My instructor after teaching Galilean transformation and showing us its flaws when velocities approach the speed of light started teaching Lorentz transformation.

He reasoned that although Galilean transformations are flawed, as they work for velocities much less than $c$, we will be able to achieve the correct transformation equations (Lorentz transformation) very easily: just by multiplying correction factor/ Lorentz factor with the transformation equations. If Galilean & inverse Galilean transformations for positions $x$ & $x'$ and for times $t$ & $t'$ are




then the Lorentz & inverse Lorentz transformation will be as follows:

$$x'=(x-vt)\gamma...(i)$$ $$x=(x'+vt')\gamma...(ii)$$

But I take issue with this. If the Lorentz transformation is $x'=(x-vt)\gamma$ then shouldn't the inverse Lorentz transformation be $$x=\frac{x'+vt\gamma}{\gamma}~?$$

How am I wrong?


2 Answers 2


The Lorentz transformation takes the coordinates $(x,t)$ which label an event in frame $S$ and maps them to new coordinates $(x',t')$, which label the same event in frame $S'$. The inverse transformation takes $(x',t')$ and spits out $(x,t)$.

As your instructor has said, a good guess for the forward transformation for the position coordinate is to let $x'=\gamma(x-vt)$. Since $\gamma\rightarrow 1$ as $v\rightarrow 0$, this reduces to the Galilean transformation in the limit of small velocities, as we know it must. The inverse transformation should just be a boost in the opposite direction for the observer in $S'$, so for symmetry reasons it would make sense that $x = \gamma(x' \color{red}{+} vt')$.

Crucially though, note the prime on the $t'$ coordinate in the inverse transformation. That prime is critical; without it, the expression $x=\gamma(x'+vt)$ is wrong. This would only be true if $t'=t$ - that is, if the Lorentz transformation did not do anything to the $t$ coordinate - which we know is not the case.

In fact, we can use our two expressions for $x$ and $x'$ to see exactly what must happen to the time coordinate: $$x'=\gamma(x-vt)$$ $$x=\gamma(x'+vt') \implies x' = \frac{x}{\gamma}-vt'$$ equating these two expressions we obtain $$\gamma(x-vt) = \frac{x}{\gamma}-vt' \implies t' = \frac{1}{v}\left(\frac{x}{\gamma}-\gamma x + \gamma vt\right) = \gamma\left(t - \frac{vx}{c^2}\right)$$ where we've used the fact that $\gamma = 1/\sqrt{1-v^2/c^2}$ and subjected ourselves to a bit of algebra.

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    $\begingroup$ Tale as old as time: an instructor doesn't write a small diacritical mark on the board clearly enough, the students write it down incorrectly in their notes without knowing that it's supposed to be there, hilarity ensues. $\endgroup$ May 18, 2021 at 15:54
  • $\begingroup$ Actually, my instructor put the prime marks correctly. It was I who messed up and couldn't put the diacritical mark in the question. Could you please answer my revised question? Should I make a new post? $\endgroup$ May 18, 2021 at 17:12
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    $\begingroup$ @user545735 My answer remains the same. Presumably your question arises because you're trying to obtain the inverse transformation by solving $x'=\gamma(x-vt)$ for $x$, but if you do this then you obtain $x =(x'-\gamma v\color{red}{t})/\gamma$. You need to additionally plug in the correct expression for $t$ in terms of $x'$ and $t'$ in order to obtain the correct inverse transformation. $\endgroup$
    – J. Murray
    May 18, 2021 at 17:17

Equation (ii) is the correct inverse Lorentz transformation. The one you wrote down after that is not.

An inverse lorentz transformation gives an expression for x in terms of x' and t'

Equation (ii) satisfies that condition. The equation you wrote after that does not. Your equation gives x in terms of x' and t .
This means it does not satisfy the requirement for being an inverse lorentz transformation


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