TcheQ said:
39:00 onwards, he deals with how you arrive at the conclusion c=c', using c=1 for simplicity
TcheQ, thanks for pointing at the relevant part of the video.
I think the lecturer himself confirms the point, although what he says does help to explain it better.
First, he considers that a light beam has, in an unprimed frame, a speed = c, whatever it is. Without special choice of units, that means that:
x (distance traversed by the beam in a given time = L = the length of a rod where that distance is marked) = ct (speed of light x that time lapse).
If we then choose to measure distance in light seconds, the expression becomes, by definition (a light-second = distance traversed by light in a second):
x = t
So far, so good. The choice of units is innocuous. No major step. Just a convenient approach.
Second, he wonders about the coordinates of that very same light beam in a primed frame, moving relative to the other. Without choice of units and without any special assumption, the equation will be:
x' = c't'
Now it’s time to make your choice of units. Can you simply say, like we did before…
x’ = t’?
Well, classical mechanics would say you can’t. Its reasoning would be:
(a) If the rod whose length we are using as reference is the same one as before, then we have x’ = x.
(b) In the primed frame the beam does not travel at c, but at (c-v) in the go trip (since the target is escaping), (c+v) in the return trip (since it is heading towards its target) and at the average between the two in the round trip.
(c) With (b)’s assumption, no matter which reference you take (light speed at the go trip, at the return trip or at the average of the round trip), if you make the calculation, you’ll conclude that the time that the beam takes to traverse the length of the rod (x=x’=L=L’) is never x light-second.
Hence with the classical assumptions you *cannot* say that x’=t’!
Of course you can change the assumptions. You can postulate that c = c’ and so x’=t’, even if as a consequence of that x’ may be different from x and t’ different from t. If you take that step (a major step, by the way), then all the rest follows. But without that postulate, you go nowhere. The lecturer somehow recognizes it when he states that, in a later part of the derivation, x^2-t^2=0 does not necessarily imply x’^2-t’^2=0, but he could have made it clearer.
To sum up: c = c’ is not a conclusion you arrive at but a postulate you derive consequences from.
Unlike what oversimplified explanations suggest, you cannot change a physical theory (you cannot shift from classical to special relativity) just with algebra or geometry, unless the original theory contained some algebraic or geometric mistake. And I wouldn’t advise anyone to tell Newton that in his face. If you wish to change, you can (and I think you must) but you need a new physical assumption for that purpose, not just math or drawings.