Large Hadron Collider finds no signatures of microscopic black holes

Dec 19, 2010
An example of a CMS event with large total transverse energy (ST=1.3 TeV) and high jet multiplicity (10 jets, denoted by yellow cones and lines), as expected from Standard Model processes. Such events are a background to the search for microscopic black holes.

The CMS experiment at CERN's Large Hadron Collider (LHC) has completed a search for microscopic black holes produced in high-energy proton-proton collisions. No evidence for their production was found and their production has been excluded up to a black hole mass of 3.5-4.5 TeV (1012 electron volts) in a variety of theoretical models.

Microscopic black holes are predicted to exist in some theoretical models that attempt to unify General Relativity and by postulating the existence of extra "curled-up" dimensions, in addition to the three familiar spatial dimensions.

At the high energies of the , such theories predict that may collide "closely enough" to be sensitive to these postulated extra dimensions. In such a case, the colliding particles could interact gravitationally with strengths similar to those of the other three fundamental forces – the Electromagnetic, Weak and Strong interactions. The two colliding particles might then form a microscopic black hole.

If it were so produced, a microscopic black hole would evaporate immediately, producing a distinctive spray of sub-atomic particles of normal matter. These would then be observed in the high-precision CMS detector that surrounds the LHC collision point. CMS has searched for such events amongst all the proton-proton collisions recorded during the 2010 LHC running at 7 TeV centre-of-mass energy (3.5 TeV per proton beam).

No experimental evidence for microscopic black holes has been found. This non-observation rules out the existence of microscopic up to a mass of 3.5–4.5 TeV for a range of that postulate extra dimensions.

The CMS results have been submitted for publication in the Physics Letters B journal. CMS will take much more data next year when the LHC resumes running in early 2011 after a brief technical stop.

Explore further: And so they beat on, flagella against the cantilever

More information:
-- Search for Microscopic Black Hole Signatures at the Large Hadron Collider, arXiv:1012.3375v1 [hep-ex] arxiv.org/abs/1012.3375
-- CERN's Scientific Summary

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Mercury_01
4.4 / 5 (10) Dec 19, 2010
The headline should read:


"Large Hadron Collider finds no signatures of microscopic black hole decay"
solar2030
Dec 19, 2010
This comment has been removed by a moderator.
Quantum_Conundrum
1.8 / 5 (20) Dec 19, 2010
yeah well, nobody has ever directly, knowingly observed so much as one particle or wave from ANY black hole's decay to confirm the theory using a known black hole.

So if they've never seen a particle from something they KNOW is a black hole, how could they be sure they have or haven't seen a black hole decay particle from something they DON'T KNOW is a black hole?

This is as absurd as two people who were born blind trying to explain a rainbow to one another, when neither has ever seen light at all, nevermind the full spectrum of color.
Noumenon
4.6 / 5 (69) Dec 19, 2010
QC, there would be no unique 'black hole particles' as no particles could ever come FROM a black hole. Micro black holes would evaporate by the virtual particle-antiparticle mechanism of hawking, and particles and anti-particles are things known.
Quantum_Conundrum
1.2 / 5 (18) Dec 19, 2010
QC, there would be no unique 'black hole particles' as no particles could ever come FROM a black hole. Micro black holes would evaporate by the virtual particle-antiparticle mechanism of hawking, and particles and anti-particles are things known.


1) I read the book. Still got a copy of it right now, in fact.

2) It's quite a different matter between seeing something and trying to make a hypothesis or theory to explain it, as compared to theorizing something ridiculously hard to find actually exists and then trying to go find it.
yyz
5 / 5 (13) Dec 19, 2010
"It's quite a different matter between seeing something and trying to make a hypothesis or theory to explain it, as compared to theorizing something ridiculously hard to find actually exists and then trying to go find it."

Like, um, neutrinos: http://en.wikiped...rguments
Noumenon
4.7 / 5 (65) Dec 19, 2010
,... and the positron, etc , etc

@QC,... The point is given the mechanism of black hole evaporation, there would result a unique detectable signature. Since it's not possible to create black holes (in our detectable 3 dim) with existing colliders the above finding places a limit on the possibility of the existence of extra curled dimensions, which is quit ironic given you last paragraph.
StillWind
1.1 / 5 (34) Dec 19, 2010
Just more BS from the guys who want more of your money to sit around and play with an enormously expensive toy that is little more useful than "the machine that goes PINGGGG" in Monty Python's Meaning of Life.
There is no science here.
Since the existence of any black hole is nothing but speculation, let alone something as exotic as "micro-black holes", how absurd to suggest that we have any idea how it would "act", or what would be produced, should it decide to "evaporate".
Millions of people are dying from want, and our society is so self-absorbed and decadent that we will pay people to play a video game on an enormously expensive game console that will produce absolutely no useful information, or anything of value.
Noumenon
4.7 / 5 (72) Dec 19, 2010
@StillWind, Your logic makes zero sense. If human progress was perpetually contingent on the non-existence of Need, we would still be living in caves.

The standard of living and life expectancy around the world has increased as the centuries advance, as a result of investing money into research, and competition and self-absorbed egoism,... the intrinsic mechanisms responsible for a successful capitalistic society.
Dummy
1.2 / 5 (22) Dec 19, 2010
The standard of living and life expectancy around the world has increased as the centuries advance...

Really? Says who? Ever see pictures of the old American Indians? They lived well into their 90's and virtually cancer-free.
lexington
5 / 5 (16) Dec 19, 2010
All of them?
Roj
3 / 5 (6) Dec 19, 2010
up to a black hole mass of 3.5-4.5 TeV

Quantum-scale black holes may require a higher critical energy, similar to the critical mass requirement for nuclear weapons detonation.

The necessity of experimentation is the mother of invention. Give the LHC a few more tries.

With enough energy the whole facility could implode revealing the spectacular signature of billions of burned-up Euros.
Noumenon
4.3 / 5 (70) Dec 19, 2010
Really? Says who? Ever see pictures of the old American Indians? They lived well into their 90's and virtually cancer-free.
Say's who. It's a plain fact that life expectancy has risen in more modern societies. We got smoking from American indians,.. I doubt they were cancer free. Savages generally don't diagnose themselves accurately.

ubavontuba
2.3 / 5 (24) Dec 19, 2010
Hawking radiation can't work, as it would be a conservation violation.

According to Hawking, as the VP pair splits, one particle escapes and one falls in. He forgot to include the GP/KE for the infalling particle in his calculation. This is real energy that must exist in the system, as described. When you include it, the violation becomes apparent.

So a lack of Hawking radiation is no surprise. I'd be more concerned with jet quenching (already reported).
beelize54
1.2 / 5 (9) Dec 19, 2010
The headline should read: "Large Hadron Collider finds no signatures of microscopic black hole decay"
If it would be true, then the string theory would be confirmed with it. But it wasn't.

http://www.math.c.../?p=3333

BTW String theory failed supersymmetry test, too.

http://www.math.c.../?p=3338
Husky
1.5 / 5 (8) Dec 19, 2010
maybe its not just making thing dense, but BIG enough as well, for example if we assume mass to displace an hypothized aetheric medium of the vacume and that this aetheric medium could interact with black hole formation, than the nucleus of a star would displace a lot of medium and be less affected by the forces of the medium, while the super tiny droplet of manmade hot quark-gluon finds itselve completely surrounded by a high grade vacume medium in the LHC tubes, wich could be the showstopper in black hole formation
Cave_Man
1.3 / 5 (13) Dec 19, 2010
id still be sleeping alot more sound if they dont run the damn thing at all in december 2012, spontaneous creation of a black hole could very well transcend causality since it distorts time so much from gravitational effects not to mention we still cant explain a black hole very well past the event horizon
sstritt
3.9 / 5 (14) Dec 19, 2010
Hawking radiation can't work, as it would be a conservation violation.

That's pretty bold of you to dismiss Hawking so matter of factly.
Raveon
1.7 / 5 (12) Dec 19, 2010
String theory is bogus, always has been. Should have been obvious the second extra dimensions were required. I understand people think it's useful but give up on the extra dimensions.

I don't even think time is a dimension. Time only exists for us. It doesn't exist for the universe because there is no past or future for it, there is only the present. The past exists for us because we can remember it and the future exists because we can imagine it. The universe has no memory or imagination, only life does.
lexington
4.1 / 5 (10) Dec 19, 2010
Hawking radiation can't work, as it would be a conservation violation.

According to Hawking, as the VP pair splits, one particle escapes and one falls in. He forgot to include the GP/KE for the infalling particle in his calculation. This is real energy that must exist in the system, as described. When you include it, the violation becomes apparent.


The energy of the infalling particle is negative.
Quantum_Conundrum
1 / 5 (6) Dec 19, 2010
maybe its not just making thing dense, but BIG enough as well


Maybe there is an incorrect interpretation of Special Relativity as it regards the mass and/or gravity of an object, or maybe there are special cases that violate relativity due to quantum effects.

If you can only have quantum units of mass, then you can only have quantum units of energy and gravity. If you can only have quantum units of gravity, then you can no longer have a black hole caused by a vertical assymptote.

Microscopic black holes come from the notion that if you have a point mass, then over a small enough distance it should have an event horizon. However real mass isn't a point, it's always associated with a particle with a width that always exceeds the limit under normal circumstances. The purpose of colliding is to try to get the energy from the velocity to count as momentum, and consequently relativistic mass.

It may be that the universe doesnt apply relativistic mass to gravity.
Quantum_Conundrum
1 / 5 (9) Dec 19, 2010
Ah, here is a possible internally consistent explaination.

It would take infinite energy to approach C from the left, that is to accelerate to speed of light.

So then to have a black hole with a gravitational acceleration equal to c per second square, would equate to infinite "force," since by the principle of equivalence Einstein said you can't tell the difference between gravity and uniform acceleration in a rocket (actually he used a train, but modern physicist often use rocket).

So force to accelerate rocket by c per second is infinite. Makes no difference whether the rocket engine does it, or whether a black hole does it. Principle of equivalent says it's the same either way.

So then in order to have an event horizon would require infinite mass, since infinite force is required to accelerate anything to c, much less have an acceleration of C per second.

Problem is people started backwards, by putting C in the gravitational formula for Force. This is nonsense.
Quantum_Conundrum
1.4 / 5 (11) Dec 19, 2010
It is nonsense to say, "If I plug in C for "F" in newtons law of gravitation, what happens?"

Why? Because by definition of the lorentz transform, "C" cannot be obtained except by infinite force.

Infinite energy or infinite force cannot exist in a finite space, and even if it could, this puny particle accelerate can't get anywhere near infinite energy.

so the micro-black hole theory is pure fantasy and a fallacy caused by a mathematical blunder: Plugging in the number "C" for "F" when "C" is outside the range of the lorentz transformation and outside the range for "real world velocity". If C is outside the range for "real world velocity" then it certain is outside the range for "real world acceleration".

It is a fallacy of interpretation combined with a fallacy of the "plugging in" a value to an equation outside it's range.

Mathematically, there is no such thing as a micro-black hole, and maybe not even a true "macro" black hole...
Quantum_Conundrum
1 / 5 (9) Dec 19, 2010
Writing:

"C = gM/(r^2)"

or

"C/s^2 = gM/(r^2)"

And saying something like "r equal some constant 'a', so solve for M." or "M equal some constant 'b' solve for r."

Is an illogical question since Lorentz transform indirectly forbids A to equal C.

F = MA

But we measure acceleration in V/s^2.

In relativity V cannot equal C.

therefore A cannot equal C/s^2.

Therefore it is not possible for "F" to equal "MC/s^2".

I don't know why I've never noticed this, but it REALLY is a mathematical, algebraic error.

Again, if Relativity is correct, you cannot put C/s^2 for acceleration, (alleged event horizon,) because it is outside the range of the lorentz transformation, which makes it nonsense.

This makes the entire science of black holes nonsense.

True black holes actually cannot exist mathematically, because V = C, and A = C/s^2 cannot exist.
Quantum_Conundrum
1 / 5 (10) Dec 19, 2010
Uh, this is "peer review" material.

Why I haven't noticed this before I got no clue, but I always catch stuff like this in other maths, butnot here.

I'm not hoaxing. I'm exactly right this time. It's been overlooked somehow.
Quantum_Conundrum
1 / 5 (8) Dec 19, 2010
Ah crap, my units are wrong, but what I'm saying is right.

A = m/s^2, or "rate of change in V".

v = at

So V cannot equal C.

c cannot equal at.

C/s^2 cannot equal a.

The value "V = C" is not in the range for lorentz transform, and so consequently all the other follows because of this.

All I know how to say is write out all of the relativistic formulas and write law of gravity out.

then recall that the range of C in the real world must be the same for any equation trying to explain the real world. Then you will see what is been overlook.

"A = C" or "A = speed of light per second" is nonsense since V cannot equal C.

You cannot make a black hole in a collider, even if you could get infinite energy, which you can't.

They aren't even CLOSE to making a black hole.

If V cant equal C, then "True" black holes do not even exist in nature. What you are seeing in center of galaxy is "close but no cigar." Ultra-heavy red shift, but no "True" event horizon.
DamienS
4.1 / 5 (10) Dec 19, 2010
I'm not hoaxing. I'm exactly right this time. It's been overlooked somehow.

Ha, ha, ha.
Mercury_01
3.9 / 5 (11) Dec 19, 2010
If you just proved that a black hole cannot exist in nature, I will personally deliver you a golden brick from my bowels. And Ill do it in a Lamborghini.
Quantum_Conundrum
1 / 5 (8) Dec 19, 2010
I'm not hoaxing. I'm exactly right this time. It's been overlooked somehow.

Ha, ha, ha.


Don't give a damn.

1/sqrt(1 - (v^2/c^2))

is UNDEFINED if v = c.

