Let's begin by discussing the source of Newtonian gravity. Later, we can also touch on the source of Einsteinian gravity. So, what substance generates a point mass? If Euclidean space were a foliation with a typical fiber in the form of a 3-dimensional sphere, then the source of gravity would be the vector (covector) field generating this foliation. Locally, such a vector field can be represented as a linear vector field orthogonal to the unit 3-sphere of 4-dimensional space. Then, the Newtonian gravitational potential is the angle (hyperbolic) of deviation of the vector field from orthogonality to the 3-sphere. In the limit, the angle tends to infinity, and therefore the source of a point mass is a vector field tangent to the 3-sphere.
What about mass M as a source of any gravitational field g with the field strength g=GM/r².
Wolfgang Konle,
Look at formula 1.1.3 where the potential inversely proportional to the distance appears
Preprint Mathematical Notes on the Nature of Things (fragment)
Igor Bayak "In the limit, the angle tends to infinity, and therefore the source of a point mass is a vector field tangent to the 3-sphere."
A 3-sphere can be described by unit quaternions multiplied by a constant R. But these unit quaternions only specify coordinate points and no vectors. How would you specify a vector field tangent to the 3-sphere?
If x,y,z is an arbitrary coordinate in the 3-sphere, what vector is a tangent in this coordinate point?
Wolfgang Konle ,
Quaternion algebra is equivalent to matrix algebra, which in turn is equivalent to the algebra of linear vector fields in space (x1,x2,x3,x4). The coordinates of Euclidean space (x,y,z) are generated by the spiral-shaped current lines of vector fields formed by the basis linear vector fields of the quaternion algebra summed with the unit vector field x1\partial x1 + x2\partial x2 + x3\partial x3 + x4\partial x4. When considering the correspondence of these vector fields with the gravitational potential, the choice of a particular element of algebra should be neglected.
Igor Bayak "Quaternion algebra is equivalent to matrix algebra..."
If (q1,q2,q3) with q1²+q2²+q3²=R² is an arbitrary coordinate in the 3-sphere, what vector is a tangent in this coordinate point?
I don't understand what your (q1,q2,q3) means. These could be coordinates on the 3-sphere or the coefficients of the basis quaternions. My example from the "notes" considers a vector field that fits onto the 3-sphere only when approaching a point mass in Euclidean space. Perhaps by (q1,q2,q3) you meant (x,y,z).
Igor Bayak "Perhaps by (q1,q2,q3) you meant (x,y,z)"
No, I can address any specific location within a 3-sphere with 3 coordinates. The range of each coordinate value is from zero to R*2π. The 3 coordinates are fully convertible. No single coordinate is special. The radius R is the radius of the four dimensional ball surrounded by the three dimensional 3-sphere.
But how are you defining the tangent vector at a specific point (q1,q2,q3)?
The coordinates of the two longitudes are in the range (R*2π), and the latitude is in the range (R*π). As for your question about the vector at this coordinate point, it depends on the choice of vector field. However, what is important for us is not the value of the vector, but its deviation at this point from the orthogonal direction.
Igor Bayak "The coordinates of the two longitudes are in the range (R*2π), and the latitude is in the range (R*π)."
In a 3-sphere, there are no longitudes and latitudes. All three coordinates are topologically exactly equal and are in the range (R*2π).
This is a consequence of the unit quaternion structure.
If this were true, we wouldn't be able to distinguish a sphere from a torus. Please see the topic "Polar coordinates of 4-dimensional Euclidean space."
Gravity can be thought of as stemming from the accelerated expansion of the universe. Consider the formation of a common polymer sponge from a catalyzed fluid that expands into a form that closely resembles the large scale structure of the universe. In this model the only driver is repulsive, yet the material aspect appears as though it is undergoing mutual attraction. Attraction -- gravity -- then, is only emergent, not fundamental in this view. https://www.linkedin.com/pulse/gravity-experimental-gap-warren-frisina-o5vqf/?trackingId=jSr64FNKHWeyms1la7Vipw%3D%3Dew.
Igor Bayak "If this were true, we wouldn't be able to distinguish a sphere from a torus."
A sphere and a torus are two dimensional and topologically different. A 3-sphere is three dimensional and absolutely symmetric in all three dimensions.
Warren Frisina "Gravity can be thought of as stemming from the accelerated expansion of the universe."
Gravity too selectively depends on gravitating objects to be dependent on such general properties of the whole universe.
