Markus Hanke 493 Posted June 9, 2020 Share Posted June 9, 2020 7 hours ago, Mordred said: This link is earlier in this thread. Unruh came up with an interesting solution. Just thought I would post here for you Joigus. https://arxiv.org/pdf/1703.00543.pdf Very interesting, thanks for sharing it! I hadn’t heard about this paper before, this seems a very reasonable approach to the issue. Link to post Share on other sites

stephaneww 19 Posted June 9, 2020 Author Share Posted June 9, 2020 (edited) 16 hours ago, swansont said: And there’s G, in equation 2. It’s related to gravity. Yes, but the vacuum catastrophe is dimensionless. 16 hours ago, swansont said: Equation 1 has no physical significance, AFAIK No, it's Planck's energy (J: kg.m^2.t^-2) I made a mistake and corrected it later : 18 hours ago, stephaneww said: ooops error 20 hours ago, stephaneww said: mp.l2pt3 (1) vaccum catastrophe = (1)/(2) (1)=mp.lp.tp2 is right (1) [math]\frac{m_p}{l_p.t_p^2}[/math] (J/m^3) is right 16 hours ago, swansont said: Show an equation that has F^2 in it, and include the derivation I don't have another equation with F^2 I only have E^2=m^2c^4+p^2.c^2 to propose as a comparable example of a square. 16 hours ago, swansont said: Where did the equation come from? from a comparison of values in a spreadsheet where I have a whole series of physical values and relationships between them. 16 hours ago, swansont said: You have to be careful assigning physical relevance when all you’re doing is rewriting constants. Yes, but I also have no idea of the physical relevance of E^2. If you do, I'm interested Edited June 9, 2020 by stephaneww Link to post Share on other sites

swansont 7484 Posted June 9, 2020 Share Posted June 9, 2020 23 minutes ago, stephaneww said: Yes, but the vacuum catastrophe is dimensionless. I don’t see why this matters. Quote No, it's Planck's energy (J: kg.m^2.t^-2) No physical significance, as in, it’s not an interaction. Not a kind of energy. It’s a scale where quantum effects of gravity would be important Quote I don't have an equation with F^2 Then why is there a question about a physical significance? Quote I only have E^2=m^2c^4+p^2.c^2 to propose as a comparable example of a square. from a comparison of values in a spreadsheet where I have a whole series of physical values and relationships between them. Yes, but I also have no idea of the physical relevance of E^2. If you do, I'm interested Sometimes you express equations in a way that makes them easier to use. You can take the Newtonian expressions and write KE = p^2/2m There’s no physical significance to p^2. It’s just how the algebra works out when you substitute some variables. In the case of the energy equation it may be to avoid using the square root Link to post Share on other sites

stephaneww 19 Posted June 9, 2020 Author Share Posted June 9, 2020 (edited) 1 hour ago, swansont said: 1 hour ago, stephaneww said: Yes, but the vacuum catastrophe is dimensionless. I don’t see why this matters. You can't compare G to another value, it's just a data in the equations, whereas you can compare energy densities by volume, and that's the problem with the vacuum catastrophe. 1 hour ago, swansont said: Then why is there a question about a physical significance? You mean it's not important ? 1 hour ago, swansont said: In the case of the energy equation it may be to avoid using the square root Thank you, I didn't know that. Edited June 9, 2020 by stephaneww Link to post Share on other sites

swansont 7484 Posted June 9, 2020 Share Posted June 9, 2020 2 hours ago, stephaneww said: You can't compare G to another value, it's just a data in the equations, whereas you can compare energy densities by volume, and that's the problem with the vacuum catastrophe. I’m not. But not having units does not detach a term from a particular phenomenon. Gamma (1/sqrt(1-v^2/c^2) is unitless. It’s still intimately tied to relativity As a general rule, if G shows up in an equation, that equation has something to do with gravity. If h-bar shows up, it’s related to quantum mechanics. G and h-bar are not “just data” Quote You mean it's not important ? Since you say it doesn’t exist, I’d guess the answer is that it’s not. Link to post Share on other sites

stephaneww 19 Posted June 9, 2020 Author Share Posted June 9, 2020 (edited) 52 minutes ago, swansont said: As a general rule, if G shows up in an equation, that equation has something to do with gravity. If h-bar shows up, it’s related to quantum mechanics. G and h-bar are not “just data” Quote You mean it's not important ? Since you say it doesn’t exist, I’d guess the answer is that it’s not. Okay, but what is your point of view about : [math]\frac{m_p}{l_p t_p^2}=\frac{c^7}{G^2 \hbar}[/math] in this case ? specifically what is G^2 or c^7 physically ? .. then the cosmological constant problem can be write as : [math]\frac{8\pi.c^3.G}{\Lambda. \hbar}=8.73*10^{122}[/math] Thank you for questioning me. It stimulates me. Edited June 9, 2020 by stephaneww Link to post Share on other sites

