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Comment: Re:H goes up, anti H goes up, unless anti-N is pre (Score 1) 255

by ByteSlicer (#43605617) Attached to: Does Antimatter Fall Up?

Back at the atomic level, particles that bounce off the bottom of the volume get an extra "push", and bounce back a bit higher.

Actually, more accurate would be to say that the particles that bounce off the bottom of the volume will not hit it hard enough and frequently enough to repel all the particles from the denser volume below. So the top boundary of that denser volume will rise, moving the bottom of the less-denser volume upwards, causing the particles in that volume to bounce slightly higher.

Comment: Re:H goes up, anti H goes up, unless anti-N is pre (Score 1) 255

by ByteSlicer (#43605391) Attached to: Does Antimatter Fall Up?

"The first mistake is to assume that helium rises. The truth is that it falls down towards the earth just like any other object. The reason for what you see is much simpler: It does not rise; it's just that everything else simply falls harder." (Freely translated from memory and German)

That is actually very misleading, to the point of being incorrect.

At the atomic level, a molecule of Oxygen falls at the same rate as an atom of Helium (about 9.8 m/(s^2), depending on altitude), so nothing really "falls harder".

What actually happens is the process of buoyancy. A He atom weighs about 16x less than an O molecule. A volume of air at specific temperature and pressure will contain the same number of particles (atoms or molecules).

If the volume contains relatively more He, the total mass of the volume will be less (lower density). If you surround this volume with volumes of higher density and place them in a gravitational field, then the sum of the pressure forces on the volume will not be in equilibrium. There will be a net force pushing the volume upwards.

At the atomic level, all the particles are falling constantly, bouncing off particles of lower layers (electrostatic repulsion), bouncing up again. At the macroscopic level, volumes of particles with lower density rise because the net forces on them are unbalanced. Back at the atomic level, particles that bounce off the bottom of the volume get an extra "push", and bounce back a bit higher.

Comment: Re:Newton? (Score 1) 231

by ByteSlicer (#43600191) Attached to: Physicists Attempting To Test 'Time Crystals'

The gravity of the body at the center is pulling the object in, the velocity vector is pushing the orbiting body out, but as they are equally matched the net result is a nice perfect circular orbit (not accounting for the fact that true circular orbits probably only occur in theory and also avoiding all those horrible to write as formulas elliptical orbits). Isn't that the same thing?

No. While the amplitude of the velocity vector remains the same, its direction changes constantly, the result of the gravitational force.

Or is the first law simply saying that a body with movement will continue to keep moving as long as nothing interacts with it?

That is exactly what it says. Or in other words: the velocity vector only changes when a net force is applied to the object.

If there are two opposite but equal forces being applied with a new force of zero, would that not be allowed to be the same thing?

That is the case, but here there is only one real force working (gravity). The centrifugal force that balances the orbit is not a real force, but a result of the object's velocity.

Comment: Re:Gravitational tides will kill you (Score 1) 412

by ByteSlicer (#43386131) Attached to: How Would an Astronaut Falling Into a Black Hole Die?

Well, these may well be the most extreme regions in space-time and the maths are some of the most difficult to solve (and honestly way over my head). Plus we would have to be able to marry GR and QM to fully describe black holes.

The way I see it, the vacuum fluctuations can form many kinds of virtual pairs, and when that pair consists of photons, then there is no problem at all since they always move at c and hence can escape the gravitational well (but they will be red shifted due to the extreme gravity). So most of the Hawking radiation will always be EM radiation.

Mass-bearing particles will form with the same small/limited energy from the vacuum, and their momentum and thus initial speed will be small too. In a normal gravitational well they would almost immediately fall back towards the event horizon, due to the extreme gravitation (it requires thrust for any mass to change/leave orbit, and these are not even in orbit).

In the ergosphere of a rotating black hole, nothing can remain stationary relative to the rest of the universe because that would require faster than light movement. This means particles are forced to move, even when initially stationary, and hence they gain momentum, which in turn can lead to the particle escaping the gravitational potential.

Of course, that is when only taking into account gravitational effects. The extreme magnetic fields surrounding black holes will also accelerate charged particles, and will probably have a bigger effect than frame dragging.

Comment: Re:Gravitational tides will kill you (Score 1) 412

by ByteSlicer (#43380443) Attached to: How Would an Astronaut Falling Into a Black Hole Die?

Instead, it can take a few swings around the black hole in a rapidly decaying orbit, until it slingshots out on a hyperbolic path. The smaller the black hole gets, the more definite the position is for every matter/antimatter particle pair, and by Heisenberg's uncertainty principle applied to position-momentum, this makes it easier for one of the two particles to escape.

Erm, that's not how orbital/slingshot mechanics work. In fact, a mass-bearing particle (from a virtual pair) could never escape a stationary black hole, because it wouldn't have enough energy to do so, and normal orbital mechanics wouldn't increase its energy.

Instead, you'd need a rotating black hole with an ergosphere. This is a weird area where space-time is dragged along the black hole faster than the speed of light relative to outer space. Here it is possible to extract energy from the black hole with what is called the Penrose process, and thus the electrons/positrons may gain enough speed to escape.

Comment: Re:Why not teach with BananaOS ? (Score 4, Interesting) 90

by ByteSlicer (#42842235) Attached to: Russian Univ. Launches Course Based On ReactOS Led By Alex Bragin

The effort invested into Reactos would make much more sense if invested into Wine instead.

ReactOS uses the higher level libraries and services from Wine, so any improvements they make in those will improve Wine too (and vice versa).

Besides that, everyone is free to invest their time and efforts in whatever they want.

... or were you driving the PONTIAC that HONKED at me in MIAMI last Tuesday?

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