Mordred Posted October 17, 2017 Posted October 17, 2017 (edited) Glad to see the trace operator above. Been a long time since I last seen a discussion mention Bose-Hubbard. Some of the other models I don't recall. Looks like your understanding of QFT based papers is sufficient that you can easily process numerous different model dynamics and get a good strong understanding of those models. Glad to see the leap and bound change in your comprehension of these papers. Most of the work you have been doing are linear treatments. A good section to study would be curvilinear treatments, in particular Bevier curve approximations. As applied to a graph, this will relate to certain operators on curvilinear treatments. Once you have curvilinear ie [latex]SU^{n+1}[/latex] there won't be a field treatment that you wouldn't be able to understand a decent portion of. You already know the SU(N) aspects with the above largely. Once you have the first group you can understand any particle physics group of the SO(10). Also good catch noticing that one state does not need to have any interaction with another state in order to have a correlation function A does not need to have anything to do with B. That's one of the tricks of statistical math. If the two have similar trends then there is a correlation. (A couple of good statistical mechanics textbooks and also on Vector Calculus is your best aid to any QFT model) Edited October 17, 2017 by Mordred
Dubbelosix Posted October 17, 2017 Author Posted October 17, 2017 (edited) We can work it out properly now. The survival probability is [math]\rho = |<\psi|U|\psi>|^2[/math] This is just [math]\rho = <\psi e^{iHt}|\psi><\psi|e^{-iHt}|\psi>[/math] As we have established before. You can expand further for (small) time intervals, it is said to be given as [math]P \approx 1 - (\Delta H^2t^2)[/math] This is the quadratic decay law also written as [math]|a_0|^2 = 1 - \omega^2 t^2[/math] Where [math]\Delta <H> = \sqrt{|H|^2 - <\psi|H|\psi>^2}[/math] That measures the uncertainty of [math]H[/math] in the state [math]\psi[/math] for one of the systems in our phase space. This can be though of in terms of the energy difference of geometries as a Hamiltonian density [math]\Delta H = \frac{c^4}{8 \pi G} (\sqrt{|R_{ij}|^2 - <\psi|R_{ij}|\psi>^2} - \sqrt{|R_{ij}|^2 - <\psi|R_{ij}|\psi>^2})[/math] (So if I have done this right) our system is one that depends on the survival probabilities of each system, while it is a statement itself about the binding energy also. I think this raises interesting situations, if the state of one system depends on the other. The survival probability https://sci-hub.bz/https://doi.org/10.1007/s11005-011-0539-0 (Mordred) I'll certainly take a look soon. Thanks. Edited October 17, 2017 by Dubbelosix
Mordred Posted October 17, 2017 Posted October 17, 2017 Mixed states will involve both linear and nonlinear treatments. Mixed states is interesting, I'm not steering you from mixed states but you will find understanding curvilinear a necessity to some of the mixed state examinations. 1
Dubbelosix Posted October 17, 2017 Author Posted October 17, 2017 It's getting late, made several typo's in the last post. Be back when I can, may not be around tomorrow. anyway, think that's the errors weeded out and fixed. My latex gets sloppy when I get tired. Good night.
Mordred Posted October 17, 2017 Posted October 17, 2017 (edited) mine too lol night mate. Here add this to your study list, you will find understanding the 4 quaturnion numbers incredibly useful to understand Hamilton and any QFT theory. https://www.google.ca/url?sa=t&source=web&rct=j&url=http://malcolmdshuster.com/Pubp_021_072x_J_JAS0000_quat_MDS.pdf&ved=0ahUKEwi5-J3R7PbWAhUH92MKHUO4DHsQFggwMAM&usg=AOvVaw2rwLoxBYHemplrpKOPlPpF Above once again can be applied to any number of dimensions including time. (basis behind the rotation matrixes) ie modelling a particle state under acceleration. Edited October 17, 2017 by Mordred 1