Warren,
It is not enough to say that matter is a form of space; it is also necessary to present this form in an explicit form.
All matter in the universe creates gravity by absorbing ether particles
Igor, For example: https://www.linkedin.com/pulse/particle-masses-revisited-warren-frisina-qzn6e/?trackingId=pF7FaFFM12EWufNTVjcjMQ%3D%3D
Igor Bayak "What I said is true for both the classical sphere and the 3-sphere."
Do you know that within a 3-sphere every point in the total volume of the 3-sphere exactly is the center of gravity of the total volume?
John-Erik Persson "All matter in the universe creates gravity by absorbing ether particles"
All forces F in our world on an object, including the gravitational force, are caused by an energy content W, depending on the positon s of the object. F=dW/ds.
In the gravitational case the position dependent energy W is the energy content of the combined gravitational field of the object g(object) and of earth g(earth).
We have g=g(object) + g(earth) and W=∫constant*g²dV, V is the volume filled with the combined field g.
Ether particles do not exist.
Do you have a formula which would quantitatively justify your absurd idea?
If ether particles do not exist there is no matter and no gravity.
John-Erik Persson "If ether particles do not exist there is no matter and no gravity."
No, matter cannot emit particles. This would exhaust matter. Matter also cannot absorb particles. This would blow up matter.
The force F between gravitating objects or charged objects is caused by the energy content W in combined force fields, which depends on the distance between s the object. F=dW/ds.
If you claim something else, then provide your alternate force law.
In the main post, I argued that the source of gravity for a point mass is a vector field in 4-dimensional space. Now, let me point out that the vector field is material, since it is the velocity field of matter particles.
Wolfgang Konle
Matter can absorb and emit ether particles.
Emission and absorption balance each other.
John-Erik
Igor Bayak is it a joke ?!
https://www.researchgate.net/publication/394250559
Pure Time Theory - Chapter I: Ontological Foundations and Spectral Dynamics of Cosmogenesis
https://www.researchgate.net/publication/394523876
Pure Time Theory - Chapter II: Modal Spectralization from Riemann Zeta Zeros - Deterministic Mass Emergence and Electromagnetic Structure
https://www.researchgate.net/publication/394523939
Pure Time Theory - Chapter III: Light as Revealer - Photons, 4D Geometry and Spatial Navigation in Pure Time Theory
https://www.researchgate.net/publication/392979130
Pure Time Theory – Chapter V - Annular Phase-Injectivity: Static RH Equivalence and Dynamic Closure
John-Erik Persson "Emission and absorption balance each other."
Why should this be balanced in every situation?
How does that lead to an attractive force?
Igor Bayak "let me point out that the vector field is material, since it is the velocity field of matter particles"
The natural force law F=dE/ds (a directional derivative) derived from E=F*s describes the source of the gravitational force F on an object in the gravitational field of earth g(earth).
We have an energy- and a force formula for the gravitational force:
E=∫constant*(g(object)+g(earth))²dV and F=dE/ds.
The constant is -1/(8πG).
The energy and force formulae are applicable to all forces exerted by force fields. In the electric case g must be replaced by the electric field vector and the constant is εₒ/2. In the magnetic case g must be replaced by the magnetic field vector H and the constant is µₒ/2.
Mr. Wolfgang Konle , it's been a long time! How are you? You're acting as if I hadn't written anything? I see that PTT still frightens you just as much!
Hi Mr. Essam Allou , it's indeed been a long time since I heard something from you in a RG forum in which I am also active.
PTT does not at all frighten me. Compared with the basic mechanic law F=dE/ds which explains all forces F on an object as caused by an energy E which depends on the position s of the object, it is extremely detached. According to your description it is even located beyond general relativity.
In our past discussions you finally admitted that your theory does not try to challenge the basics of physics. Therefore what I am telling with F=dE/ds is according to the very basic mechanics and is not touched by your PTT theory.
In fact F=dE/ds cannot be seriously challenged by PTT. It is an immediate consequence of E=F*s, energy equals force times way.
Wolfgang Konle
You are mixing up two different things, and the identity you keep repeating is simply not correct outside a very special case.
1) Your “law” F = dE/ds is not a law of mechanics.
- In 1D, for a conservative field, it is F(s) = – dV/ds, where V is the potential energy.
- The statement E = F * s only makes sense for the work of a constant force along a straight line displacement. It is not a general identity between force and “total energy”. In general: W = ∫ F·ds and F = –∇V.
