Iwonderaboutthings Posted May 14, 2014 Posted May 14, 2014 (edited) G, g, and Strong Nuclear forces.... Are these forces all the same, or are they different??? Yes I know about acceleration, but it is still a force. G from what I know is the same in all the universe yes?? Science talks about forces, but if these forces have never been seen, how then can these forces be either related, the same things, or have different values, on different planets, altitudes and etc.. As per the periodic table being involved with invisible forces, only obscures matters even more... Please pardon this, but I don't get the point, this makes no sense whats so ever... Edited May 14, 2014 by Iwonderaboutthings
ajb Posted May 14, 2014 Posted May 14, 2014 The forces of the standard model have a dependence on the energy scale, the coupling constants are running. The amazing thing is that if you extend the standard model to include supersymmetry then the three gauge couplings of the standard model converge at about 10^16 GeV. The forces appear to unify at this energy scale. Gravity is not included in this picture. Presumably we could have unification near the Planck scale. 1
Iwonderaboutthings Posted May 15, 2014 Author Posted May 15, 2014 (edited) The forces of the standard model have a dependence on the energy scale, the coupling constants are running. The amazing thing is that if you extend the standard model to include supersymmetry then the three gauge couplings of the standard model converge at about 10^16 GeV. The forces appear to unify at this energy scale. Gravity is not included in this picture. Presumably we could have unification near the Planck scale. Is Gravity " g and G" not included because unification takes place in another dimension IE ' The Speed of Light, Or the Quantum World"? Why is this? I thought that G was constant throughout the universe, similar to the speed of light? From what I am seeing then,,, "IN ATOMS THE FORCES ARE ENTIRELY DIFFERENT THAN G AND g? Residual, Force Carriers and Strong Nuclear Forces??? too?? YIkes! WAIT A MINUTE! I LOGGED BACK ON... You mean to tell me that even in the human body our atoms are not related to g and G?? Edited May 15, 2014 by Iwonderaboutthings
ajb Posted May 15, 2014 Posted May 15, 2014 G also runs and people are studying this using renormalisation group flow methods. This means that the Planck scale may not be the right scale for quantum gravity and unification with the other forces could occur at a lower energy scale. However, I am far from an expert on this.
Iwonderaboutthings Posted May 16, 2014 Author Posted May 16, 2014 (edited) G also runs and people are studying this using renormalisation group flow methods. This means that the Planck scale may not be the right scale for quantum gravity and unification with the other forces could occur at a lower energy scale. However, I am far from an expert on this. In small words, can you say the "normalization" not the---> group is: A vector? A Magnitude? or equilibrium of an entire system relative to another symmetrical system? Like a polynomial? Hymmm, how can math explain or unify such a thing? Edited May 16, 2014 by Iwonderaboutthings
ajb Posted May 16, 2014 Posted May 16, 2014 In small words, can you say the "normalization" not the---> group is: In quantum theory we formally encounter infinities. We need to deal with these and one way is to consider these infinities are really measuring our lack our lack of knowledge of the high energy regime. To remove these infinities one can by hand add a cut-off meaning we just throw away contributions to whatever we are calculating that from the sector higher than our cut-off. Loosely, the remormalisation group is a set-up of methods that allow us to investigate how the physics changes with this energy scale. In particular coupling constant vary in energy according to a group equation. Meaning that the physics at any scale can be reached by the physics at a given scale by means of a group action on the set of all couplings. 1
