# The transformations of Lorentz as a general case of the transformations of Galileo

Starting from the transformations of Lorentz,

\left\{\begin{aligned} x&=\gamma (x'+\beta ct)\\ y&=y'\\ z&=z\\ ct&=\gamma (ct'+\beta x')\\ \end{aligned}\right. \quad \tag{*}

I have thought that surely one would expect compatibility or approximation with Galileo's transformations when $$v/c=\beta \ll1 \iff v \ll c$$ and $$\gamma\simeq 1$$.

But be careful: $$\beta$$ can be negligible but the product $$\beta x'$$ cannot be overlooked because it can be a very large value. For example if it is $$\beta=0.1$$ and $$x'=10^{10}$$ then $$\beta x'=10^{9}\ggg 1$$.

\left\{\begin{aligned} x&=x'+vt'\\ y&=y'\\ z&=z'\\ t&\simeq t'+\beta x'/c\\ \end{aligned}\right. \tag{**}

My question is: starting from $$(^{**})$$ as is it possibile to arrive at the transformations of Galileo $$(^{***})$$?

\left\{\begin{aligned} x&=x+vt\\ y&=y'\\ z&=z'\\ t&=t'\\ \end{aligned}\right. \tag{***}

• It's possible I'm overlooking a subtlety in your question but isn't the Galileo transformation recovered from the Lorentz transformation in the limit $c \rightarrow \infty$? For finite $c$, you have the Lorentz transformation with a possibly wrong value for the invariant speed. – Alfred Centauri Mar 19 at 20:57
• But $c=10^8 \mathrm{m/s}$ and it is impossible that $c\rightarrow \infty$. – Sebastiano Mar 19 at 21:02
• @AlfredCentauri $c\to\infty$ means "first order in $v/c$". However, to answer the original question, since $v\ll c$ hence $v/c^2\to 0$ in the last equation, therefore you end up with $t'=t$. – gented Mar 19 at 21:07
• @gented Could I have a detailed explanation with an answer? I have guessed the solution thanks to your comment: $\beta x'/c=v/c^2$ being $\beta=v/c$. I hope, always that the question is clear and interesting. – Sebastiano Mar 19 at 21:18

## 2 Answers

The classical limit in special relativity is to be intended as expansions around $$\beta=0$$ of any function thereof, stopping at the lowest powers in $$\beta$$. For example one has: $$\gamma := \frac{1}{\sqrt{1-\beta^2}} = 1+\frac{1}{2}\beta^2 + o(\beta^2)=1+o(\beta).$$ Plugging the above into the Lorentz transformations we obtain $$x = \big(1+ o(\beta)\big)(x' + vt')$$ $$t = \big(1+ o(\beta)\big)\big(t' + \frac{\beta}{c}x'\big)$$ the argument being that $$\beta/c=o(\beta)$$, therefore at lowest order in $$\beta$$ the last piece in the right hand side of the equation for $$t=f(t')$$ vanishes after having multiplied all corresponding entries.

• Any further explanations: surely there is a series expansion. For me it is not very clear: $1+\frac{1}{2}\beta^2 + o(\beta^2)=1+o(\beta)$, $x = \big(1+ o(\beta)\big)(x' + vt')$. Similarly for the formula below? Why $\beta/c=o(\beta)?$. Now for me is +1. – Sebastiano Mar 19 at 21:42
• $o(\beta)$ is essentially any function who approaches zero faster than $\beta$: as such $\beta^2$ clearly is $o(\beta)$ and so is any higher power thereof. To understand why $\beta/c = o(\beta)$ you can just multiply and divide by $v$ in order to get $1/v \cdot \beta^2=o(\beta)$. – gented Mar 19 at 21:49

I would like to tell you, the joyous and wonderful answer is no.