So to claim “F = dE/ds explains all forces” is mathematically wrong and physically reductive (wrong sign, wrong generality, confusion between work/energy/force).
2) PTT does not “defy” the bases of physics: it generalizes and reconstructs them as limits.
PTT is a deterministic cadence grammar from which:
- GR (weak field) appears as a variational correspondence: one derives a Helmholtz–Poisson operator
(∇² – L_eff⁻²) u = (4πG / c²) ρ , with u = Φ/c²,
and Newton’s law g = –∇Φ is recovered when r
John-Erik Persson "Balanced over time and not in every moment"
What is the maximum duration of an imbalance? What does this duration depend on? What is the maximum mass percentage which can be imbalanced?
The really scientific - and so unique - answer to “What can serve as a source of gravity?” is as:
- Gravity is fundamentally nothing else than a fundamental Nature force,
- which, as that other, i.e. at least Electric, and Nuclear/Strong, Forces also do, acts by the one scheme - as that is rigorously scientifically shown in the Shevchenko-Tokarevsky’s initial Planck scale models of these Force4s, more see https://www.researchgate.net/publication/395113364_Planck_scale_informational_physical_model_and_fundamental_problems_in_physics, section 6 “Mediation of the fundamental forces in complex systems”:
The “sources of Forces”, including Gravity Force, are Forces charges, which radiate the Forces mediators that propagate [don’t move, since fundamentally aren’t particles] in the 3DXYZ space of Matter’s fundamentally absolute, fundamentally flat, fundamentally continuous, and fundamentally “Cartesian”, [4+4+1]4D spacetime with metrics (cτ,X,Y,Z, g,w,e,s,ct) only with the speed of light; intense flaws of mediators are “Forces fields”; Gravity charge is “gravitational mass”; unlike other Forces all/every particles have Gravity charges.
The radiated by a particle’s Fotce charge mediators , which, at least at statics, don’t carry energy [so at least at statics the fields don’t contain energy], when hit into other – “irradiated” – particle, transmit to this particle specific momentums, that cause some changes in the irradiated particle “own” energy E=m0c2, m0 is rest mass. If a Force is attractive, the mediators cause releasing energy portion, ∆E,in irradiated particle; and, if the particle isn’t impacted by other forces, the particle start to move in 3DXYZspace having kinetic energy Ek=∆E, the particle’s “own” energy E=m0c 2 decreases on ∆E, the m0 decreases on ∆E/c2 – that is called in physics “a Force mass defect”; in Gravity Force case – “gravitational mass defect”.
Etc., more see the link above, to read SS posts in https://www.researchgate.net/post/Field_energy_density_is_missing_in_General_Relativity_A_big_issue_preventing_to_describe_gravitation_correctly_are_you_aware_of_it#view=68cfb3952a7c84859302abc2/41/42/41/42/42/44/44/44/44/45 pages 44, 45, and links in the posts, it is useful as well.
Cheers
Essam Allou "So to claim “F = dE/ds explains all forces” is mathematically wrong and physically reductive (wrong sign, wrong generality, confusion between work/energy/force)."
You simply misunderstand. The formula F=dE/ds is a directional derivative if the position s is a vectorial entity. dE/ds=(∂E/∂sx, ∂E/∂sy, ∂E/∂sz).
It is a fundamental law as correct as E=F*s or infinitesimally dE=F*ds and the sign also is correct.
With your accusations you are completely on the wrong track.
But this is now strange, because you had admitted that you do not intend to challenge basic mechanics. But this is just what you try to do now.
In the case of an accelerated mass m we get the one dimensional equation F=dE/ds=dP/dt. With E=mc² and P=mv we get the differential equation d/ds(mc²)=d/dt(mv). The solution of this equation is just the well known relation m=mₒ/√(1-v²/c²) of the relativistic mass. This additionally confirms F=dE/ds.
But this all is absolutely basic physics knowledge and far beyond something which could be challenged. Primarily it cannot be questioned because some detached theories seem to be in contradiction to the basic facts mentioned here.
Wolfgang Konle ,
You are simply recycling the same confusion we went through months ago.
1) F = dE/ds is not a fundamental law.
- In classical mechanics, for a conservative field, the correct relation is F = –∇V, where V is the potential energy. The minus sign is essential.
- The energy of a particle is not a function of position only; the kinetic term T = ½ m v² depends on velocity. Writing ∇E as if “energy” were only a
function of s is just wrong.
- The relation dE = F·ds is nothing but the definition of work for a force along a displacement. It is not a universal differential law.
