What is really the total energy of particle? Also an interesting physical quantity f = mc^2/h

[caracal]

Hi all. 

Life is short, and i have another interesting idea regarding special relativity. 

I want to write this thread independently from the other thread i have written earlier that is about pulsating dots. Don't bring that up here in this thread. There may not be any connection between these ideas. the reason is that whatever kind of dots there may be, they may be extremely small.

I have found interesting articles that are written by E.Bakhoum starting from the year 2001 where he suggest different total energy for the particle. I want to bring it up here as i think it is an interesting idea. I don't know if it its true. I don't agree with everything and i have also own thoughts about the issue, so this thread is not just repeating what he has written. But i think it is better to introduce his texts here as a reference.

https://arxiv.org/search/physics?searchtype=author&query=Bakhoum%2C+E+G

Unfortunately, it looks like that most of the predictions from this new total energy are theoretical and does not have practical meaning.

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The usual comprehension of the subject of energies in special relativity is that the total energy, kinetic energy and momentum are

(i don't know if latex works here)

[math] E_{tot}= m \gamma c^2 [/math] 

[math] E_{kin} = m c^2 (\gamma-1) [/math]

[math] p = m \gamma v [/math] 

[math] \gamma = \frac{1}{\sqrt{1-(v/c)^2}} [/math]

,where [math] \gamma [/math] is Lorentz factor and m is mass of the particle in rest. In this notation m is always 'mass in the rest' m=m0 . 

How this kinetic energy is derived? It is derived from the situation where particle starts to accelerate from rest into velocity v and determining how much work is required to do this. I don't go there.

Where does this total energy is derived from? It is actually interpretation from the kinetic energy formula - which is coming from definite integral of work put to a particle that accelerates from the rest to velocity v. It looks like there is some kind of 'rest energy' when particle is at rest. But this definite integral is a difference of two things. You can add same thing to both of the components and still have same difference. It turns out that there is one thing that could be added to both of the components. 

First note that there is following important mathematical identity:

[math] \gamma v^{2} = c^{2}( \gamma - 1/\gamma) [/math]

E. Bakhoum suggest that there is such component in both of the terms that is [math] -mc^2(1/\gamma) [/math]  .

He suggest that the total energy would be different but kinetic energy and momentum would be same.

[math] E_{tot} = m \gamma v^2 = m c^2 (\gamma - 1/\gamma )= pv [/math]

[math] E_{kin} = m c^2(\gamma -1) [/math]

[math] p = m \gamma v [/math]

what this new total energy would mean? Particle does not have rest energy. Mass is converted to energy with ratio E=mc^2 only when particles collide in near light velocity. There is no 'relativistic mass' that is energy divided by c^2 , since the another energy component in the total energy besides kinetic energy - is not same but depends on the velocity:

[math] E_{2} = mc^2(1-1/\gamma)  [/math]  instead of [math] E_{2} = mc^2  [/math]

Now i ask question that comes clearly to my mind: How about radioactive decays? They are proven to follow formula [math] \Delta E = \Delta m c^2 [/math] ? 

I think the answer is that particles that participate on radioactive decay are accelerated before their interaction to near light velocity and decay does not happen otherwise. These accelerations may be 'hidden' under the phenomenon of quantum tunneling that is the most prominent quantum mechanical phenomenon in nuclear reactions and decays.

Therefore i don't completely agree with E. Bakhoum. I think E. Bakhoum makes mistake when he assumes that particle decays and nuclear decays and reactions can also happen when interacting particles have non-relativistic velocities at the moment of the reaction. He thinks that that there are different kind of energy distributions. I think any reaction or decay can happen only when interacting particles approach the light velocity.

If nuclear reactions and radioactive decays happen only when the participating particles are accelerated to high velocity, both total energies give same predictions for nuclear reactions and decays.

With this rule added - i think the idea might work.