Therefore v cannot equals c in any other hypothetical or model you run, as that would be a contradiction of special relativity.

If we assume relativity is correct, then at NO TIME can you EVER assume v = c for ANY massive object under any circumstances, because lorentz equation is undefined for v = c.

Nor can you ever assume "acceleration = speed of light per second."

You can laugh at me all you want.

Take Algebra 1 or a Calculus class and see what happens.

It is a fallacy.

By definition of Special Relativity, V cannot equal C.

Therefore you cannot set V = C, ever, in any equation, and then solve for something else. It violates a specific rule in algebra and calculus, but I can't remember the exact name off hand. Would require division by zero in lorentz transform, which is undefined.
DamienS
4.2 / 5 (12) Dec 19, 2010
Most cranks hang themselves with word salad, but you have excelled yourself by also serving up number salad. Bravo!
Quantum_Conundrum
1 / 5 (8) Dec 19, 2010
If you just proved that a black hole cannot exist in nature, I will personally deliver you a golden brick from my bowels. And Ill do it in a Lamborghini.


check it.

It's very simple. So easy a caveman can do it. Well, that's why it was overlooked. It's so bad...

I would discuss this on the main forum, but I'm on suspension.

I'm definitely right, shockingly right, shocking that nobody has ever noticed this before, because it's an 8th or 9th grade level algebra mistake, but it's hard to notice because of the nature of the situation.

It's a violation of basic rules of mathematics.

If you do not believe me write down all the formulas and just stare at that shit.

then like anywhere you would normally plug in a value for "v," such as in the "F" or the "A" on the left side of Newton's law when it regards a black hole, depending on how you framed it, instead plug in the expression "v = c," then stare at Lorentz transformation. v = c is undefined. c not in range of v
Quantum_Conundrum
1 / 5 (8) Dec 19, 2010
Most cranks hang themselves with word salad, but you have excelled yourself by also serving up number salad. Bravo!


It is not "number salad".

This is a matter of violation of a mathematical proof.

By defintion of Special Relativity, C is not in the range of V.

Therefore "C" is not a legal input for "V" in any formula that uses "V". Nor can any formula that uses "V" equate to a value that would require "V" to be equal to "C".

This really is correct.

I'm absolutely positive and shocked at realizing it, but it's definitely correct.

It's been a long time since I took a formal math class, so I don't remember the exact names of the rules involved, but I know I'm right.

It's kind of like how you don't forget how to differentiate a polynomial, even if you forget the proof itself.
DamienS
4.2 / 5 (10) Dec 19, 2010
This really is correct.

I'm absolutely positive and shocked at realizing it, but it's definitely correct.

It's been a long time since I took a formal math class, so I don't remember the exact names of the rules involved, but I know I'm right.

Spoken like a true crank.
lexington
5 / 5 (12) Dec 19, 2010
It is nonsense to say, "If I plug in C for "F" in newtons law of gravitation, what happens?"

Why? Because by definition of the lorentz transform, "C" cannot be obtained except by infinite force.


Uh, no, you can't replace F with c because F has to be a measure of *force* and c is only a measure of speed. This is toddler level logic, apples are not oranges.
Quantum_Conundrum
1 / 5 (8) Dec 19, 2010
Have you checked the math?

here it is for you in a more organized format.

gamma = 1/(sqrt(1 - (v^2/c^2))), v not equal c.

Even if Einstein's Theory didn't directly say V can't equal c, the equation would still forbid V = C.

However, he both said V can't equal C, and the equation mathematically forbids V = C.

Gravitation:

F = g mM/(r^2),

From above, V can never equal C.

Gravitational Acceleration:
A = gM/(r^2)

From above, V can never equal C.

F = MA

V = At, because V cannot equal C, then "At" cannot equal C.

In order to have an event horizon, i.e. a "true" black hole, you must have a situation where "At = V >= C," but this directly violates the second postulate of relativity, because once again, V cannot equal C, by definition of Special relativity.

It also directly violates the Lorentz Transformation, regardless of relativity.

At = V >= C violates the laws of physics.

By definition, this CANNOT be obtained by any massive particle.
Quantum_Conundrum
1 / 5 (8) Dec 19, 2010
It is nonsense to say, "If I plug in C for "F" in newtons law of gravitation, what happens?"

Why? Because by definition of the lorentz transform, "C" cannot be obtained except by infinite force.


Uh, no, you can't replace F with c because F has to be a measure of *force* and c is only a measure of speed. This is toddler level logic, apples are not oranges.


pay attention, it's very convoluted, but also ridiculously simple at the same time.

Force is required to have acceleration.

F = ma

At = acceleration times time = velocity

Velocity cannot equal c.

Therefore:

"At = V >= C" violates the second postulate.

Since this is a violation, you cannot get an event horizon through gravitational acceleration, because no matter how long you accelerate, V cannot and will not equal C. Even if you had an infinite mass object you could not have an event horizon, because V CANNOT equal C.

It's that simple.
Quantum_Conundrum
1 / 5 (9) Dec 19, 2010
The problem with black holes is people took Newton's law and INCORRECTLY plugged in values on the left side of his equation which are FORBIDDEN both by the second postulate of Relativity, and by the denominator of the right side of the Lorentz Transformation.

Once again, it is a violation of relativity.

Division by Zero is undefined, and V = C would require division by zero.

AND it violates at least one other axiom or postulate, etc, in mathematics, but I can't remember the name of it.
lexington
5 / 5 (10) Dec 19, 2010
Physicists aren't saying that black holes make anything move at the speed of light, so I don't see where your objection is. You're jumping between Einstein and Newton when it serves you, and then claiming that the mistake lies in the physics. The concept of surface gravity you're working with comes from Newton, it doesn't work properly with black holes and that's been known for a good ninety years now.

Like for example F=ma is incomplete in relativity so you shouldn't be using it when you talk about black holes.
Quantum_Conundrum
1 / 5 (8) Dec 19, 2010
Anyway, the notion that an object could become so massive as to warp space-time to accelerate another object to c (event horizon,) or trap a beam of light by bending it as much as 180 degrees (event horizon,) ends up being a violation of relativity. Not a consequence of relativity.

It is always said that something couldn't escape an alleged "black hole" because the instantaneous accleration at the event horizon is would allegedly require a V of greater than C to escape. This is false.

If Relativity forbids things to be accelerated to v = c, then black holes cannot exist, because relativity also forbids things accelerated by black holes to be accelerated to v = c.

Lorentz Transform require infinite force to approach C, regardless of where that "force" comes from, and you stil cannot actually reach C.
Quantum_Conundrum
1 / 5 (9) Dec 20, 2010
Like for example F=ma is incomplete in relativity so you shouldn't be using it when you talk about black holes.


The formula I used is not surface gravity. It's the gravitational formula for all distances from center of mass. There has never been a "theory of surface gravity".

In order for an object, an alleged black hole, to bend light back upon itself and trap it, as they say, "From which not even light can escape, bum, bum, bum," It would need to be accelerating photon.

Moreover, any massive particle at the exact same distance as photon would be equally accelerated.

So let's consider a "black hole candidate" and an atom in space, both initially at rest with respect to one another.

If left alone, in pre-einstein classical physics atom woudl simply accelerate inverse proportional to distance, non-stop till it hit object.

In relativity, atom will accelerate arbitrarily close to c, but never equal c.
theon
1 / 5 (1) Dec 20, 2010
Strange title: "Einstein was wrong". It was Steven Hawking who went to TV to announce they would be created. It was Hawking who is wrong. Now clear tofor everybody.
ubavontuba
2.2 / 5 (17) Dec 20, 2010
Hawking radiation can't work, as it would be a conservation violation.

That's pretty bold of you to dismiss Hawking so matter of factly.
I'll take that as a compliment. Thank you.

I'm just sad that so many have been blind to his obvious omission, for so long.
ubavontuba
2.3 / 5 (15) Dec 20, 2010
Hawking radiation can't work, as it would be a conservation violation.

According to Hawking, as the VP pair splits, one particle escapes and one falls in. He forgot to include the GP/KE for the infalling particle in his calculation. This is real energy that must exist in the system, as described. When you include it, the violation becomes apparent.


The energy of the infalling particle is negative.
Relative to a distant observer this is correct (as far as he went). However, even a distant observer will agree there is GP/KE between the infalling particle and the black hole. It's this energy that he missed. And, it's not inconsequential.
Ethelred
4.7 / 5 (12) Dec 20, 2010
Even if Einstein's Theory didn't directly say V can't equal c, the equation would still forbid V = C.
The velocity of light does equal C. So for massless particle V can and DOES equal C.
F = g mM/(r^2),

From above, V can never equal C
Yes. Especially since V isn't in that equation.
Gravitational Acceleration:
A = gM/(r^2)

From above, V can never equal C.
Yes and that is an equation has a severe shortage of V in it. However I am willing to accept the idea that nothing with mass can accelerate to C no matter how long it accelerates.
F = ma

At = acceleration times time = velocity

Velocity cannot equal c.
Yes but that isn't the right formula for relativistic velocities. Thats Newtonian.
Even if you had an infinite mass object you could not have an event horizon, because V CANNOT equal C.
You can't just claim V is where it isn't.

You are Cranking. And by Main Forum I am guessing you mean Physicsforums.com separate place. This IS the main forum here.

Ethelred
ubavontuba
2.3 / 5 (16) Dec 20, 2010
Believe it or not, Quantum_Conundrum makes a valid point. It's long been postulated (strictly in GR terms) that the boundary of the Schwarzschild radius is the boundary of a bubble in which time stops (relative to an outside observer) due to the acceleration he describes. Therefore, black holes cannot collapse into a singularity in GR spacetime. However, this solution was generally unacceptable so physicists started using other theories on black holes (like quantum field theory).

See:

http://en.wikiped...lativity
ubavontuba
2.1 / 5 (14) Dec 20, 2010
Ethelred
4.3 / 5 (9) Dec 20, 2010
It's long been postulated (strictly in GR terms) that the boundary of the Schwarzschild radius is the boundary of a bubble in which time stops (relative to an outside observer) due to the acceleration he describes.
Which is reasonable at the boundary.
Therefore, black holes cannot collapse into a singularity in GR spacetime.
Not related. That is the boundary is just that. It doesn't stop the CONTENTS of the Black Hole, that were there from the beginning, from collapsing into a singularity. What you say should only apply to the matter falling into a Black Hole. That matter never passes the boundary according to GR. So a stellar mass Black Hole should actualy have two not-really volumes where the mass is concentrated, a singularity, or whatever it is, inside and a shell at the event horizon. Super Massive Black Holes should have most of the Mass at the Event Horizon, barring Black Hole collisions where things should get even stranger.

Ethelred
ubavontuba
2 / 5 (16) Dec 20, 2010
Ha, ha, ha.
Weak.
Most cranks hang themselves with word salad, but you have excelled yourself by also serving up number salad. Bravo!
A soliloquy about ... well, nothing.
Spoken like a true crank.
I think it's rather obvious who the true cranks are here. It's the adolescent flamers like you, who have nothing real to say.
PhysicsVanAwesome
4.5 / 5 (11) Dec 20, 2010
QC, do you have any formal education in physics or advanced mathematics? Wikipedia doesn't count and neither does reading 'pop science' books.
I don't know where you get off making bold claims about subject matter that is clearly tenuously within your grasp. Hawking not taking conservation into account...absurd...the man is brilliant. If there were something, anything, not taken into consideration, I assure you, it wasn't on Hawking's end. It is better to be silent and thought a fool than open one's mouth and remove all doubt.

Once again, Ethelred ftw.

ubavontuba
1.9 / 5 (17) Dec 20, 2010
Which is reasonable at the boundary.
Which is about as far as GR (strictly speaking) can go.
Not related. That is the boundary is just that. It doesn't stop the CONTENTS of the Black Hole, that were there from the beginning, from collapsing into a singularity. What you say should only apply to the matter falling into a Black Hole. That matter never passes the boundary according to GR. So a stellar mass Black Hole should actualy have two not-really volumes where the mass is concentrated, a singularity, or whatever it is, inside and a shell at the event horizon. Super Massive Black Holes should have most of the Mass at the Event Horizon, barring Black Hole collisions where things should get even stranger.
You are sort of correct. In GR (again, strictly speaking) no singularity can exist that wasn't formed in the Big Bang. That is, the timeline for the collapse must date to the very beginning of the universe. Therefore, singularities can't spontaneously form.
ubavontuba
1.9 / 5 (17) Dec 20, 2010
QC, do you have any formal education in physics or advanced mathematics? Wikipedia doesn't count and neither does reading 'pop science' books.
I don't know where you get off making bold claims about subject matter that is clearly tenuously within your grasp. Hawking not taking conservation into account...absurd...the man is brilliant. If there were something, anything, not taken into consideration, I assure you, it wasn't on Hawking's end. It is better to be silent and thought a fool than open one's mouth and remove all doubt.

Once again, Ethelred ftw.
It appears, sir, as if you have confused QC with myself. I assure you, we are not one and the same.

What makes you think Hawking is so infallible? Is he not a man, not unlike any other?