2) Your relativistic “derivation” is circular.
- You invoke E = mc² and then differentiate “mc² with respect to s”. That makes no sense: energy is not a function of a single coordinate s.
- The actual structure of relativity is:
• p = γmv,
• E = γmc²,
• F = dp/dt.
From these definitions, one can show E² = (pc)² + (mc²)². But you cannot “derive” relativity from F = dE/ds, because you are already assuming the velocity dependence of E when you write E = mc² with m = γm₀. That is circular logic.
3) Mixing up “E = F·s” with “E = mc²” is mathematically incoherent.
- E = F·s only holds for the very special case of a constant force along a rectilinear path.
- E = mc² is a statement about rest energy and relativistic dynamics. They are not interchangeable formulas. Gluing them together is just playing with symbols, not physics.
So no, your “law” F = dE/ds is not some untouchable cornerstone “far beyond contradiction”. It is simply the work definition misapplied and mis-signed. In fact, the correct and universal relation is F = dp/dt, and from there both energy and work are defined consistently.
As for PTT: it does not need to “challenge” basic mechanics. It generalizes. In the proper limits, you recover Newton’s law, GR weak field, and even the statistical Schrödinger approximation. That is how unification works: retrieve the familiar laws as limits, and show they are not fundamental but consequences of a deeper deterministic grammar.
So please stop presenting F = dE/ds as if it were some deep universal principle. It is not. And your relativistic “confirmation” is nothing more than circular substitution.
Essam Allou You are simply recycling the same confusion we went through months ago.
No, but you are still stuck in the same errors you already made months ago.
1) F = dE/ds is not a fundamental law.
F=dE/ds is as fundamental as E=F*s.
- In classical mechanics, for a conservative field, the correct relation is F = –∇V, where V is the potential energy. The minus sign is essential.
No, this formula is incorrect. V is not an energy. It is an energy per mass.
- The energy of a particle is not a function of position only; the kinetic term T = ½ m v² depends on velocity. Writing ∇E as if “energy” were only a
function of s is just wrong.
The formula is not about the energy. It is about the change of energy caused by a force along a path s. F=dE/ds. This force is just the same force as the force related to the change of the momentum per time.
- The relation dE = F·ds is nothing but the definition of work for a force along a displacement. It is not a universal differential law.
Yes, the relation dE = F·ds is nothing but the definition of work for a force along a displacement. It is the direct consequence of the basic law E=F*s. The universal validity is a consequence of the mathematical rules of calculus.
2) Your relativistic “derivation” is circular.
- You invoke E = mc² and then differentiate “mc² with respect to s”. That makes no sense: energy is not a function of a single coordinate s.
No it is not circular because c only is an arbitrary constant in the differential equation. Only assuming a priory c=speed of light would be circular.
- The actual structure of relativity is:
p = γmv, E = γmc², F = dp/dt.
From these definitions, one can show E² = (pc)² + (mc²)². But you cannot “derive” relativity from F = dE/ds, because you are already assuming the velocity dependence of E when you write E = mc² with m = γm₀. That is circular logic.
Yes that is correct. This is the structure of relativity. But it is fact that d/ds(mc²)=d/dt(mv) just leads to m=mₒ/√(1-v²/c²) without assuming a priori m=γmₒ
3) Mixing up “E = F·s” with “E = mc²” is mathematically incoherent.
You are using the wrong terms. It is physically correct to identify the energy in E=F*s with the energy in E=mc². This identification corresponds to the situation that a mass m gets kinetic energy via the force F.
- E = F·s only holds for the very special case of a constant force along a rectilinear path.
No, you know the general law is ∆E=∫F(s)ds.
- E = mc² is a statement about rest energy and relativistic dynamics.
No, in the context of the differential equation d/ds(mc²)=d/dt(mv) c is only a constant. The relativistic interpretation is the physical consequence.
They are not interchangeable formulas. Gluing them together is just playing with symbols, not physics.
Nonsense. Nothing is glued together which does not belong together.
So no, your “law” F = dE/ds is not some untouchable cornerstone “far beyond contradiction”. It is simply the work definition misapplied and mis-signed. In fact, the correct and universal relation is F = dp/dt, and from there both energy and work are defined consistently.
That is simply wrong. F=dE/ds=(∂E/∂sx, ∂E/∂sy, ∂E/∂sz) is as basic as F=dp/dt.
