text:
Talk:Theory of Everything Project

Ideas for this research project
This research project is fine, very ambitious. There are various ways to initiate research on this subject, that do not necessarily "involving public databases for cosmological and particle physics experiments". I am at present reading Lee Smolin's book "The Trouble with Physics" (2006) where he describes the current difficulties in finding The Theory of Everything and brings it back to five major problems. I cite from his first chapter:

''One way or another, any proposed theory of fundamental physics must solve these five problems, so it's worth taking a closer look at each. ...''

Problem 1: Combine general relativity and quantum theory into a single theory...

''Problem 2: Resolve the problems in the foundations of quantum mechanics, either by making sense of the theory as it stands or by inventing a new theory that does make sense. ...''

''Problem 3: Determine whether or not the various particles and forces can be unified in a theory that explains them all as manifestations of a single, fundamental entity. ...''

''Problem 4: Explain how the values of the free constants in the standard model of particle physics are chosen in nature. ...''

Problem 5: Explain dark matter and dark energy...

These five problems are context for research as stated in the Wikiversity ressource Introduction to research. So the next thing we could do is turning these five problems into questions which we may use to address the problem of the Theory of Everything. It is worth to list some questions on the project page and develop on it in a collaborative way, examples:

 * Are there different ways to combine general relativity and quantum theory?
 * What are the similarities between GR and QT? the differences?
 * How could one make sense of Quantum Mechanics?
 * What are the foundations of Quantum Mechanics? If there are more, are they independent or is there a primary foundation?
 * What (or how) could be the single, fundamental entity if it existed?...

Arjen Dijksman 21:20, 11 September 2007 (UTC)

Science 2.0
Wolfram proclaims we need, "A New Kind of Science". Carver Mead claims the way we teach science is all wrong. I agree. In fact, most of the fundamental issues in physics are due to trying to fit what we know today into a legacy science framework. I believe we are much closer to a theory of everything than we realize. But we must abandon the postulates of quantum theory, and start fresh, not with metaphysical notions, but with empirical facts as we know them today independent of theory.

We should consider well all the well conceived objections to standard quantum theory of the great thinkers, Heisenberg, Schroedinger, Einstein, Bell, etc., and look at more sensible interpretations of the quantum such as those proposed by Bohm and Cramer.

Basic matrix mechanics is a fundamental process that accounts for all quantum phenomenon. It does not introduce classical notions of particles, fields, waves, etc. I was disappointed that the article here called "Making sense of quantum mechanics" describes standard modern quantum theory with all its excess baggage. The page tile should be changed to "Understanding Modern Quantum Theory" in my view, as this model will never "make sense". I believe we need to get down to basics of what is really known today rather than perpetuating the misunderstandings of the past if we wish to provide our students with a foundation for a theory of everything. JimScarver 09:23, 21 October 2007 (UTC)

 * For me, it is fairly clear from the introductory paragraph at Making sense of quantum mechanics what the goal of that learning resource is. This suggested name change (above) seems likely to lead to a significant change in the learning objectives and goal of that page. In my view, page name changes should not have such a large impact on a page. If there is an alternative point of view to be explored, Wikiversity tends to spawn a new page where the alternative can be explored. Why not make a new page to explore alternatives to conventional thinking about quantum mechanics? --JWS 15:09, 21 October 2007 (UTC)

 * Alternative points of view may also be listed directly at Making sense of quantum mechanics. Confronting them helps us to understand what is fundamental and what is accessory. I agree with the fact that we need to get down to basics. In my view, it is the state vector, concept which introduces the notion of a particle, system of particles, wave motion,... What do you think are the basics? Arjen Dijksman 15:48, 21 October 2007 (UTC)

 * It could always be forked too. --Remi 16:25, 21 October 2007 (UTC)

 * In the context of "Modern Physics", the page is excellent in my view. I think, as it stands, its title is inappropriate in a theory of everything context.  The Theory of Everything will most certainly over throw the existing theories of modern physics.  I concede that adding, early, links to competing theories, as needed, will indeed solve my objections to the page name.  Never mind. :-)  JimScarver 19:00, 21 October 2007 (UTC)

 * Jim, is there something particular in the rotating vector model that makes you say that this model would never "make sense"? Is there something that we can improve about it? Arjen Dijksman 20:29, 12 November 2007 (UTC)

Holistic Theory of Everything [Picture-files related to this, and the text are located at http://sites.google.com/site/theultimaterealitysite] Hasmukh K. Tank,Electronics and Communications Engineer D-12/402, Shantinagar, Sector – 7, Mira Road (E), Dist - Thane 401107 [India] Date :- 20th November 2008. Revised on : 3rd April 2009 Abstracts:This holistic theory of everything includes: the objective physical reality, emergence of living beings in it, and the subjective world of ‘mind’ and ‘consciousness’; but in this paper the subjective aspect is omitted for the ease of physicists. According to this theory, the ultimately real entity is all-pervading in space and ever-present in time. The physical world is a ‘process’ or a ‘phenomenon of fluctuations’ spontaneously generated in the ultimate reality. The continuum nature of the ultimate reality(UR)allows only spherical patterns of fluctuations to sustain, so there are integer number of spherical fluctuation-patterns, to which we perceive as fundamental ‘particles’. These spherical patterns of fluctuations interfere among themselves, linearly in the far-fields and non-linearly at near-fields. As we go towards grosser and grosser levels, one and the same process of interference shows a hierarchical order, which appears to us as ‘different’ kinds of ‘forces’. At a gross level, whenever these fluctuation-patterns form a ‘feed-back-controlled-close-loop’ then they start giving ‘response’ to their external ‘environment’, to them we consider as ‘living beings’. The interactions among the patterns is the bases for ‘mind’ and ‘consciousness’ at the higher levels. Key Words:All-pervading continuum, Integer number of spherical fluctuation-patterns, Hierarchical order of interference, Feedback-controlled-loops, Interactivity of fluctuation-patterns. Introduction: Current physics has discovered several empirical laws. Majority of physicists believe that the fundamental particles are real entities, and they are engaged in searching for a unified theory which can describe all the four kinds of particle interactions. But with the notion of ‘particles’ as ‘real’ entities, unification of empirical laws is very less likely to succeed, because the mathematical models used to describe a limited aspect cannot be extended to cover all the phenomena. So we present here a completely new approach, which starts with a postulate, derives the observed world, explains more and more observations and then tries to find proof of the initial postulate. The holistic Theory of Everything: (1)	Towards the end of his life Einstein himself realized that when the fields are continuous in space, the concept of ‘empty’ space is meaningless. There are many statements in the ancient scriptures, and in the writings of thinkers that the ultimate reality is all-pervading in space and eternal in time; and it is one without any second. So we begin here with a postulate that: space is not an ‘empty void’, the ultimately real entity (UR) is present everywhere in space, and it is eternally ever-present in time. It is a highly subtle continuum. (2)	Because of continuum nature of the ultimate reality (UR), when a small labeled dot in it moves from ‘a’ to ‘b’ as shown in fig.1, it gives rise to a chain of displacements, completing a closed path; by the dot ‘z’ occupying the place of ‘a’. (3)	There is no preferred axis about which these dots should complete a closed path; so they move partly about x-axis, partly about y-axis and partly about z-axis to complete the closed path. So this path forms a circle on an imaginary ‘spherical shell’ in the UR. (4)	This process of close-loop-displacements (or close-loop induction of fundamental field) forms a positive-feedback-loop; and so the spherical pattern of fluctuation continues for ever. The continuum nature of UR allows only spherical pattern of fluctuations to sustain, so we find integer number of such discrete fluctuation-patterns, as shown in fig.2, to which we believe as the ‘fundamental particles’ like the quarks. Here we find an explanation for why the field-concentration of the ‘particle’ does not get diluted in space. Fig.2 shows that the amplitude of waves increase with radial distance up to the quarter wavelength, and then start falling inversely with further increase of radial distance. The fluctuations in the ultimate reality get transmitted at a finite velocity of light. (5)	Fig.3 shows the peaks of wave-amplitudes of the spherical fluctuation-pattern. When two or three of such patterns are very close to each other, the wave-amplitudes interfere non-linearly, following the mathematics: 1+1=1. When two or three such patterns interfere, they give rise to radial lines of maxima and minima as shown in fig.4. When such radial lines interfere with another set of radial lines, they give birth to the magnetic lines of force. Thus it is the spherical fluctuation-pattern, representing the most fundamental field, whose hierarchical layers appear as different kinds of fields. (6)	Because these spherical patterns are of micro-microscopic size, they appear to us as ‘point particles’; so they can be mathematically represented as ‘pulse-function’ in space; and it can be Fourier transformed in to the wave-number-domain giving a wide band of wave-numbers. The pattern can also be represented in the frequency-domain, as a wide band of frequencies. Now, if the spectral components of two such spherical pattern are coherent, then they add constructively and bring the patterns closer and closer; and when the spectral components are in-coherent, then they add destructively repelling the patterns away from each other. The strength of interference depends on the coherence of spectral components. (7)	It is obvious that we can not expect any coherence between the spectral components of two independent spherical patterns; so the wide band of waves of these patterns add like the addition of ‘wideband-noise’, i.e. like the mutually perpendicular vectors. Therefore, compared to the strength of coherent superimposition of the fundamental fluctuations, the strength of incoherent superimposition of two such patterns is 1.414 times, i.e. instead of two-times it is just ‘ square-root of two’ times ; the strength of attraction of ten atoms is ‘square-root of ten ’ times to which we call molecular force; and since there are 10 raised to 80 atoms in the universe the strength of gravitational force is : square-root of 10 raised to 80, that is 10 raised to 40 times only. If one represents the wideband-noise as mutually orthogonal vectors then he would require 10 raised to 80 dimensions of space to find the vector-sum of strength of 10 raised to 80 atoms! (8)	At a macroscopic level large conglomeration of spherical fluctuation-patterns form a negative-feedback-controlled-loops. For example, when hot rays of the sun fall on the water-surface of oceans, they produce evaporation of ocean-waters, this vapor goes to the sky and form clouds, but the cloud-cover on the water-surface prevents further evaporation of ocean-waters, and provides a negative feedback to the process. It is this negative feedback which brings stability in the system of rains, otherwise oceans would have dried out completely within some years. Such negative feedback-controlled systems at molecular levels are the rudimentary forms of ‘life’. A single cell is a highly complex feedback-controlled-system, very difficult to analyze in terms feedback-controlled-system; but it is the formation of ‘system’ which brings the property of ‘response’ observed in the ‘living beings’. (9)	The ‘interaction’ of the fluctuation-patterns is at the base of ‘mind’ and ‘consciousness’; each and every pattern subjectively perceives it self as an individual, and its surroundings as the external world. With whom-so-ever we are able to establish communication, to them we believe as ‘living beings’ and we treat others as ‘dead’ and ‘inert’. The distinction of ‘living’ and ‘non-living’, and the partition of ‘I’ and ‘That’ is only a matter of convention. Actually there is only one ultimate reality, subjectively aware of its existence, and perfectly happy with itself. It is immortal and all-pervading ; dancing  out of  joy! After realizing this truth, you too are now free to dance! Conclusion Thus, according to the holistic theory of everything, the ultimate reality is all-pervading in space and ever-present in time. Because of continuum nature of UR, the fluctuations generated in it automatically assume a spherical standing-wave-pattern. Amplitude of the wave   increase directly with radial distance up to quarter wavelength, and then start falling inversely with radial distance. If we consider the strength of coherent superimposition as one, then the strength of in-coherent super imposition of two patterns is 1.414 times; the strength of ‘n’ patterns is: square-root of ‘n’, which appear to us as molecular forces; and since there are 10 raised to 80 atoms in the universe, the strength of gravitational force is 10 raised to 40 compared to the strength of coherent super imposition of 10 raised to 80. This theory explained why the field-concentration of the particles does not get diluted in space. We understood the emergence of living beings as a formation of negative-feedback-controlled-systems. References: (1)	Tank, H.K. “Some conjectures on the nature of ‘energy’ and ‘matter’, Science and Culture Volume 54, April 1988; Published by Indian Science News Association, Kolkata. I am unable to attach figures here, but I will be glad to send; my E-mail ID is: tank.hasmukh@rediffmail.com

