Widdekind Posted September 17, 2013 Posted September 17, 2013 A photon of wave-length = Planck length, has a mass-energy equivalent to the Planck-mass / energy... so, even at the smallest something can be, its energy density still doesn't dominate gravitationally... if that were not true... then wouldn't high-energy phenomena have already "twisted and contorted the fabric of space-time into donuts" and ripped out "holes to hyper-space" (for want of worthier words) ? is another way of saying "Gravity is weak", to say instead, "the fabric of space-time is strong" (and pretty impervious to (individual) particles) ? Can you construe, the "Hierarchy Problem" of particle physics, i.e. that the Planck mass-energy scale is ~17 orders of magnitude larger than typical particle mass-energies... can you construe the same, as essentially "safeguarding" the fabric of space-time, from becoming overly curved ?
ajb Posted September 17, 2013 Posted September 17, 2013 We could view the large gap between the GUT scale and the Planck scale as meaning that GR is a good theory at low energy (<< Planck) as quantum effects of space-time are hugely suppressed.
TrappedLight Posted September 20, 2013 Posted September 20, 2013 I'll just expand on what ajb said, the model you are looking for is geometrogenesis. Ajb is talking about the low energy epoch: Which is in fact when geometry and matter appeared in the universe. The Planck epoch exists in the high energy stages of our universe. General Relativity works well in a vacuum where geometry is present. Relativity is inconsistent without geometry, so it's meaningless to try and make sense of it when the universe was extremely young. No doubt this is going to require a new path of physics called ''beyond the standard model'' physics. Though it's really interesting stuff! A really strange prediction however of geometrogenesis, is that when the universe came into existence, we are led to believe there was no space to speak of, it appeared as a single point without dimensions. This must, if we take relativity seriously about it's first principles, must mean that there was no time either to speak of since space and time are considered unified as the same properties of the same space triangle. Of course, it kind of reminds us of the Wheeler de Witt equation, which has been interpreted to mean we live in a timeless universe. - you obtain this equation by quantizing the Einstein Field Equations.
Widdekind Posted September 20, 2013 Author Posted September 20, 2013 We could view the large gap between the GUT scale and the Planck scale as meaning that GR is a good theory at low energy (<< Planck) as quantum effects of space-time are hugely suppressed. is not the statement stronger still ? For, all energy densities are "low energy density" (<<Planck)... you could compress quanta, squishing them in size until their de-Broglie wavelengths were on Planck length long, and their energy density would only then begin to become "bothersome" to the fabric of space-time i perceive that the rubber-sheet analogy may mislead minds, into underestimating the enormous structural strength, of spacetime, as some sort of substance (w/ an ontological existence independent of mass-energy-density)... crudely, you could construct a bridge a million km long from space-time, and load it w/ a starmass of material, and see it sag in the center solely 3km
ydoaPs Posted September 20, 2013 Posted September 20, 2013 A photon of wave-length = Planck length, has a mass-energy equivalent to the Planck-mass / energy... so, even at the smallest something can be, its energy density still doesn't dominate gravitationally... if that were not true... then wouldn't high-energy phenomena have already "twisted and contorted the fabric of space-time into donuts" and ripped out "holes to hyper-space" (for want of worthier words) ? is another way of saying "Gravity is weak", to say instead, "the fabric of space-time is strong" (and pretty impervious to (individual) particles) ? Can you construe, the "Hierarchy Problem" of particle physics, i.e. that the Planck mass-energy scale is ~17 orders of magnitude larger than typical particle mass-energies... can you construe the same, as essentially "safeguarding" the fabric of space-time, from becoming overly curved ? Smallest something can be? IIRC, space isn't quantized.
ajb Posted September 23, 2013 Posted September 23, 2013 is not the statement stronger still ? For, all energy densities are "low energy density" (<<Planck)... you could compress quanta, squishing them in size until their de-Broglie wavelengths were on Planck length long, and their energy density would only then begin to become "bothersome" to the fabric of space-time We don't really understand what the quantum theory of gravity is, so it is impossible to give details. But as long as the energy densities are not near the Planck scale then GR will work well. It means we are lucky enough to be able to separate the quantum nature of the standard model of particle physics from gravitational physics.
Widdekind Posted September 24, 2013 Author Posted September 24, 2013 what about the generalized Klein-Gordon approach [math]D^{\mu}D_{\mu} \Psi = m^2 \Psi[/math] [math]\longrightarrow[/math] [math]D^{\mu} \eta_{\mu \nu} D^{\nu} \Psi = m^2 \Psi[/math] (inserting Minkowski metric matrix) [math]\longrightarrow[/math] [math]D^{\mu} g_{\mu \nu} D^{\nu} \Psi = m^2 \Psi[/math] (swapping out Minkowski, inserting GR metric matrix) ? The statements of AJB seem completely correct, only somewhat understated... at an age of ~1 TPL, energy densities w/in the universe were ~1 EPL, so since 1 planck time, the whole history of the complete cosmos has been "low energy"
ajb Posted September 24, 2013 Posted September 24, 2013 You can consider the KG Lagrangian on a curved background by doing exactly what you have done, replace the flat metric with the curved one. This is minimal coupling. You also have non-minimal coupling of a scalar field to the Ricci curvature, this is well discussed in the literature.
