# The Equivalence Principle And The Relativity Of Charges

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### #1 Dubbelosix

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Posted 31 January 2019 - 03:06 PM

If the rest energy of a particle changes approximately with the presence of charge, then charge becomes a correction factor within relativity.

Electromagnetic mass theories have lost it's flavor and little is written on the subject these days, but there was very convincing evidence, even supported by Feynman within his lectures, that the presence of a charge tends to make a particle a little bit heavier.

Poincare showed that if one models the electron as a hollow spherical shell with charge and a radius then the electrostatic

Indeed, from this Poincare further hypothesized there had to be stresses holding the electron together again the electrostatic repulsion of the charge on its surface! In models of electrons which are not essentially Faraday cages, this charge may be distributed evenly inside of the particle as the gravitational charge ~

$E_{charge} = mc^2 = \frac{Gm^2}{R} + \frac{e^2}{R}$

Where the rest mass contribution is associated to a gravitational charge and the last term associated to electric charge contribution also to the rest energy. The idea that the rest energy can be interpreted as a gravitational charge is related simply as:

$E_0R = mc^2R = E_{charge}R$

Where $E_{charge}$ has to be composed of two parts, as just shown.

The active $m_1$ and passive $m_2$ gravitational attraction can be given by two mass-charge terms:

$\frac{m_{act}m_{pass}}{r^2}$

And so in context of two gravitational charges we have:

$E_{charge} = \frac{\sqrt{G}m_1 \cdot \sqrt{G}m_2}{R} + \frac{e^2}{R}$

Max Planck also introduced a modified version of the mass energy equivalence, often attributed to Einstein, though a number of authors discovered this before him. The formula went like:

$E = mc^2 + PV$

where $P$ is a pressure term, including a volume, to express the relation between the mass and its latent energy (including thermodynamic contributions) within the body.

What Einstein did offer which was new, was a relativistic version of the mass-energy given as:

$E^2 = m^2c^4 + p^2c^2$

More often given as the square root, but we can write it this way:

$E = \sqrt{m^2c^4 +2 mc^2 \cdot pc + p^2c^2} = mc^2 + pc$

(we avoid writing this as a Hamiltonian, since it will not be the true total energy content)

It is completely true to talk about this equation as the relativistic counterpart to the rest energy $E =mc^2$, however, scientists tend to avoid using the word relativistic mass for a fast moving particle, as wiki explains quickly with a very nice quotation ~

Many contemporary authors such as Taylor and Wheeler avoid using the concept of relativistic mass altogether:

"The concept of "relativistic mass" is subject to misunderstanding. That's why we don't use it. First, it applies the name mass – belonging to the magnitude of a 4-vector – to a very different concept, the time component of a 4-vector. Second, it makes increase of energy of an object with velocity or momentum appear to be connected with some change in internal structure of the object. In reality, the increase of energy with velocity originates not in the object but in the geometric properties of spacetime itself."[7]

While space-time has the unbounded geometry of Minkowski-space, the velocity-space is bounded by c and has the geometry of hyperbolic geometry where relativistic-mass plays an analogous role to that of Newtonian-mass in the barycentric-coordinates of Euclidean geometry.[26] The connection of velocity to hyperbolic-geometry enables the 3-velocity-dependent relativistic-mass to be related to the 4-velocity Minkowski-formalism.[27]

It may kind of blow the mind, to think that what we seem to measure (an increase of energy) has not an origin within the structure of the particle, but instead arises from the classical concept of gravity and curvature. Wheeler of course when on to create geometrodynamics as an attempt to describe traditional phenomena in spacetime by geometry itself. If Wheeler and Taylor are correct, then Einstein's formula and the suggested corrections cannot be true since geometrodynamics explains properties like mass and charge, as not even charges or masses! It would arise from a more fundamental concept to spacetime itself and in a way, reminds me of how $P^{-4}$ propagators can be entirely described by their spacetime geometries (see Abdus Salam, Strong Gravity), apparently imitating the strong force we associate to quarks and gluons.

Let's not fret too much if Wheeler and Taylor are right - let's just settle happily that the mass-energy equivelance makes appropriate approximations within great accuracy. Keep in mind also, any objection to the true physics behind the reason for an increase of mass in a system, would also have to explain it's further success in the development of the Dirac equation which seems to have predicted the positron before experimentation was even aware of it. As you will be aware, the Dirac equation was an empirical result from the combination of quantum mechanics with that of Einstein's relativity.