general relativity and quantum theory
I've never understood why so many physicists seem to have an intuition saying that it would be neat to unify general relativity and quantum theory. Is this intuition any better than that of the ancient Greeks who imagined that the paths of celestial bodies should be perfect circles? --JWSchmidt 21:43, 11 September 2007 (UTC)

I would like to point out that we cannot presuppose that general relativity and quantum theory are incompatible just because that is the standard wisdom. There are numerous electrodynamic theories of gravity, such as Cahill's Quantum Foam Gravity that may explain gravity within the context of quantum mechanics including measured effects not accounted for by general relativity. If we consider the universe to be a quantum system, the expected increase in entropy suggested by Schroedinger equation is not incompatible with observed gravitational phenomenon. Is the purpose here to promote the current popular wisdom or what is known objectively? JimScarver 08:51, 21 October 2007 (UTC)

 * I would say that this research project offers the opportunity to critically assess popular wisdom on the hand of objective knowledge, without promoting any particular view.
 * How does Schroedinger's equation suggest increase in entropy? Arjen Dijksman 10:32, 21 October 2007 (UTC)

 * Gravitational-self-force-based explanation for the Pioneer-anomaly and the Cosmological red-shift:leading to static model of the universe;By:Hasmukh K. Tank

From: Space Applications Centre of Indian Space Research Organization, 22/695 Krishna Dham-2, Vejalpur, Ahmedabad-380051 India.E-mail: tank.hasmukh@rediffmail.com Date:20th February,2010. Abstracts: Based on the strikingly matching values of decelerations of: Pioneer-10, Pioneer-11, Galileo and Ulysses space-probes with the deceleration of the cosmologically red-shifting photons, and the 'critical acceleration' of Modified Newtonian Dynamics [MOND], this letter proposes that all these decelerations are because of 'gravitational-self-force' of the object's own gravity. This new explanation for the cosmological red-shift leads to static model of the universe. Experiments to verify this explanation are also proposed. Key Words:Pioneer anomaly, Modified Newtonian Dynamics, Gravitational-self-force Introduction: This writer has considered close to hundred alternative interpretations of the cosmological red-shift over the span of twenty years. Among them, one proposed before many years was: ‘reduction of the photon’s energy due to its own gravitational-field’. But at that time there were no other supportive-observational-evidences known to him. Recently, when I found that the Pioneer-anomaly is one of the unresolved problems of physics; ordinary explanations proposed by various authors have not succeeded; and two conferences arranged specially to solve the anomaly have remained inconclusive, I first noticed that the values of decelerations of all the four space-probes are strikingly the same, in spite of different speeds, masses and directions of the four space-probes. Since the reductions in kinetic energies of these space-probes were available in terms of their decelerations, I expressed the cosmological red-shift in terms of deceleration of the photons; and to my pleasant surprise, the value of deceleration experienced by the cosmologically red-shifting photon turned out very close to the values of decelerations of all the four space-probes. The ‘critical acceleration’ of ‘Modified Newtonian Dynamics’[MOND] is also equal to the above five decelerations, so, we have now half a dozen cases where we find acceleration of the same magnitude. This letter explains these decelerations in terms of gravitational-self-force of the moving objects themselves. This simple-looking explanation leads to static model of the universe, that there was neither any Big-Bang, nor the universe is expanding. This proposal is experimentally verifiable, and theoretically analyzable. The Details:The strikingly equal amounts of carefully measured1 anomalous decelerations of all the four space probes: For Pioneer-10, a = (8.09) x 10^-10 meters/sec2; For Pioneer-11, a = (8.56) x 10^-10 meters/sec2; For Ulysses, a = (12) x 10^-10 meters/sec2: For Galileo, a = (8.0) x 10^-10 meters/sec2; and their perfect matching with the deceleration of cosmologically red-shifting photons can not be an accidental coincidence.For Cosmologically-red-shifted-photon a = 6.87 x 10^-10 meters/sec^2 = H0 c. The reason why the deceleration of cosmologically red-shifting photon is slightly less is because: when the extra-galactic-photons enter our own milky-way-galaxy, they experience some gravitational blue-shift. So,the strikingly matching values of five different decelerations can not be ignored by a scientific mind. Moreover, this value of deceleration also matches with the ‘critical-acceleration’ of the Modified Newtonian Dynamics [MOND] devised to explain the ‘flattening of galaxies’ rotation-curves. So, this writer expressed his opinion that all these perfectly-matching-decelerations are because of ‘self-gravitational-pull’ experienced by the moving body itself. Because, every piece of matter, and every chunk of energy, produces a curvature of space-time around it; so when a body tries to move in any direction, it experiences a backward-pull, because of its own gravity. Space-time are bendable; they are not perfectly rigid like the cement-road. So, moving through space-time is like walking on a sandy path; so the moving object has to spend a part of its kinetic-energy, to keep moving. In the case of the cosmologically red-shifting-photons, the expense of energy can be expressed as follows:We can express the cosmological red-shift zc in terms of de-acceleration experienced by the photon, as: zc = (Delta f) / f = H0 D / c; i.e.(h Delta-f / h f) = H0 D / c; i.e.h Delta f = (h f / c^2) (H0 c) D. That is, the loss in energy of the photon is equal to its mass (h / c2) times the deceleration a = (H0 c),  times the luminosity-distance D  traveled by it. Here, H0 is Hubble’s constant, and c  is speed of light. i.e. a = 6.87 x 10-10 meter/sec^2. Based on my previous paper2 the acceleration a = H0 c can be expressed as: a = H0 c = G M0/R0^2 = G 10^80 mp /(10^40re)^2= G  mp / re^2; Where:  M0  was  total-mass of the universe, R0  was  radius of the universe, G  is Newton’s  gravitational-constant, mp  is mass of the proton,  and re  is  classical-radius of the electron, as defined by Dirac:  re = e^2/ me c^2. It means that the above acceleration a implies  the ‘self-gravitational-pull’ at the level of the protons and neutrons contained in the nucleus of the atoms. Works of other researchers on the Gravitational-self-force in a different context: At the time of making the above proposal, this author was unaware of the works done by other scientists. But the literature-survey shows that researchers have already done some work on‘gravitational-self-force’, in a different context. Interestingly, their findings are also favorable to the new explanation of the cosmological red-shift proposed here. For example, I quote the abstract of a paper by Eric Poisson^3: “The self-force describes the effect of a particle’s own gravitational field on its motion. While the motion is geodesic in the test-mass limit, it is accelerated to first order in the particle’s mass. In this contribution I review the foundations of the self-force, and show how the motion of a small black hole can be determined by matched asymptotic expansions of a perturbed metric. I next consider the case of a point mass, and show that while the retarded field is singular on the world line, it can be unambiguously decomposed into a singular piece that exerts no force, and a smooth remainder that is responsible for the acceleration. I also describe the recent efforts, by a number of workers, to compute the self-force in the case of a small body moving in the field of a much more massive black hole. The motivation for this work is provided in part by the Laser Interferometer Space Antenna, which will be sensitive to low-frequency gravitational waves. Among the sources for this detector is the motion of small compact objects around massive (galactic) black holes. To calculate the waves emitted by such systems requires a detailed understanding of the motion, beyond the test-mass approximation.” Conclusion:Thus, we can explain the Pioneer-anomaly and the cosmological red-shift in terms of gravitational self-force of the space-probes and the cosmologically red-shifted photons themselves. This new explanation for the cosmological red-shift will lead to static model of the universe, that there was neither any Big-Bang, nor the universe is expanding. So, we need to experimentally verify the value of deceleration caused by the ‘gravitational self-force’ by sending more and more space-probes: (i) in various different directions, (ii) of different masses and (iii) of different speeds. We should also perform experiments, in the outer space where there are no other gravitational-forces, by applying external force comparable in magnitude with the ‘gravitational self-force’. This paper should also provide motivation to theoretical physicists for working out more details. This may help us to understand flattening of galaxies’ rotation-curves and exact theoretical basis of ‘Modified Newtonian Dynamics [MOND]. References: 1.	Anderson, J.D., Laing, P.A., Lau, E.L., Liu, A.S., Nieto M. M., and Turyshev.S.G. Indication, from Pioneer 10, 11, Galileo, and Ulysses Data, of an Apparent Anomalous, Weak, Long-Range Acceleration. Phys. Rev. Letters. 81 (1998) 2858-2861 [(Comment by Katz J.I.: Phys. Rev. Lett. 83, 1892 (1999); Reply: Phys. Rev. Lett. 83, 1893 (1999) 2.  Tank H.K. “An Explanation for the Relative Strengths of ‘Gravitational’ and‘Electric’ Forces Suggesting Equality of the ‘Electrostatic-potential-energy’, ‘Gravitational-potential-energy’ and ‘Energy of Mass’ of the Universe”, Science and Culture 75, No: 9-10, Sept-Oct 2009 p. 361-363 3. Eric Poisson “ The Gravitational self-force” e-proceedings, World Scientific, DOI No:10.1142/9789812701688_0012

 * The Schroedinger's equation is a direct analog of Euler's heat flow equation. It is direct statement of the 2nd Law of thermodynamics.  It says the system tends toward equilibrium.  JimScarver 13:32, 21 October 2007 (UTC)

 * I see. I understand Schrödinger's equation as the law of conservation of energy, which is more generaly valid, whatever the variation of entropy. Arjen Dijksman 18:55, 21 October 2007 (UTC)

singularities
Einstein did not succeed in his quest for a theory of everything. He said any unified theory must be free of singularities.

a continuum
One the one hand quantum theory postulates a continuum, but at the same time its results contradict a continuum. The smallest piece of space is the Plank length, and the smallest piece of time possible is the Plank time. Many scientists today now believe that a theory of everything must be background independent, inplying there is no continuum.

determinism
Modern quantum theory presumes indeterminism. Yet there are equivalent deterministic formulations and many great scientists have never accepted this postulate. Uncertainty may be accounted for by missing information. Is it proper in a scientific theory to postulate a phenomenon that is fully accounted for?

time ordering
Einstein remarked that according to quantum theory, time ordering should be random. Clearly this is not the case according to all measurement and experience.