Mordred Posted May 17, 2014 Posted May 17, 2014 (edited) I'd like to add some details to understand GUT. First off we need to define how a force is mediated. This is done through the related bosons. -Photons are the force carriers of the electromagnetic field.-W and Z bosons are the force carriers which mediate the weak force.-Gluons are the fundamental force carriers underlying the strong force. -Higgs boson mediates mass for guage bosons and W and Z bosons(not all particles) graviton mediates gravity???? essentially what this means is the transfer the force from one particle to another. This is important. Now we need to consider the ideal gas laws in thermodynamics or specifically thermal equilibrium. Particle reactions in thermal equilibrium are essentially unstable, its a factor of temperature, density and volume, which are all related by the equation [latex]PV=nRT[/latex] The relation forms used with bosons however is Bose-Einstein statistics or distribution now to explain this is further detail. Bosons become indistinquishable from one another where N is the number of particles and V is the volume and nq is the quantum concentration, for which the interparticle distance is equal to the thermal de Broglie wavelength [latex]q=\frac{N}{V}+\ge+n_q[/latex] the number of particles of the Bose_Eintein statistics is [latex]n_i(\varepsilon_i) = \frac{g_i}{e^{(\varepsilon_i-\mu)/kT}-1}[/latex] for fermions you use the fermi-dirac statistics [latex]\bar{n}_i = \frac{1}{e^{(\epsilon _i-\mu) / k T} 1}[/latex] the De-Broglie wavelength is [latex]\frac{V}{N\Lambda^3} \le 1 \[/latex] You can google each for better information I posted those relations to show how the ideal gas laws are done in regards to fermions and bosons. as opposed to the first formula. Now when the particle species except gravity are in thermal equilbrium the types of bosons become indistinquishable from one another, hence the forces are indistinqishable as well. They would all have the same temperature and wavelength. Also any reactions that do occur such as as I said are unstable any reaction will quickly have the reverse reaction. In regards to the forces this also apply to the fundamental interactions. You can see the chart and wiki coverage here. http://en.wikipedia.org/wiki/Fundamental_interaction in terms of cosmology these two articles best cover this http://arxiv.org/pdf/hep-th/0503203.pdf "Particle Physics and Inflationary Cosmology" by Andrei Linde (this one is older may be a bit out of date) for up to date use the one below. chapter 4 covers how to use the Bose_Einstein/Fermi_Dirac distibutions (however does not explain GUT, specifically, starts at lower temperatures)http://www.wiese.itp.unibe.ch/lectures/universe.pdf:" Particle Physics of the Early universe" by Uwe-Jens Wiese Thermodynamics, Big bang Nucleosynthesis edit: forgot to mention quarks in the quark-gluon plasma is Bose-Einstein statistics for quarks, for gluons its Fermi-Dirac distributions. Higgs would AFIAK though not positive the Bose-Eintein statistics Edited May 17, 2014 by Mordred 1
Iwonderaboutthings Posted May 17, 2014 Author Posted May 17, 2014 (edited) In quantum theory we formally encounter infinities. We need to deal with these and one way is to consider these infinities are really measuring our lack our lack of knowledge of the high energy regime. To remove these infinities one can by hand add a cut-off meaning we just throw away contributions to whatever we are calculating that from the sector higher than our cut-off. Loosely, the remormalisation group is a set-up of methods that allow us to investigate how the physics changes with this energy scale. In particular coupling constant vary in energy according to a group equation. Meaning that the physics at any scale can be reached by the physics at a given scale by means of a group action on the set of all couplings. Something interesting! If g is the force due to acceleration and G is the outer limit force "IE" In Empty Space, then shouldn't these 2 be the subjects of unification?? But then wouldn't this mean that earth would be the supposed Gravitron? Earth does have an = + and - amount of electrical charges making earth an incredible reference frame in the universe right? Or am I incorrect? Re-normalization ---> equilibrium? On a different note here. There is an issue I am having on my "beliefs" of the electron orbitals though , I keep reading that electrons don't actually orbit the nucleus and the whole of the interior atom appears to be obscured with a cloud of some form restricting its actual appearance under an electron microscope.. How can science move to the plank scale and hyper gravity if this is was true? I'd like to add some details to understand GUT. First off we need to define how a force is mediated. This is done through the related bosons. -Photons are the force carriers of the electromagnetic field. -W and Z bosons are the force carriers which mediate the weak force. -Gluons are the fundamental force carriers underlying the strong force. essentially what this means is the transfer the force from one