DrmDoc Posted January 7, 2016 Posted January 7, 2016 Is dark energy considered the opposing force of both gravity and electromagnetism? As the theorized force behind the increasingly rapid expansion of our universe and eventual destroyer of atomic cohesion, can dark energy be viewed as an antimatter-like force opposite to gravity and electromagnetism?
Mordred Posted January 7, 2016 Posted January 7, 2016 (edited) No has nothing to do with electromagnetism. It also isn't a form of antimatter or antigravity. The best way to think of it is as a vacuum pressure. [latex]w=\frac{\rho}{p}[/latex] The equation of state of the cosmological constant is w=-1. https://en.m.wikipedia.org/wiki/Equation_of_state_(cosmology) Antimatter has the same equation of state as it's matter partner. So for example antiphotons has the same EoS as photons w=1/3 (ultra relativistic matter). On that same link take note of the acceleration equation Edited January 7, 2016 by Mordred
DrmDoc Posted January 7, 2016 Author Posted January 7, 2016 No has nothing to do with electromagnetism. It also isn't a form of antimatter or antigravity. The best way to think of it is as a vacuum pressure. [latex]w=\frac{\rho}{p}[/latex] The equation of state of the cosmological constant is w=-1. https://en.m.wikipedia.org/wiki/Equation_of_state_(cosmology) Antimatter has the same equation of state as it's matter partner. So for example antiphotons has the same EoS as photons w=1/3 (ultra relativistic matter). On that same link take note of the acceleration equation If I understand correctly, dark energy is conceived as a pressure-like or inflationary force filling the universe rather than some matter-intervening field of energy? I appreciate your continued indulgence.
Mordred Posted January 7, 2016 Posted January 7, 2016 (edited) More accurately it's modelled as a scalar field. For example one hypothesis suggests that the scalar field may be the Higgs field. More research is under way to validate that though. Energy being a property of particles. Doesn't exist on its own. So typically their is a particle involved. Inflation uses the inflaton. Many models of the cosmological constant borrow the inflaton to try to explain dark energy. Particles inertia interact with other particles. Those interactions causes pressure. The more energetic the particles the greater the pressure influence. This is essentially what's known as an adiabatic fluid or gas. Cosmology treats the universe and models the universe as an ideal gas. Which works extremely well. The cosmological constant is also treated as such. Even in the form of virtual particle production one can still model the system accordingly. The deeper one looks into the equations (including GR). The more aware one becomes of how intertwined the gas laws are in the FLRW and Einstein field equations. I've yet to see or read a model that's considered valid that doesn't include the ideal gas laws The scalar equation of state formulas are on that link. Those are the same equations used to model Inflation. The Higgs field and often the cosmological constant. I mentioned GR. You can see the ideal gas laws application in the energy/mass density to pressure relationships involved in the Einstein field equations. Specifically the the stress energy/momentum equation. [latex]T^{\mu\nu}=(\rho+p)U^{\mu}U^{\nu}+p\eta^{\mu\nu}[/latex] [latex]\rho[/latex] being the energy density. You can further understand the FLRW metric through the following. http://cosmology101.wikidot.com/universe-geometry This one I wrote myself it's more of a guideline training aid. These are more textbook style articles with one being a free textbook. ) http://arxiv.org/pdf/hep-ph/0004188v1.pdf:"ASTROPHYSICS AND COSMOLOGY"- A compilation of cosmology by Juan Garcıa-Bellido http://arxiv.org/abs/astro-ph/0409426An overview of Cosmology Julien Lesgourgues http://arxiv.org/pdf/hep-th/0503203.pdf"Particle Physics and Inflationary Cosmology" by Andrei Linde http://www.wiese.itp.unibe.ch/lectures/universe.pdf:"Particle Physics of the Early universe" by Uwe-Jens Wiese Thermodynamics, Big bang Nucleosynthesis The last article uses the same format and in many cases formulas that are found in Modern Cosmology by Scott Dodelson. In particular chapters 3 and 4 covering nucleosynthesis. (If I get a chance tonight I'll post how an equation of state is calculated, for matter and radiation) Edited January 8, 2016 by Mordred 1
