[stcredzero]

Mass is converted to energy as it approaches C (Light Speed).

Not so much.

https://www.youtube.com/watch?v=LTJauaefTZM

Try this thought experiment. Let's say you had a magic Bussard Ramjet rocket which scoops up energy and reaction mass from space and can accelerate forever. There are no special reference frames. So what happens if the rocket accelerates then shuts off its engine at the point where the "relativistic mass," as measured by an un-accelerated observer, should make the rocket disappear behind an event horizon?

From the POV of people on the rocket, they accelerated, then stopped accelerating. From the observer's POV, the rocket turned into a black hole? One reference frame now seems "privileged" or different somehow. How do we square this with relativity? Also, what happens if the rocket turns around, then decelerates? Wouldn't that constitute them returning from inside of an event horizon?

The answer, is that "relativistic mass" is actually just a pedagogical fiction.

(EDIT: Also, a lot of the redonkulousness in the thought experiment sneakily comes from rockets that can magically accelerate without worrying about where the fuel and energy come from. If you worked out how much fuel and reaction mass would be needed by a real rocket to perform such a feat, you'd get "unphysical" amounts of matter.)

[zspade]

I stand corrected. That's what I get for trying to make an offhand joke based on some some bad info. The rest of the time comment stands though.

[raattgift]

Don Lincoln's point boils down to momentum being a frame-dependent quantity even in plain old Newtonian mechanics, that the special-relativistic correction for momentum is simple. The point he is making about mass is, in essence, that one has to be careful in understanding equivalence relations E = m, E = p, E = hf, E = h/\lambda (here all are with c=1) as an overloading of "energy". In the last three of these there is a frame-dependent quantity: momentum, frequency, wavelength. In the first there is a non-frame-dependent quantity, the invariant mass, m_{0}. It is not the same species as a frame-dependent quantity m_{relativistic}, and at 6m57s into the video, he shows some of the dangers of "punning" them.

Your related comment says that m_{relativistic} is a pedagogical fiction. No, it is just a quantity that is non-identical to m_{0}. That students frequently forget (or do not know) that is a teaching failure.

However, you've introduced a gravitational event horizon, which is not something you can find in Special Relativity (whose spacetime is everywhere non-curved, that is, it is free of any gravitational effects at all).

If we pick out the globally flat background of Special Relativity and drop a test particle into it -- electrically neutral, low mass, and classically pointlike -- the latter generates the stress-energy tensor. Dropping indices and factors, G=T=0 -> G=T!=0, where G is the Einstein curvature tensor and T is the stress-energy tensor.

The stress-energy tensor is generally covariant: observers anywhere in spacetime, no matter how they are moving, agree on the total value of T at the point occupied by our test object. However, they are free to disagree on the quantities in the components of the stress-energy tensor.

The components of T can be written as a 4x4 matrix representing the flux of row-momentum into the column-direction. If (row,column) specifies a component of this matrix, and we count rows and columns 0..3, and we specify that our spacetime dimensions run 0...3 with 0 being the timelike dimension t, then a positive value of T (remember at a specific point in spacetime) in (0,0) and zeroes in all the other components means that momentum from the past of that point is flowing in the future of that point.

If at point p = (0,0,0,0), T^{00} = 1 and all the other components of T are zero, then at p' = (1,0,0,0), T^{00} = 1.

From Noether, a quantity that is invariant in time is conserved like the Newtonian conservation of energy.

An observer moving with our test particle generating T^{00} = 1 would relate the T^{00} quantity to the particle's rest mass (or invariant mass, if you like).

Let's call our 1 rows and columns "x" after the Cartesian direction

On a spacetime diagram, our test particle develops a worldline vertically along the t axis and not at all along the x axis.

Now let's complicate this a bit by having the test particle chuck a bit of itself out its back, engaging a classical notion of conservation of momentum. We'll call the 1 dimension backwards-and-forwards, or x, compared to our timelike 0 axis t. T^{01}, if positive, encodes the momentum coming from the past and leaving the point in the forwards direction.

Assuming coordinates (t,x,y,z) that absorb the emitted units:

At p = (0,0,0,0) we have T = 1, and T^{00} = 1.

Let's keep things normalized so that our particle is always at the origin of x.

At p' = (1,0,0,0) we have T = 1, and T^{00} = 0.9 and T^{01} = 0.1.

At P = (2,-1,0,0) we have the boosted exhaust: T^{01} = 0.1; at p'' = (2,0,0,0) we have the particle T^{00} = 0.9. Thus at p''' = (3,0,0,0) we have the particle still T^{00} = 0.9, and the exhaust P'' = (3,-2,0,0), T^{0,1} = 0.1.

But notice that this picture depends on coordinate conditions: the "rocket" particle at x=0 always, tracing out a vertical worldline on a spacetime diagram. The exhaust does not remain at x=0, and so traces out a worldline that has an angle from vertical.

If we flip this so the exhaust is at x=0, then we would say that the exhaust is at P = (2,0,0,0), P'' = (3,0,0,0) etc and its |T^{00}| remains 0.1 and it traces out a vertical worldline on the spacetime diagram. The "rocket" on the other hand is at p'' = (2,1,0,0), p''' = (3,1,0,0) etc and its |T^{01}| is 0.9.

What component(s) stress-energy appears in depends on the choice of coordinate basis.

Likewise, if we draw a spacetime diagram with the exhaust always at x=0, it traces a vertical worldline up the t axis, while the "rocket" traces out a worldline at some angle, because it is not at a constant x coordinate.

A completely different observer holding itself at x=0,y=0,z=0 would calculate different coordinate-values for particle/rocketing-particle/exhaust (changing the coordinates of p, p', p'', ... and P, P', .... However, it would agree that a normalized value of the whole T at p and p' is 1 but that the value of T at p'' is 0.9 while the value of T at P is 0.1. However, the total of 1, 0.9, or 0.1 respectively could be allocated to different components of T in 4x4 matrix form.

Technically, since T is nonzero at several points in spacetime (all those p and P points) spacetime is not flat. However, in any timelike hypervolume of this nearly empty spacetime, the total stress-energy will be 1.

Even if we move the "rocket" and its exhaust apart ultra-relativistically, the sum of their stress-energies at any time will remain 1. In Newtonian terms, what's being described is a total conservation of energy, and an unchanging centre-of-mass in a deforming system. Your rocket/bussard is similar: there is a system of ship + fuel + exhaust (+ waste heat + ...) whose observer-independent (or generally covariant) total is constant at any time. Thinking in a more field-like way, there are rocket-bits, fuel-bits, exhaust-bits (etc) which generate nonzero stress-energy at various points in the spacetime. The stress energy at each point in spacetime can be agreed by all observers, however they are free to disagree about how the stress-energy at a point is distributed among the various tensor components.

In order to form an event horizon we would need a much larger quantity of stress-energy in a small region, and that is not what is described in your posting, which is more about extremizing components of the tensor (at various points) rather than the whole tensor itself.