Furthermore, this calls into question a key assumption in Einstein’s framework: that energy gravitates. While his field equations make this a postulate by embedding energy, pressure, and stress in the energy-momentum tensor, this step is not derived from observation. Rather, it was accepted by construction. Newtonian gravity, in contrast, only allows mass to gravitate. The consequence of Einstein’s assumption leads to conceptual challenges, including the infinite regress of gravitational self-energy. The results presented here ...consistent and emerging from fundamental constants ...suggest that gravity is better understood as arising from time gradients, not from energy itself.

Energy gravitation, therefore, is questionable and may be an artifact of an incomplete theoretical construction.

Berndt Barkholz "Energy gravitation, therefore, is questionable and may be an artifact of an incomplete theoretical construction."

No, energy gravitation is needed to maintain consistence. It just reflects the relativistic mass formula m=mₒ/√(1-v²/c²). In fact it completes the theoretical construction.

Wolfgang Konle

So you claim... Wolfgang & Co, but m=mₒ/√(1-v²/c²) is nonsense ! We can only hope that you some day find out...

...and you said it "construction", but it should be a derivation !

Wolfgang Konle

According to Einstein’s relativistic framework, a mass approaching the speed of light encounters an ever-increasing momentum due to the Lorentz factor:

1. p = mv/√(1-v²/c²)

As v→c and √(1-v²/c²)→0 the momentum is said to approach infinity. This is presented as the reason why mass cannot reach the speed of light. At the same time, Einstein states that mass and energy are equivalent, via the well-known equation:

2. E=mc2

If we follow this logic honestly, then at or near the speed of light, mass "becomes" energy. In other words, if mass has transformed into pure energy, no mass remains.

This raises a fundamental contradiction:

How can momentum approach infinity if, by the same theory, the object at those extreme conditions has become energy, not mass?

Momentum, by its classical definition, requires mass:

3. p = mv

Yet, in Einstein’s model, photons (which are pure energy with no rest mass) are assigned a momentum:

4. p=E/c

This is not derived from first principles but rather imposed. It is a convenient mathematical construction, not a physical explanation. The question remains unanswered:

Why should energy without mass inherently carry momentum?

In reality, this inconsistency reveals that the relativistic treatment of mass, energy, and momentum are a carefully balanced patchwork of equations. The Lorentz factor was inserted to protect the framework from collapse, but it is not a natural outcome of a deeper principle.

If we instead consider gravity and force as the result of time gradients (as I argue) such contradictions do not arise. The infinite momentum problem disappears. Mass, energy, and momentum are not arbitrary mathematical conveniences but manifestations of temporal structure.

I ask you to question these inherited assumptions and consider whether the elegance of nature truly requires infinite quantities and imposed relationships, or whether these are signs that the theory, however celebrated, is an incomplete construct instead of a derivation.

Dear Berndt Barkholz "So you claim... Wolfgang & Co, but m=mₒ/√(1-v²/c²) is nonsense !"

No, it is not nonsense. We easily can prove it with elementary math and physics:

We begin with momentum=force*time, P=F*t, infinitesimal dP=F*dt, or F=dP/dt. Followed by

energy=force*distantance,E=F*s, infinitesimal dE=F*ds, F=dE/ds. ds is a Vector therefore we have dE/ds=(∂E/∂sx, ∂E/∂sy, ∂E/∂sz).

Physics must be consistent, therefore the force F in both equations must be the same. F=dP/dt=dE/ds.

We now apply this force equation to a particle or to any object with restmass mₒ and consider m=mₒ/√(1-v²/c²).

This leads to d/dt(mv)=d/ds(mc²). We insert mₒ/√(1-v²/c²) and apply the chain rule of differentiation:

(1) d/ds=d/dv*dv/ds, dv/ds=dv/dt*dt/ds, dt/ds=1/(ds/dt)=1/v

(2) d/dt=d/dv*dv/dt

(2) leads to (3) dP/dt=dP/dv*dv/dt

(1) leads to (4) dE/ds=dE/dv*dv/dt*(1/v)

We can cancel dv/dt in (3)=(4) an bring 1/v on the other side and get

(5) v*dP/dv=dE/dv

Now we consider the left side of (5) with the product rule and get

(6) v*dP/dv=v*d/dv(mv)=v²(dm/dv)+vmdv/dv=v²dm/dv +vm, and

(7) dE/dv=d/dv(mc²)=c²dm/dv

Now we again consider (6)=(7):

v²dm/dv+vm=c²dm/dv which leads us to dm/dv*(c²-v²)=vm. Or

(8) dm/dv=vm/(c²-v²)

Now we consider d/dv(mₒ/√(1-v²/c²)) with the chain rule, with u=(1-v²/c²), we get

d/dv(u(-1/2))=-1/2u(-3/2)*du/dv. With du/dv=-2v/c², we then get

(9) dm/dv= mₒv/c²u(-3/2). With m= mₒ u(-1/2) we get

(10) dm/dv=vm/(c²-v²)

This is exactly the same as equation (8) and proves that d/dt(mv)=d/ds(mc²) is fulfilled with m=mₒ/√(1-v²/c²).

The algebraic derivation (1)->(10) is absolutely clear and comprehensible and definitively proves m=mₒ/√(1-v²/c²) directly with absolutely basic physics and mathematics.

It is strange, that this basic and straight forward proof does not appear at uncountable places in the literature about special relativity, though it would be urgently important to calm down all the people who are having rightless doubts.

Wolfgang Konle

Since existing mass cannot evaporate the crew of a high speed spaceship including the spaceship would after a lifetime become quite heavy... don't you think ? You need to reconsider your use of the Lorentz factor or declare that mass can evaporate ! Further discussion won't make me change my mind: m=mₒ/√(1-v²/c²) is nonsense !!

Berndt Barkholz

Ignoring my proof of m=mₒ/√(1-v²/c²) is busting your head in the sand.

Wolfgang Konle

No... you need to explain what happens to the mass... where does it come from and where does it go... using math to hide black magic is not science...

Wolfgang Konle

Ich komme also zu dem Schluss: Wenn sich die Masse mit v=x Meter/Sekunde bewegt, kommt Masse an, tadaa … und bei v=0 halten wir an, um zu pinkeln, und die Masse wird unsichtbar … puff und weg … in der Tat interessante Physik!

Berndt Barkholz : Why should energy without mass inherently carry momentum?

This is known since 1894 when Poincaré derived it from the EM force and formulated the famous law dm=E/c2. The EM momentum you can find in every EM theory textbook.

But what people usually do not learn is to understand that this law is an equivalence relation between charge and B field and based on the Biot Savart force (often also called Lorenz force).

Thus E=mc2 is is an equivalence relation too, just between Energy and photon mass. Of course m=mₒ/√(1-v²/c²) does not work for real mass and

Wolfgang Konle citing existing SRT garbage is one of the family that never understood the basics.

