# Does Hawking Radiation preclude EH formation?

Discussion in 'Physics & Math' started by RJBeery, Dec 11, 2012.

1. ### AlphaNumericFully ionizedModerator

Messages:
6,697
A point of view for which you have no evidence, no formalisation, no derivation and no model to provide and explore the implications/predictions of, all based on a total lack of knowledge/experience with experimental data or mainstream models. How do you have any idea what particle dynamics are when you have absolutely no information on them? Why should anyone care what your 'theory' (what you mean is 'random supposition') says when it is based not only on no information but your demonstrably flawed grasp of what little pop science you have read.

No, I was telling you that the thing Einstein was talking about had nothing to do with frames and coordinates of space-time but rather wavefunction collapse, as you'd mistaken his discussion of the latter for discussion of the former.

You might not know the reasoning but that doesn't mean it isn't there when you actually learn and understand quantum mechanics. You don't know any of the mathematical background so how can you possibly claim there's no reasoning behind these things?

You said there was a lack of logic in the work. That is demonstrably false. Any, any, model starts with some initial postulations. Relativity has frame invariant physics and the light speed condition for instance. The axioms of quantum mechanics are perhaps less physically intuitive but, as I mentioned, they are extremely similar to those in Newtonian mechanics save for a subtle but fundamental change, namely $\{A,B\} \to ih[A,B]$, ie Poisson brackets on functions to commutator brackets on operators.

Quantum mechanics might seem illogical or hap hazard to people who don't grasp the mathematics but it is quite sound. Along with the brackets I just mentioned principles like least action, Hamiltonian mechanics, Lagrangians, Euler-Lagrange equations, Noether's theorem, Louville's theorem, functional analysis, Hilbert spaces, Lie theory. If you know how all of those are used in classical mechanics then moving into quantum mechanics is practically effortless. Sure, you have to redo many calculations to account for the subtle change in brackets but the methods are unchanged. The implications of reasonably basic assumptions and these mathematical methods can be highly counter intuitive and unexpected but that doesn't make them illogical or baseless. Human intuition didn't evolve to handle the quantum world but a classical macroscopic one, there's not only no reason to expect quantum systems to behave like everyday systems but damn good reason to expect the opposite. You previously stated you think you have good intuition for this but the evidence is clearly to the contrary.

You didn't give a sound reasoning, you spat out your unjustified supposition. Providing an answer when no one else has doesn't make it valid by default. The idea stands or falls on its merits, regardless of what others might have put forth. This is why the religion answer of "God done it" doesn't cut it even when it is about questions science cannot currently answer. Even if a guess is ultimately found to be accurate the point to believe it is when evidence is presented, not before. So until you can demonstrate your ideas/claims are capable of actually describing, in any quantitative sense, quantum systems there is no reason to think you've provided any reasoning.

On the other hand we have the mainstream models of quantum mechanics, which do have quantitative predictive/modelling power (the computer you're currently looking at is an illustration of that capability) and so if an explanation can be obtained from quantum mechanics then it is at least reasonable to think it holds some merit. In regards to your specific question you should define the question more precisely since you could be referring to general 'quantum weirdness' or you could be referring specifically to the uncertainty principle, which says simultaneous measurement of two conjugate variables, such as position x and momentum p, will never be perfectly accurate no matter the method and that the product of the uncertainties is bounded below by $\frac{\hbar}{2}$.

So can quantum mechanics explain this? Well obviously, how else that lower bound be determined so nicely! First we must formalise the definition of 'quantum uncertainty'. I'm going to do it in a minimally mathematical manner but I'll use all the relevant technical terms which I won't bother to explain because quite frankly I can't be bothered and if you're interested (which I know you won't be) you can look them up for yourself. All of this can be done in arbitrary dimensional systems with multiple particles/fields but I'll abuse notation and write things using one dimensional notation. Generalisations are straight forward to those who know how.