If you understood that of which you speak, you would know that I do not claim that Hawking didn't consider conservation in his hypothesis, but rather he missed something which destroys his conservation considerations.
PhysicsVanAwesome
4.6 / 5 (9) Dec 20, 2010
No, your name is clearly not Quantum_Conundrum. The confusion lies on your end.
PhysicsVanAwesome
4.5 / 5 (12) Dec 20, 2010
Also, I made no assertion that Hawking was infallible. He isn't the pope of physics. However, not taking conservation into account is something that one may do in physics 101, not when one is a well respected expert and pioneer in one's field.
ubavontuba
2.1 / 5 (15) Dec 20, 2010
Also, I made no assertion that Hawking was infallible. He isn't the pope of physics. However, not taking conservation into account is something that one may do in physics 101, not when one is a well respected expert and pioneer in one's field.
Quite. But even you must realize that if one's conservation considersations do not encompass the entire system in question, then these considerations may be invalid.
Ethelred
4 / 5 (8) Dec 20, 2010
In GR (again, strictly speaking) no singularity can exist that wasn't formed in the Big Bang.
Not true. GR simply can't handle a singularity at all. That doesn't mean they, or something close like a Plank length mass, can't exist. There is nothing in GR that stops the collapse withing the Event Horizon. It runs into an infinity but the collapse itself MUST occur. The only question is whether you get an actual singularity, breaking the math in GR, or does something stop it. When the collapse of a stellar mass starts the gravity that causes the effective end of time at the Event Horizon is ONLY at the Event Horizon. Within the Event Horizon gravity is LOWER. For instance the gravity of ANY hollow sphere is ZERO within the hollow. At the center of the mass the gravity is zero for ALL spheres hollow or not. Thus time does not stop within the Event Horizon.

Ethelred
Shootist
3.1 / 5 (17) Dec 20, 2010
Just more BS from the guys who want more of your money to sit around and play with an enormously expensive toy that is little more useful than "the machine that goes PINGGGG" in Monty Python's Meaning of Life.
There is no science here.
Since the existence of any black hole is nothing but speculation, let alone something as exotic as "micro-black holes", how absurd to suggest that we have any idea how it would "act", or what would be produced, should it decide to "evaporate".
Millions of people are dying from want, and our society is so self-absorbed and decadent that we will pay people to play a video game on an enormously expensive game console that will produce absolutely no useful information, or anything of value.


More left-wing nonsense.

We'd all still be sitting in drafty caves if left to your devices.
Shootist
3.9 / 5 (11) Dec 20, 2010
The problem with black holes is people took Newton's law and INCORRECTLY plugged in values on the left side of his equation which are FORBIDDEN both by the second postulate of Relativity


Tell it to Feynman.

http://www.scribd...vitation
ubavontuba
2.3 / 5 (16) Dec 20, 2010
There is nothing in GR that stops the collapse withing the Event Horizon. It runs into an infinity but the collapse itself MUST occur.
An interesting point of view, but it isn't valid in GR. It's not like a collapsing star can instantly throw out an arbitrarily large event horizon and "lasso" a bunch of mass.

Also, your lower gravity inside thing is invalid. Strictly speaking, time dilation is relative to the center of mass/gravity, not the strength of gravity in a particular place within the gravity well. For instance, time moves slower at the center of the earth (where there's no apparent gravity) than it does at the surface (where the gravitational acceleration is most strongly "felt"). Or, gravitational time dilation is related to the gravitational potential. The deeper you go into the field, the slower time goes. When you move into the center of the earth, you're moving deeper into the gravity field, hence time is slower.
Ethelred
3.9 / 5 (7) Dec 20, 2010
1/2
It's not like a collapsing star can instantly throw out an arbitrarily large event horizon and "lasso" a bunch of mass.
The mass is already there. It is where the gravity is coming from. Do you think an Event Horizon just appears in a vacuum? Mass produces it. The gravity of ANY spherical mass is strongest at the surface of the mass. Which is about where the Event Horizon forms.
Also, your lower gravity inside thing is invalid.
Sorry but it is valid.
Strictly speaking, time dilation is relative to the center of mass/gravity, not the strength of gravity in a particular place within the gravity well.
No. It IS due to strength or rather the warping of space. Space is flat in a hollow sphere for instance.

In a neutron star space is flatter towards the center of the star. The zone of maximum warping of space-time is at the surface of the neutron star. It is the same distribution when the star has accreted enough mass to collapse.

More
Ethelred
4.3 / 5 (11) Dec 20, 2010
2/2
For instance, time moves slower at the center of the earth (where there's no apparent gravity) than it does at the surface
Where did you get that idea? That denies both SR and GR. SR due to the rotation of the Earth.

(where the gravitational acceleration is most strongly "felt").
Not just felt. It is where space-time is most strongly warped. This where gravity comes from. Warped space-time. The warping of space-time is where time dilation due to gravity comes from. More strongly warped the slower time passes in that area. The flatter space is the less time slows.
When you move into the center of the earth, you're moving deeper into the gravity field, hence time is slower.
Nope. There is no deepening after the surface. The warping of space-time begins to decrease.

You need to start thinking of it as a curved field in space-time. Not a gravity well.

Ethelred
Skeptic_Heretic
4.1 / 5 (9) Dec 20, 2010
QC has no understanding of where or how the variious relativity frameworks fit in or operate.

Most certainly educated by Google and not intellectually curious enough to expose himself to criticism.

This is further evidenced on this thread.
http://www.physor...ses.html
Mr_Man
4.3 / 5 (8) Dec 20, 2010
Just more BS from the guys who want more of your money to sit around and play with an enormously expensive toy that is little more useful than "the machine that goes PINGGGG" in Monty Python's Meaning of Life.
There is no science here.
Since the existence of any black hole is nothing but speculation ... we will pay people to play a video game on an enormously expensive game console that will produce absolutely no useful information, or anything of value.


You are entitled to your opinion.

Do you pay for cable TV? Movies? etc?

There are MANY MANY ways people spend money frivolously that doesn't contribute to solving the world's problems. The LHC is providing useful information. The more we learn about our universe the further we can push technology. We may even learn how to successfully run a fusion reactor and have clean, near limitless fuel, or utilize quantum entanglement for uncrackable codes.

To say this is all just a waste is really just sad on so many levels
fmfbrestel
5 / 5 (7) Dec 20, 2010
http://hasthelargehadroncolliderdestroyedtheworldyet.com/
Modernmystic
1 / 5 (5) Dec 20, 2010
When is this thing supposed to reach its peak energies? Because up until now it's just employed a lot of engineers, construction workers, and physicists to generate press releases about things less interesting than pain drying.
Gawad
3.7 / 5 (6) Dec 20, 2010
2/2
For instance, time moves slower at the center of the earth (where there's no apparent gravity) than it does at the surface
Where did you get that idea? That denies both SR and GR. SR due to the rotation of the Earth.

Hi Eth, this is very similar to a discussion I had a while back on PO, and MaxwellsDeamon was kind enough to point me to the correct formula for calculating the correct time dilation, not just on the surface of a spherical mass, but anywhere with in. The correct formula can be found here:

http://hyperphysi....html#c4

One can see from the relationship between radius and mass that both ST curvature and position in the gravity well impact a would-be traveler trying to blast his way out of a dense mass. The center of the gravity well is where time is most dilated, the surface of the mass is where they will feel gravity most.

One thing to note, though, there is no mass shell at an event horizon, only red shifted light.
Skeptic_Heretic
4.2 / 5 (5) Dec 20, 2010
When is this thing supposed to reach its peak energies? Because up until now it's just employed a lot of engineers, construction workers, and physicists to generate press releases about things less interesting than pain drying.
I'd say the quark gluon plasma findings in relation to electromagnetic perpendicularity were rather astounding finds.
Gawad
3.3 / 5 (7) Dec 20, 2010
2/2

As M_D pointed out, in a beautiful example of symmetry between GR and SR, the GR time dilation at any point in the gravity well of an extended mass is exactly the same as the SR time dilation that would be experienced at escape velocity at the same point. Within the BH Event Horizon time dilation is such that space becomes time-like (flowing towards the singularity) but anything going down the gravity sink still feels perfectly "normal", at least until spaghettification. Whether it's the original mass or a space probe. At the singularity both GR and SR time dilation are infinite.
fmfbrestel
not rated yet Dec 20, 2010
Honest question for the physics pros on here: If a point of singularity is a place where relativity breaks down, why wouldn't proof of a singularity = disproof of relativity? It would seem that any observation of an astronomical object that cannot be explained by relativity would be a strong argument against the theory?

Despite the fact that relativity predicts them, it is still unable to deal with the actual singularity.
fmfbrestel
not rated yet Dec 20, 2010
There seem to be all these very elegant theoretical models describing what happens to the environment around a singulary, but a point of infinite density... -- wouldnt any theory which even predicts singularities be predicting its own failings as a theory?
Gawad
3.3 / 5 (7) Dec 20, 2010
Honest question for the physics pros on here: If a point of singularity is a place where relativity breaks down, why wouldn't proof of a singularity = disproof of relativity? It would seem that any observation of an astronomical object that cannot be explained by relativity would be a strong argument against the theory?

Despite the fact that relativity predicts them, it is still unable to deal with the actual singularity.
The only problem is that to observe a singularity (or whatever else would be in its stead) it would have to be "naked". While Hawking et al. have demonstrated that such "naked singularities" are not entirely theoretically ruled out in nature, the circumstances that could lead to such phenomena would be so unlikely as to probably never actually occur.
Gawad
3.4 / 5 (5) Dec 20, 2010
There seem to be all these very elegant theoretical models describing what happens to the environment around a singulary, but a point of infinite density... -- wouldnt any theory which even predicts singularities be predicting its own failings as a theory?
Indeed, GR is widely regarded as "failing" at the singularity. So does QM because the singularity is gravitational in nature (otherwise QM handles point entities just fine). This is currently our biggest scientific double whammy.
Skeptic_Heretic
4.4 / 5 (7) Dec 20, 2010
Honest question for the physics pros on here: If a point of singularity is a place where relativity breaks down, why wouldn't proof of a singularity = disproof of relativity?
Well that's a tough question to answer but I'll attempt. Relativity doesn't breakdown so much as it gives answers that we cannot comprehend. Infinity is "by definition" immesurable. Using the equations of relativity to measure infinity therefore provides a nonsensical "answer".
It would seem that any observation of an astronomical object that cannot be explained by relativity would be a strong argument against the theory?
More an indicator that we're missing something. Most relativity challenges have been found to be measurement errors of bad data thus far.
Despite the fact that relativity predicts them, it is still unable to deal with the actual singularity.
No, we're simply unable to understand the answer. The question is, is the answer wrong, or are we just too stupid.
lexington
5 / 5 (6) Dec 20, 2010
Honest question for the physics pros on here: If a point of singularity is a place where relativity breaks down, why wouldn't proof of a singularity = disproof of relativity? It would seem that any observation of an astronomical object that cannot be explained by relativity would be a strong argument against the theory?


It shows that relativity is an incomplete description of the universe but thanks to QM we've known that for almost as long as we've had relativity.
Modernmystic
2 / 5 (7) Dec 20, 2010
Something I find strange, without knowing the slightest bit about the math that "describes" a singularity is that we can't describe one no matter it's mass.

IOW our math breaks down no matter if it's a 10 solar mass black hole or a billion.

What I find even more interesting is that million solar mass black holes are VERY different from 10 solar mass ones. Why should they be? If all the matter is shrunk down to a point shouldn't they be the same "size" on the outside too?

Well we know they aren't.

I dunno, just rambling "aloud"...
Skeptic_Heretic
2.5 / 5 (4) Dec 20, 2010
What I find even more interesting is that million solar mass black holes are VERY different from 10 solar mass ones. Why should they be? If all the matter is shrunk down to a point shouldn't they be the same "size" on the outside too?
They are the same volume. The distortion of space around them is proportionate to their mass. Greater mass, greater S-T distortion. That might clear up some of the questions you're working with. If not let us know.
Modernmystic
2.4 / 5 (8) Dec 20, 2010
They are the same volume. The distortion of space around them is proportionate to their mass. Greater mass, greater S-T distortion. That might clear up some of the questions you're working with. If not let us know.


Are they? I've heard the smaller ones are about the size of the Earth, and some of the super-massive ones are about the size of the solar system. Size here being the circumference of the event horizon.

That aside for a moment, why should the mass matter? If gravity has "won" and taken the matter to it's "ultimate density" why should there be such wide variations in the size of the respective event horizons?

I always understood gravity to be very "distance dependent". I understand it's "mass dependent" too, but the actual size of the singularity is (and correct me if I'm wrong here) thought to be the same irrespective of mass.

Maybe that's telling us that the singularities in question aren't in fact points, but have some "size" to them after all?
Skeptic_Heretic
3 / 5 (4) Dec 20, 2010
Are they? I've heard the smaller ones are about the size of the Earth, and some of the super-massive ones are about the size of the solar system. Size here being the circumference of the event horizon.
Yes but the event horizon isn't the edge of the singularity, it is the after effect of the presence of the singularity. The more massive the singularity the larger the event horizon.
If gravity has "won" and taken the matter to it's "ultimate density" why should there be such wide variations in the size of the respective event horizons?
50lbs of pressure in a tire vs 40lbs of pressure. Which has more air? Did the tire change in volume, no, it did change in mass and content.
Maybe that's telling us that the singularities in question aren't in fact points, but have some "size" to them after all?
No, again, volume is the amount of space an object takes.
Mass is the content of that volume.

You can have a higher mass without a corresponding volume change.
ubavontuba
4 / 5 (12) Dec 20, 2010
Are they? I've heard the smaller ones are about the size of the Earth, and some of the super-massive ones are about the size of the solar system. Size here being the circumference of the event horizon.
As stated above, we can't observe a singularity, so we don't really know if they have a size. We only know the math leads us to certain (uncomfortable) conclusions. We do know that their mass is conserved relative to the mass which created it (and subsequently fell in), and gravity being a function of mass, the mass determines the size of the event horizon.

