Mystery111 Posted September 27, 2011 Share Posted September 27, 2011 (edited) In light of the recent developments at CERN has led me to investigate the possibility of superluminal fermionic neutrino particles in a series of modified equations from the work of Tsao Chang, a leading scientist in his area of research. First of all, I wanted to write down the equations describing how mass enters and how the yukawa coupling determines different mass sizes. Then I wanted to take it further, derive a Langrangian from the modified Dirac Equation and write it in terms of a Higgs field and then finally working out the Eigenvalues for the modified approach. The mass term is very important when speculating on the existence of a tachyonic fermion neutrino. To give a particle mass, you must assume some field [latex]\phi = \rho e^{i \alpha}[/latex]. Explaining this in terms of a mexican hat potential, you can calculate it to the first approximation in respect to [latex]f= \rho[/latex]: [latex]\phi = f e^{i \alpha}[/latex] One might know that [tex]\phi[/tex] is in fact frozen but the alpha field in the exponential is allowed to shift. To calculate the Langrangian, you first need to work out the Covariant Derivative: [latex]D\phi = \partial \phi + iA \phi[/latex] To compute the action, you must multiply this by it's conjugate: [latex] = i(\partial \alpha + A)f e^{i\alpha}[/latex] [latex] = f^2(\partial \alpha + A)^2[/latex] Doing so removes the [latex]e^{i\alpha}[/latex]. Using a special gauge transformation now, [latex]A' \Rightarrow \partial \theta + A[/latex] we should notice that in the equation [latex] = f^2(\partial \alpha + A)^2[/latex] the [latex](\partial \alpha + A)[/latex] is a gauge transformation where [latex]\partial \theta [/latex] plays the role of our alpha field. So the final expression is: [latex]f^2A'^2[/latex] [latex]f^2[/latex] term plays the role of mass and notice the alpha field has been absorbed by our object [latex]A'[/latex], that means no more Goldstone Boson, the Higgs field has provided a mass through shifting the potential from the ground state of origin. This is what a mass is. When this effects a particle like a photon, or any massless boson field it gives that quanta a mass. The mass term might be for an electron-positron pair. And analyzing how the mass term enters the Dirac Equation, permitting a positive mass for an electron and positron tells us that the Higgs field did not induce a negative matter description because there is no evidence for it in the Dirac equation. [latex]i\dot{\psi_R} = -i\partial x \psi_R + M\psi_R[/latex] [latex]i\dot{\psi_L} = i\partial x \psi_L + M\psi_L[/latex] both these equations describe right moving waves [latex]\psi_R[/latex] and left moving wave [latex]\psi_L[/latex] and they physically represent particles and antiparticles. The mass term is strictly positive, and because of the mass term, it can describe a new type of particle, the majorana particle which is a particle which is it's own antiparticle. The majorana mass term couples left and right movers together. [latex]i\dot{\psi}= -i\alpha \partial x \psi + M \beta \psi[/latex] Here, I should have noted, the [latex]f^2[/latex] term plays the role of mass. So essentially, we have a mass term which couples the fermion particle and antiparticles together. Equally on a similar note, the mass for a neutrino allows it to oscillate between flavors. The mass term in the dirac equation explicitely has no imaginary mass term, so a new equation will be required to describe it. Not only that, but mass can be measured using the Yukawa Coupling as well. keeping in mind these transformations: [latex]\psi_L \Rightarrow e^{i\phi} \psi_L[/latex] [latex]\psi_R \Rightarrow \psi_R[/latex] [latex]\phi \Rightarrow e^{i\theta} \phi[/latex] then we can see from dirac equation, that these transforms are not invariant: [latex]i(\frac{\partial \psi_L}{\partial t} + \alpha_i \frac{\partial \psi_L}{\partial x^i}) = M \beta \psi[/latex] [latex]i(\frac{\partial \psi_R}{\partial t} + \alpha_i \frac{\partial \psi_R}{\partial x^i}) = M \psi_L[/latex] [latex]i(\frac{\partial \psi_L}{\partial t} - \alpha_i \frac{\partial \psi_L}{\partial x^i}) = M \psi_R[/latex] Throw the mass term out and put in [tex]g[/tex], which is just a number, a constant: [latex]i(\frac{\partial \psi_L}{\partial t} + \alpha_i \frac{\partial \psi_L}{\partial x^i}) = M \beta \psi[/latex] [latex]i(\frac{\partial \psi_R}{\partial t} + \alpha_i \frac{\partial \psi_R}{\partial x^i}) = g \phi* \psi_L[/latex] [latex]i(\frac{\partial \psi_L}{\partial t} - \alpha_i \frac{\partial \psi_L}{\partial x^i}) = g \psi_R \phi[/latex] which makes the [latex]U(1)[/latex] allowed to make the phase changes in the equation. It's