# Is acceleration relative?

A while back in my Dynamics & Relativity lectures my lecturer mentioned that an object need not be accelerating relative to anything - he said it makes sense for an object to just be accelerating. Now, to me (and to my supervisor for this course), this sounds a little weird. An object's velocity is relative to the frame you're observing it in/from (right?), so where does this 'relativeness' go when we differentiate?

I am pretty sure that I'm just confused here or that I've somehow misheard/misunderstood the lecturer, so can someone please explain this to me.

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I find the phrase "acceleration need not be relative anything" to be awkward, but I can see where it comes from.

For the moment restrict our consideration the Galilean relativity (just to keep the math simple). Consider two frames of reference one ($S$) in which the body is at rest and another ($S'$) in which it moves with velocity $\vec{v'_i} = \vec{u} = u \hat{z}$.

So we have the initial velocity of the body in frame $S$ as $v_i = 0$, and $v' = v + u \hat{z}$

Now assume that the the body accelerates from time $t$ at acceleration $\vec{a} = a \hat{Z}$ resulting in a velocity in frame $S$ of $\vec{v_f} = a t \hat{z}$.

Compute the final velocity in frame $S'$ as $v'_f = v_f + u \hat{z} = (u + a t)\hat{z}$, and from that the acceleration in the primed frame as $a' = a$.

So the acceleration is the same in all frames (you can check the cases for $a \not{\parallel} u$ yourself), and it is reasonable to say that accelerations are not relative anything.

All of this is a consequence of the simple form of the transformation between frames:

$$\vec{x'} = \vec{x_0} + \vec{u} t$$ $$t' = t$$

Here the transformation between frames is more complicated, and the math is much more complicated resulting in observers in different frames seeing different accelerations, but they will all agree on the acceleration as measured in the body's own frame. In my opinion "the acceleration need not be relative" risks causing unnecessary confusion on these points. The magnitude and direction measured will measure in depend on the frame of the observer, which is often what is meant when people say "it's relative".

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The very simple answer to your question is that yes, acceleration is relative. Although Newtonian physics is written with respect to a preferred reference frame ("the fixed stars" as Newton said), general relativity treats all reference frames, accelerating and not, on equal footing. Newton's frame, or the "inertial frame", just means "not accelerating RELATIVE to the bulk of matter in our universe". If, locally for instance, there was enough mass moving in some direction then there would be frame dragging effects whereby the local 'inertial' reference frame (where Newton's Laws are valid) would seem to shift relative to more distant 'fixed' matter. This is basically the content of Mach's principle, which was one of Einstein's chief motives for searching for a generally covariant theory of gravitation.

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I'll sort of cheat and neglect the relativity part--instead, I'm going to focus on your mathematical confusion regarding differentiation.

What you've forgotten is that we have a constraint. Your argument is thus, right: $$\text{if } \vec v_{b}=\vec v_{a}+\vec v_{b,a}, \text{ then } \vec a_{b}=\vec a_{a}+\vec a_{b,a} \text{ by differentiation}$$ Where $b$ and $a$ are frames of reference, and $b,a$ denotes the acceleration/velocity of the frames relative to each other.

Aah, but we have a constraint: $\vec F=\frac{\rm d\vec p}{\rm d t}=m\frac{\rm d\vec v}{\rm d t}=m\vec{a} _\text{ (for constant mass)}$.

This constraint makes little difference when we switch between frames with constant relative velocities, since the derivative of velocity stays the same. But, the moment we try to switch to an accelerating frame, things get icky. We get:

$$m\vec a_{b}=m\vec a_{a}+m\vec a_{b,a}\implies \vec F_b=\vec F_a+m\vec a_{b,a}$$

Looks OK, doesn't it? We could write $m\vec a_{b,a}\to\vec F_{b,a}$, and the equation would be handy-dandy--it would show that force is relative as well. Right?

Wrong. By our assumption, acceleration and velocity are relative quantities. To measure a relative quantity, one must have a reference frame. In contrast, force is something you can measure without needing a frame-- A spring balance suffices.

This equation shows that apparent force varies with your reference frame. You probably know this, but this comes from the presence of a "psuedoforce"--the $m_{b,a}$ term. It's not a real force, but it appears whenever we try to apply Newton's laws to a noninertial{*} frame. Since it's part of the apparent force, it can be measured. Since it crops up when you change frames, you will always be able to detect that a frame change occurred. This is in contrast to switching between inertial frames--unless you have a fixed object outside, you can't tell the difference.

So basically, it's not a methematical issue but an issue with what can and can't be measured inside a frame

Actually, all of this comes from the fact that Newton's laws are only applicable in an inertial frame. When asked "which of Newton's laws is the most important?" most people would say the second and/or third law. The first law is always counted as a result of the second law. This is actually false-there is no privileged law of the three-- and they are independant. The function of the first law is to define the realm of applicability of the three. In a noninertial frame, the first law does not hold(since an object at rest still gets accelerated), so the other two do not hold either. The pseudoforce is just a way of cheating the laws into working.

So, if we have a bunch of laws that only work in a non-accelerating frame, that means that "non-accelerating" has an absolute meaning. Thus all acceleration is absolute.

In short: the "relativeness" was not there in the first place. It just appears when we take a special case of $a_{frame}=0$--useful in itself, but not generalisable.

* inertial-->constant velocity no gravity, noninertial-->acceleration and/or gravity

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You missed an opportunity to mention the Coriolis force :) –  Bernhard Mar 26 '12 at 5:41
@Bernhard: I hate the coriolis force--yes, it's another thing that can be said to pop out like a jack-in-the-box when you differentiate equations, but it's annoying. I prefer good 'ol centrifugal "force" :P –  Manishearth Mar 26 '12 at 5:58