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In Newtonian gravity, as such, there are three masses: the active gravitational mass (the mass that generates the gravitational field), the passive gravitational mass (the mass that responds to the gravitational field), and the inertial mass. It is a (more or less unexplained) fact in Newtonian gravity that all these masses are actually equal. Now, in the Einsteinian gravity, the beautiful idea of understanding the effects of gravity as those of the acceleration of a frame (with respect to a nearby freely falling frame) corresponds on the equality of the passive gravitational mass and the inertial mass. I am curious as to whether any such nice picture can be associated with the equality between the active and the passive gravitational masses in Einsteinian gravity.

An elementary explanation in Newtonian gravity for the equality between the active gravitational mass and the passive gravitational mass is based on Newton's third law of motion--that goes as follows in the familiar notations: $$|F_{12}|=G\dfrac{{m_1}^\text{passive}m_2^\text{active}}{r^2}, |F_{21}|=G\dfrac{{m_2}^\text{passive}m_1^\text{active}}{r^2}$$ For the equality of these magnitudes, it is required that $$\dfrac{{m_1}^\text{passive}}{m_1^\text{active}}=\dfrac{{m_2}^{\text{passive}}}{m_2^\text{active}}$$ Since the particles $1$ and $2$ are completely independent and arbitrary, the ratio $\dfrac{m^\text{passive}}{m^\text{active}}$ is a universal constant--and thus, in an appropriate choice of units, the active and passive gravitational masses can be made equal without any loss of generality.

But as I said, I don't know of any general relativistic (i.e. more or less geometric) explanation for this equality. So, my question is: Is there any general relativistic/geometric explanation/interpretation to the equality between the active gravitational mass and the passive gravitational mass?

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  • $\begingroup$ There is no passive gravitational mass. In both theories gravity between two bodies is a two-body problem. The difference is that in the classical theory this problem is easily solvable while the "beautiful" general covariance theory provides no analytical solution for even the simplest of the cases. $\endgroup$ – safesphere Jun 6 '18 at 17:27
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    $\begingroup$ @safesphere Can you explain what do you mean by ``there is no passive gravitational mass''? As such in standard understanding, there is--of course--a notion of the passive gravitational mass, the mass due to which a particle responds to a gravitational field, i.e. $\vec{F}=m^\text{passive}\vec{g}$. $\endgroup$ – Dvij Mankad Jun 6 '18 at 17:30
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    $\begingroup$ It's a made up concept that follows from the flawed analysis of the problem "step by step". You introduce this concept to simplify a two-body problem. Thus this concept is a mere artifact of your simplification. It is even clearer in general relativity: both masses curve spacetime and them both move in this spacetime. Nothing passive actually happens in reality, but only hypothetically in your imagination of the simplified problem. $\endgroup$ – safesphere Jun 6 '18 at 17:57
  • $\begingroup$ I share the opinion of safesphere here, if you still think there is such thing as passive gravitational mass, may I ask you for a reference where it is defined fully and a more elaborated explanation can be found? As far as my understanding goes, Newton's laws for gravitation are symmetric, it is always what you called "active" mass that appears in the equations. $\endgroup$ – ohneVal Jun 6 '18 at 18:03
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    $\begingroup$ In GR, the concept of passive mass does not arise, only that of active mass—the source of the field.... Newtonian passive gravitational mass (that which is pulled by the field) goes into banishment along with the ether, etc. Inertial mass survives in non-gravitational contexts only... Wolfgang Rindler, "Relativity: Special, General and Cosmological" pp 17, 21 (2ed). $\endgroup$ – sammy gerbil Jun 6 '18 at 19:18

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