On transfers.

[sh1pman]

1 minute ago, sevenperforce said:

Right. The answer is no; it does not. As long as you are in an inertial reference frame, your kinetic energy does not curve spacetime.

If you are in a non-inertial reference frame, all bets are off. A rotating object, for example, has centripetal acceleration and thus produces frame-dragging effects.

Ok, makes sense. By the same logic, should rotating black holes appear more massive than they are?

[sevenperforce]

3 minutes ago, sh1pman said:

Ok, makes sense. By the same logic, should rotating black holes appear more massive than they are?

Rotational energy is non-inertial, so yes.

Most mass is, in fact, relativistic kinetic mass-energy, rather than rest mass. But it is trapped inside chemical and nuclear bonds so it affects the tensor.

Rotating black holes appear smaller if you are orbiting them prograde than if you are orbiting them retrograde. For a rotating black hole, there exists orbital radii for which the orbital velocity is lower than c when orbiting prograde but greater than c when orbiting retrograde.

[YNM]

3 minutes ago, sevenperforce said:

Rotational energy is non-inertial, so yes.

That said, where does kinetic energies go on the tensor ? Is this just one "broken" parts we have no answer of yet ? (I mean, photons would be energy-less without it's speed)

Edited April 4, 2018 by YNM

[sh1pman]

1 minute ago, sevenperforce said:

Rotating black holes appear smaller if you are orbiting them prograde than if you are orbiting them retrograde. For a rotating black hole, there exists orbital radii for which the orbital velocity is lower than c when orbiting prograde but greater than c when orbiting retrograde.

So if your orbit is prograde, then you're outside of its event horizon, but if's retrograde, then you're inside the black hole? 

Mind blown.

[YNM]

2 minutes ago, sh1pman said:

So if your orbit is prograde, then you're outside of its event horizon, but if's retrograde, then you're inside the black hole? 

There isn't an "orbit", just trajectories. In these cases often referred as "worldline".

But yes, at a given (absolute metric) radius, there's going to be a future worldline outside the black hole only if you go prograde; going retrograde, all your future worldlines are inside the black hole.

Edited April 4, 2018 by YNM

[sevenperforce]

3 minutes ago, YNM said:

That said, where does kinetic energies go on the tensor ? Is this just one "broken" parts we have no answer of yet ?

Well, thanks to conservation of momentum, that kinetic energy has to have come from somewhere, so the effect on the gravitational tensor is conserved by virtue of whatever reaction mass was acted upon to produce the dV.

The relativistic mass gained by an object with an increase in kinetic energy must be defined with respect to an outside reference frame, and SR spacetime curvature effects will be present for an observer in that outside reference frame even though there is no actual change to the GR tensor.

[YNM]

2 minutes ago, sevenperforce said:

The relativistic mass gained by an object with an increase in kinetic energy must be defined with respect to an outside reference frame, and SR spacetime curvature effects will be present for an observer in that outside reference frame even though there is no actual change to the GR tensor.

What about "Kugelblitz" then ?

[sevenperforce]

4 minutes ago, YNM said:

What about "Kugelblitz" then ?

A photon has energy and momentum, which couple to gravity. Particles with rest mass (or bound relativistic mass, which functions as rest mass for all intents and purposes) couple directly to gravity, but massless particles couple to gravity through their energy-momentum-4-vector.

Interestingly enough, you could make a "dark Kugelblitz" by firing a bunch of massive particles at the same point at relativistic speeds from a bunch of different directions, and thus produce a black hole using relativistic kinetic mass-energy. But in that instance, the additional mass-energy would be the rest of timelight interactions between the particles in SR, which would produce bound states (transient or no) which will couple to the GR tensor.

[sh1pman]

13 minutes ago, sevenperforce said:

Well, thanks to conservation of momentum, that kinetic energy has to have come from somewhere, so the effect on the gravitational tensor is conserved by virtue of whatever reaction mass was acted upon to produce the dV.

What if there's was no reaction mass, and the acceleration was produced by some other means? Like a solar sail, or laser-based system like in Breakthrough Starshot project?

[sevenperforce]

4 minutes ago, sh1pman said:

What if there's was no reaction mass, and the acceleration was produced by some other means? Like a solar sail, or laser-based system like in Breakthrough Starshot project?

Photons have momentum, so momentum is conserved, and the energy-momentum-4-vector of the photon takes the place of reaction mass in the rocket equation.

This is one of the reasons why I just laugh whenever anyone claims to have a reactionless drive. You can't have one. It would break all of relativity. Mass would no longer be coupled to gravity. All the stars would explode.

"Do you see? Do you see what you do? Trying to build a reactionless drive and you destroyed the universe. Well, I hope you're happy."

[YNM]

@sevenperforce It's all better when one knows more and share it ! Thanks !

[Gargamel]

Yes, Yes, That was all very enlightening (very much so)..... But.....

22 hours ago, Gargamel said:

On 4/2/2018 at 7:31 PM, ArmchairPhysicist said:

Oh and it has about 750,000 ms D/v, nothing special.

 

Nobody asked about this!?  I'd love to see this

show me the ship dang it!

 

 

Edited April 5, 2018 by Gargamel
Push a button once, oh yes that felt good, keep pushing it.... oh no too much!