You may not be aware that the total energy is inserted to theory of quantum mechanics from outside
of the theory. If you insert this new total energy there, many things changes. For example:

1. phase velocity = group velocity = v

2. de Broglie frequency [math] f = \frac{E_{tot}}{h} = \frac{m \gamma v^{2}}{h} [/math]

3. Energy-momentum relation 

[math] E^{2} = (\frac{pv^{2}}{c})^{2} + (mv^{2})^{2} [/math] instead of [math] E^{2} = (pc)^{2} + (mc^{2})^{2} [/math]

(which may be now unpractical equation)

4. Classical radius of electron [math] r_{e} = \frac{1}{4 \pi \varepsilon_0} \frac{e^{2}}{m_e \gamma v^{2}} [/math]

The radius depend on velocity and is infinite at rest. So what does it mean, how is this interpreted then? Is this anymore useful property? I am not sure and i don't go into this here. 

All of these changes are coming from that you insert E_tot = muv^2 instead of E_tot = muc^2 during the derivation.

Also some of the four-vectors changes, such as four-momentum:

5. four-momentum

[math] P = (m \gamma v_{x} , m \gamma v_{y} , m \gamma v_{z}, i (\frac{v}{c})m \gamma v) [/math] 

(i am not sure about this)

Also Lagrangian is now different.

But i don't go into these. But i believe that everything is consistent. However most of these changes seem to be theoretical aspects without practical outcome.

The rest here is my own thinking:
-----------
An interesting physical quantity of mc^2/h 
................................................................

What i think there might be new insight that the matter wavelength depends directly
from the new total energy of the particle:

[math] \lambda = \frac{h}{m \gamma v} = \frac{hv}{m \gamma v^2} = \frac{hv}{E_{tot}} [/math]

[math] f = \frac{v}{\lambda} = \frac{m \gamma v^2}{h} = \frac{E_tot}{h} [/math]

[math] f = \frac{m \gamma v^2}{h} = \frac{mc^2}{h}(\gamma - 1/\gamma) [/math]

(note that phase velocity and group velocity are both v with this new total energy)

Where is this frequency coming from? Could it be beating between [math] m \gamma c^2 / h [/math] and [math] mc^2(1/\gamma) [/math] ?

That would mean that when particle is in rest, there is some kind of frequency that is related
to its mass:

[math] f_{rest} = \frac{mc^2}{h} [/math]

This would be a kind of "rest frequency" of the particle - whatever it is. For some reason it divides to two components when particle start to move. It could be also a maximum of some frequency distribution that changes to distribution with two maximums when particle moves. 

This is just a schematic picture of how the situation would look like if you look frequency distributions. The spikes can be different kind of:

There are two relativistic effects known to happen when particle moves: length contraction and time dilation. Maybe these components are related to these. 

This constant c^2/h has very high value:

[math] \frac{c^2}{h} = 1.356 \cdot 10^{50} [\frac{Hz}{kg}] = 1.366 \cdot 10^{32} [\frac{Hz}{eV}] [/math]

This frequency is very high even for lightweight particles such as neutrino which might have maximum mass of 0.120 eV/c^2. Therefore it is very difficult, if not impossible to observe. However the component  mc2h(1/γ)  could be observable since it goes down to zero when particle approaches light velocity.

if 

[math] m = 0.120 eV/c^{2} [/math]

=>

[math] f_{rest} =  0.120 eV * 1.366 * 10^{32} [Hz/eV] = 1.639 * 10^{31} Hz [/math] 

An effect of time dilation for solar neutrino on the frequency f_rest:

[math] f'/f_{rest} = \sqrt(1-0.99999999995^2) = 10^{-5} => f' = 1.639 * 10^{26} Hz [/math]

Wavelength of f':

[math] \lambda' = c/f = \frac{2.998*10^{8} m/s}{1.639 * 10^{26} hz}   = 1.829 * 10^{-18} m [/math]

That is still small - about 1000 times smaller than the radius of proton.

What does this frequency then cause? a second kind of interference pattern? I don't know.

