substitutematerials Posted November 1, 2016 Author Share Posted November 1, 2016 I think we may be talking about intrinsic curvature ....which does not need to be embedded in an extra dimension . Is spacetime curvature 3 or 4 dimensional?(I think it is 4 -dimensional) I think you're right, I shouldn't be invoking an extra dimension. But the mental image sort of demands another dimension for the surface to curve in. I suppose you can just imagine a distorted coordinate grid like a Mercator projection on a flat map of Earth, but this seems like a formalism and so isn't quite as satisfying. Link to comment Share on other sites More sharing options...
imatfaal Posted November 1, 2016 Share Posted November 1, 2016 ... [latex]\rho_{crit} = \frac{3c^2H^2}{8\pi G}[/latex] k = curvature constant as a dimensionless value, k = critical density. As far as we can determine the curvature constant stays constant throughout our universes history. ... Love your posts Mordred but this one has me foxed. So you give the critical density function - OK. And then define k - twice, and k is not even mentioned in the equation! [latex]R^2 \left(H^2 - \frac{8}{3} \pi G \rho \right) = -kc^2[/latex] Is the Friedman equation not the relevant one - your equation being a version of it when density is at critical level Link to comment Share on other sites More sharing options...
Mordred Posted November 1, 2016 Share Posted November 1, 2016 (edited) Sorry your correct, was having a distracted moment. Concentrating on too many things at once. Edited November 1, 2016 by Mordred Link to comment Share on other sites More sharing options...
sethoflagos Posted November 2, 2016 Share Posted November 2, 2016 (edited) This may be totally off the wall (I'm an engineer) But could the universe be approaching zero average geometrical 'curvature' asymptotically in the same sense that its thermodynamic expansion asymptotically approaches absolute zero temperature? Edit: Afterthought. Could the flattening of spacetime curvature be in some sense the work performed by the expansion of the universe on its surrounding envelope? Edited November 2, 2016 by sethoflagos Link to comment Share on other sites More sharing options...
Ophiolite Posted November 2, 2016 Share Posted November 2, 2016 This may be totally off the wall (I'm an engineer) But could the universe be approaching zero average geometrical 'curvature' asymptotically in the same sense that its thermodynamic expansion asymptotically approaches absolute zero temperature? I like that idea, but then I'm a geologist, so what do I know? Link to comment Share on other sites More sharing options...
substitutematerials Posted November 2, 2016 Author Share Posted November 2, 2016 This may be totally off the wall (I'm an engineer) But could the universe be approaching zero average geometrical 'curvature' asymptotically in the same sense that its thermodynamic expansion asymptotically approaches absolute zero temperature? Edit: Afterthought. Could the flattening of spacetime curvature be in some sense the work performed by the expansion of the universe on its surrounding envelope? I believe it's actually the opposite- flatness comes from a state of precise balance in the Friedman equations. Any tiny deviation from perfect flatness should magnify itself over cosmic history. This is part of the reason it is surprising that we observe such flatness in the present era. Link to comment Share on other sites More sharing options...