AN INSIGHT INTO THE RELATIVE STRENGTHS OF GRAVITATIONAL AND ELECTRIC FORCES Hasmukh K. Tank Electronics and Communications Engineer D-12/402, Shantinagar, Sector – 7, Mira Road (E), Dist - Thane 401107 [India] Date :- 20th November 2008. Revised on : 21st March 21, 2009 Abstracts Based on previous works of P.A.M. Dirac1,2, Eddington3,4,5, Dicke6,7,8, Hayakawa9,10, Carter11 and Adair12, this paper presents an explanation for the relative strengths of ‘gravitational’ and ‘electric’ forces, and leads to interesting conclusions that (i) sum-total of the electrostatic-potential-energy stored between the protons and electrons in the atoms, in the whole universe is equal to (ii) the gravitational potential-energy of the whole universe, is equal to (iii) the energy of mass of the universe. The strength of gravitational force depends on in-coherent superimposition of the wide-band-waves, corresponding to the ‘particles’ of ‘matter’ contained in the universe. Key Words Electromagnetic force,   Gravitational force,     Force-ratio,    Cosmological red-shift Introduction Paul Anderson Dirac won the Nobel Prize for his equation, which lead to the discovery of ‘positron’. After that, when he was on the world-tour, he thought that we measure things in our arbitrarily chosen human units of meters, kilograms and seconds; we should try to measure them in nature’s own units. So he formulated a unit of length by dividing electric-charge’s-square by energy of the electron (e2/mec2) and called it the “radius of the electron” (re). As soon as he tried to express the radius of the universe at maximum expansion (Ro) in terms of radius of the electron (re) he got the same large-number 1040, equal to the strength-ratio of electric and gravitational forces, (e2/Gmemp) = 1040! Very soon, other researchers also found that the ratio of total-mass-of the universe (M) and mass of the proton (mp); i.e. (M/mp) is also equal to 1080; the ratio of ‘radius of the electron’ and ‘Planck’s length”; i.e. re/L* is equal to 1020. Even when I divided the energy of the electron (mec2) by hHo got the same large-number 1040; here h is Planck’s constant and Ho is Hubble’s Constant. Hubble’s constant is generally expressed as 530 kilometers per seconds per megaparsac, but it has a dimension of time-1; moreover, it was also noticed by me that the product Ro Ho = c where c is the speed of light; so Ho is some sort of frequency and Ro some wavelength. So the product hHo forms the dimension of energy, and the ratio (mec2/hHo) = 1040 suggests a connection of cosmological red-shift with the strength-ratio. Moreover, all the quantities in this ratio are experimentally measurable, so by comparing it with the strength-ratio we can get the exact value of Hubble’s constant. Explanation for the relative-strength of gravitational and electric forces. In 1997, I saw in Robert K. Adair’s book ‘Concepts in Physics’ a derivation that the sum of energy of mass of the universe and gravitational potential-energy of the universe is equal to zero! That is : Mc^2 – GMM/Ro = 0. …………………….		(1) Based on this in 1997 when I expressed Ro = GM/c^2, Dirac’s ratio became expressible as^14 : Ro/re = (GM/c^2)/(e^2/mec^2). i.e. Ro/re = (GMme)/e^2 and (Ro/re) = (G me mp /e^2) (M/mp)…(2) i.e.   10^40  = 10^-40. 10^80 	Moreover, my 1997 attempt also showed that : (i) the strengths of coupling-constants-squares are directly proportional to the energy-densities of the pion, the electron and the universe; that is : Ng^2 : e^2 : G me mp : : Rho-pi : Rho-e : Rho-c 	………………	(3) Where: Rho-pi, Rho-e and Rho-c are the densities of the pion, the electron and the universe respectively. (ii) The geometric-mean of two gravitational-coupling-constants is equal to electric-coupling-constant-square ; that is :(GMme) (Gmemp)= (e2)^2 i.e. [(GMme). (Gmemp)]^1/2 = e^2 --(4) and Ro. rG(p)= re^2 --(5) Because Ro. = GM/c^2, rG(p) is gravitational-radius of the proton, rG(p) =  Gmp/c^2 and re = e^2/mec^2. The above expressions provided sufficient evidences that the strength of electric force somehow depends on total-mass of the universe. In this November 2008, while I was traveling in Mumbai’s public-transport-bus, a thought came in my mind; “What about the total electrostatic potential-energy of the universe”? That is, the energy which is stored between the electrons and protons in the atoms of the whole universe? Assuming, that this electrostatic potential energy (EU) is also equal to the energy of mass of the universe when I compared the three kinds of energies as follows: ( GMM/Ro) = Mc^2 = Eu --(6)i.e. GMM/Ro = 1836 x 10^80 mec^2.(Because M=10^80 mp and mp = 1836me). i.e. GM (1836 . 1080 me)/Ro = (1836 . 10^80) mec^2 i.e. GMme/Ro = mec^2;i.e. GMme/Ro = e^2/re;i.e. G(10^80 mp me)/ Ro = e^2/re --	(7) i.e. 1040 (Gme mp)/re = e2/re   because Dirac has already found that (Ro/re) = 10^40 so   e^2/Gmemp = 10^40--(8). This is how, the equality of (i) gravitational potential energy, (ii) electrostatic potential energy, and (iii) energy of mass, of the universe leads us to an explanation for the relative-strengths of ‘gravitational’ and ‘electric’ forces. Or, the observed relative-strengths of ‘gravitational’ and ‘electric’ forces suggest the equality of gravitational-potential-energy, electrostatic-potential-energy, and energy of mass of the universe. Now, let us think: what would be the strengths of electric and gravitational forces if the mass of the universe were just 10 raised to 40 proton-masses, or just one proton-mass, or two proton-masses? The expressions-2, 6 and 7 suggest that in that case, both: the strength-ratio and the radius-ratio would reduce to: just 10 raised to 20, one, and 1.414 respectively. This immediately leads us to an in-sight that as more and more atoms get added to the universe, the strength of gravitational force increases like the vector-sum, and not like the arithmetic-sum. This finding supports the finding reported in my earlier manuscripts that both: the rest-mass-energy and kinetic-energy of a ‘particle’ of ‘matter’ are wide-band-waves, and they are in-coherent. It means that the gravitational force among the atoms is because of in-coherent superimposition of the waves, corresponding to their masses. So, the strength-ratio changes with the increase of atoms in the universe as: ( Gravitational-force/Electric-force) = ( Radius of the electron/Radius of the Universe) = (Square-root of : n )/( n), where n is the number of atoms in the universe. Thus, the strength-ratio of gravitational force and the electric force is actually the ratio of ‘in-coherent superimposition’ and ‘ coherent superimposition’ of n number of sets of wideband waves, corresponding to ‘the fundamental particles’ ,contained in the universe. Conclusion Thus, we can explain the relative strengths of ‘gravitational’ and ‘electric’ forces based on equality of (i) the gravitational potential-energy of the universe, (ii) the total amount of electrostatic potential-energy stored in the atoms of the whole universe, and (iii) the energy of total-mass of the universe. Or, the measured relative-strengths of ‘gravitational’ and ‘electric’ forces leads us to a new finding that the gravitational potential-energy, the electrostatic potential-energy and the energy-of-mass, contained in the universe are of equal magnitudes. This finding helps us to get an insight that: the gravitational force arises from in-coherent superimposition of the wideband-waves, corresponding to all the ‘particles’ contained in the universe. References (1)	P A M. Dirac Nature 139 (1937) 323 (2)	P A M. Dirac Proc R Soc London A 165 (1938) 193-208 (3)	A S Eddington Proc Camb Phil Soc 27 (1931) 15-19 (4)	A S Eddington Relativity Theory of Protons and Electrons Cambridge England (1936) (5)	A S Eddington Fundamental Theory Cambridge Univ. Press (1946). (6)	R.H. Dicke science 129 (1954) 621-624. (7)	R.H. Dicke Nature 192 (1961) 440-441. (8)	R.H. Dicke The Theoretical signature of Experimental Relativity Gordon & Breach New York (1964). (9)	S. Hayakawa Proce. Theor. Phys 33 (1965) 538-539. (10)	S. Hayakawa Proce. Theor. Phys. Yukava Thirteenth Anniversary Issue (1965) 532-541. (11)	Carter Large Numbers in Astrophysics and Cosmology (Unpublished Preprint Institute of Theoretical Astronomy, Cambridge, Eng). (12)	Adair R.K. Concepts in Physics Academic Press, New York, (1969), pp 775. (13)	Whichmaan E.W. Quantum Physics Berkeley Physics course, Vol 4. (14)	Tank H.K. An explanation for the large number 1040 in astophysics and some insight into the nature of fundamental forces Proce-Indian. Natin. Sci, Acad, A.63, No. 6, (1997) pp 469-474.

title: Talk:Theory of Everything Project


text:
User talk:Arjen Dijksman

Hello Arjen Dijksman, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 21:32, 11 September 2007 (UTC)

Infinite Hierarchical Nesting of Matter. What do you mean about? Fedosin 02:21, 16 December 2010 (UTC)
 * Good day Arjen Dijksman. It seems you had some questions about

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User talk:Arjen Dijksman


text:
User:TwoOars

w:User:Twooars on Wikipedia

title: User:TwoOars


text:
Continuum mechanics/Entropy inequality

{| cellspacing="0" cellpadding="0" style="margin:0em 0em 1em 0em; width:80%"
 * colspan="2" style="width:80%; horizontal-align:right; vertical-align:top; border:1px solid Sienna; background-color:White;" |

Clausius-Duhem inequality
The Clausius-Duhem inequality can be expressed in integral form as

\cfrac{d}{dt}\left(\int_{\Omega} \rho~\eta~\text{dV}\right) \ge \int_{\partial \Omega} \rho~\eta~(u_n - \mathbf{v}\cdot\mathbf{n})~\text{dA} - \int_{\partial \Omega} \cfrac{\mathbf{q}\cdot\mathbf{n}}{T}~\text{dA} + \int_{\Omega} \cfrac{\rho~s}{T}~\text{dV} ~. $$ In differential form the Clusius-Duhem inequality can be written as