So if this thinking is right then there is some kind of frequency that is related to particle's mass and that divides to two components, and beating between these two components is somehow responsible for matter wave phenomenom.

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this is all. I would say that there might be something interesting in this idea of different total energy. Whether it is really the total energy of a particle, i don't know.

[swansont]

1 hour ago, caracal said:

Particle does not have rest energy.

Rest energy was derived in one of Einstein’s 1905 papers

 

1 hour ago, caracal said:

Mass is converted to energy with ratio E=mc^2 only when particles collide in near light velocity.

Inconsistent with e.g. electron-positron annihilation 

 

Quote

I think the answer is that particles that participate on radioactive decay are accelerated before their interaction to near light velocity and decay does not happen otherwise. 

Evidence? 

Some particles that “participate” don’t even exist before the decay occurs

1 hour ago, caracal said:

E. Bakhoum suggest

Citation?

[caracal]

I answer here to all of the things you pointed out.

There is actually one possibility more: Maybe there are two different kind of energies: Total energy [math] E_{tot} = m \gamma c^2 [/math] that is converted to other forms of energy in nuclear reactions and radioactive decays and "energy that is participating quantum mechanical behavior" [math] E_{qm} = m \gamma v^2 [/math] . It would be this later one that should be inserted to quantum mechanical theories instead of the total energy.

Yes, rest energy was derived by Einstein but it is coming from interpretation of the formula of kinetic energy. You can add same term to both of the energies "rest energy" and "total energy" and have still same kinetic energy.

citation:
https://arxiv.org/abs/physics/0206061
"Fundamental disagreement of wave mechanics with relativity" page 4 about. This is the first of his papers. There he argues that some equations become nicer if you add different total energy.

But by thinking that possible rule that "nuclear reactions happen only when some of the particles have high velocities",

There are three reasons that comes to mind that could accelerate particle
1. Interactions: EM interaction, strong interaction and weak interaction
2. Quantum tunneling: It may be possible that particles have high velocities during quantum tunneling.
3. Heisenberg's uncertainty principle, the energy time form of it. [math] \Delta E \Delta t = h/2\pi [/math] It allows particle to have temporarily high amount of kinetic energy but that energy is taken back also very soon.

Just before electron-positron annihilation the both particles are being accelerated towards each other. I don't know
is this enough. E.Bakhoum also writes about this in the article "Dialogue on the principle of mass-energy equivalence"

[swansont]

14 minutes ago, caracal said:

You can add same term to both of the energies "rest energy" and "total energy" and have still same kinetic energy.

Then it doesn’t matter, since the energy appears on both sides of an equation. The energy wouldn’t be something you can convert to other forms. How would you test to see if the energy existed?

16 minutes ago, caracal said:

Just before electron-positron annihilation the both particles are being accelerated towards each other. I don't know
is this enough. E.Bakhoum also writes about this in the article "Dialogue on the principle of mass-energy equivalence"

Electrons and positrons can form positronium, with a ground state of -6.8 eV, which classically corresponds to a KE of 6.8 eV and a PE of -13.6 eV. This would not give you a high-speed electron or positron.

25 minutes ago, caracal said:

https://arxiv.org/abs/physics/0206061

Published in Physics Essays, which is known to publish crackpot material. 

30 minutes ago, caracal said:

possible rule that "nuclear reactions happen only when some of the particles have high velocities",

I’ve done experiments where radioactive potassium isotopes were held in a magneto-optic trap, meaning they had almost no KE - the residual speeds would be a few cm/s - and they underwent decay.

[caracal]

Looks like he appears in arxiv. I think he is not a bad writer. But he may not have 'proofreaders' or 'moderators' so there may be mistakes. The idea is very solid - just one change and one equation.  [math] E_{tot} = m \gamma v^2  = mc^2(\gamma - 1/\gamma) [/math] and that's it. 

