Aethelwulf

I said lower.... It was a feeble demonstration to show that the workings of physics may provide new ways into this science. Yes, it's something like that. If one can get over that obstacle, there would be no problem. Personally, I don't see it as ''being outside the realms of possibilities.''

Well, I gave a demonstration that it can actually occur at lower temperatures using a tunneling processes. Certain stars work this way. For a more conventional answer, I don't have one, but we should keep our minds open for we do not know all the processes behind nature. Remarkable breakthroughs of similar nature by unlocking the secrets of the fundamental universe have been made in such ways. I'm a bit rusty on Schwingers work now, but he did work on latices for cold fusion http://www.infinite-energy.com/iemagazine/issue1/colfusthe.html

Who me? Such as would be found in the proton cycle in stars, such a barrier is penetrated by tunneling effects allowing the process to occur at much lower temperatures. So not all cases actually require extremely high thermodynamical transition phases.

''Think'' they found it and none of the results are conclusive. It will need more tests. http://www.guardian.co.uk/science/2011/dec/13/higgs-boson-glimpsed-cern-scientists

Well, let's not all be too harsh. Some scientists have seen merit in it. Schwinger, perhaps one of the most prominent scientists of all time wrote eight papers on the subjects. There is of course no reason to think it is not possible. To date, it has drawn a lot of attention, on and off.

Now, just as Motz had made clear in his paper, the Schwinger Quantization method for charged particles in fields takes on a remarkable similarity to the gravitational charge [math]Gm^2 = \hbar c[/math]. The link here will explain the charge condition: http://encyclopedia2.thefreedictionary.com/Dirac-Zwanziger-Schwinger+quantization+condition What we essentially have is [math]\frac{e\mu}{c} = \frac{1}{2}n\hbar[/math] The gravitational charge quantization is [math]\frac{Gm^2}{c}= \hbar[/math] So motz was absolutely correct in stating the importance between the two equations.

Quantum tends to be as ''elementary'' as you can get for any particle. There is such a thing as a quantum number for photons, called their spin states.

Following my own example, I end up with [math]-i \hbar (\frac{\partial \phi(t)}{\partial t}) = \lambda \phi(t)[/math] The left hand side is the energy operator [math]\hat{E} \phi(t) = -i \hbar \frac{\partial}{\partial t} \phi(t) = \lambda \phi(t)[/math] Does the fact you end up (in your simple method) using a force operator make my method wrong?

There's nothing mystical behind the use of Hookes law here. Just looking at my notes, it's just the way I defined the force. All your relativistic dynamics are still wrapped up in the [math]dp/dt[/math] (used a little p now to halt any confusion). Yes, Equation two should be written like that. The first term [math]-i\hbar \frac{\partial}{\partial x}[/math], do you recognize it? It's the momentum operator, we get this when we quantize the equation. That means what we really have [math]\frac{\partial \hat{p}}{\partial t}[/math]. This is why there is a mixture of the change in position and time on the LHS like you spotted. What's H? Anyway, assuming your approach is correct, that's fine and all dandy. But I wanted to specifically quantize the second law. Ah H is meant to be the Hamiltonian yes? Should it be defined as a Hamiltonian, I thought W=Fx?

I'm sure Hawking did not say we could exist outside the universe or see anything before the beginning of time itself.

Well, his paper defined it as the gravitational 3-force. Don't shoot the messenger, this is why I asked the questions I did. I would say, should it be surprising that a force can have a christoffel symbol in it? I mean... after all, the Christoffel Symbol (is the gravitational field) in GR. (more) I have the geodesic equation written down somewhere, I was working by memory... but if you say so. The equation however will not describe massless particles, since the equation has M defined as a gravitational charge, or passive mass in other words. So yes, s would be the proper time interval. Here it is. You where right, it is not partial derivatives, but it does use proper time. http://en.wikipedia.org/wiki/Geodesic_(general_relativity)

That doesn't make any sense to me. I take it that was an attempt of a joke.