Dubbelosix Posted October 17, 2017 Author Posted October 17, 2017 (edited) I realised while attempting to sleep there is an important interpretation in the work and once again, its because you mentioned something relevant. I remembered that the property [math]<\psi|U A U|\psi> = <\psi'|A|\psi> = <\psi'|\psi'> = 1[/math] Holds if [math]A[/math] is a linear operator. So, while in my past work I considered the geometry as the observable, it would mean we'd have to consider a linear geometry in this case. Which related back to your mentioning, it may be best to look at the non-linear cases now for the more juicy structures of the theory. I had a look at some non-linear wave equations before already... but will take a much greater look into it and see if I can see anything. Edited October 17, 2017 by Dubbelosix
Mordred Posted October 17, 2017 Posted October 17, 2017 Here this will also help to know the commutations as applied to Hamiltons. https://www.google.ca/url?sa=t&source=web&rct=j&url=http://web.science.mq.edu.au/~chris/quad/CHAP04%20Quadratic%20Algebras.pdf&ved=0ahUKEwjX-ICd9fbWAhUDzGMKHV8nBfIQFgg3MAg&usg=AOvVaw0KDOtRQEktblBG0YfpB1Us Unfortunately the better explanations are typically in textbooks so I will keep digging for you. The first link is an example applied to a spacecraft. Second link has added details 32 minutes ago, Dubbelosix said: it may be best to look at the non-linear cases now for the more juicy structures of the theory. I had a look at some non-linear wave equations before already... but will take a much greater look into it and see if I can see anything. 1
Dubbelosix Posted October 17, 2017 Author Posted October 17, 2017 Thanks! This looks valuable as reading material as well http://www.ams.org/journals/tran/1960-094-03/S0002-9947-1960-0112025-3/S0002-9947-1960-0112025-3.pdf
Dubbelosix Posted October 17, 2017 Author Posted October 17, 2017 (edited) [math]\Delta <\mathbf{H}> = \sqrt{|H|^2 - <\psi|H|\psi>^2}[/math] which measures the uncertainty in the Hamiltonian density in the state [math]\psi[/math] for the system in the phase space. Then we can implement the survival probability in the context of the binding energy equation. With [math]c = 8 \pi G = 1[/math] we have [math]\Delta \mathbf{H} = \sqrt{|R_{ij}|^2 - <\psi|R_{ij}|\psi>^2} - \sqrt{|R_{ij}|^2 - <\psi|R_{ij}|\psi>^2}[/math] So the geometries themselves depend on the survival probabilities of each system. In this case, we cannot think of the curvature tensor [math]R_{ij}[/math] as the anti-symmetric relationship - this will be studied further because we have the ground work. The uncertainties in this last equation can be though of as consisting of two terms: [math]\sqrt{|\psi|^2 \ln[|\psi(p)|^2] - |\hat{\psi}|^2 \ln[|\hat{\psi}(p)|^2]}[/math] [math]\sqrt{|\psi|^2 \ln[|\psi(q)|^2] - |\hat{\psi}|^2 \ln[|\hat{\psi}(q)|^2]}[/math] So in terms of pure statistics, the Hamiltonian density can be expressible as [math]\sqrt{|\psi|^2 \ln[|\psi|^2] - |\hat{\psi}(p)|^2 \ln[|\hat{\psi}(p)|^2]} - \sqrt{|\psi|^2 \ln[|\psi|^2] - |\hat{\psi}(q)|^2 \ln[|\hat{\psi}(q)|^2]}[/math] This application can be found from Uncertainty relations for information entropy in wave mechanics'' (linked below). Don't be put off with a wave function existing for the first terms, when there are non featured in [math]|H^2|[/math] in this part. This is because it can actually be written as [math]\Delta <\mathbf{H}> = \sqrt{<\psi|H^2|\psi> - <\psi|H|\psi>^2}[/math] This just appears to be a matter of notation. https://link.springer.com/article/10.1007/BF01608825?no-access=true Of course, it should be noted just for clarity, this does not apply to an interpretation of an antisymmetric operator, but one rather that would have to apply to linear geometric operators - we will still look into the curvilinear models like Mordred has suggested and I was already on the fence about. It's just about finding the right model (that I am happy with). Fixed the dependencies on the statistics, think I had wrong, think I have it right now. Edited October 17, 2017 by Dubbelosix
