have used my own notation. The quoted text is not exactly as it appears in the referenced sources, but it is intended to convey the same meaning. I have the sense that there is a general approach that I am not seeing which would make this discussion more coherent to me. Perhaps some application the Lagrange multiplayer method.

So an answer to my first question may obviate the subsequent question. That first question is how am I really supposed to understand the discussion in the book?

We have two Cartesian coordinate systems $\mathcal{X}$ and $\overline{\mathcal{X}}$. The general expression for the length of a 3-space interval in terms of $\mathcal{X}$ is

\begin{equation} s^{2}=\Delta x^{i}\Delta x^{j}\delta_{ij}. \end{equation}

Consider a sphere whose constant radius is given in terms of each coordinate system as

\begin{equation} \Delta x^{i}\Delta x^{j}\delta_{ij}=r^{2}, \end{equation}

\begin{equation} \Delta x^{\overline{i}}\Delta x^{\overline{j}}\delta_{\overline{i}\overline{j}}=\overline{r}^{2}. \end{equation}

Our goal is to express the $x^{\overline{i}}$ in terms of the $x^{i}$ so that the expressions for $\overline{r}^{2}$ and $r^{2}$ are equivalent. The Taylor expansion of $x^{\overline{i}}$ in terms of the $x^{i}$ for small $\Delta x^{i}$ is

\begin{equation} \Delta x^{\overline{i}}=\frac{\partial x^{\overline{i}}}{\partial x^{i}}\Delta x^{i}+\frac{1}{2}\frac{\partial^{2}x^{\overline{i}}}{\partial x^{j}\partial x^{i}}\Delta x^{i}\Delta x^{j}+\mathcal{O}^{3}\left[\Delta x^{i}\right]. \end{equation}

If we substitute $\Delta x^{\overline{i}}\Delta x^{\overline{j}}\delta_{\overline{i}\overline{j}}=\overline{r}^{2}$ in this equation and compare with $s^{2}=\Delta x^{i}\Delta x^{j}\delta_{ij}$, we see that the $x^{\overline{i}}$ must be linear functions of the $x^{i}$.

Am I correct in believing that what is really intended is that we substitute the expanded form of $\Delta x^{\overline{i}}$ into $\Delta x^{\overline{i}}\Delta x^{\overline{j}}\delta_{\overline{i}\overline{j}}=\overline{r}^{2}?$ The book indicates substitution in the opposite direction. Mine is the only interpretation that makes much sense to me. Following my interpretation produces

$$ \delta_{\overline{i}\overline{j}}\Delta x^{\overline{i}}\Delta x^{\overline{j}}=\delta_{\overline{i}\overline{j}}\left(\frac{\partial x^{\overline{i}}}{\partial x^{i}}\Delta x^{i}+\mathcal{O}^{2}\right)\left(\frac{\partial x^{\overline{j}}}{\partial x^{i}}\Delta x^{i}+\mathcal{O}^{2}\right) $$

$$ =\delta_{\overline{i}\overline{j}}\left(\frac{\partial x^{\overline{i}}}{\partial x^{i}}\frac{\partial x^{\overline{j}}}{\partial x^{j}}\Delta x^{i}\Delta x^{j}+\mathcal{O}^{3}\right), $$

which is close to what I know to be the correct result. Only $\mathcal{O}^{3}$ needs to be dealt with. We can't just neglect it since there is no assumption that our sphere is small.

But I don't see the need to go that far. It is easily shown by algebra and the established definitions that changing the length of the interval (and thereby the radius) means multiplying all components by the same number. It is therefore obvious that the terms of second order and higher must vanish for all intervals.

Indeed, I'm not coming up with a way to get rid of the $\mathcal{O}^{3}$ term that doesn't amount to the same thing.

Assuming that I have made the correct choice in my use of the Taylor expansion, how am I expected to eliminate the $\mathcal{O}^{3}$ term?

The translation of Einstein's original lecture continues:

If we therefore put

\begin{equation} x^{\overline{i}}=a^{\overline{i}}+e_{\:i}^{\overline{i}}x^{i}, \end{equation}


\begin{equation} \Delta x^{\overline{i}}=e_{\:i}^{\overline{i}}\Delta x^{i}, \end{equation}

then the equivalence of $\Delta x^{i}\Delta x^{j}\delta_{ij}=r^{2}$ and $\Delta x^{\overline{i}}\Delta x^{\overline{j}}\delta_{\overline{i}\overline{j}}=\overline{r}^{2}$ is expressed in the form

\begin{equation} \Delta x^{\overline{i}}\Delta x^{\overline{j}}\delta_{\overline{i}\overline{j}}=\lambda\Delta x^{i}\Delta x^{j}\delta_{ij}.\label{eq:2b} \end{equation}

It therefore follows that $\lambda$ must be a constant.

Since the above relationship involving $\lambda$ and the constant value $\lambda=\overline{r}^{2}/r^{2}$ follow trivially from their definitions, without any reference to the Taylor expansion, I don't understand how these follow from the intervening discussion. Would someone please explain this?

If we put $\lambda=1,$ then $\Delta x^{\overline{i}}\Delta x^{\overline{j}}\delta_{\overline{i}\overline{j}}=\lambda\Delta x^{i}\Delta x^{j}\delta_{ij}$ and $\Delta x^{\overline{i}}=e_{\:i}^{\overline{i}}\Delta x^{i}$ furnish the conditions

\begin{equation} e_{\:i}^{\overline{i}}e_{\:j}^{\overline{j}}\delta_{\overline{i}\overline{j}}=\delta_{ij}. \end{equation}

  • $\begingroup$ textbook-erratum tag? $\endgroup$ – Qmechanic Nov 26 '18 at 4:42
  • $\begingroup$ I'm not sure if the wording is correct in the book. As stated, it makes not sense to me. It's not exactly a textbook, but it may constitute an error in an important book. $\endgroup$ – Steven Thomas Hatton Nov 26 '18 at 4:49
  • $\begingroup$ I understand the relationship between $\lambda$ and the unit of measurement. I can reproduce all of conclusions of the development, but not by following the book. BTW, the assertion that $x^{\overline{i}}=a^{\overline{i}}+e_{\:i}^{\overline{i}} x^{i}$ is a linear transformation is incorrect. It is an affine transformation which is not centered. All linear transformations are centered. $\endgroup$ – Steven Thomas Hatton Nov 26 '18 at 6:14

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