Mordred

Thanks Beefpatty, just making sure I couldn't think of any reason why the B-E wouldn't be appropriate. Much like the papers you have posted the only examples I've usually read deals with the Higg's in terms of its TeV metastabiltiy above 1011 GeV. I have easily been able to find papers dealing with the Higg's at high temperature ranges. SO(10) papers have numerous examples and relies on the Higg's for the seesaw 1 and seesaw II mechanism. However I've yet to find any papers covering the Higg's below the standard model 246 VEV. So I'm not looking at the Higg's mechanism itself but rather the Higg's bosons contributions (if any) to the thermodynamics after spontaneous symmetry breaking. Which is where I'm hoping the B-E can be used to give an approximation of the energy density and pressure contributions below the VeV. (I've never read or seen any papers covering Higg's below the 246 VeV, so I'm not sure if there is any further contributions below that range in terms of thermodynamics, the fact I never see any papers dealing with the Higg's interactions temperatures of the universe from the CMB to today) makes me wonder if there is or isn't an influence in current times cosmology. I'll have to see if I can locate a copy of the paper by Krauss, thanks for these. This is specifically what I am looking for (currently studying the 39 page article) lol at least I'm not the only one that had that idea just glancing over the 39 page paper it looks like it has the information I was curious about including the related formulas much thanks for that "What is at our disposal is essentially only the value of the Higgs field at the Planck scale, since in the experimentally accessible low energy region the Higgs field is not an observable and we only know its vacuum expectation value." this statement explains, why finding papers covering current influence is so tricky. awesome paper lol (my apologies if I'm acting like a kid with a new toy hehe)

The intention of this post is not a speculation, its an ideal I wish to test out. So I don't want replies based on speculation or personal theories. Now here is the Cosmic inventory http://arxiv.org/pdf/astro-ph/0406095v2.pdf "The Cosmic energy inventory" The one value missing from this inventory is the energy contribution of the Higg's field itself. So I wish to use the Bose-Einstein statistics, based on the only confirmed Higg's Boson we have confirmed (reasonably). As the Higg's particle is a boson this statistic should work, for fermions we use the fermi-dirac statistics. The purpose of this research is to see if I can further constrain the cosmological constant itself. (possibly) The two time periods I will examine is via the temperature of the CMB and the temperature of the universe today However I have a couple of questions. 1) first does anyone know of a reason why the Bose-Einstein statistics cannot be used in this case, if not what statistic is more appropriate ? 2) The cosmological constant is a positive energy, with a negative pressure contribution determined by its equation of state. Has anyone come across or seen an equation of state relating to the 126 GeV Higg's. (or knows if it is a positive or negative pressure contributor) (I would assume the Higg's field EoS may be the same as the relativistic radiation, however thats an assumption)(may be more appropriate to use the scalar field EoS, though via the standard model the vacuum expectation value is 246 Gev considerably less than the CMB) any professional peer review papers are welcome in terms of the thermodynamic properties of the Higg's itself are welcome.

a large part of that uncertainty is due to not knowing how soon Early structure formation occurs due to inhomogeneities in the uniformity of the universe after or during inflation. There is no strong agreement on this aspect. However as you stated it does not conflict with the BB

Ok I can offhand think of a few situations that can be used to support this, so I agree with you on that aspect. Might help if you provide a few examples

roflmao

The universe from nothing model, is one of many possible beginning models. Though not the only one other models include a bounce from a previous collapsed universe. Universes from black holes etc. The universe from nothing model is one of the few that doesn't involve a previous universe. The idea has merit, but then so do the others.

I am asking you to show me how you got to your equation obviously you don't want to do that. Yes I know the above is basic level and yes I specified the above doesn't have the dimension of force I specified that in the post above thank you that was all you had to explain lol though now I feel 100% stupid for now realizing this

how is re-posting the same equation count in showing how you derived it? Newtons gravitational law Newton's law of universal gravitation states that any two bodies in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. so I have force between particle A and Earth [latex]f=\frac {GMm}{r^2}[/latex] the mass of the Earth is already combined with the mass of the test particle to determine the force between them same formula for particle B same relation now if I drop these two at the same time from the same location the two masses are essentially combined in its influence on Earth which is the same as saying [latex]f=\frac{G(m1+m2)}{r^2}[/latex] so why is the M term on the left hand side needed? in terms of acceleration? please show me how you went from this to your equation