I repeat here a little.: The derivation of equation for kinetic energy [math] E_{kin} = mc^2(\gamma - 1) [/math] is done by looking how much work has to be done for particle in order it to gain velocity v.  But after that it is interpreted this equation such that [math] E_{kin} = E_{tot} - E_{0} = m \gamma c^2 - mc^2 [/math]
But actually what you have is [math] E_{kin} = (m \gamma c^2 + X ) - (mc^2 + X ) [/math] where this X can be anything - any function of v + any constant. Looks like there is room for alternative interpretation. That would be that [math] E_{tot} = m \gamma v^2 [/math] .

This new energy term [math] E_{new} = mc^2(1-1/\gamma) [/math] appears in the equations of quantum mechanics. For example de Broglie frequency [math] f =\frac{m \gamma v^2}{h} [/math] which is now different. Most of Bakhoums writings deals with how quantum mechanics and i think QED changes when there is different total energy. But it seems that nothing practical comes out of it - equations just changes and he claims that they become better. I am not sure how it can be converted to other form of energies, but my own suggestion is that it can in particle decays and nuclear reactions but only when velocity of interacting particles approaches c.

Lets go directly to the reactions of beta decay or annihilation. From vertexes like this, without knowing the details of the reaction, i can't say whether particles have high velocities at the moment of reaction itself or not - in the places where the vertices meet or separate. It is probably quantum tunneling what makes these reactions possible. But what happen during the quantum tunneling and during the reaction itself? Can particles have high velocities during quantum tunneling?

(pictures)

The mean Z and W bosons lifetime is [math] 3 * 10^{-25}s [/math] so i guess beta decay event when W-boson is ejected from proton is even less than this.

The reaction itself where particles separate may therefore be very rapid and takes time about that much : [math] 10^{-25}s [/math] ? 

In order to estimate the event time of reaction: Radius of proton is [math] r = 0.87*10^{-15}m [/math]. If some particle is moving at light speed, it goes across proton in time T : [math] T = 2R/c = \frac{1.74*10^{-15}m}{2.998*10^{8} m/s} = 5.804*10^{-24} s [/math]

In such very short time scales the Heisenberg uncertainty principle may also play bigger role: Particles may have for short period of time high kinetic energies such that [math] \Delta E \Delta T = \frac{h}{2 \pi} [/math]

If [math] \Delta T = 10^{-25}s [/math] , then

[math] \Delta E = \frac{6.582 * 10^{-16}eVs}{10^{-25}s} = 6.582 * 10^{9} eV = 6582 MeV [/math] 

Maybe particles could have high velocities during very short moment of tunneling and after that lose the velocity, possibly both due to Heisenberg uncertainty principle? In that case they could have also high velocities at the moment of reaction. 

[swansont]

2 hours ago, caracal said:

Looks like he appears in arxiv.

Which is a preprint server (not peer-reviewed) and the paper was published in Physics Essays, which does not instill much confidence.

2 hours ago, caracal said:

I think he is not a bad writer. But he may not have 'proofreaders' or 'moderators' so there may be mistakes. The idea is very solid - just one change and one equation.  Etot=mγv2=mc2(γ−1/γ) and that's it. 

I repeat here a little.: The derivation of equation for kinetic energy Ekin=mc2(γ−1) is done by looking how much work has to be done for particle in order it to gain velocity v.  But after that it is interpreted this equation such that Ekin=Etot−E0=mγc2−mc2
But actually what you have is Ekin=(mγc2+X)−(mc2+X) where this X can be anything - any function of v + any constant. Looks like there is room for alternative interpretation. That would be that Etot=mγv2 .

 

If it can be anything, that’s not meaningful. But all models must be confirmed by experiment. What experiment would confirm this additional energy - which can’t be converted into other forms, if it always appears in the equations on both sides.

 

2 hours ago, caracal said:

This new energy term Enew=mc2(1−1/γ) appears in the equations of quantum mechanics. For example de Broglie frequency f=mγv2h which is now different.