\rho~\dot{\eta} \ge - \boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right) + \cfrac{\rho~s}{T} ~. $$
 * }

Proof:

Assume that $$\Omega$$ is an arbitrary fixed control volume. Then $$u_n = 0$$ and the derivative can be taken inside the integral to give

\int_{\Omega} \frac{\partial }{\partial t}(\rho~\eta)~\text{dV} \ge -\int_{\partial \Omega} \rho~\eta~(\mathbf{v}\cdot\mathbf{n})~\text{dA} - \int_{\partial \Omega} \cfrac{\mathbf{q}\cdot\mathbf{n}}{T}~\text{dA} + \int_{\Omega} \cfrac{\rho~s}{T}~\text{dV} ~. $$ Using the divergence theorem, we get

\int_{\Omega} \frac{\partial }{\partial t}(\rho~\eta)~\text{dV} \ge -\int_{\Omega} \boldsymbol{\nabla} \cdot (\rho~\eta~\mathbf{v})~\text{dV} - \int_{\Omega} \boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right)~\text{dV} + \int_{\Omega} \cfrac{\rho~s}{T}~\text{dV} ~. $$ Since $$\Omega$$ is arbitrary, we must have

\frac{\partial }{\partial t}(\rho~\eta) \ge -\boldsymbol{\nabla} \cdot (\rho~\eta~\mathbf{v}) - \boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right) + \cfrac{\rho~s}{T} ~. $$ Expanding out

\frac{\partial \rho}{\partial t}~\eta + \rho~\frac{\partial \eta}{\partial t} \ge -\boldsymbol{\nabla} (\rho_\eta)\cdot\mathbf{v} - \rho~\eta~(\boldsymbol{\nabla} \cdot \mathbf{v}) - \boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right) + \cfrac{\rho~s}{T} $$ or,

\frac{\partial \rho}{\partial t}~\eta + \rho~\frac{\partial \eta}{\partial t} \ge -\eta~\boldsymbol{\nabla} \rho\cdot\mathbf{v} - \rho~\boldsymbol{\nabla} \eta\cdot\mathbf{v} - \rho~\eta~(\boldsymbol{\nabla} \cdot \mathbf{v}) - \boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right) + \cfrac{\rho~s}{T} $$ or,

\left(\frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \rho\cdot\mathbf{v} + \rho~\boldsymbol{\nabla} \cdot \mathbf{v}\right) ~\eta + \rho~\left(\frac{\partial \eta}{\partial t} + \boldsymbol{\nabla} \eta\cdot\mathbf{v}\right) \ge -\boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right) + \cfrac{\rho~s}{T} ~. $$ Now, the material time derivatives of $$\rho$$ and $$\eta$$ are given by

\dot{\rho} = \frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \rho\cdot\mathbf{v} ~; \dot{\eta} = \frac{\partial \eta}{\partial t} + \boldsymbol{\nabla} \eta\cdot\mathbf{v} ~. $$ Therefore,

\left(\dot{\rho} + \rho~\boldsymbol{\nabla} \cdot \mathbf{v}\right)~\eta + \rho~\dot{\eta} \ge -\boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right) + \cfrac{\rho~s}{T} ~. $$ From the conservation of mass $$\dot{\rho} + \rho~\boldsymbol{\nabla} \cdot \mathbf{v} = 0$$. Hence,

{  \rho~\dot{\eta} \ge -\boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right) + \cfrac{\rho~s}{T} ~. } $$

{| cellspacing="0" cellpadding="0" style="margin:0em 0em 1em 0em; width:80%"
 * colspan="2" style="width:80%; horizontal-align:right; vertical-align:top; border:1px solid Sienna; background-color:White;" |

Clausius-Duhem inequality in terms of internal energy
In terms of the specific entropy, the Clausius-Duhem inequality is written as

\rho~\dot{\eta} \ge - \boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right) + \cfrac{\rho~s}{T} $$ Show that the inequality can be expressed in terms of the internal energy as

\rho~(\dot{e} - T~\dot{\eta}) - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \le - \cfrac{\mathbf{q}\cdot\boldsymbol{\nabla} T}{T} ~. $$
 * }

 Proof:

Using the identity $$ \boldsymbol{\nabla} \cdot (\varphi~\mathbf{v}) = \varphi~\boldsymbol{\nabla} \cdot \mathbf{v} + \mathbf{v}\cdot\boldsymbol{\nabla} \varphi$$ in the Clausius-Duhem inequality, we get

\rho~\dot{\eta} \ge  - \boldsymbol{\nabla} \cdot \left(\cfrac{\mathbf{q}}{T}\right) + \cfrac{\rho~s}{T} \qquad\text{or}\qquad \rho~\dot{\eta} \ge - \cfrac{1}{T}~\boldsymbol{\nabla} \cdot \mathbf{q} - \mathbf{q}\cdot\boldsymbol{\nabla} \left(\cfrac{1}{T}\right) + \cfrac{\rho~s}{T} ~. $$ Now, using index notation with respect to a Cartesian basis $$\mathbf{e}_j$$,

\boldsymbol{\nabla} \left(\cfrac{1}{T}\right) = \frac{\partial }{\partial x_j}\left(T^{-1}\right)~\mathbf{e}_j = -\left(T^{-2}\right)~\frac{\partial T}{\partial x_j}~\mathbf{e}_j = -\cfrac{1}{T^2}~\boldsymbol{\nabla} T ~. $$ Hence,

\rho~\dot{\eta} \ge - \cfrac{1}{T}~\boldsymbol{\nabla} \cdot \mathbf{q} + \cfrac{1}{T^2}~\mathbf{q}\cdot\boldsymbol{\nabla} T       + \cfrac{\rho~s}{T} \qquad\text{or}\qquad \rho~\dot{\eta} \ge -\cfrac{1}{T}\left(\boldsymbol{\nabla} \cdot \mathbf{q} - \rho~s\right) + \cfrac{1}{T^2}~\mathbf{q}\cdot\boldsymbol{\nabla} T ~. $$ Recall the balance of energy

\rho~\dot{e} - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} + \boldsymbol{\nabla} \cdot \mathbf{q} - \rho~s = 0 \qquad \implies \qquad \rho~\dot{e} - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} = - (\boldsymbol{\nabla} \cdot \mathbf{q} - \rho~s) ~. $$ Therefore,

\rho~\dot{\eta} \ge \cfrac{1}{T}\left(\rho~\dot{e}-\boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v}\right) + \cfrac{1}{T^2}~\mathbf{q}\cdot\boldsymbol{\nabla} T   \qquad \implies \qquad \rho~\dot{\eta}~T \ge \rho~\dot{e}-\boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} + \cfrac{\mathbf{q}\cdot\boldsymbol{\nabla} T}{T} ~. $$ Rearranging,

{    \rho~(\dot{e} - T~\dot{\eta}) - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \le - \cfrac{\mathbf{q}\cdot\boldsymbol{\nabla} T}{T} ~. } $$

title: Continuum mechanics/Entropy inequality


text:
Continuum mechanics/Clausius-Duhem inequality for thermoelasticity

Proof: Since $$e = e(\boldsymbol{F}, T)$$, we have

\dot{e} = \frac{\partial e}{\partial \boldsymbol{F}}:\dot{\boldsymbol{F}} + \frac{\partial e}{\partial \eta}~\dot{\eta} ~. $$ Therefore,

\rho~\left(\frac{\partial e}{\partial \boldsymbol{F}}:\dot{\boldsymbol{F}} + \frac{\partial e}{\partial \eta}~\dot{\eta}    - T~\dot{\eta}\right) - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \le - \cfrac{\mathbf{q}\cdot\boldsymbol{\nabla} T}{T} \qquad\text{or}\qquad \rho\left(\frac{\partial e}{\partial \eta} - T\right)~\dot{\eta} + \rho~\frac{\partial e}{\partial \boldsymbol{F}}:\dot{\boldsymbol{F}} - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \le - \cfrac{\mathbf{q}\cdot\boldsymbol{\nabla} T}{T} ~. $$ Now, $$\boldsymbol{\nabla}\mathbf{v} = \boldsymbol{l} = \dot{\boldsymbol{F}}\cdot\boldsymbol{F}^{-1}$$. Therefore, using the identity $$\boldsymbol{A}:(\boldsymbol{B}\cdot\boldsymbol{C}) = (\boldsymbol{A}\cdot\boldsymbol{C}^T):\boldsymbol{B}$$, we have

\boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} = \boldsymbol{\sigma}:(\dot{\boldsymbol{F}}\cdot\boldsymbol{F}^{-1}) = (\boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T}):\dot{\boldsymbol{F}} ~. $$ Hence,

\rho\left(\frac{\partial e}{\partial \eta} - T\right)~\dot{\eta} + \rho~\frac{\partial e}{\partial \boldsymbol{F}}:\dot{\boldsymbol{F}} - (\boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T}):\dot{\boldsymbol{F}} \le - \cfrac{\mathbf{q}\cdot\boldsymbol{\nabla} T}{T} $$ or,

\rho~\left(\frac{\partial e}{\partial \eta} - T\right)~\dot{\eta} + \left(\rho~\frac{\partial e}{\partial \boldsymbol{F}} - \boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T}\right):\dot{\boldsymbol{F}} \le - \cfrac{\mathbf{q}\cdot\boldsymbol{\nabla} T}{T} ~. $$


text:
Continuum mechanics/Balance of energy for thermoelasticity

{| cellspacing="0" cellpadding="0" style="margin:0em 0em 1em 0em; width:80%"
 * colspan="2" style="width:80%; horizontal-align:right; vertical-align:top; border:1px solid Sienna; background-color:White;" |

Balance of energy for thermoelastic materials
Show that, for thermoelastic materials, the balance of energy

\rho~\dot{e} - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} + \boldsymbol{\nabla} \cdot \mathbf{q} - \rho~s = 0 ~. $$ can be expressed as

\rho~T~\dot{\eta} = - \boldsymbol{\nabla} \cdot \mathbf{q} + \rho~s ~. $$
 * }

Proof:

Since $$e = e(\boldsymbol{F}, T)$$, we have

\dot{e} = \frac{\partial e}{\partial \boldsymbol{F}}:\dot{\boldsymbol{F}} + \frac{\partial e}{\partial \eta}~\dot{\eta} ~. $$ Plug into energy equation to get

\rho~\frac{\partial e}{\partial \boldsymbol{F}}:\dot{\boldsymbol{F}} + \rho~\frac{\partial e}{\partial \eta}~\dot{\eta} - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} + \boldsymbol{\nabla} \cdot \mathbf{q} - \rho~s = 0 ~. $$ Recall,

