Trading in the Real Space for Real Time

Idiot. I've said that the singularity is by definition a point of infinite density and zero volume. Please find where I've said otherwise.

I've also said that when your equations yield infinity as an answer,it's a pretty sure sign that the equations have exceeded their realm of appliciblity.

Do some reading other than wiki. Maybe even study it for a while. But actual education would be asking too much, wouldn't it?

 

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******* ROFL

First page http://sciforums.com/showthread.php?t=104927

“ Originally Posted by AlexG
The singularity is not massless, and it does not consist of dense matter. Nor is it formed by the accretion of matter.

Please do some reading on black holes and astronomy in general before you shoot your mouth off again. ”

 

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See the other thread. I see no reason to point out you're an idiot in two places, when one is sufficient.

 

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Yes, please, anyone is invited to see this. It's dense, but not with matter.

Please Register or Log in to view the hidden image!

 

How long will you keep up this pretense? Nobody with even casual acquaintance of Hilbert spaces believes you, and frankly, anybody who's liable to be fooled is also liable to think TimeCube is the pinnacle of human thought.

I ask you: What do you think you stand to gain by keeping this up?

Furthermore it has been amply demonstrated that you do not know the preliminaries, either. You've repeatedly made grave mistakes with basic linear algebra, but worse, also in such stuff as sets, intervals, simple arithmetic, etc. You don't even know the preliminaries to the preliminaries, and you're either lying or deluded when you say you know it.

 

But you obviously think your work good enough to be evaluated by such a person.

I've 'tinkered' in a few things, come up with minor results which I haven't read anywhere, but I wouldn't email them to a friend with a PhD in that area for their comments. I might, during a conversation at the pub, discuss the area with them but even then the discussion wouldn't go down to the level of "Here's a print out of my work, can you read it".

I said that often 0, as an index, can refer to time. In the sense that a 4 vector is of the form \(v = v^{a}\frac{\partial}{\partial x^{a}} = v^{0}\frac{\partial}{\partial t} + v^{1}\frac{\partial}{\partial x} + v^{2}\frac{\partial}{\partial y} + v^{3}\frac{\partial}{\partial z}\). That is often used in relativity. If you'd read much relativity (not wordy explanations, actual relativity) you'd know that my use of the word 'often' is entirely valid.

Thanks, I'm glad I got your okay to do so....

I gauge it from your posts, just as people gauge my level of understanding from my posts and everyone else does for everyone else's posts. If I couldn't add up and didn't know what a function was few people would think I'd spent much time reading mathematics. You wouldn't have to know precisely what I had or hadn't read, you'd be able to tell that if I haven't heard of one of the most fundamental concepts in mathematics then I can't have gotten far in my reading of it. Or if I couldn't add up then I am unlikely to be able to do more advanced stuff.

You can't add vectors. You can't even divide by 2 or take square roots properly!! And you expect us to think you're familiar with more advanced stuff?

You just like to talk about vorticity and about how you've come up with a result yourself?

You link to wordy essays. The rare occasion you do link to something with an ArXiv or journal reference you don't show any understanding of it.

Clearly my explanation went over your head. If you had a little more experience with basic differential equations and operators you'd have understood why there isn't a unique bijection between the wave function and the Hamiltonian. This isn't even quantum mechanics, its analysis 101.

For instance, consider the particle trapped in a box of constant potential. You end up with a system of the form \(H = \frac{\partial^{2}}{\partial x^{2}} + k\). If you solve for the wave-function the equation \(H \psi(x) = 0\) then you have TWO linearly independent solutions and thus infinitely many possible solutions, ie \(\psi(x) = A\psi_{1}(x) + B \psi_{2}(x)\) (where the individual solutions happen to be sin and cos). Given just the Hamiltonian you cannot reconstruct the wave function uniquely. Basically the equation \(H\psi(x) = 0\) is a second order differential equation, which means that you're going to need initial conditions to uniquely determine the wave function. That's something true of any kind of differential system, you have to have initial/boundary conditions in order to extract out a specific solution from the (typically) infinite choices you have.