Mordred Posted January 8, 2016 Posted January 8, 2016 (edited) First take the first law of thermodynamics. [latex]dU=dW=dQ[/latex] U is internal energy W =work. As we dont need heat transfer Q we write this as DW=Fdr=pdV Which leads to dv=-pdF. [latex]V=r^3[/latex] Through several steps we will end up with. [latex]\rho=-3(\rho+p)\frac{\dot{r}}{r}[/latex] We will use the last formula for both radiation and matter. Assuming density of matter [latex]\rho=\frac{M}{4/3\pi r^3}[/latex] [latex]\rho=\frac{dp}{dr}[/latex] [latex]\dot{r}=3\rho \frac{\dot{r}}{r}[/latex] Using the above equation the pressure due to matter gives an Eos of Pressure=0. Which makes sense as matter doesn't exert a lot of kinetic energy/momentum. For radiation we will need some further formulas. Visualize a wavelength as a vibration on a string. [latex]L=\frac{N\lambda}{2}[/latex] As we're dealing with relativistic particles [latex]c=f\lambda=\frac{2L}{N}[/latex] [latex]U=\hbar w=hf[/latex] [latex]U=\frac{Nhc}{2}\frac{1}{L}\propto V^{-\frac{1}{3}}[/latex] Using the formula above. [latex]p=1/3\rho[/latex] for ultra relativistic radiation. Those are examples of how the first law of thermodynamics fit within the equations of state. There is more intensive formulas involved. In particular the Bose-Einstein statistics and Fermi-Dirac statistics but the above serves as a good approximation. Edited January 8, 2016 by Mordred 3
DrmDoc Posted January 8, 2016 Author Posted January 8, 2016 (edited) Most eloquently explained; I have a lot to study and I greatly appreciate all the insight you've shared, thank you. NOTE TO MODERATORS: Please remove the negative rating on the previous post. I mistakenly selected the wrong arrow to register my approval of Mordred's post. I don't normally rate posts but was moved to do so because of his very helpful efforts. Thank you. Edited January 8, 2016 by DrmDoc
Carrock Posted January 21, 2016 Posted January 21, 2016 This post was quoted on another thread by Mordred Beyond a demonstration that the 'right' answer need not depend on correct equations I don't see its relevance there so I'm responding to it here before looking at the rest of his post. It's quite simple to spot many of the errors but I haven't made much attempt to spot the more subtle ones. First take the first law of thermodynamics. [latex]dU=dW=dQ[/latex] U is internal energy W =work. As we dont need heat transfer Q we write this as DW=Fdr=pdV Which leads to dv=-pdF. dv=-pdF fails dimensional analysis. [latex]V=r^3[/latex]Implicitly requiring the volume to be cubic or irregular is problematic initially and would require your next correct equation to be changed. Perhaps you meant [latex]V={\frac{4}{3}}\pi r^3[/latex]? Why bother defining it when you don't bother defining most of your other terms? Through several steps we will end up with. [latex]\rho=-3(\rho+p)\frac{\dot{r}}{r}[/latex] [latex]\rho=-3(\rho+p)\frac{\dot{r}}{r}[/latex] fails dimensional analysis. We will use the last formula for both radiation and matter. Assuming density of matter [latex]\rho=\frac{M}{4/3\pi r^3}[/latex] [latex]\rho=\frac{dp}{dr}[/latex] [latex]\rho=\frac{M}{4/3\pi r^3}[/latex] I expect it's OK but I've never been that strong on explicit and implicit operator precedence. Why not put [latex]\rho=\frac{3M}{4\pi r^3} [/latex] or at least avoid using different styles in the same equation [latex]\rho=\frac{M}{\frac{4}{3}\pi r^3} [/latex]? [latex]\rho=\frac{dp}{dr}[/latex] looks wrong but with no context it's not obvious. [latex]\dot{r}=3\rho \frac{\dot{r}}{r}[/latex][latex]\dot{r}=3\rho \frac{\dot{r}}{r}[/latex] fails dimensional analysis. Using the above equation the pressure due to matter gives an Eos of Pressure=0. Which