swansont 7484 Posted June 9, 2020 Share Posted June 9, 2020 41 minutes ago, stephaneww said: Okay, but what is your point of view about : mplpt2p=c7G2ℏ in this case ? specifically what is G^2 or c^7 physically ? There is no physical significance that I’m aware of. You have mass/length*time^2 What is that supposed to represent? What physics equation does that come from? Link to post Share on other sites

stephaneww 19 Posted June 9, 2020 Author Share Posted June 9, 2020 (edited) 25 minutes ago, swansont said: You have mass/length*time^2 What is that supposed to represent? What physics equation does that come from? It is the energy density per volume in Planck units (Ep/lp^3 , J/m^3) in the QM This is a theoretical vacuum energy in QM [math]\frac{c^4 \Lambda}{8\pi .G}[/math] is the observed value, The cosmological constant problem is the mismatch between the two with a ratio of 10^122. https://en.wikipedia.org/wiki/Cosmological_constant_problem Edited June 9, 2020 by stephaneww Link to post Share on other sites

Mordred 1372 Posted June 9, 2020 Share Posted June 9, 2020 4 hours ago, stephaneww said: Okay, but what is your point of view about : mplpt2p=c7G2ℏ in this case ? specifically what is G^2 or c^7 physically ? .. then the cosmological constant problem can be write as : 8π.c3.GΛ.ℏ=8.73∗10122 Thank you for questioning me. It stimulates me. I really don't know how your getting your numbers particularly on the powers. No cosmological equation I have ever encountered gives [math] c^7[/math] for example. I highly recommend step back and refresh your memory on the relevant cosmological constant equations. It looks to me as your following the wrong garden path on key relations but I cannot discern where the error is arising from. Link to post Share on other sites

stephaneww 19 Posted June 9, 2020 Author Share Posted June 9, 2020 (edited) Hi Mordred, Apart from the dimensional problems encountered and attempts to introduce the notions of force or electromagnetism (I'll come back to that later) , we can be satisfied with this reminder: https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1115799 [math]c^7[/math] appears naturally when we rewrite mp/(lp.tp^2) with the fundamental constants ______________________________________________ [math]l_p^2=\frac{G\hbar}{c^3}[/math] [math]m_p^2=\frac{c\hbar}{G}[/math] [math]t_p^2=\frac{G\hbar}{c^5}[/math] ( mp/(lp.tp^2) )^2 = [math]\frac{c^{14}}{G^4.\hbar^2}[/math] _______________________________________________ I have a begining of idea : in [math] A=\frac{ m_p}{l_p.t_p^2}=\frac{c^7}{G^2.\hbar}[/math] we have [math]G^2[/math], a QM approch of energy density with value of gravitation ? in [math]B=\frac{\hbar.\Lambda_{m^{-2}}^2.c}{(8\pi)^2}[/math] we have [math]\Lambda^2[/math], a QM approch of energy density with value of dark energy ? [math]B[/math] came from the same message where I suggest a mathematic solution of the cosmological constant problem https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1115799 or is ot only a circular raisonnement ? Edited June 9, 2020 by stephaneww Link to post Share on other sites

Mordred 1372 Posted June 10, 2020 Share Posted June 10, 2020 (edited) Well quite frankly it doesn't make sense to mp/(lp.tp^2). Let's start with the first two terms. How does it make sense to take Planck mass ( mass is resistance to inertia change) regardless of being a Planck unit in this case and divide that by a minimal Planck length ? Not to mention the Planck time unit. ? Time being a rate of change. Sorry but that simply doesn't make any applicable sense. PS I'm assuming your using the [math]\cdot[/math] for the multiply operand rather than the cross product. As I am aware your not familiar with inner and cross products in vector notation though that wouldn't make a difference in this case. The logic in either case escapes me. It makes zero sense to take mass and divide or even multiply by length even with time as part of the operation. Edited June 10, 2020 by Mordred Link to post Share on other sites

stephaneww 19 Posted June 10, 2020 Author Share Posted June 10, 2020 (edited) I relied on the equality (1) of this document: https://arxiv.org/ftp/physics/papers/0611/0611115.pdf https://royalsocietypublishing.org/doi/full/10.1098/rspa.2007.0370 is that a nonsense? Edited June 10, 2020 by stephaneww Link to post Share on other sites