Going the other way, if you give me a wave function then the question "What is the Hamiltonian for this wave function which satisfies \(H\psi = 0\)?" is akin to asking "Give me the unique second order differential operator for which this wave function is a zero eigenvalue eigenfunction.". There's infinitely many such operators, trivially so since if H satisfies that condition then so does kH.

So your "They uniquely encode one another" isn't a matter of arguing physics, this is fundamental mathematics. The issue of uniqueness of solutions to equations is one 'even physicists' should know.

Applying the Schrodinger equation and actually understanding what the states in the Hilbert space can be isn't 'my concept', its simply an application of definitions. The 0 cannot literally be 'time' since the Hilbert space the equation is defined upon doesn't include that.

I've explained many times and the fact you still don't get it demonstrates that, despite your claims, you don't understand even the simplest thing about Hilbert spaces.

Are you hoping if you keep saying it it'll make it so?

And yet you maintain the 0 is literally time.

The 0 means there's no time dependence. I never said it means there's time dependence.

When you can grasp something the first time you're told it then maybe I'll stop having to repeat myself.

Evidence says otherwise.

No, you failed to understand what I said, which is because you don't understand the Schrodinger equation. Though its possible to formulate a time dependent Hamiltonian (time dependent operators are in the Heisenberg picture, compared to the Schrodinger one), I explained that the Schrodinger equation relates the time dependency of the wave function to the state obtained by acting the Hamiltonian on the wave function. If the Hamiltonian acts on the wave function to give zero then it means the wave function has no time dependency. This is nothing but a trivial application of the Schrodinger equation and yet you need to have it explained again.

No, its not some 'remedial sense', if something satisfies \(\partial_{t}\psi = 0\) then it has no explicit time dependency. Remember what you were saying about agreed upon definitions? Twisting them to suit your purpose does you no favours.

Actually I covered the different 'pictures' in my 3rd year in the 'Principles of Quantum Mechanics' course. And I love how you complain when I supposedly make assumptions about what you have or haven't read yet you have no problem assuming what I have or haven't read.

You seem to be under the misapprehension that you're treading in areas which are advanced and stretching the bounds of what I've done myself. Like I said, the different pictures and time dependency of operators and states is in one of the opening sections of one of the first course in QM I took... *counts quickly in head* ..... 6 years ago.

If you hadn't just jumped ahead and decided to copy/paste big words from papers you don't understand and instead you'd actually learnt quantum mechanics step by step you'd have known about the different pictures too. Instead you take a pot shot at me for the supposedly silly notion that operators can be time independent (ironically I never actually referred to such a concept, you misunderstood me), only for your Googling to turn it up a little while later.

I know that threads like this push your Googling skills to the max and you often come across concepts a few days after they've come up and then you realise they are common place but some of us don't have that issue. Some of us have this thing called memory where we can recall almost instantly things we've previously learnt, it makes understanding and discussion so much easier when you have more bits of the puzzle in your head and don't have to use Google every 5 seconds. Try it some time.

So you've gone from not knowing what 'covariant' means and not being able to add vectors to having a sufficient grasp of the work of Penrose and Hawking that you can tell other people "Do some reading!"?! And you seriously think anyone is going to swallow that load of crap? You claim you're modest and not trying to be a scientist but you want people to think you're well read, ie about Penrose, Hawking, quantum mechanics, Hilbert spaces etc, and you're willing to tell other people "Do your homework first!" when its painfully clear you haven't yourself.

Its comments like that which make me feel entirely justified in being short with you. You don't know a damn thing about these areas of physics, other than what you can parrot back off Wikipedia, and yet you want people to think you do and you act as if you do. No one buys it. What modicum of scientific merit this thread might have had goes right out the window when you come up with things like that because no one believes you understand those things. If you're going to be a hypocritical hack then I'll treat you as such.

 

Oh - you are starting to infuriate me alphanumeric. You are starting to steal my arguements. I call Alex a hypocrite, because its obvious hes not a scientist, but he goes about telling people to study physics, then you call me a hypocrite for telling another person they are a hypocrite.

