How does the Lorentz factor (gamma) affect Lorentz transformations?

Question

Does the lorentz factor in lorentz transformations of space and time between reference frames mean that time and space increase the amount that they shift at an exponential rate as v approaches c? Does this shift grow at an exponentially higher or lower rate as v approaches c? does the lorentz factor being in both the space and time transformations somehow cancel so that the increase in the shifting of spacetime stays constant as v approaches c? What is the effect of the lorentz factor on the shifting of spacetime as v approaches c? Image credit to: http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/veltran.html

Answer 1

Well, let's look at the math: Clearly when $\vec{v} = 0$, then $\gamma =1$ and so we expect neither a shift in time or space. That's pretty self-explanatory, after all, it's just a stationary frame isn't it?

What about when we start increasing $\vec{v}$? Well,

$\frac{d\gamma}{d\vec{v}} = \frac{d}{d\vec{v}}\frac{1}{\sqrt{1-\frac{\vec{v}^2}{c^2}}} = - \frac{\vec{v}}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{v^2c^2}{c^2-v^2}. $

The point here being that as $v$ approaches $c$, then the change in $\gamma$ becomes assymtotic. Plugging in a $v$ close to $c$ produces a near 0 denominator, and thus $\gamma \rightarrow \infty.$ Thus it's fair to claim that $\gamma$ increases at an increasing rate, and thus time dilation and length contraction will as well. Finally, let us look at velocity.

if $t' = \gamma(t-\frac{xv}{c^2})$, then $t = \frac{t'}{\gamma} + \frac{xv}{c^2}$.

Plugging this in, $x'=\gamma(x-vt)=\gamma(x-\frac{t'v}{\gamma}-\frac{xv^2}{c^2})$. After some further simplification, you will arrive at the conclusion that $\gamma$ is not necessary for this velocity equation. That having been said, there is still a $\frac{v^2}{c^2}$ term in the velocity equation. As such there are still relativistic effects even without the Lorentz coefficient.