Mordred 1372 Posted June 10, 2020 Share Posted June 10, 2020 Your application is incorrect neither of those documents tells you mass is divisible by (length multiplied by time). You need to study and pay attention to what each of those terms represent and how they are defined. Mass isn't associated with any geometric length nor time. Link to post Share on other sites

stephaneww 19 Posted June 10, 2020 Author Share Posted June 10, 2020 (edited) Uh-huh: ε_{p}=m_{p }c^2/l_{p}^3 c^2=l_{p}^2/t_{p}^2 so m_{p} c^2/lp^3 = m_{p}/ (l_{p} t_{p}^2) , right? thank you for your requests, it obliges me to bring clarifications that should have been made from the start 1 hour ago, Mordred said: PS I'm assuming your using the ⋅ for the multiply operand rather than the cross product. As I am aware your not familiar with inner and cross products in vector notation though that wouldn't make a difference in this case. The logic in either case escapes me. It makes zero sense to take mass and divide or even multiply by length even with time as part of the operation. Yes, I use multiplication. Sorry I didn't say it before. I'll avoid the "." in the future so there won't be any confusion. Edited June 10, 2020 by stephaneww Link to post Share on other sites

stephaneww 19 Posted June 10, 2020 Author Share Posted June 10, 2020 (edited) sorry DeepL gave me a mistranslation for "Uh-huh:", I meant "Uh... (Clears throat)" Edited June 10, 2020 by stephaneww Link to post Share on other sites

swansont 7484 Posted June 10, 2020 Share Posted June 10, 2020 15 hours ago, stephaneww said: It is the energy density per volume in Planck units (Ep/lp^3 , J/m^3) in the QM This is a theoretical vacuum energy in QM That does not follow. Planck units are not a QM prediction. Link to post Share on other sites

stephaneww 19 Posted June 10, 2020 Author Share Posted June 10, 2020 (edited) 13 hours ago, swansont said: That does not follow. Planck units are not a QM prediction. Same answer as for Mordred : 22 hours ago, stephaneww said: I relied on the equality (1) of this document: https://arxiv.org/ftp/physics/papers/0611/0611115.pdf https://royalsocietypublishing.org/doi/full/10.1098/rspa.2007.0370 is that a nonsense? (Ep=mp c^2) Edited June 11, 2020 by stephaneww Link to post Share on other sites

Mordred 1372 Posted June 11, 2020 Share Posted June 11, 2020 (edited) In the first article do you understand what a mathematical coincidence is ? Let's define that term. "A mathematical coincidence is said to occur when two expressions with no direct relationship show a near-equality which has no apparent theoretical explanation " now given that definition does the first article make more sense when the author keeps referring to the pure numbers and the coincidence between the pure numbers ? https://en.m.wikipedia.org/wiki/Mathematical_coincidence Edited June 11, 2020 by Mordred Link to post Share on other sites

stephaneww 19 Posted June 11, 2020 Author Share Posted June 11, 2020 (edited) which article are you referring to, please? https://arxiv.org/ftp/physics/papers/0611/0611115.pdf https://royalsocietypublishing.org/doi/full/10.1098/rspa.2007.0370 are the same paper I understand the definition of mathematical coincidence. It's not the question of numbers in 10^120 wich interrested me but the way to calculate the cosmolgical constant problem do by the equalty (1) The definition of wikipedia being too vague, I looked for a more precise definition. Besides, I couldn't find any scientific discussion on the rest of the paper. (the coincidence between these numbers) a priori this is probably not relevant while the subject of the cosmological constant problem remains open Edited June 11, 2020 by stephaneww Link to post Share on other sites

Mordred 1372 Posted June 11, 2020 Share Posted June 11, 2020 (edited) Well your going to have to be clearer on which cosmological problem your addressing. The cosmological constant has several problems associated with it. The problem you have been working on in the past is the problem of why the observed values are so small. Much smaller than the 10^120 originally predicted value. The other problem is the Coincidence problem which amounts to "why now" do the DM and DE values seem so close. The latter related to the fine tuning problem. The first link doesn't address either of these two problems. The second link is the same paper just on a different site. Thise links does you no good for the first problem. It doesn't even attempt to address the problem described in the wiki link https://en.wikipedia.org/wiki/Cosmological_constant_problem Edited June 11, 2020 by Mordred Link to post Share on other sites

stephaneww 19 Posted June 11, 2020 Author Share Posted June 11, 2020 (edited) 1 hour ago, Mordred said: Well your going to have to be clearer on which cosmological problem your addressing. The cosmological constant has several problems associated with it. The problem you have been working on in the past is the problem of why the observed values are so small. Much smaller than the 10^120 originally predicted value. It is indeed this latter question that I am trying to answer. I don't know what DM and DE is (edit: Dark Matter and Dark Ernergy ?), probably, it doesn't matter in the context of what I'm trying to address. The paper I have linked is not trying to solve it, he's just stating it in simple terms with equality (1). That's what I thought. Can the problem be approached in this way, which also poses the terms of the problem: https://en.wikipedia.org/wiki/Vacuum_energy with [math]E =\frac{1}{2} h \nu = E_p=m_p c^2[/math] if I don't make a mistake (I'm not sure about these equalities) and [math] E_p/l_p^3=m_p c^2/l_p^3[/math] to obtain the value of [math]10^{113} J/m^3[/math] That's what I was trying to say before. I hope it will be clearer Edited June 11, 2020 by stephaneww Link to post Share on other sites