Fuck off.

 

Actually I covered the different 'pictures' in my 3rd year in the 'Principles of Quantum Mechanics' course. And I love how you complain when I supposedly make assumptions about what you have or haven't read yet you have no problem assuming what I have or haven't read.

You seem to be under the misapprehension that you're treading in areas which are advanced and stretching the bounds of what I've done myself. Like I said, the different pictures and time dependency of operators and states is in one of the opening sections of one of the first course in QM I took... *counts quickly in head* ..... 6 years ago.

If you hadn't just jumped ahead and decided to copy/paste big words from papers you don't understand and instead you'd actually learnt quantum mechanics step by step you'd have known about the different pictures too. Instead you take a pot shot at me for the supposedly silly notion that operators can be time independent (ironically I never actually referred to such a concept, you misunderstood me), only for your Googling to turn it up a little while later.

I know that threads like this push your Googling skills to the max and you often come across concepts a few days after they've come up and then you realise they are common place but some of us don't have that issue. Some of us have this thing called memory where we can recall almost instantly things we've previously learnt, it makes understanding and discussion so much easier when you have more bits of the puzzle in your head and don't have to use Google every 5 seconds. Try it some time.


Except, there is such a picture where you can treat the operator as time-dependant, something which you managed to completely ignore when you told me the operator is not.

 

Why don't you put me in my place by showing your amazing grasp of the physics you profess you understand? I'm more than happy to overlook any clashes of personality I might have with someone if they obviously know their shit and they are honest. Mathematicians aren't known for their social skills and while I might not like someone I can still respect them for their abilities.

When you show you know your shit I'll stop saying "You don't understand that".

Actually I said that \(H\psi = 0\) means that wave function is time independent. Via the Schrodinger equation the time dependence of a wave function can be related to the Hamiltonian acting on that wave function. The issue of the time dependence of \(\psi\) is different from the time dependence of \(H\), which is different from the expression \(H\psi\). You seem to think I refer to one when I refer to the other.

I suggest you get your head around the concept of operators and states and how they relate to one another. Until then you're seriously hindered in how much of any quantum mechanics you can grasp.

 

I'm sure you said that 'i was mistaken' that the operator was time-dependant. I'll look back.

 

http://sciforums.com/showthread.php?t=104881&page=5

yes you say time is not a state in the Hilbert space - I do beleive you meant the Hamiltonian as we are not discussing the Hilbert space. It's post 92.

 

No, I meant precisely what I said. The WdW written as \(H|\Phi\rangle = 0\) is expressed as an operator, H, acting on a state, \(|\Phi\rangle\). It is phrased in the bra-ket notation and in order to have such notions as 'expectation value' for the operator (which is associated to energy in regards to the Hamiltonian) you need an inner product as \(E_{\Phi} = \frac{\langle \Phi|H|\Phi\rangle}{\langle \Phi|\Phi\rangle}\). This (along with other technical points) means that \(|\Phi\rangle\) is an element in a Hilbert space. So is \(H|\Phi\rangle\). Thus anything equal to it must also reside in the same space, else its mathematically gibberish. Thus since \(H|\Phi\rangle = 0\) then the 0 is in the same Hilbert space. Thus, in this case, the 0 is an element in the Hilbert space of quantum states.

To give a simple example, suppose we consider 'states' which are vectors, \(\underline{v} \in \mathbb{R}^{N}\). H then has an NxN matrix representation, it maps vectors in \(\mathbb{R}^{N}\) to vectors in \(\mathbb{R}^{N}\). Thus if \(|\Phi\rangle \sim (\phi_{1},\ldots,\phi_{N}) \sim \phi_{i} \sim \underline{\phi}\) then the expression \(H|\Phi\rangle\) becomes vector valued and in terms of components we get \(H_{ij}\phi_{j}\). This it a vector in \(\mathbb{R}^{N}\), so its equal to \(H_{ij}\phi_{j} = \xi_{i}\) for some vector \(\underline{\xi}\). If \(H|\Phi\rangle = 0\) then we have that \(\xi_{i} = 0\) for all indices, thus \(\underline{\xi} = \underline{0}\), thus \(\underline{H\cdot \underline{\phi}} = \underline{0}\). So in actual fact the zero in this example is not just zero like the 0 which is 1 less than 1 in the integers, its the zero of the Hilbert space \(\underline{\phi}\) belongs to.