It is obvious that basic mechanics is not your preferred domain. You make too many primitive errors and you follow fatally wrong concepts. You’d better stay in your nebulous PTT theory where nobody can show you specific errors.
Wolfgang Konle ,
I will prove to you below who knows physics better between us (regarding PTT, if you are not able to criticize it, that is not my fault. Everything is derived and can be reconstructed).
1) “F = dE/ds” is not a universal vector law.
- Decompose total energy: E = T + U.
• For conservative forces, F = –∇U and dU = –F·ds.
• Along a path with unit tangent t̂, one has dU/ds = –F·t̂ = –F_parallel.
• For the total energy, the kinetic part T = ½ m v² depends on velocity; ∇E is not defined purely in space.
=> The correct general identity is:
dE/dt = F·v (power),
and therefore dE/ds = (1/v) dE/dt = F·t̂.
This is a **scalar identity along the path**, not a universal vector equation.
2) “E = F·s” is not a basic law.
The general statement is the work integral:
ΔE = ∫ F·ds.
The shortcut E = F s holds only for constant force along a straight line. Presenting it as a cornerstone law is wrong.
3) Potential vs potential energy.
- U(x) is potential energy (Joules).
- φ(x) is potential per unit mass (J/kg).
Mixing the two and saying “V is not an energy” is just a confusion of notation. The correct law is:
F = –∇U, with U = m φ.
4) Your “relativistic derivation” is tautological.
You write: d/ds(mc²) = d/dt(mv).
But note: d/ds = (1/v) d/dt. Substituting gives
(1/v) dE/dt = d(mv)/dt.
Since power is P = F·v = dE/dt, the left-hand side is P/v = F_parallel. The right-hand side is also F_parallel.
So the equation just reduces to F_parallel = F_parallel.
It does not derive relativity; it is only a restatement of the definition of power. The appearance of γ in relativistic mass requires starting from Lorentz invariance, not this identity.
Conclusion.
- The universal dynamical law is F = dp/dt.
- Energy/work relations follow: dE = F·ds and P = F·v.
- For conservative fields: F = –∇U and along a path F_parallel = –dU/ds.
Everything else you wrote (treating F = dE/ds or E = F s as fundamental) are misuses of special cases.
I see you haven’t changed! Absolute confidence in form and a completely empty substance.
- Absorption of ether particles is the cause of gravity.
- Absorption affects ether. Matter causes the force of gravity to emerge.
Essam Allou I will prove to you below who knows physics better between us (regarding PTT, if you are not able to criticize it, that is not my fault. Everything is derived and can be reconstructed).
There is no need to criticize PTT. Ignoring it is sufficient.
1) “F = dE/ds” is not a universal vector law.
- Decompose total energy: E = T + U.
• For conservative forces, F = –∇U and dU = –F·ds.
• Along a path with unit tangent t̂, one has dU/ds = –F·t̂ = –F_parallel.
• For the total energy, the kinetic part T = ½ m v² depends on velocity; ∇E is not defined purely in space.
=> The correct general identity is:
dE/dt = F·v (power),
and therefore dE/ds = (1/v) dE/dt = F·t̂.
This is a **scalar identity along the path**, not a universal vector equation.
Sorry that is wrong because generally dE/ds is the vector (∂E/∂sx, ∂E/∂sy, ∂E/∂sz) the inversion of dE=F*ds according to the rules of calculus leads to the force vector F=(∂E/∂sx, ∂E/∂sy, ∂E/∂sz). But in a scalar situation d/dt(mc²)=v*d/dt(mv) is also correct and also leads to m=mₒ/√(1-v²/c²) (see below)
Do you really think, you can show with such primitive errors that you really know physics better?
2) “E = F·s” is not a basic law.
The general statement is the work integral:
ΔE = ∫ F·ds.
Yes, that is the general statement. But F*ds still is a scalar product and the integral generally is a line integral over any selected path.
The shortcut E = F s holds only for constant force along a straight line. Presenting it as a cornerstone law is wrong.
E=F*s is the basic expression for the definition of energy.
3) Potential vs potential energy.
- U(x) is potential energy (Joules).
- φ(x) is potential per unit mass (J/kg).
Mixing the two and saying “V is not an energy” is just a confusion of notation. The correct law is:
F = –∇U, with U = m φ.
This is not a natural law because U does not exist in reality. Or can you tell where U is located?
The location of U is the critical point. Because m is the mass of the subject of the force F, U must be located in the object itself. But what about the gradient of U? How can there be a gradient if U is located in the object?