These fringe SRT claims came into live because electron's did cheat the scientists, that did not know the Biot Savart dependency and the close (99.8%) photon nature of the electron....

The real (exact) proton structure you can find in : Data The proton and its resonances RII.pdf

Jürg Wyttenbach "Thus E=mc2 is is an equivalence relation too, just between Energy and photon mass. Of course m=mₒ/√(1-v²/c²) does not work for real mass and Wolfgang Konle citing existing SRT garbage is one of the family that never understood the basics."

But you are understanding the basics?

If you are understanding the basics, then you should also understand the following proof, I had presented to Berndt Barkholz , which proves that m=mₒ/√(1-v²/c²) applies to any real mass with a restmass.

Photons have resmass mₒ=0, they are not subject of this relation.

We begin this proof with momentum=force*time, P=F*t, infinitesimal dP=F*dt, or F=dP/dt. Followed by

energy=force*distantance,E=F*s, infinitesimal dE=F*ds, F=dE/ds. ds is a Vector therefore we have dE/ds=(∂E/∂sx, ∂E/∂sy, ∂E/∂sz).

Physics must be consistent, therefore the force F in both equations must be the same. F=dP/dt=dE/ds.

We now apply this force equation to a particle or to any object with restmass mₒ and consider m=mₒ/√(1-v²/c²).

This leads to d/dt(mv)=d/ds(mc²). We insert mₒ/√(1-v²/c²) and apply the chain rule of differentiation:

(1) d/ds=d/dv*dv/ds, dv/ds=dv/dt*dt/ds, dt/ds=1/(ds/dt)=1/v

(2) d/dt=d/dv*dv/dt

(2) leads to (3) dP/dt=dP/dv*dv/dt

(1) leads to (4) dE/ds=dE/dv*dv/dt*(1/v)

We can cancel dv/dt in (3)=(4) an bring 1/v on the other side and get

(5) v*dP/dv=dE/dv

Now we consider the left side of (5) with the product rule and get

(6) v*dP/dv=v*d/dv(mv)=v²(dm/dv)+vmdv/dv=v²dm/dv +vm, and

(7) dE/dv=d/dv(mc²)=c²dm/dv

Now we again consider (6)=(7):

v²dm/dv+vm=c²dm/dv which leads us to dm/dv*(c²-v²)=vm. Or

(8) dm/dv=vm/(c²-v²)

Now we consider d/dv(mₒ/√(1-v²/c²)) with the chain rule, with u=(1-v²/c²), we get

d/dv(u(-1/2))=-1/2u(-3/2)*du/dv. With du/dv=-2v/c², we then get

(9) dm/dv= mₒv/c²u(-3/2). With m= mₒ u(-1/2) we get

(10) dm/dv=vm/(c²-v²)

This is exactly the same as equation (8) and proves that d/dt(mv)=d/ds(mc²) is fulfilled with m=mₒ/√(1-v²/c²).

The algebraic derivation (1)->(10) is absolutely clear and comprehensible and definitively proves m=mₒ/√(1-v²/c²) directly with absolutely basic physics and mathematics.

Wolfgang Konle : Photons have restmass mₒ=0, they are not subject of this relation.

This is exactly the base of your ein Einstein's fallacy!

Nobody can convert real mass "m" into energy according E=mc2!! Only E=dmc2 is/has been proven in experiments. Further you cannot make a physical derivative for gamma (v/c)2 as this expression has no physical units!!!!!!!!! Thus (v/c) is neither a time like nor a space like quantity!!

This is and was just "Einstein garbage math". You also permanently forget that m (mass) can rotate and hence can have kinetic energy with out any vector velocity. This again shows that your derivation is just garbage!

Wolfgang Konle

This wall of text might look impressive to the untrained eye, but once again, you’re trying to “derive” a relativistic formula by assuming the very equation you claim to prove.

That’s not physics, that’s circular reasoning.

In theoretical physics, a derivation means starting from a set of independent postulates or assumptions, and logically deducing consequences.

If your proof starts by inserting m = m₀ / √(1 - v² / c²)​​, and ends up “proving” that same formula, you haven’t shown anything, you’ve just looped.

And more importantly:

Relativistic mass is not something that’s “deduced” from Newtonian mechanics with chain rules and algebra. It replaces Newtonian assumptions, based on new postulates (like the invariance of the speed of light and the relativity of simultaneity), which lead to tested and verified predictions.

You're not defending physics here. You're misusing equations outside their conceptual context, and presenting that as "proof." It doesn't work that way.

You should seriously consider stopping your habit of correcting others and instead focus on catching up with modern physics yourself. Because the way you're doing things now, you're only polluting the conversation and spreading confusion

Berndt Barkholz "...you need to explain what happens to the mass... where does it come from and where does it go... using math to hide black magic is not science..."

dE/ds=dP/dt explicitly describes and explains what happens to the entity, we are calling "mass". Mass can have kinetic energy and momentum and its rest mass. But this is all what mass can have. Or do you know something else?

The other point is that only the rest mass is coming from nowhere and it is going to nowhere. Restmass is constant and it is simply given as it is. There is absolutely no black magic involved. The math, I am using in the proof is basic algebra, which is easily understandable, and does not hide anything.

Einstein now has discovered that kinetic energy has a mass eqivalent. This mass equivalent of the kinetic energy is expressed in Ekin=(m-mₒ)/c². And just this mass equivalent makes the relation between energy and momentum consistent as I am showing in my proof.

You also can verify that Ekin=(m-mₒ)/c² is just equivalent to Ekin=mv²/2 in the limit of small velocities. (We know that 1/√(1-v²/c²) ≈ 1/(1-v²/(2c²)) for v

Wolfgang Konle

Repeating equations doesn’t make them physical explanations. Berndt asked you a clear and simple question: where does the mass come from, and where does it go?

You responded with algebra, but no physics.

You claim there’s “no black magic,” but your entire reasoning relies on inserting mathematical identities without ever addressing the real conceptual question. Just saying “rest mass is constant and just is” is not an answer, it’s evasion.

This is exactly why you should stop lecturing others and start reflecting more critically on the foundations of your own arguments. Equations are only meaningful when grounded in physics, not when used to pretend that you’ve solved something you haven’t even addressed.

Essam Allou "If your proof starts by inserting m = m₀ / √(1 - v² / c²)​​, and ends up “proving” that same formula, you haven’t shown anything, you’ve just looped."

My proof is not starting with the insertion of m = m₀ / √(1 - v² / c²). It starts from dE=F*ds and dP=F*dt and the consistency requirement that the force in both equations is the same force. This leads to dE/ds=dP/dt, the real start of my proof.

With your primitive insinuation of circularity in that proof you only have demonstrated a fatal ignorance.

Wolfgang Konle

Before rushing into new “proofs” and corrections addressed to others, perhaps take a moment to address the serious contradictions and misunderstandings we uncovered yesterday, the ones you have quietly chosen to ignore.

