@ Akira Kanda - By now I have become familiar with your perspective on Schrödinger, illogical people and the like. But I hope you can appreciate that I do not share your views and can possibly not condone them. If that means that I take myself "too seriously as the king", so be it; it is certainly not my intention to impose my views on others; as so many others in RG, I write my views on matters that I deem to know something about, without any claim to be right or have the last word; the only thing implicit in my comments is that they are right to my best knowledge, which of course is no guarantee that they also right. Those who disagree with my views should certainly feel free to reject my views. I have no special privileges in RG and am subject to the same system of being voted up and voted down as others. In the end, it is up to each individual to decide what is right and what is wrong.

This is my last comment on this issue on this page. 

States in classical mechanics are completely characterized by the (parameterized) position and momentum as functions of time, or more simply by position and momentum. Momentum is the scalar value of mass times the velocity vector. In short, we can describe quantum mechanically some classical system via some equation with h-bar or we can take Hamilton's equations and replace pwith mv but doing the former is usually pointless and doing the latter is what one does anyway as one can't know the momentum without knowing the mass.

But perhaps I am misunderstanding your question.

I strongly disagree with the assertion made in the main question on this page. What is the Lagrangian of a classical free particle if not L = m v2/2 ≡ p2/2m? Similarly as regards the Hamiltonian H of a classical free particle. We could consider m as merely a constant of proportionality, or mass. However, irrespective of what we call m, it is exactly the same constant that one encounters in the quantum description of a free particle. Note that using the correspondence principle in quantizing the Hamiltonian of the classical problem, we do not change the constant m.

I totally agree with Professor B. Farid. Also, the rate of change of the expectation value of position and the expectation value of the momentum are related to each other via the mass of the particle. Furthermore, you compare between equation of the path in classical dynamics with energy equation in quantum mechanics.

Dear Juan, I cannot quite capture your point. Lagrangian has the dimension of energy, but if I take your Lagrangian as you mean to present it, it has the dimension of length squared. If you attempt to convert this Lagrangian into a form with proper dimension, then you will necessarily have to have a constant in there which has the dimension of mass. But this is a distraction. If you consider the main question on this page, on which my above response has bearing, it explicitly refers to a specific Lagrangian, or Hamiltonian, which both classically and quantum-mechanically contains a constant m called mass. My argument was that in going from the classical description of a point particle with mass m to its quantum-mechanical description, we do not dispense with mass.

Dear Yuri Rylov,

The answer is very simple. Because Schroedinger equation expresses about the wave-particle duality, while   in classical physics is not. As I explained to you before the problem is in the interpretation of Lorentz transformation in SRT depending on objectivity in classical physics. Lorentz transformation will lead to the wave-particle duality and Heisenberg uncertainty principle by removing objectivity in Lorentz transformation in a simple way. And then Lorentz transformation will express about Schroedinger equation, where, in case of non-inertial frame we get four vector, the source of which is the four-current, a first-rank tensor which is related to vacuum fluctuations described by Schroedinger equation which means photons mediates gravitation. 

Review my last comments in my discussion below. I told you my theory solved all the problems in physics. It is the unified theory.

https://www.researchgate.net/post/Why_must_graviton_have_a_spin_of_2_while_a_photon_has_a_spin_of_1

Sir Isaac Newton first presented his three laws of motion in the "Principia Mathematica Philosophiae Naturalis" in 1686. Newton's first law states that 

every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. 

This classical law is indeed independent of the mass of the particle.

But Schrodinger formulated the wave equation as an eigenvalue equation, which is of the form, H psi= E psi. Here H is an operator and E is the energy eigenvalue whereas  psi is the wave function. To know the space part of psi,  the most straight forward way is to  solve the Time Independent Schrodinger Equation (TISE). In TISE, E appears as a separation constant. Time dependent part of psi is known. The free particle wave function  (combining space part and time part) is what is known as a plane wave solution

psi=A ei(kx - wt)   Here w=E / hbar, We can explicitly check  that psi is an eigenfunction of the momentum operator.

In general there is a subscript of psi as psin, where the subscript signifies that this eigenfunction corresponds to energy En. The particle can be in more than one state depending on the value of n. For a free particle H is the operator corresponding to the kinetic energy of the particle as this is the only form of energy a free particle can possess. The operator for K.E. is inversely proportional to the mass of the particle. This is the reason why, even for a free particle, the eigenvalue equation stated above depends on the mass of the particle.

