This is false. A logical contradiction supersedes any "deep dives" into the mathematics of a theory. Making a proclamation that the problem doesn't exist is not addressing the problem.

There is no contradiction in GR. Rather, you seem to be mixing GR with your own assumptions that contradict it. Specifically, you seem to be under the impression that there is some simple relation in GR that always applies along the lines "if the energy density is higher than some threshold then you will always have a black hole". That is wrong.

First, the gravitational field in GR does not depend on the "energy" or "energy density" but on the stress-energy tensor, which is more general. Like the name implies, the stress-energy tensor is a

*tensor*. If you don't know what that is, you can think of it as a kind of generalised vector. Like a vector in Euclidean geometry, it is an invariant object. It has

*components* and those individually depend on the reference frame, much like the $$x$$, $$y$$, and $$z$$ components of a vector depend on the axes you choose, but the tensor as a whole is a geometrical object defined on the spacetime manifold and can be understood as such independently of any choice of coordinate system.

The stress-energy tensor has ten components in any given coordinate system. In an inertial reference frame,

*one* of those components is the local energy density. The other nine consist of the $$x$$. $$y$$, and $$z$$ components of the momentum density and six momentum fluxes.

Another thing about the stress-energy tensor is that it is a

*field*, which means it is a function of position in spacetime. So if you were comparing a moving mass to a stationary one then not only are the momentum components now nonzero but also the tensor and its components will now be changing over time, since the place where the energy and momentum are concentrated is moving, and you would need to take

*all* of that into account if you were attempting to calculate what gravitational field a moving mass will produce. It is not like the "energy" of a moving mass is a bigger number and everything else is the same.

Second, the relation between the stress-energy tensor and the gravitational field itself is not simple. If there were a simple rule that always tells you when you would have a black hole, it would be pretty amazing given that nobody even knows how to accurately predict the gravitational field in general in GR. Unlike in Newtonian gravity, GR does not have a simple rule that directly tells you if the stress-energy tensor is such-and-such then the gravitational field will be such-and-such. GR instead only says that the gravitational field and the stress-energy tensor must be related in such a way that they satisfy the Einstein field equation. So predicting the gravitational field (including whether there is a black hole or not) in general requires solving the Einstein field equation for a given stress-energy tensor or integrating it starting from some known initial conditions. The problem here is that the Einstein field equation is a complicated nonlinear differential tensor equation. Solving it is

*very* hard. Nobody knows how to do it accurately in general.

There are some predictions in GR that do give relations between matter/energy density and getting a black hole, but you need to keep in mind that they are

*not* general and it is not justified to assume whatever relation they predict will apply outside of the specific conditions they are derived for. For example, the Schwarzschild geometry tells us that you can't compress a sphere of mass smaller than a certain radius without getting a black hole, but that relation only necessarily applies under the conditions and assumptions that the Schwarzschild geometry is derived for as a solution to the Einstein field equation. Importantly, one of those assumptions is that the gravitational field is static and the source producing it is at rest, so you can't just take the result and assume it will be the same for a moving mass.

Or to summarise all of the above: things are not as simple as you are assuming.