§ An incorrect derivation of special relativity in 1D

I record an incorrect derivation of special relativity, starting from the single axiom "speed of light is constant in all inertial reference frames". I don't understand why this derivation is incorrect. Help figuring this out would be very appreciated.

§ The assumption

We assume that the velocity of light as measured by any inertial frame is constant. Thus if x,x′x, x' are the locations of light as measured by two inertial frames, and t,t′t, t' is the time elapsed as measured by two inertial frames, we must have that dx/dt=dx′/dt′dx/dt = dx'/dt'. This ensures that the speed of light is invariant.

§ The derivation

  • Our coordinate system has an xx space axis and a tt time axis.
  • We have observer (1) standing still at the origin, and measures time with a variable tt.
  • We have observer (2) moving to the left with a constant velocity vv.
  • Observer (1) who is at rest sees a photon starting from the origin travelling towards the right with constant velocity cc. The position of the photon at time tt is x=ctx = c t.
  • Observer (2) also sees this photon. At time tt, he sees the position of the photon as x′=vt+ctx' = vt + ct.
  • From our rule of invariance, we have that dx/dt=c=dx′/dt′dx/dt = c = dx'/dt'.
We calculate dx′/dt′=(dx′/dt)(dt/dt′)dx'/dt' = (dx'/dt)(dt/dt') [chain rule ], giving:
c=dxdt=dx′dt′=dx′dtdtdt′c=d(vt+ct)dtdtdt′c=(v+c)dtdt′cv+c=dtdt′dt′=(v+c)dt/ct′=(v+c)t/c=(1+v/c)t \begin{aligned} &c = \frac{dx}{dt} = \frac{dx'}{dt'} = \frac{dx'}{dt} \frac{dt}{dt'} \\ &c = \frac{d(vt + ct)}{dt}\frac{dt}{dt'} \\ &c = (v + c) \frac{dt}{dt'} \\ &\frac{c}{v+c} = \frac{dt}{dt'} \\ &dt' = (v+c)dt/c \\ &t' = (v+c)t/c = (1 + v/c) t \end{aligned}
So we get the relation that time elapsed for observer (2) is related to observer (1) as t′=(1+v/c)tt' = (1 + v/c) t.
  • This checks out: Assume our observer is moving leftward at v=cv = c. He will then see the photon move rightward at x′=2ctx' = 2ct. So if his time slows down to have t′=2tt' = 2t, we will have that x/t′=2ct/2t=cx/t' = 2ct/2t = c.
  • However, this forumla allows us to go faster than the speed of light with no repercurssions! (It is also not the correct formula as anticipated by the usual derivation). This can be fixed.
  • Now assume that Observer 2 was moving rightward , not leftward . That is, we simply need to set v=−vv = -v, since this change of sign accomplishes flipping the direction of motion in 1D. This gives us the equation t′=(1−v/c)tt' = (1 - v/c)t.
  • According to this new equation, we are not allowed to approach the speed of light. If we attempt to do so, we will need to elapse zero time; and if we exceed the speed of light, we will need to elapse negative time .
  • However, these formulae lead to an absurdity. If our observer 1 and observer 2 witness two photons, one moving leftward and one moving rightward, one will need to write down the equations t′=(1±v/c)tt' = (1 \pm v/c)t, which plainly leads one to contradiction.

§ What's the issue?

The issue is the equation x′=vt+ctx' = vt + ct.
  • It is true that as per observer (1) standing at the origin , the distance between observer (2) and the photon is vt+ctvt + ct.
  • It is not true that observer(2) sees the distance between them and the photon as vt+ctvt + ct.
  • Intuitively, this equation of x′=vt+ctx' = vt + ct completely ignores length contraction, and hence cannot be right.
  • Alternatively, the equation of x′=vt+ctx' = vt + ct imposes galilean relativity, where I am attempting to naively connect reference frames, which cannot be correct.