Derivation of time transformation in Lorentz transformations Consider the Lorentz transformation
$$\Delta x'=\gamma(\Delta x-v\Delta t)\tag1$$
Now let a particle move with speed $u'$ and $u$ in prime and unprimed systems respectively. Thus by using $\Delta x=u\Delta t$ we get
$$\frac{\Delta t'}{\Delta t}=\gamma\frac{u-v}{u'}\tag 2$$
Since we don't know the relation between $u$ and $u'$ in general we let $u=c$ thus $u=u'=c$
This reduces to
$$\Delta t'=\gamma\bigg(\Delta t-\frac vc\Delta t\bigg)\tag{3}$$
But since $\Delta t=\Delta x/c$ we get
$$\Delta t'=\gamma\bigg(\Delta t-\frac{v\Delta x}{c^2}\bigg)\tag{4}$$
Can we say that this result holds true in general solely based on the above derivation which was a special case of the equation $1$?
 A: Not quite, which I will explain in the second paragraph. First, let's get a sense of what you are doing. You are assuming the Lorentz transformation rule for $\Delta x$ and trying to derive the transformation rule for $\Delta t$. Note that in doing this you are assuming that if $u=c$, then $u'=c$. This is one of the postulates of special relativity so this is a fine assumption but it's good to explicitly know which assumptions you are using.
Up till equation (2) your derivation is general; it would work for any (sensical) values of $\Delta x,\Delta t$. After this you take $\Delta x=c\Delta t$. This means you are restricting yourself to intervals in spacetime which move at the speed of light. What you have derived should only be expected to work for intervals moving at the speed of light and this becomes clear in the last step. In the last step you go from $\Delta t'=\gamma\left(\Delta t-\frac v c\Delta t\right)$ (3) to $\Delta t'=\gamma\left(\Delta t-\frac {v\Delta x} {c^2}\right)$ (4). You should be able to do this step in reverse, i.e. derive (3) from (4) for general intervals, but you can't because (3) is less general and only works for paths moving at the speed of light.
