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In general relativity, gravity is seen as the result of spacetime curvature. This gives a automatically encodes the equivalence principle, explaining why all objects fall at the same rate and removes the coincidence that inertial and passive gravitational mass are the same.

But I think there is another coincidence involving mass which isn't explained by this:

The motion of a particle in curved spacetime and subject to a force field (e.g. a charge in EM) is given by:

$$ m_{I} \frac{Du_μ}{d\tau}​=qF_{μν}​u^ν .$$

We see a mass term appear as a "resistance to being deflected off a geodesic trajectory".

While in the Einstein field equations, the stress energy tensor couples to the 4-momentum of the particles:

$$G_{\mu\nu} = kT_{\mu\nu} $$

Where: $$T_{\mu\nu} = T^{EM Field}_{\mu\nu} + m_{A} u_\mu u_\nu.$$

So we see a mass term as causing curvature of spacetime.

The equivalence principle explains why inertial and passive gravitational mass, but I see no reason why we would expect ${m_I = m_A}$.

Why is this the case? What happens in a metric theory of gravity in which these two do not coincide?

Is it logically possible for an object to fall like ordinary matter (mass-independent geodesic motion) while curving spacetime more or less than its inertial mass would suggest - where inertial mass is measured by interactions with other forces?

Attempted solution:

I assume it would violate local 4-momentum conservation on some level.

If you have two like-charged particles initially at rest in some IRF, with the same active mass but different inertial masses, you could put them at a distance so that their EM force of one particle matches the acceleration it feels inwards. But since the other particle has a different charge to inertial mass ratio, it will start accelerating, so the system's total momentum has changed.

I am struggling to make this rigorous. I would also expect violation of momentum conservation be a consequence of broken translation symmetry, but I can't see how having two distinct types of mass does this!

I have found similar questions on this site, but they look to be focused on the equivalence principle - which I think only explains equality of passive and inertial mass (not active mass).

Is there a fundamental reason why gravitational mass is the same as inertial mass?

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    $\begingroup$ Your expressions are wrong. The correct equations do not involve $m_I$, and you are ignoring that $u^\nu$ heavily modify whatever appearance of mass that you think you have. $\endgroup$ Commented yesterday
  • $\begingroup$ @naturallyInconsistent I have taken the relativistic Lorentz force law from here: en.wikipedia.org/wiki/… - I don't think equation is wrong? m appears in the same place as $m_I$ which I denote as such purely to distinguish from the m appearing in the stress energy tensor. What do you mean my $u^\nu$ modifies the apperance of mass? $\endgroup$ Commented yesterday
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    $\begingroup$ I am not denying that plenty of people express the GR Lorentz force law that way, because people are extremely glued to Newton's 2nd Law, even when it really ought to be retired from modern physics. The correct Equations of Motion have 4-momentum derivative in it; what is really happening is that EVERY appearance of mass in Newtonian mechanics is actually coming from a sensitivity to absolute amount of total energy, of which in Newtonian mechanics the rest mass energy contribution dominates. That's why in the EFE mass ought to never appear, but rather the stress-energy-momentum tensor appears. $\endgroup$ Commented yesterday
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    $\begingroup$ Therein also lies the correct resolution to the fundamental "coincidence"; all the old arguments about the coincidence that gravitational mass and inertial mass are the same, is resolved in GR by them both being the measure of one single thing, the total stress-energy-momentum tensor, which in Newtonian mechanics is dominated by rest mass energy. It is from when you understand that it is really just energy, then you will see that there is no more confusion about "coincidences". $\endgroup$ Commented yesterday
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    $\begingroup$ Your stress-energy-momentum tensor also doesn’t account for the pressure terms that appear in realistic circumstances. It is valid for an electromagnetic field plus a dust, but not for a star for example. $\endgroup$ Commented yesterday

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I can show that your equations are inconsistent unless $m_I=m_A$. Indeed your intuition that energy-momentum is not conserved otherwise is correct. In order to complete your Einstein-Maxwell-dust system we have to give both Einstein and Maxwell's equations

$$G_{\mu\nu}=T_{\mu\nu}$$ $$\nabla_\mu F^{\mu\nu}=-J^\nu$$

with the relevant stress-energy tensor and electromagnetic currents

$$T_{\mu\nu}=T_{\mu\nu}^{EM}+m_A u_\mu u_\nu$$ $$J_\mu=q u_\mu$$ However the equations of motion for your dust fluid are not independent from the equations above, they are given by the Bianchi identity applied to the Einstein field equations above. Indeed we have

$$\nabla_\mu T^{\mu\nu}=0$$ which is equivalently $$\nabla_\mu (m_A u^\mu u^\nu)=-\nabla_\mu T^{\mu\nu}_{EM}$$ Maxwell's equations then imply that

$$\nabla_\mu T^{\mu\nu}_{EM}=-q F^{\mu\nu} u_\mu$$ If we reinsert this in the equation above and project orthogonally to $u^\mu$ (by contracting both sides with the projector $h^\mu_{\ \,\nu}=\delta^\mu_\nu+u^\mu u_\nu$) we get

$$m_Au^\mu\nabla_\mu u^\nu=qF^{\mu\nu}u_\mu$$ which is simply the Lorentz equation you wrote with $m_A=m_I$. This equation comes from the conservation of energy-momentum and is only consistent if inertial mass is equal to active gravitational mass.

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  • $\begingroup$ Thanks, really nice answer - I haven't seen the trick involving the projection onto the orthogonal compliment of u before. So summarising, the total stress-energy tensor must be divergence free to locally conserve 4-momentum. And you have explicitly shown for EM & dust in curved spacetime, 4-momentum in the stress energy tensor must match the 4-momentum in the Lorentz law, otherwise 4-momentum is transferred between dust and EM field at different rates. $\endgroup$ Commented 17 hours ago
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I see no reason why we would expect $m_I=m_A$.

Why is this the case?

It isn’t the case. In GR, $m_A$ does not exist. The source of gravity is not some active gravitational mass, a scalar, but the full stress energy tensor.

However, in the weak field limit GR reduces to Newtonian gravity. So the question can be asked of Newtonian gravity, and the result can be applied to GR specifically in the Newtonian limit.

For Newtonian gravity the equality of $m_I$ and $m_A$ is buried in the gravitational constant $G$. The most that you can actually say physically is that they are proportional. This is in the linearity of Newtonian gravity. And in the Newtonian limit GR is also linear.

Violations of $m_A=m_I$ would show up as variations of $G$. It could vary over space or time or as a function of mass or some other quantity. There are alternative theories that allow for such scalar fields, but so far there is no data that supports them over GR.

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