Jump to content

Dubbelosix

Senior Members
  • Posts

    518
  • Joined

  • Last visited

Everything posted by Dubbelosix

  1. Please don't start posting diagrams which are unintelligible without some explanation or equations backing them. And I have seen your equations, you might think I am being harsh, but they equally are unintelligible - No one can properly follow what is being suggested, but even if I could read it, you are not careful with dimensional analysis which is very important.
  2. I don't know what you mean by cycles a such - though there are cycles in the Berry phase which has possible implications to this geometric approach of gravity. The Eigenstates are calculated in the Planck phase space as [math]R_{[\mu, \nu]} \equiv [\nabla_x\nabla_0 - \nabla_0 \nabla_x] \geq \frac{(\ell^{2})^{-1}}{\sum_i \sqrt{n_i(n_i + 1)}}[/math]
  3. what did you mean by ''all possible states''....? The phase space I am working in involves higher orders that involves the Eigenstates of the system.
  4. Yes. Anything in nature that has a motion is translated into, say linear or rotational kinetic energies.
  5. That's not what I got from the paper, what I got was that gravity acts on all energy states in the same way. That didn't mean gravity has no effect on superpositioned particles. You're not understanding what was being discussed. If you read it again, you will (maybe) understand, my toy model is a theory of the Planck Space in terms of non-commutative phase space.
  6. I don't doubt more work needs to be done, but understanding rotation as part pf the Poincare group, it feels natural to suggest this as a theory and maybe even taken seriously in some circles that understand a more global impact for the dark flow phenomenon.
  7. Notice from my opening post, there is a force associated to the equation [math]\ddot{R} = \frac{8 \pi G R}{3}\rho - \frac{d\omega}{dt} \times R - 2\omega \times \frac{dR}{dt} - \omega \times (\omega \times R)[/math] In fact, this last equation is very much like the absolute acceleration (or four component acceleration equation) or an ''apparent acceleration,'' in the rotating frame. Because Newtons laws apply to the absolute acceleration, the effect that arises is a ficticious force which can be understood when mass is involved in the picture: [math]m\ddot{R} = \frac{8 \pi Gm R}{3}\rho - m\frac{d\omega}{dt} \times R - 2m\omega \times \frac{dR}{dt} - m\omega \times (\omega \times R)[/math] [math]F = m\ddot{R}[/math] What you are looking at is the ''force'' which is pushing the universe away from its centre, due to a proposed rotary property of the universe. Already, the acceleration term has been shown by other scientists https://arxiv.org/ftp/arxiv/papers/1111/1111.3873.pdf to satisfy the required acceleration to satisfy dark energy.
  8. being edited Note, even though I recite a paper, I have actually corrected an equation that may be a printing error Spiral Trajectories and Extra Background Radiation Source A quick explanation first why that third derivative in time in the Friedmann equation leads to non-conservation. The time derivative of the Hubble radius is [math]\frac{dR}{dt} = v = \dot{R}[/math] Second derivative in time leads to acceleration (as would be expected say, in Friedmanns acceleration equation) [math]\frac{dv}{dt} = a = \ddot{R}[/math] Third derivative in time leads to chaotic systems and is denoted as the jerk [math]\frac{da}{dt} = j = \dddot{R}[/math] The suggested equation for a non-conservation in particle number located in the effective density was suggested in a form (with rotation): [math]\frac{\dot{R}}{R}(\frac{\ddot{R}}{R} + \frac{kc^2}{a})= \frac{8 \pi G}{6c^2}(\frac{\rho + P}{n})\dot{n} + \omega^2 \frac{\dot{R}}{R}[/math] The rotating universe (at least in the early cosmology case) coupled to the dust inside of it strongly. This causes the charged particles in spacetime to experience a circular trajectory (in which they lose energy through the loss of radiation) which is known as a cyclotron radiation, similar to how we view charged particles accelerating in spacetime giving rise to Larmor radiation - in the case of gravity, this would be due to the weak equivalence principle. [math]m\dot{R}^2 = \frac{e^2}{6 \pi c^3} \dddot{R} + eV[/math] In which [math]\ddot{R} \propto \dddot{R}[/math]. As noted by Arun and Sivaram, this leads to a path that is an exponentially increasing logarithmic spiral. Of course, in the context of a rotating expanding spacetime, the decaying rotational properties means that the logarithmic path too is overcome by expansion in the bigger picture. So instead of an exponential increase, the coupling of rotation to matter requires also that the coupling fall off as rotation equally decays. Such a logarithmic spiral would instead follow an exponential decay rule in accordance to the rotation which decays ~ [math]\omega = \omega_0 e^{-\lambda t}[/math] We can see how this relates to the third derivative directly. Differentiation leads to in the spiral equation, terms that will fit the expanding and rotating Friedmann model [math]2m\dot{R}\ddot{R} = \frac{e^2}{6 \pi c^3} \ddddot{R} + e\dot{V}[/math] Notice, the potential difference, also known as the voltage [math]V[/math] has picked up a charge to mass ratio coefficient, [math]\dot{R} \ddot{R} = \frac{e^2}{6 \pi c^3} \dddot{R} + (\frac{e}{m})V[/math] (As it is, this matches the non-conserved form of the Friedmann equation). We can replace the charge to mass ratio with a gyromagnetic ratio because the universes rotation, is also a classical property. This term that can replace the charge to mass ratio works only if charges in spacetime are distributed evenly. Due to spacetime homogeneity, this seems to be a fitting case. The interesting thing, the additional rotational radiation coming from these charged particles in the early universe can contribute to an exotic zoo. The high radiation densities would lead to new particles of various types. It also stands as a contributor to the background temperatures. The differentiation of both the spiral trajectory equation and the Friedmann Langrangian (an equation I derived some time back) we can see how they relate as power equations [math]m\dot{R}^2 = \frac{e^2}{6 \pi c^3} \dddot{R} + eV[/math] [math]\rightarrow m\dot{R}\ddot{R} = \frac{e^2}{12 \pi c^3} \ddddot{R} + \frac{1}{2}e\dot{V}[/math] In which the Friedmann Langrangian and the Friedmann power equations where identified as: [math]\mathcal{L} = m\dot{R}^2 - \frac{8 \pi Gm R}{3c^2}(\rho + 3P) + mR \omega^2[/math] [math]\rightarrow \mathcal{P} = m\dot{R}\ddot{R} - \frac{8 \pi Gm R}{6c^2}(\rho + 3P)\frac{\dot{R}}{R} + mR \omega \dot{\omega}[/math] The rotating universe is compatible with the spiral paths taken giving rise to the extra radiation. Notice also, the differentiation of the spiral equation yields the jolt a rare symbol ever if there was one in physics [math]\ddddot{R}[/math]. Very rarely do we have to consider such derivatives but in this model, it cannot be avoided. ref: https://arxiv.org/ftp/arxiv/papers/1402/1402.5071.pdf
  9. no it can't - do you understand, for instance why [math]mc^2 \ne ma[/math] If m is the mass, c is the speed of light and a is acceleration?
  10. Speaking with you was arduous, because even when I corrected you, you continued to make the same mistakes - in other words, you where blatantly ignoring what was being said to you, such as the dimensional consistency of equations. I appreciate you have an fascination for physics, but along the way, who are you trying hustle?
  11. Where did you get this link from? I have quoted the link before, I am just wondering how you came across it? I did actually find correlations between black hole dynamics and the evolution and stability of galactic structures.
  12. Mordred, you are the only one here who makes coherent sense. Vmedril, a word of warning, I am not even a mod, but even I can see through the bullshit. You have done this to me on another site, you inundated a previous thread of mine with complete nonsense. This is complete trash you are posting and it is actually annoying me... which takes quite a bit. It annoys me because you become part of the new age anti-scientific regime. The world is crumbing to pseudo scientists and here you are posting rubbish that doesn't even make sense when you study it closely. Just like your manipulations of my equations on another site without regard about complexities involving dimensional consistency.
  13. Do you like it? Do you think this geodesic equation is simple enough to be possibly right? 

