Peter Strohmayer said:
I never denied that Einstein accurately derived the L-T from an introductory "Gedankenexperiment". I only pointed out that it is an abstract-mathematical derivation (cf. #75), where his thought experiment is not very helpful (#99), because it reminds too much of tennis balls (of Newtonian mechanics) (#89).
This is not true. You've said things like "the emission events can only be determined with the help of the L-T" (in
post #106), which is a lot stronger than the claim you now say you were making, and is also incorrect. It is perfectly possible to derive the Lorentz transforms from Einstein's version of the thought experiment. I don't personally find his exposition particularly clear, so here is my version.
A train travelling at speed ##v## passes a stationary embankment of length ##2l##. As the nose and tail of the train pass the end and start of the embankment, respectively, they are struck by lightning. An observer standing at the center of the embankment sees the flashes arrive at his location simultaneously and concludes that the strikes were simultaneous. Can we deduce the Lorentz transforms?
Let there also be an observer on the train. She measures the train's length to be ##2l'##. Both observers zero their clocks when they pass one another, and both measure distance from their own locations. Since both frames are global inertial frames the transform (whatever it is) between the ##x,t## coordinates of the embankment observer and the ##x't'## coordinates of the train observer must take the form$$\left(\begin{array}{cc}A&B\\C&D\end{array}\right)$$
The embankment observer assigns coordinates ##x=l, t=0## to the front strike. The train observer assigns coordinates ##x'=l', t'=T_+'##. Applying our matrix yields:$$\left(\begin{array}{c}T_+'\\l'\end{array}\right)=\left(\begin{array}{cc}A&B\\C&D\end{array}\right)\left(\begin{array}{c}0\\l\end{array}\right)$$from which we deduce that ##B=T_+'/l## and ##D=l'/l##. The rear strike is at ##x=-l,t=0## for the embankment observer and ##x'=-l', t'=T_-'## for the train observer, and the equivalent calculation merely adds that ##T_-'=-T_+'##.
Now we can turn our attention to the events where the train observer receives light from the flashes. A straightforward intercept calculation tells us that the embankment observer assigns coordinates ##x=lv/(c+v), t=l/(c+v)## to the arrival of the pulse from the front while the train observer assigns ##x'=0, t'=l'/c+T_+'##. Again using the matrix on the embankment observer's coordinates and equation this with the train observer's coordinates gives us that ##A=\frac{(c+v)l'+c^2T_+'}{cl}## and ##C=-l'v/l##. For the arrival of the flash from the rear, another intercept calculation tells us that the embankment observer assigns coordinates ##x=lv/(c-v), t=l/(c-v)## and the train observer assigns ##x'=0,t'=l'/c-T_+'## (where we have used our result for ##T_-'##). Again we apply our matrix to the embankment coordinates, recovering that ##T_+'=-l'v/c^2##.
Putting all this together, we have found that the Lorentz transforms expressed as a matrix are$$\left(\begin{array}{cc}l'/l&-l'v/c^2l\\-l'v/l&l'/l\end{array}\right)$$We only need to work out what is the relationship between ##l## and ##l'##. We can do this by noting that the principle of relativity says that the inverse transform must be the same except for the sign of ##v##. Inverting our matrix gives us$$\gamma^2\left(\begin{array}{cc}l/l'&lv/c^2l'\\lv/l'&l/l'\end{array}\right)$$where ##\gamma=\left(1-\frac{v^2}{c^2}\right)^{-1/2}## is the Lorentz gamma factor, and the condition that this is the same except for the sign of ##v## tells us that ##l'=\gamma l## and hence that the Lorentz transform as a matrix is$$\left(\begin{array}{cc}\gamma&-\gamma v\\-\gamma v&\gamma\end{array}\right)$$And multiplying this by the generic coordinates ##(x,t)^T## gives us the usual form.
That is a nothing more than a process of writing down coordinates and deducing a relationship. If that's abstract maths, all of physics is abstract maths.
