Does relative velocity time dilation occur at any relative velocity?

Question

The reason I ask this question is due to the fact that every example I've found is presented such that the relative velocity is "near the speed if light", or that it happens as the object " approaches the speed of light". If I've understood anything that I've learned recently, wouldn't time dilation and length contraction occur at any relative velocity? Perhaps it's a layman level of question but you can't fault me for trying to transcend that level of knowledge.

I've assumed thusfar that such high velocities are used in examples because the effect is easier to observe and explain in those regimes.

Answer 1

What the commenters are referring to is that the Lorentz factor $\gamma$ really only deviates significantly from $1$ at speeds that are fairly close to $c$, the speed of light.

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

With $v$ the relative speed between observers.

The Lorentz factor pops up in relativistic calculations all the time and is a measure of how much the relative speed affects the magnitude of time or momentum.

Just try a few values of $v$:

For $v=0.1 c$, $\gamma=1.005$

For $v=0.5 c$, $\gamma=1.155$

For $v=0.9 c$, $\gamma=2.294$

So even for a very respectable speed of $v=0.1c \approx 30000 \text{ km/s}$, $\gamma$ is still very close to $1$ and has little effect on observables like time or momentum. For now, significant relativistic effects like time dilation remain outside the domain of personal human experience.

Answer 2

The basic answer is yes, you are understanding things right.

The "low-speed effect" is that if a ruler of size $2L$ with a set of clocks is moving past you at low speed $v$, and the clock at the center shows time $t$, and those clocks are in sync in the moving frame, then they appear out-of-sync in your frame: if the moving ruler measures $x$ relative to the center (with the positive sign being in the forwards direction from your persective) then the clocks on the ruler measure $t - v x / c^2.$ In other words, the clock at the "front" of the ruler shows time $t - vL/c^2$ and the clock at the "back" shows time $t + vL/c^2.$ (These are actually relativistically valid too, as long as $2L$ is the proper length of the ruler.)

This "desynchrony" effect, and the "classical" effect that their coordinates travel through yours approximately like $x' = x - v t,$ (not relativistically valid!) are the only effects of the Lorentz transform which are linear in $v.$ Everything else (length contraction, time dilation) is an effect of a lot of these little desynchronies adding up.

When they add up, they build up the Lorentz transform, which can be written concisely by defining $w = c t,\;\beta = v/c,\;\gamma = 1/\sqrt{1 - \beta^2},$ as $$\begin{align}w' =&~ \gamma~(w - \beta~x)\\ x' =&~ \gamma~(x - \beta~w)\\ y' =&~ y\\ z' =&~ z\end{align}$$(assuming both coordinate systems have the same origins.) The coefficient $\gamma$ is the origin of the time dilation and length contraction.

As you can see, for small velocities $\gamma \approx 1 + \frac 12 \beta^2$ and for small $v$ these effects disappear; then they reappear by summing lots of the linear-effects.

So, the length-contraction and time-dilation effects are not tied to any particular "critical velocity" directly; they just disappear in the noise threshold (where you measure effects of size ($1 \pm \epsilon$) for lengths and durations) unless $\beta^2 > \epsilon,$ which happens before you think because of the squaring. Like, if you want to calculate a trip to a nearby star at $\beta = 0.1$ then you don't care about the length-contraction or time-dilation effects unless you're measuring things to one part in a hundred, because that is $\beta^2.$ So you can just calculate for example that the kinetic energy you need for a 1000-ton spaceship to make the journey is something like $4\cdot10^{20}\text{ J}$ of energy, classically, and you can know that the correction is only on the order of 1% of that energy.

There is no finite cutoff, just that it gradually becomes too weak of a correction to measure.