The state of a classical system is described by the variable q and its dynamical properties by the time derivative, $\dot{q}$. From this we define the kinetic energy $T = T(q,\dot{q})$ and the potential energy $V = V(q)$. The Lagrangian is then $L = T-V$. By the principle of least action we obtain the Euler-Lagranges $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} - \frac{\partial L}{\partial q} = 0$ and we define the momentum conjugate partner of q to be $p = \frac{\partial L}{\partial \dot{q}}$. From this we can define the Hamiltonian as the Legrendre transform of the Lagrangian, $H = H(q,p) = \dot{q}p - L = T+V$. All classical fields are now upgraded to quantised operators, $q \to \hat{q}$ and $p \to \hat{p}$ (likewise for anything built out of them such as the Lagrangian and Hamiltonians), through the enforcement of a quantisation condition on the operators $[\hat{q},\hat{p}] = i\hbar$. We now drop the hats for notational convenience. The state of the system is now written as an element of a Hilbert space upon which the operators act, $|\psi\rangle \in \mathcal{H}$, $\mathcal{O} : \mathcal{H} \to \mathcal{H}$, $\langle \;\cdot | \cdot \; \rangle : \mathcal{H} \otimes \mathcal{H} \to \mathbb{R}$. Observables are the expectations of Hermitian operators with respect to the state, $\langle \mathcal{O} \rangle = \frac{\langle \psi | \mathcal{O} \psi \rangle}{\langle \psi | \psi \rangle}$. We assume all states are normalised from here on. The variance of an observable follows the same definition as in statistics, $\mathbb{V}(\mathcal{O}) \equiv (\Delta \mathcal{O})^{2} \equiv \langle (\mathcal{O} - \langle \mathcal{O} \rangle)^{2}\rangle = \langle \mathcal{O}^{2} \rangle - \langle \mathcal{O} \rangle^{2}$

Now we can define quantum uncertainty, namely if q,p are conjugate then $\Delta q \,\Delta p \geq \frac{\hbar}{2}$. Using the definition of the standard derivation we get $(\Delta q \,\Delta p)^{2} = \langle (q - \langle q \rangle)^{2} \rangle\langle (p - \langle p \rangle)^{2} \rangle$. By the Cauchy-Schwarz inequality we get $\langle (q - \langle q \rangle)^{2} \rangle\langle (p - \langle p \rangle)^{2} \rangle \geq |\langle (q - \langle q \rangle) (p - \langle p \rangle) \rangle|^{2}$. It is simple to show $\langle (q - \langle q \rangle) (p - \langle p \rangle) \rangle = \langle qp \rangle - \langle q\rangle\langle p\rangle$. Further restructuring leads to $\left(\frac{1}{2} \langle qp+pq \rangle - \langle q \rangle \langle p \rangle \right)^{2} + \left \frac{1}{2i}(qp-pq) \right)^{2}$. Since $qp-pq = [q,p] = i\hbar$ we get $\left \frac{1}{2i}(qp-pq) \right)^{2} = \left \frac{\hbar}{2} \right)^{2}$. Therefore $(\Delta q \,\Delta p)^{2} \geq \left(\frac{1}{2} \langle qp+pq \rangle - \langle q \rangle \langle p \rangle \right)^{2} + \left\frac{\hbar}{2} \right)^{2}$. It remains to show that $\langle qp+pq \rangle \in \mathbb{R}$ which follows from the fact it is Hermitian, $q^{\dagger} = q$ and $p^{\dagger} = p$ implies $(qp+pq)^{\dagger} = qp+pq$. Done.

There, with a bit of formalisation the link between the quantisation of a relationship between infinite dimensional operators on a Hilbert space and the observation fact we cannot measure position and momentum conjugate pairs to arbitrary position is made. To someone who doesn't know the mathematics this might as well be written in hieroglyphics but to anyone with even a rudimentary grasp of linear algebra it is straight forward. A seemingly illogical and highly counter intuitive result explained in an utterly logical manner. That's the power of formalising ideas, of being rigorous and of actually learning things rather than pulling them out of your backside because you're too lazy to find out properly and too enamoured with yourself to consider perhaps you don't have divine knowledge about things you have no experience or actual knowledge of.

And before anyone (I'm looking at you Farsight) tries the "Oh you're just off in maths land!" or "Are you trying to throw out really complicated stuff to hammer PL into submission", firstly this stuff accurately describes physical systems, including the piece of electronics you're currently sitting in front of and secondly if you think this is complicated you haven't seen anything. I wish physics stayed this simple but reality isn't so.