particle to another. This is important. Now we need to consider the ideal gas laws in thermodynamics or specifically thermal equilibrium. Particle reactions in thermal equilibrium are essentially unstable, its a factor of temperature, density and volume, which are all related by the equation [latex]PV=nRT[/latex] The relation forms used with bosons however is Bose-Einstein statistics or distribution now to explain this is further detail. Bosons become indistinquishable from one another where N is the number of particles and V is the volume and nq is the quantum concentration, for which the interparticle distance is equal to the thermal de Broglie wavelength [latex]q=\frac{N}{V}+\ge+n_q[/latex] the number of particles of the Bose_Eintein statistics is [latex]n_i(\varepsilon_i) = \frac{g_i}{e^{(\varepsilon_i-\mu)/kT}-1}[/latex] for fermions you use the fermi-dirac statistics [latex]\bar{n}_i = \frac{1}{e^{(\epsilon _i-\mu) / k T} 1}[/latex] the De-Broglie wavelength is [latex]\frac{V}{N\Lambda^3} \le 1 \[/latex] You can google each for better information I posted those relations to show how the ideal gas laws are done in regards to fermions and bosons. as opposed to the first formula. Now when the particle species except gravity are in thermal equilbrium the types of bosons become indistinquishable from one another, hence the forces are indistinqishable as well. They would all have the same temperature and wavelength. Also any reactions that do occur such as as I said are unstable any reaction will quickly have the reverse reaction. In regards to the forces this also apply to the fundamental interactions. You can see the chart and wiki coverage here. http://en.wikipedia.org/wiki/Fundamental_interaction I think this gives me 95% clarity, so its heat interaction? the charges have thermal energy.. I don't know why I never figured this out, but you explaining it makes sense! I have always heard that thermodynamics had issues with equilibrium and balances, I can see why things can get " extremely difficult" considering all the constituents at hand... YIKES!!!!!!!!!!!!!!!! I Don't even see how calculus nor maths can even be of use with something so complex here.. But I am excited about this new information and will read it over and over again. thanks! Edited May 17, 2014 by Iwonderaboutthings
Mordred Posted May 17, 2014 Posted May 17, 2014 yes the simple way to look at it is at extreme temperatures, as volume increases pressure and temperature drops. So at dense volume and high density you have high temperatures. The energy-density has an equation of state to relate to its pressure. For cosmology applications use the radiation EoS http://en.wikipedia.org/wiki/Equation_of_state_%28cosmology%29 When temperatures are that high you simply cannot distinquish one particle from another, and like I stated any reactions that do occur are unstable. As the universe cools particles drop out of thermal equilibrium. Reactions of that particle species become stable
ajb Posted May 17, 2014 Posted May 17, 2014 essentially what this means is the transfer the force from one particle to another. It is better to think of exchange particle as exchanging energy, momentum and sometimes charge between particles rather than "transferring force". Now when the particle species except gravity are in thermal equilbrium the types of bosons become indistinquishable from one another, hence the forces are indistinqishable as well. Is this really the case? You are describing a Bose-Einstein condensate, right? If I have a box containing two different species of boson then under the right conditions I will not be able to tell there are two different species? This sounds strange to me as the assumption for Bose-Einstein condensates requires that I have a system of identical particles. If i had a mix of two species that are non-interacting then won't I effectively see two condensates? Can you give me a reference on this? They would all have the same temperature and wavelength. What would have the same temperature? If g is the force due to acceleration and G is the outer limit force "IE" In Empty Space, then shouldn't these 2 be the subjects of unification?? I understand g here to the the acceleration due to gravity at the Earth's surface. It is about 10ms^-2 and varies slightly depending on your location. It is not a fundamental parameter of the Universe. G however is, it is Newton's constant and tells us about the strength of gravity. Re-normalization ---> equilibrium? It means removal of troublesome infinities in our theory.
