Mystery111

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. 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

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

I think the consensus implies an open universe.

I forward it as a conjecture to be thought upon swansont.

The good news is, if the neutrino is a superluminal fermion, then the idea it may interact with an electromagnetic field could be drawn out.

No, read the first link swanson. You are like others here, seem to be taking the EDM as the main part of the investigation. The first link explains that they have measured the electron to be like a sphere. The first experiment is much more decisive than the first: '' What we find is, as best we can tell and we've looked really very carefully, there's no wobble. The electron shows all signs of being round at our current sensitivity.'' I was wrong. The consensus in 2011 may I think, the experiment measured the interaction of the field created by the particle, which was spherical. This did not imply the actual structure.

As a non-zero dimensional particle.

Either way ajb, we could be technical to every point. But I saw people arguing my interpretation of their work does not imply what I said, but to be honest, you cannot have both a pointlike particle and something which has a sphere! The two identities are totally different with the latter containing dimension. So my interpretation cannot surely be argued with. If their experimentation is correct, then the electron must have a structure, call it the classical radius, or simply a radius. The size of an electron matters not much, so long as we remember it is not as small enough to say it doesn't even have a body-structure!

negative and positive what? Are we talking about EM charge, mass-charge? My instincts indicate EM charge, but we've been talking about negative mass signs as well, so I need you to be specific. Let me try this again. Your confusion seems to lye, correct me if I am wrong, in the original state of the antiparticle existing with a negative energy in the vacuum. While it is generally true that particles have negative descriptions before they become real particles existing in the vacuum, implies that the correct equation to represent energy and mass in a Hamiltonian is [math]E= \pm Mc^2[/math]. However, the Mass term does not pick up a negative sign when it becomes a real particle. Recall the equations I showed you concerning the Dirac Equation. The mass term in the equations are identical for those which are particles and antiparticles. What is a mass term? If theory is correct, it is a symmetry-breaking in the equations. To give a particle mass, you must assume some field [math]\phi = \rho e^{i \alpha}[/math]. Explaining this in terms of a mexican hat potential, you can calculate it to the first approximation in respect to [math]f= \rho[/math]: [math]\phi = f e^{i \alpha}[/math] One might know that [math]\phi[/math] 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: [math]D\phi = \partial \phi + iA \phi[/math] To compute the action, you must multiply this by it's conjugate: [math] = i(\partial \alpha + A)f e^{i\alpha}[/math] [math] = f^2(\partial \alpha + A)^2[/math] Doing so removes the [math]e^{i\alpha}[/math]. Using a special gauge transformation now, [math]A' \Rightarrow \partial \theta + A[/math] we should notice that in the equation [math] = f^2(\partial \alpha + A)^2[/math] the [math](\partial \alpha + A)[/math] is a gauge transformation where [math]\partial \theta [/math] plays the role of our alpha field. So the final expression is: [math] = f^2A'^2[/math] Notice the alpha field has been absorbed by our object [math]A'[/math], 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. Here, I should have noted, the [math]f^2[/math] term plays the role of mass. The [math]A[/math] the electromagnetic potential just comes from the equations of electrodynamics. Note that in the electromagnetic equation [math]F_{\mu \nu} = \frac{\partial A_{\mu}}{\partial X^{\nu}} - \frac{\partial A_{\nu}}{\partial X^{\mu}}[/math] where you might see derivates in respect of the [math]A[/math] only, there is no mass term. Also don't ever as a warning mix up the idea of [math]F_{\mu \nu}F^{\mu \nu}[/math] with the [math]f^2[/math] term above. They are different.

lol

I don't think you can say it doesn't imply what I said. Pointlike particles have no structure - it cannot be described as a sphere. Before the release of the information in the OP, we must have treated the electron; an object which has no structure, pointlike, dimensionless. Now we can prove that wrong. evidence suggests the electron does have a structure after all and thus can be described by a classical radius, which is simply a line seperating the radius of curvature, or the Compton Wavelength from it's center. How more classical can you get? And why would it not mean what I've said? I think it's all very clear. (I was naturally buzzed when I read the article in the OP. I had been saying the electron, indeed all particles must have a structure for while.)

This is why experiments need to be conducted time and time again to find any flaws.

Hmmmm turns out the nuetrino possessing a magnetic moment could in fact interact electromagnetically according to wiki ;The neutrino has half-integer spin (½ħ) and is therefore a fermion. Neutrinos interact primarily through the weak force. The discovery of neutrino flavor oscillations implies that neutrinos have mass. The existence of a neutrino mass strongly suggests the existence of a tiny neutrino magnetic moment[12] of the order of 10−19 μB, allowing the possibility that neutrinos may interact electromagnetically as well. Dipole I meant.

