Mordred 1372 Posted February 7, 2020 Share Posted February 7, 2020 (edited) Your still missing a vital point, you will never get one master oscillator equation to describe all particles and their interactions. Every particle and every interaction involves their own specific formulas. If you ignore the math you will miss that detail. For example invariant quantities are not ocsillating. All wavefunctions are also not oscillations. Edited February 7, 2020 by Mordred Link to post Share on other sites

Kartazion 3 Posted February 7, 2020 Author Share Posted February 7, 2020 14 minutes ago, Mordred said: Your still missing a vital point, you will never get one master oscillator equation to describe all particles and their interactions. However ... But it would be like calculating all the interactions of particles at the same time. 15 minutes ago, Mordred said: Every particle and every interaction involves their own specific formulas. If you ignore the math you will miss that detail. No, no I know. 22 minutes ago, Mordred said: For example invariant quantities are not ocsillating. All wavefunctions are also not oscillations. This is where it gets interesting. This amounts to drawing the wave function with the particle. Link to post Share on other sites

swansont 7465 Posted February 7, 2020 Share Posted February 7, 2020 1 hour ago, Kartazion said: The oscillator explains a lot. But the oscillator has a trajectory. A trajectory of what? Or at least a periodic charge signal. No. A particle in a potential well does not have a trajectory. Maybe stay away from QM if you haven’t learned about it (for real, not pop-sci summaries) 1 hour ago, Kartazion said: If you only have one particle in action, it is sure that there can be no interactions and interference of itself. single particles through a double-slit will interfere, so no. 1 hour ago, Kartazion said: Matter is made up of vacuum. also no. 1 hour ago, Kartazion said: In the context of the SM, and for the boson, the principle of reiteration makes it possible to superimpose the "particle" several times in the same place, or at least increase its level of density, at a point. Principle of reiteration? Does this principle exist outside of your own pet theory? Link to post Share on other sites

Kartazion 3 Posted February 7, 2020 Author Share Posted February 7, 2020 (edited) 1 hour ago, swansont said: No. A particle in a potential well does not have a trajectory. Maybe stay away from QM if you haven’t learned about it (for real, not pop-sci summaries) I understand. There is no particle in the quantum oscillator. Or there is a particle without motion? 1 hour ago, swansont said: single particles through a double-slit will interfere, so no. It's the wave that interferes, right? 1 hour ago, swansont said: Principle of reiteration? Does this principle exist outside of your own pet theory? Yes, a principle of me. What does mean pet theory? Edited February 7, 2020 by Kartazion Link to post Share on other sites

swansont 7465 Posted February 7, 2020 Share Posted February 7, 2020 2 hours ago, Kartazion said: I understand. There is no particle in the quantum oscillator. Or there is a particle without motion? It's the wave that interferes, right? Quantum particles exhibit wave behavior 2 hours ago, Kartazion said: Yes, a principle of me. What does mean pet theory? An individual conjecture, untested and not mainstream science. Speculation. Also not to be used as a basis for further speculation. IOW, if you have an idea based on some conjecture, you can’t build a greater conjecture on it. One per thread, and limited to discussion here in speculations. Your original discussion was based in classical physics. What’s the point of moving into QM? Link to post Share on other sites

Kartazion 3 Posted February 7, 2020 Author Share Posted February 7, 2020 16 minutes ago, swansont said: Also not to be used as a basis for further speculation. IOW, if you have an idea based on some conjecture, you can’t build a greater conjecture on it. One per thread, and limited to discussion here in speculations. Your original discussion was based in classical physics. What’s the point of moving into QM? Yes my oscillator remains classic. But then my time is limited for the quantum oscillator? Link to post Share on other sites

swansont 7465 Posted February 8, 2020 Share Posted February 8, 2020 42 minutes ago, Kartazion said: Yes my oscillator remains classic. But then my time is limited for the quantum oscillator? I don’t know what this means. Link to post Share on other sites

Mordred 1372 Posted February 8, 2020 Share Posted February 8, 2020 (edited) 3 hours ago, Kartazion said: Yes my oscillator remains classic. But then my time is limited for the quantum oscillator? Neither do I understand what this means. You have looked at numerous QM relations with regards to oscillator behavior under QM through the course of this thread but very little of this thread has dealt with classical. For example the uncertainty principle is quantum in nature. The statement above doesn't make sense how do you get classical only while discussing the quantum oscillator All physics apply time Edited February 8, 2020 by Mordred Link to post Share on other sites

Kartazion 3 Posted February 8, 2020 Author Share Posted February 8, 2020 3 hours ago, Kartazion said: Yes my oscillator remains classic. But then my time is limited for the quantum oscillator? My speaking time is limited in quantum oscillator since my oscillator is classic? Link to post Share on other sites

Kartazion 3 Posted February 8, 2020 Author Share Posted February 8, 2020 Maybe l should split this thread into one title like: Classical vs. Quantum Harmonic Oscillator No? Link to post Share on other sites

