Mystery111

Sure it can be viewed this way. So what is it rotating in? I've pondered this question. Suppose we get really exotic and steal an idea from certain string theories suggesting our universe is floating about in a multidimensional pool. It could be an eleven dimensional pool. 26 dimensions for some boson theories. Or maybe even a simple 5 dimensional vacuum where our branch is rotating relative to other branches floating about. But this goes against what I am comfortable with. I don't believe in parallel universes, (or at that a multiverse) and I certainly don't believe in any dimensions existing in this universe that is higher than four. So obviously I am at a problem with what relativity could say about a rotating machian universe.

I got somewhat interested today in how the quantum dispersion relation for a photon was originally derived and the reason why was because I accidently derived it today, and through a series of simple gestures. But because the dispersion relation is quite an easy equation for photons, there is doubt in my mind my derivation was somewhat easier at all. [math]\hbar c = Gm^2[/math] Rearranging for the angular momentum component: [math]\hbar = \frac{Gm^2}{c}[/math] Then multiply [math]\frac{\partial}{\partial t}[/math] on both sides gives: [math]\hbar \partial_t = \frac{GM^2}{c} \cdot \partial_t[/math] Hit this with [latex]\psi[/latex] and it makes [math]\hbar \cdot i\omega = \frac{Gm^2}{c} \cdot i \omega[/math] Cancel out the [math]i[/math]'s this gives and replace the [math]\omega[/math] with [math]ck[/math] on the right [math]\hbar \omega = \frac{Gm^2}{c} \cdot ck[/math] clearly has units of energy. From now on, we will call [math]\frac{Gm^2}{c} = \beta[/math]. Since [math]\omega=ck[/math] then we can see this equation as freely exchanging the terms [math]\hbar ck = \beta \cdot \omega[/math] At first, I wasn't sure what to make of this equation, if indeed anything at all. So I carried on: turns out that a series of proportionalities could be measured from altering the equation: [math]\frac{E}{(\beta \cdot \omega)} = k[/math] because [math]E= \hbar c k [/math] so [math]k = \frac{E}{\hbar c}[/math] It then struck me I'd seen the expression [math]\frac{E}{\hbar c}[/math] before. It is the case of electromagnetic radiation as a linear "dispersion relation" given as [math]E=\hbar ck[/math] with [math]k = \frac{E}{\hbar c}[/math] This is obviously strictly a case only for a photon. I certainly derived a form of the photon dispersion relation but with a pesky mass sqaured term in there. As elegent it seemed deriving a dispersion relation by messing about with some terms, I wanted to have the equation suit a dispersion relation for a particle with mass... let us assume the simple case of an electron. To derive the right one, I needed to go back to the derivations. [math]\hbar ck = \beta \cdot \omega[/math] Divide [math]c[/math] off the left hand side gives you a momentum expression [math]p = \hbar k = \frac{\beta \cdot \omega}{c}[/math] square [math]\hbar^2 k^2 = (\frac{\beta \omega}{c})^2[/math] Divide the left by [latex]2m[/latex] and you get the dispersion relation for an electron [math]\frac{\hbar^2 k^2}{2m} = E[/math] Has anyone got any links to how the dispersion formula was derived for a photon? I bet it was easier on second thoughts... I did a lot of fiddling about...

It's nice.

I believe the universe spins. If we find enough evidence suggesting that the universe spins, then there is a new problem. What is it spinning relative to? The only way to answer that is by saying there must be something outside the universe, which goes directly against what relativity has to say about space and time in general.

Except this is used as a reason that it cannot be physical, at least this is the general consensus. Just because relativity - or our application of relativity cannot handle them, should and I refine the arguement again, should not indicate a breakdown of physical theory. Our general understanding of relativity will break down, but there most certainly will be a theory which will deal with it, and if I am right, will help pave a more physical understanding of singular regions of spacetime.