Which means that anything that depends on the deBroglie wavelength is wrong. How do you reconcile this with experiments where the mainstream equation works?

 

 

2 hours ago, caracal said:

 

Most of Bakhoums writings deals with how quantum mechanics and i think QED changes when there is different total energy. But it seems that nothing practical comes out of it - equations just changes and he claims that they become better. I am not sure how it can be converted to other form of energies, but my own suggestion is that it can in particle decays and nuclear reactions but only when velocity of interacting particles approaches c.

Lets go directly to the reactions of beta decay or annihilation. From vertexes like this, without knowing the details of the reaction, i can't say whether particles have high velocities at the moment of reaction itself or not -

As I noted, I’ve done beta decay experiments where the parent is at rest. 

 

[caracal]

Reconciling this new total energy to quantum mechanical theories is what Bakhoum has done in his writings. I don't go to check them now. But what it seemed to be is that It just clears out some things in the theories of quantum mechanics. The simplest is that there is not anymore super luminal phase velocity. 

...

Yes i can see what kind of experiment that is. A sample of radioactive material contain an atom nuclei that undergo beta decays and that happens in rest. Energy,momentum,electric charge and spin are conserved. Proton has less mass than neutron and therefore it is energetically possible for neutron to decay to proton, electron and antineutrino. Down quark emits W- boson and changes to up quark. W- boson decays after some time into electron and antineutrino.

In fact it does not matter whether the parent is in rest or in constant velocity - the situation would be same, only the sample would be rest in other inertial reference frame. That is because of Relativity principle. 

What i mean by this suggestion is high relative velocity between interacting particles + powerful interaction.

But what happens at the moment when down quark changes to up quark and emit W- boson? That is a moment that takes about [math]N * 10^{-25}s [/math] or less since the lifetime of W boson is already [math] T = 3 * 10^{25}s [/math] And also what happens at the moment when W- boson decays into electron and antineutrino? I don't know. Particles just seem to appear or change to some other particles.

There are three reasons why there might be high velocities present:
- Quarks already have high kinetic energies
- Potential energies
- Heisenberg uncertainty principle

Two last ones can temporarily give more kinetic energy to particles. This kinetic energy does not show up in the outcome of the reaction.
 

[Genady]

8 minutes ago, caracal said:

what happens at the moment when down quark changes to up quark and emit W- boson?

The down quark field interacts with the up quark field and the W-boson field transferring to them its energy and momentum.

[caracal]

On 9/22/2023 at 8:18 PM, swansont said:

If it can be anything, that’s not meaningful. But all models must be confirmed by experiment. What experiment would confirm this additional energy - which can’t be converted into other forms, if it always appears in the equations on both sides.

It may not be just anything, but i would need more information. I don't know which total energy would be better. Bakhoum just treats the definite integral differently to get his total energy. Here is reference: https://arxiv.org/pdf/physics/0206061.pdf (page 6). But this is just mathematical trick. This does not prove that his suggestion is the right total energy. 

Work integral [math] E_{k} = \int_{0}^{s} F dx = \int_{0}^{p}v dp [/math] , is mathematically speaking always a difference between two things : [math] E_{k} = E(s) - E(0) [/math]. You can add same function to E(s) and E(0) and still get same [math] E_{k} [/math]. In other words, you don't know how much energy particle has at rest and how does the total energy vary between 0 and s. 

Bakhoum suggested that [math] E(s) = m \gamma v^2 = m c^2 (\gamma - 1/\gamma) [/math] and [math] E(0) = m c^2(1-1/\gamma) [/math]. 

I think Nuclear reactions support [math] E_{tot} = m \gamma c^2 [/math], but Bakhoum states that [math] E_{tot} = m \gamma v^2 [/math] reconcile some equations in relativistic quantum mechanics better. 

Bakhoum studied Dirac equation and Hydrogen atom, how different total energy change them.

