Time component in Lorentz transformation

Question

I am slightly confused about the Lorentz transformation for time

$t=\gamma (t'+\frac {vx'}{c^2}) $

I have seen the derivation and understand it (maybe there are several and some are able to explain my question better than others? The derivation I have seen is considering how the x components transform using a simple geometric/algebraic approach of reference frames coinciding, and then using the results for the x transformation to obtain the time transformations)

What I do not understand is why the time is not simply

$t'=t/{\gamma} $

which is the simple result obtained at the beginning of any special relativity book for the time intervals on a moving and stationary clock evaluated by an observer. In essence, why do we have the Lorentz transformations if we have the previous easy result, and how are the Lorentz transformations consistent with the previous results? They clearly are not algebraically equivalent...

Answer 1

You are mixing up two different things.

Lorentz transformation is a one-to-one mapping between two reference frames, each of which is a collection of all spacetime points. That is, Lorentz transformation uniquely maps each $(t, x, y, z)$ to some $(t^\prime, x^\prime, y^\prime, z^\prime)$.

On the other hand, the time dilation equation $\Delta \tau = \Delta t / \gamma$ is defined for two points in spacetime (also called events) if it is possible to make their spatial coordinates (but not time) coincide by Lorentz-transforming to an appropriate reference frame. The time interval measured in this reference frame is the proper time $\Delta\tau$, and $\Delta t$ is the time interval measured in some other reference frame.

The result $\Delta \tau = \Delta t / \gamma$ can be derived from Lorentz transformation by demanding that in one of the two reference frames related by the transformation, the time interval be measured at a fixed point in space.