I'm asking for clarity about how you derived the above equation I can't figure out where you got the extra M term

can you clarify statement one.? As far as I know EP always applies for falling bodies. (this is rigorously tested and is in every textbook and related article I have ever read in over 20 years (non controversial that is) mi =mg I don't need to prove this as it is well known and established. now statement 2 I agree with, take 2 objects at rest. object A exerts a force upon object b , however that force is also applied to object a (forces are applied in equal and opposite directions) however f=ma so the the math will show that mass A's rate of acceleration will be different than mass B now statement 3 and 4 is where I have a problem, [math]m_1+m_2[/math] is combined to determine [latex]\mu=G(m1+m2)[/latex] where [latex]\mu [/latex] is the standard gravitational parameter [latex]\mu =GM[/latex] [latex]\mu=G m_1+G m_2 =\mu_1+\mu_2 [/latex] now here is my problem the right hand of your equations uses the law of universal gravitation [latex] f_1=f_2=G\frac{m1m2}{r^2}[/latex] http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation so therefore your equation should work for gravitational bodies.. [latex]M\frac{d^2X}{dt^2}=+(\frac{GMm_1}{(X-x_1)^2}+\frac{GMm_2}{(X-x_2)^2})[/latex] [latex]\frac{d^2 x}{dt^2} =\frac{dv}{dt} =\frac{dx}{dt}\frac{dv}{dx} =v\frac{dv}{dx} =\frac{GM}{x^2}[/latex] if [latex]\frac{d^2 x}{dt^2} =\frac{GM}{x^2}[/latex] why do you have [latex]M\frac{d^2 x}{dt^2}[/latex] ? in the above equation?

I have a bit of an unusual story about dreams in terms of nightmares. The last nightmare I ever recall having, when I was roughly 6 or 7. In that nightmare I recall running away from a brown bear then having my head bitten off, when I woke up. The nightmare in and of itself wasn't particularly unusual. The unusual part is that I am now close to 50 years old and its the only nightmare I can recall even to this day. I've often wondered why that is the case, I can recall numerous other dreams etc but I can honestly say I do not recall any nightmares since that last time. Its quite possible I may have had some and never known, after all I've heard the claims that we have roughly 200 dreams a night. However as stated I do not recall any

quite honestly the calculation is already in the references I pointed out in the article. I do not need to merely copy or paste them. The related formulas are already derived. 17.2 clearly shows how a 2 body problem can be reduced to a one body problem. "We shall begin our solution of the two-body problem by showing how the motion of two bodies interacting via a gravitational force (two-body problem) is mathematically equivalent to the motion of a single body with a reduced mass given by" [latex] \mu=\frac{m_1m_2}{m1+m2} [/latex] the steps on how to derive from this formula to arrive at the formula at 17.3.12. which all conform to Newtons 3 laws. 17.3.12 is the force between two bodies When I have such a clear reference I will post that reference and point out the related sections. In this case there is nothing that needs to be further derived or reduced.this article clearly shows all the textbook related formulas in how to derive both the force between gravitational bodies, the centrifugal force between those bodies, including the elliptical case. It clearly takes you from the force related calculations to arrive at Kepler's laws of rotation. by the method commonly found in textbooks. textbooks such as, Extragalactic Astronomy and Cosmology by Peter Schneider though this one doesn't go into detail in this area Introduction to Astronomy and Cosmology by Ian Morison chapter 1.9.1 Derivation of Kepler’s third law I am aware that the Kepler solutions has limitations,(particularly in 3 body problems) however its covered extensively in textbooks. I do have 2 other Astronomy textbooks. my main interest is in the field of Cosmology though, in that arena I own 12 textbooks, In the related metrics such as relativity, I own Ronald M Wald's General relativity book (excellent highly recommend it) and lecture notes on General Relativity by Mathius Blau. (handy 925 page article. (its not a textbook as per se but close enough) http://www.blau.itp.unibe.ch/newlecturesGR.pdf "Lecture Notes on General Relativity" Matthias Blau I don't believe there is a completely reliable solution to the 3 body problem, I have some papers covering the circular restricted 3 body problem, and the Poincare solutions, as well as how N-body can be used to treat the 3 body problem. (including one textbook on N-body Gravitational treatments, can't recall the authors name though) However Astronomy is not particularly my primary interest. (except in its relation to cosmology applications lol) by the way I do follow your formula's my problem was in misinterpreting how I thought you were describing it in terms of the equivalence principle but I'm clear on that now

