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I was reading a recent article on Mach's Principle. In it, the author talks about inertia in an empty universe. I'll quote some lines from the article:

Imagine a single body in an otherwise empty universe. In the absence of any forces, (Newton's second law gives): $$m\mathbf{a} = 0$$ What does this equation imply? Following Newton we would conclude from that $\mathbf{a} = 0$, that is, the body moves with uniform velocity. But we now no longer have a background against which to measure velocities. Thus $\mathbf{a} = 0$ has no operational significance. Rather, the lack of any tangible background for measuring motion suggests that $\mathbf{a}$ should be completely indeterminate. And it is not difficult to see that such a conclusion follows naturally, provided we come to the remarkable conclusion that $$m = 0$$ In other words, the measure of inertia depends on the existence of the background in such a way that in the absence of the background the measure vanishes!

I don't see how the argument is complete. For example, in an empty universe, how is it possible to assign a value of 0 to a force? And wouldn't the existence of mathematics in such an empty universe be questionable?

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    $\begingroup$ "But we now no longer have a background against which to measure velocities." You don't need a background to measure acceleration, it can be performed in a closed lab (like an elevator). Since this is a hypothetical empty universe, you can make the lab infinitesimally small. $\endgroup$
    – Marton Trencseni
    Jul 31, 2011 at 21:38
  • $\begingroup$ FYI philosophy questions are off topic here - though I don't think this one is actually about philosophy, just the mathematical basis of some rather speculative physics. $\endgroup$
    – David Z
    Jul 31, 2011 at 22:25
  • $\begingroup$ related: physics.stackexchange.com/q/5483/4552 $\endgroup$
    – user4552
    Jun 28, 2013 at 20:20

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The author appears to be assuming that $a$ is indeterminate in an empty universe. That assumption fits in nicely with some people's philosophical preconceptions (including Mach's), but of course we don't know it to be true. In particular, in general relativity, one can have an empty universe described by good old special-relativistic Minkowski spacetime. In such a universe, a test particle would "know" whether it was accelerating or not. In this sense, general relativity does not appear to be Machian.

Something like Mach's principle is in the end an axiom, which one can choose to assume or not depending on one's taste.

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    $\begingroup$ Mach's principle is not really an axiom, it's just a philosophical worldview because it's only vague qualitive statement. It states that inertia of a local body depends on the rest of the universe but as soon as one tries to make this statement precise one finds that the theory is inconsistent (either mathematically or with experiment). The idea is inspiring and has some relation to nature via frame dragging but that's all there is to it, IMHO. $\endgroup$
    – Marek
    Jul 31, 2011 at 14:23
  • $\begingroup$ @Marek: I would consider the following to be a pretty reasonable way of formulating Mach's principle precisely: Brans-Dicke gravity holds, with a value of the $\omega$ parameter that is of order unity. See physics.stackexchange.com/q/5483/4552 $\endgroup$
    – user4552
    Jun 28, 2013 at 20:25
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I think it is straight forward: assuming a lack of a force means no acceleration, so the equation $m \cdot a = 0$ is satisfied. There is no information or assumptions about the mass, nor about the metric - so you may not conclude that $m=0$ without further information.

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First of all, i apologize for my english. The equation F=ma is correct only from the point of view of an inertial system. What an inertial system actually is: any system which is in agree with the statement that if no force is acting on a body (your example is the case) then the body moves in a straight line and whit constant velocity. So, you are right, there is always an observer wich measures a nonvanishing acceleration (no matter you are in an empty universe or not), but all this observers are not inertial frames, so they have no right to say F=ma. Hope i have been of help

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The argument fails when he attempts to prove that mass=0.
IMO the particle preserves the mass.

The author Narlikar is correct here:

in the absence of the background the measure vanishes!

My bold measure.
The eq. F=m.a is a relation to measure any of "force, mass or acceleration" when at least 2 of the 3 terms are present.
In Static we have forces and we have no acceleration and the mass is still present. Mass is a property of matter and is not defined with the help of that equation, although measured.
IMO the particle preserves the mass.
First attempt:
If the universe contains only 1 particle we have no means to define an external referential (except in a mathematical sence wich is useless), neither to exert a force. The center of mass of the system is always centered with that single particle that is to say: the particle is immobile.

Second attempt:
Lets try with a universe with two particles where 1 of them is reducing its mass until 0:
The gravitational interaction is present, a center of mass frame (COM) is present and both particles contribute to the overall mass. We can attribute mass to each one as a part of the overall mass.
As the mass of particle 1 is evolving to 0 the other particle is migrating to the COM. The remaining particle retains its mass and stay immobile irt the COM (centered in itself) in the limit of mass=0.
There is a profound reason to preserve mass: the force and the acceleration are consequence of a interaction, and it takes at least two objects to interact (and a time basis). The mass is an intrinsic property of matter.

The COM velocity is zero: $V_{CM}=\frac{v_1m_1+v_2m_2}{m_1+m_2}=0=v'_1\frac{m_1}{m_1}$ and $m_1$ <> 0;

Your sentence:

existence of mathematics in such an empty universe be questionable?

Math is a human mental construct. Only.

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And wouldn't the existence of mathematics in such an empty universe be questionable?

I don't see why. A implies B and A together imply B whether the universe contains matter or not. $3$ (or, if you prefer, $\mathtt{plus.one}(\mathtt{plus.one}(\mathtt{plus.one}(0)))$) doesn't divide $5$ regardless of whether the universe exists or not. And so on.

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