Mordred Posted November 3, 2016 Share Posted November 3, 2016 (edited) Love your posts Mordred but this one has me foxed. So you give the critical density function - OK. And then define k - twice, and k is not even mentioned in the equation! [latex]R^2 \left(H^2 - \frac{8}{3} \pi G \rho \right) = -kc^2[/latex] Is the Friedman equation not the relevant one - your equation being a version of it when density is at critical level I like the form you use of that equation, there is another form from Ryden that I find more useful. For other readers I will detail the equations with some explanation. First thing to understand is that the critical density formula and the one posted by Imatsfaal is the GR aspects of the FLRW metric, Other key aspects is the acceleration equation and the ds^2 line element of the metric. However for this post I'm just going to focus on the two equations posted above. Essentially those two equations are derived by inserting the Einstein field equations into the FLRW metric. This is for all contributors (photons, matter, radiation etc). So first we replace [latex]\rho(t)[/latex] mass density with energy density in the form [latex]\epsilon(t)/c^2[/latex] the GR form of the Freidmann equations is in the Newton limit in GR, this is low gravity such as stars, galaxies, LSS etc. It is a specific class solution in GR. This gives the form of [latex](\frac{\dot{a}}{a})^2[/latex][latex]=\frac{8\pi G}{3}\frac{\epsilon(t)}{c^2}[/latex][latex]-\frac{kc^2}{R_0^2}\frac{1}{a^2(t)}[/latex] If [latex]k\le0[/latex] and the energy density is positive, then the R.H.S of the last equation is always positive. This is an expanding universe that will expand forever. If matter is the dominant form of energy, as opposed to radiation this implies [latex]\epsilon\propto \frac{1}{a^2(t)}[/latex]. If k=+1 then the R.H.S must eventually reach 0, after which the universe will contract. To get to the density parameter we can substitute [latex]H(t)=(\frac{\dot{a}}{a})^2[/latex] and we can rewrite the above equation into the Hubble parameter. (note I hate calling it constant, as its only constant at a particular moment in time) [latex]H(t)=\frac{8\pi G}{3}\frac{\epsilon(t)}{c^2}[/latex][latex]-\frac{kc^2}{R_0^2}\frac{1}{a^2(t)}[/latex] if k=0 then [latex]\rho_c(t)=\frac{e_c(t)}{c^2}=\frac{3H^2(t)}{8\pi G}[/latex] with the following density parameter relations [latex]\Omega=\frac{\epsilon}{\epsilon_c}=\frac{\epsilon}{c^2}*\frac{8\pi G}{3H^3}[/latex] note how we correlate the constant c in the the above. The cosmological constant isn't included in the above, essentially the Cosmological constant leads to an increase rate of expansion from the above relations. Also as it is constant as far as we can tell, this universe will continue to expand. This is what I should have taken the proper time to post. Again thanks Imatsfaal for catching the above. Busy work week lol This may be totally off the wall (I'm an engineer) But could the universe be approaching zero average geometrical 'curvature' asymptotically in the same sense that its thermodynamic expansion asymptotically approaches absolute zero temperature? Edit: Afterthought. Could the flattening of spacetime curvature be in some sense the work performed by the expansion of the universe on its surrounding envelope? The thermodynamic details take a bit to explain but from the above and using the equations of state for each contributor one can determine the deceleration equation. Wiki has a decent enough coverage. https://en.m.wikipedia.org/wiki/Equation_of_state_(cosmology) https://en.m.wikipedia.org/wiki/Friedmann_equations Edited November 3, 2016 by Mordred 1 Link to comment Share on other sites More sharing options...
Quantum321 Posted December 2, 2016 Share Posted December 2, 2016 General relativity predicted the BB. We think we know this part of his theory is "incomplete" He predicted infinite density at a singularity. We also know that when the equations result in infinities it means science can't explain it. Its off our charts. While it seems there is no center of the universe and the current thinking that the BB happened (is happening) everywhere I think it premature to call his classical theory incomplete. There is something we just don't understand yet. His theory has explained how the particles where created. Everything in the early universe was governed by temperature. There had to be something like a BB happen somewhere or everywhere. I don't believe there was infinite mass in a singularity. Why? I personally believe the universe is finite not infinite. Infinite mass would create an infinite universe. Perhaps the James Webb telescope will give us a clearer understand if the universe has boundaries. Hubble's deep field image tells us galaxies exist beyond Hubble's capabilities. I am hoping the James Webb telescope will help science answer the finite or infinite universe question. Or perhaps the answer to that question will lay beyond that telescopes capabilities. Link to comment Share on other sites More sharing options...
StringJunky Posted December 2, 2016 Share Posted December 2, 2016 I don't believe there was infinite mass in a singularity. The singularity of GR is infinitely dense not massive. Link to comment Share on other sites More sharing options...
Quantum321 Posted December 2, 2016 Share Posted December 2, 2016 This is an interesting question. What exactly does infinite density mean? Link to comment Share on other sites More sharing options...
