Natural Science Forum / Physics / Relativity / November 2005

Both. Sort of. One can in principle distinguish "active gravitational
mass" and "passive gravitational mass". The former is the m in
    Del^2 \phi = -4 pi G \rho
    m = integral \rho d^3x
The latter is the m in
    f = -m grad \phi
But Newton's third law guarantees these two are the same, even though
they appear so different -- of course they aren't different at all in
    f = GMm/r^2

BTW in Newtonian physics "inertial mass" is the m in
    f = ma
Suitable choice of the gravitational constant G ensures that this is the
same m as in the previous equations.

In GR there are no such things as "gravitational mass" or "inertial
mass", there is just mass. Mass does not "generate the curvature of
spacetime", that role is played by the energy-momentum tensor T in the
Einstein field equation
    G = T
But, of course, mass contributes to T. The G in that equation is the
Einstein curvature tensor.

Note that vacuum regions can have nonzero curvature (even though T=0),
because G is not the complete Riemann curvature tensor; continuity of
the fields at the boundary between matter and vacuum requires nonzero
curvature in the vacuum region. The Bianchi identities put constraints
on the Riemann curvature tensor at such boundaries (and throughout the
manifold).

Tom Roberts    tjroberts@lucent.com