Dubbelosix Posted October 17, 2017 Author Posted October 17, 2017 (edited) On 04/10/2017 at 9:52 PM, Dubbelosix said: With the past post help from Mordred, I believe I can sum this up in the following: von Neumann introduced S=−Tr(ρlogbρ) where we have used the base power of the logarithm. The Shannon entropy has a relationship to this H=−∑i(pilogbpi) Which works, when we consider a mixture of orthogonal states. In this case, the density matrix does contain classical probabilities on the diagonal. Quantum mechanical density matrices in general though, have off-diagonal terms, which for pure states, reflect the quantum phases in superpositions. So... we know where we want to go with this. A useful equation I came across in information theory was the Shannon entropy in terms of correlation: H(A)=H(A|B)+H(A:B) How do you read this strange equation? Well, the H(A|B) is the entropy of A after having measured the systems that become correlated in B . And H(A:B) is the information gained about A through measuring the system B . (Apparently), as is well known, these two quantities complement each other such that H(A) will remain unchanged to satisfy the conservation of the second law. While this was interesting, I realised my understanding of the context of which I wanted to construct this theory relies on an interpretation of information theory that can contain an uncertainty principle - since my investigations have been primarily involved in understanding a possible non-trivial spacetime relationship, this shouldn't be too suprising. A paper by I. Białynicki-Birula, J. Mycielski, on ''Uncertainty relations for information entropy in wave mechanics'' (Comm. Math. Phys. 1975) contains a derivation of an uncertainty principle based on an information entropy and features here as −∫|ψ|2ln[|ψ(q)|2] dq−∫|ψ^|2ln[|ψ^(p)|2] dp≥1+lnπ Where |ψ(q)|2=∫W(q,p) dp using generalized coordinates just in case anyone wonders what p and q is. And, |ψ^(p)|2=∫W(q,p) dq I have a bit of interest in this method, I generally get a feeling if I know I can do something with something else, like this above. It seems to have all the physics I need to try and piece this puzzle together. I do seem to be getting the impression that if the system is in equilibrium, it calculates as S=−∑i1Nln(1N) =−N1Nln(1N) =ln(N) Which just looks like the Boltzmann entropy. While this last bit wasn't pertinent to the work, it is interesting for educational purposes. ref: http://www.cft.edu.pl/~birula/publ/Uncertainty.pdf The paper by I. Białynicki-Birula, J. Mycielski shows that there are ways to describe my model with gravitational logarithmic nonlinear wave equations. More* In particular, I want to see if these two equations can be merged into some unified definition, by merging the physics somehow to make sense of each other in context of my investigation. −∫|ψ|2ln[|ψ(q)|2] dq−∫|ψ^|2ln[|ψ^(p)|2] dp≥1+lnπ Which is the equation we just featured, and an equation I derived was: ΔE=c48πG∫<ΔRij> dV=c48πG∫<ψ|Rij−<ψ|Rij|ψ>)|ψ> dV =c48πG∫(<ψ|Rijψ>−<ψ|Rij|ψ>) dV Which was the difference in quantum geometries which as was already established, related to the uncertainty principle in the antisymmetric indices of the curvature tensor Rij . We never proved by any means, that this is how an uncertainty principle should be interpreted with gravity - that is a hard thing to do without a full understanding of gravity as it is. There are disagreements on how to approach a quantum theory, right down to vital questions about whether gravity is even the same as the other fundamental forces in nature. If it lacks a graviton, then you can count large portions of gauge theory will become questionable. I didn't note this before, but these are Wigner functions. It has relationships with the density operator [math]\int W(q,p)\ dp = <q|\rho|q>[/math] [math]\int W(q,p)\ dq = <p|\rho|p>[/math] [math]\int W(q,p)\ dq\ dp = Tr(\rho) = \mathbf{1}[/math] There is also a property which satisfies [math]|W(q,p)| \geq \frac{1}{\pi \hbar}[/math] Edited October 17, 2017 by Dubbelosix
Vmedvil Posted October 17, 2017 Posted October 17, 2017 15 minutes ago, Dubbelosix said: I didn't note this before, but these are Wigner functions. It has relationships with the density operator ∫W(q,p) dp=<q|ρ|q> ∫W(q,p) dq=<p|ρ|p> ∫W(q,p) dq dp=Tr(ρ)=1 There is also a property which satisfies |W(q,p)|≥1πℏ Still at it I see, Dubbelosix you never stop do you? When you solve it just make sure that I get a copy, it is going to be something good.