You still haven’t responded to:

• Your misuse of “SR applies everywhere” and the false generalization it implies.
• Your failure to clarify how your so-called gravitational equation fits within any coherent spacetime geometry.
• Your repeated avoidance of deriving your core formula from any Lagrangian, variational principle, or consistent field framework.
• Your blatant confusion between local applicability and global validity in GR.

And now you add a “proof” that is, in essence, a disguised loop.

Let’s be clear:

If you begin by inserting the relativistic mass formula m = m₀ / √(1 - v² / c²)​​ into both sides of your derivation and then arrive at that same formula, you have not derived anything, you've just confirmed an assumption you made at the start.

That is, by definition, circular reasoning.

Claiming it starts from “dE=F⋅ds ” and “dP=F⋅dt” doesn't change that fact if the actual crux of your derivation still requires assuming the relativistic mass formula before “proving” it.

And calling this “fatal ignorance” only reveals your discomfort with being corrected on basics, something that happens often when one refuses to listen.

Please. One thing at a time. Start by addressing the pile of unresolved fallacies you’ve left behind in the other threads or eventually, others will expose them for you.

Essam Allou

Your whataboutism is funny:

"Before rushing into new “proofs” and corrections addressed to others, perhaps take a moment to address the serious contradictions and misunderstandings we uncovered yesterday, the ones you have quietly chosen to ignore."

But this is not the right place to discuss your ridiculous and irrelevant accusations.

Wolfgang Konle

What you call "whataboutism" is actually accountability.

You can’t pretend to build serious physics by accumulating half-baked claims in every thread, while refusing to clean up the mess you've left behind in your own.

When multiple foundational contradictions have been exposed, including your failure to derive or validate your key equations, it is not “irrelevant”, it’s the core of scientific integrity.

So no, it’s not just about yesterday. It’s about everything you still haven’t answered. All while polluting other discussions with nothing of actual value.

We’ll let the readers decide which one of us is running from the fact

Essam Allou

You are claiming that deriving the relativistic mass formula from dE/ds=dP/dt is circular because this relation also could be derived from the relativistic mass formula.

But energy E is defined by "energy equals force times distance", E=F*s and momentum P is defined by "momentum equals force times time", P=F*t.

Without this definition the entities energy and momentum would not exist.

But a not existing entity cannot appear in any relativistic formula like E=mc².

With your claim about circularity in the derivation of the relativistic mass formula you are reverting causality in all the relations, which constitute theoretical physics. This reversion is alarming.

Wolfgang Konle

You’ve already left an entire trail of unanswered contradictions, many of which, as this very thread shows, stem from your persistent misunderstanding of the most basic meaning of the word relativity.

This is not a rhetorical insult; it’s a demonstrable fact, visible to anyone reading your exchanges across threads.

So it is genuinely baffling to see you return here, once again posturing as if you’re in a position to teach basic physics, when your previous statements have already revealed you don't even grasp the role of reference frames, covariance, or the core motivations behind modern frameworks like GR.

That alone should invite humility, not arrogance.

Now let’s briefly address your latest confusion:

You say: “E = F·s and P = F·t are the definitions of energy and momentum. Therefore, deriving anything from them is valid and non-circular.”

But here is where you still miss the point:

You can’t derive a velocity-dependent expression for mass using only those definitions.

Why?

Because F = dE/ds and F = dP/dt are kinematic identities: they describe how force relates to changes in energy and momentum, not the functional form of mass as a function of velocity.

To go from: F = dP/dt = d/dt [m(v) · v]

to a result like: m(v) = m₀ / √(1 - v² / c²)

you must assume either:

• A relativistic energy-momentum relation
• A Lagrangian or Hamiltonian framework
• Lorentz invariance

None of which appear in your derivation.

So yes, when you start your “proof” with expressions involving m(v), and then later say “see, this confirms the relativistic mass formula,” you are implicitly assuming what you’re supposed to derive. That’s textbook circular reasoning.

Your reply once again avoids the real issue: you keep trying to rebuild modern physics from intuitive Newtonian-style formulas, refusing to engage with the mathematical structures that actually govern relativistic systems.

And every time you’re shown the limits of your reasoning, you resort to blaming “theoretical physics” for being too abstract.

But the truth is simpler: It’s not the theory that’s upside down, Wolfgang. It’s your understanding of it.

We tried to help clarify that for your own sake. But if you insist on lecturing others while ignoring the most basic conceptual corrections, we will continue to point out the disconnect between your confidence and your comprehension.

Essam Allou "So yes, when you start your “proof” with expressions involving m(v), and then later say “see, this confirms the relativistic mass formula,” you are implicitly assuming what you’re supposed to derive. That’s textbook circular reasoning."

This statement is blatantly wrong and without any logic.

The complete derivation of the relativistic mass (not only a consistency check) is as follows:

We begin with momentum=force*time, P=F*t, infinitesimal dP=F*dt, or F=dP/dt. Followed by energy=force*distantance,E=F*s, infinitesimal dE=F*ds, F=dE/ds. ds is a Vector therefore dE/ds=(∂E/∂sx, ∂E/∂sy, ∂E/∂sz).

Physics must be consistent, therefore the force F in both equations must be the same. F=dP/dt=dE/ds.

We now apply this force equation to an object with mass m and apply E=mc² and P=mv.

This leads to d/dt(mv)=d/ds(mc²). We apply the chain rule of differentiation:

(1) d/ds=d/dv*dv/ds, dv/ds=dv/dt*dt/ds, dt/ds=1/(ds/dt)=1/v

(2) d/dt=d/dv*dv/dt

(2) leads to (3) dP/dt=dP/dv*dv/dt

(1) leads to (4) dE/ds=dE/dv*dv/dt*(1/v)

We can cancel dv/dt in (3)=(4) an bring 1/v on the other side and get

(5) v*dP/dv=dE/dv

Now we consider the left side of (5) with the product rule and get

(6) v*dP/dv=v*d/dv(mv)=v²(dm/dv)+vmdv/dv=v²dm/dv +vm and

(7) dE/dv=d/dv(mc²)=c²dm/dv

Now we consider (6)=(7):

v²dm/dv+vm=c²dm/dv which leads us to dm/dv*(c²-v²)=vm. Or

(8) dm/dv=vm/(c²-v²)=>m‘/m=vm/(c²-v²) in mathematical notation: y'/y=x/(c²-x²)

We now can formally solve the equation y'/y=x/(c²-x²). This is a differential equation of type Bernoulli. Its solution is

(9) ln(y)=-ln((c²-x²)/2) which leads to

(10) y=C/√(c² - v²).

Selecting the constant C as C=m₀c then formally leads to

(11) m=m₀/√(1-v²/c²).

With the solution of the differential equation I have derived m=m₀/√(1-v²/c²) from dE/ds=dP/dt and E=mc².

