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I'm certain I'm just under a grave misapprehension. I really just don't understand how one can conclude that inertial mass is equal to gravitational mass. It seems to me that we've simply scaled our "fundamental" constants so that it just so happens that we can set $m_i=m_g$.

Here's exactly what I mean. Consider a point-particle of inertial mass $m_i$ and gravitational mass $m_g$. What this means is, under the classical theory of mechanics and gravitation,

$$m_i\mathbf{a}=m_g\nabla\phi$$

where, as usual, $\mathbf{a}$ is the measurable acceleration, and $\phi$ is some scalar field. Suppose the weak equivalence principle is false, and that $m_g=3m_i$. Then, by plugging it in,

$$\mathbf{a}=3\nabla\phi$$

Now we aren't able to say that the gradient of the scalar potential we've previously defined is equal to the measurable acceleration, but hey, if we just redefine our scalar potential $\phi$ to be $\phi/3$, then we can stick with our previous definition of the scalar field and everything will work out "fine".

In a nutshell, I think that it's only reasonable to say that there is a linear relationship between the gravitational force and a certain "gravitational-mass" parameter. My question is, how is our definition of the gravitational constant devoid of the possibility of the inertial-mass and gravitational mass being related by arbitrary and immeasurable constant, rather than just $1$? Please tell me where I am going wrong.

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I'd say you have a grave apprehension of the equivalence principle: you've thought about it thoroughly. And what you say is absolutely true: the independence of the motion of an interaction-free point mass in spacetime is independent of its mass (in the small mass limit[1]) is the essence of the equivalence principle and, for its fulfilling, it only requires the inertial mass $m_i$ should be proportional to the "gravitational coupling" $m_g$. With a nonunity scaling constant, you'll still get the same equivalence classes of inertial motion states that cannot be told apart from one another by any experiment wholly from within one of the inertial frames, as described by Galileo's allegory of Salviati's Ship: general relativity simply describes for us the equivalence classes of "Saliviati Ship" motions and their relatively accelerated (in the sense of having certain accelerometer readings) cosets given a stress-energy distribution and boundary conditions. None of these conclusions change if you add a scaling constant between $m_i$ and $m_g$. It's simply that, given we find the equivalence principle to be true, we choose to make our unit system as simple as we can by setting the scaling constant to unity.

[1]: i.e. so that its own contribution to the stress-energy tensor can be taken to be negligible so that its presence does not change the spacetime geometry.


Question from OP

Why is it then, that I hear so many times people infer that because all matter responds to the curvature of space-time the same (i.e. equal acceleration), $m_i=m_g$?

Because it's a hypothesis; $m_i=m_g$ is a sufficient condition, but it's stronger than needed to explain that "gravity's effect is independent of composition". Often hypotheses are made in physics that are stronger than really needed to explain experimental results, usually on the grounds of simplicity or the so-called "Occam's Razor". General Relativity is the classic example: it explains Galileo's experiment perfectly and it gets rid of the weird "conspiracy" of a gravity force that is complicated or "intelligent" enough to work out what the inertial mass $m_i$ of a body is and accelerate with a force proportional to $m_i$. But it is only one possible explanation: we (or Einstein) chose it for its amazingly simplicity, not for its uniqueness. It reduces the theory of gravitation to a small clutch of axioms: (1) Spacetime is a metrical manifold, (2) a body feeling no interaction follows geodesics in this manifold and (3) the Einstein field equations and relevant boundary conditions define the metric that arises from matter distributions. Misner, Thorne and Wheeler in their book "Gravitation" do the same (or rather, they replicate Élie Cartan's treatment) for Newtonian Gravity, i.e. show how to describe Newtonian gravity as a local, geometrical theory and it becomes monstrously complicated compared to GTR in this form.

However once you choose the geometric hypothesis - that gravity is not a force and freefall motion is inertial motion - then the unity constant is indeed fixed as follows. On the one hand, there is no interaction between an observer in freefall and anything else; it feels no force whatsoever. Therefore, the notion of a "coupling constant" to the gravitational field has no meaning and there is no such thing as $m_g$. But suppose we have a clutch of objects "resting" on the surface of a planet, say Earth to be concrete. Then each, through the relevant solid state physics (certain kinds of stuff can't pass through other kinds of stuff), are constrained to accelerate at the same rate relative to the local freefalling frame. So now ask, "what force on each of these objects is needed from the ground to make them accelerate thus?". By Newton's second law, it is $m_i\,g$. Notice how the notion of $m_g$ doesn't even enter the discussion. The situation is altogether analogous to the situation where you put a bunch of different mass boxes of things on your car's back seat and then accelerate off at acceleration $g$. You conclude, from Newton's second law, that each object must be being pushed by the back seat forwards with a force $m_i\,g$.

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  • $\begingroup$ Why is it then, that I hear so many times people infer that because all matter responds to the curvature of space-time the same (i.e. equal acceleration), $m_i=m_g$? $\endgroup$ – Arturo don Juan Oct 17 '15 at 2:35
  • $\begingroup$ @ArturodonJuan See my updated discussion. I've just realized that I've banged on a bit more than I meant to, so hopefully I am still understandable. $\endgroup$ – WetSavannaAnimal Oct 17 '15 at 3:37

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