stephaneww Posted October 17, 2017 Posted October 17, 2017 (edited) Hello Dubbelosix, Your last quote have eat the fraction [latex] \frac{c^4}{8 \pi G}[/latex] in [latex] \Delta E [/latex] Edited October 17, 2017 by stephaneww latex
Dubbelosix Posted October 17, 2017 Author Posted October 17, 2017 (edited) Yes the sites latex is sensitive to quotes it seems. 1 hour ago, Vmedvil said: Still at it I see, Dubbelosix you never stop do you? When you solve it just make sure that I get a copy, it is going to be something good. No I don't give up. I don't think I'll ever become bored of physics to be honest. Right now, I am writing up on why curvature in the Hilbert space is not very well defined - though I will offer suggestions. It became clear to me, it wasn't a simple topic in just adding it curvature to Hilbert space, but I did come across a great paper that will be linked after that helped this post. Does a Hilbert space have a curvature? No it doesn't - not intrinsically, but it does actually have a geometric structure, simply because it does possess a scalar product. I've shown the Hilbert space in the context of a geometric uncertainty relationship for space, known as the Cauchy Schwarz space. What is important, though not intrinsically possessing curvature, it may indeed have one and I see no reason why it can't. The linear Hilbert space will only satisfy the natural flat geometry. There is in fact, no easy way to give ''curvature'' to a Hilbert space, There is a way to describe that through the flat metric [math]d(\psi, \phi) = |\psi - \phi|[/math] There is a projective Hilbert space [math]\mathbf{C}P^n[/math] then we can consider what is known as the distance function [math]d_{proj}(\psi, \phi) = \min_{\alpha}|\psi - e^{i\alpha}\phi| = \sqrt{2 - 2|<\psi,\phi>|}[/math] This is often used to find the ground state of some Hamiltonian. Interestingly (something I learned) the distance function contains a singularity at a quantum phase transition. The distance function can be thought of as a metric - this may be a better alternative to the Bure's metric (which I have read some authors) claim that it is not actually a true metric: Neverthless, we have looked at a statistical theory of gravity and the Bure's metric has implication for quantum geometry information theory. So there are links here that should not really be ignored without investigation. How true the statement that the Bure metric is not a true metric is, I am unsure. What I can say is that the Bure metric is a ''metric'' which measures the infinitesimal distance between density operators which define the quantum states. The Bure's metric, can also be thought of as a ''statistical distance.'' This has formal similarity to a distance function of the form [math]d(\psi,\phi) = \arccos(|<\psi|\phi>|)[/math] This measures also, the shortest geodesic between any two states! It is an angle formula, we may consider it also in the form [math]d(\rho_1,\rho_2) = \arccos \sqrt{F(\rho_1,\rho_2)}[/math] Of course, (iff) the Bure's metric is not a true metric, then neither is this ''shortest geodesic distance'' equation. It will be interesting to read vigorous arguments for and against the Bure's metric as being a real metric or not. Now... we must go back quickly, and go back to the distance function. Using just single notation now, we have [math]|\psi - \psi'|[/math] This still doesn't yet describe curvilinear space - this is actually a Euclidean measure of the distance between two states. The length of a curve however can be given as: [math]\frac{ds}{dt} = \sqrt{<\dot{\psi}|\dot{\psi}>}[/math] This involves the tangent vector [math]\dot{\psi}[/math] has lengths that is the belocity which travels in the Hilbert space. I can construct a Schrodinger (like) equation from this. [math]\frac{ds}{dt} = \frac{1}{\hbar} \frac{c^4}{8 \pi G}\sqrt{<\psi|R_{ij}|\psi>} = \frac{1}{\hbar} \sqrt{<\psi|H|\psi>}[/math] and the metric of the solution is [math]ds = \frac{1}{\hbar}(t_1 - t_2)\sqrt{<\psi|H|\psi>}[/math] Anandan, a physicist whome I used an equation that described the difference of geometries (but made it within the context of the curvature tensor) has proposed that the Euclidean length is an intrinsic parameter of the Hilbert space. references http://www.physik.uni-leipzig.de/~uhlmann/PDF/UC07.pdf CURVATURE OF HILBERT SPACE AND q-DEFORMED 'QUANTUM MECHANICS' BINAYAK DUTTA-ROY Saha Institute of Nuclear Physics, Calcutta 700 064, India Edited October 18, 2017 by Dubbelosix