• Using E=mc² does not imply what m exactly is. This finally is derived with m=m₀/√(1-v²/c²).
• The proof initially states that m=m(s,t), which is a consequence of the differential equation. m=m(v) then has been derived mathematically as a consequence of the chain rule in calculus.
• The entire logic is based on mathematical rules and physical laws.

This is exactly how deductions in physics work.

However your not comprehensible and false conclusions only show that logic is not your friend.

Wolfgang Konle

I’ve already addressed your latest “contribution” in the main thread, which, as usual, isn’t new in any meaningful way. But since you insist on repeating yourself here as well, I’ll repeat my response too. Still, for your own sake, you might want to stop posting, all you're doing is giving more people the chance to read the same nonsense and watch your credibility sink further.

Or maybe you’ve already realized there’s nothing left to lose. 😊

You say:

“Using E = mc² does not imply what m exactly is. This finally is derived with m = m₀ / √(1 - v² / c²).”

Let’s stop right here. This is the core of your misunderstanding, and the reason your argument has failed since the very first post.

You assume E = m·c², a relation that is not Newtonian, but intrinsically relativistic, and then use that together with P = m·v to extract a velocity dependence for m.

But here’s the catch: the functional form of E = m·c² already encodes the assumption that energy and mass are related via relativistic invariance.

So when you manipulate that equation to recover m(v), you’re not deriving anything, you’re just unfolding what was already built in.

That’s why every physicist with proper training will tell you: your “derivation” is just a consistency loop, not a deduction from first principles.

And repeating it over and over again, without ever addressing this central flaw, doesn’t make it stronger, it just makes it louder and more embarrassing for anyone watching.

If you still can’t grasp that assuming a relativistic postulate and “deriving” a relativistic result is circular, then this isn’t a physics discussion anymore , it’s just an exercise in self-hypnosis.

The sad part is, we’ve tried to explain this multiple times, clearly and patiently. But since you’re not here to understand, only to preach, there’s no point in engaging further until you start showing some intellectual honesty.

With each reply, you're not advancing the discussion, you're merely chiseling away what's left of your credibility. You may not see it, but to any serious reader, your stubbornness isn’t resilience, it’s the slow self-inflicted collapse of whatever authority you thought you had.

Essam Allou "But here’s the catch: the functional form of E = m·c² already encodes the assumption that energy and mass are related via relativistic invariance.

So when you manipulate that equation to recover m(v), you’re not deriving anything, you’re just unfolding what was already built in."

Your so called catch is fundamentally wrong and without any logic because the relation of energy and mass is defined with the relation between energy and momentum and not via relativistic invariance.

Your fundamental error is misinterpreting the basic character of the relation between energy and momentum. Obviously this error fundamentally compromises your general view on causal relations in physics.

In fact it is the force equation which basically defines the relation between energy and mass. The force equation relates energy and momentum. But the entity "momentum" is directly proportional to mass. Therefore the force equation is the root cause for the proportionality between mass and energy. This proportionality relation does not at all depend on a relativistic invariance.

Because the momentum is proportional to the mass, the force relation explicitly also requires that the energy as well must be proportional to the mass. The only open question then is why the proportionality constant just is c². But it is obvious that the proportionality constant must be a velocity squared.

Obviously you have missed or deliberately ignored that in the whole derivation of the relativistic mass the actual value of c does not explicitly occur or does not explicitly matter. According to that derivation c only is the constant which occurs in E=mc². In this derivation of m=m₀/√(1-v²/c²) c can be any velocity, just the constant, which appears in E=mc².

Again, the proportionality of energy and mass in dE/ds is entirely a consequence of the force equation because the momentum in dP/dt by definition already is proportional to the mass.

Not even the requirement that for velocities v

Wolfgang Konle

Why you're wrong, point by point:

1. E = mc² is a direct consequence of relativistic invariance.

It is rigorously derived from the conservation of energy and momentum within the relativistic framework, using the invariance of the norm of the energy-momentum four-vector:

E² − (pc)² = (m₀c²)²

Which leads to:

E = γ·m₀·c² ⟹ E = m·c²

This equation does not exist in classical mechanics. In Newtonian physics, kinetic energy is given by Ek = ½·mv², and there is no intrinsic link between mass and energy.

2. The “force equation” does not automatically produce E = mc².

The identity F = dP/dt = dE/ds is kinematic. It expresses a consistency between changes in momentum and energy, but it does not define the functional forms of E or m(v).

If you write:

dE/ds = dP/dt ⇒ d(mc²)/ds = d(mv)/dt

…you’ve already assumed that E = mc².

You’re using the equation, not deriving it.

3. Claiming that “c” could be any velocity is absurd.

In a world where c = 3 km/h, E = mc² would give you an absurdly small energy for any mass.

In reality, c has been measured and, more importantly, the constancy of c is the fundamental postulate of special relativity. It’s not a tunable parameter, it is the keystone of the structure of spacetime.

Essam Allou “1. E = mc² is a direct consequence of relativistic invariance.

It is rigorously derived from the conservation of energy and momentum within the relativistic framework, using the invariance of the norm of the energy-momentum four-vector:

E² − (pc)² = (m₀c²)²

Which leads to:

E = γ·m₀·c² ⟹ E = m·c²

This equation does not exist in classical mechanics. In Newtonian physics, kinetic energy is given by Ek = ½·mv², and there is no intrinsic link between mass and energy.”

Without acknowledging the definition of energy and momentum the energy-momentum four-vector is pointless. Therefor using it here even in competition to the original definition also is pointless.

“2. The “force equation” does not automatically produce E = mc².

The identity F = dP/dt = dE/ds is kinematic. It expresses a consistency between changes in momentum and energy, but it does not define the functional forms of E or m(v).

If you write:

dE/ds = dP/dt ⇒ d(mc²)/ds = d(mv)/dt

…you’ve already assumed that E = mc².

You’re using the equation, not deriving it.“

No, the force equation explicitly requires/produces E=mc² with an arbitrary constant c² because the momentum is defined as being proportional to the mass. The force equation only can be fulfilled if the energy also is proportional to the mass.

3. Claiming that “c” could be any velocity is absurd.

In a world where c = 3 km/h, E = mc² would give you an absurdly small energy for any mass.

In reality, c has been measured and, more importantly, the constancy of c is the fundamental postulate of special relativity. It’s not a tunable parameter, it is the keystone of the structure of spacetime.

You are completely misinterpreting the statement that in the equation E=mc² which is used in the derivation of the relativistic mass, c is an arbitrary constant. Identifying c as the velocity of light only is a consequence of the compatibility of the relativistic mass to the Lorentz transform. In the entire derivation of the relativistic mass, c definitively is an arbitrary constant.

With your absurd conclusions you only again show your incompetence.

Wolfgang Konle

If you still feel the need to continue this discussion, let’s move it elsewhere, there's no reason to keep polluting Berndt Barkholz’s thread with repeated misconceptions.

Once again, you're mistaking algebraic consistency for physical derivation.