The only existing gradient is the gradient of φ. But the gradient of U does not exist.
4) Your “relativistic derivation” is tautological.
You write: d/ds(mc²) = d/dt(mv).
But note: d/ds = (1/v) d/dt. Substituting gives
(1/v) dE/dt = d(mv)/dt.
What is here tautological ?
dE/dt=vd(mv)/dt with E=mc² and P=mv leads to d/dt(mc²)=v²dm/dt+vmdv/dt.
dm/dt(c²-v²)=vm*dv/dt
with dm/dt=dm/dv*dv/dt we get:
dm/dv*dv/dt(c²-v²)= vm*dv/dt, we cancel dv/dt and get:
dm/dv=vm/(c²-v²) and m‘/m=v/(c²-v²).
ln(m)=-ln((c²-v²)/2)+C
We select C=mₒc and get m=mₒ/√(1-v²/c²)
Since power is P = F·v = dE/dt, the left-hand side is P/v = F_parallel. The right-hand side is also F_parallel.
So the equation just reduces to F_parallel = F_parallel.
It does not derive relativity; it is only a restatement of the definition of power. The appearance of γin relativistic mass requires starting from Lorentz invariance, not this identity.
Conclusion.
- The universal dynamical law is F = dp/dt.
- Energy/work relations follow: dE = F·ds and P = F·v.
- For conservative fields: F = –∇U and along a path F_parallel = –dU/ds.
Everything else you wrote (treating F = dE/ds or E = F s as fundamental) are misuses of special cases.
I see you haven’t changed! Absolute confidence in form and a completely empty substance.
Real conclusion:
You are only repeating arguments you have derived from erroneous assumptions.
Wolfgang Konle ,
You confirm yourself that you do not master the basics you pretend to defend.
1) F = dE/ds.
You keep saying “dE/ds is a vector”. No. This is first–year material:
– ∇E = (∂E/∂x, ∂E/∂y, ∂E/∂z) is a vector.
– dE/ds is a scalar directional derivative, equal to ∇E·t̂.
So your statement “dE/ds is a vector” is simply false. You are confusing gradient with directional derivative.
2) E = F s.
Again, an elementary confusion: E = F s is valid only for the very special case of a constant force along a straight line.
The general law is ΔE = ∫ F·ds.
Trying to elevate E = Fs to a universal cornerstone law is like confusing a classroom shortcut with the general definition.
3) Potential U.
Saying “U does not exist” is nonsense. All of conservative mechanics is built on the existence of a scalar potential U(x) with F = –∇U.
You even add: “the location of U is the critical point”. No: U is a state function, not a little object you store somewhere.
Your argument is equivalent to saying “temperature does not exist because we don’t know where it sits inside the matter”. Seriously?
4) Your relativistic derivation.
Your equation “d/ds(mc²) = d/dt(mv)” only rewrites P = F·v. It is a tautology: you end up with F_parallel = F_parallel. Congratulations.
And your integration “ln(m) = –ln((c²–v²)/2)+C” is dimensionally absurd: you cannot take the logarithm of a dimensional quantity. This is a high–school level error.
In the end, yes, you can juggle the algebra until you recover m = m₀/√(1–v²/c²), but only because you already slipped in the assumption m = m(v). You derive nothing, you just re–package what was assumed.
Conclusion.
– You confuse gradient and directional derivative.
– You present a special case (E = Fs) as a universal law.
– You deny the existence of potential U while it is central to conservative mechanics.
– You attempt a relativistic derivation but write a tautology and even a dimensionally inconsistent equation.
You needed 7 hours to reply with this? You just wanted to post nonsense at a time when I might be asleep so you could have, as you like, the empty “last word”. But it is shameful, your level is seriously low, and you still try to lecture others. Stop it! This is really bad.
PS: And stop this habit of copying all my message in italics when you have no real substance in your reply, just to give the illusion of a long answer when it is empty. It makes reading and following very annoying! You only try to drown nonsense under a fake defense.
Essam Allou "You confirm yourself that you do not master the basics you pretend to defend."
1) F = dE/ds.
dE/ds resolves F*ds in respect to F. We have dE=(Fx*dsx+Fy*dsy+Fz*dsz)
This leads us to Fx=dE/dsx, Fy=dE/dsy,Fz=dE/dsz.
Put together in vector notation we get F=(dE/dsx,dE/dsy,dE/dsz) or abbreviated to (∂E/∂sx, ∂E/∂sy, ∂E/∂sz).