Dubbelosix Posted October 18, 2017 Author Posted October 18, 2017 (edited) Assuming I have done all this right so far, I did find an interesting continuation which uses the Wigner function, which is why I explained them in more depth not long ago. The curve of the length can be related to an inequality by pulling out the terms to write a Wigner function. [math]\frac{ds}{dt} \equiv |W(q,p)| \sqrt{<\psi|R_{ij}^2\psi>}\ \geq \frac{1}{\hbar} \sqrt{<\psi|H^2|\psi>}[/math] due to the inequality relationship: [math]|W(q,p)| \geq \frac{1}{\pi \hbar}[/math] There are other ways to argue this inequality. For instance, some standard formula [math]\int <q|\rho|q>\ dq = \int <p|\rho|p>\ dp = Tr(\rho) = \mathbf{1}[/math] and [math]\int <q|\rho|q> <p|\rho|p>\ dqdp = Tr(\rho)^2 = \mathbf{1}[/math] Because our model deals with the phase space, you can argue there exists a Cauchy Schwarz inequality [math]2 \pi \hbar W(q,p)^2\ \leq\ <q|\rho|q><p|\rho|p>[/math] or [math]W(q,p)^2\ \geq\ \frac{1}{2 \hbar} <q|\rho|q><p|\rho|p>[/math] Again, to satisfy our Cauchy Schwarz space. The inequality satisfies the norm through [math]\int W(q,p)^2\ dqdp \geq \int \int <q|\rho|q><p|\rho|p>\ dqdp = Tr(\rho)^2 = \mathbf{1}[/math] There is an error in the last two equations of not the last post (which will merge with this) but the one before it, the Hamiltonian should be squared, which when [math] c=8 \pi G = 1[/math] [math]\frac{ds}{dt} = \frac{1}{\hbar} \sqrt{<\psi|R_{ij}^2|\psi>} = \frac{1}{\hbar} \sqrt{<\psi|H^2|\psi>}[/math] and so the length is [math]ds = \frac{1}{\hbar}(t_1 - t_2) \sqrt{<\psi|R_{ij}^2|\psi>} = \frac{1}{\hbar} (t_1-t_2)\sqrt{<\psi|H^2|\psi>}[/math] And again, all updates are wrote down here as well for a more compact collection of the work. http://www.physicsgre.com/viewtopic.php?f=10&t=127412&p=198855#p198855 Edited October 18, 2017 by Dubbelosix
Mordred Posted October 18, 2017 Posted October 18, 2017 (edited) Good on the above but lets be careful here. One can linearize nonlinear systems to good approximation which is essentially what you are doing above. In this there is huge variety of techniques. Two primary classes of field treatments apply to this. Conformal and canonical. Canonical it is the distances via end points that is the priori. On conformal it is the angles. So be careful to identify the treatment involved when jumping theories or mixing them... Also remember [latex]\mathcal{H}×\mathcal{H}=\mathbb{C}[/latex] "× " is The cross product. So these two Hilbert spaces. The two vector fields are perpindicular to one another in cross products. Angular momentum is a Cross product space https://en.m.wikipedia.org/wiki/Angular_momentum Side note your primarily using ODEs observable differential equations, the other being partial Pde,s Edited October 18, 2017 by Mordred 1
Dubbelosix Posted October 18, 2017 Author Posted October 18, 2017 ''Side note your primarily using ODEs observable differential equations, the other being partial Pde,s'' Well spotted, I will need to tackle this. Thanks for the rest.
Mordred Posted October 18, 2017 Posted October 18, 2017 (edited) Np it was also for the benefict of other readers to help them follow the thread somewhat without derailing it. Lets summarize heuristically for other readers benefict. The OP is studying different techniques and specifics on how to mathematically define an object and system under Vector fields via symmetry relations that apply to Noethers theorem. First he went through the proofs defining a vector itself, ie the requirement of 2 units to describe. (magnitude and direction). Magnitude is a scalar quantity. Then he established that these two units are independant quantities. One can change in value without changing the other. Then he estsblished the boundary conditions of the magnitude as a normalized unit under a coordinate basis. ie the scale of the graph, to the ratio of change to the length of the vector 1 to 1 ratio to axis coordinates. Then he went further and applied Cauchy inequality, to a plane, which also shows the triangle inequality. This relates to i,j. Thus establishing an orthonormal and orthogonal basis. In order to define Hilbert space he had to apply all the above to apply the outer products of two vectors to close the Hilbert group. Now he is looking at how these two Hilbert spaces are applied under treatments. So here is a practicum question. Take right angle triangle ABC with identity connection i,j. Is C an independent unit ? Is it a Hilbert space? Edited October 18, 2017 by Mordred 1
Dubbelosix Posted October 19, 2017 Author Posted October 19, 2017 Good summary and worded in a fresh way.