a nice trick for the mass term in the Dirac Equation which does not violate charge conservation. Describing the equation is easy. [latex]i(\frac{\partial \psi_R}{\partial t} + \alpha_i \frac{\partial \psi_R}{\partial x^i}) = g \phi* \psi_L[/latex] The left describes a right handed particle propogating which splits into a phi-particle and a left handed particle. Since there is discussion of the neutrino mass, the coupling which allows a certain handedness for neutrons will be important when concerning a mass. If anyone followed the link I gave, Tsao Chang gave a modified dirac equation suitable for a tachyonic neutrino. He never worked out a langrangian for his equation, which would be the langrangian for the neutrino particle. We begin by taking his equation: [latex]-i(\alpha k)c\psi + \beta_s M_s c^2 \psi = i\hbar \partial_t \psi[/latex] [latex]g_{\gamma} \bar{\psi}\psi = (g_{\gamma}\psi^{\dagger}_L \phi \psi_R + g_{\gamma}\psi^{\dagger}_R \psi_L \phi^{\dagger})[/latex] Where [tex]g_{\gamma}[/tex] is our Yukawa mass term coupling, removing the phi and replacing for our mass term f, we rewrite this equation as: [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] Where [latex]H[/latex] is the Higgs Field.So the entire equation to the point at which I derived and altering with our yukawa couplings and Higgs Field we have: [latex]\gamma^{0}\cdot i\hbar \partial_t(\psi^{\dagger}_L \psi_R + \psi^{\dagger}_R \psi_L) + \gamma^{i} \cdot \hat{p} c(\psi^{\dagger}_L \psi_R + \psi^{\dagger}_R \psi_L) + M_s(\psi^{\dagger}_L \psi_R + \psi^{\dagger}_R \psi_L) = \mathcal{L}[/latex] and move everything to the left [latex]i\hbar \partial_t \psi + i(\alpha k)c\psi - \beta_s M_s c^2 \psi = 0[/latex] To calculate the tachyonic neutrino Tsao Langrangian, you need to multiply this by psi dagger: [latex]\psi^{\dagger}(i\hbar \partial_t \psi + i(\alpha k)c\psi - \beta_s M_s c^2 \psi) = \mathcal{L}[/latex] In covariant language, if you multiply beta by psi dagger, you get psi bar. Simplifying the equation, you end up with [latex]\bar{\psi}\beta (i \hbar \partial_t) + \bar{\psi} \beta (\alpha \cdot k)c\psi + M\bar{\psi} \psi = \mathcal{L}[/latex] This can further be simplified as: If anyone followed the link I gave, Tsao Chang gave a modified dirac equation suitable for a tachyonic neutrino. He never worked out a langrangian for his equation, which would be the langrangian for the neutrino particle. We begin by taking his equation: [latex]-i(\alpha k)c\psi + \beta_s M_s c^2 \psi = i\hbar \partial_t \psi[/latex] and move everything to the left [latex]i\hbar \partial_t \psi + i(\alpha k)c\psi - \beta_s M_s c^2 \psi = 0[/latex] To calculate the tachyonic neutrino Tsao Langrangian, you need to multiply this by psi dagger: [latex]\psi^{\dagger}(i\hbar \partial_t \psi + i(\alpha k)c\psi - \beta_s M_s c^2 \psi) = \mathcal{L}[/latex] In covariant language, if you multiply beta by psi dagger, you get psi bar. Simplifying the equation more by using gamma notation [latex]\bar{\psi}\gamma^{0} (i \hbar \partial_t) + \bar{\psi} (\gamma^{i} \cdot k)c\psi + M\bar{\psi} \psi = \mathcal{L}[/latex] This makes a nice compact four vector covariant tachyonic neutrino equation. So I have went from deriving the Tsao Langrangian [latex]\psi^{\dagger}(i\hbar \partial_t \psi + i(\alpha \hat{p})c\psi - \beta_s M_s c^2 \psi) = \mathcal{L}[/latex] to allowing a higgs field interaction [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] which allowed me to create the equation: [latex]\gamma^{0}\cdot i\hbar \partial_t(g_{\gamma}\psi^{\dagger}_L \phi \psi_R + g_{\gamma}\psi^{\dagger}_R \psi_L \phi^{\dagger}) + \gamma^{i} \cdot \hat{p} c(g_{\gamma}\psi^{\dagger}_L \phi \psi_R + g_{\gamma}\psi^{\dagger}_R \psi_L \phi^{\dagger}) + M_s(\gamma^{0}\psi^{\dagger}_L \phi \psi_R + \gamma^{0}\psi^{\dagger}_R \psi_L \phi^{\dagger}) = \mathcal{L}[/latex] Now I'm going to do something different. To calculate the Tsao-Dirac Hamiltonian, you can state: It would simply be [latex]i\hbar \partial_t\psi = H\psi[/latex] since we are speculating neutrino's it would be a small number. This is not the same as finding a small number of the yukawa coupling. It has a small energy because it travels so fast. Allow the wave function to be expressed more compactly as: [latex]\Psi = \begin{pmatrix} \psi_1(\vec{x}, t) \\ \chi_2(\vec{x}, t) \end{pmatrix}[/latex] Then we can write the Eignevalues of the Tsao-Dirac Hamiltonian [latex]\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] which allows you to write: [latex]\beta_s M_sc^2\psi(\vec{x}, t) - (\sigma \cdot \hat{p})c\chi(\vec{x}, t) = 0[/latex] Done, sorry about that Edited September 27, 2011 by Mystery111 Link to comment Share on other sites More sharing options...