I know the neutrino would contain a magnetic moment with the advent of it having a mass, so the particle still has a charge. I take it, it would also require an electric moment then?

y They don't leave a boom, but they do leave a trail called '' Cherenkov radiation.'' some have likened this to the sonic boom analogy.

I remembered doing a little research on faster than light neutrino's 6 years back. I found that it was possible to describe nuetrino's this way because of the negative mass squared term which enters the equations describing it. It is possible to observe this from the decay of tritium. Also, the idea that the Neutrino is a tachyonic fermion has been investigated by this physicist: http://arxiv.org/PS_...1/0011087v4.pdf He proposes a superluminal tachyonic-fermion dirac equation. If the neutrino is moving at superluminal speeds, then why won't they use some other method to test this prediction, like an observable Cherenkov radiation? We see the stuff from test reactors, why wouldn't we notice it from neutrinos? There are effectively billions passing through my body at this very moment in time.

In what sense are you using ''polarity''?

It is possible now (with the recent evidence of the first link) that the electron could really be a classical object - it is a sphere and it might entail the idea of some substructure. The idea should not be beyond physics at all. Glaswegian scientists have even shown that it is possible to explain the electron in terms of a trapped photon following a toroidal knot.

http://www.thenakedscientists.com/HTML/content/interviews/interview/1711/ http://www.sciencenews.org/view/feature/id/69229/title/Sizing_up_the_Electron Good news for anyone who said mainstream science was wrong in it's contentions about zero dimensionless objects. This evidence suggests that an electron actually does have a classical radius [math]e^2/ Mc^2[/math].

1) The mass term for all particles we work with have a positive mass. The appearance of a negative charged particle does not imply changing the positive sign of attraction in the mass term. The Dirac Equation describes particles and antiparticles of fermions. The mass term in that equation does some interesting things... Two specific wave equations, which come to mind are: [math]i\dot{\psi_R} = -i\partial x \psi_R + M\psi_R[/math] [math]i\dot{\psi_L} = i\partial x \psi_L + M\psi_L[/math] both these equations describe right moving waves [math]\psi_R[/math] and left moving wave [math]\psi_L[/math] 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. [math]i\dot{\psi}= -i\alpha \partial x \psi + M \beta \psi[/math] 2) This question is hard to understand... what is the gravitational sign of the mass terms for matter and antimatter? They will still attract each other, because they have a real mass. 3) No

Gravitational waves are none trivial in Einsteins Theory, meaning that you can have a curvature of spacetime but not the presense of matter. This is of course the presence of gravitational waves. Gravitational waves would be the curvature itself.