Mordred Posted May 17, 2014 Posted May 17, 2014 (edited) It is better to think of exchange particle as exchanging energy, momentum and sometimes charge between particles rather than "transferring force". Is this really the case? You are describing a Bose-Einstein condensate, right? If I have a box containing two different species of boson then under the right conditions I will not be able to tell there are two different species? This sounds strange to me as the assumption for Bose-Einstein condensates requires that I have a system of identical particles. If i had a mix of two species that are non-interacting then won't I effectively see two condensates? Can you give me a reference on this? What would have the same temperature? I understand g here to the the acceleration due to gravity at the Earth's surface. It is about 10ms^-2 and varies slightly depending on your location. It is not a fundamental parameter of the Universe. G however is, it is Newton's constant and tells us about the strength of gravity. It means removal of troublesome infinities in our theory. In thermal equilibrium all particle species have the same temperature, particle species in thermal equilibrium are described by its temperature only see the physics of the Early universe article chapter 4 as far as the indistinquishable, its a term I borrowed from David Griffiths Elementary to particle physics textbook. A lot of the various GUT (older articles use this descriptive. Keep in mind GUT articles are typically older, its incredibly difficult to find recent coverage of specifically GUT. So quite frankly its questionable if its even supported currently. Weinberg's first three minutes have the particles essentially vanish, at each step. I also included the caveat that the reaction rates of the reaction rates and stability. Some articles describe it as below.... "Early Universe is a soup of particles and anti-particle that under go frequent interactions with each other .Consider a single particle species (for example,electrons,or neutrinos,or protons...),and let the rate of the main reaction that keeps the species in thermal equilibrium with other species and the radiation field" see attachment EarlyU. the particle physics and Inflationary Cosmology by Linde posted above has this statement "The basic idea underlying unified theories of the weak, strong, and electromagnetic interactions is that prior to symmetry breaking, all vecto rmesons (which mediate these interactions) are massless, and there are no fundamental differences among the interactions" the Indistinquishable in David Griffiths book has a further caveat. "Nevertheless, it is an exciting idea, for it means that the observed difference in strength among the three interactions is an “accident” resulting from the fact that we are obliged to work at low energies, where the unity of the forces is obscured. If we could just get in close enough to see the “true” strong, electric, and weak charges, without any of the screening effects of vacuum polarization, we would find that they are all equal." page 77 In the particle physics by Uwe Gen Weise posted above The fact that the early Universe was in thermodynamical equilibrium leads to a tremendous simplification compared to the present epoch, because the equilibrium state is characterized completely by the temperature. further on a particle that has frozen out of equilibrium would have a different temperature and not be in thermal equilibrium. Particle Physics of the Intergalactic Mediums states that any temperature variations of different particles is negligable when in thermal equilibrium. http://arxiv.org/abs/0711.3358 now here is Wiki's caveat note the distinct ? "A Grand Unified Theory (GUT) is a model in particle physics in which at high energy, the three gauge interactions of the Standard Model which define the electromagnetic, weak, and strong interactions, are merged into one single interaction characterized by one larger gauge symmetry and thus one unified coupling constant. If Grand Unification is realized in nature, there is the possibility of a grand unification epoch in the early universe in which the fundamental forces are not yet distinct." take your pick which way you would prefer to describe it, some articles have the particles disappear, some say the particles stay but are the same temperature and any anistrophy is negligible . Which is essentially the same thing as indistinguishable when you think about it. Mosat GUT articles simply state the forces are unified however they are usually questionable articles when they don't clarify what they mean by Unified so I didn't include them. Keep in mind at sufficiently high temperatures all particle species are relativsistic and its total energy is dominated by its kinetic energy so its rest mass is essentially meaningless. As such their momentum is all the same at c (so momentum and frequency is the same), Their energy is all the same at T, so all you have left is spin. Or in thermal dynamics its spin is its entropy, which contributes to the temperature. (see Uwe gen weisse article above). A particle is determined by its energy, momentum and spin so what do you have left? http://galileo.phys.virginia.edu/classes/252/energy_p_reln.html