Black holes at CERN?, Full story at http://www.physorg.com/news12171.html

I think, that electron and positron can be some kind of rotations... And then if they two anihilate then this rotations energy became wave energy.

Technically incorrect. It's the total energy of all the energy/mass absorbed into the nano-black-hole. However, it isn't/wouldn't be at rest.

You mean to say:
As I have already said, if we were to measure the rest MASS of a Proton with a total energy of 1150 TeV, we would find that it only weighed 4.7 TeV. This includes the other nonsense that you keep blithering about.

It's kinetic.

Presuming you meant "proton" and not "photon," you are incorrect (not withstanding the point I made above). Forget the earth for a moment. Consider two colliding protons. Our observer is at rest with one of them. Where is the center of gravity for the collision. Does it simply stop relative to our observer? What about the (not yet mentioned) observer riding along with the other proton. Does the center of gravity zing away at the speed of light relative to him? How can that be?

I don't dispute that.

Slower, but not stopped.

Your momentum is relative to only one observer, thereby suggesting a preferred reference frame.

I never said that. I simply said it wouldn't stop (relative to the earth).

They are conserved separately, but they must be fully conserved within the isolated system. One cannot be transformed into the other.

This post has been edited by ubavontuba on Oct 28 2007, 12:09 AM

You only think you're making a point because you don't understand it.

One photon? You're funny.

Matter/antimatter annihilation is relatively the single most powerful release of energy possible.

This post has been edited by ubavontuba on Oct 28 2007, 12:05 AM

You mean to say:
As I have already said, if we were to measure the rest MASS of a Proton with a total energy of 1150 TeV, we would find that it only weighed 4.7 TeV. This includes the other nonsense that you keep blithering about.

No, because I understand the difference between measuring the mass of an object travelling at relativistic speeds, and measuring the rest mass of an object.

You, however, have just demonstrated, once again, that you have no clue what you're talking about.

I said mass, and I meant mass. The rest mass of a proton is constant, and it's 938 MeV (The rest mass of a Neutron is 939 MeV).

The mass of a proton travelling at 0.9999c is 4.7 TeV.
The total energy of a Proton travelling at 0.9999c is 1150 Tev.

Savy?

No, I doubt that you do.

And you apparently still have no clue what Rest energy is.

No, because I understand the difference between measuring the mass of an object travelling at relativistic speeds, and measuring the rest mass of an object.

You, however, have just demonstrated, once again, that you have no clue what you're talking about.

I said mass, and I meant mass. The rest mass of a proton is constant, and it's 938 MeV (The rest mass of a Neutron is 939 MeV).

The mass of a proton travelling at 0.9999c is 4.7 TeV.
The total energy of a Proton travelling at 0.9999c is 1150 Tev.

Savy?

No, I doubt that you do.

And you apparently still have no clue what Rest energy is.

Fine, relativistic mass then. I was just trying to keep it simple for you.

Are you arguing the point about kinetic energy?

Are you arguing the point about the preferred reference frame?

Are you arguing the point about conservation?

This post has been edited by ubavontuba on Oct 28 2007, 01:19 AM

You arrogant jackass.

And you wonder why I loose my temper with you sometimes?

WHAT point about Kinetic energy?

NOWHERE has anything I said implied a preferred refference frame. I've been through this with you, but apparently you still can not, or will not understand.

WHAT point about conservation.

EVERYTHING I have said conserves momentum before and after the collision.
EVERYTHING I have said conserves Total energy before and after the collision.

Presumably we're creating a Kerr black hole so Charge and Spin are conserved.

And that's all there really is.

Again. You're failing to grasp the simple concept of what rest energy actually is.

Ubavontuba.

It has occured to me why you think that what I have said violates causality, and I'll admit, it took me sitting down and working it out on paper to figure it out.

You're ignoring the Oxygen atom that the cosmic ray is colliding with.

An observer in a reference frame that is co-moving with the cosmic ray (let's call it a proton as that's what we've been dealing with) observes the Proton to be stationary relative to themselves, and so measures the protons rest mass to be equivalent to it's mass, and all the rest of that.