\frac{\partial e}{\partial \eta} = T \qquad\text{and}\qquad \rho~\frac{\partial e}{\partial \boldsymbol{F}} = \boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T} ~. $$ Hence,

(\boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T}):\dot{\boldsymbol{F}} + \rho~T~\dot{\eta} - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} + \boldsymbol{\nabla} \cdot \mathbf{q} - \rho~s = 0 ~. $$ Now, $$\boldsymbol{\nabla}\mathbf{v} = \boldsymbol{l} = \dot{\boldsymbol{F}}\cdot\boldsymbol{F}^{-1}$$. Therefore, using the identity $$\boldsymbol{A}:(\boldsymbol{B}\cdot\boldsymbol{C}) = (\boldsymbol{A}\cdot\boldsymbol{C}^T):\boldsymbol{B}$$, we have

\boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} = \boldsymbol{\sigma}:(\dot{\boldsymbol{F}}\cdot\boldsymbol{F}^{-1}) = (\boldsymbol{\sigma}\cdot\boldsymbol{F}^{-T}):\dot{\boldsymbol{F}} ~. $$ Plugging into the energy equation, we have

\boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} + \rho~T~\dot{\eta} - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} + \boldsymbol{\nabla} \cdot \mathbf{q} - \rho~s = 0 $$ or,

{    \rho~T~\dot{\eta} = - \boldsymbol{\nabla} \cdot \mathbf{q} + \rho~s ~. } $$

{| cellspacing="0" cellpadding="0" style="margin:0em 0em 1em 0em; width:80%"
 * colspan="2" style="width:80%; horizontal-align:right; vertical-align:top; border:1px solid Sienna; background-color:White;" |

Rate of internal energy/entropy for thermoelastic materials
For thermoelastic materials, the specific internal energy is given by

e = \bar{e}(\boldsymbol{E}, \eta) $$ where $$\boldsymbol{E}$$ is the Green strain and $$\eta$$ is the specific entropy. Show that

\cfrac{d}{dt}(e - T~\eta) = - \dot{T}~\eta + \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} \qquad\text{and}\qquad \cfrac{d}{dt}(e - T~\eta - \cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{E}) = - \dot{T}~\eta - \cfrac{1}{\rho_0}~\dot{\boldsymbol{S}}:\boldsymbol{E} $$ where $$\rho_0$$ is the initial density, $$T$$ is the absolute temperature, $$\boldsymbol{S}$$ is the 2nd Piola-Kirchhoff stress, and a dot over a quantity indicates the material time derivative.
 * }

Taking the material time derivative of the specific internal energy, we get

\dot{e} = \frac{\partial \bar{e}}{\partial \boldsymbol{E}}:\dot{\boldsymbol{E}} + \frac{\partial \bar{e}}{\partial \eta}~\dot{\eta} ~. $$ Now, for thermoelastic materials,

T = \frac{\partial \bar{e}}{\partial \eta} \qquad \text{and} \qquad \boldsymbol{S} = \rho_0~\frac{\partial \bar{e}}{\partial \boldsymbol{E}} ~. $$ Therefore,

\dot{e} = \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} + T~\dot{\eta} ~. \qquad \implies \qquad \dot{e} - T~\dot{\eta} = \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} ~. $$ Now,

\cfrac{d}{dt}(T~\eta) = \dot{T}~\eta + T~\dot{\eta} ~. $$ Therefore,

\dot{e} - \cfrac{d}{dt}(T~\eta) + \dot{T}~\eta = \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} \qquad \implies \qquad {  \cfrac{d}{dt}(e - T~\eta) = -\dot{T}~\eta + \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} ~. } $$ Also,

\cfrac{d}{dt}\left(\cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{E}\right) = \cfrac{1}{\rho_0}~\boldsymbol{S}:\dot{\boldsymbol{E}} + \cfrac{1}{\rho_0}~\dot{\boldsymbol{S}}:\boldsymbol{E} ~. $$ Hence,

\dot{e} - \cfrac{d}{dt}(T~\eta) + \dot{T}~\eta = \cfrac{d}{dt}\left(\cfrac{1}{\rho_0} \boldsymbol{S}:\boldsymbol{E}\right) - \cfrac{1}{\rho_0}~\dot{\boldsymbol{S}}:\boldsymbol{E} \qquad \implies \qquad {  \cfrac{d}{dt}\left(e - T~\eta - \cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{E}\right) = - \dot{T}~\eta - \cfrac{1}{\rho_0}~\dot{\boldsymbol{S}}:\boldsymbol{E} ~. } $$

{| cellspacing="0" cellpadding="0" style="margin:0em 0em 1em 0em; width:80%"
 * colspan="2" style="width:80%; horizontal-align:right; vertical-align:top; border:1px solid Sienna; background-color:White;" |

Energy equation for thermoelastic materials
For thermoelastic materials, show that the balance of energy equation

\rho~T~\dot{\eta} = - \boldsymbol{\nabla} \cdot \mathbf{q} + \rho~s $$ can be expressed as either

\rho~C_v~\dot{T} = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s +\cfrac{\rho}{\rho_0}~T~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}:\dot{\boldsymbol{E}} $$ or

\rho~\left(C_p - \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}\right) ~\dot{T} = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s -\cfrac{\rho}{\rho_0}~T~\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}:\dot{\boldsymbol{S}} $$ where

C_v = \frac{\partial \hat{e}(\boldsymbol{E},T)}{\partial T} \qquad \text{and} \qquad C_p = \frac{\partial \tilde{e}(\boldsymbol{S},T)}{\partial T} ~. $$ For the special case where there are no sources and we can ignore heat conduction (for very fast processes), the energy equation simplifies to

\rho~\left(C_p - \cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{\alpha}\right) ~\dot{T} =     -\cfrac{\rho}{\rho_0}~T~\boldsymbol{\alpha}:\dot{\boldsymbol{S}} $$ where $$\boldsymbol{\alpha} := \frac{\partial \tilde{\boldsymbol{E}}}{\partial T}$$ is the thermal expansion tensor which has the form $$\boldsymbol{\alpha} = \alpha\boldsymbol{1}$$ for isotropic materials and $$\alpha\,$$ is the coefficient of thermal expansion. The above equation can be used to calculate the change of temperature in thermoelasticity.
 * }

Proof:

If the independent variables are $$\boldsymbol{E}$$ and $$T$$, then

\eta = \hat{\eta}(\boldsymbol{E}, T) \qquad \implies \qquad \dot{\eta} = \frac{\partial \hat{\eta}}{\partial \boldsymbol{E}}:\dot{\boldsymbol{E}} + \frac{\partial \hat{\eta}}{\partial T}~\dot{T} ~. $$ On the other hand, if we consider $$\boldsymbol{S}$$ and $$T$$ to be the independent variables

\eta = \tilde{\eta}(\boldsymbol{S}, T) \qquad \implies \qquad \dot{\eta} = \frac{\partial \tilde{\eta}}{\partial \boldsymbol{S}}:\dot{\boldsymbol{S}} + \frac{\partial \tilde{\eta}}{\partial T}~\dot{T} ~. $$ Since

\frac{\partial \hat{\eta}}{\partial \boldsymbol{E}} = -\cfrac{1}{\rho_0}~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}   ~; \frac{\partial \hat{\eta}}{\partial T} = \cfrac{C_v}{T} ~; \frac{\partial \tilde{\eta}}{\partial \boldsymbol{S}} = \cfrac{1}{\rho_0}~\frac{\partial \tilde{\boldsymbol{E}}}{\partial T} ~;\text{and} \frac{\partial \tilde{\eta}}{\partial T} = \cfrac{1}{T}\left(C_p - \cfrac{1}{\rho_0}    \boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}\right) $$ we have, either

\dot{\eta} = -\cfrac{1}{\rho_0}~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}:\dot{\boldsymbol{E}} + \cfrac{C_v}{T}~\dot{T} $$ or

\dot{\eta} = \cfrac{1}{\rho_0}~\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}:\dot{\boldsymbol{S}} + \cfrac{1}{T}\left(C_p - \cfrac{1}{\rho_0}    \boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}\right)~\dot{T} ~. $$ The equation for balance of energy in terms of the specific entropy is

\rho~T~\dot{\eta} = - \boldsymbol{\nabla} \cdot \mathbf{q} + \rho~s ~. $$ Using the two forms of $$\dot{\eta}$$, we get two forms of the energy equation:

-\cfrac{\rho}{\rho_0}~T~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}:\dot{\boldsymbol{E}} + \rho~C_v~\dot{T} = - \boldsymbol{\nabla} \cdot \mathbf{q} + \rho~s $$ and

\cfrac{\rho}{\rho_0}~T~\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}:\dot{\boldsymbol{S}} + \rho~C_p~\dot{T} - \cfrac{\rho}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}~\dot{T} = - \boldsymbol{\nabla} \cdot \mathbf{q} + \rho~s ~. $$ From Fourier's law of heat conduction

\mathbf{q} = - \boldsymbol{\kappa}\cdot\boldsymbol{\nabla} T ~. $$ Therefore,

-\cfrac{\rho}{\rho_0}~T~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}:\dot{\boldsymbol{E}} + \rho~C_v~\dot{T} = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s $$ and

\cfrac{\rho}{\rho_0}~T~\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}:\dot{\boldsymbol{S}} + \rho~C_p~\dot{T} - \cfrac{\rho}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}~\dot{T} = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s ~. $$ Rearranging,

{ \rho~C_v~\dot{T}  = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s +\cfrac{\rho}{\rho_0}~T~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}:\dot{\boldsymbol{E}} } $$ or,

{ \rho~\left(C_p - \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}\right) ~\dot{T} = \boldsymbol{\nabla} \cdot (\boldsymbol{\kappa}\cdot\boldsymbol{\nabla T)} + \rho~s -\cfrac{\rho}{\rho_0}~T~\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}:\dot{\boldsymbol{S}} ~. } $$


text:
Continuum mechanics/Stress-strain relation for thermoelasticity

Proof:

Recall that the Cauchy stress is given by

\boldsymbol{\sigma} = \rho~\frac{\partial e}{\partial \boldsymbol{F}}\cdot\boldsymbol{F}^T \qquad \implies \qquad \sigma_{ij} = \rho~\frac{\partial e}{\partial F_{ik}}F^T_{kj} = \rho~\frac{\partial e}{\partial F_{ik}}F_{jk} ~. $$ The Green strain $$\boldsymbol{E} = \boldsymbol{E}(\boldsymbol{F}) = \boldsymbol{E}(\boldsymbol{U})$$ and $$e = e(\boldsymbol{F},\eta) = e(\boldsymbol{U},\eta)$$. Hence, using the chain rule,

