What's the opposite of Absolute Zero?

Infinity + 1 :rolleyes:

What could make the number of particles exiting a black holes event horizon equal to the number of virtual particles entering it?

Is that even possible in theory? “Empty” space is full of virtual particles, after all.

Yes, but the distribution of virtual particles is such that they don’t specify a reference frame. So being “at rest” in a sea of virtual particles is the same as “zooming through” the same sea: You’ll interact with them in the same way in either case. They do actually make a difference if you’re accelerating, but that’s a different topic.

Opposite of absolute zero? Why, molten cheese, particularly fondue, applied generously to one’s soft palate :eek:

Pleonast said:

Read the link, find the system completely preposterous.

First off, positive and negative zero? At the ends of the scales?

Second, infinity in the middle, and the transition runs from positive infinity to negative infinity as energy is applied?

If there were anything more designed to sound preposterous and create confusion, I find it hard to fathom - and that includes quantum issues like wave-particle duality.

Irishman, yep, it’s hard to understand. Unfortunately that math is the simplest way we have to describe those systems. Personally, I think about it terms of a two’s complement representation, but that is probably more confusing to most people.

Sometimes in science the hardest part is coming up with the math used to describe a phenomenon.

Ok…this is confusing.

E=MC[sup]2[/sup]

Mass is interchangeable with energy.

So, keep adding energy to a system and its mass perforce increases.

I thought this was one of the sticking points of traveling at light speed. An object with mass would have infinite mass at light speed which is a problem.

So how is it these energetic particles have no more mass than less energetic ones?

The full equation is E[sup]2[/sup] = m[sup]2[/sup]c[sup]4[/sup] + p[sup]2[/sup]c[sup]2[/sup]. M is the rest mass, which is an invariant quantity–all observers will agree on what a particle’s rest mass is. P is momentum. In the case of zero momentum (that is, in the particle’s rest frame of reference), you get the famous equation E = mc[sup]2[/sup].

When a particle is moving, its rest mass* does not change, but it’s energy E does. Note that a particle’s energy is not invariant–different observers will see a different momentum, and thus disagree about the energy.

*Physicists only use rest mass, or simply “mass”. The concept of “relativistic mass” (the “mass” that increases with a particle’s speed) is not a useful one. Confusion as we see here is a result of using it.

Why? (genuinely curious…particularly when we are talking about energies where relativistic effects would seem to apply)

Trying to read up on it some I found the bit below which seems to argue for using relativistic mass:

Interesting linked article. I encourage others to read it in full.

I think the reason I prefer to use rest mass only, and not think in terms of relativistic mass, is that rest mass is a nice invariant property of a particle, like charge or spin. I put all the relativistic accounting into energy and momentum. It’s basically a choice about where to place the complications of relativity. It seems more natural to put the speed-variable variation in the quantities that vary the most with speed (including slow non-relativistic speeds).

And as we’ve seen in this thread, relativistic mass engenders the misconception that a fast-moving object can collapse into a black hole due to it’s increased “mass”.

Compare mass to length. While mass is a fundamental property of a particle, length is not. It is a macroscopic property of a system of particles, like temperature. Actually I have a similar complaint about relativistic length. It encourages some people to think that objects get compressed (as in forces squeezing the objects to make them flatter) when moving relativistically fast. They don’t; the foreshortening is a geometrical effect of spacetime.

Not trying to be a pain…trying to understand so bear with me if you will:

So, what happens if I take a kilo of anti-hydrogen and a kilo of hydrogen and bring them together (apart from a spectacular explosion)?

I thought we go to E=MC[sup]2[/sup] again for the answer.

We are taking the mass (2 kilos) and converting it all to energy. No more mass, lots and lots of energy.

Are we converting the rest mass or relativistic mass using that equation? Where does the rest mass go if that equation only deals with relativistic mass?

Wouldn’t the opposite of absolute zero be “burning with the intensity of 1000 suns”?

The inside of my car at 3:00pm on any given day in July or August.

What would you heat it with?

For matter-antimatter annihilation, the energy available is the combined total energies of the matter and antimatter. Most physicists would compute it using the equation I gave in Post 29. That uses the rest mass (invariant) and relativistic momentum (varies with speed).

Instead, one could use E = mc[sup]2[/sup], where the mass is the relativistic mass. Of course, you need to know the rest mass and speed to compute that, so it’s no different than the other formulation in terms of information. And the result will be the same.

In case it isn’t clear, the total energy of the matter and antimatter is converted.

If you take a kilogram of matter and a kilogram of antimatter and bring them together (gently, so there’s no kinetic energy to worry about), then you’ll have a total of 2 kg of mass, both before and after the Earth-shattering kaboom. The key here is that the mass is a property of the whole system, not of the particles that compose it. A single photon has no mass, but a collection of photons (like what you’ll get out of that antimatter reaction) can have mass. It only looks like you’re losing mass if you look at the individual pieces of the system.

So if particles were speeding toward each other at the very very very (etc) high speeds Cecil mentions in his column about miniature black holes, when they get close to each other, an observer from outside should find that the total mass of the system exceeds the necessary mass to create a black hole out of the two particles. Is there a reason this should not happen (or would it happen)?