Let’s settle this clearly, one last time:

You say “the force equation produces E = mc² with an arbitrary constant c².”

That is flatly false.

Let’s walk through it:

• You start from the identity F = dP/dt = dE/ds, fine.
• You define P = m·v and E = m·c², okay, but stop here: You already inserted E = m·c², which means you've assumed the proportionality between energy and mass with c².

So yes, you can algebraically verify that if E = m·c², it fits into dE/ds = dP/dt. But you did not derive E = m·c², you used it.

Your entire "proof" is built on an assumption you then pretend to deduce. That’s textbook circularity.

2. Your idea that c is an “arbitrary constant” is deeply flawed.

You write:

“In the entire derivation of the relativistic mass, c definitively is an arbitrary constant.”

No. In physics, constants have meaning. c is not just a mathematical placeholder. It is a measured universal constant, tied to the structure of spacetime, and its invariance is a postulate of special relativity.

Inserting a “generic velocity” squared into E = m·c² means you are no longer doing physics, just dimensional bookkeeping.

3. Your resistance to the energy-momentum 4-vector reveals a deeper misunderstanding.

You dismiss the 4-vector as “pointless.” Yet it is the very foundation of relativistic mechanics. The invariant relation:

E² − (pc)² = (m₀c²)²

is what allows us to derive E = γ·m₀·c² and thus connect mass, energy, and momentum coherently, without circularity.

Rejecting this isn’t an argument, it’s just refusing to learn.

You haven’t proven anything new. You’ve just:

• Assumed E = mc²
• Called c² “arbitrary” to pretend you’re not assuming relativity
• Then used those assumptions to “derive” a result already built into them.

It’s not derivation. It’s algebraic recycling dressed as insight.

Please stop calling it a proof and stop confusing readers who genuinely seek clarity.

Essam Allou

OK, continuing here would be unfair. You will find my answer on the thread

Is there a reasonable alternative to the theory of the expanding universe?

Essam Allou : E² − (pc)² = (m₀c²)²

As said: This equation is incomplete and only works for totally rigid point objects/field sources.

In all collisions of real non point, not rigid objects p is not conserved except for some very special cases. Of course energy still is conserved! We always will see that part of the energy/momentum is passed to the S(J) terms.

This is why all classic equations of standard model physics do break down in the real world and fantastic (e.g. vacuum) couplings must be introduced etc...

Of course Wolfgang Konle still wrongly believes that field energy exists even for rotating bodies.... What a nonsense.

As said: most should go back to school and study Sommerfeld.

Jürg Wyttenbach

Let’s go through your statements with precision and see what stands the test of physics:

1. “E² − (pc)² = (m₀c²)² only works for rigid point-like objects.”

That is incorrect.

This equation is a direct consequence of special relativity, derived from the invariant norm of the energy-momentum four-vector:

E² - (p·c)² = (m₀·c²)²

It applies to any isolated system, whether point-like, extended, rigid, rotating, or not, as long as all energy contributions are correctly accounted for (including kinetic, internal, rotational, vibrational, thermal, etc.).

This relation is not limited to idealized systems. It is used and experimentally confirmed in particle physics daily. For example, when dealing with unstable particles, composite systems, or even resonances, physicists apply this same relation with effective mass m₀ that includes the internal energy of the system.

2. “p is not conserved in real collisions.”

Momentum conservation is a fundamental symmetry tied to translational invariance via Noether's theorem.

If you claim p is not conserved in "non-pointlike collisions," you are confusing internal redistribution of energy/momentum with a breakdown of conservation laws.

Momentum is still globally conserved, even when part of the energy goes into deformation, heating, or radiation. The only requirement is that the system is closed and isolated.

3. “Wolfgang Konle still believes in field energy for rotating bodies.”

Let’s be clear: Wolfgang’s main issue is not whether field energy exists, it’s that he abuses classical reasoning and then claims he has derived relativistic mass or E = mc² from Newtonian logic. That’s the real nonsense.

So yes, Jürg, I do confirm that Wolfgang is not the only one who needs to reopen the books.

But let’s be honest: I’m not interested in debating this with someone who throws shade while presenting half-truths wrapped in condescension. As long as you’re not polluting the threads like Wolfgang, you’re free to think what you want.

Just don’t pretend it’s the new physics.

Please stop misusing my thread ! -> The smallest quantum of gravitational mass

Berndt Barkholz

May be we can continue our discussion based on mutual respect before it got rudely intercepted by Essam.

Are you still thinking that the concept of the relativistic mass is for the birds?

Jürg Wyttenbach "Of course Wolfgang Konle still wrongly believes that field energy exists even for rotating bodies.... What a nonsense."

What exactly do you mean? Moving or rotating masses only contain kinetic energy. Do you see that differently?

Wolfgang Konle You bet... but the problem is much deeper down... a theory must not be constructed, but exactly that is the main problem. First there must be a foundation that can carry the building, so you better look at that instead of trying to repair Einstein's hut, which doesn't fit properly on the existing foundation...

Essam Allou : derived from the invariant norm of the energy-momentum four-vector:

As you say it: Its just correct for this kind of simple minded physics problems. The 4-vetor momentum is not a conserved quantity in real world mechanics!!! Only in fantasy physics and highly simplistic models like Dirac/QED etc. this is claimed and just a precondition for century long wrong physics.

IF a real (non point) object owns a vector momentum of inertia and a force (in a collision) acts not centrally or a slight compression changes the surface and the reflection angle, then a part of the vector momentum gets transferred to inertia hence inertial momentum. So the total collision vector momentum sum is not zero! (You cant simply add vector momentum and inertial momentum vector like!)

Real physics knows to types of momentum and dumb standard model physics only knows one type. (Except some simplistic quantum numbers)

Check out the Goos Haenchen effect and you will immediately learn that not even photon reflection follows the classic laws and there no collision point 4 vector pair does even exist!

I only recommend you to once watch a snooker game and learn how humans make use of the non conservation of the vector momentum vector.

Wolfgang Konle

What means E=mc2 ? Doesn't it mean that at the speed of light mass EQUALS (IS) energy, that means Joule ?! You write a momentum as m*v/√(1-v²/c²) right ? Now how can the momentum grow when the mass is gone, and presents now energy. Energy can deliver a momentum to a mass, yes, but can't have momentum self... because the kilogram is gone. So the right momentum should be m*v*√(1-v²/c²), that means at the speed of light the momentum is zero... and I am very serious about that ! Now you may laugh...

Berndt Barkholz

Einstein's theory has problems with gravitational field energy and with gravitational radiation. These problems leads to misconceptions like dark matter and strange properties of black holes.

But the concept of relativistic mass is ok. The properties of mass must be consistent for any observer with an arbitrary velocity. This is only possible if mass transforms with the same Lorentz transform, which applies to space and time.

Dear Wolfgang Konle

Gravitational radiation is just as nonsensical as the ever increasing momentum on the way to the speed of light... you need to think !