Forget technical terms. Most of the times, scientists dispute the existence of these exotic regions simply because we cannot fathom them. I sware this is true. No one in the scientific community, apart from a few, actually can appreciate regions of spacetime which have singular applications. No one likes infinities, never mind the vacuum! Also, rocket said a singularity in GR is a failure of spacetime to be timelike geodesically complete. ...is a statement itself wrapped up in some rather complicated physics. To understand the swap between spacelike and timelike conditions rely on an understanding of the mathematics behind black holes themselves, which I would wager, not many here can appreciate. ps.... in fact if you are really nice to me, I Might even take you through the maths of black holes leading to where space time flip

I don't. I think a singularity is breakdown of our understanding, rather than the physical theory. I like singularities. I don't see why infinities should always be frowned upon.

I said the existence of these particles four years ago. Today, only the most prominent papers which speculate their existences are coming to light... http://www.stardrive.org/index.php?option=com_content&view=article&id=5377:a-1997-paper-published-on-cornells-aixvorg-neutrinos-must-be-tachyons&catid=43:science&Itemid=82 This is not a matter of relativity being wrong, but is rather a methodology of bad science.

That theory has not yet been decided. The most promising attempt is quantum cosmology, the application of quantum rules on cosmological scales.

I propose [latex]\bar{\psi}c(\gamma^i \cdot \hat{p})\chi + \bar{\psi}(\Box \phi \ell^3) c^2\psi = \bar{\psi} \gamma^0 (i\hbar \partial_t) \psi[/latex] where the rest part is derived from the mass dependance of the gravitational field [latex]\phi[/latex] in four dimensions. This an analogue of the dirac equation in graviattional interaction terms. The part [latex]\gamma^i[/latex] determines whether to express [latex]\hat{p}[/latex] as either positively or negatively and for rest energy particles travelling lower than the speed of light is given as [latex]\bar{\psi}c(\gamma^i \cdot \hat{p})\chi > c[/latex] (rest) This component [latex]\bar{\psi}(\Box \phi \ell^3) c^2\psi[/latex] is the intrinsic rest energy. It is non-relativistic on it's own. Of course, the full equation [latex]\bar{\psi}c(\gamma^i \cdot \hat{p})\chi + \bar{\psi}(\Box \phi \ell^3) c^2\psi = \bar{\psi} \gamma^0 (i\hbar \partial_t) \psi[/latex] Makes up the Hamiltonian for a free particle when relativistic momentum effects are reduced. If we take normal mathetical convention we have [latex]f=0[/latex] [latex]\bar{\psi}(\Box \phi \ell^3) c^2\psi = \bar{\psi} \gamma^0 (i\hbar \partial_t) \psi[/latex] If [latex]f[/latex] has dimensions of mass, then we can see that if [latex]\bar{\psi}c(\gamma^i \cdot \hat{p})\chi = fc^2\bar{\psi}\psi[/latex] You can insert some yukawa couplings in here, expand this equation out in consideration of a Higgs field with terms satisfying: [latex]g_{\gamma} \bar{\psi}\psi = g_{\gamma}f(\psi^{\dagger}_L \psi_R + \psi^{\dagger}_R \psi_L) + g_{\gamma}H(\psi^{\dagger}_L \psi_R + \psi^{\dagger}_R \psi_L)[/latex] the rest energy actually depends variationally on the state of the momentum component in the sense that the rest energy part requires the idea that mass requires the dependance on the gravitational field as Nordstrom classically proved. [latex]g_{\gamma}\bar{\psi}(\Box \phi \ell^3) c^2\psi =g_{\gamma}(\psi^{\dagger}_L \psi_R + f\psi^{\dagger}_R \psi_L) + H(\psi^{\dagger}_L \psi_R + \psi^{\dagger}_R \psi_L) )[/latex] This would mean that mass is a relativistic phenomenon; then that would mean that the momentum which directly influences variantional energy in the rest energy is provided to the system. It becomes covariant under the same translation you would give a fermion with a mass under the symmetry breaking of a [latex]U(1)[/latex] group.

There is evidence suggesting the early universe possessed a spin http://www.stardrive.org/index.php?option=com_content&view=article&id=4596:the-universe-may-have-been-born-spinning-according-to-new-findings-on-the-symmetry-of-the-cosmos&catid=43:science&Itemid=82

It's still an important property. I wonder how long it will take to collect the relevant data. I've heard a year... I've heard three.

