Does the near-C mass increase create gravitaton too?

[Agent86]

As an object approaches the speed of light and its mass increases, does its gravitation increase as well? As in, if a vehicle approaches 99.99% of C, would its gravity come to eqhal that of a planet or spar, based on its increased mass?

[K^2]

First of all, you have to be careful when you say that mass in creases. Inertial, or relativistic mass increases. Invariant, or rest mass is going to remain exactly the same. In modern physics, the word "mass" typically means the later.

Now, gravitational mass is, indeed, identical to the relativistic mass. But it's still a little more complicated than that. Source of gravity isn't just mass, but the stress-energy tensor. For a point object, it depends not only on object's mass, but also on its momentum. An object moving really close to speed of light is going to have a different gravitational field around it, but not in such a trivial way.

Unfortunately, this is way outside of what you can describe with classical physics. This really has to be treated from perspective of GR, where the answer is going to be governed by a Lorentz-boosted Scwarzschild metric. Qualitative effect, however, is going to be a lot like magnetism. You will experience stronger pull if you are moving in direction opposite to the object, but a lesser pull if you are traveling in the same direction.

[Dispatcher]

Thank goodness that the gravitation doesn't become nearly infinite!

[Guest]

It doesn't become infinite, but might approach infinity (along with relativistic mass) as the object approaches c. So it might become nearly infinite, but that would require a rather massive object at a near-c speed. An unlikely occurrence, to say the least.

[LextheNonTrivial]

How about this for those familiar with only special relativity...

Simply consider the Sun from a frame moving at, say, c - .0003 m/s , so that its mass increases and its Schwarzschild radius is greater than its physical radius (by a factor of about 5 at that speed, if my maths are good). It's a black hole now. As a black hole is a black hole in any frame, we know that the Sun is actually a black hole in its own rest frame, regardless of appearances...

In other words, no- gravity isn't simply proportional to relativistic mass.

Yes, I know this is technically, or even wildly, inaccurate, given GR. I feel it's a fun refutation nonetheless.

@K^2 : If you're familiar enough with GR maths, try contracting indeces on Einsteins equations (by way of a locally Minkowskian metric). It's kind of pertinent to this.

[K^2]

@K^2 : If you're familiar enough with GR maths, try contracting indeces on Einsteins equations (by way of a locally Minkowskian metric). It's kind of pertinent to this.

Hm? Why? I can easily compute the proper acceleration vector for an object at rest in the moving Schwarzschild metric using Christoffel symbols.

[LextheNonTrivial]

Hm? Why? I can easily compute the proper acceleration vector for an object at rest in the moving Schwarzschild metric using Christoffel symbols.

Mathematically: It shows that the Ricci scalar is proportional to the rest mass density- after all scalar equations can't reference energy or momentum as those concepts are undefined without a metric. It's relevant to the question about whether gravity is proportional to relativistic mass or rest mass.

Philosophically: I don't know, curiosty?

[K^2]

Ah, I see what you mean. Get scalar curvature to be equal to trace over the stress-energy tensor. But yeah, as you point out yourself, that's going to be a coordinate-dependent relationship. Which, I suppose, is what OP is asking about in the first place, but it's still an oversimplification. Yeah, curvature due to a moving object is higher, but the real fallout is in how that effects geodesics, and it's a very different effect from simply making the source heavier.

[LextheNonTrivial]

You seem to have misread that a bit- I was trying to point out that the scalar curvature is proportional to the invariant rest mass, not relativistic mass. The whole idea behind using a scalar equation is that they don't depend on the metric/frame/coords used.

Going faster doesn't have any affect on how much scalar curvature something causes; at best the curvature is just redistributed through spacetime.