[math]F=Ma[/math] is not true in relativity. The force equation which remains unchanged in relativity is [math]F=d(m_{rel}v)/dt[/math] The reason why is because in [math]F=Ma[/math], mass is generally a constant which is not true in relativity. I had a little look on line and there is a wiki article which mentions it: http://en.wikipedia.org/wiki/Mass_in_special_relativity Oh, and -kx is just hookes law for force.

I decided to quantize Newtons 2nd Law [math]-kx = \frac{\partial P}{\partial t}[/math] today. Has anyone done this before? I need some help. I start with quantizing the equation, naturally; [math]-i \hbar \frac{\partial}{\partial x} (\frac{\partial}{\partial t}) = -kx[/math] Hitting it with a wave function [math]\Psi = \psi(x)\phi(t)[/math] gives [math]-i \hbar \frac{\partial}{\partial x} (\frac{\partial \phi(t)}{\partial t}\psi(x)) = -kx \psi(x)\phi(t)[/math] To solve it we are therefore going to use the separation of variables method. (But I need you guys to make sure I am doing this right) Divide through by [math]\Psi[/math] gives [math]-i \hbar \frac{\partial}{\partial x} (\frac{\partial \phi(t)}{\partial t}\frac{1}{\phi(t)}) = -kx[/math] So this is as far as I have got. Should I move the [math]\frac{\partial}{\partial x}[/math] term to the right as [math]-i \hbar (\frac{\partial \phi(t)}{\partial t}\frac{1}{\phi(t)}) = \frac{-kx}{(\frac{\partial}{\partial x})}[/math] So that on the left I purely have variables of t and on the right variables of x? I think this would mean that both sides are independent, meaning that they equal a constant? [math]-i \hbar (\frac{\partial \phi(t)}{\partial t}\frac{1}{\phi(t)}) = \lambda[/math] [math]\frac{-kx}{(\frac{\partial}{\partial x})} = \lambda[/math] Then I would like to concentrate on the time-dependant case [math]-i \hbar (\frac{\partial \phi(t)}{\partial t}) = \lambda \phi(t)[/math] Will now be [math](\frac{\partial \phi(t)}{\phi(t)}) = \frac{-i\lambda}{\hbar}\partial t[/math] [math]\int (\frac{\partial \phi(t)}{\phi(t)}) = \int \frac{-i\lambda}{\hbar}\partial t \rightarrow \frac{-i\lambda}{\hbar} t +C[/math] Then the solution would be [math]C e^{\frac{-i \mathcal{A}}{\hbar}}[/math] where I have let [math]\lambda t = \mathcal{A}[/math]. Does this seem right, or have I messed up somewhere?

I thought you were replying to me. I apologize.

I never said it was an illusion. You need to start reading more carefully Swansont. You have a record with me now of not reading people correctly. I said time was not a physical phenomenon.

Pmb, where was the thread you posted your work on mass? Was it this one, I had been looking for it because I had a question. In your paper, you define the gravitational charge component in a gravitational 3-force equation. I believe it had the form [math]f^k = m\Gamma_{ij}^{k}v^i v^j[/math] I just wanted to know, how one would derive this equation. Presumably the connection is playing your usual role of the gravitational field yes? I see there is a dependance on the velocity. The [math]\Gamma_{ij}^{k}v^i v^j[/math] part looks a little bit like it comes from the geodesic equation of motion which involves usually a term of second derivatives. The equation I had in mind was this one [math]\frac{\partial^2 x^{\mu}}{\partial s^2} + \Gamma^{\mu}_{ij} \frac{\partial x^i}{\partial s} \frac{\partial x^j}{\partial s}=0[/math] Ok, they look quite different but if you rearrange it [math] \Gamma^{\mu}_{ij} \frac{\partial x^i}{\partial s} \frac{\partial x^j}{\partial s} = \frac{\partial^2 x^{\mu}}{\partial s^2}[/math] The left hand side of the equation does look a little like [math]\Gamma_{ij}^{k}v^i v^j[/math] So I am just wondering how the force equation ''gets'' the terms it has. I'm especially interested in the dependence of the velocity and how this comes about. I'm actually starting to think its definitely from the geodesic equation. The s is just 'proper time'. ct = x and c = x/t. [math] \Gamma^{\mu}_{ij} \frac{\partial x^i}{\partial s} \frac{\partial x^j}{\partial s} = \Gamma^{\mu}_{ij} v^i v^j[/math] I think that works out. This adds up to the same dimensions as you find in that force equation. But then I start thinking about the dimensions of the equation. [math]m\Gamma^{\mu}_{ij} v^i v^j[/math] How does this equate to a 3-force? If we just take it at face value, the part described as [math]\Gamma^{\mu}_{ij} v^i v^j[/math] would have to have dimensions of acceleration.

