In mechanics there is a relative concept of "inertial frame": frame A is inertial with respect to frame B if A moves uniformly with respect B. That concept is easy to understand.
There also seems to be an absolute concept of "inertial frame". I keep reading things like "A is an inertial frame", without specifying with respect to which other frame B. Every time I read that kind of statements I get stuck. I cannot see how A can be "inertial". I can only see how it can be "inertial with respect to B".
Related to this, I keep reading things like "the solar system is accelerating" or "an object is moving" (for example here and here). Those statements I simply can't understand, unless they specify with respect to what frame (or object) that movement is defined.
I suspect my inability to understand the absolute concept of inertial frame is related to my inability to understand the statement "an object is moving". I only keep wondering: "with respect to what?".
So, my question is: can you really say "A is inertial" or "B is moving" in an absolute sense? (i.e without adding "with respect to C"). If so, how is that interpreted?
Best Answer
You really have two questions here
The first one is harder than the second.
Identifying inertial frame
We define an inertial frame as one in which the laws of physics take on their usual (simple) form, and identify non-inertial frame by the need to impose (pseudo-)forces like the Coriolis Force that depend explicitly on place or direction to explain the behavior of objects.
The definition feels a little circular, but it does give rise to a simple method of finding additional inertial frames once you have identified a single one: any frame moving at constant velocity relative an inertial frame is also inertial.
Acceleration not relative?
Allow that in some inertial frame $S$ we have identified all the forces acting on a body, applied newton's law and computed the acceleration $a = F_{net}/m$ of the body. The equation of motion of the body is (using one dimension only to simplify the mathematical expressions) $$ x(t) = x_0 + v_0 t + \frac{1}{2} a t^2 \,. $$
Now we observe the same action from a second frame $S'$ moving at velocity $u$ as seen from frame $S$. For simplicity (again) we'll assume that at $t = t' = 0$ the origins of $S$ and $S'$ coincided. That means that we can compute positions as measured in either frame from the position as measured in the other by $$ \begin{align*} x' &= x - ut & x &= x' + ut \,. \end{align*} $$ In particular it means that $x_0 = x'_0$ We also know that the initial velocity of our object in frame $S'$ was $v'_0 = v_0 - u$.
Taking these facts together we write the equation of motion in frame $S'$ by transforming that in $$ \begin{align*} x' &= x - u t \\ &= (x_0 + v_0 t + \frac{1}{2} a t^2) - u t \\ &= x'_0 + (v_0 - u) t + \frac{1}{2} a t^2\\ &= x'_0 + v'_0 t + \frac{1}{2} a t^2 \,. \end{align*} $$ Two things are immediately apparent: (1) that the algebraic form of the equation of motion is identical in the primed frame as that in the unprimed frame and (2) that the acceleration is also the same.
If you are of a particularly skeptical bent you may not quite believe this, but try choosing a set of values for $x_0$, $v_0$ and $a$; tabulating the motion for a while, converting the tabulation to the primed frame and computing the acceleration in the primed frame from the table.