The Lorentz Transformation are derived by scalling the following terms by $\gamma$.

$$x' = (x - ct)\gamma$$ $$x = (x' + ct')\gamma$$

And then by considering the conditions that:

$$c = \frac{x}{t}$$ $$c = \frac{x'}{t'}$$

The $x'$ and $t'$ are derived.


Why don't we scale the $t=t'$ in the same way by some factor other than $\gamma$. Say $\lambda$: $$t' = t\lambda$$ $$t = t'\lambda$$

  • 1
    $\begingroup$ Your last two equations (question v1) imply $\lambda=1$. $\endgroup$
    – Ruslan
    Dec 19 '16 at 14:11
  • $\begingroup$ @Anonymous I don't understand the "Question" part of your question. Also, $x'$ and $t'$ are not derived from imposing the condition that $c=\dfrac{x}{t}$ should map to $c=\dfrac{x'}{t'}$ and vice-versa on the first set of equations you wrote down. In fact, these conditions are already built-in those formulae. $\endgroup$
    – Dvij D.C.
    Dec 19 '16 at 14:55
  • $\begingroup$ @Electrodynamist Why is $x$ part of the gallilean transformation special to get scaled and not the $t$ part? $\endgroup$
    – Anonymous
    Dec 19 '16 at 15:20
  • 1
    $\begingroup$ @Anonymous Do you mean by "special to get scaled" that writing $x'=\gamma(x-ct)$ is something like $x'=x-vt$ from Galilean transformation up to a scaling factor namely, the $\gamma$? $\endgroup$
    – Dvij D.C.
    Dec 19 '16 at 15:33
  • $\begingroup$ @Electrodynamist exactly. $\endgroup$
    – Anonymous
    Dec 19 '16 at 16:23

I am basing this answer on some of the clarification I obtained in the comments on the OP apart from the OP.

The reason we take the form of the transformation to be


is simply that we know that the events happening at (according to .the unprimed frame) time $t$ and the spatial coordinate $x$ are certain to be recorded at the spatial origin of the primed coordinate system if and only if $x-vt=0$.

It is not that we are giving any privilege to the spatial component over the temporal component. Using the 'Einsteinian' definition of simultaneity along with the assumption of the invariance of the speed of light, we do formulate a similar relation for the temporal coordinate as well. We deduce that the events happening at (according to .the unprimed frame) time $t$ and the spatial coordinate $x$ are certain to be recorded at the temporal origin of the primed coordinate system if and only if $t-\dfrac{vx}{c^2}=0$. Thus, we write down

$t' = \gamma_2\big(t-\dfrac{vx}{c^2}\big)$

One should keep in mind that the assumption of homogeneity and isotropy of space and time is inherent to all of the above arguments. I have used different factors - $\gamma_1$ and $\gamma_2$ because we don't know a priori that they are going to be the same. It is only later that we find out, through further exploitation of the axioms of special relativity (laws of Physics are same in all the frames and speed of light is frame-invariant), that


Coming back to the original question that is posed by the OP, the reason for the equation coming from the knowledge about the spatial origin of the primed frame in terms of the unprimed frame's coordinates is similar to the one in Galilean transformation but the corresponding equation for time is very different from its Galilean counterpart is that the Galilean transformations were written down on the basis of crude observational instincts. Since all the relative speeds we experience are very small as compared to the speed of light, we got the equation for spatial part better than the temporal part because the factor $vt$ in $x-vt$ is always significant but the factor $\dfrac{vx}{c^2}$ in $t-\dfrac{vx}{c^2}$ is not significant for small speeds. Anyway, the point is that we don't write down our equations for SR by cherry-picking some parts of Galilean equations. We derive SR equations simply from the axioms of SR and some of them happen to have similarities with some Galilean equations and the others don't. Also, we can explain all these similarities or their absence on the basis of the value of the speed of light being high.

  • $\begingroup$ +1 for a carefully worded answer, taking nothing , except the axioms for granted. But if you ever write a textbook, it will be big :) I have a feeling the OP still wants a special place for time. $\endgroup$
    – user139561
    Dec 20 '16 at 14:56

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