Mordred Posted October 19, 2017 Posted October 19, 2017 Thanks, granted any heuristic explanation is an oversimplification. Quite frankly the detail you placed in this thread has been an excellent in quality and paid close attention to all the essential aspects. 1
Dubbelosix Posted October 20, 2017 Author Posted October 20, 2017 (edited) Thank you Mordred. I have tried. It's a combination of idea's heavily laden with knowledge from papers and additional help from posters like you. Just a small note, not really for your benefit as I am sure you know how I have gone around most of these things, but for other posters, a quick explanation of the last inequalities.We used the form: [math]\frac{ds}{dt} = |W(q,p)| \sqrt{<\psi|R_{ij}^2|\psi>}\ \geq \frac{1}{\hbar} \sqrt{<\psi|H^2|\psi>}[/math] But when expressed fully in Dirac notation, I prefer it in the form: [math]\sqrt{<\dot{\psi}|\dot{\psi}>} = |W(q,p)| \sqrt{<\psi|R_{ij}^2|\psi>}\ \geq \frac{1}{\hbar} \sqrt{<\psi|H^2|\psi>}[/math] You can argue it satisfies the Schrodinger equation: [math]H |\psi> = i \hbar |\dot{\psi}>[/math] I'll be clear how the inequality arises also from the dimensions, the equation given in the reference below is [math]\frac{ds}{dt} = \frac{1}{\hbar} \sqrt{<\psi|H^2|\psi>}[/math] We noted in my work that [math]\frac{c^4}{8 \pi G}<\psi|R_{ij}|\psi> = <\psi|E|\psi>[/math] square the operators and take the square root of the components like : [math]\sqrt{\frac{c^8}{16 \pi^2 G^2}<\psi|R_{ij}^2|\psi>} = \sqrt{<\psi|E^2|\psi>}[/math] and so. [math]\frac{ds}{dt} = \frac{1}{ \pi \hbar} \sqrt{\frac{c^8}{16 G^2}<\psi|R_{ij}^2|\psi>} = \frac{1}{\hbar}\sqrt{<\psi|H^2|\psi>}[/math] Pull the pi out of the denominator and into the definition of the first inverse hbar and use natural units after identifying [math]\frac{1}{\pi \hbar}[/math] as the mean Wigner function |W(q,p)| [math]\frac{ds}{dt} = |W(q,p)| \sqrt{<\psi|R_{ij}^2|\psi>}[/math] and the inequality holds because of the quantization of hbar after identifying [math]\frac{ds}{dt} = \sqrt{<\dot{\psi}|\dot{\psi}>}[/math] we simply get, [math]\sqrt{<\dot{\psi}|\dot{\psi}>} = |W(q,p)| \sqrt{<\psi|R_{ij}^2|\psi>}\ \geq \frac{1}{\hbar} \sqrt{<\psi|H^2|\psi>}[/math] http://www.physik.uni-leipzig.de/~uhlmann/PDF/UC07 Edited October 20, 2017 by Dubbelosix
Dubbelosix Posted October 20, 2017 Author Posted October 20, 2017 (edited) ( really theoretical post) Let's go back for a moment because I want to apply some ''older knowledge'' I have on a possible continuation for the following (what I call) Hilbert-Wigner Distribution Inequality: [math]\sqrt{<\dot{\psi}|\dot{\psi}>} = |W(q,p)| \sqrt{<\psi|R_{ij}^2|\psi>} \geq \frac{1}{\hbar}\sqrt{<\psi|H^2|\psi>}[/math] Under the Jacobi formulation of the Maupertuis principle involving an action of the form [math]\int\ p \cdot dq[/math], the kinetic energy is related to the generalized velocities, [math]T = \frac{1}{2} \frac{dq}{dt} \cdot \mathbf{M} \cdot \frac{dq}{dt}[/math] where the not-so-frequently encountered mass tensor [math]\mathbf{M}[/math] features here. The key point is that it relates the temperature with the action in the following way [math]2T = p \cdot \dot{q}[/math] Alternative to the first equation, we can also find arguements for the metric of the form [math]ds^2 = dq \cdot \mathbf{M} \cdot dq[/math] and even maybe an argument for the kinetic energy written with a mass term (where now I fix the dimensions) [math]k_B T = \frac{1}{2}m(\frac{ds}{dt})^2[/math] Actually according to the Virial theorem or even from equipartition, you could even argue the quantity is more true for the kinetic motion of systems: [math]k_B T = \frac{3}{2}m(\frac{ds}{dt})^2[/math] Notice we have a curve argument in the term [math](\frac{ds}{dt})^2[/math]? From here we will use this physics to implement into our model, for a purely theoretical approach. Remember, the Wigner function plays a possible fundamental role in the notion of the curve in the Hilbert space, at least in theory, since that curve could be adjusted to suit a definition of the form [math]|W(q,p)| \sqrt{<\psi|R_{ij}|\psi>}[/math] - this curve was also set equal to the primary definition: [math]\sqrt{<\dot{\psi}|\dot{\psi}>}[/math]. It is also possible to retrive a half factor on the terms for the first equation, I just haven't derived it exlicitely but will do if asked. I just haven't to save time. So for equation 1, we have a number of simple procedures to get to where we want to be. Square the equation [math]<\dot{\psi}|\dot{\psi}> = |W(q,p)^2| <\psi|R_{ij}^2|\psi>[/math] Let the mass tensor now play a role and apply one half to the right hand side [math]<\dot{\psi}|\mathbf{M}|\dot{\psi}> = \frac{1}{2}|W(q,p)^2| <\psi|R_{ij}^{2'}|\psi>\ \geq \frac{1}{2(L \cdot S)}<\psi|H^{2'}\psi>[/math] Where we have used for the square of the action, the spin-orbit coupling factor. Where I have used the notation [math]R_{ij}^{2'}[/math] to say that the mass tensor is acting on this curvature tensor. I think of this latter relationship as (maybe being tied) to some way to describe the correlations between matter and geometry, a subject that has been taken of interest in emergent gravity theories (see Fotini Markoupolou) - in her theory it is an important feature that matter can entangle with geometry. Nothing in the laws of physics forbids it - in fact, matter can entangle with matter. Space can even entangle with space! And so, the laws of physics definitely shows it is possible to link the two through a dynamic entanglement of matter to geometry. I won't lie, there are some interesting but very important questions from this last equation, like, is the spin-orbit actually quantum in nature [math](L \cdot S)[/math] as it give rise to the inequality like we have done? Well I have known for a while that the mathematics of classical spin are actually identical the quantum spin dynamics, there is no