Messages:
982

5. ### SyneSine qua nonValued Senior Member

Messages:
3,515
No, proper time is used in SR/GR. QM utilizes the absolute time of Newton. Thus you continue to prove your ineptitude at every turn.

You said:
1. Point particles, by definition, cannot contract.
2. Length contraction is only observed parallel to the direction of travel.

7. ### Prof.Laymantotally internally reflectedRegistered Senior Member

Messages:
982
Does the lorenzt tranformation ring any bells?

I never said the particles contract. There frame of refernce contracts if they assume they are at rest. Your right it is only in the direction of travel because if you consider the value of contraction in the derivation it is the same distance in the other directions.

8. ### AlphaNumericFully ionizedModerator

Messages:
6,697
You provide no formalisation at all so at best you can claim that you are consistent with it but then it is easy to say "My supposition matches that mathematics", showing it is something else.

If you claim it does then prove it. Let's see the formalisation of your work. Or do you not actually have any?

What would 'more accurately' mean? Predict numerical values more precisely. You have none of that. If someone can construct a quantitative model of quantum phenomena which can do such a thing, great. But if all they do is arm wave and fail to provide any formalism then they never will.

A superficial idea which you like the sound of. You have no idea if it could lead to be a better description. Using words is easy, you say assert anything you like. For example, quantum uncertainty is due to torsion of space-time causing an inductance in the permeability of the gauge field of electromagnetism. I claim that explains quantum uncertainty better than you.

Or rather I don't because it is just words and assertions. Without details I have no reason to think what I just said has any relevance to reality. Just as you don't. If you cannot see this then you have a fundamental lack of understanding about the scientific method.

How can I add it to the mathematics of quantum mechanics, you cannot formalise it? Saying "Here is an idea, you fill in the details" isn't how science is done. In fact it is one of the points on the crackpot index. As for drinking beer and watching TV I'm a professional researcher, including in the domain of quantum mechanics for multi-billion dollar industrial giants. I spend more than half my waking life doing science.

If you do then you're in for one hell of a shock when you realise you know nothing about science and all the arm waving in the world won't help you solve real world problems.

Free advice? You think what you're providing has any merit? Coming up with wordy superficial explanations for bits of physics is easy. You have shown you have no idea how the scientific method works, the formal details of mainstream models, you're deluded enough to think you have an understanding of phenomena you have zero information on and no experience with and you think you're giving me advice?

And people call me egotistical! At least I can do the science I claim!

9. ### GrumpyCurmudgeon of LucidityValued Senior Member

Messages:
1,876
Prof.Layman

Virtual particles do not create energy, nor do they destroy energy. If they remain "virtual" they are not directly observable because they are created and destroyed in Planck time and space. Their annihilation does not result in a photon being released to the Universe, as they are a completely neutral system with no net energy change(which the release of a photon would negate). Only if the two particles are separated prior to annihilation do they become "real" particles in our Universe, they then interact like any other particle. Particle pair production creates the particles using the energy available in the environment and then immediately returns it. Between those two points the internal energy of the two particles can be at any level, but most are close to the energy level available. This is not a violation of conservation because either the energy is returned or, if one or both particles survive by being separated, the energy is lost to the environment. There is no free lunch(as Feineman liked to say).

Sciency sounding gobbledegoop is still gobbledegoop, you know?

Grumpy

10. ### GrumpyCurmudgeon of LucidityValued Senior Member

Messages:
1,876
AlphaNumeric

There, in a nut shell, is why I taught High School physics rather than continuing in my studies and doing research. Those who can(as you have demonstrated you can)do, those who cannot, teach. Both are necessary, but teaching at least kept me from the headaches advanced math always gave me. I don't grok math, though I do understand most of the concepts(or at least as well as we did in the 70s I should say). But I do recognize that my understanding rests on the work of many mathematicians being explained to me, not the other way round as Prof.Layman thinks.

Grumpy

11. ### SyneSine qua nonValued Senior Member

Messages:
3,515
Sure, it figures prominently in SR, as it describes relations between coordinate systems within Minkowski space. The Standard Model of QM is framed in absolute space and time. Any of that ring a bell?

