Smaller mass in gravity well? When sitting in a gravity well, as we do on earth, does our effective mass become smaller than our rest mass due to having negative potential energy? Correspondingly, does a free falling mass (from infinity, radial path) have an effective mass equal to its rest mass, as here potential and kinetic energies are ballanced? The question is based on the concept of equivalence of energy and mass.
 A: In most cases, it doesn't really make sense to talk about a lowered effective mass caused by sitting in a gravitational potential well, since the equivalence principle says that locally the spacetime looks flat, and hence it looks like the gravitational field vanishes.  
However, in certain special cases, there is a sensible notion of energy that is different from your rest mass.  This is when you have a timelike Killing vector, which means there is a preferred time coordinate under whose flow the metric is invariant.  The existence of time translation symmetry leads to a conserved energy.  If $\xi^a$ is the Killing vector representing the time flow, and $p^a$ is the 4-momentum of an object, the conserved energy is 
$$E = -g_{ab}\xi^a p^b$$
while the rest mass is 
$$m = (g_{ab}p^a p^b)^{1/2}. $$
The Schwarzschild spacetime is a classic example of this.  The metric is
$$ds^2 = -\left(1-\frac{2GM}{r}\right)dt^2+\frac{dr^2}{\left(1-\frac{2GM}{r}\right)}+r^2d\Omega^2$$
Since the metric is independent of time, it has a Killing vector $\xi^\alpha = (1,0,0,0)$.  
Let's first consider the 4-momentum of an observer initially at rest at some radius $r_0$.  Initially at rest here means they start of with $p^a \propto \xi^a$, and the normalization of $p^a$ tells us that it is
$$p^\alpha = m\left(\left(1-\frac{2GM}{r_0}\right)^{-1/2},0,0,0\right),$$
Then the energy for this particle is
$$E = -g_{0\alpha}p^\alpha=m\left(1-\frac{2GM}{r_0}\right)^{1/2}$$
As long as we are outside the event horizon $r=2GM$ (which is the only place where the energy really makes sense), this shows that the Killing energy $E$ is less than the rest mass.  And if you are far away from $r=2GM$, you can expand to first order in $GM/r$ to get
$$E\approx m - \frac{GMm}{r_0}$$
Which is the rest mass minus the Newtonian gravitational potential energy.  So in this sense the potential energy of the particle is negative, since it's Killing energy is less than its rest mass energy.  
Finally, for a particle falling in from at rest at infinity, we use the fact that Killing energy is conserved along all points along the geodesic.  At infinity, the metric is asymptotically Minkowski, and being initially at rest means $p^a\propto\xi^a$, hence
\begin{align}
p^\alpha &= m(1,0,0,0)\\
E &= m
\end{align}
Since $E$ is conserved, we see that in this case it is always equal to the rest mass energy.  You can sort of see this as a cancellation between kinetic energy and potential energy: to get the kinetic energy you need to specify who your observer is, so lets say it is the observers at constant radius.  Their 4-velocity is 
$$u^\alpha=\left(\left(1-\frac{2GM}{r}\right)^{-1/2},0,0,0\right)$$
and they would define the total energy (rest mass plus kinetic, but not including potential energy) as
$$T = -g_{ab}u^a p^b = \frac{m}{1-\frac{2GM}{r}} \approx m + \frac{GMm}{r}.$$
To derive this, we used the fact that $E=m$ is conserved, which means that $p^t = \dfrac{m}{\left( 1-\frac{2GM}{r}\right)^{1/2}}$.  If we continue to assign the potential energy $V = \frac{GMm}{r}$, then we get
$$E=T-V = m + K-V = m \implies K=V $$
So in some sense the relations you postulated hold when there is a well-defined "effective mass" i.e. Killing energy, but in a general spacetime with no timelike Killing vector, you won't be able to make a sensible definition of such a thing.