difference between the two except notation, so yes, this is a quantum property if we wish to assume it is, like we do. From this I deduct we now have a reason why the curvature of the Hilbert space must be implied: Through the spin-orbit coupling, specifically which obviously must follow a curvilinear trajectory. This means, the curvature in the Hilbert space is not an observable of time per se, but can be though of that, but in this one, it seems the curve is brought on by the spin coupling of the system with the orbit. Sorry typos fixed. Also you can argue the mass tensor transforms the curvature tensor is such a way you end up with the modified Hamiltonian [math]H^{2'}[/math]. The spin-orbit coupling will also allow for energy shifts. On 18/10/2017 at 4:01 AM, Mordred said: Np it was also for the benefict of other readers to help them follow the thread somewhat without derailing it. I didn't realise you said this earlier, I think you are far too modest.You are in my opinion (one of the least likely) posters here who could accidentally add knowledge in any hope of derailing the thread. I am also going to apply my knowledge of electroweak symmetry breaking to the context above and see if there are any obvious reasons why mass should appear in the theory - for instance, can we work in a massless understanding of the theory as well? This will take longer than my usual posts and may be some time before I come to a conclusion. Edited October 20, 2017 by Dubbelosix
Dubbelosix Posted October 20, 2017 Author Posted October 20, 2017 (edited) After (what in hindsight) appears to be some trivial misunderstandings, it seems that ultmately my friend Matti P. (PhD) has finally accepted this space theory is acceptable (in context of the last posts concerning the action) . I asked his permission to relay a certain quote of his that I found, interesting and insightful: ''OK. We come back where we started. This is non-relativistic case and one considers kinetic energy. One can introduce in condensed matter physics mass tensor when the effective kinetic energy is not anymore mv^2/2 but more complex function m_ijv^i v^j/2. This is just effective description and non-relativistic. I would not introduce it in GRT since the mass is by definition Lorentz scalar in this context. I would speak about energy momentum tensor and introduce to it this kind of terms.'' I consider Matti, not just a friend, but also a mentor. He has been influential. I have made it clear to him, if he was within distance, I'd properly pay him for his assistence to take me on as a pupil. Matti is very intelligent, and while I do not agree with every aspect of his topological geometrodynamics theory, I think its one of the most promising theories on the market that really values an investigation. Ah, I just looked at the curve equation, it also predicts that temperature arises from the curvature of space. I think I just found my new favourite investigation. That really puzzled me, but can you guess which equation can be understood this way? Why should the temperature arise from the curvature of space? Well I will tell you why, because with curvature comes acceleration and with acceleration as a parameter to a condensed collection of systems results in increased thermodynamic properties. This feels promising. Thermodynamics is about the movements of the constituents it is made of. It seems curvature at the quantum scale could provide a thermodynamic translation into gravity (or geometry). This is totally unexpected, but I admit, actually makes sense for once. Edited October 20, 2017 by Dubbelosix
Mordred Posted October 20, 2017 Posted October 20, 2017 (edited) I was going to suggest the stress tensor approach as well. The time time component T^00 describes the massless stress components. ie pc^2. This is a scalar value as in the line element for seperation distance, a massless particle will have [math] ds^2=0.[/math]. If you like tonight I can post some of the essentials to the stress tensor for the other non reducible components. On this a also agree with Matti in applying the stress tensor. In the above via Kronecker delta connection, which you have already modelled, this is Euclidean and will follow the principle of equivalence. An easy way to understand Kronecker is parallel transport of two vectors. In Euclid geometry that parallel transport is preserved. Now under curvature of spacetime you will get a screw symmetry from the simultaneous affects upon the x and ct axis. (length contraction, time dilation.) However barring relativity simple curvature will cause a loss of parallel transport, the two vectors will either converge or diverge, Example Newtons laws with a central potential. So one can assign a vector mapping the seperation distance between those previous two parallel paths. This is the essence of the Principle of Covariance, Yes indeed curvature has definitive affects on temperature. Edited October 20, 2017 by Mordred 1
Dubbelosix Posted October 21, 2017 Author Posted October 21, 2017 Many of us know of John A. Wheeler --- what many do not know, is that he approached de Witt in the 60's ... possibly the year of 68' .... in which he asked for his direct help in the quantization of gravity. In their early approach, which used early methods, they arrived at the Wheeler de Witt equation, which has caused much controversy since. Like Wheeler, I want to reach out to people like Mordred who puts time into contemplating theories, they may not even agree in. A wise man once said, it takes an educated mind to appreciate a theory you may not even agree with. (Mordred) Once again, thank you for your in-put and I will contemplate it for a bit before responding. I am meticulous this way, you know me Matti will be pleased you agree with him (ps.) I'm trying to write a thermodynamic theory now based on the above. We'll see how it goes.