Berndt Barkholz "You write a momentum as m*v/√(1-v²/c²) right ?"

Momentum P=m₀*v/√(1-v²/c²). m₀ is the rest mass, which is constant.

"Now how can the momentum grow when the mass is gone, and presents now energy."

The mass is not gone. We have E=m₀*c²/√(1-v²/c²). The remarkable point is that kinetic energy has a mass equivalent Ekin/c²=m-m₀. This mass eqivalent is an inertial mass and a gravitational mass.

Wolfgang Konle

...and YOU can't see what nonsense that is ? Oh my...

Dear Berndt Barkholz ,

Congratulations on your work! I find it interesting, in a good way!

However, you wrote something in your first comment.

Electromagnetic gravity overcomes the difficulty that energy gravity has. I think your current work can also be accepted as evidence. Newton's gravity, only applies mechanically to the surface of the Earth.

Regards, Laszlo

László Attila Horváth

Newtonian gravity has no difficulties... the mainstream has difficulties with Newton's gravity, but there is no Electromagnetic gravity, sorry.

regards, Berndt

Berndt Barkholz

It seems that you are having a quite different understanding of what is called "relativistic mass" in physics.

Please don't tell only "thats nonsense". Instead please try to explain how you see the relation between mass and kinetic energy and momentum.

Wolfgang Konle

Sorry Wolfgang, but I have the understanding that relativistic mass is nonsense ! The Lorentz factor is working with meter and seconds and is derived that way... that is without using any mass in the process, so you are not allowed to use that factor on the mass !

The standard relativistic formulation of momentum, given by: p = m v / √(1 - v²/c²) leads to contradictions at v = c, where mass is said to become infinite while also transforming into pure energy. This inconsistency suggests the standard formula is a mathematical artifact rather than a physically sound relationship. I propose a corrected formulation: p = m v √(1 - v²/c²) In this form, the momentum naturally reduces to zero as velocity approaches the speed of light, consistent with mass having transitioned into energy and no longer contributing classical momentum. When integrating this corrected momentum to determine the corresponding energy, the result is: E = ∫ p dv = (m c²) / 3 Remarkably, this result is not just a mathematical curiosity. It aligns with physical observations from the mass-limit collapse phenomena (such as supernova explosions), where the released energy fraction corresponds closely to this one-third relation. This suggests that the famous E = m c² (still valid in some applications) may represent a theoretical limit or idealization, whereas nature operates on a more refined law where the effective usable energy is given by E = (m c²) / 3. The consistency between this corrected momentum formulation, its derived energy relationship, and astrophysical evidence strongly suggests that this framework is physically correct and not merely theoretical.

I know you won’t accept this, but I focus on results ...not on preserving the fantasies of celebrity scientists. My own reasoning leads me down a different path than Einstein’s did. Einstein was an architect, skillfully constructing ideas, but in his curved spacetime {where no forces act} nothing can fall. Absolutely nothing. Some "mind-creatures" are simply born dead, and we have to accept that.

Berndt Barkholz "I know you won’t accept this, but I focus on results ...not on preserving the fantasies of celebrity scientists. My own reasoning leads me down a different path than Einstein’s did. Einstein was an architect, skillfully constructing ideas, but in his curved spacetime {where no forces act} nothing can fall. Absolutely nothing. Some "mind-creatures" are simply born dead, and we have to accept that."

I thank you for your detailed description of your motives and your respectful tone.

Yes indeed I can't accept your proposal to replace m=m₀/√(1-v²/c²) by m=m₀√(1-v²/c²). I know that you are strongly convinced that your proposal is justified. Nevertheless I propose to compare the approximation v

Berndt Barkholz : whereas nature operates on a more refined law where the effective usable energy is given by E = (m c²) / 3.

This is quite reasonable as we know that the acting energy of dense mass (atoms) maximally is 1/3 of the total mass as always 2/3 are conform with the strong force flux bond.

In the famous H*-H* bond the free acting flux reduces to 2/3 and the total potential of the first electron gets reduced by 1/3... (as experiment with Rydberg matter do show).

So definitely the factor of 1/3 is well known in real (serious) nuclear physics.

Not a single person had anything to say about the main issue: "The smallest quantum of gravitational mass" ??!

Berndt Barkholz "Not a single person had anything to say about the main issue: "The smallest quantum of gravitational mass" ??!"

Because of E=mc² and "inertial mass=gravitational mass" there is no "smallest quantum of gravitational mass". Any continuous amount of energy gravitates.

Wolfgang Konle Why do you always first talk... and then maybe think... just a little would do it !

Since when mans E=mc2 that E=m ?

Berndt Barkholz "Since when mans E=mc2 that E=m ?"

It means m=E/c² and not m=E.

BULLSHIT !

Berndt Barkholz "B...t!"

Could you be more polite and more specific!

If you are thinking differently you should be able to express what you are thinking. Otherwise your whole thread does not make any sense.

You didn't once explain what happened to the mass after the mass (the spaceship) is back to speed zero... is the the mass just evaporating or is the mass now greater than before, according to Lorentz ? I am sure you have a very exotic answer to that !

At the speed of light the momentum (m*v) can only be zero ! How can E*v be a momentum ?

The fact that E/v and E/c represent momentum has nothing to do with whether E*c is a momentum ...because at that speed, mass becomes energy. You can accelerate mass, but how do you accelerate energy? Joule * speed gives units of kg·m³/sec³—does that look like momentum to you? Now, kindly spare me the Einstein-worship. Ergo, the true momentum of a mass is given by m * v * sqrt(1 - v²/c²). The only way to deny this is by invoking Einstein… and you're more than welcome to do so!

Wolfgang Konle

"Otherwise your whole thread does not make any sense."

Until now you didn't even touch the issue of my PDF file...

Berndt Barkholz "back to speed zero"

The rule is m=mₒ/√(1-v²/c²). This rule does mention any speed history.

"The fact that E/v and E/c represent momentum has nothing to do with whether E*c is a momentum".

E/v is not correctly representing a momentum, and E*c is not at all a momentum. You must consider the mass equivalent of energy m=E/c² and then momentum P=mv.

What you are doing is simply confusing physical units. Of course you also can accelerate the mass equivalent of energy.

If you are having two blocks of exactly 1kg of iron with a temperature of 0° Celsius. Now you are heating up one block to 1000°C, the energy of the hot block increases by Q=0,439J/(kgK)*1000K*1kg=439J. This increases the mass of the block by ∆m= 439J/c²=439/(9E+16) kg=4,88E-15kg.

The mass equivalent is extremely small. It does not lead to a noticeable difference if you accelerate the hot or the cold block of iron.

Wolfgang Black Magic is NOT physics ! The good thing is that this alleged mass difference is not possible to measure... so there is no possibility to prove you wrong... what's wrong is that you believe ...