But the energy and velocity of the Oxygen atom are no longer insignificant.

In the reference frame of the co-moving observer, the Oxygen atom has a velocity of (about) 0.9999 c, a mass of about 7.5 TeV, and a total energy of god only knows what, because I can't be bothered working it out. (Give or take binding energy).

So... Once you take the whole system into account...

Believe me or don't, it's up to you, but from the point of view of the co-moving observer you can no longer ignore the total energy of the Oxygen atom that the Proton collides with, which you, in your protestations that I'm wrong, and violating casuality have been doing.

What ubavontuba surely lacks is primary level education, and sadly demonstrates the incapacity of sound reasoning. Most normal people instictively 'weigh up data' in a pefectly sensible manner, whilst typically brain damaged hapless, unfortunate people trend towards the ubavontuba mind-set.

It's very, very sad really.

nothing personal ubavontuba.

Take heart, you're never alone whilst the likes of kaneda, Mr.Robin Parsons exist.

Thanks for the encouragement bunny.

Keep this up and I'll begin blocking you again.

Why do you lose it at all? Have you no self-control?

You asked me where the other 1145TeV was. I answered.

If you forget the earth for a moment. Your model is the same as this:
Two protons approach each other at relativistic speeds. Our observer is at rest with one of them. WHAMMO! They collide. Where is the center of gravity for the collision? Does it simply slow to less than 11km/s relative to our observer (as you contend)? What about the (not as yet mentioned) observer riding along with the other proton. Does the center of gravity zing away at the speed of light relative to him? How can that be?

Why does the earthbound observer see a "slowing" (note: "slowing" is relative to him), but the other does not? You're preferring one observer over the other (they don't both see the same thing). The only way they could see the same thing in your case is if the collision result somehow cloned itself and both observers had their own little collision result to watch.

The truth is:
Both observers will see their respective particle mass' accelerated (rearward) by the collision and the incoming mass is respectively decelerated. The inertia of the observers will keep them moving along uniformly, but the collision result will appear to fall behind them (literally accelerate away from them).

There's a strange relativistic effect here though (Walter L. Wagner saw it). It's pretty complicated so you might want to skip this part.

The collision result can appear to move rearward at near light velocity (each observer might think it's following along with the other observer). This is a time dilation/relativistic effect. It's dependent on the relative energy/velocity of the observers and particles before the collision.

The point that each form of conservation must be conserved within the isolated system, together.

No it doesn't. You keep ignoring the momentum of the kinetic energy. You think it only converts to mass (an energy/mass transformation). You don't understand that its momentum is conserved separately (it does not contribute to the mass). The momentum of the whole system (including the KE that converts to mass) is conserved.

No it doesn't. The energy/mass conversion doesn't change the total mass of the system (even though you think it does). The mass is always there (largely in the form of KE). There is no increase in mass (increasing the mass would be a violation of the mass/energy conservation).

Maybe not.

"Rest" is relative.

This post has been edited by ubavontuba on Oct 28 2007, 11:32 PM

Now you're just being silly. A cosmic ray proton cannot collide with a whole oxygen atom, all at once.

Sure.

Right. Our observer sees the oxygen atom as moving relativistically toward his particle!

Sure.

Sure I can. It's only the part of the oxygen atom that the proton can collide with that's relevant.

Do you have anything relevant to add to the discussion?

More importantly, can you add anything relevant to the discussion?

Ubavontuba.

The nucleon the proton is colliding with is part of a whole atom, you can not simply ignore the rest of the atom.

At work I have a really nifty video of a fighter jet rocketing on a rail sled and hitting a concrete wall (it's probably available on YouTube). The wall is a little narrower than the jet's wingspan.

In slow-motion, you can see the jet vaporize against the concrete, but the wingtips carry onward past the concrete barrier, without apparent affect. No jerk, no apparent reaction.

Lesson: Energy absorption cross sections are narrowed by relative momentum.

As the concrete barrier was unable to affect the wingtip's momentum, so would most of the oxygen atom be unaffected by the cosmic ray proton.





This post has been edited by ubavontuba on Oct 29 2007, 05:45 AM