Oh found another way of describing it lol http://www-ekp.physik.uni-karlsruhe.de/~deboer/html/Forschung/review3.pdf link may or may not work seems to be intermittant so I included the attachment GUTreview3.pdf " the well known forces, strong and electroweak are assumed to be equally strong." later on "At high energy, all forces are equally strong" took me a bit to dig this article out of my archives, had to locate it, its one of the better ones dealing specifically with GUT what I described above is how to model the tthermodynamics. The last article covers what that means in terms of symmetry breaking and guage theories some other articles related. These two include SU(10) http://arxiv.org/pdf/0904.1556.pdf http://pdg.lbl.gov/2011/reviews/rpp2011-rev-guts.pdf EarlyU.pdf GUTreview3.pdf Edited May 17, 2014 by Mordred
IsiacTorres Posted May 18, 2014 Posted May 18, 2014 When you think about it, if everything is always made up of some smaller unit of matter, then at the smallest level wouldn't ALL forces be Kinetic energy? Now it might not seem so, but to primitive civilizations fire and the power of movement were two completely different things. Now we know that Fire is really kinetic vibrations at the atomic level. You see were i am going? 1
Mordred Posted May 18, 2014 Posted May 18, 2014 (edited) No, you seem to misunderstand what kinetic energy is. In physics, the kinetic energy of an object is the energy which it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest. http://en.wikipedia.org/wiki/Kinetic_energy in other words you need the motion of particles or an object. Energy also requires particles In physics, energy is a property of objects, transferable among them via fundamental interactions, which can be converted in form but not created or destroyed. The joule is the SI unit of energy, based on the amount transferred to an object by the mechanical work of moving it 1 metre against a force of 1 newton. In other words energy is not a standalone entity Edited May 18, 2014 by Mordred
ajb Posted May 18, 2014 Posted May 18, 2014 @Mordred okay I have a better understanding of what you are saying. You are talking about thermal equilibrium in the early Universe and how this was "broken" by the expansion of the Universe leading to massive particles freezing out, which is of course very important for many things. I don't think that it is correct to say that all the particle species are indistinguishable when in thermal equilibrium alone. When people discuss freeze-out and the thermal history of the Universe they are probably talking about epochs after 100GeV and so we have already experiences the EW phase transition. We have the different species of particle as found in the standard model. Anyway, I am sure this by itself is not really at the heart of unification, the fact that the coupling constants are running is key. In other words energy is not a standalone entity This is very important. There is no such thing as pure energy in just the same way there is no such thing as pure momentum. Both a properties of a physical configuration and not, as you say standalone entities.
Iwonderaboutthings Posted May 18, 2014 Author Posted May 18, 2014 (edited) It is better to think of exchange particle as exchanging energy, momentum and sometimes charge between particles rather than "transferring force". Is this really the case? You are describing a Bose-Einstein condensate, right? If I have a box containing two different species of boson then under the right conditions I will not be able to tell there are two different species? This sounds strange to me as the assumption for Bose-Einstein condensates requires that I have a system of identical particles. If i had a mix of two species that are non-interacting then won't I effectively see two condensates? Can you give me a reference on this? What would have the same temperature? I understand g here to the the acceleration due to gravity at the Earth's surface. It is about 10ms^-2 and varies slightly depending on your location. It is not a fundamental parameter of the Universe. G however is, it is Newton's constant and tells us about the strength of gravity. It means removal of troublesome infinities in our theory. ajb, g is not a fundamental perimeter of the universe??? All known science functions uses the evolution of time from this planet.. NOTE: It may then appear that Physics and QM are then two totally complete different subjects??? So, then you are saying "no" not all forces are the same including those found at the atomic scale then??? I hope you understand what I am asking... I Want to know if: Gravity G, is the same at the atomic scale..... This includes all the constituents of the atom, it would not hurt to know where G plays a role in this if it does at all.. The reasoning behind this is because it appears " maths " does not really allow solutions to infinities nor equilibrium, however given the phenomenon of "Forces" what would science be able to do with boundaries of controlled limits of infinity?? Isn't pi ratio an infinity??? I see it all the time in equations... Would it be wrong to think that "if forces are not the same at the atomic scale" then: Physics and QM are two " ENTIRELY" separate subjects?? We live in a Quantum world and a Physical world?? Edited May 18, 2014 by Iwonderaboutthings