the asinine cretin Posted September 27, 2011 Share Posted September 27, 2011 :applause: Link to comment Share on other sites More sharing options...

ajb Posted September 27, 2011 Share Posted September 27, 2011 Not that I have gone through your workings very closely, but don't we have issues with Hermitian properties of the action or Hamiltonian for tachyonic fermions? (This might not really be a problem in light of PT symmetry etc.) Link to comment Share on other sites More sharing options...

Mystery111 Posted September 28, 2011 Author Share Posted September 28, 2011 (edited) Is there a mod here who can alter my post for me... I wrote this the other day, then after reading it over, I noticed a few mistakes, like as always happens with me when I go to write out lengthly passages of latex. I picked up a superfluous gamma^0 term in the Higgs representation of g \bar{\psi}\psi and I am missing a big part of my matrix. In light of the recent developments at CERN has led me to investigate the possibility of superluminal fermionic neutrino particles in a series of modified equations from the work of Tsao Chang, a leading scientist in his area of research. First of all, I wanted to write down the equations describing how mass enters and how the yukawa coupling determines different mass sizes. Then I wanted to take it further, derive a Langrangian from the modified Dirac Equation and write it in terms of a Higgs field and then finally working out the Eigenvalues for the modified approach. The mass term is very important when speculating on the existence of a tachyonic fermion neutrino. To give a particle mass, you must assume some field [latex]\phi = \rho e^{i \alpha}[/latex]. Explaining this in terms of a mexican hat potential, you can calculate it to the first approximation in respect to [latex]f= \rho[/latex]: [latex]\phi = f e^{i \alpha}[/latex] One might know that [latex]\phi[/latex] is in fact frozen but the alpha field in the exponential is allowed to shift. To calculate the Langrangian, you first need to work out the Covariant Derivative: [latex]D\phi = \partial \phi + iA \phi[/latex] To compute the action, you must multiply this by it's conjugate: [latex] = i(\partial \alpha + A)f e^{i\alpha}[/latex] [latex] = f^2(\partial \alpha + A)^2[/latex] Doing so removes the [latex]e^{i\alpha}[/latex]. Using a special gauge transformation now, [latex]A' \Rightarrow \partial \theta + A[/latex] we should notice that in the equation [latex] = f^2(\partial \alpha + A)^2[/latex] the [latex](\partial \alpha + A)[/latex] is a gauge transformation where [latex]\partial \theta [/latex] plays the role of our alpha field. So the final expression is: [latex]f^2A'^2[/latex] [latex]f^2[/latex] term plays the role of mass and notice the alpha field has been absorbed by our object [latex]A'[/latex], that means no more Goldstone Boson, the Higgs field has provided a mass through shifting the potential from the ground state of origin. This is what a mass is. When this effects a particle like a photon, or any massless boson field it gives that quanta a mass. The mass term might be for an electron-positron pair. And analyzing how the mass term enters the Dirac Equation, permitting a positive mass for an electron and positron tells us that the Higgs field did not induce a negative matter description because there is no evidence for it in the Dirac equation. [latex]i\dot{\psi_R} = -i\partial