\frac{\partial e}{\partial \boldsymbol{F}} = \frac{\partial e}{\partial \boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial \boldsymbol{F}} \qquad \implies \qquad \frac{\partial e}{\partial F_{ik}} = \frac{\partial e}{\partial E_{lm}}~\frac{\partial E_{lm}}{\partial F_{ik}} ~. $$ Now,

\boldsymbol{E} = \frac{1}{2}(\boldsymbol{F}^T\cdot\boldsymbol{F} - \boldsymbol{\mathit{1}}) \qquad \implies \qquad E_{lm} = \frac{1}{2}(F^T_{lp}~F_{pm} - \delta_{lm}) = \frac{1}{2}(F_{pl}~F_{pm} - \delta_{lm}) ~. $$ Taking the derivative with respect to $$\boldsymbol{F}$$, we get

\frac{\partial \boldsymbol{E}}{\partial \boldsymbol{F}} = \frac{1}{2}\left(\frac{\partial \boldsymbol{F}^T}{\partial \boldsymbol{F}}\cdot\boldsymbol{F} +      \boldsymbol{F}^T\cdot\frac{\partial \boldsymbol{F}}{\partial \boldsymbol{F}}\right) \qquad \implies \qquad \frac{\partial E_{lm}}{\partial F_{ik}} = \frac{1}{2}\left(\frac{\partial F_{pl}}{\partial F_{ik}}~F_{pm} +      F_{pl}~\frac{\partial F_{pm}}{\partial F_{ik}}\right) ~. $$ Therefore,

\boldsymbol{\sigma} = \frac{1}{2}~\rho~\left[\frac{\partial e}{\partial \boldsymbol{E}}: \left(\frac{\partial \boldsymbol{F}^T}{\partial \boldsymbol{F}}\cdot\boldsymbol{F} +      \boldsymbol{F}^T\cdot\frac{\partial \boldsymbol{F}}{\partial \boldsymbol{F}}\right)\right]\cdot\boldsymbol{F}^T \qquad \implies \qquad \sigma_{ij} = \frac{1}{2}~\rho~\left[\frac{\partial e}{\partial E_{lm}} \left(\frac{\partial F_{pl}}{\partial F_{ik}}~F_{pm} +      F_{pl}~\frac{\partial F_{pm}}{\partial F_{ik}}\right)\right]~F_{jk} ~. $$ Recall,

\frac{\partial \boldsymbol{A}}{\partial \boldsymbol{A}} \equiv \frac{\partial A_{ij}}{\partial A_{kl}} = \delta_{ik}~\delta_{jl} \qquad \text{and} \qquad \frac{\partial \boldsymbol{A}^T}{\partial \boldsymbol{A}} \equiv \frac{\partial A_{ji}}{\partial A_{kl}} = \delta_{jk}~\delta_{il} ~. $$ Therefore,

\sigma_{ij} = \frac{1}{2}~\rho~\left[\frac{\partial e}{\partial E_{lm}} \left(\delta_{pi}~\delta_{lk}~F_{pm} +      F_{pl}~\delta_{pi}~\delta_{mk}\right)\right]~F_{jk} = \frac{1}{2}~\rho~\left[\frac{\partial e}{\partial E_{lm}} \left(\delta_{lk}~F_{im} +      F_{il}~\delta_{mk}\right)\right]~F_{jk} $$ or,

\sigma_{ij} = \frac{1}{2}~\rho~\left[\frac{\partial e}{\partial E_{km}}~F_{im} + \frac{\partial e}{\partial E_{lk}}~F_{il}\right]~F_{jk} \qquad \implies \qquad \boldsymbol{\sigma} = \frac{1}{2}~\rho~\left[\boldsymbol{F}\cdot\left(\frac{\partial e}{\partial \boldsymbol{E}}\right)^T + \boldsymbol{F}\cdot\frac{\partial e}{\partial \boldsymbol{E}}\right]\cdot\boldsymbol{F}^T $$ or,

\boldsymbol{\sigma} = \frac{1}{2}~\rho~\boldsymbol{F}\cdot\left[\left(\frac{\partial e}{\partial \boldsymbol{E}}\right)^T + \frac{\partial e}{\partial \boldsymbol{E}}\right]\cdot\boldsymbol{F}^T ~. $$ From the symmetry of the Cauchy stress, we have

\boldsymbol{\sigma} = (\boldsymbol{F}\cdot\boldsymbol{A})\cdot\boldsymbol{F}^T \qquad \text{and} \qquad \boldsymbol{\sigma}^T = \boldsymbol{F}\cdot(\boldsymbol{F}\cdot\boldsymbol{A})^T = \boldsymbol{F}\cdot\boldsymbol{A}^T\cdot\boldsymbol{F}^T \qquad \text{and} \qquad \boldsymbol{\sigma} = \boldsymbol{\sigma}^T \implies \boldsymbol{A} = \boldsymbol{A}^T ~. $$ Therefore,

\frac{\partial e}{\partial \boldsymbol{E}} = \left(\frac{\partial e}{\partial \boldsymbol{E}}\right)^T $$ and we get

{ \boldsymbol{\sigma} = ~\rho~\boldsymbol{F}\cdot\frac{\partial e}{\partial \boldsymbol{E}}\cdot\boldsymbol{F}^T ~. } $$


text:
Continuum mechanics/Maxwell relations for thermoelasticity

{| cellspacing="0" cellpadding="0" style="margin:0em 0em 1em 0em; width:80%"
 * colspan="2" style="width:80%; horizontal-align:right; vertical-align:top; border:1px solid Sienna; background-color:White;" |

Maxwell relations between thermodynamic quantities
For thermoelastic materials, show that the following relations hold:

\frac{\partial \psi}{\partial \boldsymbol{E}} = \cfrac{1}{\rho_0}~\hat{\boldsymbol{S}}(\boldsymbol{E},T) ~; \frac{\partial \psi}{\partial T} = -\hat{\eta}(\boldsymbol{E},T) ~; \frac{\partial g}{\partial \boldsymbol{S}} = \cfrac{1}{\rho_0}~\tilde{\boldsymbol{E}}(\boldsymbol{S}, T) ~; \frac{\partial g}{\partial T} = \tilde{\eta}(\boldsymbol{S}, T)   $$ where $$\psi(\boldsymbol{E},T)$$ is the Helmholtz free energy and $$g(\boldsymbol{S},T)$$ is the Gibbs free energy.

Also show that

\frac{\partial \hat{\boldsymbol{S}}}{\partial T} = - \rho_0~\frac{\partial \hat{\eta}}{\partial \boldsymbol{E}} \qquad\text{and}\qquad \frac{\partial \tilde{\boldsymbol{E}}}{\partial T} = \rho_0~\frac{\partial \tilde{\eta}}{\partial \boldsymbol{S}} ~. $$
 * }

Proof:

Recall that

\psi(\boldsymbol{E}, T) = e - T~\eta = \bar{e}(\boldsymbol{E}, \eta) - T~\eta ~. $$ and

g(\boldsymbol{S}, T) = - e + T~\eta + \cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{E} ~. $$ (Note that we can choose any functional dependence that we like, because the quantities $$e$$, $$\eta$$, $$\boldsymbol{E}$$ are the actual quantities and not any particular functional relations).

The derivatives are

\frac{\partial \psi}{\partial \boldsymbol{E}} = \frac{\partial \bar{e}}{\partial \boldsymbol{E}} = \cfrac{1}{\rho_0}~\boldsymbol{S} ~;\qquad \frac{\partial \psi}{\partial T} = - \eta ~. $$ and

\frac{\partial g}{\partial \boldsymbol{S}} = \cfrac{1}{\rho_0}~\frac{\partial \boldsymbol{S}}{\partial \boldsymbol{S}}:\boldsymbol{E} = \cfrac{1}{\rho_0}~\boldsymbol{E} ~;\qquad \frac{\partial g}{\partial T} = \eta ~. $$ Hence,

{ \frac{\partial \psi}{\partial \boldsymbol{E}}  = \cfrac{1}{\rho_0}~\hat{\boldsymbol{S}}(\boldsymbol{E},T) ~; \frac{\partial \psi}{\partial T} = -\hat{\eta}(\boldsymbol{E},T) ~; \frac{\partial g}{\partial \boldsymbol{S}} = \cfrac{1}{\rho_0}~\tilde{\boldsymbol{E}}(\boldsymbol{S}, T) ~; \frac{\partial g}{\partial T} = \tilde{\eta}(\boldsymbol{S}, T)   } $$

From the above, we have

\frac{\partial^2 \psi}{\partial T\partial\boldsymbol{E}} = \frac{\partial^2 \psi}{\partial \boldsymbol{E}\partial T}   \qquad\implies\qquad -\frac{\partial \hat{\eta}}{\partial \boldsymbol{E}} = \cfrac{1}{\rho_0}\frac{\partial \hat{\boldsymbol{S}}}{\partial T} ~. $$ and

\frac{\partial^2 g}{\partial T\partial\boldsymbol{S}} = \frac{\partial^2 g}{\partial \boldsymbol{S}\partial T}   \qquad\implies\qquad \frac{\partial \tilde{\eta}}{\partial \boldsymbol{S}} = \cfrac{1}{\rho_0}\frac{\partial \tilde{\boldsymbol{E}}}{\partial T} ~. $$ Hence,

{   \frac{\partial \hat{\boldsymbol{S}}}{\partial T} = - \rho_0~\frac{\partial \hat{\eta}}{\partial \boldsymbol{E}} \qquad\text{and}\qquad \frac{\partial \tilde{\boldsymbol{E}}}{\partial T} = \rho_0~\frac{\partial \tilde{\eta}}{\partial \boldsymbol{S}} ~. } $$

{| cellspacing="0" cellpadding="0" style="margin:0em 0em 1em 0em; width:80%"
 * colspan="2" style="width:80%; horizontal-align:right; vertical-align:top; border:1px solid Sienna; background-color:White;" |

More Maxwell relations
For thermoelastic materials, show that the following relations hold:

\frac{\partial \hat{e}(\boldsymbol{E},T)}{\partial T} = T~\frac{\partial \hat{\eta}}{\partial T} = -T~\frac{\partial^2 \hat{\psi}}{\partial T^2} $$ and

\frac{\partial \tilde{e}(\boldsymbol{S},T)}{\partial T} = T~\frac{\partial \tilde{\eta}}{\partial T} + \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T} = T~\frac{\partial^2 \tilde{g}}{\partial T^2} + \boldsymbol{S}:\frac{\partial^2 \tilde{g}}{\partial \boldsymbol{S}\partial T}~. $$
 * }

Proof:

Recall,

\hat{\psi}(\boldsymbol{E},T) = \psi = e - T~\eta = \hat{e}(\boldsymbol{E}, T) - T~\hat{\eta}(\boldsymbol{E}, T) $$ and

