Lorentz Transformation Proof - Special Relativity

This is from A.P. French Special Relativity book, Chapter 3 (page 78)

Setup of the proof: $$S$$ and $$S'$$ be inertial reference frame. $$S'$$ move to the right with respect to $$S$$ at velocity $$v$$.

Let co-ordinates in $$S$$ be $$(x,t)$$ and co-ordinates in $$S'$$ be $$(x',t')$$

Equation (3-8) in the book, he writes that transformation will be of the form:

$$x = ax' + bt'$$ and by symmetry of the reference frames as implied by relativity principle, $$x' = ax - bt$$

My question:

How does symmetry of reference frame argument lead to the above conclusion? For e.g. why can't the second equation above be of the form $$x' = -ax - bt$$ or maybe $$x' = -ax + bt$$. These equations look as symmetric to me (mathematically!) as the one author uses. (I know I'm wrong but want to understand more clearly why am I wrong)

Thanks

Without loss of generality, let's assume $$v$$ is positive. From the viewpoint of the S' frame, the S frame is moving to the left with velocity $$v$$. (Equivalently, S moves to the right with velocity $$-v$$). Now, if we make a video of this, and play the video backwards, it will look like S is moving to the right with velocity $$v$$.
Playing the video backwards is equivalent to replacing $$t$$ with $$-t$$. So the transformation from S coordinates to S' coordinates is the same as the transformation from S' coordinates to S coordinates, but with time reversed.
• @aman_cc Maybe this will help. If you start at position O and face left and walk at 1 m/s for 3 seconds you end up 3 m to the left of O. And if you start at position O and face right but walk backwards at 1 m/s for 3 seconds you also end up 3 m to the left of O. That is, walking backwards at $v$ is the same as walking forwards at $-v$. And vice versa. May 24 '20 at 8:55
• @aman_cc Ok. For the moment, let's assume that $v\ll c$, so we can ignore time dilation & length contraction, and see what happens in Galilean relativity. If I'm at rest in the S frame and you're at rest in the S' frame, we agree that the magnitude of our relative speed is $v$. But I say that you are moving with a speed of $v$ to the right in my frame, and you say that I'm moving to the left at a speed of $v$ in your frame. Do you agree? May 24 '20 at 12:09
$$b$$ has to have the dimension of a velocity to be consistent. When the reference frame $$S$$ looks at the reference frame $$S^\prime$$, he sees it has moving to the right with velocity $$+v$$. But if you swap the frames, in the $$S^\prime$$ frame, if you look at the $$S$$ frame, you'll see it moving with the same velocity as before but in the opposite direction, so $$-v$$. There's no reason to let even the $$a$$ change sign since swapping refrence frames just swaps the velocity of the other frame in that reference.