We now only can discuss if the vector component ∂E/∂sx is formally the same as the component dE/dsx. But even this discussion is pointless because in that expression it is defined as being the same.
2) E = F s.
The scalar product E=F*s simply is the definition of mechanical energy in physics. Every natural law or every basic definition in physics is written in that notation. P=F*t is not wrong because in fact the general law also is ΔP=∫ F(t)·dt.
3) Potential U
A scalar potential U=m φ which leads to the force F on the mass m itself is physically inconsistent. It in the first grade implies that m has an internal energy supply which according to external needs can be converted to kinetic energy. But m cannot also have a momentum supply which also according to external needs can be converted to the momentum the mass always gains together with the kinetic energy.
This self-related property of m of being the cause of a force proportional to m is the reason why the potential φ is not accepted of a plausible cause for the gravitational force.
The real cause is F=dE/ds, the cause for all forces in our world.
4) Your relativistic derivation.
The equation d/ds(mc²) = d/dt(mv) is compatible with P=F*t or F=dP/dt. P=F*v is wrong.
It is compatible to all physically correct equations which involve v,s,t,E,P. How could you expect to be taken seriously if you insert such primitive errors like P=F*v.
About the mathematical correctness of physical units within expressions, we can say that this correctness is maintained by the selection of the integration constant C as a constant with units. (C=mₒc)
ln(m)=-ln((c²-v²)/2)+C has been derived by integrating m‘/m=v/(c²-v²) which is dimensionally correct because m’=dm/dv. The dimension of m’/m is 1/velocity like the dimension of v/(c²-v²).
Insinuating a principal violation of physical consistency rules within standard algorithmic operations only reveals serious knowledge gaps.
You needed 7 hours to reply with this?
Do you really think I have nothing else to do than replying on your unqualified arguments?
As you have requested I have abbreviated your text to the absolute minimum without losing the context. But each of the four points contains the proof that your arguments are blatantly wrong. Your trivial principal errors concerning P=F*v, F=(dE/dsx,dE/dsy,dE/dsz), and the physical correctness of the units in ln(m)=-ln((c²-v²)/2)+C reveal your severe and embarrassing knowledge gaps in the fundamentals of mathematics and physics.
Misinterpreting the unfortunate self-reflexive property of the potential U in respect of the mass m is not your fault. It is a general and principal fault in standard physics.
As a summary I would say that your insolent claim of mastering the basics better than I do, only shows that your self-confidence has lost the contact to your actual capabilities.
You obviously have spent too much time and intellectual energy in irrelevant and detached domains of physics which had obscured your view on the pragmatic and simple logic of basic mechanics and math.
Wolfgang Konle ,
You keep piling elementary confusions on top of each other. Last pass from me, point by point, with standard definitions anyone can check in a first-year text.
1) “F = dE/ds” and what dE/ds is
• Gradient vs directional derivative:
∇E = (∂E/∂x, ∂E/∂y, ∂E/∂z) is a vector.
dE/ds is the directional derivative along the unit tangent t̂, and equals ∇E · t̂. It is a scalar.
Writing “F = (dEx/dsx, dEy/dsy, dEz/dsz)” is not a definition of anything standard; it just re-labels components without a coherent limit process. In a conservative field the correct relation is
F = −∇U and dE/ds = F · t̂ (work per arclength).
Elevating dE/ds to a vector is a category error.
• Special case vs general law:
E = F s holds only for constant force along a straight line. The general work–energy relation is
ΔE = ∫ F · ds (line integral along the path).
Claiming “E = F s is the definition of energy” is simply false.
2) On potentials
• Conservative mechanics uses a scalar potential U(x) with
F = −∇U and ΔU = −∫ F · ds (path independent).
U is a state function, not a little “thing” located somewhere. Saying “U=m φ implies the mass powers itself” is a misunderstanding. In Newtonian gravity the field has energy; the potential energy U is an interaction energy that depends on the configuration in the external field. None of this implies the test mass carries an “internal fuel.”
3) Your use of P, F, v, t
You mix three different quantities:
p = m v (momentum, vector)
P = F · v (power, scalar)
F = dp/dt (force, vector)
They are not interchangeable. When you write “P = F * t” you are confusing power with impulse. When you write “P = F * v is wrong,” you are denying the definition of power. This is not a debate point; it’s a definition you can verify in any reference.