Dubbelosix Posted October 23, 2017 Author Posted October 23, 2017 (edited) If we identify [math]\frac{1}{2}<\dot{\psi}|\mathbf{M}|\dot{\psi}>[/math] (1) with wave functions [math]|\psi> = e^{iHt}|q>[/math] [math]<\psi| = <q| e^{-iHt}[/math] In these last two equations, though the wave function will give rise to a probability of the wave function with dimensions of either [math]1/L[/math]. [math]1/L^2[/math] or cubed density (volume) [math]1/L^3[/math] we will not renormalize it in this approach, we'll leave that to the reader to assume there is some differential attached in there to remove the density. ie. [math]<\psi|\psi>\ dV[/math] for a wave function in three dimensions. In the two equations, we will also assume for the largest part there is encoded a generalized position [math]q[/math]. You'll see why when we reach equations 5 and 6. and identify it as ''being akin'' to the following equation [math]k_BT = \frac{1}{2}mc^2 =\frac{1}{2} \frac{dq}{dt} \cdot \mathbf{M} \cdot \frac{dq}{dt}[/math] (2) where [math]q[/math] is the generalised position and so [math]\dot{q}[/math] is the generalized velocity and knowing that [math]k_BT\ dt^2 = \frac{1}{2}mc^2\ dt^2 = \frac{1}{2}dq \cdot \mathbf{M} \cdot dq[/math] (3) When considering the Hertz principle of least curvature, the mass is often set equal to 1 without loss of generality in the theory - you may do such a thing if all the masses are equal, or if you wish to continue with a ''massless theory'' - whatever the motive, keep in mind that a massless theory would not have the mass tensor. This is why you will come to encounter some authors write it in the form: [math]ds^2 \equiv c^2\ dt^2 = \frac{1}{2}dq \cdot \mathbf{M} \cdot dq[/math] (4) According to the almost similar nature of the first expressions relationship with equation 2, we can theorize some relationships, one related to the kinetic energy and another related to the metric [math]k_BT = \frac{1}{2}<\dot{\psi}|\mathbf{M}|\dot{\psi}>[/math] (5) featuring the mass tensor on the RHS but not the mass term on the LHS, it would also hold that [math]ds^2 = \frac{1}{2}<\psi|\mathbf{M}|\psi>[/math] (6) The mass tensor in this model, appears to be a metric tensor. I'll even propose the possible quantization: [math]\hbar = \frac{1}{2}t^{-1} <\psi|\mathbf{M}|\psi>[/math] The temperature was once again, related to the curve, [math]k_BT = \frac{1}{2}m(\frac{ds}{dt} \cdot \frac{ds}{dt})[/math] and the curve in our Hilbert space was identified using the Wigner function, here as the square of the metric [math]ds^2[/math], [math]<\dot{\psi}|\dot{\psi}>\ = |W(q,p)^2|\ <\psi|R^2_{ij}|\psi>\ \geq \frac{1}{\hbar^2}\ <\psi|H^2|\psi>[/math] and as noted, will satisfy a Schrodinger equation [math]\frac{1}{ i \hbar}H|\psi> = |\dot{\psi}>[/math] remember, [math]ds = cdt[/math]. From this under the Jacobi formulation there exists an action related to the temperature: [math]k_BT = \frac{p \cdot \dot{q}}{2}[/math] In which [math]\frac{1}{2}p \cdot \dot{q}[/math] is the minimized action. Further under the Jacobi formulation, the action is related to the following relationships for the temperature [math]k_BT\ dt = \int p \cdot dq[/math] So the interesting thing now which is different to my last thoughts, I thought the mass tensor could be just a mass term acting on a matrix, but it is actually identified as a more complicated object which appears to be a [metric tensor]. I need more literature on this to understand it. Edited October 23, 2017 by Dubbelosix
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