swansont 7484 Posted June 11, 2020 Share Posted June 11, 2020 The problem I see is that Planck units are a unit system https://en.wikipedia.org/wiki/Planck_units#Definition “the system is internally coherent. For example, the gravitational attractive force of two bodies of 1 Planck mass each, set apart by 1 Planck length is 1 coherent Planck unit of force. Likewise, the distance traveled by light during 1 Planck time is 1 Planck length.” Notice how it’s not referencing any actual particles, or measured phenomena I don’t understand why anyone is ascribing more significance. It doesn't seem to justified anywhere. You can do your exercise with other units. The SI energy density is 1 Joule/m^3.The cgs energy density is 1 erg/cm^3, which is not the same - the differ by a factor of 10. That’s OK, though, because there’s no physical significance here. I have seen reference to the vacuum energy calculation from the Casimir effect, and from a particle in a box. Those have QM significance. Link to post Share on other sites

Mordred 1372 Posted June 11, 2020 Share Posted June 11, 2020 I agree with Swansont on needing the SI units to delve into the cosmological problem. You should also avoid normalized units. You need the higher precision to equate to the extremely small value of the cosmological constant. Link to post Share on other sites

Endy0816 454 Posted June 11, 2020 Share Posted June 11, 2020 2 hours ago, swansont said: The problem I see is that Planck units are a unit system https://en.wikipedia.org/wiki/Planck_units#Definition “the system is internally coherent. For example, the gravitational attractive force of two bodies of 1 Planck mass each, set apart by 1 Planck length is 1 coherent Planck unit of force. Likewise, the distance traveled by light during 1 Planck time is 1 Planck length.” Notice how it’s not referencing any actual particles, or measured phenomena I don’t understand why anyone is ascribing more significance. It doesn't seem to justified anywhere. You can do your exercise with other units. The SI energy density is 1 Joule/m^3.The cgs energy density is 1 erg/cm^3, which is not the same - the differ by a factor of 10. That’s OK, though, because there’s no physical significance here. I have seen reference to the vacuum energy calculation from the Casimir effect, and from a particle in a box. Those have QM significance. Yeah, often incorrectly thought of as the be all end all, when in some cases you can have even smaller measurements. Link to post Share on other sites

stephaneww 19 Posted June 11, 2020 Author Share Posted June 11, 2020 (edited) 4 hours ago, swansont said: The problem I see is that Planck units are a unit system The problem goes beyond that for all I understand. But first of all, there may be a Pi factor missing from my proposal, or it needs to be added for it to be correct: [math]E_\text{Vacuum Energy}=1/2 h\nu=???? =E_p=m_pc^2[/math] I think so, but I have a doubt about the value of [math]\nu[/math] that I don't know. Can you tell me the value, formula, and unit of [math]\nu[/math] please? https://en.wikipedia.org/wiki/Vacuum_energy In wikipedia link you also find the definition of the problem : Quote However, in both quantum electrodynamics (QED) and stochastic electrodynamics (SED), consistency with the principle of Lorentz covariance and with the magnitude of the Planck constant suggest a much larger value of 10^113 joules per cubic noted by me : (I) and : Quote Using the upper limit of the cosmological constant, the vacuum energy of free space has been estimated to be 10^−9 joules per cubic meter. noted by me : (II) Quote This huge discrepancy is known as the cosmological constant problem. ____________________________________ You can change the value of [math]h[/math] and [math]\nu[/math] if you want, you'll have always a spread (that you'll call cosmologycal constant problem with another value) between (I) and (II), excepted in rare cases. The problem is that (I) don't match (II) I have checked, indeed, my proposal works in all cases as long as the values given to Planck's units are consistent with the formulas that govern them. It makes sense when my proposal is formulated in terms of fundamental constants as I do in this quote below : On 6/10/2020 at 1:09 AM, stephaneww said: I have a begining of idea : in A=mplp.t2p=c7G2.ℏ we have G2 , a QM approach of energy density with value of gravitation ? in B=ℏ.Λ2m−2.c(8π)2 we have Λ2 , a QM approach of energy density with value of dark energy ? B came from the same message where I suggest a mathematic solution of the cosmological constant problem https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1115799 or is ot only a circular raisonnement ? [math]C=\sqrt{A}\sqrt{B}[/math] = value of density energy of cosmological constant. (note hbar disappears with the multiplication A B square root) 3 hours ago, Mordred said: I agree with Swansont on needing the SI units to delve into the cosmological problem. it's what I've always done. Edited June 11, 2020 by stephaneww Link to post Share on other sites

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