x \psi_R + M\psi_R[/latex] [latex]i\dot{\psi_L} = i\partial x \psi_L + M\psi_L[/latex] both these equations describe right moving waves [latex]\psi_R[/latex] and left moving wave [latex]\psi_L[/latex] and they physically represent particles and antiparticles. The mass term is strictly positive, and because of the mass term, it can describe a new type of particle, the majorana particle which is a particle which is it's own antiparticle. The majorana mass term couples left and right movers together. [latex]i\dot{\psi}= -i\alpha \partial x \psi + M \beta \psi[/latex] Here, I should have noted, the [latex]f^2[/latex] term plays the role of mass. So essentially, we have a mass term which couples the fermion particle and antiparticles together. Equally on a similar note, the mass for a neutrino allows it to oscillate between flavors. The mass term in the dirac equation explicitely has no imaginary mass term, so a new equation will be required to describe it. Not only that, but mass can be measured using the Yukawa Coupling as well. keeping in mind these transformations: [latex]\psi_L \Rightarrow e^{i\phi} \psi_L[/latex] [latex]\psi_R \Rightarrow \psi_R[/latex] [latex]\phi \Rightarrow e^{i\theta} \phi[/latex] then we can see from dirac equation, that these transforms are not invariant: [latex]i(\frac{\partial \psi_L}{\partial t} + \alpha_i \frac{\partial \psi_L}{\partial x^i}) = M \beta \psi[/latex] [latex]i(\frac{\partial \psi_R}{\partial t} + \alpha_i \frac{\partial \psi_R}{\partial x^i}) = M \psi_L[/latex] [latex]i(\frac{\partial \psi_L}{\partial t} - \alpha_i \frac{\partial \psi_L}{\partial x^i}) = M \psi_R[/latex] Throw the mass term out and put in [tex]g[/tex], which is just a number, a constant: [latex]i(\frac{\partial \psi_L}{\partial t} + \alpha_i \frac{\partial \psi_L}{\partial x^i}) = M \beta \psi[/latex] [latex]i(\frac{\partial \psi_R}{\partial t} + \alpha_i \frac{\partial \psi_R}{\partial x^i}) = g \phi* \psi_L[/latex] [latex]i(\frac{\partial \psi_L}{\partial t} - \alpha_i \frac{\partial \psi_L}{\partial x^i}) = g \psi_R \phi[/latex] which makes the [latex]U(1)[/latex] allowed to make the phase changes in the equation. It's a nice trick for the mass term in the Dirac Equation which does not violate charge conservation. Describing the equation is easy. [latex]i(\frac{\partial \psi_R}{\partial t} + \alpha_i \frac{\partial \psi_R}{\partial x^i}) = g \phi* \psi_L[/latex] The left describes a right handed particle propogating which splits into a phi-particle and a left handed particle. Since there is discussion of the neutrino mass, the coupling which allows a certain handedness for neutrons will be important when concerning a mass. If anyone followed the link I gave, Tsao Chang gave a modified dirac equation suitable for a tachyonic neutrino. He never worked out a langrangian for his equation, which would be the langrangian for the neutrino particle. We begin by taking his equation: [latex]-i(\alpha k)c\psi + \beta_s M_s c^2 \psi = i\hbar \partial_t \psi[/latex] [latex]g_{\gamma} \bar{\psi}\psi = (g_{\gamma}\psi^{\dagger}_L \phi \psi_R + g_{\gamma}\psi^{\dagger}_R \psi_L \phi^{\dagger})[/latex] Where [tex]g_{\gamma}[/tex] is our Yukawa mass term coupling, removing the phi and replacing for our mass term f, we rewrite this equation as: [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] Where [latex]H[/latex] is the Higgs Field.So the entire equation to