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I am currently reading Carroll's GR book p. 48. There he says that

If you like, $m_g/m_i$ can be thought of as the "gravitational charge" of the body

with the gravitational mass $m_g$ and the inertial mass $m_i$. Why? Wouldn't it make more sense to say that just $m_g$ is the gravitational charge? After all, we don't call $q/m_i$ is the electric charge of a body.

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  • $\begingroup$ $q/m$ is called the specific charge. $\endgroup$
    – Farcher
    Commented Sep 21, 2021 at 8:02

1 Answer 1

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Imagine there is a uniform electric field $\vec{E}$ then for a particle with charge $q$ the force acting on that particle is $\vec{F}=q\vec{E}$. So the acceleration will be $$\vec{F}=\frac{q\vec{E}}{m}$$ Notice that if we double the charge the acceleration will also be doubled.

Imagine there is a uniform newtonian gravitaional field $\vec{g}$ then for a particle with inertial mass $m_i$ and gravitational mass $m_g$ the force acting on that particle is $\vec{F}=m_g\vec{g}$. So the acceleration will be $$\vec{F}=\frac{m_g\vec{g}}{m_i}=\vec{g}\text{ since }m_i=m_g$$ Notice that if we double the $m_g$ the acceleration will not be doubled.

In that sense, all particles have the same gravitational charge which is +1. The main difference compared to the electric field case is that q and m are independent, but $m_i$ and $m_g$ are not independent.

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