# Does general relativity (GR) solve mass-equivalence problem or just renames it?

It is usually said that GR solved problem of equivalence of gravitational and inertial mass. Yet, my objection is:

OK, there is no gravitational mass in GR, but Einstein had to introduce mass as "spacetime curving constant".

So the question

A: Why is the gravitational and inertial mass the same?

is replaced by the question:

B: Why is inertial mass and spacetime-curving constant the same?

I am honestly unable to clearly solve my objection, I have troubles thinking abstract enough. Is or is not my objection correct?

Is is possible to imagine a universe, where, for example, electron has two different masses: one (inertial) appears in EM interactions (and other non-gravitational interactions), other appears in spacetime-curving equations.

• GR does not solve the the question "Is gravitational mass equal to inertial mass?", it assumes that they are equal. en.wikipedia.org/wiki/Equivalence_principle – Natanael Feb 22 at 9:31
• It certainly depends on the point of view: GR says bodies move on straight lines, so there is no need at all of gravitation and consequently no need of gravitational mass. Then the assumption does not make sense: one cannot assume A being equivalent to B if B does not exist. – F. Jatpil Feb 22 at 9:36
• GR says how matter and energy curve space-time. The curved space-time tells matter and energy how to move, this is what we call gravity. How the curved space-time tells matter to move, depends on the matters gravitational mass. – Natanael Feb 22 at 9:42
• I am almost sure you are wrong. Once the S-T is curved it tells nothing to a body. Body moves straight (on geodesics) in that curved S-T, which is a natural generalization of first Newton law. GR was build having in head (assuming) equivalence principle, but once it is completed and written down on the paper the equivalence principle is its (i.e. GR's) consequence. – F. Jatpil Feb 22 at 9:49
• Rather than simply renaming, GR makes a stronger statement inspired by the weak equivalence principle and tells us that the effects of gravity are purely determined by our deviation from the local inertial frame. – Dvij Mankad Feb 22 at 9:55

The "why" questions about physical laws are rarely questions that can actually be answered by physics. I can just tell you that Einstein gravity was constructed so that there is no difference between inertial and gravitational mass. You can see this from the Einstein field equations which read $$R^{\mu\nu} - \frac{1}{2} R g^{\mu\nu} = \frac{8 \pi G}{c^2} T^{\mu\nu}$$ where the left hand side corresponds to the geometry of the space-time, and the right-hand side to the matter-energy sources.
For instance, for a coherent cloud of dust, we would just have $$T^{\mu\nu} = \rho u^\mu u^\nu$$, where $$\rho$$ is the matter density and $$u^\mu$$ the four-velocity field. (Roughly speaking, in the Newtonian limit only the $$\mu\nu=00$$ component is relevant and $$\rho u^0 u^0$$ is the matter density.) So the field equations tell you "space-time geometry (gravitational field) is sourced by matter-energy".
But now consider the following. The left-hand side automatically fulfills $$\left(R^{\mu\nu} - \frac{1}{2} R g^{\mu\nu}\right)_{;\nu} = 0$$ where $$;\nu$$ means a space-time-covariant gradient and I use the Einstein summation convention. That is, the covariant divergence of the left-hand side of the Einstein equations automatically vanishes. So, if the Einstein equations are fulfilled, this also implies $$T^{\mu\nu}_{\;\;\;;\nu} = 0$$ If you aply this to dust and combine it with the conservation of matter (continuity equation) $$(\rho u^\mu)_{;\mu} = 0$$, you just get the geodesic equation $$u^\mu_{;\nu}u^\nu = 0$$! (In the Newtonian limit this just gives you that all particles experience a universal gravitational acceleration $$\vec{a}$$.)