Classical and quantum cases, however, can be reconciled using Ehrenfest's theorem, using which we can prove that mass times the time derivative of the expectation value of position is the expectation value of  momentum.

@ Akira Kanda - The main question on this page concerns "the Schroedinger equation for a free particle", and hence my particular response to it. Responses on the questions in RG are not supposed to encyclopaedic, covering an entire field. 

@ Akira Kanda - By now I have become familiar with your perspective on Schrödinger, illogical people and the like. But I hope you can appreciate that I do not share your views and can possibly not condone them. If that means that I take myself "too seriously as the king", so be it; it is certainly not my intention to impose my views on others; as so many others in RG, I write my views on matters that I deem to know something about, without any claim to be right or have the last word; the only thing implicit in my comments is that they are right to my best knowledge, which of course is no guarantee that they also right. Those who disagree with my views should certainly feel free to reject my views. I have no special privileges in RG and am subject to the same system of being voted up and voted down as others. In the end, it is up to each individual to decide what is right and what is wrong.

This is my last comment on this issue on this page. 

Dear Yuri, the Schrodinger equation is a mathematical model in which a massive particle with kinetic energy and momentum is considered like a wavefunction. In all different expressions of the Schrodinger equation for the description of motion of the elementary particle there is mass because mass is present whether in the expression of the kinetic energy or in the expression of De Broglie's equivalent wavelength of particle.

In dynamic equation of motion of a classic body a different equation is considered and  mass there is always also here. Only in motion of a body in a gravitational field, under convenient physical conditions, body mass is absent.

Dear All,

In classical motion in macro world, quantum effect is not appeared why? Most of physicists think because h (Planck's constant) approaches to zero. But that is not the real cause.

The real cause is because in case of low velocity in classical motion the length contraction and time dilation are negligible and are not appeared. Length contraction and time dilation are appeared in case of high speed near the speed of light , and thus in this case quantum effect will appear. where according to my theory quantum effect produced by the new interpretation and understanding to the length contraction and time dilation according to refusing objectivity and then the Lorentz symmetry in the Lorentz transformation where in this case Lorentz transformation is vacuum energy dependent instead of relative velocity. In this case Lorentz symmetry is not affected in the Lorentz invariance, where we keep on Lorentz invariance while refusing Lorentz symmetry. Lorentz symmetry is only to keep on objectivity.

In this case in case of relativistic speeds for macro world (high speeds) the wave-particle duality and the Heisenberg uncertainty principle appear also. Then by refusing Lorentz symmetry it is disappeared all the paradoxes in SRT the Twin paradox, Ehrenfest paradox, Ladder paradox and Bell's spaceship paradox. So relativistic motion for micro world or macro world will lead to quantum effect. Where quantum and relativity are one theory leading to each other by my new equivalence principle, where in my new equivalence principle the escape velocity of the free fall object in gravity is defined as relativistic escape velocity not classical as in Einstein's GR. GR is not completely relativistic. Because of that entropy and temperature are related to time dilation in relativity, which related to vacuum energy according to my new transformation equations.

http://dx.doi.org/10.14299/ijser.2014.10.002

https://www.researchgate.net/post/Can_zero_point_energy_explain_the_rotation_curve_of_a_typical_spiral_galaxy_the_flat_appearance_of_the_velocity_curve_out_to_a_large_radius2

https://www.researchgate.net/post/Why_must_graviton_have_a_spin_of_2_while_a_photon_has_a_spin_of_1

Why does the Schroedinger equation for a free particle contain the particle mass?

This question is asking for an explanation of Schroedinger dynamics. It could be compared with asking `why is a billiard ball rigid?'. This question cannot be answered by the classical theory of rigid bodies. You need a different theory for that, either a

classical or quantum mechanical atomic theory of the solid state, explaining the observed rigidity of the billiard ball by pointing to the tight bindings of the ball's atoms. Rigid body theory just describes rigidity, it does not explain it.

By the same token quantum mechanics does not explain, it just describes what we see when interfering with microscopic objects by carrying out measurements. It seems to me that for explanations we have to turn to sub-quantum (hidden-variables) theories. The Schroedinger equation just describes the way a wave packet changes,

it does not explain it. But its mass-dependence may tell us something about the sub-quantum process changing the shape of the wave function in the way described by it, this process probably encompassing a kind of diffusion process having a mass-dependent diffusion rate.