Now back to my question: So you chose to use a heated one kilogram Iron mass and of course the heat is after a while disappeared since the Iron-mass is cooling, loosing energy, but not mass !! But that mass equals energy/c2 is true but not relevant in this connection... nothing converted the energy to mass.

...but maybe your frying pan became too heavy for you after years of use ?

Berndt Barkholz "nothing converted the energy to mass."

A conversion of energy to mass happens in the synthesis of elements heavier than iron. In the synthesis of elements below iron from lightweighter elements, mass gets converted to energy. Also nuclear fission of heavy elements like Uranium converts mass to energy.

The synthesis of heavy elements converts energy to stable mass. But the equivalence of mass and energy does not require such a conversion. It is a duality.

The actual form of energy or mass does not matter for the gravitational impact. The gravitational impact of energy is exactly equal to the gravitational impact of the mass=energy/c².

But that's all included in E=mc². Why are you doubting?

Wolfgang Konle You are really joking now...

What is the number of nucleons before and after fission of Uranium 235 ?

I come again tomorrow... in the mean time please explain...

Berndt Barkholz "What is the number of nucleons before and after fission of Uranium 235 ?"

The number of nucleons did not change. That is baryon number conservation. But the sum of the nuclear binding energy in the fission products exceeds the binding energy in Uranium 235.

The binding energy is the opposite of the field energy. Therefore, there is less nuclear field energy in the fission products than in Uranium. The difference in field energy is the energy which has been generated with the fission of Uranium.

You also can see this in considering the inverse fission reaction in supernovae which produce Uranium from those fission products in a reaction which consumes energy.

Are you having another theory how fission reactors can produce energy?

Wolfgang Konle

As long as you see energy as gravitating we have only a "yes-no-yes-no" discussion... we wouldn't go anywhere... Instead you should have a look at the file I send with this thread.

Berndt Barkholz

What do you think is the relation of proton attributes, whatever they are, to the energy content in nuclear fields, called quark fields, which interact whithin atomic nuclei? Considering a proton does not tell us enough about the interaction of nucleons.

The only thing we know is that these quark fields contain a lot more energy density than the electric field in the nucleus. We also know that combining different quark charges, reduces the energy density in quark fields, similar to the reduction of electric field energy by combining different charges.

For very large nuclei, the combination of electric charges in the nucleus leads to a high electric energy density which partially compensates the energy gain of combining nucleons.

That is or current knowledge about binding effects in atomic nuclei. Speculations about the structure of protons, based on semi-classic Bohr physics, are not really helpful in that context.

I knew it... you were flying through the file without having the internal computer turned on in order to understand what you are reading.. You are one big dogma-crystal unable to correct your memory and unable to understand... how could I ever have an intelligent discussion with you ?! ...and you still don't know that "quark" is a German sour-milk product...

Berndt Barkholz "and you still don't know that "quark" is a German sour-milk product..."

I am talking about the interaction between nucleons, and you are insisting on looking into details of your semi-classic description of protons.

Just considering

"I propose that the Proton mass {mP} is given by the atomic mass divided by the mass number. ", immediately rings the bell "irrelevant", because it does not consider the interaction between nucleons via force fields, which contain an energy density, and it ignores the existence of neutrons.

I don't think that quantum-chromo-dynamics is the final answer of physics to the properties of atomic nuclei. But I see that this answer is much closer to reality than your semi-classic approach.

Wolfgang Konle

Explain to me why my results are 100% consistent. Explain why the difference between the Proton and the Hydrogen atom is about 19 times the Electron mass ...just another lucky coincidence, right? You dismiss it as "semi-classical nonsense", but what you fail to consider is that your entire framework could be wrong. You ignore the numbers when they tell a different story ...how can mathematical consistency be irrelevant? A consistency that, by the way, your ‘banana quark’ can’t reproduce!

You fire electrons at Hydrogen atoms, assuming a fixed Proton mass, yet Bohr’s equations predict the Proton weight as 1/M of every atom. Am I crazy for pointing that out, or are the equations themselves wrong? And if they are wrong, why do they yield exact results? You speak before you think, and I’m sorry, but YOU can’t just wave away my calculations as irrelevant ...you can only shield yourself behind your everything-controlling dogma!

Berndt Barkholz "Explain to me why my results are 100% consistent."

Consistent to what?

"Explain why the difference between the Proton (mass) and the Hydrogen atom (mass) is about 19 times the Electron mass."

Your difference is derived from the mass of an iron nucleus and not from the mass of a free proton. If you compare the mass of a free proton and the mass of a hydrogen atom the difference is much smaller.

You really don't know what you talk about... it just shows a mass 1/56 of the Iron atom, or 1/7 of the Lithium atom... you didn't really read that file... you have no idea about the mass of a free Proton... you are a believer of dogma, nothing else... lets stop here. YOU WANT TO LECTURE, BUT NOT LISTEN TO OTHERS !

IF you did read the file with an open mind you would have noticed that I do not derive this Proton mass from the Iron atom !

You just refuse to listen !!!!

Wolfgang Konle

On the Misinterpretation of Photon Momentum

It is widely stated in modern physics that light, despite being massless, carries momentum. This claim is typically justified through the relation

p=E/c

where E is the energy of the photon and c is the speed of light. However, upon closer examination, this interpretation raises a serious conceptual problem.

From dimensional analysis:

Energy has units of kg⋅m2/sec2...

Momentum has units of kg⋅m/sec

While the expression p=E/c yields the correct units for momentum, it does not follow that energy as such intrinsically carries momentum. In Newtonian mechanics, momentum arises from mass in motion,

p=m⋅v

and without mass, there is no classical momentum. A photon has no rest mass; hence, assigning it momentum requires deeper scrutiny.

The confusion stems from conflating energy transfer with momentum possession. Photons indeed interact with matter in ways that cause recoil (as in the photoelectric effect or Compton scattering), and in those interactions, momentum appears in the system. But that momentum is not inherent to the energy itself; it is the result of energy being absorbed by a mass and converted into kinetic motion.

Thus, energy alone does not possess momentum. It can result in momentum only when interacting with mass. Energy is scalar and lacks direction; momentum is a vector quantity defined by motion and mass. The presence of momentum, therefore, implies a carrier with mass and a specific direction in spacetime.

The claim that "light has momentum" is at best a useful mathematical abstraction for balancing energy-momentum equations, and at worst, a conceptual shortcut that violates physical logic. In truth, energy as such is incapable of momentum ...it merely creates the conditions for momentum to arise when absorbed by a mass-bearing system.

This distinction is not trivial. It reflects a deeper misunderstanding in how physics treats massless fields and should be carefully revisited, especially in contexts where mathematical formalisms are taken as physical realities.

Berndt Barkholz "The claim that "light has momentum" is at best a useful mathematical abstraction for balancing energy-momentum equations, and at worst, a conceptual shortcut that violates physical logic. In truth, energy as such is incapable of momentum ...it merely creates the conditions for momentum to arise when absorbed by a mass-bearing system."