It is an old descriptive of how Newton reconciled, gravity with orbiting bodies, I got the descriptive from my 1923 1st year University physics textbook. (lol coincidentaly how they modeled the atom didn't include the neutron, just a side note) All bodies are falling due to gravity. Their angular momentum keep them in orbit. Your right on the nearer of two bodies however if you look at say for example 17.5.1 page 15 the center of mass of a system is not at an actual body. Its between body A and body B. so both A and B will orbit the effective center of mass.. In the case of our solar system though or center of mass coincides near or at the sun, However local objects to a stronger mass source will orbit the local strong mass source over the solar systems center of mass. example planets and its moons "Thus each body undergoes a motion about the center of mass in the same manner that the reduced body moves about the central point" The reduced body has a similar geometry application figure 17.2 page 3 http://web.mit.edu/8.01t/www/materials/modules/guide17.pdf

The question boils down to a matter if practicality. For every day circumstances the Earth reaction is not a factor. With Kepler's law all bodies are understood in GR to be considered falling towards the effective center of mass. Their angular momentum offsetting their rate of fall. However this is more a statement that mass affects the location of the effective center of mass. Not that different mass objects fall at different rates. So in this aspect I agree with you

this is correct in certain circumstances, 1) the mass of the falling object must be large enough to have a measurable influence upon the effective center of gravity, (must be sufficient to move the Earth appreciably)

the newton mass is cancelled by the weight can't you see that?? are you that blind???? =equivalence principle now I am not about to go through all these equations for you in how to PROPERLY use the 2 body problem in regards to gravitational bodies as the 2 body problem applies to 2 bodies with similar mass ie they each have their own gravity, for object falling within Earths atmosphere the Euclidean forms are more than adequate where the mass of the falling obect do not significantly offset the effective common gravity. http://web.mit.edu/8.01t/www/materials/modules/guide17.pdf its up to you to read them Kepler’s Laws: Each planet moves in an ellipse with the sun at one focus. •The radius vector from the sun to a planet sweeps out equal areas in equal time. •The period of revolution T of a planet about the sun is related to the major axis of the ellipse by T2 =kA3 =where k is the same for all planets "We shall begin our solution of the two-body problem by showing how the motion of twobodies interacting via a gravitational force (two-body problem) is mathematically equivalent to the motion of a single body with a reduced mass given by" [latex]\mu=\frac{m_1m_2}{m1+m2}[/latex] this statement is also wrong "These results were important to Newton's analysis of planetary motion; they are not immediately obvious, but they can be proven with calculus. (Alternatively, Gauss's law for gravity offers a much simpler way to prove the same results.)" http://en.wikipedia.org/wiki/Shell_theorem

try it in straight newton physics mass a=1,000 kg mass b=10 kg force of gravity at Earths surface= 9.8 N/KG lets just round it off at 10 newtons/kg f=ma therefore object a=10,000 netwons object b= 100 newtons f=ma therefore a=f/m object a acceleration=10,000newtons/1000kg =10 m/s2 object b acceleration=100N/10kg=10 m/s2 wow they have the same acceleration therefore they land at the same time

fine where in any of your terms or math did you apply acceleration? or did you forget that step

well some computer codes lol, PLC ladder logic work a bit different as each is specific to the machine, in order to copyright the code in this case I would also have to copyright the machine itself. Trust me on this I write PLC code all the time, plus if your working for the company. The company itself has the rights as they paid you to produce it

