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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 $\beta \ll 1$.

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  • $\begingroup$ Lorentz transformations are $x' = \gamma ( x - v t)$ and $t' = \gamma ( t + \beta \frac{x}{c} )$. I don't see why you can't just take $\beta \to 0$ keeping $v$ fixed which gives $x' = x- vt$ and $t' = t$. $\endgroup$
    – Prahar
    Commented Feb 7, 2021 at 14:06
  • $\begingroup$ @PraharMitra: $\beta \rightarrow 0$ with $v$ constant means $c \rightarrow \infty$. This is an irreal assumption. $v$ could tend to zero (event be equal to zero), but $c$ can not tend to $\infty$. $\endgroup$ Commented Feb 7, 2021 at 14:09
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    $\begingroup$ @PraharMitra: "trend" is not "equal". Trend is, in practical sense, a decision of how may terms you take in series and how these terms are hidden by other elements in the equations. $\endgroup$ Commented Feb 7, 2021 at 14:15
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    $\begingroup$ Why can't $c \to \infty$?? That is precisely the non-relativistic limit! If you're suggesting that in real life, $c$ is a constant and such a limit is not possible, then I agree with you because in real life, you also cannot have a non-relativistic setting. If you're not happy with taking a limit, you can do a Taylor expansion in $v/c$. $\endgroup$
    – Prahar
    Commented Feb 7, 2021 at 14:24
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    $\begingroup$ @GiuseppeNegro: done $\endgroup$ Commented Feb 7, 2021 at 16:36

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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.

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  • $\begingroup$ Thanks for your answer. That means that at a speed of 10 m/s (=36km/h) and distance of 30 m, I can experiment relativistic effects if measurement of $1 \mu s$ are done ? $\endgroup$ Commented Feb 7, 2021 at 14:04
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    $\begingroup$ I mean for fixed time $t$, $x$ has to be large enough. I fixed the formula, how large actually depends on $\beta$. For $t=1\mu s$ and $v=8\text{km/s}$ (speed of circular orbit near Earth), $t'$ gets noticeably different from $t$ for coordinates $x$ comparable and greater than $ct/\beta = 11300 ~\text{km}$, which is around diameter of the Earth. So for an observer on ISS, the mismatch in times of events that happen at the same time on opposite sides of the Earth is around $1\mu s$. $\endgroup$ Commented Feb 7, 2021 at 14:05
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Yes, you're right about that. Here is a picture of the Lorentz vs. Galilean transformations to illustrate:

Lorentz vs. Galilean transformations

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.

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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)
robphy-STTrig-1

E=0 (Galilean)
robphy-STTrig-2

E=-1 (Euclidean)
robphy-STTrig-3

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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.

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    $\begingroup$ Yes. There is a point P (0,0) that corresponds to P' (0,0). I will edit. $\endgroup$ Commented Feb 6, 2021 at 21:55
  • $\begingroup$ Do you assume that all events that observer O measures must have a spacetime trajectory that passes through (0,0) ? $\endgroup$ Commented Feb 7, 2021 at 9:47
  • $\begingroup$ No, but if one of the events is taken as happened in (0,0) the calculations are simplified. $\endgroup$ Commented Feb 7, 2021 at 15:53

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