However, the event horizon isn't really a thing. It's a non-physical, unidirectional barrier which is really just an effect of spacetime curvature at a specified distance from the center of mass. More succinctly, it's not really a "size," per se. It's a distance.
Gawad
3.4 / 5 (5) Dec 20, 2010
Maybe that's telling us that the singularities in question aren't in fact points, but have some "size" to them after all?
Actually, virtually all realistic BH solutions show that the singularity in fact has "size", though not volume. The singularity is only a point (and always a point regardless of mass) in an idealized BH solution. Virtually all BH will have spin (and might have a very small charge, the combination of which kind of makes them like a giant fundamental partical—spin, mass, charge ;) resulting in an inner event horizon, an outer event horizon and an ergosphere, and at the centre an annular, one-dimensional ring singularity through which something could pass if it is small enough and light enough that its very presence does not mess up the BH geometry too badly. The size of the ring depends on the speed of the rotation rather than the mass. For additional info, search for Kerr black hole.
Aristoteles
1 / 5 (5) Dec 20, 2010
Probably the strength of gravity in nD increases like in 3D with decreasing distance ! ? This mean that
gravity is a quantuym-process not "field-interacting"
and also, that there isnt any taste of "antigravitation" ( hey Sabine from Tucson-Arizona uniwersity !...). [ Sabine invent "anticovariant derivative" and "antigeodescic equations"...] .
TabulaMentis
1 / 5 (3) Dec 20, 2010
Now we know where not to look.
fmfbrestel
5 / 5 (1) Dec 20, 2010
Thanks for the answers guys. hopefully as LHC ramps up to full power we still start to get some better observations for the theorists to play with.
Mr_Man
not rated yet Dec 20, 2010
http://hasthelargehadroncolliderdestroyedtheworldyet.com/


that is excellent! the site was probably put up by a physicist.
fmfbrestel
5 / 5 (1) Dec 20, 2010
here is how I imagine black holes, feel free to correct.

infinite density, + any amount of mass = 0 volume point

adding mass increases the size of the event horizon but aught to do nothing to the infinitely dense point at the center.

past the even horizon no amount of energy can resist the gravitational pull of the singularity. (so can things still orbit the singularity once inside the event horizon?)
fmfbrestel
5 / 5 (1) Dec 20, 2010
well, if a singularity can change rotational speeds, something would have to impact it with angular momentum... If so, then things must still be able to spiral down inside the event horizon. But i doubt anyone has observed anything that could be interpreted as a singularity changing rotational speed, so that seems totally hypothetical.
fmfbrestel
5 / 5 (1) Dec 20, 2010
ok, so a spinning toroidal singularity: zero volume in the toroid (so its more of a one dimensional ring) but the mass of the singularity would have to be equally distributed along the singularity (if its infinitely dense, you cant have part that is more dense then another part). so all points along the ring would have mass and be infinitely dense. --- how in the heck does the ring stay a ring? wouldnt it need to spin infinitely fast to maintain the minimum (infinite) energy required to remain a stable ring?
fmfbrestel
5 / 5 (4) Dec 20, 2010
mental math involving infinite sums hurts the head. You show me someone who can divide by zero, and I'll show you someone who can understand a singularity.
Modernmystic
3.8 / 5 (10) Dec 20, 2010
I'm getting it now.

Yes gravity is heavily distance dependent, but again also mass dependent. Therefore the event horizon is larger because the mass is larger irrespective of distance from the singularity.

IOW it's not the size (as in volume) of the singularity which makes for big event horizons, but the mass.
mojination
5 / 5 (2) Dec 20, 2010
The standard of living and life expectancy around the world has increased as the centuries advance...

Really? Says who? Ever see pictures of the old American Indians? They lived well into their 90's and virtually cancer-free.


And if they had invented vaccines they might be here today to help back your claim.
Terrible_Bohr
3 / 5 (2) Dec 20, 2010
To all who find the LHC's experimentation wasteful and boring: why are you bothering to read a physics website? Your "holier than thou" attitudes and lack of foresight suggest you would all fit right in at the nearest church.
ubavontuba
2.3 / 5 (8) Dec 21, 2010
mental math involving infinite sums hurts the head. You show me someone who can divide by zero, and I'll show you someone who can understand a singularity.
What's so hard about that? It's Simple Simon. If I divide a number by zero, then I haven't done anything, so the answer is there is no answer ...because I didn't do any math from which to derive an answer! See? Piece o' cake! LOL!

It's the fancy-pants pocket-protector wielding guys, with their insistence that there must be a solution to everything, that make a mess of it for everyone. ;D

Reference:

http://en.wikiped..._by_zero

ubavontuba
2.6 / 5 (5) Dec 21, 2010
ok, so a spinning toroidal singularity: zero volume in the toroid (so its more of a one dimensional ring) but the mass of the singularity would have to be equally distributed along the singularity (if its infinitely dense, you cant have part that is more dense then another part). so all points along the ring would have mass and be infinitely dense. --- how in the heck does the ring stay a ring? wouldnt it need to spin infinitely fast to maintain the minimum (infinite) energy required to remain a stable ring?

Naw. Here's a Space Daily article which explains it pretty well:

http://www.spaced...999.html
ubavontuba
2.3 / 5 (9) Dec 21, 2010
here is how I imagine black holes, feel free to correct.

infinite density, + any amount of mass = 0 volume point

adding mass increases the size of the event horizon but aught to do nothing to the infinitely dense point at the center.

past the even horizon no amount of energy can resist the gravitational pull of the singularity. (so can things still orbit the singularity once inside the event horizon?)

There are two answers. In a Schwarzschild (point singularity) black hole: No, as all paths point to the singularity.

However, in a Kerr (rotating) black hole, or a Reissner-Nordstrom (charged) black hole, the paths are smeared and all kinds of crazy things might happen.

See:

http://en.wikiped...gularity

Skeptic_Heretic
not rated yet Dec 21, 2010
However, in a Kerr (rotating) black hole, or a Reissner-Nordstrom (charged) black hole, the paths are smeared and all kinds of crazy things might happen.
I thought this model had been mathematically invalidated.
ubavontuba
2.6 / 5 (5) Dec 21, 2010
However, in a Kerr (rotating) black hole, or a Reissner-Nordstrom (charged) black hole, the paths are smeared and all kinds of crazy things might happen.
I thought this model had been mathematically invalidated.
What, would you so willingly lose the romance of time travel?
Rohitasch
1 / 5 (3) Dec 21, 2010
They haven't reached anywhere near the Planck mass yet! Absence of BH decay in the infra-Planck mass spectrum shouldn't be a suprise.
Rohitasch
1 / 5 (1) Dec 21, 2010

They haven't reached anywhere near the Planck mass yet! Absence of BH decay in the infra-Planck mass spectrum shouldn't be a suprise.
savroD
1 / 5 (1) Dec 21, 2010
There is a 4th spatial dimension that physics has missed!
suntraider
1 / 5 (7) Dec 22, 2010
If colleges is where speculations of democracy, socialism, and the general polectics of economics came from.

Then there exists a platform where the American Revolution, French Revolution, and Russian Revolution started from.

Now if the students for some reason saw the function of integrative university research to city process control technologies. Then that is natural to the flow of things mind then matter.

But if the students saw the direction as differentiated and loss of education but regardless research for urbanization must be scaled back to the social sciences and nessessary control.

Could this mechanism be a black hole? European colleges held in suspension while the US has rescaled the military and colleges open to the DREAM ACT. The british students creating defense to parliament that relates to schools. Rather than democracy creating defense relative to the american revolution. Then russia had some kind of issue relating to Vladamir Putin and France as well
Skeptic_Heretic
3.7 / 5 (3) Dec 22, 2010
However, in a Kerr (rotating) black hole, or a Reissner-Nordstrom (charged) black hole, the paths are smeared and all kinds of crazy things might happen.
I thought this model had been mathematically invalidated.
What, would you so willingly lose the romance of time travel?
Because reality isn't what I want it to be, it is what it is.

suntraider appears to be a Turing test.
suntraider
1 / 5 (7) Dec 22, 2010
If colleges is where speculations of democracy, socialism, and the general polectics of economics came from.

Then there exists a platform where the American Revolution, French Revolution, and Russian Revolution started from.

Now if the students for some reason saw the function of integrative university research to city process control technologies. Then that is natural to the flow of things mind then matter.

But if the students saw the direction as differentiated and loss of education but regardless research for urbanization must be scaled back to the social sciences and nessessary control.

Could this mechanism be a black hole? European colleges held in suspension while the US has rescaled the military and colleges open to the DREAM ACT. The british students creating defense to parliament that relates to schools. Rather than democracy creating defense relative to the american revolution. Then russia had some kind of issue relating to Vladamir Putin and France as well
yoron
2.3 / 5 (3) Dec 23, 2010
Hawking radiation postulates a radiation on the EV (event horizon) Due to the energy involved you see a lot of spontaneous 'pair productions'. The idea is that the the negative 'twin' annihilate the positive particle it meets inside the EV, and the positive then 'wander' in SpaceTime. If a positive particle falls in the negative can't survive outside since 'a negative-energy particle can't continue to exist outside the horizon for a time longer than h/E. So the black hole lose energy to vacuum fluctuations, but can't gain energy.'

And that's also the idea of 'information' communicating through the Event horizon as pair production implies an entanglement which the make the 'negative' particle communicate its state to the positive as it annihilates inside the EV.

You can see this two ways. Either an Entanglement has no limitations as we are discussing 'singularities'. And there you can as easily consider worm holes. Or the information is not 'meaningful' other than as 'energy quanta'.
yoron
1 / 5 (1) Dec 23, 2010
What you might argue against it is that an particle anti particle annihilation is expected to leave positive rest products. That should mean that the Black hole won't decrease proportionally to SpaceTime gaining? But I'm not sure on that one as the same could be said for all particle/anti particle annihilations inside SpaceTime too?

Anyone that have a opinion on that one?
yoron
1 / 5 (2) Dec 23, 2010
One reasonable approach would be to assume that the information lifted out is not 'meaningful' that is, leaves no information per se. Instead it 'sets' the energy quanta, spin etc. and then you need another concept defining 'meaningful information' as emerging only when the energy quanta interacts in complex patterns enough.

And yes, that would bring us to 'chaos theory'.
fmfbrestel
not rated yet Dec 23, 2010
Wow suntrader, have you taken your meds recently?

Anyway after doing some reading (thanks guys) ring singularities seem to be possible, but only if the ring has it's mass perfectly evenly distributed, which is why they are unstable. Because if any mass disturbed the ring, the gravitational equilibrium would collapse, and so would the ring.

Is that about right?
yoron
1 / 5 (2) Dec 24, 2010
And anyone thought of this. If we don't see them created,'evaporating' maybe they don't? So now they are down at the middle, eating silently, burping now and then?

What a exiting Xistmas :)

Naaah.

Joking, and a merry Xmas to ye all.
Skeptic_Heretic
not rated yet Dec 24, 2010
That should mean that the Black hole won't decrease proportionally to SpaceTime gaining? But I'm not sure on that one as the same could be said for all particle/anti particle annihilations inside SpaceTime too?
Actually it should leave a small positive quantity of rest products. EM energy is positive while gravity is negative energy. When the particle anti-particle pair splits and the negative particle encounters a positive particle inside the ev, the release of positive energy reduces the gravitational potential of the mass of the blackhole, thereby balancing in whole or in part for the release of gamma upon annihilation. I'd hypothesize that the balance for the resultant energy balance should follow the perceived asymmetry of matter and antimatter creation in the universe, leaving an infinitely small but positive potential force upon spacetime.

Been working on this one for a while, not ready for publication yet.
yoron
1 / 5 (2) Dec 24, 2010
Good Sceptic, tell me when you're ready. Also try to make it simple enough for me :)

I don't like people engaging in superfluous learning. Makes for a hard stomach, and, a probably not thought through proposition, well as I see it :)

But you're on it then?
Cool.
yoron
1 / 5 (1) Dec 24, 2010
Sorry, Skeptic not Sceptic. My English sux, especially when fighting with the elixir of life.

No, it's not water, although it takes part :)
yoron
1 / 5 (1) Dec 24, 2010
Rereading you, interesting idea :)

To me gravity is no energy. On the other hand you can release more energy than you otherwise would have, through it. But that energy is 'positive energy', not negative.

So you will have to be very clear to make me understand how you think there.
Husky
5 / 5 (1) Dec 24, 2010
yes, i suspect that the a black hole is indeed a bubble with a dense compressed shell and inside is a true timeless spaceness nothingness not even a vacume, in fact the inside would only exist as a conceptual idea, because the blackbubble would only exist in space/time by its outer shell
Husky
not rated yet Dec 24, 2010
quantum pertubations in the surface of the bubble shell would want to shrink the bubble eventually releasing hawking radiation in the process, thats why a micro black hole won't work, sure you got the density, but you need the sheer mass of a star nucleas to generate enough surface tension in the shell to keep the bubble from popping immediately, just like you need a minimum size for soap bubbles before you can blow them, its easier to make bigger soapbubbles
Husky
4 / 5 (1) Dec 24, 2010
you could do away with a density singularity approaching infinity at the center of a black hole if you assume that there is no center, no space at all,past the dense shell at the event horizon, this would do away with the infinite entropy problem stored in the classical black hole as well
Husky
5 / 5 (1) Dec 24, 2010
in fact one should be able to calculate the density of the shell and how this relates to the Plank scale by oberving the merging of a black hole with another black hole and the difference in radius of the old event horizon vs the new horizon radius, will tell you how thick of a layer was added by the merge, knowing the thickness of the layer and the mass contributed by the swallowed object you can derive the density
Skeptic_Heretic
4.3 / 5 (3) Dec 24, 2010
Rereading you, interesting idea :)

To me gravity is no energy. On the other hand you can release more energy than you otherwise would have, through it. But that energy is 'positive energy', not negative.

So you will have to be very clear to make me understand how you think there.

