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pervect
#28
Jan19-06, 06:46 AM
Emeritus
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P: 7,594
Quote Quote by ubavontuba
Sure, but spaceship 1's mass will appear to change (relative to spaceship 2) as it approaches the black hole. But, I'm not talking about black holes (in general) here anyway.
I will reiterate as delicately as I can. I am not talking about black holes. I am talking about an apparent effect of relativity. There is no real black hole in question to consider.
You've totally missed the point, partially because of my choice of example.

Spaceship 1 is approaching a planet or a black hole - it doesn't matter. (I said black hole in the first post, but it doesn't really matter, black holes are just convenient to calculate with - and the example in my textbook that this is drawn from involves a black hole).

From the frame of spaceship 1, the planet is more massive. (Not its invariant mass, which is invariant, but it has a greater energy, and hence a greater "relativistic mass".)

(Technical note: to define the mass at all, you need an asymptotically flat space-time.)

However, the gravity of the planet/black hole is not spherically uniform in the moving frame. Thus when spaceship1, directly approaching (or fleeing) the planet, fires up its Forward mass detector (a real device, which however measures tidal forces, and not mass directly), it finds that the gravity (specifically, tidal gravity) of the planet has not increased _even though_ the relativistic mass (energy) of the planet has increased.

The same thing happens if the spaceship is moving away from the planet. The same thing happens if the planet is moving away from the spaceship.

Motion towards or away from a planet (or black hole) does not affect tidal gravity.

The same thing does _not_ happen if the spaceship is moving perpendicularly to the line connecting it to the planet - in that case, the tidal gravity actually does incease.

This can be understood (in an approximate sense) by thinking of the gravitational field of a moving object being compressed, much like the electric field of a moving charge is. (The anology is not exact, mainly because gravity isn't really a field in GR).

The bottom line is your idea does not work - there is no "infinite mass" due to the expanding universe, even when the velocity of recession is very large. The velocity of recession is on a direct line, and thus it does not affect the gravitational forces.

There's another way of saying something similar that is more in the spirit of true GR and perhaps simpler. The net gravitational effect of a spherically symmemtric expanding shell of matter can be shown to be zero inside the shell.

The cosmological horizon (assuming it exists, which means assuming that the universe's expansion is accelerating and will continue to accelerte) is thus not due to the effects of the matter outside - it is due to the effects of the matter (and dark energy) inside the horizon.

[add]
Here's a reference (unfortunately it's in postscript - but with google you can view it as text, though it won't be quite as intelligible). The theorem is called Birkhoff's theorem.

www.ucolick.org/~burke/class/grclass.ps

What this means, roughly, is that spherical symmetry is enough to ensure that the spacetime is static, without separately requiring it. Another way to appreciate the content of the theorem is to consider any spherical object. If you go inside it and do anything you wish that does not disturb the spherical symmetry, then the external gravitational field cannot change. This is consistent with our earlier result that there is no monopole gravitational radiation. Yet another meaning for the theorem is than inside of a spherically symmetric distribution of mass there are no gravitational effects, just as in Newtonian theory, despite the nonlinearities.