What does the following mean with respect to special relativity?
$$\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$
Answer
Colin McFaul's answer is accurate. If you're wondering why it's in that weird form, it has to do with a fundamental postulate of special relativity.
In 2D, say you're given the endpoint of a line that you know starts at the origin, and you want to find its length. According to the diagram below, if you're given a vector in $xy$ coordinates, $(a, b)$ (represented by the red line), you can look at it in an arbitrary coordinate system by rotating your head, but physically, the length of the arrow must remain unchanged. If the coordinates in this other coordinate system are $(c,d)$, we can phrase this with the pythagorean theorem: $a^2+b^2=c^2+d^2=\mbox{distance}$
If we have a special coordinate system drawn, $x'y'$, then $d=0$ and we just have $a^2+b^2=c^2$. In this new frame, its length is just the position on one axis.
In relativity, we have one axis $ct$ (where $c$ is the speed of light), and another axis $x$. $ct$ and $x$ are both measured in meters. Instead of a Euclidean distance relation, the factor that remains unchanged from "rotations" is $(ct)^2-x^2$. This is the crux of relativity. It means you can "rotate" space into time and vice versa.
In one frame, the particle may have a constant velocity, so $x=v t$ and the law is conservation of $(ct)^2-(v t)^2=(c^2-v^2)t^2$. If we choose a primed reference frame so that the speed is zero, then we must have $(c^2-v^2)t^2=(ct')^2$. With some algebra we find that this equation implies that $$t=t' \frac{1}{\sqrt{1-\left( \frac{v}{c}\right)^2 }}$$
This is written as $t=\gamma t'$. It all stems from the invariance of $(ct)^2-x^2$. You can do analogous things when talking about $x$ and $x'$.
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