The ability to have a pressure term for the internal dynamics of the system could hold importance for the concept of Poincare stresses. Casimir himself wasn't daft. He did believe the electron was a conducting sphere of radius and associated zero point energy as

$E = -C \frac{\hbar c}{2R}$

The term $\hbar c$ is a charge also and is dimensionally equivalent to $Gm^2$, where C is a dimensionless constant that if positive implied an inward force. That inward force balances the outward Coulomb force (as Poincare stress) when the magnitudes of the corresponding energies are equal ~

$\frac{e^2}{2r} = C \frac{\hbar c}{2r}$

In fact, you could argue for a case in which we identify $\hbar c$ associated to an electromagnetic part, in which case the total charge $Q$ is argued:

$Q = Gm^2 + \hbar c$

A detailed  calculation  of  C by Boyer, however, shows that Casimir’s intuitive approach was off the mark: The constant is negative, equal to about −0.09; that is, the stress on a conducting sphere tends to make it expand. Boyer’s result remains of interest because it highlights the geometry dependence of the Casimir force, a subject that has received considerable attention. Some papers in academia is of particular interest here, in that it shows the calculations of the Casimir electron in a new way to avoid the errors.

Point particles are a problem for physics and definitely problematic to our understanding of them for a number of reasons, but the biggest problem arises from classical physics, which says singularities appear when we make the electron radius go to zero.

References:

http://vixra.org/pdf/0712.0003v1.pdf

http://earthtech.org...ir_electron.pdf

http://www.casimir-n...f/Lamoreaux.pdf

Edited by Dubbelosix, 01 February 2019 - 09:25 AM.

### #2 exchemist

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Posted 31 January 2019 - 03:45 PM

E = mc² +pc is wrong:

E= √(m²c⁴ +p²c²)  =>  E² = m²c⁴ +p²c²

Whereas  (mc² +pc)² = m²c⁴ +2mpc³ + p²c² = E² + 2mpc³

Edited by exchemist, 31 January 2019 - 03:53 PM.

### #3 exchemist

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Posted 01 February 2019 - 02:45 AM

The correct form is  E= (pc)2 + (mc2)is it not

Exactly, i.e. as in the 2nd line of my post.

In the case where one or other term is zero, i.e. p=0 (for an object at rest relative to the observer), or m=0 (for a photon, for instance)  the expression reduces to E=mc², or E=pc, respectively.

In the latter case, applying de Broglie's relation, p=h/λ, this becomes E=hc/λ, and since c=νλ, this is equivalent to E=hν, which is Planck's relation. Rather neat, I always think.

But if you have an entity with both mass and momentum, then you need the full expression in terms of E².

Edited by exchemist, 01 February 2019 - 03:08 AM.

### #4 Dubbelosix

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Posted 01 February 2019 - 02:58 AM

So long as neither side of the equation is negative, then E = mc^2 + pc is acceptable. Squaring the formula leads to the ordinary

$E^2 = m^2c^4 + p^2c^2$

It's only a square difference away.

### #5 Dubbelosix

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Posted 01 February 2019 - 03:00 AM

The correct form is  E= (pc)2 + (mc2)is it not

Certainly the form we encounter the most, it doesn't make it any more ''correct'' as far as I am aware.

### #6 Dubbelosix

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Posted 01 February 2019 - 03:01 AM

A good example why it is fine, is just take the statement:

$\sqrt{m^2c^4 + p^2c^2} = mc^2 + pc$

and ask yourself, is this true? Mathematically, it is absolutely fine. In fact the left implies the right.

### #7 OceanBreeze

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Posted 01 February 2019 - 03:15 AM

A good example why it is fine, is just take the statement:

$\sqrt{m^2c^4 + p^2c^2} = mc^2 + pc$

and ask yourself, is this true? Mathematically, it is absolutely fine. In fact the left implies the right.

are you serious?

${ E }^{ 2 }={ (m{ c }^{ 2 }) }^{ 2 }+{ (pc) }^{ 2 }$

So then

$E=\sqrt { { m }^{ 2 }{ c }^{ 4 }+{ p }^{ 2 }{ c }^{ 2 } }$

Note that:

$\sqrt { { m }^{ 2 }{ c }^{ 4 }+{ p }^{ 2 }{ c }^{ 2 } } \neq \quad m{ c }^{ 2 }+pc\quad$

Because the laws of exponents do not allow for a power to be distributed across a sum or a difference.

The binomial expansion needs to be used in those cases.