makes sense as matter doesn't exert a lot of kinetic energy/momentum. For radiation we will need some further formulas. Visualize a wavelength as a vibration on a string. [latex]L=\frac{N\lambda}{2}[/latex] As we're dealing with relativistic particles [latex]c=f\lambda=\frac{2L}{N}[/latex] [latex]c=f\lambda=\frac{2L}{N}[/latex] fails dimensional analysis. [latex]U=\hbar w=hf[/latex] [latex]U=\frac{Nhc}{2}\frac{1}{L}\propto V^{-\frac{1}{3}}[/latex] Using the formula above. [latex]p=1/3\rho[/latex] for ultra relativistic radiation. Those are examples of how the first law of thermodynamics fit within the equations of state. There is more intensive formulas involved. In particular the Bose-Einstein statistics and Fermi-Dirac statistics but the above serves as a good approximation. With so many errors and so few definitions and intermediate steps it's very hard to follow this post. To quote the expert (lol this post will likely cause difficulty)I only used dimensional analysis and only for the obvious errors so there may well be other errors. (Having dimensional homogeneity is a necessary but not sufficient requirement for an equation's validity.) The thought of having to fully check every relevant equation in the rest of Mordred's post on the other thread is rather daunting and I am wondering if there's any point. Halfway through this post I lost the will to post but I eventually decided to finish it so I wouldn't have wasted my time completely. If the errors here are not fixed and much more detail provided I would recommend reading these cosmology lectures which cover extremely similar ground in a much clearer and more accurate way. Ok lets go through this. First I posted pv=nRt. It's a basic (very basic thermodynamic equation). It Doesn't get used much in Cosmology papers on arxiv simply because we use another form. I tend to use pv=nRt as many that post on this forum are weak in thermodynamic laws. Let alone mathematical Cosmology.(lol this post will likely cause difficulty) (The formula shows the relevant relations without too much complexity)Somewhere along the way I forgot that my objection was the inappropriate use of classical physics to derive cooling of the CMBR. Using pv=nRt is perfectly valid. So in terms of admitted mistakes Mordred is winning with fewer than me. 2
Mordred Posted January 21, 2016 Posted January 21, 2016 (edited) Carrock you obviously have some difficulty seeing the ideal gas law relations involved. Might I suggest picking up a textbook.... In one of the other posts I posted you a few references. You obviously didn't read... Instead you choose to bounce through posts with a rather rude format. So I will respond in kind. 1) is not the term equation of state not a thermodynamic (statistical mechanic terminology? "In physics and thermodynamics, an equation of state is a relation between state variables.[1] More specifically, an equation of state is a thermodynamic equation describing the state of matter under a given set of physical conditions. It is a constitutive equation which " https://en.m.wikipedia.org/wiki/Equation_of_state Why do you think the terms "perfect fluid and ideal gas laws are used in Cosmology applications? They are after all classical statistical mechanic terminology So you explain why Cosmology involves equations of state without classical thermodynamic relations. Go on give it a whirl I could use a good laugh. https://en.m.wikipedia.org/wiki/Equation_of_state_(cosmology) I explained in that other thread that I often post pv=nRt as it's a simple thermodynamic relationships. I choose the simple form as many cannot comprehend the advanced forms. Personally what mathematical forms I choose to use is my choice provided its accurate. Instead of responding in the other thread my reply to you. You choose to attack a difference thread. PS there is nothing wrong with posting a previous post in another thread. It's a time saver so I don't have to do a bunch of latex I posted you numerous formulas showing the terms pressure and energy-density. Whenever you have ANY system with pressure you WILL have temperature. The More energetic or kinetic energy the particles has the greater it's energy density to pressure ratio This is basic physics. No matter how many references I've posted to you in the past you've ignored seeing the ideal gas law relations. So do your own search. Google "Ideal gas laws cosmology pdf" Then Google "thermodynamics cosmology pdf" Edited January 21, 2016 by Mordred 1