\tilde{g}(\boldsymbol{S},T) = g = - e + T~\eta + \cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{E} = -\tilde{e}(\boldsymbol{S}, T) + T~\tilde{\eta}(\boldsymbol{S}, T) + \cfrac{1}{\rho_0}~\boldsymbol{S}:\tilde{\boldsymbol{E}}(\boldsymbol{S}, T)~. $$ Therefore,

\frac{\partial \hat{e}(\boldsymbol{E},T)}{\partial T} = \frac{\partial \hat{\psi}}{\partial T}     + \hat{\eta}(\boldsymbol{E}, T) + T~\frac{\partial \hat{\eta}}{\partial T} $$ and

\frac{\partial \tilde{e}(\boldsymbol{S},T)}{\partial T} = - \frac{\partial \tilde{g}}{\partial T}     + \tilde{\eta}(\boldsymbol{S},T) + T~\frac{\partial \tilde{\eta}}{\partial T}     + \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T} ~. $$ Also, recall that

\hat{\eta}(\boldsymbol{E},T) = -\frac{\partial \hat{\psi}}{\partial T}    \qquad \implies \qquad \frac{\partial \hat{\eta}}{\partial T} = -\frac{\partial^2 \hat{\psi}}{\partial T^2}  ~, $$

\tilde{\eta}(\boldsymbol{S}, T) = \frac{\partial \tilde{g}}{\partial T}    \qquad \implies \qquad \frac{\partial \tilde{\eta}}{\partial T} = \frac{\partial^2 \tilde{g}}{\partial T^2}  ~, $$ and

\tilde{\boldsymbol{E}}(\boldsymbol{S}, T) = \rho_0~\frac{\partial \tilde{g}}{\partial \boldsymbol{S}} \qquad \implies \qquad \frac{\partial \tilde{\boldsymbol{E}}}{\partial T} = \rho_0~\frac{\partial^2 \tilde{g}}{\partial \boldsymbol{S}\partial T}~. $$ Hence,

{ \frac{\partial \hat{e}(\boldsymbol{E},T)}{\partial T} =  T~\frac{\partial \hat{\eta}}{\partial T} = -T~\frac{\partial^2 \hat{\psi}}{\partial T^2} } $$ and

{  \frac{\partial \tilde{e}(\boldsymbol{S},T)}{\partial T} = T~\frac{\partial \tilde{\eta}}{\partial T} + \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T} = T~\frac{\partial^2 \tilde{g}}{\partial T^2} + \boldsymbol{S}:\frac{\partial^2 \tilde{g}}{\partial \boldsymbol{S}\partial T}~. } $$


text:
Continuum mechanics/Specific heats of thermoelastic materials

{| cellspacing="0" cellpadding="0" style="margin:0em 0em 1em 0em; width:80%"
 * colspan="2" style="width:80%; horizontal-align:right; vertical-align:top; border:1px solid Sienna; background-color:White;" |

Relation between specific heats - 1
For thermoelastic materials, show that the specific heats are related by the relation

C_p - C_v = \cfrac{1}{\rho_0}\left(\boldsymbol{S}-T~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}\right): \frac{\partial \tilde{\boldsymbol{E}}}{\partial T} ~. $$
 * }

Proof:

Recall that

C_v := \frac{\partial \hat{e}(\boldsymbol{E},T)}{\partial T} = T~\frac{\partial \hat{\eta}}{\partial T} $$ and

C_p := \frac{\partial \tilde{e}(\boldsymbol{S},T)}{\partial T} = T~\frac{\partial \tilde{\eta}}{\partial T} + \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T} ~. $$ Therefore,

C_p - C_v = T~\frac{\partial \tilde{\eta}}{\partial T}       + \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}         - T~\frac{\partial \hat{\eta}}{\partial T} ~. $$ Also recall that

\eta = \hat{\eta}(\boldsymbol{E}, T) = \tilde{\eta}(\boldsymbol{S}, T) ~. $$ Therefore, keeping $$\boldsymbol{S}$$ constant while differentiating, we have

\frac{\partial \tilde{\eta}}{\partial T} = \frac{\partial \hat{\eta}}{\partial \boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial T} + \frac{\partial \hat{\eta}}{\partial T} ~. $$ Noting that $$\boldsymbol{E} = \tilde{\boldsymbol{E}}(\boldsymbol{S},T)$$, and plugging back into the equation for the difference between the two specific heats, we have

C_p - C_v = T~\frac{\partial \hat{\eta}}{\partial \boldsymbol{E}}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}       + \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \tilde{\boldsymbol{E}}}{\partial T}  ~. $$ Recalling that

\frac{\partial \hat{\eta}}{\partial \boldsymbol{E}} = - \cfrac{1}{\rho_0}~\frac{\partial \hat{\boldsymbol{S}}}{\partial T} $$ we get

{   C_p - C_v = \cfrac{1}{\rho_0}\left(\boldsymbol{S} -T~\frac{\partial \hat{\boldsymbol{S}}}{\partial T}\right): \frac{\partial \tilde{\boldsymbol{E}}}{\partial T} ~. } $$

{| cellspacing="0" cellpadding="0" style="margin:0em 0em 1em 0em; width:80%"
 * colspan="2" style="width:80%; horizontal-align:right; vertical-align:top; border:1px solid Sienna; background-color:White;" |

Relation between specific heats - 2
For thermoelastic materials, show that the specific heats can also be related by the equations

C_p - C_v = \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \boldsymbol{E}}{\partial T} + \frac{\partial \boldsymbol{E}}{\partial T}:\left(\frac{\partial^2 \psi}{\partial \boldsymbol{E}\partial\boldsymbol{E}}:        \frac{\partial \boldsymbol{E}}{\partial T}\right) = \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \boldsymbol{E}}{\partial T} + \cfrac{T}{\rho_0}~ \frac{\partial \boldsymbol{E}}{\partial T}:\left(\frac{\partial \boldsymbol{S}}{\partial \boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial T}\right) ~. $$ We can also write the above as

C_p - C_v = \cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{\alpha}_E + \cfrac{T}{\rho_0}~\boldsymbol{\alpha}_E:\boldsymbol{\mathsf{C}}:\boldsymbol{\alpha}_E $$ where $$\boldsymbol{\alpha}_E := \frac{\partial \boldsymbol{E}}{\partial T} $$ is the thermal expansion tensor and $$\boldsymbol{\mathsf{C}} := \frac{\partial \boldsymbol{S}}{\partial \boldsymbol{E}}$$ is the stiffness tensor.
 * }

Proof:

Recall that

\boldsymbol{S} = \rho_0~\frac{\partial \psi}{\partial \boldsymbol{E}} = \rho_0~\boldsymbol{f}(\boldsymbol{E}(\boldsymbol{S},T),T)~. $$ Recall the chain rule which states that if

g(u,t) = f(x(u,t), y(u,t)) $$ then, if we keep $$u$$ fixed, the partial derivative of $$g$$ with respect to $$t$$ is given by

\frac{\partial g}{\partial t} = \frac{\partial f}{\partial x}~\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}~\frac{\partial y}{\partial t} ~. $$ In our case,

u = \boldsymbol{S}, t = T, g(\boldsymbol{S}, T) = \boldsymbol{S}, x(\boldsymbol{S},T) = \boldsymbol{E}(\boldsymbol{S},T), y(\boldsymbol{S},T) = T, \text{and} f = \rho_0~\boldsymbol{f}. $$ Hence, we have

\boldsymbol{S} = g(\boldsymbol{S}, T) = f(\boldsymbol{E}(\boldsymbol{S},T), T) = \rho_0~\boldsymbol{f}(\boldsymbol{E}(\boldsymbol{S},T),T)~. $$ Taking the derivative with respect to $$T$$ keeping $$\boldsymbol{S}$$ constant, we have

\frac{\partial g}{\partial T} = \frac{\partial \boldsymbol{S}}{\partial T} = \rho_0~\left[\frac{\partial \boldsymbol{f}}{\partial \boldsymbol{E}}: \frac{\partial \boldsymbol{E}}{\partial T} +\frac{\partial \boldsymbol{f}}{\partial T}~\frac{\partial T}{\partial T}\right] $$ or,

\mathbf{0} = \frac{\partial \boldsymbol{f}}{\partial \boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial T} + \frac{\partial \boldsymbol{f}}{\partial T}~. $$ Now,

\boldsymbol{f} = \frac{\partial \psi}{\partial \boldsymbol{E}} \qquad \implies \qquad \frac{\partial \boldsymbol{f}}{\partial \boldsymbol{E}} = \frac{\partial^2 \psi}{\partial \boldsymbol{E}\partial\boldsymbol{E}} \quad \text{and} \quad \frac{\partial \boldsymbol{f}}{\partial T} = \frac{\partial^2 \psi}{\partial T\partial\boldsymbol{E}} ~. $$ Therefore,

\mathbf{0} = \frac{\partial^2 \psi}{\partial \boldsymbol{E}\partial\boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial T} + \frac{\partial^2 \psi}{\partial T\partial\boldsymbol{E}} = \frac{\partial }{\partial \boldsymbol{E}}\left(\frac{\partial \psi}{\partial \boldsymbol{E}}\right): \frac{\partial \boldsymbol{E}}{\partial T} + \frac{\partial }{\partial T}\left(\frac{\partial \psi}{\partial \boldsymbol{E}}\right) ~. $$ Again recall that,

\frac{\partial \psi}{\partial \boldsymbol{E}} = \cfrac{1}{\rho_0}~\boldsymbol{S} ~. $$ Plugging into the above, we get

\mathbf{0} = \frac{\partial^2 \psi}{\partial \boldsymbol{E}\partial\boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial T} + \cfrac{1}{\rho_0}~\frac{\partial \boldsymbol{S}}{\partial T} = \cfrac{1}{\rho_0}~\frac{\partial \boldsymbol{S}}{\partial \boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial T} + \cfrac{1}{\rho_0}~\frac{\partial \boldsymbol{S}}{\partial T} ~. $$ Therefore, we get the following relation for $$\partial \boldsymbol{S}/\partial T$$:

\frac{\partial \boldsymbol{S}}{\partial T}      = - \rho_0~\frac{\partial^2 \psi}{\partial \boldsymbol{E}\partial\boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial T}       = - \frac{\partial \boldsymbol{S}}{\partial \boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial T} ~. $$ Recall that

C_p - C_v = \cfrac{1}{\rho_0}\left(\boldsymbol{S}-T~\frac{\partial \boldsymbol{S}}{\partial T}\right): \frac{\partial \boldsymbol{E}}{\partial T} ~. $$ Plugging in the expressions for $$\partial \boldsymbol{S}/\partial T$$ we get:

C_p - C_v = \cfrac{1}{\rho_0} \left(\boldsymbol{S}+T~\rho_0~\frac{\partial^2 \psi}{\partial \boldsymbol{E}\partial\boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial T}       \right): \frac{\partial \boldsymbol{E}}{\partial T}      = \cfrac{1}{\rho_0} \left(\boldsymbol{S}+T~\frac{\partial \boldsymbol{S}}{\partial \boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial T}\right) :\frac{\partial \boldsymbol{E}}{\partial T} ~. $$ Therefore,

C_p - C_v = \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \boldsymbol{E}}{\partial T} + T~\left(\frac{\partial^2 \psi}{\partial \boldsymbol{E}\partial\boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial T}       \right): \frac{\partial \boldsymbol{E}}{\partial T}      = \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \boldsymbol{E}}{\partial T} + \cfrac{T}{\rho_0}~\left(\frac{\partial \boldsymbol{S}}{\partial \boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial T}\right) :\frac{\partial \boldsymbol{E}}{\partial T} ~. $$ Using the identity $$ (\boldsymbol{\mathsf{A}}:\boldsymbol{B}):\boldsymbol{C} = \boldsymbol{C}:(\boldsymbol{\mathsf{A}}:\boldsymbol{B})$$, we have

{  C_p - C_v = \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \boldsymbol{E}}{\partial T} + T~\frac{\partial \boldsymbol{E}}{\partial T}:\left(\frac{\partial^2 \psi}{\partial \boldsymbol{E}\partial\boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial T}       \right) = \cfrac{1}{\rho_0}~\boldsymbol{S}:\frac{\partial \boldsymbol{E}}{\partial T} + \cfrac{T}{\rho_0}~ \frac{\partial \boldsymbol{E}}{\partial T}:\left(\frac{\partial \boldsymbol{S}}{\partial \boldsymbol{E}}:\frac{\partial \boldsymbol{E}}{\partial T}\right) ~.  } $$ {| cellspacing="0" cellpadding="0" style="margin:0em 0em 1em 0em; width:80%"
 * colspan="2" style="width:80%; horizontal-align:right; vertical-align:top; border:1px solid Sienna; background-color:White;" |

Specific heats of Saint-Venant–Kirchhoff material
Consider an isotropic thermoelastic material that has a constant coefficient of thermal expansion and which follows the Saint-Venant–Kirchhoff model, i.e,

\boldsymbol{\alpha}_E = \alpha~\boldsymbol{\mathit{1}} \qquad\text{and}\qquad \boldsymbol{\mathsf{C}} = \lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2\mu~\boldsymbol{\mathsf{I}} $$ where $$\alpha$$ is the coefficient of thermal expansion and $$3~\lambda = 3~K - 2~\mu$$ where $$K, \mu$$ are the bulk and shear moduli, respectively.

Show that the specific heats related by the equation

C_p - C_v = \cfrac{1}{\rho_0}\left[\alpha~\text{tr}{\boldsymbol{S}} + 9~\alpha^2~K~T\right]~. $$
 * }

Proof:

Recall that,

C_p - C_v = \cfrac{1}{\rho_0}~\boldsymbol{S}:\boldsymbol{\alpha}_E + \cfrac{T}{\rho_0}~\boldsymbol{\alpha}_E:\boldsymbol{\mathsf{C}}:\boldsymbol{\alpha}_E ~. $$ Plugging the expressions of $$\boldsymbol{\alpha}_E$$ and $$\boldsymbol{\mathsf{C}}$$ into the above equation, we have

\begin{align} C_p - C_v & = \cfrac{1}{\rho_0}~\boldsymbol{S}:(\alpha~\boldsymbol{\mathit{1}}) + \cfrac{T}{\rho_0}~(\alpha~\boldsymbol{\mathit{1}}): (\lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2\mu~\boldsymbol{\mathsf{I}}): (\alpha~\boldsymbol{\mathit{1}}) \\ & = \cfrac{\alpha}{\rho_0}~\text{tr}{\boldsymbol{S}} + \cfrac{\alpha^2~T}{\rho_0}~\boldsymbol{\mathit{1}}: (\lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2\mu~\boldsymbol{\mathsf{I}}): \boldsymbol{\mathit{1}} \\ & = \cfrac{\alpha}{\rho_0}~\text{tr}{\boldsymbol{S}} + \cfrac{\alpha^2~T}{\rho_0}~\boldsymbol{\mathit{1}}: (\lambda~\text{tr}{\boldsymbol{\mathit{1}}}~\boldsymbol{\mathit{1}} + 2\mu~\boldsymbol{\mathit{1}})\\ & = \cfrac{\alpha}{\rho_0}~\text{tr}{\boldsymbol{S}} + \cfrac{\alpha^2~T}{\rho_0}~ (3~\lambda~\text{tr}{\boldsymbol{\mathit{1}}} + 2\mu~\text{tr}{\boldsymbol{\mathit{1}}})\\ & = \cfrac{\alpha}{\rho_0}~\text{tr}{\boldsymbol{S}} + \cfrac{3~\alpha^2~T}{\rho_0}~ (3~\lambda + 2\mu)\\ & = \cfrac{\alpha~\text{tr}{\boldsymbol{S}}}{\rho_0} + \cfrac{9~\alpha^2~K~T}{\rho_0}~. \end{align} $$ Therefore,

{  C_p - C_v = \cfrac{1}{\rho_0}\left[\alpha~\text{tr}{\boldsymbol{S}} + 9~\alpha^2~K~T\right]~. } $$


text:
BCP/Medeola virginiana

Medeola virginiana

title: BCP/Medeola virginiana


text:
BCP/Medeola virginiana/Logs

2008 Logs

 * --Jomegat 22:31, 3 June 2008 (UTC)
 * --Jomegat 21:18, 21 June 2008 (UTC)
 * --Jomegat 23:16, 1 July 2008 (UTC)

2009 Logs

 * Jomegat 00:21, 25 May 2009 (UTC)
 * Jomegat 00:19, 29 May 2009 (UTC)
 * Jomegat 00:31, 2 June 2009 (UTC)
 * Jomegat 00:27, 9 June 2009 (UTC)
 * Jomegat 00:19, 30 June 2009 (UTC)

Recent Logs

 * Jomegat 00:27, 30 May 2010 (UTC)

title: BCP/Medeola virginiana/Logs


text: Continuum mechanics/Tensor algebra identities


text: Continuum mechanics/Tensor-vector identities


text: Continuum mechanics/Relations between surface and volume integrals


text: Continuum mechanics/Curl of a gradient of a vector


text:
Asian film studies

Short Response 1
Due: 9/19/07 in Class

Please answer the following questions with examples from the movie. Your answers should be typed. Handwritten answers will NOT be accepted. No late assignments will be accepted either.

1) A common theme in Japanese movies and dramas is the value of perseverence. How does the movie, Waterboys, demonstrate this value?

2) By having boys participate in synchronized swimming, this movie tackles the issues of social expectations and gender stereotypes. What do you think that this movie is trying to say about these issues?

title: Asian film studies


text:
User talk:Azeba356

Hello Azeba356, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:02, 12 September 2007 (UTC)

title: User talk:Azeba356


text:
User talk:Jteja242

Hello Jteja242, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:04, 12 September 2007 (UTC)

title: User talk:Jteja242


text:
User talk:Nryan849

Hello Nryan849, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:04, 12 September 2007 (UTC)

title: User talk:Nryan849


text:
User talk:Ghona906

Hello Ghona906, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:08, 12 September 2007 (UTC)

title: User talk:Ghona906


text:
User talk:Nacun212

Hello Nacun212, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:08, 12 September 2007 (UTC)

title: User talk:Nacun212


text:
User talk:Wkeyt078

Hello Wkeyt078, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:09, 12 September 2007 (UTC)

title: User talk:Wkeyt078


text:
User talk:Luren135

Hello Luren135, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:09, 12 September 2007 (UTC)

title: User talk:Luren135


text:
User talk:Skamp305

Hello Skamp305, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:10, 12 September 2007 (UTC)

title: User talk:Skamp305


text:
User talk:Sdima334

Hello Sdima334, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:11, 12 September 2007 (UTC)

title: User talk:Sdima334


text:
User talk:Aguil989

Hello Aguil989, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:11, 12 September 2007 (UTC)

title: User talk:Aguil989


text:
User talk:Jgill391

Hello Jgill391, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:12, 12 September 2007 (UTC)

title: User talk:Jgill391


text:
User talk:Jkamp405

Hello Jkamp405, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:12, 12 September 2007 (UTC)

title: User talk:Jkamp405


text:
User talk:Clitt149

Hello Clitt149, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:14, 12 September 2007 (UTC)

title: User talk:Clitt149


text:
User talk:Rdavi872

Hello Rdavi872, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:15, 12 September 2007 (UTC)

title: User talk:Rdavi872


text:
User talk:Ireid767

Hello Ireid767, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:15, 12 September 2007 (UTC)

title: User talk:Ireid767


text:
User talk:Showe778

Hello Showe778, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:15, 12 September 2007 (UTC)

title: User talk:Showe778


text:
User talk:Slace043

Hello Slace043, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:22, 12 September 2007 (UTC)

title: User talk:Slace043


text:
User talk:Jmcne852

Hello Jmcne852, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:23, 12 September 2007 (UTC)

title: User talk:Jmcne852


text:
User talk:Rsiva031

Hello Rsiva031, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:23, 12 September 2007 (UTC)

title: User talk:Rsiva031


text:
User talk:Kbhag494

Hello Kbhag494, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:23, 12 September 2007 (UTC)

title: User talk:Kbhag494


text:
User talk:Rclar450-b

Hello Rclar450-b, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) --JWSchmidt 03:24, 12 September 2007 (UTC)

title: User talk:Rclar450-b


text:
User:Elsbeth~enwikiversity

Learner: Elsbeth Interests: Culture, Language and Literature Location: Philippines

title: User:Elsbeth~enwikiversity


text:
User talk:Janreddam

Hello Janreddam, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) Erkan Yilmaz ( evaluate me!, discussion ) 07:51, 12 September 2007 (UTC)

title: User talk:Janreddam


text:
User talk:Elsbeth~enwikiversity

Hello Elsbeth, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) Erkan Yilmaz ( evaluate me!, discussion ) 07:52, 12 September 2007 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Elsbeth. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Elsbeth~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:35, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:07, 19 April 2015 (UTC)

title: User talk:Elsbeth~enwikiversity


text:
User talk:Asianfilm

Hello Asianfilm, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

If you're already excited about editing, our sandbox is the perfect place for all test edits. Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity!

P.S. Do you want live feedback ? You can find us in #wikiversity-en (see Chat) Erkan Yilmaz ( evaluate me!, discussion ) 07:56, 12 September 2007 (UTC)

title: User talk:Asianfilm