'Time' is a parameter, not a state or a vector or an element of any vector space (all Hilbert spaces are vector spaces). States in the Hilbert space and operators which act on them can be dependent upon time but they can't equal time in the sense you're claiming.

Would you also care to comment on my explanation as to why the wave function and Hamiltonian aren't uniquely encoded into one another, that you need additional information to pin down one given the other?

 

The \(\psi\)-function describes all the possible physical states in the equation \(H|\psi>= 0\). The classical Hamiltonian constraint provides \(H= 0\), and should be quantised so that every physical state given by \(\psi\) satisfies the WDW equation, hence why I kept saying the \(\psi\)-function encodes all the necessery information. I said, if you like, it shows in the Schrodinger equation that time does not depend on \(\psi\) but in all respects of the interpretation of the WDW equation, frozen time models, and the universe where internal energy changes cease to exist, are all a reality of the equation. In every respect of the WDW equation, time certainly ceases to exist, and is purely an interpretational method to whether see the zero in the equation refer to this. In every sense, it does, since the right handside of the schrodinger equation \(\hat{H}|\psi>= i\hbar \partial_t \psi\) but it's not clear because of this why classical quanties would depend on time. The encoding I explained however, could be as trivial from Tomonaga-Schwinger approach, where both the operator and the wave function have (parts) which refer to the existence of time.

 

I suspect you've been reading Wiki or the like and not understood what it says well enough to articulate a coherent explanation.

The wave function of the WdW equation is formed by summing over space-time geometries. It is in this sense that it can be said to describe 'everything', it includes contributions from all viable space-time configurations. This doesn't mean it encodes the Hamiltonian, in that you cannot be given a wave function, asked "What operator is this the zero eigenvalue eigenfunction of?" and then provide a unique answer. Similarly, given a Hamiltonian there is no unique wave function which has zero eigenvalue wrt the Hamiltonian.

The whole "The wave function encodes everything" is in reference to space-time geometries and matter distributions, not the Hamiltonian.

You mean \(\psi\) on time.

When summing over all space-time geometries you're implicitly including variation in time. A single space-time configuration with no time dependency can be viewed as 'frozen' but if you're considering an entire ensemble of space-time configurations then you can't immediately conclude they are separately time independent, as they could mix and cancel one another's time dependences.

A wave front like \(e^{i(kx-wt)}\) is time dependent but its possible to form time independent things from normalised linear combinations of waves. Standing waves are an example, where its only their amplitude which varies in time and that can be removed by normalisation.

I can't help but feel you're back peddling. Before you were very adamant that it equalled time and no amount of 'interpreting' would make that statement correct.

The wave function isn't describing one space-time, it describes a slew of them as a single entity, so just because such an entity overall has no time dependence doesn't mean there isn't some kind of variation occuring within it.

For instance, steady state fluid flows can be time independent, ie all physical quantities satisfy \(\partial_{t}f = 0\) and yet something is still flowing, the system isn't frozen.

Explain, in your own words and carefully, what the 'Tomonaga-Schwinger approach' is and what relevance it has.

 

I am not the best at articulating sometimes, but it's certainly not wiki.

The Hamiltonian simply refers to the total energy of the system. The wave function of that system however must have some characteristic to the equation. It's not enough to know the classical limit, which is why the wave function encodes information about the energy limits. There are some applications I have learned about this, for instance, by reconfiguring the equation as an average \(<\psi|\hat{H}|\psi>= 0\).

Actually I do believe the problem is that shuffling spacetime coordinates do not change anything, as found in GR, the problem of the WDW equation is discovered by Diffeomorphism invariance on the theory.

Ok

I am not back peddling as such. The Time is a factor on the right hand side of the equation. If it vanishes then so must the time description.

I'm not sure how that would apply to the WDW equation.