    [math]\nabla_n \dot{\gamma}(t) = \nabla_n\frac{dx^{\mu}}{d\tau} \equiv\ min\ \sqrt{<\dot{\psi}|[\nabla_j,\nabla_j]|\dot{\psi}>} = 0[/math]

    1. Dubbelosix

      Dubbelosix

      I hold that the RHS must [always] equal zero because the curve component is squared, meaning by definition that the covariant derivatives have to be the same. I defined the time derivative early on with a j-subscript. This necessary application fundamentally implements that the geodesic always equals zero. The fact the connections are acting classically could also be a hint for classical gravity at the quantum scale? Certainly, Penrose has suggested classical gravity in the phase space.

  14. I'll be closing my studies on gravity in the Hilbert space soon now. I will finalise a more comprehensible paper from it as well, but i wanted to reserve for this post, the most interesting conclusions I made. There were many things I found while investigating this model, but the following will only be concerned with the dynamics of [math]L^2[/math] Cauchy-Schwarz space and the Hilbert space formalism. The ultimate goal was to find a Hilbert space theory of gravity which seemed (at least plausible) by linking it dynamically with the Schrodinger evolution. I achieved this but it needs so much more work that (maybe) someone else will take up as I will be going back into cosmology investigations. Working (back from where I ended up) actually would make more sense to the reader as I can keep it short and simple. The analogue of the geodesic equation found in GR: [math]\frac{d^2x^{\mu}}{d\tau^2} + \Gamma^{\mu}_{\nu \lambda} \frac{dx^{\nu}}{d \tau} \frac{dx^{\lambda}}{d\tau} = 0[/math] was in fact, satisfied in a very concise way using the bra-ket notation, [math]\nabla_n \dot{\gamma}(t) = \nabla_n\frac{dx^{\mu}}{d\tau} \equiv\ min\ \sqrt{<\dot{\psi}|[\nabla_j,\nabla_j]|\dot{\psi}>} = 0[/math] There is classical commutation going on, on the RHS [math][\nabla_j, \nabla_j] = 0[/math] which just means in the usual sense, that the order of the operators does not matter, for obvious reasons. This is why it should be noted, that space time non-commutivity is subtly hinted at when you consider connections with derivatives in both space and time. Anyway, moving on, from the second equation we can construct two solutions for the time-dependent Schrodinger equation which satisfies [math]\frac{1}{ i \hbar}H|\psi>\ = |\dot{\psi}>[/math] When you consider the geometry related to the Hamiltonian in such a way: [math]\sqrt{<\dot{\psi}|\dot{\psi}>} = \int \int\ |W(q,p)^2| \sqrt{<\psi|\Gamma^2|\psi>}\ dqdp \geq \frac{1}{\hbar}\sqrt{<\psi|H^2|\psi>}[/math] Which was derived by myself using the Wigner function, then it becomes more apparent that it was possible to construct a wave equation encoding the rank 2 tensor dynamics of the connection and the stress energy tensor in the following way [math]\nabla_n|\dot{\psi}>\ = \frac{c^4}{8 \pi G} \int \int\ |W(q,p)^2|\ (\frac{d}{dx^n}\Gamma^{ij} + \Gamma^{i}_{n\rho} \Gamma^{\rho j} + \Gamma^{j}_{n \rho}\Gamma^{\rho}_{i})|\psi>\ dqdp \geq \int\ \frac{1}{\hbar}(\frac{d}{dx^n}T^{ij} + \Gamma^{i}_{n\rho} T^{\rho j} + \Gamma^{j}_{n \rho}T^{\rho}_{i})dV|\psi>[/math] This is in fact, totally analogous to an acceleration/curve in the general theory of relativity, except this time, it satisfies an equality bound and it also satisfies the wave dynamics of a Schrodinger equation, albeit, a non-linear one. It is not immediately obvious there are terms in this last equation which can satisfy an inequality bound, but in my early investigation, the bounds where found as: [math]\sqrt{|<\nabla^2_i><\nabla^2_j>|} \geq \frac{1}{2}<\psi|[\nabla_i, \nabla_j] |\psi>\ = \frac{1}{2} <\psi|R^2_{ij}| \psi>[/math] [math]<\psi|[\nabla_i, \nabla_j] |\psi>\ =\ <\psi| R_{ij}| \psi>\ \leq 2 \sqrt{|<\nabla^2_i><\nabla^2_j>|}[/math] Which is basically a mean deviation that can reach twice the classical upper bound. This bound holds importance for our equation which satisfied the geodesic equation [math]\nabla_n \dot{\gamma}(t) = \nabla_n\frac{dx^{\mu}}{d\tau} \equiv\ min\ \sqrt{<\dot{\psi}|[\nabla_j,\nabla_j]|\dot{\psi}>} = 0[/math] Which can have a Berry curve definition [math]\gamma = i \oint <n(\mathbf{R})\nabla_{\mathbf{R}}|n(\mathbf{R})>\ d\mathbf{R}[/math] This formulation links the wave dynamics with the geometry and the energy associated to the geometry of the system. This geometric look at the Hilbert space has allowed us to find solutions to the geodesic equation, which appear at the surface, a simple argument. This ability to give a vector space curvature was really the most enlightening key of the investigation. The Hilbert space, even though is an abstract space, is one that actually works very well with quantum mechanics, for whatever unseen intrinsic reason there is for this. Equally, in such an abstract mathematical space, you can create mathematical objects to describe the necessary curvature required to describe the accelerations in the phase space. Since gravity is a pseudo force from the first principles of relativity, then it is unlikely quantization of the field into the usual spin 2 graviton gauge theory is a bit extreme - why should gravity get a mediator particle when the Coriolis force and the Centrifugal force do not require one? How is gravity a pseudo force? Gravity is a pseudo force that can be understood in the following (neat) and (concise and short) way: [math]\frac{d^2x^{\mu}}{d\tau^2} + \Gamma^{\mu}_{\nu \lambda} \frac{dx^{\nu}}{d \tau} \frac{dx^{\lambda}}{d\tau} = 0[/math] where [math]\Gamma^{\mu}_{\nu \lambda} = \frac{\partial x^{\mu}}{\partial \eta^{a}}\frac{\partial^2 \eta^a}{\partial x^{\nu}\partial x^{\lambda}}[/math] or more compactly [math]\Gamma^{\mu}_{\nu \lambda} = J^{\mu}_{a} \partial_{\nu} J^{a}_{\lambda} = J^{\mu}_{a} \partial_{\lambda} J^{a}_{\nu} \equiv J^{\nu}_{a} J^{a}_{\nu \lambda}[/math] which represents a pseudo force for gravity which makes it in the same league as the Coriolis and the Centrifugal forces.The previous equations also expose the inner structure of our investigation. Keep in mind for the Schrodinger Curve equation we derived, the covariant derivative acts on rank 2 tensors in the following way: [math]\nabla_n\Gamma^{ij} = \frac{d}{dx^n}\Gamma^{ij} + \Gamma^{i}_{n\rho} \Gamma^{\rho j} + \Gamma^{j}_{n \rho}\Gamma^{\rho}_{i}[/math] [math]\nabla_nT^{ij} = \frac{d}{dx^n}T^{ij} + \Gamma^{i}_{n\rho} T^{\rho j} + \Gamma^{j}_{n \rho}T^{\rho}_{i}[/math] Where the Hamiltonian has been 'naively(?)' replaced with the stress energy tensor. In a much later study when I get interest in this again, I will try and formulate a theory of how temperature and gravity translate into each other, since it was an observation from early investigations that the curve of a metric is in fact related to the temperature of the system given as (and as you will see, the definition on the left imples a factor of one half attached to our metric): [math]K_BT = \frac{1}{2}(\frac{dx^{\mu}}{d\tau} \cdot \frac{dx^{\mu}}{d\tau}) \equiv\ \frac{1}{2}<\dot{\psi}|\dot{\psi}>[/math] This extra factor of [math]\frac{1}{2}[/math] is acceptable since one such term exists in the classical upper bound equation and [math]m=1[/math] for a constant mass. That's the latex fixed, sorry about that. I'd also like to say, Mordred helped me along with this. He/she did direct my in the right way a few times. One such example was finding the bound which I had not considered until they offered it.
  15. Actually, I knew there had to be an inconsistency with setting the geodesic equation to zero but finding that it could satisfy non-commutation. The answer I have figured out relied on the fact both connections have derivatives in time. I have been so fixated on the non-commutative properties of spacetime, I forgot in this one simple case, it doesn't hold. The question was, was the geodesic equation a case where non-commutivity means it is non-zero? The answer it seems, there is actually commutation between them since they are both covariant derivatives in (time). In the case of a covariant derivative with both space and time indices, there would be spacetime non-commutivity. So in this case, the order of the covariant derivatives do not matter and they actually commute which preserves the definition of the geodesic being zero [math][\nabla_j,\nabla_j] = 0[/math]
