I was told that i could ask new question instead of continuing old question. I hope this new question has better clarity.

(I have completely rewrited this question what i asked 8 years ago. It needed details or clarity and was closed then. Despite minus votes there(-4) i still think there is something worth of thinking here.)

E.Bakhoum has published several papers such as http://arxiv.org/pdf/physics/0206061.pdf "Fundamental Disagreement of Wave Mechanics with Relativity" where he assumes different total energy for particle.

You don't need to go to these papers. I am not advocating his other possible claims there. I want to go only to his main suggestion.

The only thing he there suggest is a different total energy :

$$ E_{tot} = \gamma mv^2 $$

,where $ \gamma = \frac{1}{\sqrt{1-(\frac{v}{c})^2}} $

while keeping the kinetic energy exactly same:

$$ E_{kin} = (\gamma - 1)mc^2 $$

and also keeping the relativistic momentum exactly same:

$$ p(v) = \gamma mv $$

I think it his suggestion is just as simple as this.

Kinetic energy can be easily proven by looking how much energy is needed to put into particle that is accelerating from rest to velocity v. But Bakhoum argues that the total energy $E= \gamma mc^2 $ comes actually from simple assumption that particle has rest energy $E_0 = mc^2$ And he says it is an assumption that is not proven.

I think this is reasonable argument. (But What do you think?)

There is following mathematical identity:

$$ \gamma v^2 = c^2(\gamma - \frac{1}{\gamma}) $$

Therefore if you reduce from total energy $E_{tot} = \gamma mv^2$ the kinetic energy, the leftover term is:

$$ E_m = mc^2 (1 - \frac{1}{\gamma}) $$

This term becomes $ E_m = mc^2 $ only if particle velocity approaches c. And it becomes zero when relative velocity of particle goes to zero.

(This following is not a part of original question)

I know that many particle decays such as beta decay would point out that the second term would be $E_m = mc^2$ because there is following relation:

$$ \Delta E = \Delta m c^2 $$

That is i think proven experimentally.

Therefore i think this different kind of second term can be possible only if particles that are involving decay happen to accelerate high relative velocities that are close to c just a very short moment before the decay takes place. And after the decay they possibly decelerate again.

In particle collisions experiments the relative velocities are usually close to c so the second term is then close to $E_m \rightarrow mc^2$

What would be the reason why particles would accelerate just before decay, i don't know, could it be quantum fluctuations or Heisenberg's uncertainty principle and possibly strong force fields? I don't know.

Starting from this different total energy, the energy momentum relation becomes different:

Edit: i corrected this equation


$$\gamma = \sqrt{1+(\frac{p}{mc})^2} $$


$$ \gamma v^2 = c^2(\gamma - 1/\gamma) $$

$$ \rightarrow E = mc^2(\sqrt{1+(\frac{p}{mc})^2}- \sqrt{1-\frac{p^2}{m^2c^2+p^2}})$$

Edit: (alternatively this E could be 'energy that participate QM' instead of 'total energy')

Also the phase velocity of particle becomes v if i remember it correctly.

$$ v_{phase} = v $$

unfortunately i can't prove this later one at the moment. Now the phase velocity is not anymore superluminal.

How this works with photon:

The momentum for photon is $ p = (h/\lambda) $ (that is coming from Quantum Mechanics) but the energy-momentum relation is $ E(v) = pv $ and the following holds also between frequency and velocity: $ f(v) = (v/\lambda) $ . therefore i get energy for photon $\rightarrow E(c) = pc = hc/\lambda = hf $.

My question is, can this kind of total energy $E_{tot}=\gamma mc^2$ be possible taking also possibly into concideration also that in particle decays or interactions the involving particles could accelerate to high relative velocities when the total energy becomes $E_{tot}\rightarrow \gamma mc^2$ and the second term becomes $E_{m} \rightarrow mc^2$ and therefore following equation can hold:

$$\Delta E = \Delta m c^2 $$


  • $\begingroup$ Possible duplicate by OP: physics.stackexchange.com/q/221493/2451 $\endgroup$
    – Qmechanic
    Commented May 19 at 20:28
  • 1
    $\begingroup$ Non-mainstream physics is off-topic here. $\endgroup$
    – Ghoster
    Commented May 19 at 20:50
  • $\begingroup$ This question is correctly closed as being non-mainstream physics, but we can do a far better job at explaining why this is definitely never going to work. The issue is that a physics theory must have consistency in and of itself. The Special Theory of Relativity had already explained that the only acceptable energy-momentum relation is $$E=+\sqrt{(m_0c^2)^2+(pc)^2}\qquad\text{with}\qquad\frac{\vec v}c=\frac{c\vec p}E$$ with the result that if $\vec v\to0$ or $\vec p\to0$ (same thing), then $E\to m_0c^2$, a fact that is extremely well-established in experiments. So his suggestion is wrong. $\endgroup$ Commented May 20 at 4:57
  • $\begingroup$ I see. I have one question. But is it certain that this energy-momentum relation is applicable to relativistic quantum mechanics? The reason for different E-p relation for QM would be that not all of the total energy take apart in quantum mechanics (for some reason). But i guess different e-p relation would lead to wrong experimental results. $\endgroup$
    – Sami M
    Commented May 25 at 12:06

1 Answer 1


$E_{tot}=\gamma m v^2$ is clearly wrong. A series expansion in $v$ gives the low-velocity limit of $$E_{tot}=mv^2 + O(v^4)$$ This is clearly wrong because it does not have any rest mass (the $v^0$ term is $0$) and the kinetic energy has the wrong constant ($m$ instead of $m/2$).

  • $\begingroup$ Yes, in the low velocity limit $ E_{tot} = mv^2 $. But $ E_m = \frac{1}{2} mv^2 $ and $ E_{kin} = \frac{1}{2} mv^2$ also. This $ E_m $ is no longer "rest energy". There would not be any "rest energy", only some kind of energy that is $E_m = mc^2(1-1/\gamma) $ . that term is zero when velocity is zero. What kind of energy this $ E_m = \frac{1}{2} mv^2 $ is then i don't know. $\endgroup$
    – Sami M
    Commented May 19 at 21:40
  • 3
    $\begingroup$ @SamiM that is nonsense to try and split it out this way. With this wrong approach there is a term in the total energy proportional to $v^2$. When a force does work on an object the total energy increases by $mv^2$, which is twice as much as the work done. Trying to split it in half and call one half $E_{kin}$ and the other half $E_{m}$ doesn't fix anything. You still have twice as much energy increase as work done. The problem isn't the labeling, it is the wrong function. This idea just doesn't work. $\endgroup$
    – Dale
    Commented May 20 at 0:20
  • $\begingroup$ Yes that $E_m = \frac{1}{2} mv^2$ would appear to be "free energy" that would come into the bargain just by doing the half of the work to push particle into movement. because of energy conservation principle, the only reason for that to happen could be that there is some kind of energy reservoir that has maximum energy $ E=mc^2 $ where this energy is coming. Maybe that this another energy term does not change to kinetic energy or other forms of energy when the relative velocities are low? It just stands there. I suggested possibility that in particle reactions particles accelerate. $\endgroup$
    – Sami M
    Commented May 20 at 18:28
  • $\begingroup$ @SamiM that is just nonsense. It is not a mystery reservoir. It is kinetic energy, and it is the wrong value. The proposal is just wrong. I am done wasting my time telling you that it is wrong. Best of luck $\endgroup$
    – Dale
    Commented May 20 at 18:59
  • $\begingroup$ oh i forgot to say thank you. $\endgroup$
    – Sami M
    Commented May 20 at 19:30

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