ajb Posted May 18, 2014 Posted May 18, 2014 (edited) Then you are saying "no" not all forces are the same including those found at the atomic scale then??? The forces of nature are for sure not the same at the energy scale of atomic physics. The forces of the standard model, so not including gravity may be the same at a higher energy level called the GUT scale which is about 10^16 GeV. I hope you understand what I am asking... I Want to know if: Gravity G, is the same at the atomic scale..... Truthfully, gravity has not been studied well below the sub-millimeter scale. It is possible that something different happens with gravity at the atomic scale, but of course as gravity is weak we just don't see any effect on atomic physics. Personally, I doubt anything changes until we are closer to the Planck scale, but gravity needs to be properly tested at the sub-millimeter and atomic scales. This includes all the constituents of the atom, it would not hurt to know where G plays a role in this if it does at all.. Well as I have said, gravity is weak at the atomic scale and so does not play any real role in atomic or molecular physics. In fact we ignore its effects in particle collisions at CERN as they are just swamped by the forces of the standard model. Gravity only plays an important role on macroscopic objects or at very high energies that we just can't probe. Would it be wrong to think that "if forces are not the same at the atomic scale" then: Physics and QM are two " ENTIRELY" separate subjects... We live in a Quantum world and a Physical world... The physical Universe dances to the tune of quantum mechanics for sure. It is just that this may not be evident in our human world. Edited May 18, 2014 by ajb 1
Mordred Posted May 18, 2014 Posted May 18, 2014 (edited) @Mordred I don't think that it is correct to say that all the particle species are indistinguishable when in thermal equilibrium alone. When people discuss freeze-out and the thermal history of the Universe they are probably talking about epochs after 100GeV and so we have already experiences the EW phase transition. We have the different species of particle as found in the standard model. Anyway, I am sure this by itself is not really at the heart of unification, the fact that the coupling constants are running is key. yes and no, thermodynamics and freeze-out is also involved when discussing GUT. Thermal equilibrium is involved in GUT. You are right there is more involved than just thermo equilibrium in regards to the coupling constants. however lets use the GUTreview.pdf as a reference. look at the temperature, Energy(Gev) vs time graph. on page 6. The time component is determined by the expansion history of the universe, Essentially as the volume increases (expansion), energy-density (Gev) decreases, as well as the temperature. Just as I described above. You will notice on page 6 they also use the terms freeze out. So from that you can easily see that GUT does include thermal equilibrium within its calculations. Here is one sample statement on the same page. "After spontaneous symmetry breaking, The Guage bosons freeze out" , However how you define thermal equilibrium depends on the particles being examined. You can describe a system at thermal equilibrium at any stage, and exclude any particles not in thermal equilibrium. when symmetry breaking occurs is when the temperature drops enough to allow coherence or freeze-out, in simpler words stability. In other words you cannot exclude the thermodynamic state in regards to the coupling constants in GUT it would be like trying to describe pressure without temperature and volume. as an example In regards to coupling constant "IN QFT the coupling constants are only effective at a certain energy, they are energy or equivalently distance dependent through virtual corrections" the section continues on the corrections specific to various models till you hit this line "the strong coupling constant decreases with higher energy, while the electromagnetic coupling constant increases with energy, so that at some point they are equal" this is equal to saying the strong coupling constant and electromagnetic coupling constant are in thermal equilibrium. section 3.3.2 page 32. later on in the same paragraph "and if the process goes through a state of non equilibrium" PS (Its too bad I could never find any one article that has a complete picture on GUT. They usually leave out the thermodynamic calculations. at least the GUTreview sort of covers that in the Cosmology section at least as far as the temperature vs expansion corelations using the equations of state), Its not bad for approximations, however doing the same with Gibbs law or the Boltzmann statistics (Bose_Einstein and Fermi_Dirac statistics) is more accurate. Edited May 18, 2014 by Mordred