Mordred Posted January 22, 2016 Posted January 22, 2016 (edited) Maybe if you see the pv=nRt used in a point slide lecture note you will learn to listen. http://www.google.ca/url?sa=t&source=web&cd=2&rct=j&q=equation%20of%20state%20cosmology%20pv%3DnRt%20pdf.&ved=0ahUKEwjt3tGxnrzKAhXrn4MKHd8KC9QQFggbMAE&url=http%3A%2F%2Fwww.ir.isas.jaxa.jp%2F~cpp%2Fteaching%2Fcosmology%2Fdocuments%2Fcosmology01-05.pdf&usg=AFQjCNFB61v07XFKiEK2Oe9wYZmwX8d0xg&sig2=_7QjNO2IeKppzI6jnDMhBw "Fundamentals of Cosmology 5, the equations of state" page 7. By the way in case your not aware [latex]pv=nRt=nkt[/latex]. The latter being an adiabatic fluid. Edited January 22, 2016 by Mordred
Mordred Posted January 22, 2016 Posted January 22, 2016 (edited) You know the funny part is you kept posting fails dimensional analysis with a 57 page book on dimensional analysis without reference to any specifics. Nor did you state why it fails dimensional analyses. However no matter there was some mistakes. First equation noted should have been [latex]dU=-pdf[/latex] I did make a mistake on this equation in forgetting an overdot. [latex]\rho=-3(\rho+p)\frac{\dot{r}}{r}[/latex] Should be [latex]\dot{\rho}=-3(\rho+p)\frac{\dot{r}}{r}[/latex] So how was this derived? [latex]dU=-\rho dV[/latex] [latex]U=\rho V[/latex] [latex]\dot{U}=\dot{\rho}V+{\rho}\dot{V}=-p\dot{V}[/latex] [latex]V\propto r^3[/latex] [latex]\frac{\dot{V}}{V}=3\frac{\dot{r}}{r}[/latex] Which leads to [latex]\dot{\rho}=-3(\rho+p)\frac{\dot{r}}{r}[/latex] If I choose to latex this as [latex]\rho=\frac{M}{4/3\pi r^3}[/latex] that's my choice not yours. Anyone can see it's the same as [latex]\rho=\frac{M}{\frac{4}{3}\pi r^3}[/latex] Missed a term [latex]\rho=\frac{dp}{dr}\dot{r}[/latex] and missed the minus sign [latex]\dot{r}=-3\rho \frac{\dot{r}}{r}[/latex] Last noted equation I missed a term. Comes from posting from a phone at 2am. [latex]c=f\lambda=f\frac{2L}{N}[/latex] Next time try exercising a little diplomacy. I don't mind mistakes pointed out in my latex or posted formulas. However you can keep the attitude about it at home. Thanks for pointing out the errors. Next try a little diplomacy. PS I'm sorry you have a problem with thermodynamic applications in Cosmology. However that is the way it is. Edit: seeing as you did find the errors I'll grant a +1 Edited January 22, 2016 by Mordred 3
Carrock Posted January 23, 2016 Posted January 23, 2016 (edited) You know the funny part is you kept posting fails dimensional analysis with a 57 page book on dimensional analysis without reference to any specifics. Nor did you state why it fails dimensional analyses. However no matter there was some mistakes.I only intended the reference once, but used cut and paste without checking. In one of the other posts I posted you a few references. You obviously didn't read...As you expect me to read your references without specifying which part, I thought you would be happy to do the same or at least google "dimensional analysis". I looked at a few references and didn't find any suitable examples. I only use it as an informal quick check of equations and I thought you would want a formal description. I was taught it in an off the syllabus end of term maths class at school and don't know or care if I'm using it correctly as long as it gets results. This is problematic when I refer to it here. First equation noted should have been [latex]dU=-pdf[/latex] Informal worked examples of dimensional analysis using all my knowledge:-) .. There may be errors. [latex]dU=-pdf[/latex] Since dy/dt in this context is equivalent to y/t, multiply both sides by t; set sign positive (also ok). then L.H.S. = energy R.H.S.