12. ### OnlyMeValued Senior Member

Messages:
3,914
Are you sure?

QM and the standard model do incorporate SR and Lorentz transformations.

There is no such thing as an absolute space and time, frame of reference (velocity or acceleration etc.). Me thinks you have been jousting with Motor Daddy, too long.

13. ### SyneSine qua nonValued Senior Member

Messages:
3,515
I never said there was an absolute time and space, only that QM is framed in such, although I did misspeak about the Standard Model, as it is a relativistic quantum theory.

14. ### Prof.Laymantotally internally reflectedRegistered Senior Member

Messages:
982
I don't think I could ever be able to become a happy scientist until we could show how relativity applies to all the basic priciples of quantum mechanics. It wouldn't be until then I would be able to suffer from any deliousions that physics is complete.

The Lorentz transformations is just the equation for proper time, just in a different form. Getting to the equation of the proper time has nothing to do with how things in a clock are veiwed to move at different speeds. It only has to do with how real distances relate to each other from two different coordinate systems (the relation of their "measuring rods", as put by Einstein). I have made multiple threads that clearly show this, and it only requires the most basic algebra.

15. ### SyneSine qua nonValued Senior Member

Messages:
3,515
Who said physics was complete? People have erroneously claimed so several times, only to be proven wrong. Most real scientists in the field are happy to have such a puzzle as quantum gravity to aim at solving, with the promise of new physics to puzzle over upon its resolution.

16. ### AlphaNumericFully ionizedModerator

Messages:
6,697
... you realise your random suppositions based on almost complete lack of information or experience is not how you do science.

We can never show such a thing. You can never prove something is fundamental true like that, you'd have to test every possible system. We can construct models which mix special relativity and quantum mechanics, it is known as quantum field theory, and it is demonstrably superior to non-relativistic quantum mechanics.

Who said it was complete? We know there is a fundamental gap in our understanding of the subatomic, in that we don't know how quantum gravity works. It's funny you accuse physicists of such a view, you seem to suffer from delusions of competency though, given what you've said about your ability to grasp subatomic physics and just pluck personal 'explanations' out of nowhere.

No, they aren't and no, you haven't. Lorentz transformations are so much more. Constructing the expression $t' = \gamma t$, which is the expression you're referring to, is a special case of part of a Lorentz transform. That expression alone is not enough to have physics independent of inertial frame choice. What about how length changes? Well even if you also consider $x' = \gamma x$ that still isn't enough. Lorentz transforms are defined as the transforms which lead the Minkowski metric invariant, $ds^{2} = \eta_{ab}dx^{a}dx^{b} = -dt^{2} + \sum_{i=1}^{3}(dx^{i})^{2}$ where $\eta = \textrm{diag}(-1,+1,+1,+1)$ (for 1+3 dimensional space-time). A Lorentz transform should change the coordinates (t,x,y,z) to (t',x',y',z') such that $-dt^{2} + d\mathbf{x} \cdot d \mathbf{x} = -dt'^{2} + d\mathbf{x}' \cdot d \mathbf{x}'$. The transform you have stated doesn't do that. Lorentz transforms also have a group structure (look up the mathematical meaning of 'group', don't assume it means what it normally means!), where the compositions of Lorentz transforms are Lorentz transforms since they sequentially preserve the metric.