An interesting paper come to light: http://arxiv.org/PS_cache/arxiv/pdf/1109/1109.6562v1.pdf Of course, I actually believe that one generation of nuetrino particle was observed to go at superluminal speeds, but apparently this paper admits that if the velocity anomaly was indeed true, then it would effect all three generations. The paper also makes use of an analogue cherenkov radiation. There is a great deal of talk on the cherenkov radiation from very notable scientists, some even using it as a basis for the experiment to be wrong. Maybe swansont you have neglected the idea to quickly.... http://arxiv.org/abs/astro-ph/9505117

Look at that, I rearranged my langrangian wrong, when moving everything to left hand side of the dirac equation, it should be given as i(ħ∂/∂t)ψ - i(αk)cψ - β(s)M(s)c²ψ = 0 not as i(ħ∂/∂t)ψ + i(αk)cψ - β(s)M(s)c²ψ = 0 I got confused because you can write it like this i(αk)cψ + β(s)M(s)c²ψ - i(ħ∂/∂t)ψ = 0 But obviously I arranged it differently.

... doesn't matter.

Then say a positive or negative field. I assumed a charge because all you have me was + and a - .... anyway... ...no it does not interchange the sign of the field. Nothing in the equations will determine that.

Actually it doesn't violate relativity so long as the particle did not accelerate to or past the speed of light. If the nuetrino is superluminal, it most likely moves superluminally at all times.

Looky at what I found http://iopscience.iop.org/1742-6596/134/1/012010/pdf/1742-6596_134_1_012010.pdf

I don't want this included as some part of the OP because I don't know how to interpret it fully. There is an interesting way to express the dirac equation treating mass against Nordstrom's relativistic mass invariance expressed in four dimensions so using the d'Alembertian [latex]i\hbar \partial_t \psi + c(\sigma \cdot \hat{p})\chi - \beta M_sc^2\psi = 0[/latex] We are going to introduce the expression for mass M_s from now on, given as: [latex]\Box \phi \ell^3 = \frac{M_s}{\ell^3}\ell^3 = M_s[/latex]. [latex]i\hbar \partial_t \psi + c(\sigma \cdot \hat{p})\chi - \beta (\Box \phi \ell^3)c^2\psi = 0[/latex] Multiply by psi dagger to retreive our hermitian Langrangian and simplify using covariant gamma notation [latex]\bar{\psi}(i\hbar \partial_t) \psi + c(\sigma \cdot \hat{p})\chi - \beta (\Box \phi \ell^3) c^2\psi = \mathcal{L}[/latex] [latex]=\bar{\psi} \gamma^0 (i\hbar \partial_t \psi + \bar{\psi}c(\gamma^i \cdot \hat{p})\chi - \bar{\psi}(\Box \phi \ell^3) c^2\psi[/latex] Rearranging this for a different form of a Dirac Equation, where the relativistic solution of the mass that depends on the gravitational field [latex]\phi[/latex] we have: [latex]\bar{\psi}c(\gamma^i \cdot \hat{p})\chi + \bar{\psi}(\Box \phi \ell^3) c^2\psi = \bar{\psi} \gamma^0 (i\hbar \partial_t) \psi[/latex] The non-relativistic limit is [latex] \bar{\psi}(\Box \phi_i \ell^3) c^2\psi = \bar{\psi} \gamma^0 (i\hbar \partial_t) \psi[/latex] These equations wills suit a particle for a neutrino if one can determine what role the [latex]\ell^3[/latex] plays. Conventially, one can say a particle which is a sphere can be expressed in terms of any four coordinates of Minkowski space. There is some evidence (mathematical and some experimental) suggesting that there is a slim possibility that the nuetrino as possessing a non-zero magnetic moment could actually interact with an electromagnetic field. In that case, you simply take the Eigenvalue matrix and solve for [latex]q\phi[/latex] instead of [latex]M\phi[/latex]. Interestingly, dividing both sides by [latex]\ell^3 c^2[/latex] we find a density relationship: [latex] \bar{\psi}(\Box \phi_i)\psi = \frac{\bar{\psi} \gamma^0 (i\hbar \partial_t)\psi}{\ell^3 c^2}[/latex] Which uses the same Nordstrom classical relativistic relation [latex]\Box \phi_i = \rho_i[/latex] using mass-inertial subscript relationship. More compactly this would be: [latex]\bar{\psi}\rho_i{\psi} = \frac{\bar{\psi} \gamma^0 \hat{E}_i \psi}{\ell^3c^2}[/latex]