Willem de Muynck

You can read 

http://mbrausen.blogspot.mx/2013/05/la-generalizacion-de-la-ecuacion-de-de.html

Sorry! is in spanish but you could use google translate. This is the historical derivation involving the de Broglie length and the classical motion equation. Also comes a critical to its derivation.

Oh Akira, meanwhile we all learned how deep and brilliant your understanding of physics is. Please spare us more examples.

Manuel:

   En cuanto a que la velocidad de la onda sea doble de la partícula, te mando este artículo mío que quizá te sirva.

   Saludos. 

As Yuri says, the motion of classical noninteracting particles (or non-relativistic particles in a gravity potential) does not depend on their masses. This is not the case in the real world, which is better described by quantum mechanics (for instance by the Schrödinger equation).

Crudely, the mass in the Schrödinger equation tells us when a classical description of the motion is sufficient (very likely for a billiard ball on a snooker table) or not (as for the electron in a hydrogen atom). For a particle in free motion the mean position will still follow the classical description (Ehrenfest theorem), but the quantum uncertainty around the mean will increase faster for a particle with smaller mass. The mass term in the Schrödinger equation tell you that.

If Einstein had grown up with an education in quantum mechanics, could he have  come to the equivalence principle, and later general relativity?

Hello, everyone.My apologies for writing in Spanish; I got carried away. Here is a link to the  same paper in English.

@ Akira Kanda - If you had paid attention, I closed my last comment on this page (the one before this one) with the announcement that it would be my last response to your points on this page. I emphasise once more, that I did not, and do not, say that your points are necessarily invalid, only that given what I personally believe to be correct, I could not, and cannot, condone them. Further, you raise a host of issues which are not relevant to the main question on this page. I responded to this very question, and not to all possible questions that may be related to it. I hope you can appreciate that it becomes a full-time job to respond to each and every question on pages of RG in such a way that all related questions are also thereby answered. Please accept my apologies, but time limitations do not permit me to respond to your above points. If this implies my defeat in your eyes, I accept the defeat. 

Akira, I am not obliged to argue; life is simply too short to take up any single challenge offered. As I wrote earlier, I have my views and you yours, and I am comfortable with that.

Akira, nowhere have I stated or implied that my particular response to you was in accordance with scientific methods or principles. As an individual, I have to divide my available time between a number of responsibilities, which implies that there is only so much time that I can devote to my activities on RG. I cannot possibly fit the task of answering all possible questions on the pages of RG into the limited time available to me, not to say anything about the essential fact that my expertise does not extend to all fields of science. I think I am entitled to decide on the way I spend my time. So, let us part in peace.

Point particle dynamics can neglect the "m" because of the equivalence principle as mentioned above.  This works to the extent that the spatial extent of the body and any radiation reaction effects are small.  

When we build wave equation theories which give some sort of "classical limit" in an Ehrenfest fashion the mass is needed to modulate the oscillation in the packets that form the classical particles in the limit.  With out it, any packet would have to have the same frequency of oscillations if it had the same velocity.  This is problematic if we want to include particles with different mass (hence momentum and energy for the same velocity).  

I've never accepted complimentarity but I do believe that a continuum description is the correct (or best we can currently have) description for microscale physics.  Akira, the equivalence principle is an experimental result to the best of our measurements.  I think it will ultimately fail especially when there is acceleration and radiation reaction-like effects.  You can't add accelerations in relativity even when one can construct a global frame.  

I am very interested in the cases where continuum mechanics fails especially in regards to rheology.  

Akira> "e=mv2/2 is not the energy needed to accelerate from m*0 to m*v unless the acceleration is uniform."

That statement is a blatant misunderstanding. Any textbook on elementary mechanics will carefully explain that. It follows from F=m(dv/dt) (Newton 2), and

(de/dt) = Fv = d (mv^2/2)/dt 

Akira> "I do not think"

It is difficult to disagree with that. But you should, before you post.

Akira,

since in your opinion there is hardly anything in physics on which physicists have a correct view, I extrapolate that you will be able to produce hundreds of pages with ignorant and arrogant statements that will disgust people and make them adding their own hundreds of pages. So, I kindly ask you to discontinue this unworthy game.

Raul Simon. Gracias, lo lei. 

