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## Main Question or Discussion Point

I'm trying my hand at deriving Lorentz transformations using 3 postulates - it's a linear transformation, the frames are equivalent, so they see the same speed of each other's origins and that the speed of light is the same. Let's say frame ##S## is moving at velocity ##v## in the ##x##-direction w.r.t. ##S'##. From linearity, we have

$$t'=a_1t+a_2x+a_3\\x'=b_1t+b_2x+b_3$$

Let the position of the origin of ##S## in the ##S'## frame at ##t'=0## be ##x'_{O}[t'=0]##. Since ##x=0##, we get ##t=-a_3/a_1## by substituting ##t'=0,x=0## in the first equation. So ##x'_O[t'=0]=-\frac{b_1a_3}{a_1}+b_3##. Similarly at ##t'=T'##, we have for the ##S## origin ##t=\frac{T'-a_3}{a_1}##. So ##x'_O[t'=T']=\frac{b_1(T'-a_3)}{a_1}+b_3##. Using ##x'_O[t'=T']-x'_O[t'=0]=vT'##, we get ##b_1=va_1##. Now we have

$$t'=a_1t+a_2x+a_3\\x'=va_1t+b_2x+b_3$$Now consider a photon ##P## that starts from the origin in ##S## at ##t=0##. In the ##S## frame, ##x_P[t=0]=0## and ##x_P[t=T]=cT##. The starting time in frame ##S'## will be ##t'[t=0,x=0]=a_3##. The ending time will be ##t'[t=T,x=cT]=a_1T+a_2cT+a_3##. Similarly, ##x'[t=0,x=0]=b_3## and ##x'[t=T,x=cT]=va_1T+b_2cT+b_3.## This gives

$$va_1T+b_2cT=ca_1T+a_2c^2T\\\implies va_1+b_2c=a_1c+a_2c^2$$

Finally, let the position of the origin of ##S'## in the ##S## frame at ##t=0## be ##x_{O'}[t=0]##. Substituting ##t=0,x'=0## in the second equation, ##x_{O'}[t=0]=-b_3/b_2##. Similarly ##x_{O'}[t=T]=\frac{-b_1T-b_3}{b_2}##, and so

$$x_{O'}[t=T]-x_{O'}[t=0]=-\frac{b_1T}{b_2}=-vT\implies b_1=vb_2\implies b_2=a_1

\\\implies va_1=a_2c^2\implies a_2=\frac{va_1}{c^2}$$

The transformation can thus be restated as (on the RHS I'll replace ##a_1## by ##\gamma##):

$$t'=a_1t+\frac{va_1}{c^2}x+a_3=\gamma(t+\frac{vx}{c^2})+a_3\\x'=va_1t+a_1x+b_3=\gamma(x+vt)+b_3$$

And now beyond this, I'm really not sure how to proceed. I'm lost on how to derive the value of ##\gamma##. Can anyone please help? Thanks (I've looked at the Lorentz transformation derivation Wikipedia page but haven't found a resolution to my query)

$$t'=a_1t+a_2x+a_3\\x'=b_1t+b_2x+b_3$$

Let the position of the origin of ##S## in the ##S'## frame at ##t'=0## be ##x'_{O}[t'=0]##. Since ##x=0##, we get ##t=-a_3/a_1## by substituting ##t'=0,x=0## in the first equation. So ##x'_O[t'=0]=-\frac{b_1a_3}{a_1}+b_3##. Similarly at ##t'=T'##, we have for the ##S## origin ##t=\frac{T'-a_3}{a_1}##. So ##x'_O[t'=T']=\frac{b_1(T'-a_3)}{a_1}+b_3##. Using ##x'_O[t'=T']-x'_O[t'=0]=vT'##, we get ##b_1=va_1##. Now we have

$$t'=a_1t+a_2x+a_3\\x'=va_1t+b_2x+b_3$$Now consider a photon ##P## that starts from the origin in ##S## at ##t=0##. In the ##S## frame, ##x_P[t=0]=0## and ##x_P[t=T]=cT##. The starting time in frame ##S'## will be ##t'[t=0,x=0]=a_3##. The ending time will be ##t'[t=T,x=cT]=a_1T+a_2cT+a_3##. Similarly, ##x'[t=0,x=0]=b_3## and ##x'[t=T,x=cT]=va_1T+b_2cT+b_3.## This gives

$$va_1T+b_2cT=ca_1T+a_2c^2T\\\implies va_1+b_2c=a_1c+a_2c^2$$

Finally, let the position of the origin of ##S'## in the ##S## frame at ##t=0## be ##x_{O'}[t=0]##. Substituting ##t=0,x'=0## in the second equation, ##x_{O'}[t=0]=-b_3/b_2##. Similarly ##x_{O'}[t=T]=\frac{-b_1T-b_3}{b_2}##, and so

$$x_{O'}[t=T]-x_{O'}[t=0]=-\frac{b_1T}{b_2}=-vT\implies b_1=vb_2\implies b_2=a_1

\\\implies va_1=a_2c^2\implies a_2=\frac{va_1}{c^2}$$

The transformation can thus be restated as (on the RHS I'll replace ##a_1## by ##\gamma##):

$$t'=a_1t+\frac{va_1}{c^2}x+a_3=\gamma(t+\frac{vx}{c^2})+a_3\\x'=va_1t+a_1x+b_3=\gamma(x+vt)+b_3$$

And now beyond this, I'm really not sure how to proceed. I'm lost on how to derive the value of ##\gamma##. Can anyone please help? Thanks (I've looked at the Lorentz transformation derivation Wikipedia page but haven't found a resolution to my query)