the point at which I derived and altering with our yukawa couplings and Higgs Field we have: [latex]\gamma^{0}\cdot i\hbar \partial_t(\psi^{\dagger}_L \psi_R + \psi^{\dagger}_R \psi_L) + \gamma^{i} \cdot \hat{p} c(\psi^{\dagger}_L \psi_R + \psi^{\dagger}_R \psi_L) + M_s(\psi^{\dagger}_L \psi_R + \psi^{\dagger}_R \psi_L) = \mathcal{L}[/latex] and move everything to the left [latex]i\hbar \partial_t \psi + i(\alpha k)c\psi - \beta_s M_s c^2 \psi = 0[/latex] To calculate the tachyonic neutrino Tsao Langrangian, you need to multiply this by psi dagger: [latex]\psi^{\dagger}(i\hbar \partial_t \psi + (\alpha k)c\psi - \beta_s M_s c^2 \psi) = \mathcal{L}[/latex] In covariant language, if you multiply beta by psi dagger, you get psi bar. Simplifying the equation more by using gamma notation [latex]\bar{\psi}\gamma^{0} (i \hbar \partial_t) + \bar{\psi} (\gamma^{i} \cdot k)c\psi + M\bar{\psi} \psi = \mathcal{L}[/latex] This makes a nice compact four vector covariant tachyonic neutrino equation. So I have went from deriving the Tsao Langrangian [latex]\psi^{\dagger}(i\hbar \partial_t \psi + (\alpha \hat{p})c\psi - \beta_s M_s c^2 \psi) = \mathcal{L}[/latex] to allowing a higgs field interaction [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] which allowed me to create the equation: [latex]\gamma^{0}\cdot i\hbar \partial_t(g_{\gamma}\psi^{\dagger}_L \phi \psi_R + g_{\gamma}\psi^{\dagger}_R \psi_L \phi^{\dagger}) + \gamma^{i} \cdot \hat{p} c(g_{\gamma}\psi^{\dagger}_L \phi \psi_R + g_{\gamma}\psi^{\dagger}_R \psi_L \phi^{\dagger}) + M_s(g_{\gamma}\psi^{\dagger}_L \phi \psi_R + g_{\gamma} \psi^{\dagger}_R \psi_L \phi^{\dagger}) = \mathcal{L}[/latex] Now I'm going to do something different. To calculate the Tsao-Dirac Hamiltonian, you can state: It would simply be [latex]i\hbar \partial_t\psi = H\psi[/latex] since we are speculating neutrino's it would be a small number. This is not the same as finding a small number of the yukawa coupling. It has a small energy because it travels so fast. Allow the wave function to be expressed more compactly as: [latex]\Psi = \begin{pmatrix} \psi_1(\vec{x}, t) \\ \chi_2(\vec{x}, t) \end{pmatrix}[/latex] Then we can write the Eignevalues of the Tsao-Dirac Hamiltonian [latex]\begin{pmatrix} i\hbar\partial_t\psi & 0 \\ 0 & i\hbar\partial_t\psi \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] If there are any other mistakes, please point them out! if as moderator could take the contents of this post and place them in the OP, it would be appreciated. Not that I have gone through your workings very closely, but don't we have issues with Hermitian properties of the action or Hamiltonian for tachyonic fermions? (This might not really be a problem in light of PT symmetry etc.) Are you meaning that hermition operators spit out positive real numbers and so the mass operator of your tachyonic fermion field equation is non-hermitian? Well, yes.... normally this is the route which one will come face to face with. However, you can deal with the spacelike four momentum of the equation and ridding yourself of any negative terms arising from the timelike nature of your equation using the proper mass as given by Tsao. In fact I dare say there maybe a number of solutions. If this is not what you meant, could you underline the discrepency? Thanks Edited September 28, 2011 by Mystery111 Link to comment Share on other sites More sharing options...