your welcome

basic physics also teaches you that inertial mass is equivalent to gravitational mass the article I posted above shows that test of this are accurate to 1 part in 1013 (this just allows for possible systematic errors) http://www.damtp.cam.ac.uk/user/tong/concepts/gr.pdf in other words how objects fall depend on its gravitational acceleration not upon its mass the problem is your not considering the effective center of mass. yes the mass of the 3 objects is important but in the sense its needed to calculated the effective center of mass, not in the sense of heavier objects falling faster than lighter ones here is the proper 2 body problem equations http://web.mit.edu/8.01t/www/materials/modules/guide17.pdf all objects in a 2,3,x number of objects always fall towards an effective center of mass. the reaction of [latex]M\frac{d^2X}{dt^2}[/latex] is a consequence of an offset center of mass relative to the Earths center. it does not mean that the heavier object fall faster, it means the Earth also has to fall towards the center of mass just an amusing side note here is a brief NASA coverage for the feather and hammer test on the moon. http://science1.nasa.gov/science-news/science-at-nasa/2007/18may_equivalenceprinciple/ also [latex]f=\frac{GMm_2}{r^2}[/latex] is a newtonian law in terms of instantaneous acceleration your usage doesn't consider all the factors. (this is more for others reading the post informational purpose) [latex]\frac{d^2 x}{dt^2} =\frac{dv}{dt} =\frac{dx}{dt}\frac{dv}{dx} =v\frac{dv}{dx} =\frac{GM}{x^2}[/latex] Newtonian law works great at short distances not so great at longer ones such as spacecraft. (granted you probably wasn't trying to include the varying gravity due to distance from the gravitational source, and for the purpose of your equation assumed constant gravity)(this is more for others reading the post informational purpose) http://en.wikipedia.org/wiki/Equations_for_a_falling_body "A set of dynamical equations describe the resultant trajectories when objects move owing to a constant gravitational force under normal Earth-bound conditions. For example, Newton's law of universal gravitation simplifies to F = mg, where m is the mass of the body. This assumption is reasonable for objects falling to earth over the relatively short vertical distances of our everyday experience, but is very much untrue over larger distances, such as spacecraft trajectories. Please note that in this article any resistance from air (drag) is neglected." please note none of these equations at the various gravity levels include a mass term for the object falling on that wiki link wonder why that is?? http://www.npl.washington.edu/eotwash/sites/www.npl.washington.edu.eotwash/files/webfiles/publications/pdfs/schlamminger_AAPT07.pdf http://www.umich.edu/~mctp/SciPrgPgs/events/2008/SS08/EP_Mich.pdf "On a Theoretical Proof of the Weak Equivalence Principle from within the Confines of Newtonian Gravitation" http://vixra.org/pdf/1111.0082v1.pdf "the weak equivalence principle is the statement that bodies of different mass and composition will – as first demonstrated by Galileo Galilee; fall at the same rate in a gravitational field. Current measurements on the equality of gravitational and inertial mass indicate that this equality holds on a level of one part to 1013 "

your interpretation of that equation is wrong period

equivalence principle http://www.damtp.cam.ac.uk/user/tong/concepts/gr.pdf the problem is your trying to state the equivalence paper is wrong, the equivalence principle clearly states 2 objects of differing mass removing all friction will fall to the center of mass at the same rate. your statement that the heavier mass will fall faster is plain wrong. but instead of listening to the truth you misinterpret your formula and state every textbook and test in GR is wrong. You even go so far as to call a well established and rewarded physicist a crack pot. [latex]M\frac{d^2X}{dt^2}=+(\frac{GMm_1}{(X-x_1)^2}+\frac{GMm_2}{(X-x_2)^2})[/latex] but its form is more correctly [latex]M\frac{d^2X}{dt^2}=+(\frac{GMm_1}{r^2}+\frac{GMm_2}{r^2})[/latex] this formula may work great where the 3 bodies are of similar mass to each other as in 3 bodies orbiting each other, however IT does not mean the equivalence principle does not apply. if you took particle a and place it near to particle b so they drop at the same time from the same location their mass is effectively combined. If the center of mass is still the center of the Earth due to the size of mass being so insignificant to the mass of the Earth. the two will land at exactly the same time. if mass a is 4kg and mass b is 10 kg and they are dropped from opposite ends of the Earth whose mass is far greater the center of mass is the center of the Earth effectively. the two object will land at the same time. hence equivalence principle here is your 2 body problem with the center of mass correlations http://en.wikipedia.org/wiki/Gravitational_two-body_problem notice where your center of mass is?

link isn't working for me can you repost it? or is that the Mario Rabinowitz article? yeah works now

Quite frankly I'm going to pay more attention to what the textbooks teach, and what someone as distinquished as Mario Rabinowitz is over your opinion no offense, but quite frankly you started insulting his paper just from the letterhead initially. In case you didn't notice he stated in the paper that all 3 views in a certain sense is correct. Depending on the treatment. In case you haven't figured it out I don't need the mathematics as I'm not claiming either of the 3 views are more correct than the other. I'm not the one making claims you are. I merely posted the paper showing that depending on its treatment all 3 views can be accurate