You are addressing a general widespread misinformation about the relationship between energy and momentum. However, the ratio is clearly determined by the force in the mechanics. In mechanics, we have the definitions energy = force times displacement, E=F*s and momentum = force times time. P=F*t

This leads to a differential equation F=dE/ds=dP/dt, which we can apply to anything in physics where forces occur.

You also correctly say that we cannot derive momentum from energy, since energy is a scalar, while momentum is a vectorial quantity. Therefore, we must also understand dE/ds as a directional derivation. dE/ds=(∂E/∂sx, ∂E/∂sy, ∂E/∂sz).

By the way, your remarks are highly topical, because in standard physics this basic differential equation is practically not used.

For a photon, however, the relationship between energy and momentum is not a real problem, since the speed of the photon is in fact already a vector.

P=E/c then becomes P=E*ec/c. ec is the unit vector in the c-direction.

But this is only valid for Photons or generally for particles without a restmass. For everything, which has a restmass mₒ we have E=mc² and P=mv with m=γmₒ. m=γmₒ, by the way, with an arbitrary constant c is a direct solution of the differential equation d/ds(mc²)=d/dt(mv).

Wolfgang Konle

I have more complains:

On Spherical Emission, the Nature of Photons, and the Absurdity of Force-Carriers

If we accept the atom as a spherically symmetric physical system, then we must also accept that any radiation it emits—such as a photon—must initially propagate spherically. There are no directional apertures in the atomic structure, no built-in mechanism to "aim" a photon. Thus, the common image of a photon being emitted like a tiny bullet in a specific direction collapses under scrutiny.

In this view, the so-called photon is not a particle at all. It is the localized result of a spherically emitted energy wave interacting with a detection surface. What we call "a photon" is simply where the energy from a spherical wave happens to be absorbed. There is no need for a photon "trajectory," no quantum "decision" made by the atom. The wave exists until it is measured. The particle is the byproduct of that interaction ...not a physical object flying through space.

This alone casts doubt on the particle interpretation of light, but the issue becomes even more tangled when we consider the force-carrier model used in quantum field theory. In this model, forces ...like attraction and repulsion ...are said to occur through the exchange of virtual particles (e.g., photons for electromagnetism, gluons for the strong force, etc.).

But this raises an obvious and rarely addressed question:

Who throws the particle first?

If two particles are interacting, do they coordinate? If both emit force-carrying particles simultaneously, do the exchanged particles collide mid-space? And in the case of attractive forces, are we expected to believe that both particles are somehow firing particles at each other's backs to pull one another closer? This is absurd. The logic of momentum exchange breaks down in this context. Momentum transferred in a repulsive collision makes some sense; but a pulling effect via exchanged particles is completely non-intuitive and conceptually unjustified.

Furthermore, if we accept that all emitted quanta are spherical energy waves, then the idea of a directional exchange becomes meaningless. There is no well-defined trajectory, only a field of influence that grows outward until an interaction occurs.

Conclusion

The notion that all interactions occur via the exchange of force-carrying particles collapses under physical reasoning. It assumes coordination where none exists, directionality where there is none, and particles where there are only waves.

A more coherent view is that:

1. Atoms emit spherical energy.

2. Momentum and force effects arise from interactions with this wave, not from bullet-like particles.

3. Attraction and repulsion are time-gradient phenomena, not mechanical particle exchanges.

These assumptions restore logical clarity to a subject that has, under modern interpretation, become clouded by metaphor and abstraction.

Now I killed my original thread...

Dear Berndt Barkholz : If we accept the atom as a spherically symmetric physical system, then we must also accept that

Unluckily not even particles are of spherical nature. Some large atoms look more like peanuts than spheres...

In this view, the so-called photon is not a particle at all....

Of course one can call a photon particle, because its extension is finite contrary to an EM wave that expands to infinity.

Absorption of a photon simply is adding its EM flux to the atoms surface EM flux. In rare cases total absorption happens that is restricted to so called resonant states of the atom's spectrum.

The adsorbed EM flux no longer acts particle like and is transported along a torus trajectory, that shows different coupling for the involved radii, what causes shifted reflections.

Of course you are correct: Any (G,N etc..) force exchange via particles is a brain dead idea of some clueless standard model fantast.

Berndt Barkholz If we accept the atom as a spherically symmetric physical system, then we must also accept that any radiation it emits—such as a photon—must initially propagate spherically. There are no directional apertures in the atomic structure, no built-in mechanism to "aim" a photon. Thus, the common image of a photon being emitted like a tiny bullet in a specific direction collapses under scrutiny.

An excited atom, which actually emits a photon is not spherically symmetric. Only some atoms like hydrogen and helium have a spherically symmetric ground state. As soon as as we have a state with the angular momentum quantum number l ≠ 0, the atom is not spherically symmetric.

Haven't you ever seen visualisations of the shape of different atomic states?

They are not at all generally spherically symmetric.

Wolfgang Konle Jürg Wyttenbach

If a "photon" is ejected from an atom where is the tiny (ass)hole releasing this photon ? ...or is this photon particle "materializing" somewhere outside the atom. What direction has this photon... arbitrary, or a fixed, but not specifiable direction ? You are both good at telling us what you think isn't true... so how about telling us for a change what is true. I have problems with a tiny particle flying around and calling himself for a photon... and how can a particle have a frequency (color) ? Sorry, but the narrative about photons as particles stinks...

Berndt Barkholz "What direction has this photon... arbitrary, or a fixed, but not specifiable direction ?"

The photon exactly carries the momentum and energy difference of the atomic or molecular phase transition.

"how can a particle have a frequency (color) ?"

A photon has a certain length corresponding to the duration of the phase transition, which had created the photon. This length comprises many wave lengths. The width of the photon corresponds to the width of the transition zone which emitted the photon.

Jaja... das ist so wie das berühmte Seemannsgarn...

Berndt Barkholz

So much for sailor's yarn! Do you have a better explanation?

Berndt Barkholz : If a "photon" is ejected from an atom where is the tiny (ass)hole releasing this photon ?

Most people mix up photons with EM radiation. You can only detect photons at interaction. Just use a LASER pointer that emits zillions of photons in just one direction. EM radiation is difficult to focus and you usually must use higher order resonances and complex focusing senders but the field is always a cone. Not so for a bunch of photons that only become a cone when they pass through a media.

Physically a photon is a self contained higher dimensional packet of EM energy. To be self contained internally charge must be produced that makes the packet stable.

To find the photon structure some more modelling is needed of the type I did for the proton.

Jürg Wyttenbach said: "Just use a LASER pointer that emits zillions of photons in just one direction."

Jürg... but a laser pointer needs a collimating lens that transforms a divergent light beam into a parallel beam...

Berndt Barkholz : but a laser pointer needs a collimating lens...

Cheap ones yes. But LED based emitter are almost plane... Key is that a single photon does not form a cone.