MigL Posted May 19, 2014 Posted May 19, 2014 (edited) I do remember reading about this years ago but I can't give a source after all this time. Basically at low energies particles are not energetic enough to get close to other particles such that they 'see' each other's true field strength. This happens because particles surround themselves with a multitude of virtual particles which pop in and out of existence.. An electron for instance, would be surrounded by positive and negative charged virtual particles, and for a brief instant, the positive charges would move closer to the real electron before vanishing. This migration of positive virtual particles towards the electron would 'screen', or mediate, some of its true charge, and we would detect it to be less than it actually is. At higher energies such as the GUT scale ( about 10^15 GeV IIRC ), particles are energetic enough to reach inside the virtual particle screen and see the true strength of that particle's field. At this temp the particle's field strength are all equal ( if you include supersymmetric virtual particle contributions of course ). Edited May 19, 2014 by MigL
Mordred Posted May 19, 2014 Posted May 19, 2014 (edited) Not sure I've ever heard it described in quite this manner, however your descriptive bears some similarities to the regions in Yukawa couplings. Or at least thats what it reminds me of from the way you described it page 98, GUTreview.pdf. Yukawa couplings has 4 regions. edit actually scholarpedia has a similar descriptive, "They predict that the (distance-dependent) interaction strengths of the known interactions should become equal at short distance scales, lead to partially successful relations between quark and lepton masses, may be associated with small neutrino masses, may have implications for cosmology, and may lead to new gauge interactions that survive to low energies." http://www.scholarpedia.org/article/Grand_unification Edited May 19, 2014 by Mordred
Iwonderaboutthings Posted May 20, 2014 Author Posted May 20, 2014 (edited) The forces of nature are for sure not the same at the energy scale of atomic physics. The forces of the standard model, so not including gravity may be the same at a higher energy level called the GUT scale which is about 10^16 GeV. Truthfully, gravity has not been studied well below the sub-millimeter scale. It is possible that something different happens with gravity at the atomic scale, but of course as gravity is weak we just don't see any effect on atomic physics. Personally, I doubt anything changes until we are closer to the Planck scale, but gravity needs to be properly tested at the sub-millimeter and atomic scales. Well as I have said, gravity is weak at the atomic scale and so does not play any real role in atomic or molecular physics. In fact we ignore its effects in particle collisions at CERN as they are just swamped by the forces of the standard model. Gravity only plays an important role on macroscopic objects or at very high energies that we just can't probe. The physical Universe dances to the tune of quantum mechanics for sure. It is just that this may not be evident in our human world. aYyyyySssee. WOW! this now makes me understand the things that don't make much sense in mathematics and maybe life in general...WOW! thanks! Question: IF THE sub-millimeter and atomic scales WERE TO BE CALCULATED WITH PURE DIMENSIONAL NUMERICAL VALUES TO THEIR EXACT "PRECESSION" HOW WOULD THESE SOUND VALUES BE TESTED?? AND OF COARSE WHAT COPY RIGHTS, INSURES THE " FINDER" CREDIT IN THE WORLD OF SCIENCE?? IS THEIR A NUMERICAL EXAMPLE YOU CAN GIVE ME OF THE ISSUES OF sub-millimeter and atomic scales?? BOUNDARY A AND BOUNDARY B ??? I ASSUME THEM TO BE SIMILAR TO INTRIGALS, ! Sorry I am too lazy to remove the caps and retype, its been a long night for me no jocking Thanks for the reply... Edited May 20, 2014 by Iwonderaboutthings
ajb Posted May 20, 2014 Posted May 20, 2014 IF THE sub-millimeter and atomic scales WERE TO BE CALCULATED WITH PURE DIMENSIONAL NUMERICAL VALUES TO THEIR EXACT "PRECESSION" HOW WOULD THESE SOUND VALUES BE TESTED?? People are devising tests of gravity on small scales. You should try a quick google as this is well outside my area of expertise. Basically at low energies particles are not energetic enough to get close to other particles such that they 'see' each other's true field strength.... This is the notion of bare and dressed particles. The cloud of virtual particles around a given particle contribute to its properties. This is important in interpreting what is going in remormalisation.
MigL Posted May 20, 2014 Posted May 20, 2014 You're right AJB, this would be the same mechanism which gives us the excuse to use the ( terrible ) mathematical tool known as re-normalization ( although it does seem to work just fine ). It seems that a particle's field strength is then temperature ( energy ) dependant, as a more energetic test particle will penetrate further into the surrounding virtual particle cloud. What I find interesting is that without supersymmetric virtual particles the field strength converges to two distinct points, not at the GUT field strength. The calculation for the mass of the Higg's boson also uses this mechanism and without supersymmetric virtual particles, the calculated mass is huge. With supersymmetric virtual particles the mass turns out to be approx. the experimentally found ( LHC ) mass. So where are these supersymmetric particles ???