Einstein derived what is called the Einstein tensor which is: [math]\nabla_{\mu} G^{\mu \nu}=0[/math] and [math]G^{\mu \nu}[/math] is just the Ricci tensor minus [math]\frac{1}{2}g^{\mu \nu}R[/math]. These equations express the local continuity of energy and momentum. As a side note, it could be imagined for a universe to be devoid of mass, so this indiates the right hand side of [math]\nabla_{\mu} R^{\mu \nu} = \frac{1}{2 \nabla_{\mu}g^{\mu \nu} R[/math] But does the absence of matter imply zero curvature for a metric? The answer is no. Gravitational waves are not trivial, where the Ricci tensor is zero everywhere, but the Reimann tensor is not. Let us now concentrate on a specific wave equation, quite a famous one: [math]\frac{\partial^2 \phi}{\partial t^2} = c^2\frac{\partial^2 \phi]{\partial x^2}[/math] This describes two kind of waves, one wave which moves to the left, another to the right. From now on, we will use natural units. To express this equation in three dimensions [math]\frac{\partial^2 \phi}{\partial t^2} = [\frac{\partial^2 \phi]{\partial x^2} + \frac{\partial^2 \phi]{\partial y^2} + \frac{\partial^2 \phi]{\partial x^2}[/math] We can rewrite this as [tex]\eta^{\mu \nu} \frac{\partial^2 \phi}{\partial X^{\mu} \partial X^{\nu}=0[/tex] To make it into a tensorial equation, we can take [tex]\eta^{\mu \nu}[/tex] to be [tex]g^{\mu \nu}[/tex] thus we can state that [tex]g^{\mu \nu} \frac{\partial \phi}{\partial X^{\nu}[/tex] and differentiate as: [math]\frac{\partial}{\partial x^{\mu} g^{\mu \nu} \frac{\partial \phi}{\partial X^{\mu}[/math] We need to bring in the Covariant derivative, and make: [math]\frac{\partial}{\partial x^{\mu} g^{\mu \nu} \frac{\partial \phi}{\partial X^{\nu}+ \Gamma_{\mu \alpha}^{\mu} g^{\nu \beta} \frac{\partial \phi}{\partial X^{\beta}}=0[/math] [math] \nabla g^{\mu \nu} \frac{\partial \phi}{\partial x^{\nu}}[/math] To work out the covariant derivative involves Christoffel Symbols. This is by definition, the wave equation in curved coordinates. The equations coeffients depend on position, and describes a wave in curvilinear coordinates. Curvature varies throughout space, and understanding the relativity of photons, for instance, you disover that photons couple to spacetime - the intrinsic relationship between curvature, matter and energy are numerous. Of course, you wind the clock back, then the curvature becomes stronger and stronger, it should get to a point where there is a perfect curvature, i.e (an object of infinite curvature, with infinite mass and energy). The parameters are defined in a negative spacetime region. Could it be possible, to have a lot of curvature which typically emanates mass from it's geometry? The answer might be yes, because you can deal with [math]\nabla_{\mu} R^{\mu \nu} = \frac{1}{2 \nabla_{\mu}g^{\mu \nu} R[/math] Where you can remove all the mass from the universe, but still have a non-zero curvature. Part Two [math]R_{\mu \nu} - 1/2g_{\mu \nu}R + \Lambda g_{\mu \nu} = kT_{\mu \nu}[/math] The Newtonian analogue is: [math]\nabla^2 \phi + \Lambda = 4 \pi G \rho[/math] where [math]\Lambda[math] is our energy density, the cosmological constant. I ran into a problem with this, because the 00-component tells me that the [math]g_{\mu \nu}[/math] part is close to the limits of special relativity, and as we know, special relativity is a flat spacetime. The idea that this describes our early universe very well seems bleak, but let's work with it for now. Suppose this is taken as a uniform energy density, then we would have [math]\nabla \phi + \Lambda = 0[/math] [math]\nabla \phi = \Lambda[/math] we must assume that we cannot deal with [tex]\phi[/tex] properly as we did if we have the metric of spacetime any significant time past the first instant of the universe. We would differentiate [tex]\phi[/tex] with the function [math]x^2 + y^2 + z^2[/math] but that is an solution for three dimensional spacetime - this kind of representation for [tex]\phi[/tex] must become obsolete. Of course, normally you do this by saying the del of our function is 6, since differentiating [math]x[/math] twice gives you [math]2[/math]], same with the [math]y[/math] and [math]z[/math] components; you simply end up with [math]\phi = \frac{\Lambda}{6}[x^2+y^2+z^2][/math] But of course, the spatial dimensions must vanish to leave a point on the metric - we would also find [math]\Lambda|_{\infty}[/math] since the singularity can permit an infinite volume of energy. This must mean that the curvature is also infinite. Then again, if we wanted to draw out the use of this energy density when the universe becomes sufficiently large enough, then you simply differentiate the gravitational potential to find the force, and these equations let you retrieve the important terms involving the cosmological constant as a driving force for continued expansion. Using the cartesian coordinates, we can understand for instance that the x-component of force is given as: [math]\frac{\partial \phi}{\partial x} = \frac{\Lambda}{6}2x = \frac{\Lambda x}{3}[/math] This tells us there is a component of force along the x-direction. This is just basic relativity, but in order to understand my theory, we need a new understanding of how to treat [math]\phi[/math] as a potential in zero dimensions, with an infinitely large [math]\Lambda[/math] which would be related to the identity of either the stress energy or the curvature; you can swap the constant term around to either the right or the left of Einsteins field equation describing how space tells matter how to move, and how matter tells space how to bend, by understanding that the constant can be seen in terms of either it's geometry or by the energy-distribution of spacetime - of course, I have surmized it is more rewarding to think of the geometry in this case, where it contributes to the potential of new particles which are created as you highly stress spacetim.

Energy in Einstein's Relativity is simple a diffused state of matter. That is to say, Einstein's Relativity makes energy and matter as different fascets of the same thing. So energy is a fluctuation and it is a diffused state of matter.

Not just this surely? Every experimentally confirmed value in physics holds a mystery. The [math]g[/math] in the Yukawa Coupling, with different values for every mass on the standard model is one such example. What about the fine structure constant? In fact there are many precise values in the standard model just begging to be answered for and by another theory. It may just happen that our theory is a mathematical monstrocity.

The universe does not have an exact defined energy. For that to be so, someone would need to be sitting outside of the universe to measure the energy.