ajb Posted September 28, 2011 Share Posted September 28, 2011 Not quite, let me be a bit more careful. Is the Lagrangian real? I ask this as I think it could be very problematic if the action is not real, or more technically Hermitian. I believe there are issues with the S-matrix if not? Please correct me if I am wrong on that. Link to comment Share on other sites More sharing options...

Mystery111 Posted September 28, 2011 Author Share Posted September 28, 2011 Not quite, let me be a bit more careful. Is the Lagrangian real? I ask this as I think it could be very problematic if the action is not real, or more technically Hermitian. I believe there are issues with the S-matrix if not? Please correct me if I am wrong on that. 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. Link to comment Share on other sites More sharing options...

ajb Posted September 28, 2011 Share Posted September 28, 2011 There is a redefinition in order to avoid [math]m[/math] being complex? I mean taychons have negative mass squared, you can modify the mass shell-condition to get a "new real mass". I think Wikpedia says something about this. Link to comment Share on other sites More sharing options...

Mystery111 Posted September 28, 2011 Author Share Posted September 28, 2011 (edited) 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 Edited September 28, 2011 by Mystery111 Link to comment Share on other sites More sharing options...

ajb Posted September 28, 2011 Share Posted September 28, 2011 Tachyons are generally unstable and would be thought to decay very quickly, I have no idea about using them in cosmology for inflation. But it sounds a fun idea. Link to comment Share on other sites More sharing options...

Mystery111 Posted September 28, 2011 Author Share Posted September 28, 2011 (edited) Tachyons are generally unstable and would be thought to decay very quickly, I have no idea about using them in cosmology for inflation. But it sounds a fun idea. 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] Edited September 28, 2011 by Mystery111 Link to comment Share on other sites More sharing options...

Mystery111 Posted September 28, 2011 Author Share Posted September 28, 2011 (edited) 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] Edited September 29, 2011 by Mystery111 Link to comment Share on other sites More sharing options...

Mystery111 Posted September 29, 2011 Author Share Posted September 29, 2011 Tachyons are generally unstable and would be thought to decay very quickly, I have no idea about using them in cosmology for inflation. But it sounds a fun idea. Looky at what I found http://iopscience.iop.org/1742-6596/134/1/012010/pdf/1742-6596_134_1_012010.pdf Link to comment Share on other sites More sharing options...

Mystery111 Posted October 2, 2011 Author Share Posted October 2, 2011 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. Link to comment Share on other sites More sharing options...

Mystery111 Posted November 18, 2011 Author Share Posted November 18, 2011 Well, I was right... atleast I think I was. The experiment has now been replicated showing the same faster-than-light results. I am very happy. I have had debates with several scientists stating that the experiment should be taken as a true result, that we should believe in science and our ability to test nature. One scientist even called me a crackpot for denying the paper written by Sheldon Glashow on an analogue Cherenkov Radiation. I have been made a very happy man! Link to comment Share on other sites More sharing options...

Mystery111 Posted December 23, 2011 Author Share Posted December 23, 2011 A Link to comment Share on other sites More sharing options...

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