= pressure x force In S.I. units L.H.S. = kg*m2*s-2 R.H.S. = (kg*m-1*s-2) * (kg*m*s−2) = kg2*s-4 The dimensional inequality is sufficient to show the equation is not valid. Also useful: exponents must be dimensionless. Worked example: [latex]V=V_0 e^{-\frac{t}{RC}}[/latex] [latex]\frac{t}{RC}[/latex] ie [latex]\frac {time}{resistance*capacitance}[/latex] must be dimensionless. In S.I. units [latex]\frac {time}{resistance*capacitance} = \frac{s}{(kg*m^{2}*s^{-3}*A^{-2})*(kg^{-1}*m^{-2}*s^4*A^2)} =1[/latex] Exponent is dimensionless. I did make a mistake on this equation in forgetting an overdot. [latex]\rho=-3(\rho+p)\frac{\dot{r}}{r}[/latex] Should be [latex]\dot{\rho}=-3(\rho+p)\frac{\dot{r}}{r}[/latex] So how was this derived? [latex]dU=-\rho dV[/latex] [latex]U=\rho V[/latex] [latex]\dot{U}=\dot{\rho}V+{\rho}\dot{V}=-p\dot{V}[/latex] [latex]V\propto r^3[/latex] [latex]\frac{\dot{V}}{V}=3\frac{\dot{r}}{r}[/latex] Which leads to [latex]\dot{\rho}=-3(\rho+p)\frac{\dot{r}}{r}[/latex] If I choose to latex this as [latex]\rho=\frac{M}{4/3\pi r^3}[/latex] that's my choice not yours. Anyone can see it's the same as [latex]\rho=\frac{M}{\frac{4}{3}\pi r^3}[/latex] Missed a term [latex]\rho=\frac{dp}{dr}\dot{r}[/latex] and missed the minus sign [latex]\dot{r}=-3\rho \frac{\dot{r}}{r}[/latex] It's often possible to find dimensional errors by inspection if the equation is simplified e.g. [latex]\dot{r}=-3\rho \frac{\dot{r}}{r}[/latex] solving for [latex]\rho [/latex]: [latex]\rho = -\frac{r}{3}[/latex] which 'looks' as well as is wrong. Perhaps you meant [latex]\dot{\rho}=-3\rho \frac{\dot{r}}{r}[/latex] Edited January 23, 2016 by Carrock 1
Mordred Posted January 23, 2016 Posted January 23, 2016 (edited) You didn't catch the full correction on the first equation. Completely replace dv=pdf with DU=pdv. It's understandable your confused there. (For some dumb reason I typed f again Grr).{dumb spell check on phone} [latex]DU=pdV[/latex]. Here it should read as follows with the corrections. I will add a few details. First take the first law of thermodynamics. [latex]dU=dW=dQ[/latex] U is internal energy W =work. As we dont need heat transfer Q we write this as [latex]DW=Fdr=pdV[/latex] Which leads to [latex]dU=-pdV.[/latex]. Which is the first law of thermodynamics for an ideal gas. [latex]U=\rho V[/latex] [latex]\dot{U}=\dot{\rho}V+{\rho}\dot{V}=-p\dot{V}[/latex] [latex]V\propto r^3[/latex] [latex]\frac{\dot{V}}{V}=3\frac{\dot{r}}{r}[/latex] Which leads to [latex]\dot{\rho}=-3(\rho+p)\frac{\dot{r}}{r}[/latex] We will use the last formula for both radiation and matter. Assuming density of matter [latex]\rho=\frac{M}{\frac{4}{3}\pi r^3}[/latex] [latex]\rho=\frac{dp}{dr}\dot{r}=-3\rho \frac{\dot{r}}{r}[/latex] Using the above equation the pressure due to matter gives an Eos of Pressure=0. Which makes sense as matter doesn't exert a lot of kinetic energy/momentum. For radiation we will need some further formulas. Visualize a wavelength as a vibration on a string. [latex]L=\frac{N\lambda}{2}[/latex] As we're dealing with relativistic particles [latex]c=f\lambda=f\frac{2L}{N}[/latex] substitute [latex]f=\frac{n}{2L}c[/latex] into Plancks formula [latex]U=\hbar w=hf[/latex] [latex]U=\frac{Nhc}{2}\frac{1}{L}\propto V^{-\frac{1}{3}}[/latex] Using [latex]dU=-pdV[/latex] using [latex]p=-\frac{dU}{dV}=\frac{1}{3}\frac{U}{V}[/latex] As well as [latex]\rho=\frac{U}{V}[/latex] leads to [latex]p=1/3\rho[/latex] for ultra relativistic radiation. Those are examples of how the first law of thermodynamics fit within the equations of state. There is more intensive formulas involved. In particular the Bose-Einstein statistics and Fermi-Dirac statistics but the above serves as a good approximation. That should make more sense now again I apologize for the errors above and thanks again on the assistance in correcting. (You will note I added some missing details to assist) A further+1 for the last post. It was well thought out and polite. Thanks Edited January 23, 2016 by Mordred
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