Mordred 1372 Posted February 8, 2020 Share Posted February 8, 2020 (edited) Likely a good idea, that way you can focus on specific applications. When you get into QM those applications can get rather complex. Edited February 8, 2020 by Mordred Link to post Share on other sites

Kartazion 3 Posted February 20, 2020 Author Share Posted February 20, 2020 Split reference: Conclusion: The oscillator that I am trying to exploit is classical and not quantum. Indeed in my case the oscillator allows simply to alternate a particle from point A to point B at high speed. Ultimately, however, it would be difficult to locate the particle at a given fixed point, since the particle is represented on several positions. It is up to me to find the link and the mathematical equation between the classical alternation of a particle and the HUP and the probability of distribution of the particle. Spoiler _ Link to post Share on other sites

swansont 7465 Posted February 20, 2020 Share Posted February 20, 2020 14 minutes ago, Kartazion said: Conclusion: The oscillator that I am trying to exploit is classical and not quantum. Indeed in my case the oscillator allows simply to alternate a particle from point A to point B at high speed. What is the equation of motion of the system? i.e. the solution to the oscillator equation. Link to post Share on other sites

Kartazion 3 Posted February 21, 2020 Author Share Posted February 21, 2020 12 hours ago, swansont said: What is the equation of motion of the system? i.e. the solution to the oscillator equation. As I am a beginner in mathematics I do not really know. I thought maybe having help here. Here is how I think to identify the values: Axa (t1) + xb (t2) + Bxc (t1) t1 = 1ns t2 = 1μs xa = -1 xc = 1 xb = -0.99 to 0.99 If we neglect any form of damping the equation of the movement is: m.d2x / dt2 + F.x = 0 (1) There is no analytical solution to the problem and digital integration is required. We use here the Runge-Kutta method at order 4. Link to post Share on other sites

swansont 7465 Posted February 21, 2020 Share Posted February 21, 2020 9 minutes ago, Kartazion said: As I am a beginner in mathematics I do not really know. I thought maybe having help here. Here is how I think to identify the values: Axa (t1) + xb (t2) + Bxc (t1) What is this equation supposed to mean? 9 minutes ago, Kartazion said: t1 = 1ns t2 = 1μs xa = -1 xc = 1 xb = -0.99 to 0.99 If we neglect any form of damping the equation of the movement is: m.d2x / dt2 + F.x = 0 (1) No, that doesn’t work. m.d2x / dt2 = F is almost F=ma. So there’s a minus sign missing With Fx, it has the wrong units. Link to post Share on other sites

Kartazion 3 Posted February 21, 2020 Author Share Posted February 21, 2020 18 minutes ago, swansont said: What is this equation supposed to mean? The movement of the particle along x and t. 13 minutes ago, swansont said: No, that doesn’t work. m.d2x / dt2 = F is almost F=ma. So there’s a minus sign missing With Fx, it has the wrong units. The equation comes from a university. 11 th line: ressources.univ-lemans.fr/AccesLibre/UM/Pedago/physique/02/meca/oscilanhar.html Link to post Share on other sites

swansont 7465 Posted February 21, 2020 Share Posted February 21, 2020 21 minutes ago, Kartazion said: The movement of the particle along x and t. No, it makes no sense 21 minutes ago, Kartazion said: The equation comes from a university. 11 th line: ressources.univ-lemans.fr/AccesLibre/UM/Pedago/physique/02/meca/oscilanhar.html Still wrong, and as I don’t speak French, I don’t know what they’re claiming Link to post Share on other sites