Going purely by the equations established by QM and Relativity, Gravity is always seen as a negative energy in the balance of the equations for spacial geometry. Since we're zeroing in on a value of Omega=1, gravity is required to be negative under the Lamba CDM model of cosmology.
yoron
1 / 5 (1) Dec 24, 2010
Ok, I will have to read up on that Skeptic :)

As for a center, that a definition we use inside an 'arrow of time'. without it, or even if only 'reversed' it becomes meaningless, I think?

Awh :)
theknifeman
not rated yet Dec 25, 2010
Ok, so I'm not one of you scientists. But I do have a serious question. If so much of this is based on the speed of light and particles moving at near the speed of light, why is it that two beams moving toward each other at near the speed of light aren't colliding at the equivalent of near twice the speed of light? Like two cars traveling at 100 MPH head on is the same energy as one car hitting a stationary object at 200 MPH.
theknifeman
5 / 5 (1) Dec 25, 2010
Ok, so I'm not one of you scientists. But I do have a serious question. If so much of this is based on the speed of light and particles moving at near the speed of light, why is it that two beams moving toward each other at near the speed of light aren't colliding at the equivalent of near twice the speed of light? Like two cars traveling at 100 MPH head on is the same energy as one car hitting a stationary object at 200 MPH. Then haven't you functionally exceeded the speed of light relative to colliding particles?
Skeptic_Heretic
4 / 5 (4) Dec 25, 2010
Ok, so I'm not one of you scientists. But I do have a serious question. If so much of this is based on the speed of light and particles moving at near the speed of light, why is it that two beams moving toward each other at near the speed of light aren't colliding at the equivalent of near twice the speed of light?
They are.
Like two cars traveling at 100 MPH head on is the same energy as one car hitting a stationary object at 200 MPH.
That's actually exactly how this is measured out. The top power for LHC collisions is a sum of each of the beams' individual powers. So a 5TeV collision is really two 2.5 TeV beams colliding.
Then haven't you functionally exceeded the speed of light relative to colliding particles?
No. It's a function of special relativity. What happens is the particles actually get heavier and time passes more slowly for them, making the speed of light appear as a constant.
ubavontuba
2.6 / 5 (5) Dec 25, 2010
Actually it should leave a small positive quantity of rest products. EM energy is positive while gravity is negative energy. When the particle anti-particle pair splits and the negative particle encounters a positive particle inside the ev, the release of positive energy reduces the gravitational potential of the mass of the blackhole, thereby balancing in whole or in part for the release of gamma upon annihilation.
Obviously based on thermodynamics, not GR.

Anyway,how does the resultant energy escape the ev?
I'd hypothesize that the balance for the resultant energy balance should follow the perceived asymmetry of matter and antimatter creation in the universe, leaving an infinitely small but positive potential force upon spacetime.
Why would it have to favor spacetime, and not the black hole?
Been working on this one for a while, not ready for publication yet.
yoron
1 / 5 (1) Dec 25, 2010
The resultant energy is the rest of the pair production, after the negative particle is 'eaten' by a positive inside the EV. And the information 'leaked' is the 'spin' that then will have to be set as it's 'other half' got 'eaten'.

Makes me head spin that one does:)

Because if it's true then I can't see any problem sending energy between 'parallel universe'. We already know that plants do it, by entanglement and somewhere I saw a proposition for testing sending energy through entanglements.
yoron
1 / 5 (3) Dec 25, 2010
It builds on the concept of a black hole as a 'singularity' of course. But if that is true, then we have a way to send 'energy' through the Event Horizon, and also a 'worm hole' if we like. Furthermore it's possible to assume such radiation already existing if so, maybe made by some more advanced culture.

And that's another reason for why we might need to split the concept of 'information' and 'meaningful information' just as we assumed that entanglements couldn't carry 'information' before.

It's interesting.

Maybe it can carry 'information' if it can carry 'energy'? But maybe not making it 'meaningful', depending on how we want to define that?

Still?
Isn't energy 'meaningful'??
Ask a plant :)
Skeptic_Heretic
5 / 5 (2) Dec 25, 2010
Obviously based on thermodynamics, not GR.
No, it is based on the balance equations for the Lamba CDM model of the universe, which is QM and GR.
Anyway,how does the resultant energy escape the ev?
It doesn't. E=mC^2 the escaping positive particle in the particle antiparticle pair is energy. Effectively you've created a particle outside of the EV by destroying a particle inside the EV with an antiparticle that has fallen into the EV. Returning the universe to energy balance has allowed you to utilize gravity and QM to "escape" a blackhole.
dtxx
3.4 / 5 (5) Dec 25, 2010
2) It's quite a different matter between seeing something and trying to make a hypothesis or theory to explain it, as compared to theorizing something ridiculously hard to find actually exists and then trying to go find it.


Oh you mean like the christian god?
ubavontuba
2.6 / 5 (5) Dec 25, 2010
No, it is based on the balance equations for the Lamba CDM model of the universe, which is QM and GR.
Well the Lambda CDM model certainly, I was referring more specifically to the process.

Anyway, as dark energy (DE) has an anti-gravity effect, wouldn't DE therfore have to be a form of negative mass then? How does all that fit in?
It doesn't. E=mC^2 the escaping positive particle in the particle antiparticle pair is energy. Effectively you've created a particle outside of the EV by destroying a particle inside the EV with an antiparticle that has fallen into the EV. Returning the universe to energy balance has allowed you to utilize gravity and QM to "escape" a blackhole.
Why would particle pair splitting favor antiparticles falling in, and particles escaping? Just because the energy is negative (from a distant observer's point of view), doesn't mean only antiparticles are falling in. And you still haven't explained how the particles escape (unless they're massless).
Skeptic_Heretic
4.4 / 5 (5) Dec 25, 2010
Anyway, as dark energy (DE) has an anti-gravity effect, wouldn't DE therfore have to be a form of negative mass then? How does all that fit in?
No, DE is positive energy, ie: emission, similar to electromagnetism.
Why would particle pair splitting favor antiparticles falling in, and particles escaping?
Never said it would.
Just because the energy is negative (from a distant observer's point of view)
Gravity is always negative.
doesn't mean only antiparticles are falling in.[q/] Again, didn't say that.
And you still haven't explained how the particles escape (unless they're massless).
The particle doesn't escape, it is annihilated and replaced by a different particle that already exists outside of the event horizon.
ubavontuba
2.3 / 5 (8) Dec 26, 2010
No, DE is positive energy, ie: emission, similar to electromagnetism.
Really? I didn't know we knew that much about it. AFAIK, it supposedly doesn't interact through any of the fundamental forces, other than gravity.

See: http://en.wikiped...k_energy
Never said it would.
As you described the process, it seems dependent on it.
Gravity is always negative.
It looks like you're simply trying to reinvent the wheel (so to speak). How is this fundamentally different than Hawking radiation?
The particle doesn't escape, it is annihilated and replaced by a different particle that already exists outside of the event horizon.
So are you advocating a form of quantum tunneling then?

And, are you suggesting that simply being outside the event horizon, is a guarantee of escape?
Skeptic_Heretic
3.7 / 5 (3) Dec 26, 2010
Really? I didn't know we knew that much about it. AFAIK, it supposedly doesn't interact through any of the fundamental forces, other than gravity.

No, that'd be dark matter, not dark energy.
It looks like you're simply trying to reinvent the wheel (so to speak). How is this fundamentally different than Hawking radiation?
It isn't.
So are you advocating a form of quantum tunneling then?
Yep.
And, are you suggesting that simply being outside the event horizon, is a guarantee of escape?
Nope.
yoron
1 / 5 (3) Dec 26, 2010
Thinking of my last comments. They are wild speculations only. Also, as positive energy get killed so quickly inside that Event Horizon?
So, Maybe not?

But if we then found a source of positive energy constantly expiring? Sounds very alike 'quantum fluctuations' if so :)

Wild wild speculations, a speciality of mine :)
ubavontuba
2.3 / 5 (6) Dec 26, 2010
No, that'd be dark matter, not dark energy.
Having trouble with the references, are we? My reference was about dark energy, not dark matter. Check it again.
Skeptic_Heretic
4.2 / 5 (5) Dec 26, 2010
No, that'd be dark matter, not dark energy.
Having trouble with the references, are we? My reference was about dark energy, not dark matter. Check it again.

I'm telling you that you're wrong. I'm not confused by your statement.
Negative energy exerting negative pressure upon baryonic matter is positive energy output. You're going to need to understand what these terms mean, and yes it gets confusing if you don't look at the equations.

Secondly, when it comes to conceptual physics, wikipedia gets it wrong more often than right.
Daan
3 / 5 (2) Dec 26, 2010
I'd be more concerned with jet quenching (already reported).

that is the first thing that came to mind here. Any more info on the lack of energy conservation in jet quenching? it reminds me of the discovery of neutrinos... the things that dont quite add up are the research fields of tomorrow.
ubavontuba
2.6 / 5 (5) Dec 27, 2010
Negative energy exerting negative pressure upon baryonic matter is positive energy output. You're going to need to understand what these terms mean, and yes it gets confusing if you don't look at the equations.
Are you aware that the quantum energy calculations for spacetime in regards to dark energy are about 10^120 times too big (a 1 with 120 zeros after it!)? How do you reconcile that?

The "quintessence" model I touched on previously (my "negative matter" reference) doesn't really make any more sense, I suppose. However, at least it isn't based on a gross miscalculation!
Secondly, when it comes to conceptual physics, wikipedia gets it wrong more often than right.
I disagree, as it's been independently rated as being quite accurate in regards to its science articles, but we'll move on...

If you're going QM, perhaps spacetime itself is gravitationally attractive - via VP's spitting out gravitons?
Egleton
1 / 5 (2) Dec 27, 2010
Black holes are an affront to my aesthetic. I do not approve.
Skeptic_Heretic
3.7 / 5 (3) Dec 27, 2010
Are you aware that the quantum energy calculations for spacetime in regards to dark energy are about 10^120 times too big (a 1 with 120 zeros after it!)? How do you reconcile that?
Are you aware that the vaccuum catastrophy only occurs when you're using old formulae without Dark Energy? Assuming that the cosmological constant, and Omega_lambda are the same is the reason for vaccuum catastrophy, however there is no logical reason why you would assume the two are the same.

You're making the same mistake that many physicists in the past have made by not reading forward on the issue.

Read: Weinberg, S (1987). Anthropic Bound on the Cosmological Constant. Phys. Rev. Lett. 59 (22):2607-610. doi:10.1103/PhysRevLett.59.2607. PMID 10035596. http://prl.aps.or.../p2607_1

for starters.
ubavontuba
2.6 / 5 (5) Dec 27, 2010
Are you aware that the vaccuum catastrophy only occurs when you're using old formulae without Dark Energy? Assuming that the cosmological constant, and Omega_lambda are the same is the reason for vaccuum catastrophy, however there is no logical reason why you would assume the two are the same.

You're making the same mistake that many physicists in the past have made by not reading forward on the issue.

Read: Weinberg, S (1987). Anthropic Bound on the Cosmological Constant. Phys. Rev. Lett. 59 (22):2607-610. doi:10.1103/PhysRevLett.59.2607. PMID 10035596.

for starters.
Well now you're digressing into anthropic philosophy. Variable cosmological constant. Why not?

Or, maybe it's magic!
ubavontuba
2.4 / 5 (9) Dec 27, 2010
Applying Occam's razor, what's most likely?

1. The cosmological constant, isn't constant.

2. The universe is made up mostly of stuff we can't see, which passes light unperturbed.

3. Something is critically wrong with our theories, or interpretations therof.

Personally, I vote for number three. Particularly, the latter part.

Skeptic_Heretic
3.8 / 5 (4) Dec 27, 2010
Applying Occam's razor, what's most likely?
You make a lot of assumptions in very few words.
1. The cosmological constant, isn't constant.
You mean that we're wrong?
2. The universe is made up mostly of stuff we can't see, which passes light unperturbed.
Which we have evidence for.
3. Something is critically wrong with our theories, or interpretations therof.
So number 1 again.

Personally, I vote for number three. Particularly, the latter part.
Same, and we know number 3 will come to pass, because it always does in physics.
ubavontuba
3 / 5 (4) Dec 27, 2010
Same, and we know number 3 will come to pass, because it always does in physics.
Well, as I stated above, that's what gets my vote.

I still like my semi-gravitationally attractive vacuum. What do you think?
Skeptic_Heretic
4.3 / 5 (3) Dec 28, 2010
Well, as I stated above, that's what gets my vote.

I still like my semi-gravitationally attractive vacuum. What do you think?
Eh, observation indicates that it is gravitationally repulsive. It's not a bad idea, just doesn't have much in the way of support.
ubavontuba
3 / 5 (4) Dec 28, 2010
Well, as I stated above, that's what gets my vote.

I still like my semi-gravitationally attractive vacuum. What do you think?
Eh, observation indicates that it is gravitationally repulsive.
Well, then you're talking about fantastical concepts like negative mass and anti-gravity.
It's not a bad idea, just doesn't have much in the way of support.
Time will tell.

Anyway, if it's semi-attractive (by this I mean it attracts mass, but mass can't attract it), it would have the effect of appearing to be repulsive on large scales (by dragging galaxies apart as the galaxies ever attempt to fill larger voids). It would also account for the accelerating expansion.

It's originally based on QM (supposing gravitons, like photons, are their own antiparticle).

It works in GR too, by supposing each VP pair creates a tiny gravity wavelet.

It's certainly highly speculative, but fun.
Skeptic_Heretic
3.7 / 5 (3) Dec 28, 2010
well, then you're talking about fantastical concepts like negative mass and anti-gravity.
Not necessarily, we could also be talking spacial geometry, or other forces entirely. DE isn't well fleshed out. The math indicates negative energy with a negative coefficient, ie: anti-gravity, but not necessarily what you or I would call "anti-gravity".
Solkar
4.2 / 5 (5) Dec 28, 2010
@ubavontuba

ubavontuba wrote
I still like my semi-gravitationally attractive vacuum.