Special relativity in fact asserts something new about the world. It asserts that if you are accelerating forwards (for some definition of “forwards”) with acceleration $$a$$ then your best guess about what is happening ahead of you (independent of Doppler effect) is that the clocks at a distance $$L$$ away from you are ticking faster by a rate of $$a L/c^2$$ and clocks behind you are similarly ticking slower by the same amount, with ticking stopping altogether at an $$x = -aL/c^2.$$

Indeed, the whole Lorentz transform can be correctly inferred from this low-velocity limit, and this low-velocity limit can be understood in the following way: suppose someone is moving with velocity $$v$$ relative to some “ground,” and emits a light pulse in all directions, expanding at speed $$c$$ as seen by an observer stationary on the ground. After some time $$t$$, the bubble has radius $$L = c t$$ and the traveler is off-center by some small-by-comparison distance $$v t$$. The question is, how do we modify Galilean relativity so that they see themselves at the center of the bubble in their reference frame? And the answer that special relativity gives us is called “relativity of simultaneity,” where the traveler does not think that those edges of the light bubble are being measured at the same time: they would prefer you to have chosen an earlier instant for the measurement of the light ‘behind’ them and a later instant for the measurement of the light ‘ahead’ of them, so that they think that the point we're saying is the light ahead of them is actually at their time $$t - vt/c = t - vL/c^2.$$ Clocks that we think are in-sync at those two ends must appear to therefore be out-of-sync to the traveler by this factor, and we thus have the low-velocity limit \begin{align} t' &= t - vx/c^2,\\ x' &= x - v t,\\ y' &= y,\\ z' &= z. \end{align}To infer the whole Lorentz transform, simply define $$w = c t$$ and build a big transform out of a limit of little ones, $$\begin{bmatrix}w'\\x'\\y'\\z'\end{bmatrix} = \lim_{N\to\infty} \begin{bmatrix}1 & -\phi/N & 0 & 0\\ -\phi/N & 1 & 0 & 0\\ 0&0&1&0\\ 0&0&0&1\end{bmatrix}^N \begin{bmatrix}w\\x\\y\\z\end{bmatrix}$$and after diagonalizing and exponentiating and un-diagonalizing this gives the Lorentz boost matrix there as, $$\begin{bmatrix}\cosh\phi&-\sinh\phi&0&0\\ -\sinh\phi&\cosh\phi&0&0\\ 0&0&1&0\\ 0&0&0&1\end{bmatrix}$$which after defining $$\gamma = \cosh\phi$$ and $$v = c~\tanh\phi$$ is the familiar expression, with an identity $$\cosh^2 - \sinh^2 = 1$$ giving us $$1 - (v/c)^2 = 1/\gamma^2$$ for the familiar expression $$\gamma = 1/\sqrt{1-\beta^2}.$$ (One may also cheat and simply observe that the composition of one "forwards" and "backwards" transform is $$\operatorname{diag}(1-\beta^2,1-\beta^2, 1, 1)$$ and come up with the idea of dividing the 2x2 sub-block by $$\sqrt{1-\beta^2}$$ of course, but that just feels like cheating.)

So really, relativity of simultaneity is the only fundamental physical thing that special relativity contributes, and it is indeed nontrivial, and as long as it is present in the equations the equations compose together to force the Lorentz transformation with its additional length contraction and time dilation mechanics. Those are simply aggregated effects of lots of little relativity-of-simultaneities in a Galileian world; they have no independent physical content.

So then, when we are trying to take a Galileian approximation, we must be operating under the assumption that clocks are sufficiently nearby that given our typical speeds and accelerations, we never see them ticking fast or slow, so that the relativity of simultaneity never kicks in. This is not just a constraint on $$v/c\ll 1$$ but as you have noticed it is something different, stating that in addition the time for light to propagate through the system $$L/c$$ is small enough that we don't need to consider these sorts of things, so that $$vL/c^2$$ is still a very small time offset.

• Very nice answer. I understood the concept because I know the hyperbolic transformations of Lorentz :-) I thank you so much for your kindness and I also approve your answer. My grades are always positive because I appreciate those who have dedicated time to me. – Sebastiano Mar 20 at 20:53