...messed the latex here.

I've made a full mathematical theory against a modified version of the Dirac Equation permitting a hermitian solution to finding the Langrangian of the equations describing tachyonic neutrino fermions. http://www.scienceforums.net/topic/60141-studying-tachyonic-neutrino-fermion-equations-against-a-modified-dirac-equation/page__pid__628878#entry628878

Exactly! In fact, I've heard quite the opposite, that consciousness requires every cell the human body is composed of, not only the brain (Candice Pert, nuerobiologist). Bottom line is, there is no part of the brain which we can pin-point and call the origin of consciousness. It is a sum collection of all the dynamics in our brain and the collective nervous system most likely.

That it does! Just been thinking about inflationary implications. If [latex]M\phi[/latex] was a potential energy attributed to the Hubble parameter which involves the transition temperature phases, then the matrix derived would be written as: [latex]\begin{pmatrix} i\hbar\partial_t\psi + M\phi & 0 \\ 0 & i\hbar\partial_t\psi + M\phi \end{pmatrix} \begin{pmatrix} \psi_1(\vec{x}, t) \\ \chi_2(\vec{x}, t) \end{pmatrix} - \begin{pmatrix} 0 & (\sigma \cdot \hat{p})c \\ (\sigma \cdot \hat{p})c & 0 \end{pmatrix} \begin{pmatrix} \psi_1(\vec{x}, t) \\ \chi_2(\vec{x}, t) \end{pmatrix} - \begin{pmatrix} M_sc^2 & 0 \\ 0 & -M_sc^2 \end{pmatrix}\begin{pmatrix} \psi_1(\vec{x}, t) \\ \chi_2(\vec{x}, t) \end{pmatrix} = 0[/latex] This yields: [latex](\epsilon + M \phi)\psi(\vec{x},t) - c(\sigma \cdot \hat{p})\chi(\vec{x},t) = 0[/latex] where [latex]\epsilon = i\hbar \partial_t - M_sc^2[/latex] Just an added dynamic, an interesting layout though. I mean, I don't know how the tachyon neutrino has implications for the cosmological side of inflation, but if a M\phi term was to enter the matrix, it would enter it like this, and the resulting equation would be: [latex](\epsilon + M \phi)\psi(\vec{x},t) - c(\sigma \cdot \hat{p})\chi(\vec{x},t) = 0[/latex]

Yup. Don't know about the wiki link, but in a recent discussion with a scientist, ''real on-mass shell tachyonic neutrino fermions are thought to destabilize the vacuum causing inflation of baby universes. The Higgs-Goldstone phase transition is when there is a ground state phase transition to a coherent order parameter that is the expectation value of a non-Hermitian destruction field operator of spin 0 in the usual model.'' Not my area of expertize, but the redefinition is certainly there to avoid M being complex

Yes... !! The Langrangian will be real, because the Langrangian is derived by simplyy multiply it by \psi(dagger) from the Dirac Equation. Before deriving any of the later equations, remember that we are specifically taking the Dirac Equation and making it suitable for a nuetrino with a positive mass, meaning the modified dirac equation which is then multiplied by psi(dagger) producing the langrangian, which will remain real as well. We are dealing with a langrangian for a tachyon fermion in special conditions treating the mass as proper in four momentum spacelike configuration. If Tsao had not defined the proper mass as being positive and real, then the Langrangian for a nuetrino could not be derived from a Dirac Equation because fundamentally the dirac equation would have been obsolete to begin with.