Preaching to someone who knows near nothing about fluid mechanics. I will just take your word for it.

\(i \hbar \frac{\partial}{\partial t}A_I |\psi_I(t)>= H_I(t)|\psi_I(t)>\)

The 'Tomonaga-Schwinger approach' gives a time evolution to the operator in the schrodinger equation along with the psi-function. The relevence is unique. I've heard one unique factor is it gives a ''many fingered time''.

 

Just a matter of notation here, but I take it the ij-components refer to matrix rows and columns yes?

 

The Hamiltonian allows you to work out the energy for a wave function in a given system but without the wave function there's no 'energy of the system'.

Firstly, as a point on LaTeX (a professor I had used to do the same and it makes me grit my teeth) the < > are not 'less than' and 'great than', their LaTeX codes are \langle and \rangle, 'left angle bracket' and 'right angle braket'.

Secondly \(\langle\psi|\hat{H}|\psi\rangle\) is the expectation value of the energy of wavefunction. That isn't something unique to the Hamiltonian, the expectation value of any operator with respect to a particular state is written as \(\langle\psi|\hat{\mathcal{O}}|\psi\rangle\). For non-normalised states it is \(\frac{\langle\psi|\hat{\mathcal{O}}|\psi\rangle}{\langle\psi|\psi\rangle}\), precisely the thing I typed before.

What I said has nothing to do with diffeomorphisms. The wave function in the WdW equation is obtained by doing a summation over all possible space-time and matter geometries, not a single geometry. If someone says "The metric is time invariant" then that's more restrictive than the statement "The ensemble of metrics is time invariant". Individual metrics might be time dependent but when you combine them in a certain way you get a time independent quantity.

An example, which I'm sure you're unfamiliar with, is gauge invariance. A quantity might not be gauge invariant, say \(\partial_{\mu}A_{\nu}\) under \(A_{\mu} \to A_{\mu} + \partial_{\mu}\Lambda\). Neither is the quantity \(\partial_{\nu}A_{\mu}\). But put them together to make \(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}\) and you do get a gauge invariant quantity. Individual components need not be time invariant in order for the entire system to be time invariant overall. This is particularly so if you're dealing with not just a single space-time but entire families of space-times. Multi-dimensional geometry is difficult when the dimensions are just those of space-time, adding in additional parametrisations in the metric makes things even less intuitive.

Which is quite different from the fervent statement you were making about the 0 being time, not just something related to time dependence.

Because you skipped the basics, the required fundamentals, and jumped into things you have no grounding in.

And the \(A_{I}\)? Remember, when you're asked to explain something if you introduce notation not previously discussed then the onus is on you to define it. Its a problem you have in previous threads, you're asked to explain something because the notation seems dubious and you explain one undefined thing in terms of another.

And what's the I subscript for? I'm not asking in the sense of "I don't know", I'm asking because I want to see if you do and I want you to explain why you're using it.

The time evolution of the wave function is determined by the Hamiltonian but that doesn't mean that given the wave function you can determine the Hamiltonian from the WdW equation. Besides, aren't you making the point that the WdW removes time dependency? If it does that then the 'T-S approach' becomes trivial, nothing happens.

 

The subscript \(I\) refers to interaction, I think.

 

Whilst I have little time reading these massive posts you are making and equally the patience to reply to every bit, I will be ceasing interaction here after this next post. You say I am not familiar, or atleast you think I am not familiar with this particular gauge. I think you should start giving me a little credit. I can identify many things you post, such as the Reimann Tensor, which I am sure you would have bet on that too.

It is true that quantity is not gauge invariant. The \(A_{\mu} = \eta_{\mu \nu}A^{\nu}\) and is a vector (t-0,1) tensor field. Your \(\partial_{\mu}\) is simply \(\frac{\partial}{\partial x^{\mu}}\). To be gauge invariant there needs to be a local symmetry of the action, which is why the first example you gave \(A_{\mu} \rightarrow A_{\mu}+\partial_{\mu} \Lambda\) is not.

 

I want you to answer my question though, on the ij components.