Orion1 10 Posted February 21, 2020 Share Posted February 21, 2020 (edited) 3 hours ago, swansont said: No, that doesn’t work. m.d2x / dt2 = F is almost F=ma. So there’s a minus sign missing With Fx, it has the wrong units. Still wrong, and as I don’t speak French, I don’t know what they’re claiming Translated from Google translate: Quote The potential energy of a harmonic oscillator is of the form Ep (x) = ½.Kx 2 . For this type of oscillator, the movement is sinusoidal, the phase portrait v = f (x) is an ellipse and the period is independent of the value of the total energy of the system. We consider here an oscillator (for example a mass m fixed to a special spring) for which the expression of the potential energy is Ep (x) = ½.Kx 2 - 1 / 3.AKx 3 . A is a dimensionless coefficient which will characterize the anharmonicity of the oscillator. To this potential corresponds the force F (x) = - dEp / dx = - Kx + AKx 2 . This force is zero for x = 0 and for x = 1 / A which are equilibrium positions. The expression of the second derivative of Ep is d 2 Ep / dx 2 = - K + 2.AKx For x = 0 it is positive and this position is a stable equilibrium. On the other hand, for x = 1 / A, the second derivative is negative and the equilibrium unstable. The corresponding value of Ep is Ep (1 / A) = Epmax = K / 6A 2. If we neglect any form of damping the equation of motion is: md 2 x / dt 2 + Fx = 0 (1) There is no analytical solution to the problem and digital integration is required. We use here the Runge-Kutta method at order 4. For this integration, the initial conditions must be specified. We take at the moment t = 0, v = 0. The total energy of m being E = Ep + Ec = Ep(x) + ½.mv 2 , at t = 0, we have E = Ep. To find the initial value of x, we must solve E = Ep(x) by a method of zero. It is also possible to take x = 0 for t = 0. As Ep(0) = 0, we determine v(0) from E = Ec = ½.mv 2 . The right part of the animation shows the curve Ep(x) = ½.Kx 2 - 1 / 3.AKx 3 (in red), the curve Ep (x) = ½.Kx 2 (in gray), the movement of the mass m. The orange line corresponds to potential energy and the cyan line to kinetic energy. The horizontal green line corresponds to the value of the total energy. At the top left we find the curve v = f(x) (phase portrait) and at the bottom right the curve x = f (t). The program also displays the value of the period T 0 = 2π / (M / K) ½ of the harmonic oscillator. The page appears to be modelling a classical harmonic oscillator for a mass m fixed to a special spring. 2 hours ago, Kartazion said: The equation comes from a university. 11 th line This formula appears to be attempting to utilize Leibniz's notation for Newtons second law of motion: [math]m \frac{d^2 x}{dt^2} + F \cdot x = 0[/math] However, Newtons second law of motion in Leibniz's notation is: [math]F = m \frac{d^2 x}{dt^2}[/math] Subtracting force from both the equation left hand side and right hand side results in: [math]m \frac{d^2 x}{dt^2} - F = 0[/math] However, note the equation definition for the potential force on line 5: [math]F(x) = -\frac{dE_p}{dx}[/math] And the resulting potential energy: [math]E_p = -F(x) \cdot dx[/math] So, The equation on line 11 is mixing systeme internationale units of force (newtons) and energy (joules), which is mathematically incorrect. Kartazion, I recommend purchasing a university level physics textbook and study the section on Newtonian mechanics for classical harmonic oscillators. University level physics textbooks already provide the required level mathematics in Leibniz's notation, as well as numerical integration for models such as these, instead of landing in the middle of other online models that display crude and incorrect formulas and with insufficient modeling experience. Some experience in Latex modelling would be beneficial also. Reference: Anharmonic oscillator - Lemans university: https://bit.ly/2SKdIJX Edited February 21, 2020 by Orion1 source code correction... Link to post Share on other sites

studiot 2148 Posted February 21, 2020 Share Posted February 21, 2020 (edited) 1 hour ago, Orion1 said: Kartazion, I recommend purchasing a university level physics textbook and study the section on Newtonian mechanics for classical harmonic oscillators. University level physics textbooks already provide the required level mathematics in Leibniz's notation, as well as numerical integration for models such as these, instead of landing in the middle of other online models that display crude and incorrect formulas and with insufficient modeling experience. Some experience in Latex modelling would be beneficial also. Several members have already remarked on this. Additionally the need for a crash course in A level and post A level Maths has already been admitted by the OP. Forcing equations of non linear oscillators is definitely a post graduate subject, let alone post A level. Edited February 21, 2020 by studiot Link to post Share on other sites

Kartazion 3 Posted February 21, 2020 Author Share Posted February 21, 2020 But you, the experts, do you have a mathematical approach to give me for my equation, namely the alternation of a particle from point A to point B? Or does it not exist at all? Link to post Share on other sites

swansont 7465 Posted February 21, 2020 Share Posted February 21, 2020 24 minutes ago, Kartazion said: But you, the experts, do you have a mathematical approach to give me for my equation, namely the alternation of a particle from point A to point B? Or does it not exist at all? Without existing in between those points? Link to post Share on other sites

Kartazion 3 Posted February 21, 2020 Author Share Posted February 21, 2020 2 minutes ago, swansont said: Without existing in between those points? The speed between two is almost instantaneous, even instantaneous. Link to post Share on other sites

swansont 7465 Posted February 21, 2020 Share Posted February 21, 2020 10 minutes ago, Kartazion said: The speed between two is almost instantaneous, even instantaneous. I don’t know what the function would look like for “almost instantaneous” You could start with a simple oscillator using Hooke’s law and tinker from there For “instantaneous” I doubt there is a solution. Link to post Share on other sites

Kartazion 3 Posted February 21, 2020 Author Share Posted February 21, 2020 (edited) 1 hour ago, swansont said: For “instantaneous” I doubt there is a solution. It can be of the order of: 1x10 ^{-1000000 }s for a trip from A to B or from B to A. Edited February 21, 2020 by Kartazion Link to post Share on other sites

swansont 7465 Posted February 21, 2020 Share Posted February 21, 2020 8 minutes ago, Kartazion said: It can be of the order of: \tag && 1 \times 10^-^1^0^0^0^0^0^0 s && for a trip from A to B or from B to A. That’s not helpful Link to post Share on other sites

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now