## § 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 $x, x'$ are the locations of light as measured by two inertial frames, and $t, t'$ is the time elapsed as measured by two inertial frames, we must have that $dx/dt = dx'/dt'$. This ensures that the speed of light is invariant.

#### § The derivation

• Our coordinate system has an $x$ space axis and a $t$ time axis.
• We have observer (1) standing still at the origin, and measures timewith a variable $t$.
• We have observer (2) moving to the left with a constant velocity $v$.
• Observer (1) who is at rest sees a photon starting from the origintravelling towards the right with constant velocity $c$. The position of thephoton at time $t$ is $x = c t$.
• Observer (2) also sees this photon. At time $t$, he sees the positionof the photon as $x' = vt + ct$.
• From our rule of invariance, we have that $dx/dt = c = dx'/dt'$.
We calculate $dx'/dt' = (dx'/dt)(dt/dt')$ [chain rule], giving:
\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 $t' = (1 + v/c) t$.
• This checks out: Assume our observer is moving leftward at $v = c$. He willthen see the photon move rightward at $x' = 2ct$. So if his time slows down to have $t' = 2t$, we will have that $x/t' = 2ct/2t = c$.
• However, this forumla allows us to go faster than the speed of light withno repercurssions! (It is also not the correct formula as anticipated bythe usual derivation). This can be fixed.
• Now assume that Observer 2 was moving rightward, not leftward. That is,we simply need to set $v = -v$, since this change of sign accomplishesflipping the direction of motion in 1D. This gives us the equation$t' = (1 - v/c)t$.
• According to this new equation, we are not allowed to approach the speed oflight. If we attempt to do so, we will need to elapse zero time; and ifwe 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 2witness two photons, one moving leftward and one moving rightward, one willneed to write down the equations $t' = (1 \pm v/c)t$, which plainly leadsone to contradiction.

#### § What's the issue?

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