?

"semi-gravitational" ???

What does that mean?

@Quantum_Conundrum
- It is quite clear that just taking constness of c and Newton's law into account will not yield a GR-black hole. GR is formulated in terms of curvature of spacetime-manifolds, whereas classical mechanics is not.
- Even your classical eq of motion are flawed
v = at is true if and only if a is const, but a isn't const on trajectories in the vicinity of massive body but ∝ to 1/r^2.
- What you apparently wanted to show is that for energy preservation on inbound trajectories E_kin at the EH would be or exceed 1/2mc^2, thus v >= c. But your approach is wrong:

Aside of the fact that your approach even disgards SR it suffers from a misinterpretation of Schwarzschild coordinates, namely "r"

@ubavontuba:

Your proposal about potential effects *inside* massive bodies is wrong because it violates Gauss's div. theorem
ubavontuba
3 / 5 (4) Dec 28, 2010
@ubavontuba
I still like my semi-gravitationally attractive vacuum.
?

"semi-gravitational" ???

What does that mean?
I described it above. Did you miss it? Basically, I propose that the virtual particle flux may have a gravitational effect on spacetime.
@ubavontuba:

Your proposal about potential effects *inside* massive bodies is wrong because it violates Gauss's div. theorem
What proposal about what "effects *inside* masssive bodies" are you talking about?

Perhaps you're referring to my use of the term, "large scales?" If so, then you have woefully misinterpreted what I was writing about.

I was using "large scales" in relation to astronomical voids.

ubavontuba
3 / 5 (4) Dec 28, 2010
well, then you're talking about fantastical concepts like negative mass and anti-gravity.
Not necessarily, we could also be talking spacial geometry, or other forces entirely. DE isn't well fleshed out. The math indicates negative energy with a negative coefficient, ie: anti-gravity, but not necessarily what you or I would call "anti-gravity".
So now you admit that it's currently perceived to react with matter gravitationally and not through other fundamental forces? Isn't that what I said?
Skeptic_Heretic
3.5 / 5 (2) Dec 29, 2010
So now you admit that it's currently perceived to react with matter gravitationally and not through other fundamental forces? Isn't that what I said?
No, I'm saying we don't know how it interacts.

"DE is not well fleshed out."
Solkar
3.7 / 5 (3) Dec 29, 2010
@ubavontuba:

What does that mean? I described it above. Did you miss it?


No, what I'm missing is consistency in your proposals.

Basically, I propose that the virtual particle flux may have a gravitational effect on spacetime.


And I'm also missing precision in the formulation of your "theory".

- Which "effect"?
- How large is it?
- "flux" through which surface or volume?

And last not least what are the observables, what is the prediction and what are falsification criteria?

What proposal about what "effects *inside* masssive bodies" are you talking about?


So you cannot get the connection between your own hypotheses in this thread and Gauss's div. theorem yourself?

Very well; that explains a lot, esp. why you did not understand Ethelreds remarks.
ubavontuba
3 / 5 (4) Dec 29, 2010
No, I'm saying we don't know how it interacts.

"DE is not well fleshed out."
Um, didn't I say that too?
ubavontuba
3 / 5 (4) Dec 29, 2010
@ubavontuba:

No, what I'm missing is consistency in your proposals.
Or in your interpretations thereof...
And I'm also missing precision in the formulation of your "theory".
When did I call it a "theory?" I called it "highly speculative" and "fun." Get a grip man!
- Which "effect"?
Uh, a gravitational effect.
- How large is it?
Relative to what?
- "flux" through which surface or volume?
You never heard of flux density in QM?
And last not least what are the observables, what is the prediction and what are falsification criteria?
More on this next...
So you cannot get the connection between your own hypotheses in this thread and Gauss's div. theorem yourself?
You don't even understand my speculation, so how can you even assert it's a relevant factor?
Very well; that explains a lot, esp. why you did not understand Ethelreds remarks.
He was wrong! Another poster (Gawad - Dec 20, 2010) even provided to him the correct formula!
ubavontuba
3 / 5 (4) Dec 29, 2010
And last not least what are the observables, what is the prediction and what are falsification criteria?
Oh, I don't know. It's just a musing, a speculation, a wild notion...

But if it's true, the prediction is mass will tend to accelerate toward voids.

In relation to Gauss's theorem, I haven't really worked that out. I'm speculating that where gravity abounds (in high mass densities) it simply lends its support (so to speak) and acts toward the center of mass, and in very low densities, it acts toward the center of voids (essentially, the void becomes a very low density mass). So, the demarcation would have to have something to do with the volume of the voids versus the size/density of the mass.

The effect would be vanishingly weak locally, and only be apparent on extremely large scales (like astronomical voids).

I don't think it'd be detectable in a lab. Therefore, I can only think of observation as a means to falsify it.

continued...
Skeptic_Heretic
not rated yet Dec 29, 2010
But if it's true, the prediction is mass will tend to accelerate toward voids.
Yeah but what is a void? Are you referring to false vaccuum, true vaccuum, etc? That's a little too loose to even be a musing you can play much with. I know I'm iunterrupting a multi post so feel free to ignore if you subsequently answered it.
ubavontuba
3 / 5 (4) Dec 29, 2010
What you'd look for are Doppler signals which indicate large filaments/structures are being gravitationally affected by voids. But even then, I think it'd be very difficult to confirm or falsify, as the signal noise would tend to be very high from all the ordinary mass effects.

Anyway, don't get too excited by it. It's only a speculation, a conjecture, a supposition, a musing, a contemplation, a tentative insight...

ubavontuba
3 / 5 (4) Dec 29, 2010
But if it's true, the prediction is mass will tend to accelerate toward voids.
Yeah but what is a void? Are you referring to false vaccuum, true vaccuum, etc? That's a little too loose to even be a musing you can play much with. I know I'm iunterrupting a multi post so feel free to ignore if you subsequently answered it.
Well by definition, a void is the empty space between filaments, so I think it's best described as a relatively low mass density.
Skeptic_Heretic
5 / 5 (1) Dec 29, 2010
Well by definition, a void is the empty space between filaments, so I think it's best described as a relatively low mass density.
Ok, but then we know what the gravitational attraction would be and it wouldn't be a "void" it would be the virtual particles within said volume of space.
ubavontuba
3 / 5 (4) Dec 29, 2010
Ok, but then we know what the gravitational attraction would be and it wouldn't be a "void" it would be the virtual particles within said volume of space.
Right. That's what I'm saying. However, there's a time dependent twist. The gravitational density is higher than the mass volume density at any given moment.
Skeptic_Heretic
5 / 5 (1) Dec 29, 2010
Right. That's what I'm saying. However, there's a time dependent twist. The gravitational density is higher than the mass volume density at any given moment.
That's established. Virtual particles are 90% of the weight of a proton.

editted for undue flippancy.
yoron
1 / 5 (1) Dec 29, 2010
The energy density of a vacuum is, as I understand it, a 'negative' one, also called the cosmological constant as that is what is thought to define this 'negative density', and also called Einsteins biggest blunder. Maybe it isn't though, a blunder I mean. QM 'zero point energies' if added up differs from the measured cosmological observation by a magnitude of one hundred and twenty orders, making the discrepancy very large and unexplainable. It is also thought of as one of the, or the, reason(s) why we see an expansion as it corresponds mathematically to a gravitational repulsion.

But, being 'negative', the reason two plates join I do not see as 'space' having a pressure. Space have a 'zero' pressure. I see the real effect as disallowing certain 'virtual wave lengths' by having the plates extremely close, creating an local 'unbalance' that SpaceTime rectify by forcing the plates together.

Anyone want to define how the expansion come to be, creating 3-D space?
Skeptic_Heretic
5 / 5 (1) Dec 29, 2010
Yoron,

What?

Energy density of a "vacuum" is negative because there is mass present in the form of virtual particles, which would mean there's a small gravitational attraction to "nothing". This is why dark energy is positive.

As for the plate statement, are you referring to the Casimir effect?
yoron
1 / 5 (1) Dec 29, 2010
Well, I see it a little differently :)

I see it as a result from our 'arrow of time'. A 'unfettered space' have all those wavelengths outside Plank time as I see it. When we introduce the plates we disallow some of them, creating a effect inside our arrow of time, as SpaceTime equilibrium is disturbed. So yes, to me the Casimir effect has to do with time. And so has Rindler observers and Unruh radiation.

As for the first of my statements you can look at *** books.google.com/books?id=5dryXCWR7EIC&pg=PA187&hl=en#v=onepage&q&f=false

The virtual particles we talk about is 'energy carriers', existing inside us, as well as all around us. Without them we have no idea for how 'energy' gets transformed and distributed. My view isn't that strange I think :)
ubavontuba
3 / 5 (4) Dec 29, 2010
That's established.
Sure, when defining the spacetime curvature of the universe. Here's a reference:
http://math.ucr.edu/home/baez/vacuum.html
I'm suggesting there could be a peculiar local effect which compounds the gravity of the void relative to the size of the void (beyond the mass/energy density of the vacuum).

It's probably an implausible and perhaps inconsistent idea, as it implies the universe's expansion (from the big bang and onward) is the result of a (higher dimensional?) void pulling it apart (it works best with the concept of "bubble universes").

Anyway, like I said before: It's highly speculative, but fun. I'm not claiming anything more.
ubavontuba
3.4 / 5 (5) Dec 29, 2010
It looks like it'll now accept a link in text form.

http//math.ucr.edu/home/baez/vacuum.html
yoron
1 / 5 (2) Dec 29, 2010
As I think of it 'SpaceTime' exist in a 'size' of its own, defined by its dimensions, and there we have some different 'dimensional topologies' to choose between, times arrow being a really important one. Could we place ourselves outside that arrow I doubt a 'SpaceTime' would be seen. Naively seen my 'SpaceTime' is infused with a lot of other possible 'realities', one that we call 'virtual'. If there are more I don't know, so far we have 'virtual particles' and gluons as the biggest contenders, but I expect both to be under what is 'meaningful' for us, impossible to observe inside our arrow.
ubavontuba
Dec 29, 2010
This comment has been removed by a moderator.
yoron
1 / 5 (1) Dec 29, 2010
Okay I'll try again.

General relativity: an introduction for physicists by Michael Paul Hobson, George Efstathiou, Anthony N. Lasenby

http//books.google.com/books?id=5dryXCWR7EIC&pg=PA187&hl=en#v=onepage&q&f=false
yoron
1 / 5 (2) Dec 29, 2010
Thanks ubavontuba, it seemed to do the trick:)
Density is a concept we use, remember that this is my view now:), inside SpaceTime. Outside an arrow of time it becomes different. That's why Space is nulled, becoming classically 'zero' as I see it.

You can discuss 'negative pressure' as a concept but to me it's more of a 'time reversal' well maybe not reversal, more like opening a 'rift' in SpaceTime. If we look at a Black Hole it seems as SpaceTime have one answer to singularities, close it.

So when those plates are brought together, disallowing the wavelengths, we create a disturbance and SpaceTime 'closes it'.
yoron
1 / 5 (2) Dec 29, 2010
Hawing radiation is a 'communicative' process only if seen from the viewpoint of its entanglement as I understands it. You can not speak of it as anything 'moving' from the inside of the EV to the outside.

What we see is to me a self regulatory expression of a state of SpaceTimes equilibrium, and if you like 'entropy'. It's like a game, if you stop looking at motion and distance, it has some very peculiar rules that will make things happen inside our arrow, where the idea of entanglements for making a SpaceTime work seems really important to me.
yoron
1 / 5 (1) Dec 29, 2010
And furthermore, if we define Hawking radiation as a 'communicative process' we introduce a communication from a 'singularity' making my idea more plausible as I see SpaceTime as a place where we only can see a part of the 'real thing'. The 'arrow of time' is what creates it for us, making linearity work, without an arrow it would all be chaos. It seems as 'SpaceTime' uses linearity inside non-linearity, inside linearity insi.. ad infinitum, as one of its main 'rules'.

But we haven't realized that particular concept until recently. And now we are trying to make sense of it, and it may make sense if linearity is one of the peculiarities you get with an arrow.
yoron
1 / 5 (2) Dec 29, 2010
You can see a Lorenz contraction two ways as I understand. As a geometrical 'twist' fooling our senses and measurements or as 'real'. If you see it as real, and you really should if you believe in a 'time dilation' then distance is a function of SpaceTime and your 'room time geometry'.

I've used 'frames of reference' before, the problem being that it is a very neutral definition, you can talk about your own, then about someone else's in the same breath fooling yourself to see both as being of the same 'value'.

They're not, no matter what you do, how fast you travel, if you're at a neutron star, you have the same expiration date. But all other frames you ever will notice, or as I call it 'room time geometries' are relative you open for 'change', and by the same 'experiences' that you will find 'not' changing your duration. If you see how I think.
yoron
1 / 5 (2) Dec 29, 2010
When I use my description I know where I stand with it, I'm looking at the universe from my 'room time geometry' and mine is unique. If discussing yours then that one also will be unique, solely yours.

There is a subtle difference in it to me :)

Helping me to see what I think is important. People seem to think that this universe is a seamless 'whole', and I agree in one way, but if you look at it as 'room time geometries' they seem to come in all 'sizes' from a pebble on the beach to a electrons 'orbital' to, whatever I can think up i guess :) And they all have to be slightly different I think, presenting a different SpaceTime, but still giving us all this feeling of 'sharing' the exact same universe.

And how they do that is radiation. There are three things SpaceTime is to me.

A topology defined by your 'room time geometry' (gravity/space)

Radiation.

Matter.