Therefore:

$E\quad \neq \quad m{ c }^{ 2 }+pc$

### #8 exchemist

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Posted 01 February 2019 - 03:16 AM

Thus illustrating that Dubblesox cannot do GCSE level algebra.......

√(a² + b²)  is not a + b.

Example: a = 2, b=3.

a + b = 5.

√(a² + b²) =  √(4 + 9) = √13

Last time I checked, 5 was √25

### #9 Dubbelosix

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Posted 01 February 2019 - 03:21 AM

### #10 Dubbelosix

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Posted 01 February 2019 - 03:39 AM

Thus illustrating that Dubblesox cannot do GCSE level algebra.......

√(a² + b²)  is not a + b.

Example: a = 2, b=3.

a + b = 5.

√(a² + b²) =  √(4 + 9) = √13

Last time I checked, 5 was √25

I had to refresh my memory on this. It is true that:

$m^2c^4 + p^2c^2$

$(mc^2 + pc)(mc^2 + pc)$

$m^2c^4 + 2mc^2 \cdot pc + p^2c^2$

and so

$\sqrt{m^2c^4 + 2mc^2 \cdot pc + p^2c^2} = mc^2 + pc$

or

$\sqrt{a^2 + 2ab + b^2} = a + b$

### #11 exchemist

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Posted 01 February 2019 - 03:42 AM

I had to refresh my memory on this. It is true that:

$m^2c^4 + p^2c^2$

$(mc^2 + pc)(mc^2 + pc)$

$m^2c^4 + 2mc^2 \cdot pc + p^2c^2$

and so

$\sqrt{m^2c^4 + 2mc^2 \cdot pc + p^2c^2} = mc^2 + pc$

or

$\sqrt{a^2 + 2ab + b^2} = a + b$

Yes, that is what I wrote in post 2.

I have got my 15 yr old son past this within the last 3 months, for his impending GCSE.

### #12 Dubbelosix

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Posted 01 February 2019 - 03:49 AM

Yes, that is what I wrote in post 2.

I have got my 15 yr old son past this within the last 3 months, for his impending GCSE.

The attitude you can drop. Exchemist, I don't know what kind of world you live in, but people do make mistakes. I've seen Susskind get his units wrong on the board, mix up things that are not true. I don't care how accomplished your son is, you reveling in someone's genuine error is pretty ugly. But you are an ugly person anyway. It can't be about correction for the sake of correction, but for the sake of belittling. You'd make a lousy teacher.

### #13 exchemist

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Posted 01 February 2019 - 04:04 AM

The attitude you can drop. Exchemist, I don't know what kind of world you live in, but people do make mistakes. I've seen Susskind get his units wrong on the board, mix up things that are not true. I don't care how accomplished your son is, you reveling in someone's genuine error is pretty ugly. But you are an ugly person anyway. It can't be about correction for the sake of correction, but for the sake of belittling. You'd make a lousy teacher.

But I got you to see reason here, did I not, after you had initially insisted you were right?

My point, which I will continue to illustrate for readers from time to time, is that you, Reiku, have no business writing grandiose mathematical screeds when your basic algebra is at this level.

Everyone make mistakes of course. But, if you knew your algebra, you would have realised your error immediately it was pointed out. Instead, you foolishly insisted, Trump-like, that you were right, until a simple enough example was given to make it blindingly obvious to everyone that you were wrong.  Susskind would never have done that.

### #14 Dubbelosix

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Posted 01 February 2019 - 04:26 AM

Yes, you did make me see reason, and it took very little you will have noticed. When you don't use algebra a lot, you do tend to forget things over time. I think what annoys you though, judging you by the way you pounced on the error, is that you saw an opportunity to attack, and you are still following the attack in the last post just now, saying I have no right to do this or that, when basic errors can be made.

Well, if you are such a genius yourself, why don't you find errors in the ''grandiose'' posts of mine? Secondly, if I was such a dunce, how was I able to recognize the mistake even when pointed out? You yourself said I gave in pretty easy, ... that's because I too notice mistakes when they are made, sometimes by myself or another has to point it out.

You need a reality check on how you approach things, or how you speak to people. You saw an opportunity to desperately jump on a mistake and you come across as a bit of a d_ick in the way you do it. I have pointed out mistakes of yours in the past, but I only got personal when you could not concede your own mistakes.

Edited by Dubbelosix, 01 February 2019 - 04:36 AM.