También yo te mando el URL de uno de mis  artículos que es más técnico que pone matemáticamente para mí la concepción de que se puede re formular la teoría cuántica con éxito simplemente no olvidando que en realidad las partículas no son corpúsculos sino en realidad campos de fuerza en movimiento. Muy en la filosofía inicial de Faraday descrita en el libro Fields of Force de William Berkson. Editorial Cambridge

El segundo documento es para entender las matemáticas desplegadas en el primer artículo.

https://www.researchgate.net/publication/260825909_A_modication_of_classical_electrodynamics_with_quantized_levels_of_energy._A_new_quantum_theory

https://www.researchgate.net/publication/260918114_A_general_solution_for_Navier-Stokes_radial_pressure_equation._Case_of_frictionless_movement_of_fluids

Raul Simon

Also I command the URL one item that is more technical, it's to me the mathematical formulation of the idea that you can reformulate quantum theory successfully simply not forgetting that in reality the particles are not corpuscles but actually moving force fields. In the initial philosophy of Faraday described in the book Fields of Force by William Berkson. Cambridge

The second document is to understand the mathematics deployed in the first article.

Working Paper A modication of classical electrodynamics with quantized lev...

Working Paper A general solution for Navier-Stokes radial pressure equatio...

@Akira

On your mv²/2 problem: If you accelerate (i.e.  (a|v) >0 with obvious notation for the scalar product and a for acceleration) then you add energy to the mass and if you decelerate (i.e.(a|v)

Dear all,

If one reads the question carefully, he realizes that Yuri assumed that the mass m is constant. So, please leave relativity and all other answers aside and only remember that Yuri is comparing between equation of the path in classical mechanics and equation of energy in linear quantum mechanics.   

Dear RG-ers,

The answer to my question does not lie in classical dynamics and in quantum mechanics. It lies in the space-time geometry. The boundary between the particle dynamics and the space-time geometry is mobile. The best example of this mobility is a motion of a charged particle in the given electromagnetic field, which can be described as a free motion of a  particle in the 5D space-time of Kaluza-Klein. In this example the electromagnetic field is included in the space-time geometry, and the boundary is shifted in the direction of dynamics. Note, that the inertial mass is a dynamic mass. The gravitational mass may be treated as a geometric mass, because the gravitational field is essentially a geometric field.

Unfortunately, we do not use coordinateless description of the space-time geometry. We cannot imagine that the space-time geometry may have an indefinite dimension, because the dimension is considered as a fundamental quantity, describing geometry. Introduction of the Riemannian geometry starts by the words: “Let us consider a manifold of dimension n and a coordinate system on it”. We cannot imagine such a situation, when some properties of the proper Euclidean geometry GE are special properties of GE. We believe that all properties of GE are the general geometric properties, i.e. properties of any space-time geometry. The curvature is considered as unique special property of Riemannian geometry. This property is absent in GE.

In reality, there are only two general geometric properties (i.e. properties, which are common for all possible space-time geometries). These properties (1) scalar product of two vectors and (2) linear dependence of n vectors are expressed only via metric \rho, or via world function \sifma =1/2\rho2. All other properties of GE connected with special form of Euclidean world function \sigmaE. The dimension of GE is defined via world function \sigmaE as maximal number of linear independent vectors in GE.

Scalar product (AB.CD) of two vectors AB and CD is defined by the relation

(AB.CD)=\sigma(A,D) +\sigma(B,C) -\sigma(A,C) -\sigma(B,D)

n vectors P0P1, P0P2,…P0Pn are linear dependent if Fn(Pn+1) = 0, where Fn(Pn+1) is the Gram’s determinant

Fn(Pn+1) = det||(P0Pi.P0Pk)||, i,k = 1,2,…n

In the n-dimensional Euclidean geometry GE in the following conditions take place

There are such Pn+1 = {P0,P1,…Pn}. that Fn(Pn+1) does not vanish, whereas Fk(Pk+1)=0 for k>n.                                                                                     (*)

Condition (*) determines dimension n(P0) at the point P0, which is maximal number of linear independent vectors at the point P0. As far as GE is uniform, the dimension n(P0) does not depend on the point P0. One can see that (*) form a lot of constraints on the world function \sigmaE of GE.

In the discrete space-time geometry Gd, whose metric \rhod satisfies the relation

|\rhod(P,Q)| does not belong to interval (0,\lambda)             (**)

where \lambda is a minimal length in the space-time. Usually one considers (**) as a condition on the pointset \Omega, where geometry is given. As a result one obtains a geometry on a lattice. The geometry on a lattice is anisotropic and not uniform. Such a geometry cannot be considered as a geometry of the space-time.