For the moment that is, I'm open for change :)
yoron
2.3 / 5 (3) Dec 29, 2010
And if we accept that 'the arrow of time' just as your 'room' is a function of that 'room time geometry' then what you call 'distance' I prefer to look at as a 'sliding relation, magnifying and contracting' your 'reality'. How that works is hard to see but it have to be a function of our arrow, well maybe we need 'it all' to get to that 'arrow' probably it is so.

Speed is a distance measured in time. So speed is relative your 'room time geometry.'

With one exception.

Light.
yoron
1 / 5 (2) Dec 29, 2010
So what is motion?
If it can change with 'potential gravity' or 'stress'.

And does light really move?

Think of it. I know, it sounds like a rather weird question, doesn't it :)
yoron
1 / 5 (2) Dec 29, 2010
I find light easier to understand if I go after its definitions, instead than after what I think I observe 'normally'.

Then light will have the same 'speed' in all 'frames of reference', be that your own, or any other 'frame' you measure (like that attacking space fleet coming at Earth, and no, they're not, just an example:)

And light only exist in its interactions.

Weak observations will still build on 'interacting' even if removed from what you observe. To prove that wrong you need to observe without observing at all.

(It makes most sense if you remember that I see SpaceTime as a 'game', the rules doesn't necessarily have to make any 'sense' to me if I compare it to mine 'immediate reality' writing here)
yoron
1 / 5 (2) Dec 29, 2010
You can look at space as a 'spring system' where the density will express itself as springs, resting against springs, resting against spr.. Ad infinitum.

Your light will then represent a 'oscillating motion' having a velocity in that mass of springs where the light-corn/wave :) is represented by the oscillation 'propagating' in the medium. The density of the medium it travels through may be thought of as the rigidity of those springs, or their 'inertia'.

Or as a single extremely long spring being 'SpaceTime'. And doing so the oscillation/light still can be seen to travel through that spring, but the 'spring' never moves.
yoron
1 / 5 (2) Dec 29, 2010
A third possibility is to let those springs 'pop up' from nowhere, making indentions very close to each other creating an illusion of movement for those looking, like a sheet with someone doting it from behind with a marker. Like our 3-D was sheet upon sheet upon sheet, with a he* of a lot of markers everywhere :) Then you also would have our 'arrow'.

It's all a question about what 'times arrow' really is.
yoron
2.3 / 5 (3) Dec 29, 2010
But if I'm correct in assuming that 'SpaceTime' is related to unique 'room time geometries', and that everything you observe, have it's own.

And that light will in and from all those 'room time geometries' only present us with that same one speed?

Why?
And how?

In 'frames of reference' I find a difference between what I see as my 'time' relative all other frames, just as you can do comparing your to all others. I don't see how I can do the same with the speed of light?

Can you see a way?

To me it seems as if this 'light' always is 'the same' no matter how I choose to look at it?

And why do we only observe its 'interactions'?

And 'virtual light' is real 'light' for a Rindler observer and as Unruh radiation.

And, how did it become 'real'?
yoron
1 / 5 (2) Dec 29, 2010
You do it by 'compressing' your 'room time geometry'.
There are two ways I know of.

Invariant proper mass (restmass)
Motion.
yoron
1 / 5 (2) Dec 29, 2010
Imagine it as a 'still frame.'
Let light exist in 'propagation'

Mark out where those light-corns will be on your frame.

Now compress the frame speeding away.

What happened with your coordinate system?
What coordinate system will your neighbor see from home?
yoron
1 / 5 (2) Dec 29, 2010
The question could be stated as which frame is the 'correct' one?

And to see its inherent weirdness: Remember that your neighbor never lost touch with you 'speeding away' looking in his telescope.

And, what the he* happened with our 'propagating light'?

Is it in two places?
yoron
1 / 5 (2) Dec 29, 2010
Eh, its' real, no joke. There are two coordinate systems there, each one having its own 'time dilation' & Lorentz contraction as I see it. You can't just expect one object motion to create it. There is no absolute 'frame of rest' inside SpaceTime. The best we can do is to define them relative each other. And if you succeeded in placing two non uniformly accelerating rockets in absolute sync, we could define those too as being 'at rest' relative each other

Lets back up a little. Entanglements?

What about 'ordinary light'.
Can that be entangled too?

I don't know, do you?
If it was, would that make a difference?

To me, a entanglement makes a difference, by just existing. We just need to know that the concept exist.

And light, how can it 'communicate' in 'no time', over singularities, and as far as I understand, having no limit to their distance as they do so.

'Communicate' here, I see as setting a opposite spin for its other half, by you interacting with one of them.

yoron
1 / 5 (2) Dec 29, 2010
I better stop there.
It feels like I'm hijacking the thread :)

But those are the things I wonder about. And now you do it too :) Heh.

And 'room time Dimensions/topology' includes string and loop quantum theory too. Although I look at dimensions as 'properties', not singular 'thingies', but I might be wrong in it all :)

Life huh :)
DamienS
5 / 5 (1) Dec 30, 2010
Jeez. Have you tried the ATM forum at bautforum.com?
Solkar
3 / 5 (2) Dec 30, 2010
Hi yoron!

I better stop there. It feels like I'm hijacking the thread :)


There are some really interesting concepts in your posts, so I dont think anyone will mind the clustering of your posts! ;)

Some remarks of mine:

You wrote:
Although I look at dimensions as 'properties', not singular 'thingies', but I might be wrong in it all :)


That's not as strange as it looks on a first glance; vectors of QM-Hilbert-space can be represented in a multitude of bases, the space-representation just being one possibility amongst others; another one e.g being the spin-up/spin-down basis. Thus the concept of space being a "property" amongst others is quite plausible.

You wrote
As I think of it 'SpaceTime' exist in a 'size' of its own, defined by its dimensions,


Riemannian Geometry is indeed an *intristic* formulation of geometry thus "size" may solemnly depend on the unit of the metric compared to other quantities existing "in" that very spacetime.
Solkar
2.3 / 5 (3) Dec 30, 2010
Yoron wrote

Okay I'll try again.

General relativity: an introduction for physicists by Michael Paul Hobson, George Efstathiou, Anthony N. Lasenby


There is also a very good video lecture on GR including Powerpoint-docs available from Uni Tuebingen:

www dot tat dot physik dot uni-tuebingen dot de slash ~kokkotas slash Teaching slash GTR dot html

It's a German site, but the docs and vids are English.

ubavontuba
3 / 5 (4) Dec 30, 2010
Hi yoron!
I better stop there. It feels like I'm hijacking the thread :)
There are some really interesting concepts in your posts, so I dont think anyone will mind the clustering of your posts! ;)
I'll second that. It looks like you have both a good and unique understanding of relativity. I like your concepts.

However, I feel there is always an intrinsic frame of rest which is always the first person, observer's "room."

And, I've always had difficulties with time, myself. I tend to think of time as a dimension (not unlike any other) and time dilation is a result of perspective within that dimension, relative to your "room."

Solkar:

Here's a copy and paste version of your excellent reference:
http://www.tat.physik.uni-tuebingen.de/~kokkotas/Teaching/GTR.html
yoron
1 / 5 (2) Dec 30, 2010
Thanks, I really like the Internet. When we share here we all get ideas. My ideas comes from your ideas, same as yours comes from what you find. I was worried that you might find me up the walls :)
yoron
1 / 5 (2) Dec 30, 2010
"I've always had difficulties with time, myself. I tend to think of time as a dimension (not unlike any other) and time dilation is a result of perspective within that dimension, relative to your "room.""

I agree :) it's hard to 'divide' what we call 'SpaceTime' We need to see the rules making it a 'whole' like the Feigenbaum constant and the Bekenstein bound. I'm not saying that they are absolutes, but I think we will find more constants if we just start to look, and we need to too :)

So maybe this universe of ours is even weirder than we thought huh :)
yoron
1 / 5 (2) Dec 30, 2010
And now I got all those nice references to look at too :)"vectors of QM-Hilbert-space" and "Riemannian Geometry" Thanks for those.

Don't know how easy they will be to understand but I will at least try to. And then those excellent links too :)

After fire and the wheel I believe Internet to be the most important 'invention' ever made. We may go towards a 'poorer Earth' but we have a unique way of communicating now. And to me it seems that is what we really like, most of us at least :)

As long as it is kept free that is.
yoron
1 / 5 (2) Dec 30, 2010
Rereading you ubavontuba. Maybe you meant that there is two 'dimensions'? Namely the 'room 3-D' and 'times arrow'? I think so at times too, it depends on my mood :). It's very hard to ignore the arrow, and I don't agree with placing entropy before the arrow. To me it you can't even discuss 'entropy' and neither 'discrete events' without implicatively assuming a 'arrow' resting behind it. To put it bluntly, there is a need of a 'glue' for your causality chains to exist.

Then, on the other hand, we have the concept of a 'whole' 'SpaceTime'. Looking at it that way your 'arrow' becomes a property of the whole 'shebang' impossible to free from the idea of a 'SpaceTime'.

There is a way to solve this though, at least as I see it, and that is to look at the way our 'universe' changes appearance through what chaos math calls 'emergences'. Water becoming ice, gaining totally new properties (ever tried to skate on water?:)
yoron
1 / 5 (2) Dec 30, 2010
Some look at that as 'hand waving' but I don't. To me this is no stranger argument than when you define what you call your 'system'. The idea of a 'system' is to me a equivalence to 'emergences', meaning that, you or the 'universe', defines an 'importance'.

The difference being that the 'universe' was first doing it :) It's either that or walking a lonely path to smaller and smaller 'thingies' and then walking back trying to see how they build this 'SpaceTime' and the 'states' we associate with different 'sizes'. Macroscopically, QM, Plank Size and under. Because we see that the 'universe' behave differently, don't we?
yoron
1 / 5 (2) Dec 30, 2010
In a really good game you should have several ways to solve your goal. And I think that this is exactly what SpaceTime offers us. There might be a possibility to prof SpaceTime the 'lonely way'? Another might be through 'emergences'? The worst thing we can do is to assume that there exist an 'objective truth' a linear process, simple to see that will define it all. On the other hand I do believe there is one (not necessarily 'linear' in the way we used to think of it though:) But I also expect us to have to redefine a lot of 'self evident' truths and ideas we have first to see it. And that's where my interest lies. To see what is possible to redefine.
yoron
1 / 5 (2) Dec 30, 2010
And now I'm sounding both pompous and self-contradictory. It's not what I set out to do :) But I do think we will find a 'causality chain' making 'sense'. And it seems like a lot of guys, & gals, are doing just what I think is necessary to get there. although the chain will seem to have very little with what we call 'reality' in our daily lives. I don't know if it's true but I've seen some suggest that with new words comes new ways to see. So I'm hoping we will find those words.

(Like Emmy Noether did)
ubavontuba
3 / 5 (4) Dec 31, 2010
yoron:

It looks like we're on the same page. I view time in lots of ways, depending on the application.

Like how spatial dimensions hold mass/energy distributions, time holds event distributions.

Time is also the river which passses through the spatial dimensions and changes the spatial mass/energy distributions.

By doing so, time effectively shatters the Bekenstein bound - by allowing the defined space to hold more things, but over time.

So time can be viewed as both a dimension, and an effect.

It's also part of the very topology of space.

And, "the arrow of time" has many definitions, as seen here:

http:/en.wikipedia.org/wiki/Arrow_of_time#Arrows

How marvelous is time, that it's so difficult to describe, yet so seemingly ordinary!
yoron
1 / 5 (2) Dec 31, 2010
Yeah :) but in the definition of 'entropy', we can also find a reversibility as well as 'islands' that stays the same, according to several sources. Entropy is not a desert of sand growing, killing us off. It's more like an ocean in where you can find different salinity and spots that stays the same. So in that motto it does not fit the unique arrow you have, no matter what others might think of your 'age'. Maybe there is some way to adapt 'entropy' to that fact though?
yoron
1 / 5 (2) Dec 31, 2010
The reversibility of entropy comes from biology, but the 'spots that stays the same' is actually an accepted definition amongst those working with 'entropy' since long time past.

We only need to accept one of them to see that your unique 'room time geometry' isn't 'entropic'. To be so it should be possible for your arrow to stop in your own 'frame of reference'. And that one I've never heard about? I would like too though, especially when meeting those lovely ladies :)
ubavontuba
2.7 / 5 (7) Jan 01, 2011
Well, entropy is certainly observed.

I wouldn't suggest that life is a reversed entropy, as the energy of the entire system which supports life certainly is dissipating. Rather life is the result of the useful energy which remains in the system.

It's been suggested there may be a sort of entropic "reset" mechanism, which would be the cyclic universe model, but it seems unlikely. However, these musings boil down to philosophy, as we simply don't know from whence the energy of the universe originates.

Perhaps universes spring into being as a regular occurrence. Perhaps there's only the one universe, dying a slow death. Perhaps new energy is being introduced into our universe from sources we simply haven't yet observed. Perhaps...

You see? Nothing but speculation and philosophy.

So what we know is entropy is real locally, and locally is all we really have.

As for staying young, I think that's a matter for the microbiologists and geneticists to sort out.
yoron
1 / 5 (2) Jan 01, 2011
Yeah, it's interesting. But what I'm boiling down too is that entropy when defined as I do, in unique 'room time geometries' don't fit in. Not if I assume that those laboring with it are right in that you can have 'isolated islands' where the energy don't transform into 'energy done' at the same rate as seen from a larger perspective.

That is if one assume that ones personal arrow constantly tick with the same rate measured inside. And I can't find any way to disprove that notion? And no, to me that one isn't philosophy, rather an unavoidable fact of relativity.

Maybe someone already have an answer to that question?
yoron
1 / 5 (2) Jan 01, 2011
Another thing, defining times arrow. If I assume that what 'communicates' is radiation, what does that make 'the arrow of time'? What I notice is that lights speed is a constant. Your time as measured from/in your own 'room time geometry' can also be seen as a 'constant'.