  16. Minimum Distance in Relativity and the Non-Commutation of the Phase Space First... a blow by blow account of what led up to the proposal of the curve equation: Anandan proposes an equation [math]E = \frac{k}{G} (\Delta \Gamma)^2[/math] and I offer also true [math]E = \frac{c^4}{G} \int \Delta \Gamma^2\ dV = \frac{c^4}{G} \int \frac{1}{R^2} \frac{d\phi}{dR}(R^2 \frac{d\phi}{dR})\ dV[/math] That was after correcting the constant of proportionality, and after I derived a following inequality by making use of the Wigner function - The Mandelstam-Tamm inequality for instance I have shown can be written in the following way, with [math]c = 8 \pi G = 1[/math] (as usual), we can construct the relationship: (changing notation only slightly) [math]|<\psi(0)|\psi(t)>|^2 \geq \cos^2(\frac{[<\Gamma^2> - <\psi|\Gamma^2|\psi> ]\Delta t}{\hbar}) = \cos^2(\frac{[<H> - <\psi|H|\psi> ]\Delta t}{\hbar}) = \cos^2(\frac{\Delta H \Delta t}{\hbar})[/math] This linking of geometry to the energy of the system can be understood through a curve equation I derived using the same principles [math]\frac{ds}{dt} \equiv \sqrt{<\dot{\psi}|\dot{\psi}>}\ = |W(q,p)| \sqrt{<\psi|\Gamma^4|\psi>} \geq \frac{1}{\hbar} \sqrt{<\psi|H^2|\psi>}[/math] By understanding that the curve equation when squared provides solutions to the time-dependent Schrodinger equation (in the following way) ~ [math]\frac{1}{ i \hbar}H|\psi>\ = |\dot{\psi}>[/math] Then you can construct a more serious equation that may be seen as a gravitational analogue to the Schrodinger equation when the covariant derivative acts on the tensor components and I calculate it as: [math]\nabla_n|\dot{\psi}>\ = \frac{c^4}{8 \pi G} \int \int\ |W(q,p)^2|\ (\frac{d}{dx^n}\Gamma^{ij} + \Gamma^{i}_{n\rho} \Gamma^{\rho j} + \Gamma^{j}_{n \rho}\Gamma^{\rho}_{i})|\psi>\ dqdp \geq \int\ \frac{1}{\hbar}(\frac{d}{dx^n}T^{ij} + \Gamma^{i}_{n\rho} T^{\rho j} + \Gamma^{j}_{n \rho}T^{\rho}_{i})dV|\psi>[/math] Also keep in mind, when you take the bra solution [math]<\dot{\psi}| \nabla[/math] and it;s product with its conjugate above, you could understand the product as producing an object like [math]<\dot{\psi}|[\nabla,\nabla]|\dot{\psi}>[/math] which shows the non-commutation between covariant derivatives - the wave functions, in its most simplest form can be understood as [math]|\psi>\ = e^{iHt}|q>[/math] [math]<\psi| =\ <q| e^{-iHt}[/math] When we take the product of the bra and ket solutions, we get an analogous identity found in General relativity - we simply take the tangent vector [math]\frac{dx^{\mu}}{d\tau}[/math] and allow the covariant derivative to act on this (and further set it to zero) and defines the minimum curve, or better yet, as a geodesic for a minimum distance, [math]\nabla_n\frac{dx^{\mu}}{d\tau} \equiv\ min\ \sqrt{<\dot{\psi}|[\nabla,\nabla]|\dot{\psi}>}[/math] We know how the covariant derivative acts on the curve from the following equation [math]T_{nm}(y) = \nabla_n V_m = \frac{\partial V_m}{\partial y^{n}} + \Gamma_{nm} V_{r}(x)[/math] The interesting thing about setting this to zero and using this as the definition of the minimum distance is that non-commutation between the covariant derivatives in a phase space is generally not zero [math][\nabla_i \nabla_j] \ne 0[/math]! The Von Neumann algebra insists that deviation from the classical vacuum relies on the non-commutative properties of the quantum phase space. The covariant derivative also acts on rank 2 tensors in the following way The covariant derivative acting on our connection must obey a rank 2 tensor, [math]\nabla_n\Gamma^{ij} = \frac{d}{dx^n}\Gamma^{ij} + \Gamma^{i}_{n\rho} \Gamma^{\rho j} + \Gamma^{j}_{n \rho}\Gamma^{\rho}_{i}[/math] It also follows then the stress energy tensor responds in much the