I'm sure you didn't understand that and I'm sure you won't understand what I'm about to say but I do it to illustrate how staggeringly far from grasping Lorentz transforms you are if you think what you're said about proper time essentially covers it. Lorentz transforms are formed from boosts and rotations. Clearly the 3 dimensional spatial part of the 1+3 dimensional metric has Euclidean structure and thus inherits all of its symmetries, namely the space of rotations SO(3). In addition to this 3 dimensional space of transforms (the dimensionality of the SO(N) group is N(N-1)/2 so in this case N=3 gives a 3 dimensional rotation group) there is a 3 dimensional space of boost transforms (in this case the 3 is because of 3 spatial dimensions). Due to rotational invariance we can define the structure of the boost in the x direction and the other two are obtained by rotation conjugation. The Lorentz boost in the (t,x) plane is $(t,x) \to (t',x') = \gamma(t-vx,x-vt)$. Note how this is more general than your expression. To confirm it does leave $ds^{2}$ unchanged we do the necessary algebra : $dt' = \gamma(dt - v dx)$ and $dx' = \gamma(dx - v dt)$ gives $-(dt')^{2} + (dx')^{2} = -\gamma^{2}(dt^{2} - 2v\,dx\,dt + v^{2} dx^{2}) + \gamma(dx^{2} - 2v\,dt\,dx + v^{2}dt^{2}) = \gamma^{2}\Big( -(1-v^{2})dt^{2} + (1-v^{2})dx^{2} \Big) = \gamma^{2}(1-v^{2})(-dt^{2} + dx^{2}) = -dt^{2} + dx^{2}$ and so we indeed have a Lorentz boost. A general Lorentz transform is expressible in terms of a sequence of rotations and boosts. We can prove the group structure of the Lorentz transform by composing two of them and rearranging. It is a bit tedious to do it in full generality so I'll prove it for the combination of two boosts in the x direction to illustrate, one by speed v and another by speed v', with associated Lorentz factors $\gamma,\gamma'$ respectively. If $(t',x') = \gamma(t-vx,x-vt)$ and $(t'',x'') = \gamma'(t'-v'x',x'-v't')$ then we have $(t'',x'') = \gamma'\gamma\Big( (t-vx) - v'(x-vt) , (x-vt) - v'(t-vx) \Big) = \gamma\gamma' \Big( (1+vv')t - (v+v')x , (1+vv')x - (v+v')t \Big)$. Pulling out a factor of 1+vv' gives $(t'',x'') = (1+vv') \gamma\gamma' \Big( t - \frac{v+v'}{1+vv'}x , x - \frac{v+v'}{1+vv'}t \Big)$. The terms in the brackets match the Lorentz boost form if the composed velocity is $\frac{v+v'}{1+vv'}$. With some algebra it is simple to show that if $\gamma(v) = \frac{1}{\sqrt{1-v^{2}}}$ then $\gamma(\frac{v+v'}{1+vv'}) = (1+vv')\gamma(v)\gamma(v')$ as required and so the composition of a Lorentz boost with speed v with one with speed v' is the same as a Lorentz boost by speed $\frac{v+v'}{1+vv'}$. There, just proven the 1 dimensional relativistic velocity addition formula.

As with your previous glaring gaps in understanding none of this is particularly advanced, it is something covered in a 1st year course in special relativity, something any introductory book on relativity would cover. But since you've never read any such thing you have zero idea what special relativity actually involves on anything even remotely like a working level. Instead you come out with this ridiculous (ie deserving of ridicule!) statements about how you've done some deep result or you've grasped something well or how really its all pretty straight forward. No you haven't, no you haven't and no it isn't, at least not for you.

How many more times do we have to do this little dance? How many more times do I have to explicitly walk you through rudimentary physics you claim to grasp but don't? How many more times will be necessary before you realise what you do know, on a qualitative arm wavy half misunderstood level, is just a tiny tiny tip of an enormous iceberg of what relativity or quantum mechanics really involve? If you'd even read the first chapter of a special relativity textbook or set of lecture notes you'd never have made the laughable statement you did. Come on, think about it. You know you haven't read such material so you know you don't have very much knowledge about what the quantitative side of special relativity actually involves so why do you persist in trying to present yourself as if you do? It's just like how you claimed to have a good intuition for quantum systems, how can you possibly say such a thing when you have no experience with any working models of quantum mechanics or any experience with proper experimental data? You're asserting you know things you have no real knowledge of! It would be like me claiming to know what a Japanese newspaper says when I've never seen one and even if I did I cannot read the language. Farsight does the same and you really don't want to be emulating him.

17. ### Prof.Laymantotally internally reflectedRegistered Senior Member

Messages:
982
I considered the first experrion here, and I think I found a problem. You say, $t' = \gamma t$ , but then what if you just try and go ahead and solve for $\gamma$ ? Then you would get something like, $\gamma = t' / t$ , then you could say that $\gamma = sqrt{1- \frac{v^{2}}{c^{2}}}$ , by canceling out time in the equation; but then you say that $\gamma(v) = \frac{1}{\sqrt{1-v^{2}}}$