Mordred Posted May 20, 2014 Posted May 20, 2014 (edited) now that is a good question lol, one of the major hurdles for the MSSM (minimum supersymmetric models) GUT models. Those are covered in the GUTreview. Minimum supersymmetric SU(5) guage symmetry group. If I remember correctly.. in that I have one question is the SU(10) guage symetry group just the Higg's sector? edit never mind that question the Higgs sector is needed to break SU(5) to Su(3)*SU(3)*U(1) which means the Higg's sector needs 12 Goldstone bosons. As SU(5) has 24 guage bosons and standard model has 12. including antiparticles If I understand that's for the SU(5) to work the SU(10) GUT, from what I understand is a strong candidate from this paper, as it does not involve the supersymmetric particles http://www-f1.ijs.si/~ziherl/Greljo12.pdf man I really need a more modern particle physics textbook The two I have from David Griffith just aren't cutting it, any recommendations? Edited May 20, 2014 by Mordred
Iwonderaboutthings Posted May 21, 2014 Author Posted May 21, 2014 now that is a good question lol, one of the major hurdles for the MSSM (minimum supersymmetric models) GUT models. Those are covered in the GUTreview. Minimum supersymmetric SU(5) guage symmetry group. If I remember correctly.. in that I have one question is the SU(10) guage symetry group just the Higg's sector? edit never mind that question the Higgs sector is needed to break SU(5) to Su(3)*SU(3)*U(1) which means the Higg's sector needs 12 Goldstone bosons. As SU(5) has 24 guage bosons and standard model has 12. including antiparticles If I understand that's for the SU(5) to work the SU(10) GUT, from what I understand is a strong candidate from this paper, as it does not involve the supersymmetric particles http://www-f1.ijs.si/~ziherl/Greljo12.pdf man I really need a more modern particle physics textbook The two I have from David Griffith just aren't cutting it, any recommendations? Not sure if this link is of any help, BUT! reading it now places the things that confused me on a better path of perspective. http://profmattstrassler.com/articles-and-posts/particle-physics-basics/virtual-particles-what-are-they/ At-least as a start for me, I read this on Wikipedia: Goldstone bosons http://en.wikipedia.org/wiki/Goldstone_boson In fluids, the phonon is longitudinal and it is the Goldstone boson of the spontaneously broken Galilean symmetry. Insolids, the situation is more complicated; the Goldstone bosons are the longitudinal and transverse phonons and they happen to be the Goldstone bosons of spontaneously broken Galilean, translational, and rotational symmetry with no simple one-to-one correspondence between the Goldstone modes and the broken symmetries. I find this " Galilean symmetry" very interesting, would anyone know " why " this is mentioned? Does this have anything to do with Cartesian Space??? Basically Is time the issue?? You're right AJB, this would be the same mechanism which gives us the excuse to use the ( terrible ) mathematical tool known as re-normalization ( although it does seem to work just fine ). It seems that a particle's field strength is then temperature ( energy ) dependant, as a more energetic test particle will penetrate further into the surrounding virtual particle cloud. What I find interesting is that without supersymmetric virtual particles the field strength converges to two distinct points, not at the GUT field strength. The calculation for the mass of the Higg's boson also uses this mechanism and without supersymmetric virtual particles, the calculated mass is huge. With supersymmetric virtual particles the mass turns out to be approx. the experimentally found ( LHC ) mass. So where are these supersymmetric particles ??? energy dependant?? Is this the same as dependant on " time" as in cycles much like wave phenomena? As in frequencies and cycles, periods and etc?
Mordred Posted May 21, 2014 Posted May 21, 2014 Thanks for the attempt I'm more interested in the SU(10) portion, David Griffith didn't cover this group, or at least my copy didn't. No worries though. I found some material on the seesaw mechanism type I and Type II. I already have numerous Higg's articles as well as several Higg's inflationary model articles, and a couple of CERN Higg's thermodynamic papers. Just need to fit the pieces together. Along with the textbooks I do have. My problem is none of my textbooks cover S0(10) I would much rather have a textbook that does, than a scattering of potentially misleading articles
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