That makes them interesting as an equivalence. You could also argue that radiation is the best 'clocks' existing. Getting closer to 'times arrow' than using radiation I don't think is possible? And it works when measuring other 'frames of reference' from your own too. When you do you use the 'arrow of time' you have locally, aka 'room time geometry'.

So, maybe it's right? I don't really know, but maybe?
To define radiation as our arrow of time?
yoron
1 / 5 (2) Jan 01, 2011
One problem being that, if so, it contradict my earlier question about 'lights motion', doesn't it :)

But maybe both ideas works together? It's just us not seeing it? In defense of my notion I will spell out two 'magic words'..

Lorentz contraction
Time dilation.

And one more.
Constants.
yoron
1 / 5 (2) Jan 01, 2011
I'm going for one more proof why I think light 'doesn't move', risking to bore you to death.

To explain how 'photons' can be of different strengths I've seen some suggest that they come in 'higher concentrations' per time unit. That as they are thought of as invariant light-quanta. Not waves now, 'photons'.

Think of the sun as a sun-hose streaming out 'photons' at a black hole. Put a solar panel between the black hole and the sun. What do you expect to happen as we move that detector closer to the black holes EV (Event horizon)?

Will the photons deliver more energy?
Why?
yoron
1 / 5 (2) Jan 01, 2011
If they do, can I assume that to be a consequence of them getting 'compacted' bunched together by 'gravity'?

Nope. If we assume gravity to 'accelerate' them they should be spaced out:) Not 'bunched together' as they close in. So that one didn't hold water, did it?
yoron
1 / 5 (2) Jan 01, 2011
Well they can't 'accelerate' you point out. Instead they 'change' energy. Okay, but then they can't close in on each other either, right?

And what do we say they do instead of accelerating?
yoron
1 / 5 (2) Jan 01, 2011
Change energy? By themselves, intrinsically you mean?
But they're timeless? And only defined in their 'interactions'? Where do you expect them to do this amazing feat?

And if they did?
Why isn't that an 'interaction'?
yoron
1 / 5 (2) Jan 01, 2011
It's no different from a 'photon' bouncing near the Event horizon. As it 'climbs up' it will become red shifted relative the 'far observer' and so 'lose energy'.

And it's a very good argument of it not existing until in its 'interaction' to me:) As we otherwise would have to find an explanation to why it 'loses energy' when it's expected to be of a defined 'light quanta'. To see what I mean there you have to remember that light is 'time less' intrinsically, and only 'existing' in its interaction.

If you want to define it as it changes 'energy' as it climbs you will have defined an 'interaction'. That's not possible, if so, all light would annihilate as soon as it meet another gravitational potential, and it doesn't.

That's where its 'timelessness' comes in too as that is what we assume to make it possible to 'propagate' vast 'distances' without losing 'energy', as I understands it.
yoron
1 / 5 (2) Jan 01, 2011
To see it my way you need to see it as a game, nothing more but nothing less either. Another proof is that this photon climbing if measured outside that gravity well will be found to have gotten all its 'strength' back, telling us that it expended no energy climbing, no matter what we would have measured it to be if inside that gravitational potential. So looking at it as a game helps one accepting the rules. Looking at it as we observe it here in our daily life won't.
ubavontuba
3 / 5 (4) Jan 02, 2011
If they do, can I assume that to be a consequence of them getting 'compacted' bunched together by 'gravity'?

Nope. If we assume gravity to 'accelerate' them they should be spaced out:) Not 'bunched together' as they close in. So that one didn't hold water, did it?
It looks like you're forgetting to apply time dilation to your analogies. That is, your solar panel isn't going to experience time the same in the two positions.

As the time is slower closer to the black hole (relative to the sunward position), the solar panel will indeed perceive the photons as having higher energy. Do you see? The photons aren't changing, it's that the clocks of the observers are different.

More succinctly: The energy perceived is relative to the clock.
yoron
1 / 5 (2) Jan 02, 2011
A very nice twist ubavontuba :)

If I was the solar panel, my time would be 'as always' as per my definition of every objects unique and, in some truly weird way, equivalent 'room time geometry'. As in that one your time always will 'tick as usual' according to your wristwatch.

So why would I define the photon as being 'stronger' due to that when I according to my wristwatch don't have any 'time dilation' to notice?

The only way to define a time dilation that I can see is when I return to my 'origin'. You can as you know of it theoretically define it mathematically due to velocity, or potential gravity. but that's a theoretic exercise.

That's how I look at it for the moment, I differ between what I call 'room time geometries' having an 'equivalent time' as seen from their own frames against what I might call 'second hand observations' like someone telling me their 'clock' at some other 'frame of reference'. I will think about it though:) And see what more arguments I can muster.
yoron
1 / 5 (2) Jan 02, 2011
Gravity is equivalent to a uniform acceleration right?
Does that mean that when you turn of your engine the infalling photons measured will fall back to some 'original energy level' as before that acceleration?

I don't think so myself, and if they don't you now are in a uniformly moving frame, equivalent to all other uniformly moving frames, like earth. You can pick any other frame to define your velocity from now, without finding the infalling photons energy to change.
yoron
1 / 5 (2) Jan 02, 2011
So it can't be velocity per se that defines the 'strength' of that photon. And if it isn't velocity then the equivalence only exist in the acceleration, but as I think I showed here the photon still keeps its 'strength' relative you when uniformly moving.
==

It's a really good argument though, let me sleep on it and I we discuss it tomorrow, or today as it is :)

After I've woken up and got me some coffee..

Sweet thinking nevertheless ubavontuba, I need to get my ideas straight to make them make sense :)
ubavontuba
3 / 5 (4) Jan 02, 2011
Does that mean that when you turn of your engine the infalling photons measured will fall back to some 'original energy level' as before that acceleration?
Generally, no. The perceived energy will stay the same.

Your clock rate/relative energy level was initially set by the position you held in the gravity well. Going into free fall from that point will cause you to accelerate toward the black hole. The acceleration will reduce the relative photon energy, but this is countered by the depth you fall into the gravity well, which slows your clock and increases the relative energy.

Or more succinctly: You free-fall in with the relative energy perception you started with.
yoron
1 / 5 (2) Jan 02, 2011
That one is really nice. Acceleration in the manner of a 'free fall' towards a gravitational potential aka 'matter'.

I'm of two minds there, nothing unusual with that:) one shouting. "Huh, what acceleration?:)" The other not as sure.

What are the definitions of a free fall?

And are all 'free falls', aka following a geodesic, 'equivalent'?

I hope I'm looking at it from the perspective of relativity now. You can look at it as 'forces' too, but then you will use the Newtonian definitions, well, as I see it. Sometimes I find it quite hard to differ them, as I'm so used to the idea of 'forces' defining my world.

And I'm still not fully awake here.
Hope I will be later :)
yoron
1 / 5 (2) Jan 02, 2011
There is one thing to notice though, differing between a uniform motion and an acceleration. And I'm still wondering about that one?

In a uniform motion two observers in relative motion will see different frequencies for the same source, but they can not change the sign if you go by Lorentz transformations. That is not true in a acceleration though there, if i got it right, the sign can be interchangeable and you will see what the rocket 'coasting' won't, and nothing to do with what velocity/speed you have.

And that's the Unruh radiation. I've seen it explained as your engine becoming your 'detector', but I have difficulties turning my head around this 'sign switch' possible here?

To me that seems like a reversal of 'flow'?
Time?

Anyone that can define how it's possible?
yoron
1 / 5 (2) Jan 02, 2011
Let's go back to entropy. The reversibility of our arrow is a thing defined in a possible two object scenario. Only there will you be able to define it as possible right? As you go up in numbers you can't any more.

And that's how things arrange them selves from orderly to disorderly. You can have a orderly system containing several object, neatly laid out, but it will change into a disorderly system as 'time' goes on.

But it's also true that there are so many more ways of leaving them disorderly than orderly. So why do we see a system consistently working from order to a greater disorder? What made the state of order the one we started from?
yoron
1 / 5 (2) Jan 03, 2011
So how should I define my 'room time geometry'. According to me both time dilation and Lorenz contraction is real. But they are not noticeable as changing that room time geometry, not by me measuring inside it at least, are they?

I'm not talking about 'remembering how it used to be' and then 'compare the difference' here. I'm talking about taking out my clock and yard stick and then start to measure.

And that must mean that it all 'changes together' as a 'whole experience' for me and my 'SpaceTime'. So what do I have if looked at like that, one object or several? I would say one, but my arrow still exist inside it?
yoron
1 / 5 (2) Jan 03, 2011
People love to make statements that are not correct it seems, and we all do it from time to to time. One of the worst is to assign plus and minus to a so called 'gravitational acceleration' like throwing a ball in the air. Worthless and superfluous to me, and only confounding the reader (me:).

But what I'm wondering of here is not that. As far as I understand the 'sign switch' is expected to be 'real' and a part of the explanation to why we see a Unruh radiation. so how is it possible? What would we see.
==

Or is it the same?
I've seen some putting Hawking radiation and Unruh radiation as one and the same? And then, if so, I will ignore the statement of reversing signs as that seems to be a mathematical joke with no relation to the factual effect.
==

And then we should have an extreme pair production in a Unruh radiation which if the idea was right should cancel itself out as the particles and 'anti particles' react with each other.
yoron
1 / 5 (2) Jan 03, 2011
And then we should have an extreme pair production in a Unruh radiation which if the idea was right should cancel itself out as the particles and 'anti particles' react with each other.
But there are some weird effects to it.

1. We shouldn't see it. Not if it cancels itself out, if it doesn't though you can expect radiation to escape, hitting your retina.

2. it will leave a positive 'rest product' whatever that is thought to be. And so, once more, we meet that scarlet pimpernel 'Energy'.

Hmm?

To see the inherent weirdness of 2. You need to remember that Unruh radiation doesn't exist for our uniformly moving observer. So he won't observe any such 'effect'. But what about that 'positive rest product' then?
yoron
1 / 5 (2) Jan 03, 2011
There is a clear difference between Hawking radiation and Unruh radiation.

In a Hawking radiation you will have the so called 'negative' particle-half inside a event horizon negating the 'positive' particles, and even if leaving a positive rest product not being inside our SpaceTime. That will also slow down the Black hole relative our universe and so have an uneven equivalence.

I forgot to mention that here you will see a radiative effect, with all right too as we will have a surplus. But in a Unruh radiation I do not expect you too, that is, if it's expected to be the same?

In a Unruh radiation it all happens in 'SpaceTime'. and if you want to assume that we all share the same, then the rest product should 'exist' for us all.

Even though the radiation didn't :)
yoron
1 / 5 (2) Jan 03, 2011
From the idea of unique 'room time geometries' I might be able to argue that it's not necessary for all accelerated effects to be shared over our common 'SpaceTime'.

Don't know how though :)
ubavontuba
3 / 5 (4) Jan 03, 2011
And are all 'free falls', aka following a geodesic, 'equivalent'?
Generally and locally, yes. However optically, you might see some peculiarities.
And that's the Unruh radiation.
AFAIK, Unruh radiation has never been observed.
So why do we see a system consistently working from order to a greater disorder?
Entropy is the dissipation of, or saturation of (in a closed system), energy. Not all systems are necessarily subject to it (unless the universe is open and dark energy eventually rips everything to shreds).

http:/en.wikipedia.org/wiki/Ultimate_fate_of_the_universe#Open_universe

...not by me measuring inside...?
Correct. Local time is always perceived as a steady "flow," but it isn't necessarily so.

yoron
1 / 5 (2) Jan 03, 2011
"Correct. Local time is always perceived as a steady "flow," but it isn't necessarily so."

Can you give me a example on how you think there?
ubavontuba
3 / 5 (4) Jan 03, 2011
"Correct. Local time is always perceived as a steady "flow," but it isn't necessarily so."

Can you give me a example on how you think there?
Sure.

You can only experience time in situ. So, if it stops and starts, you stop and start - at the same rate. You won't notice the stopping and starting, because you're part of it. To you, time will always feel smooth and steady.
yoron
1 / 5 (2) Jan 03, 2011
That's a possibility, but, as you say impossible to notice. When I define it I do it from what I am able to notice, and then I come to this conclusion :)

But entropy and the arrow of time seems different to me. Which shouldn't be seen as they do not co-exist, as they go into each other. But the arrow is like radiation, steady and unchanging in duration for you, inside your own frame of reference.
yoron
1 / 5 (2) Jan 03, 2011
The biggest point I make here is that there are observations and then there are conceptual thinking.

Conceptually I can define a blue-shift both as a time dilation, and a Lorentz contraction, or either of them. But observationally I will only use one 'clock'. The one inside my own 'frame of reference', aka 'room time geometry'.

To use any other frames time measuring becomes a academic exercise, and you can see how wrong that one 'ticks' any time you compare it against your own frames 'clock'.
ubavontuba
3 / 5 (4) Jan 04, 2011
Conceptually I can define a blue-shift both as a time dilation, and a Lorentz contraction, or either of them. But observationally I will only use one 'clock'. The one inside my own 'frame of reference', aka 'room time geometry'.
Sure, but there's nothing special about your clock, other than it being yours.

It's only natural to relate everything to yourself, as you need a rest frame from which to make comparisons Your "room" is as good as any, and the most convenient.
yoron
1 / 5 (2) Jan 04, 2011
Well, there I go the other way in fact :)
Because what I notice is that my 'room time geometry' also will be yours, seen from your eyes that is, or as we 'share' it.

But I don't need to 'share' to see the equivalence.

In all frames imaginable I know that light will have the same 'speed' and that my clock will 'tick' at the same duration as compared to my heartbeats.