I read everything, but this part struck me the most. I agree it was an absurdity, but I think it came about because of the unification of space with time. I think, because fields could create a gravitational distortion in space, then time somehow was physical. Which is a load of rubbish. I think today, time is one of the most misunderstood concept there is - and a much abused one, where people often I read equate motion with change and a change with time, or that time is something which can be observed because there is a clock on the wall... these things are just ridiculous. Time does not mean change nor is time an observable, just as much time is not a real physical entity.

Yes, it is called the Null Energy Condition. [math]Mc^2 - \frac{GM^2}{2R} = E[/math] When [math]M=0[/math] it is said you are left with the metric.

Can you explain to me, in your own words, the reason for a gravitional 3-force. I see it in your paper, but you define mass as the gravitational charge - this is not the same kind of charge I think of or another author... can you elaborate on your christoffel symbol and its use, thank you. Regards.

Well, no one moved my post; It seems outside those who can answer it. I want to know what the speculators say. http://www.scienceforums.net/topic/67240-flat-geometry-and-the-wdw-equation/page__pid__685565#entry685565 Speculate away.

While on a train today, I came to a question I could not answer. It grew from recent talks on the Wheeler-deWitt (WDW) equation ... whilst I have pretty much advocated the idea over many years that we may actually live in a timeless universe (as many scientists have), it did cross my mind to believe that perhaps I have had it all wrong and that I must at least entertain the idea that the WDE-equation has it all wrong. I considered some reasons why this approach could be wrong and I came across a question I couldn't answer within myself. The WDW-equation is obtained from quantizing the Einstein Field Equations - the general relativistic equations which describe curvature in the universe - noting this in my mind, I reminded myself that the wave function of the WDW-equation was a global case, describing the entire universe. However, I have known for a while that the Wilkinson Microwave Anisotropy Probe has determined that we live in a universe which is more or less flat, which begs to question whether General Relativity would effectively break down on large enough levels. If the universe is flat therefore, it seems unreasonable to attend idea's about it from equations which satisfy curvature [math]g_{\mu \nu} \ne 0[/math]. Instead it seems only reasonable to use the Newtonian Limit [math]E = \int_v T^{00} dV[/math] The WDW-equation is derived from the equations which describe curvature, but if this is not the true configuration of the universe at large, then we might assume that many of the problems which have arisen from the WDW-equation, such as a timelessness and even the non-complexifying wave function (real) have arisen from faulty premises to begin with. Instead, should we not be trying to quantize an approach which actually fits the universe at large, one that is pretty much flat in all directions we look? (On a separate note) It also occurred to me that the universe would appear mostly flat if it was rotating (So in its natural form, it would be sphere like, if it was rotating, it would be flat-disk like). But that's for another discussion. Is this thread a matter of no one can, or no one will answer it?

The Chaldeans had in fact placed a lot of numerical significance to words. This does not prove the case of the OP however.

Yes, of course I realize that, which is why I said you need to adjust the frequency as I said. When you do, no particle of energy can adjust to ''its'' size. We are talking about a very simple photon. All particles have sizes, but without the splitting of hairs, the photon can be the smallest and is the smallest.

Quite possibly so. Bolded by me. How so? Also you make a point saying the SS equation is defined in a Hilbert Space beyond the scope of the WDW equation. It is true the wave function of the WDW equation is no longer a spatial function, however, the operator H is a relativistic case which does act on the Hilbert Space. You've made no comments on the complexifying of the SS equation, or the deep phyiscal meaning I purported to when physicists view quantum gravity under the non-complex wave function. ( I realize in post 14 I said the time dependant SE is the analogue of the WDW equation - this is a mistake, I meant the time independEnt case).