There is another possibility. One should consider the pointset \Omega as a manifold \OmegaM, where the geometry of Minkiwski is given. In this case the simplest solution of (**) may be taken in the form of a discrete geometry Gd with the world function

\sigmad =\sigmaM + \lambda2/2 sgn(\sigmaM)                                (***)

where \sigmaM is the wotld function of the geometry of Minkowski. It is easy to verify that (***) satisfies (**). However, the discrete geometry Gd is uniform and isotropic, because \sigmad is a function of \sigmaM. Condition (*) is not fulfilled in Gd, and one cannot introduce concept of dimension in Gd. We have the following situation. In the discrete space-time geometry for vectors P0Pi, whose length is of the order l>>\lambda, the Grams determinant Fn(Pn+1)

is of the order 4\lambdan-4 for n>4.

Of course in Gd there are no smooth world lines. World lines are replaced by world chains, which are broken lines, whose links are segments of straight lines of equal lengths \mu. Let ..P0,P1,…Ps,Ps+1,…be break points of the world chain C. The link length |PsPs+1|=\mu is the geometric mass of the particle which is connected with the conventional mass m by the relation m = b\mu, where b is an universal constant. Thus, in Gdthe particle mass is geometrized. In Gd the world chain wobbles, because Gd is a miltivariant space-time geometry. As a result a random walking is added to the progressive motion of breaking points of the world chain C. Amplitude of this walk depends on relation between \lambda and \mu. The less \mu, the greater the amplitude of random wolk. Nature of the mass m in the Schroedinger equation is the mass geometrization together with the discreteness of the space-time geometry. See for details in “ Metrical conception of the space-time geometry” Int. J. Theor, Phys. 54, iss.1, 334-339, (2014). Electronic version: http://gasdyn-ipm.ipmnet.ru/~rylov/mcstg2e.pdf, and in “Different conceptions of Euclidean geometry and their application to the space-time geometry” , Available at http://arXiv.orgperceived/abs/0709.2755v4.

I should note, that such things as uniform discrete space-time geometry and multivarint geometry are perceived hardly, because they contradict  to those representation of a geometry, which is taught in the school.

in my previous post there is a mistake in the last reference

http://arXiv.org/abs/physics/0603237v6

Schroedinger equation for a free particle does not contain the particle mass.It only contains the sum of kinetic energy and the potential energy. This sum does not include the rest mass energy. Following article describes relativistic version of the Schrodinger equation (eq.2.5) which yields Dirac energy levels. This equation (2.5) includes the rest mass energy.

Article Quarkonium and hydrogen spectra with spin-dependent relativi...

Sorry Yuri, your last explanation indicates that the assertion made in your question was incomplete as B. Farid demonstrated earlier.

Yuri.

So, in the end it becomes clear that this was not a serious question at all! It was just a lure to attract buyers to the shit you want to sell. I think such questions, and the people who asks them, should be banned from ResearchGate. This should be a place for serious science and sincere questions.

I am sorry to disagree with Kåre Olaussen. In my view the question ``Why does the Schroedinger equation for a free particle contain the particle mass?'' is a legitimate one if quantum mechanics is an incomplete theory, which it is if there is a sub-quantum dynamics (to be described by a sub-quantum theory). An answer like Kåre Olaussen's one makes sense only on the basis of a belief in the completeness of quantum mechanics. Notwithstanding the Copenhagen interpretation's assertion of completeness, we have no reason to believe in such a completeness.

Willem de Muynck

Willlem.

I don't oppose the question.

What I strongly oppose is the falsehood under which the question was posted, which is completely destructive to the Q&A section of ResearchGate. Which serious scientist will waste time answering questions when they discover that the questioner were totally dishonest? Unfortunately, Yuri is not the only one behaving this way on Q&A. Those of us who wants Q&A (and in the long run ResearchGate) to survive as a serious entity, must act strongly against such antisocial behavior.

Kåre,

Thanks for your reaction. Of course I agree with your intention to keep discussion as rational as possible. In my view the best thing to do is to ignore contributions of which you fear that they lead to a waste of time.

Willem de Muynck

Kare, you make an interesting point.  Online forums can degenerate into narcissism and rubbish.  If this was the author's intention he probably should have started the thread differently.  I wonder about the value of very long technical posts versus short summaries and references.