Widdekind Posted September 25, 2013 Author Posted September 25, 2013 You can consider the KG Lagrangian on a curved background by doing exactly what you have done, replace the flat metric with the curved one. This is minimal coupling. You also have non-minimal coupling of a scalar field to the Ricci curvature, this is well discussed in the literature. minimal coupling would seem appropriate, for single particle wave-function solutions, given the "low energy" limit ? if the KG generalizes, in the presence of EM fields, to [math]D^{\mu} \rightarrow D^{\mu} - e A^{\mu}[/math]... then would any energy potential, be treated, similarly, inserted into the time component of the generalized differential operator ?
ajb Posted September 26, 2013 Posted September 26, 2013 if the KG generalizes, in the presence of EM fields, to [math]D^{\mu} \rightarrow D^{\mu} - e A^{\mu}[/math]... Right, so you want to consider a complex scalar field coupled to a classical EM background on a curved space-time? Don't forget that we have a scalar field here and so we don't need the covariant derivative with respect to the Christoffel symbols. ...then would any energy potential, be treated, similarly, inserted into the time component of the generalized differential operator ? You can find a description here; aesop.phys.utk.edu/qft/2004-5/4-2.pdf
Widdekind Posted October 4, 2013 Author Posted October 4, 2013 Right, so you want to consider a complex scalar field coupled to a classical EM background on a curved space-time? Don't forget that we have a scalar field here and so we don't need the covariant derivative with respect to the Christoffel symbols. You can find a description here; aesop.phys.utk.edu/qft/2004-5/4-2.pdf first, if i understand the Lagrangian formalism, then any potential, of any form or origin, would wind up incorporated into the time component, of the (generalized) derivative... e.g. a hypothetical exponential-well, for a mathematical model, of a strong-force potential... second, i perceive a potential problem, in properly interpreting the symbols, since [math]|| d^4x \sqrt{det(g)} ||[/math] = length4 so, seemingly, for a Schwarzschild metric, [math]det(g) = r^4 sin^2(\theta)[/math], so that [math]d^4x \sqrt{det(g)} = c \; dt \; dr \; d\theta \; d\phi \; r^2 \; sin(\theta)[/math] so, similarly, for the KGE, paying particular attention to units, the inverse metric is appropriate, for the "inverse differentials" of the derivatives -- the determinant is of the metric matrix, the derivatives multiply the inverse metric matrix (??) looks like the Laplacian is reconstituted from the formulas, with the exception of the Schwarzschild metric terms, in the time and radius double derivatives moreover, looks like a hypothetical strong-force exponential scalar potential, to try to model quarks in nucleons, would fold into the time derivative component, exactly as an EM scalar potential i want to ask about "minimal coupling" First, MC usually refers to the generalization, of the derivative operators, to include EM scalar & vector potentials, "imposed" upon the particle's wave-function, w/o any feedback from the particle to the field (apparently). So, seemingly, MC of particle <----> EM fields is distinguished, from MC of particle <----> space-time curvature, where the space-time curvature is "imposed", w/o any feedback from the particle to the SET <----> Ricci tensor. Second, starting from the Schrodinger equation, for the H atom, for simplicity's sake, and observing that the proton's implicitly-present wave-function is basically [math]\Psi_p^* \Psi_p \approx \delta^3(0)[/math], the generalization seems sound: [math]V( r ) \rightarrow \int d^3x \Psi_p^*(x) \Psi_p(x) V(r-x)[/math] Thus, turning probability amplitudes into probability (and mass, charge) densities, seemingly allows one to calculate the fields (and SE / curvatures) caused by particles. That approach would be an "integro-differential" equation approach -- particles' wave-functions (amplitudes) are evolved thru a differential time-step, knowing the fields & curvatures; and then the updated densities determine the updated fields & curvatures. Prima facie, such a two-step integro-differential approach seems sound, w/o any obvious reason why such would not work well
ajb Posted October 5, 2013 Posted October 5, 2013 first, if i understand the Lagrangian formalism, then any potential, of any form or origin, would wind up incorporated into the time component, of the (generalized) derivative... e.g. a hypothetical exponential-well, for a mathematical model, of a strong-force potential... Classically you can add any potential you like to the scalar field on flat space-time. The trouble is you require this to be renormalisable and this will put constraints on the potential that depend on the space-time dimensions. I guess that as locally we have a flat space-time on a general space-time then we must have similar constraints on the form of the potential. I don't know the details here and studying interacting fields on curved space-times properly is very difficult. ...the determinant is of the metric matrix, the derivatives multiply the inverse metric matrix (??) I am not quite sure what you are asking here. However remember the action is built from a Lagrangian which is an object that is intergarted over space-time. Thus the Lagrangian must be a density. First, MC usually refers to the generalization, of the derivative operators, to include EM scalar & vector potentials, "imposed" upon the particle's wave-function, w/o any feedback from the particle to the field (apparently). So, seemingly, MC of particle <----> EM fields is distinguished, from MC of particle <----> space-time curvature, where the space-time curvature is "imposed", w/o any feedback from the particle to the SET <----> Ricci tensor. Minimal coupling means that when you set the connections to zero, but not nessisarily their derivates you end up with the same physics as if the connections were never there in the first place. So on a curved space-time you want things to reduce to the flat space-time physics when in normal coordinates, for example. ... such a two-step integro-differential approach seems sound, w/o any obvious reason why such would not work well I don't know.
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