But Bakhoum also claims that nuclear fission would have different energy spectrum, and beta decay happens differently. I tend to think that he might be wrong there. I think the energy spectrum and beta decay could be explained well without this new total energy.

I suggest two different kind of things to explain energies in nuclear reactions:

1. there are actually two different kind of total energies - energy that participates to relativistic QM and energy that participates to mass to energy conversions

or

2. All nuclear reactions and decays just happen to involve high velocities + powerful interaction of the mass that is converting (to particles, kinetic energy or photons) , when [math] m \gamma v^2 [/math] will aprroach [math] m \gamma c^2 [/math]. 

For example in beta decay that seem to happen in rest - I listed these possibilities that could make option 2. possible - Heisenberg uncertainty principle, quark kinetic energies, potential energies and quark field interactions could make the reaction involve something that is accelerating from low velocities to high velocities - therefore mass to energy conversion look like to be form [math] \Delta E = \Delta mc^2 + \Delta E_{kin} + \Delta E_{binding} + etc [/math]. 

Actually equation [math] \Delta E = \delta mc^2 [/math] in nuclear reactions could be still right. That would mean that Mass is still the measure of the energy content of matter. But particles just have this different total energy [math] E = (\gamma - 1/\gamma ) mc^2 [/math] That would mean that the energy budget of protons mass or nucleus mass could be slightly different. There is that 'could' because my thinking is not working now.

 

On 9/22/2023 at 8:18 PM, swansont said:

Which means that anything that depends on the deBroglie wavelength is wrong. How do you reconcile this with experiments where the mainstream equation works?

I don't know where de Broglie frequency is needed, if is it needed anywhere. If it is not needed, then different de Broglie frequency does not change any practical calculations.

 

Looks like total energy of particle is used in relativistic quantum mechanics such as quantum field theory. I didn't manage to find information where it is needed exactly. But i understood that it is an assumption that is inserted to the theory. (It is assumed that total energy would be [math] m \gamma c^2 [/math].) 

Edited October 7, 2023 by caracal

[swansont]

15 minutes ago, caracal said:

It may not be just anything, but i would need more information. I don't know which total energy would be better. Bakhoum just treats the definite integral differently to get his total energy. Here is reference: https://arxiv.org/pdf/physics/0206061.pdf (page 6). But this is just mathematical trick. This does not prove that his suggestion is the right total energy. 

Work integral Ek=∫s0Fdx=∫p0vdp , is mathematically speaking always a difference between two things : Ek=E(s)−E(0) . You can add same function to E(s) and E(0) and still get same Ek . In other words, you don't know how much energy particle has at rest and how does the total energy vary between 0 and s. 

Bakhoum suggested that E(s)=mγv2=mc2(γ−1/γ) and E(0)=mc2(1−1/γ) . 

I think Nuclear reactions support Etot=mγc2 , but Bakhoum states that Etot=mγv2 reconcile some equations in relativistic quantum mechanics better. 

Bakhoum studied Dirac equation and Hydrogen atom, how different total energy change them.

The spectrum of the hydrogen atom is well-known, and agrees with mainstream theory.
If they are changed and no longer agree with experiment, then the new treatment is wrong.

 

15 minutes ago, caracal said:

But Bakhoum also claims that nuclear fission would have different energy spectrum, and beta decay happens differently. I tend to think that he might be wrong there. I think the energy spectrum and beta decay could be explained well without this new total energy.

The spectrum is an observed phenomenon, and does not change.

 

15 minutes ago, caracal said:

I don't know where de Broglie frequency is needed, if is it needed anywhere. If it is not needed, then different de Broglie frequency does not change any practical calculations.

The wavelength is used in e.g. diffraction. If you change the wavelength, then your predicted pattern disagrees with observation, and is therefore wrong.

Models must agree with experiment. 

 

[kenjimckinstry]

The kinetic energy of something makes it come out. If you look at the math one can note by the change in energy is done by many variances. Every change causes the change in energy. All the deltas are one. Everybody can do this math. So you have to put it together.