Galilean transform as limit of Lorentz one

Question

Galilean transform:

$$\begin{pmatrix} ct' \\x' \end{pmatrix} = \begin{pmatrix} 1 & 0 \\-\beta & 1 \end{pmatrix}\begin{pmatrix} ct \\x \end{pmatrix}$$

Lorentz transform:

$$\begin{pmatrix} ct' \\x' \end{pmatrix} = \gamma\begin{pmatrix} 1 & -\beta \\-\beta & 1 \end{pmatrix}\begin{pmatrix} ct \\x \end{pmatrix}$$

where $\gamma^{-2} = 1-\beta^2$ and $\beta=v/c$.

It is always said that Lorentz is equivalent to Galilean in case of small speed ($v \ll c$ or equivalent $\beta \ll 1$). I've no doubt that in this case $\gamma \simeq 1$, thus:

$$\begin{pmatrix} ct' \\x' \end{pmatrix} = \gamma\begin{pmatrix} 1 & -\beta \\-\beta & 1 \end{pmatrix}\begin{pmatrix} ct \\x \end{pmatrix} \simeq\begin{pmatrix} 1 & -\beta \\-\beta & 1 \end{pmatrix}\begin{pmatrix} ct \\x \end{pmatrix}$$

However I do not see how to continue without add the assumption $|\beta x| \ll ct$ (equivalent to $\beta \ll \left|\frac{ct}{x}\right|$ or $v \ll c^2\left|\frac{t}{x}\right|$). It seems an assumption stronger than $\beta \ll 1$.

Answer 1

You are correct, and that's why the Galilei transformation is the limit of the Lorentz transformation when $c\to\infty$ rather than $v\to 0$. Limit $v\to 0$ can't produce useful transformation because we get rid of $v$ that way.

Lorentz transformation has, as you have found, a different behaviour at low $\beta$ from the Galilei transformation in that it manifests relativity of simultaneity. The coordinates have to be large enough, $x \approx ct/\beta$ to notice this.

Answer 2

Yes, you're right about that. Here is a picture of the Lorentz vs. Galilean transformations to illustrate:

LEFT: The purple arrow is a pure temporal displacement (the trajectory of the observer themself), and the red is spatial. RIGHT: According to Galileo (solid arrows), the primed observer sees that same purple displacement as taking the same amount of time (dashed orange line), but shifting in space -- but the spatial displacement is unchanged. But according to relativity (dashed arrows), any displacement slides along the hyperbola with asymptote $c$ (red & blue curves) -- specifically, all displacements sweep out the same amount of area under that hyperbola (the area between the $t'$ axis and the dashed purple arrow).

Based on that, you can see that $\gamma$ really just tells you how the hyperbola is stretched away from the horizontal line. For low speed between the two observers, these displacements stay close to the hyperbola vertex, so the $\gamma$ effect dies out. But both displacements are still rotated. It is only because $\beta \ll 1$ that you can say the rotation of the spatial displacement is negligible if you want. And by not applying that same approximation to the time displacement, since it doesn't matter if you apply it or not, you end up back at the Galilean transform.

Answer 3

A unified expression for these boosts is $$\begin{pmatrix} ct' \\x' \end{pmatrix} = \gamma\begin{pmatrix} 1 & -E\beta \\-\beta & 1 \end{pmatrix}\begin{pmatrix} ct \\x \end{pmatrix},$$ where $$\gamma=\frac{1}{\sqrt{1-E \beta^2}}.$$

When $E=1$, you have the Minkowski case.
When $E=0$, you have the Galilean case.
When $E=-1$, you have the Euclidean case (a rotation).
(These are associated with the Cayley-Klein geometries,
and lead to something I call "Spacetime Trigonometry".)

The "unit circles" that go along these boosts are $$(ct)^2-E\cdot x^2=1$$

The $c$ above plays the role of a convenient conversion constant, say the speed of light $c=c_{light}=3\times 10^8{\rm\ m/s}$, so that the coordinates have the same units and so that light travels along lines of slope $\pm 1$... even in the Galilean case.

The speed that is associated the various $E$-cases is (what I call) $$c_{max}=\mbox{``the maximum signal speed''},$$ so that $$E=\left(\frac{c_{light}}{c_{max}}\right)^2.$$

Thus, the Galilean case $E=0$ corresponds to an infinite maximum signal speed $c_{max}=\infty$ ,
whereas the Minkowski case $E=1$ corresponds to a certain finite maximum signal speed $c_{max}=c_{light}$ (since light in vacuum seems to travel at the maximum signal speed).

Visit my "spacetime diagrammer" at https://www.desmos.com/calculator/kv8szi3ic8 to play with some of these ideas. Try the E-slider. You can also open the "BOOST" folder and change the lab-frame.

E=1 (Minkowski)

E=0 (Galilean)

E=-1 (Euclidean)

Answer 4

The equations for Lorentz transformation for $1+1$ spacetime supposes an origin $(0,0)$.

For any timelike distance between the origin and an event: $|\frac{ct}{x}|>1$. So $\beta \ll 1 \implies \beta \ll |\frac{ct}{x}|$

The problem is the spacelike distances when $|\frac{ct}{x}|<1$. In that case, even for low velocities, strange conclusions like the Andromeda paradox results from SR, that are not present in the Galilean transform.