### #15 Dubbelosix

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Posted 01 February 2019 - 04:29 AM

With the pressure term we have:

$E = mc^2 + pc + pV = \sqrt{m^2c^4 + p^2c^2 + P^2V^2 + 2mc^2 \cdot pc + 2pc \cdot PV + 2mc^2 \cdot PV}$

And so I define the total charge of the system related to the rest mass as

$E_0 = \frac{Gm^2}{R} + \frac{\hbar c}{R} = mc^2$

and so the proposed relativistic case would be:

$E = \sqrt{\frac{G^2m^4}{R^2} + \frac{\hbar^2 c^2}{R^2} + p^2c^2 }$

But a pressure correction as noted before took the form on the rest mass as:

$E = mc^2 + PV$

If we take the pressure as the squared form:

$E = \sqrt{(mc^2 + pc + pV)^2} = \sqrt{\frac{G^2m^4}{R^2} + \frac{\hbar^2 c^2}{R^2} + p^2c^2 + P^2V^2}$

( as in similar form to Einstein's ordinary)

$E = \sqrt{m^2c^4 + p^2c^2}$

Then its expansion is

$E = (mc^2 + pc + pV) = \sqrt{m^2c^4 + p^2c^2 + P^2V^2 + 2mc^2 \cdot pc + 2pc \cdot PV + 2mc^2 \cdot PV}$

But this is a really messy equation, to know the physics I propose all we need to know about is:

$E = \sqrt{(mc^2 + pc + pV)^2} = \sqrt{\frac{G^2m^4}{R^2} + \frac{\hbar^2 c^2}{R^2} + p^2c^2 + P^2V^2}$

Feynman has said himself in his lectures, that there is strong evidence to suggest that electromagnetic effects actually do contribute to an over all mass; comparing the uncharged particles with charged particles of almost same magnitudes, shows that the presence of a charge seems to make a particle heavier! The only two he compared that did not share this property was a tight squeeze between the proton and neutron values: this probably has to do with the fact that while the net charge is zero for a neutron, it does in fact have a charge distribution which is not zero - It's only that its net charge is zero and so calculating the electromagnetic mass is a fair bit more complicated.

He also makes a point there are other electromagnetic energies to take into account, such as the non*zero magnetic moment. Funnily enough, as Feynman has also shown, the neutron sometimes looks like a proton with a negative *meson charge cloud surrounding it" As far as we can tell, the muon differs only very slightly from the electron. It acts very similar to an electron and because of this it has no nuclear forces and interacts with the electromagnetic field. Feynman went as far to say that the muon is actually just a heavier electron - Feyman has rightly noted that whoever can explain the mass of an electron, will have to then deal with the muon! the reason why is because their behaviour is indistinguishable and so the mass, as Feynman stated, ''ought to come out the same!''

So here we can deduce some things from the equation: If two particles of mass ought to come out the same then their magnitude will be made of a reduced rest mass term:

$E = \sqrt{(mc^2 + pc + pV)^2} = \sqrt{\frac{G^2m^4}{R^2} + p^2c^2 + P^2V^2}$

Since the simple rest mass associated to rest gravitational charge is the same:

$E_0 = \frac{Gm^2}{R}$

But if there is the presence of a charge, the rest mass gets a correction:

$E_0 = \frac{Gm^2}{R} + \frac{e^2}{R}$

I do not see this correction as a trivial matter, taking into respect the contributions of electromagnetic energy to the system. The pressure describing latent heat and internal thermodynamics is why the pressure term also becomes relevant for the rest system:

$E_0 = \frac{Gm^2}{R} + \frac{e^2}{R} + PV$

But we speculated it's squared form not long ago.

Edited by Dubbelosix, 01 February 2019 - 04:40 AM.

### #16 Dubbelosix

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Posted 01 February 2019 - 04:31 AM

And by the way exchemist, I totally agree it is a simple error... except even top scientists make simple errors from time to time. Maybe you shouldn't gauge the nature of an error but by the persons ability to present what he can. I remember reading about David Hilbert... you know, that brilliant mathematician? Turns out a lot of his work, recognized after death, contained many errors, some even simple. Einstein too, had to have adjustments to his papers because they contained errors. Just some perspective.

### #17 Dubbelosix

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Posted 01 February 2019 - 04:54 AM

The interesting thing to note, is that all particles in nature that have mass, contain a charge - except for one - the neutrino. It has according to models, a non-zero but very small mass with which we assume the electric charge is zero. I still think it is possible to understand where you have mass you must have a charge and in the case of a neutrino, it would show that the magnitude of the charge has to be proportional to its mass content - in other words, if the neutrino has a vanishingly small mass, then maybe its charge is equally weak enough not to couple to other charged objects in a general sense.