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I know this is gonna sound backward reasoning. But still, just for the sake of curiosity:

Is it possible to use just Lorentz Contraction and Time Dilation along with the lorentz factor $\gamma$ to derive the Lorentz transformations from one inertial frame to another? If not, what other assumptions are necessary?

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  • $\begingroup$ What would you want to assume about the contraction / dilation? Would you want to simply allow their possibility and see what's compatible with other basic assumptions or begin with the Lorentz factor $\gamma = \sqrt{1-v^2}^{-1}$ as a given? $\endgroup$ – WetSavannaAnimal Feb 13 '18 at 11:46
  • $\begingroup$ @WetSavannaAnimal aka Rod Vance Yep, assuming the value of $\gamma$. $\endgroup$ – PhyEnthusiast Feb 13 '18 at 11:48
  • $\begingroup$ You probably need to assume that directions perpendicular to the velocity are not contracted. $\endgroup$ – lalala Feb 13 '18 at 11:56
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We assume that Lorentz transformations are linear and that the origin in one frame maps to the origin in another frame (we do this in the usual derivation of Lorentz transformations as well!). Then, the general form is $$ t' = a t + b x\,,\qquad x' = c t + d x \,. $$ Since the transformations are linear, this also implies $$ \Delta t' = a \Delta t + b \Delta x\,,\qquad \Delta x' = c \Delta t + d \Delta x \,. $$

Time Dilation

This states that if $\Delta x = 0$, then $\Delta t' = \gamma \Delta t$. Thus, $a=\gamma$. Inversely, if $\Delta x' = 0$ then $\Delta t = \gamma \Delta t'$. This implies $d = \gamma ( ad - b c )$.

Length Contraction

This states that if $\Delta t' = 0$, then $\Delta x' = \frac{ \Delta x}{ \gamma }$. This implies $a=\gamma(ad-bc)$. Inversely, if $\Delta t = 0$, then $\Delta x = \frac{ \Delta x'}{\gamma}$. This implies $d=\gamma$. Using these equations, we can obtain $$ a = d = \gamma \,,\qquad b c = \gamma^2 - 1 = \frac{\gamma^2 v^2}{C^2} $$ EDIT - I'm using $C$ for the speed of light so as not to confuse with the coefficient $c$.

With this, we can write a generic Lorentz transformation as $$ t' = \gamma t - \frac{\gamma v}{C^2 } f(v) x \,,\qquad x' = \gamma x - \frac{\gamma v}{f(v)} t\, . $$ for some function $f(v)$. We can now study some properties of this function.

First, we note the inverse Lorentz transform takes the form $$ t = \gamma t' + \frac{\gamma v}{C^2} f(v) x' \,,\qquad x= \gamma x' + \frac{ \gamma v }{ f(v) } t' $$ The two must be related via $v \to - v$. Thus, we must have $f(v) = f (-v)$.

I don't think we can say more about $f(v)$ without additional input! Let me add one additional input and complete the derivation.

Constancy of Light Speed

This requires that if $x=Ct$ then $x'=Ct'$. This immediately implies $f(v)=1$.

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  • $\begingroup$ What do you mean "Solving these equations" before the last step? $\endgroup$ – PhyEnthusiast Feb 13 '18 at 13:35
  • $\begingroup$ @PhyEnthusiast - That was misleading wording. I have edited it. $\endgroup$ – Prahar Feb 13 '18 at 13:47
  • $\begingroup$ But how do you use the product bc to find what b and c are seperately? $\endgroup$ – PhyEnthusiast Feb 13 '18 at 13:49
  • $\begingroup$ I cannot fix $b$ and $c$ separately. That's what the function $f(v)$ was introduced for. $\endgroup$ – Prahar Feb 13 '18 at 13:51
  • $\begingroup$ yeah but how do you know $b = -\frac{\gamma v}{c^2} f(v)$ $\endgroup$ – PhyEnthusiast Feb 13 '18 at 13:55

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