same way [math]\nabla_nT^{ij} = \frac{d}{dx^n}T^{ij} + \Gamma^{i}_{n\rho} T^{\rho j} + \Gamma^{j}_{n \rho}T^{\rho}_{i}[/math] which were the primary tools we used to construct our form of the non-linear Schrodinger equation. Also, it has also been established that the square of the curve can be seen as related to the kinetic energy: [math]K_BT = (\frac{dx^{\mu}}{d\tau} \cdot \frac{dx^{\mu}}{d\tau}) \equiv\ <\dot{\psi}|\dot{\psi}>[/math] Where again, this is a massless form (m = 1) for a constant mass. This linking of temperature to geometry could be lucrative. https://en.wikipedia.org/wiki/Maupertuis'_principle
  17. The Christoffel symbol can be 'loosely' though of as being analogous to a force in Newtons equations (where mass has been set to 1 to denote that it is a constant in this formulation): [math]\Gamma = \frac{1}{2} \frac{\partial g_{00}}{\partial x}[/math] Newtonian formulation of this acceleration is [math]F = -\frac{\partial \phi}{\partial x}[/math] However as mentioned, the gravitational force is not actually a true definition of a force as we come to expect say, in the proposed fundamental fields of nature, which are inherently complex (when quantum gravity is not) and that require quantization of field particles acting as mediators of the force (something which gravity is expected to use to form the unification theory in the opinions of many scientists). It's actually a crucial component of many theories, most notably string theory. Gravity is a pseudo force and can be understood in the following (neat) and (concise and short) way: [math]\frac{d^2x^{\mu}}{d\tau^2} + \Gamma^{\mu}_{\nu \lambda} \frac{dx^{\nu}}{d \tau} \frac{dx^{\lambda}}{d\tau} = 0[/math] where [math]\Gamma^{\mu}_{\nu \lambda} = \frac{\partial x^{\mu}}{\partial \eta^{a}}\frac{\partial^2 \eta^a}{\partial x^{\nu}\partial x^{\lambda}}[/math] or more compactly [math]\Gamma^{\mu}_{\nu \lambda} = J^{\mu}_{a} \partial_{\nu} J^{a}_{\lambda} = J^{\mu}_{a} \partial_{\lambda} J^{a}_{\nu} \equiv J^{\nu}_{a} J^{a}_{\nu \lambda}[/math] which represents a pseudo force for gravity which makes it in the same league as the Coriolis and the Centrifugal forces.
  18. The Christoffel symbol can be 'loosely' though of as being analogous to a force in Newtons equations (where mass has been set to 1 to denote that it is a constant in this formulation): [math]\Gamma = \frac{1}{2} \frac{\partial g_{00}}{\partial x}[/math] Newtonian formulation of this acceleration is [math]F = -\frac{\partial \phi}{\partial x}[/math] However as mentioned, the gravitational force is not actually a true definition of a force as we come to expect say, in the proposed fundamental fields of nature, which are inherently complex (when quantum gravity is not) and that require quantization of field particles acting as mediators of the force (something which gravity is expected to use to form the unification theory in the opinions of many scientists). It's actually a crucial component of many theories, most notably string theory. Gravity is a pseudo force and can be understood in the following (neat) and (concise and short) way: [math]\frac{d^2x^{\mu}}{d\tau^2} + \Gamma^{\mu}_{\nu \lambda} \frac{dx^{\nu}}{d \tau} \frac{dx^{\lambda}}{d\tau} = 0[/math] where [math]\Gamma^{\mu}_{\nu \lambda} = \frac{\partial x^{\mu}}{\partial \eta^{a}}\frac{\partial^2 \eta^a}{\partial x^{\nu}\partial^{\lambda}}[/math] or more compactly [math]\Gamma^{\mu}_{\nu \lambda} = J^{\mu}_{a} \partial_{\nu} J^{a}_{\lambda} = J^{\mu}_{a} \partial_{\lambda} J^{a}_{\nu} \equiv J^{\nu}_{a} J^{a}_{\nu \lambda}[/math] which represents a pseudo force for gravity which makes it in the same league as the Coriolis and the Centrifugal forces. Also keep in mind, when you take the bra solution [math]<\dot{\psi}| \nabla[/math] and it;s product with its conjugate above, you could understand the product as producing an object like [math]<\dot{\psi}|[\nabla,\nabla]|\dot{\psi}>[/math] which shows the non-commutation between covariant derivatives
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.