# Neutron Star going BH

• B
timmdeeg
Gold Member
The energy would not be the same in all frames. Energy is a frame variant quantity, even in Newtonian physics. Energy is conserved, but not invariant.
But the mass of the fuel is invariant. Could one argue that this mass is converted into proper acceleration of a rocket such that escape velocity is achieved?

Dale
Mentor
But the mass of the fuel is invariant. Could one argue that this mass is converted into proper acceleration of a rocket such that escape velocity is achieved?
Yes, that is the type of argument one would need to make. Everything in terms of invariants like invariant mass and proper acceleration

Nugatory
Mentor
But the mass of the fuel is invariant. Could one argue that this mass is converted into proper acceleration of a rocket such that escape velocity is achieved?
Sure, but to do the calculation properly you'll have to account for all the energies involved, including the kinetic and potential energy of the rocket exhaust which does not escape and the rocket which does escape. These energies are all frame dependent, but when you add them all up you'll find that the difference between the pre-launch energy and the post-launch energy is frame-independent and what you'd expect from burning the invariant amount of fuel consumed.

Before you follow this line of thought further, you'll want to be sure that you have a solid understanding of a much simpler classical "paradox": a gun fires a two kilogram projectile with a muzzle velocity of 1000 m/sec. We use the classical ##E=mv^2/2## to calculate the before and after energies. If we're at rest relative to the gun, we calculate that the explosive charge released ##1\times{10}^6## joules (velocity of projectile went from 0 to 1000 m/sec). If the gun is approaching us at 1000 m/sec we calculate that the explosive charge released ##3\times{10}^6## joules (velocity of projectile went from 1000 to 2000 m/sec). But it's the same amount of the same explosive, so the energy released has to be the same.

(It turns out that the first calculation is correct, and the second one is an example of how to manufacture a paradox by choosing coordinate systems that obscure the physics).

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• Dale
Dale
Mentor
Actually, it would, as long as it is isolated. The symmetry would not be apparent in coordinates in which it is moving, but could be recovered by analyzing the killing vector fields.
Thanks for pointing this out to me. The more I think about it the more powerful a concept this becomes.

timmdeeg
Gold Member
Sure, but to do the calculation properly you'll have to account for all the energies involved, including the kinetic and potential energy of the rocket exhaust which does not escape and the rocket which does escape. These energies are all frame dependent, but when you add them all up you'll find that the difference between the pre-launch energy and the post-launch energy is frame-independent and what you'd expect from burning the invariant amount of fuel consumed.
I had the idealized assumption to neglect the mass of the rocket in my mind but should have mentioned that.

Thank you for all the replies and insight.