Why does isotropy principle require existence of inertial transformation when axes are reversed?

Question

Assuming one spatial and one termporal dimension, let's assume an intertial transformation $A(v)$ as follows: $$ \begin{pmatrix} t' \\ x' \\ \end{pmatrix} = A(v) \begin{pmatrix} t \\ x \\ \end{pmatrix} $$ where $$ A(v) = e^{\sigma v} \begin{pmatrix} 1 & 0 \\ -v & 1 \\ \end{pmatrix} $$ for some constant $\sigma \ge 0$ (we get the Galileo transformation for $\sigma=0$ as a special case). It is simply a Galileo transformation combined with a velocity-dependent (and not reflection invariant, as shown below) global dilatation.

It can be easily shown, that $A(0) = 1$, $A^{-1}(v) = A(\bar v)$ with $\bar v=-v$ and $A(u)A(v)=A(w)$ with $w=u+v$, so $A(v)$ forms a group. The transformation is linear, as required by space homogeneity.

Following [1], if the space is isotropic, then $(t, -x)$ and $(t, -x')$ qualify as well for equivalent coordinates. Introducing the space reflection (parity) by the matrix $T=diag(1, -1)$, we can see, that $T A(v) T$ does not belong into the group (i.e. there is no $\bar v$ so that $T A(v) T = A(\bar v)$) unless $\sigma=0$ (then $\bar v=-v$), because $$ T A(v) T = e^{\sigma v} \begin{pmatrix} 1 & 0 \\ +v & 1 \\ \end{pmatrix} $$

In other words, if an event has coordiantes $(t, x)$ and $(t', x')$ in the two inertial frames, then for $\sigma > 0$ there is no inertial transformation (of the above form), that would connect $(t, -x)$ and $(t, -x')$, so the transformation $A(v)$ does not respect isotropy.

What is the physical meaning of the above $A(v)$ transformation for $\sigma > 0$? Does it mean that it depends on the axes orientation? How could an experiment be constructed to show that the space is not isotropic? This transformation must be preferring a space direction, but I would like to see this explicitly.

[1] Levy-Leblond, J.-M. (1976). One more derivation of the Lorentz transformation. American Journal of Physics, 44(3), 271–277.

Answer 1

Adding the factor of $\exp(\sigma v)$ to the Galilean transformation matrix simply means that every infinitesimal Galilean transformation is "decorated" by an extra uniform infinitesimal rescaling of $t,x$ by the same factor. Such a "velocity-dependent scaling of time (and space)" has of course never been considered in classical mechanics where time was universal and absolute (including the normalization).

But you may consider it. When you do, the very form of the factor $\exp(\sigma\cdot v)$ makes it obvious that $\sigma$ is really a spatial vector (and not a scalar), too. If there's one spatial dimension only, spatial vectors and scalars may get confused but it's easy to unconfuse oneself, especially if you try to construct higher-dimensional generalizations of this construction (higher than 1+1).

Because $\sigma$ is a spatial vector, it's obvious that under the spatial reflection $x\to -x$, $\sigma$ goes to $-\sigma$, too. That's why you can't construct a parity-symmetric theory with a fixed nonzero $\sigma$: the parity transformation simply wants to revert the sign of $\sigma$ but it isn't allowed because the theory wants a fixed $\sigma$. I find the ordering of the description of the problems of this theory you offered kind of upside down because you are trying to "impose" the parity symmetry and claim that the Galilean transformations won't exist to relate two events in spacetime. It seems more logical to me to assume/demand the more nontrivial symmetry, the Galilean one, and then you may simply see that the parity symmetry is broken for $\sigma\neq 0$.

Answer 2

I'll try to answer my own question. Here is the experiment which shows the space anisotropy. The above transformation implies the following time dilatation: $$ \Delta t' = e^{\sigma v} \Delta t $$ which depends on the direction of $v$. So we consider two synchronized clocks A and B at x=0, then move A to $x=1$ and B to $x=-1$ with velocities $\pm v_1$. Then we exchange A, B with velocities $\pm v_2$. Finally we move A, B back to $x=0$ with velocities $\pm v_1$. The idea is to use $v_1 \ll v_2$, so that $v_1$ virtually doesn't cause any dilatation, while $v_2$ causes almost all the dilatation. As such, the clock B must show more time.

Now in equations (I hope I didn't make a mistake below). Let's start with t=t'=0. For the clock A, by moving from $x=0$ to $x=1$ it will show the time $t'=t_1'$: $$ \begin{pmatrix} t_1' \\ 0\\ \end{pmatrix} = A(v_1) \begin{pmatrix} t_1 \\ 1 \\ \end{pmatrix}\,, $$ from which $t_1 = \frac{1}{v_1}$ and $t_1' = e^{\sigma v_1} t_1 = e^{\sigma v_1}\frac{1}{v_1} $. Now we move to $x=-1$, the clock A will show time $t'=t_1'+t_2'$, where $t_2' = e^{-\sigma v_2}\frac{2}{v_2}$. Finally, by moving back to $x=0$, the clock will show the time $t'=t_1' + t_2' + t_3'$ where $t_3' = e^{\sigma v_1}\frac{1}{v_1}$. All together then: $$ t' = t_1' + t_2' + t_3' = e^{\sigma v_1}\frac{1}{v_1} + e^{-\sigma v_2}\frac{2}{v_2} + e^{\sigma v_1}\frac{1}{v_1} = e^{\sigma v_1}\frac{2}{v_1} + e^{-\sigma v_2}\frac{2}{v_2} $$ In the same way, the clock B will show: $$ t' = e^{-\sigma v_1}\frac{2}{v_1} + e^{\sigma v_2}\frac{2}{v_2} $$ Using for example $v_1=1$ and $v_2=10$ and $\sigma=0.1$ we obtain for clock A: $$ t' = 2.2839... $$ and clock B $$ t' = 2.3533... $$ as can be calculated by this simple Python script:

In [1]: from math import exp                                                   
In [2]: v1 = 1.
In [3]: v2 = 10.                                                               
In [4]: sigma = 0.1                                                            
In [5]: exp(sigma*v1) * 2 / v1 + exp(-sigma*v2) * 2/v2                         
Out[5]: 2.2839177243855837
In [6]: exp(-sigma*v1) * 2 / v1 + exp(sigma*v2) * 2/v2                         
Out[6]: 2.3533312017637282

This experiment is symmetrical, but we still got different times for A, B. This means that we can use this experiment to measure if time goes faster when moving to the left or to the right.