Strange Posted December 2, 2016 Share Posted December 2, 2016 I think it premature to call his classical theory incomplete. There is something we just don't understand yet. So you don't think it is incomplete but you do think there are things we don't understand? That doesn't make much sense. Everyone else thinks it is incomplete because there are things we don't understand. His theory has explained how the particles where created. GR says nothing about the creation of particles. Link to comment Share on other sites More sharing options...
Delta1212 Posted December 2, 2016 Share Posted December 2, 2016 This is an interesting question. What exactly does infinite density mean? Density is mass/volume. To get infinite density, you need infinite mass and/or infinitesimal volume. So: infinite mass/finite volume infinite mass/infinitely small volume finite mass/infinitely small volume Any of those three situations will give you infinite density. Link to comment Share on other sites More sharing options...
Airbrush Posted December 2, 2016 Share Posted December 2, 2016 Density is mass/volume. To get infinite density, you need infinite mass and/or infinitesimal volume. You need infinite mass and/or zero volume. "Infinitesimal" is not small enough. Link to comment Share on other sites More sharing options...
StringJunky Posted December 2, 2016 Share Posted December 2, 2016 (edited) You need infinite mass and/or zero volume. "Infinitesimal" is not small enough. No, it needs to be a non-zero value which is what it means; vanishingly small but not zero. Edited December 2, 2016 by StringJunky Link to comment Share on other sites More sharing options...
Airbrush Posted December 2, 2016 Share Posted December 2, 2016 (edited) I don't understand what you mean. Density becomes infinite, for a finite amount of mass, only when the volume equals zero. What am I missing? I'm not an expert. Edited December 2, 2016 by Airbrush Link to comment Share on other sites More sharing options...
Delta1212 Posted December 2, 2016 Share Posted December 2, 2016 I don't understand what you mean. Density becomes infinite, for a finite amount of mass, only when the volume equals zero. What am I missing? I'm not an expert. A finite value divided by zero is not infinity. It is undefined. You need an infinitesimal divisor in order to yield infinity. Link to comment Share on other sites More sharing options...
Airbrush Posted December 2, 2016 Share Posted December 2, 2016 (edited) But isn't an infinitesimal divisor a non-zero value? As the divisor approaches zero, the fraction's value approaches infinity. But a fraction's value using a zero divisor is undefined, so infinity is also undefined. Infinite density is undefined, right? Edited December 2, 2016 by Airbrush Link to comment Share on other sites More sharing options...
StringJunky Posted December 2, 2016 Share Posted December 2, 2016 (edited) But isn't an infinitesimal divisor a finite value? You can always go smaller whatever value you set just as you can always use a higher number in infinity. Edited December 2, 2016 by StringJunky Link to comment Share on other sites More sharing options...
Quantum321 Posted December 2, 2016 Share Posted December 2, 2016 So you don't think it is incomplete but you do think there are things we don't understand? That doesn't make much sense. Everyone else thinks it is incomplete because there are things we don't understand. GR says nothing about the creation of particles. I stand by my original statement. I am not ready to throw the baby out with the bath water. To say any part of his theory is incomplete is to say it is wrong. I am just saying just because we may not understand his singularity doesn't mean his theory is incomplete (wrong) Link to comment Share on other sites More sharing options...
StringJunky Posted December 2, 2016 Share Posted December 2, 2016 (edited) To say any part of his theory is incomplete is to say it is wrong. I am just saying just because we may not understand his singularity doesn't mean his theory is incomplete (wrong) You are wrong in conflating 'incomplete' with 'wrong'. All theories are incomplete in some way. The most important aspect of a theory is it's utility; complete or otherwise. Edited December 2, 2016 by StringJunky 2 Link to comment Share on other sites More sharing options...
Strange Posted December 2, 2016 Share Posted December 2, 2016 I stand by my original statement. I am not ready to throw the baby out with the bath water. To say any part of his theory is incomplete is to say it is wrong. I am just saying just because we may not understand his singularity doesn't mean his theory is incomplete (wrong) Newtonian gravity is incomplete, but it isn't wrong. Link to comment Share on other sites More sharing options...
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