4) Your “relativistic derivation”
• You wrote: d/ds (m c^2) = d/dt (m v). The left is a scalar derivative, the right is a vector derivative. You’re equating different objects.
• You then integrate “m’/m = v/(c^2 − v^2)” and claim “ln(m) = −ln((c^2 − v^2)/2) + C”. Two problems:
(i) you are taking ln of a dimensional quantity (c^2 − v^2), not allowed; you must work with a dimensionless ratio,
(ii) even fixing units, your manipulation tacitly assumes m = m(v), i.e., the conclusion is smuggled into the hypothesis.
• The standard, correct route is:
Work–energy: d(γ m0 c^2)/dt = F · v
Momentum: p = γ m0 v, F = dp/dt
Dispersion: E^2 = (p c)^2 + (m0 c^2)^2
From there you get γ = 1/√(1 − v^2/c^2) without dimensional nonsense and without equating scalars to vectors.
5) Bottom line on your “F = dE/ds explains all forces”
No. In general:
• In conservative fields: F = −∇U. Along a path, dE/ds = F · t̂ (scalar).
• With dissipation: work splits into potential change plus heat; there is no path-independent U but F is still not “the vector dE/ds”.
• In EM: forces come from fields (q(E + v × B)), not from “dE/ds”.
• In GR: free fall is geodesic motion; “force” is geometric, and potential is a metric component in weak field. Your slogan does not survive any modern framework.
Given the volume of basic mistakes (gradient vs directional derivative; E = F s vs ∫ F · ds; power P = F · v; logs of dimensional quantities; mixing vectors and scalars), there is no productive path here. If you want to keep exploring your private notation, do it, but please stop presenting it as “fixing basic mechanics.” It isn’t.
I stop here with you, and everyone will be able to see the difference between your answers and mine. In any case, I will learn nothing from you, and you are not humble enough to learn or to recognize your mistakes. BUT stop trying to "correct" others!
Essam Allou "dE/ds is the directional derivative along the unit tangent t̂, and equals ∇E · t̂. It is a scalar."
No, F=dE/ds resolves dE=F*ds in respect to F. The scalar product F*ds is "Fx*dsx+Fy*dsy+Fz*dsz". This leads to Fx=dE/dsx, Fy=dE/dsy, Fz=dE/dsz.
in vector notation we get F=(Fx,Fy,Fz)=(∂E/∂sx, ∂E/∂sy, ∂E/∂sz).
Let us first focus on just this point because it is essential also for all other points.
Wolfgang Konle ,
Well, we really have to stop here. You keep insisting that
F = dE/ds somehow means "F = (∂E/∂sx, ∂E/∂sy, ∂E/∂sz)".
That is simply not standard mechanics:
– E(x,y,z) is a scalar. Its gradient is ∇E = (∂E/∂x, ∂E/∂y, ∂E/∂z), a vector.
– The directional derivative along a path s is dE/ds = ∇E · t̂, a scalar.
Confusing these two (gradient vs directional derivative) is the basic mistake in all your replies.
In conservative mechanics the correct, universal relation is
F = –∇U, and along a trajectory ΔE = ∫ F · ds.
Not “E = F s” and not “F = dE/ds (vector)”.
This is first–year material. You can check in any standard text, e.g.:
– Goldstein, *Classical Mechanics*, Ch. 1–2.
– Taylor, *Classical Mechanics*, Ch. 4.
So let’s stop here: continuing is pointless if even these elementary definitions are not accepted.
Essam Allou "E(x,y,z) is a scalar. Its gradient is ∇E = (∂E/∂x, ∂E/∂y, ∂E/∂z), a vector."
Just E(x,y,z) is your misconception. E is not a scalar field with a gradient. It is an energy assigned to an object, dependent on the position s of the object.
E=E(s), not E=E(x,y,z).
Wolfgang Konle ,
Nothing to say except :
- H. Goldstein, Classical Mechanics (3rd ed.), §1.5–1.7 (potentials as scalar fields).
- J.R. Taylor, Classical Mechanics (2005), §4.2 (potential energy functions U(x,y,z) and forces as gradients).
Essam Allou "potential as scalar fields", "potential energy functions U(x,y,z) and forces as gradients"
Yes, a "potential" and an "amount of energy", are different things.
But F=dE/ds just considers a concrete amount of energy, which depends on the position s of the object which is subject of the force F.
If you cannot grasp this fact, it's your problem.
Wolfgang Konle ,
Really, final word from me. You keep confusing two different notions:
1. A potential energy field U(x,y,z), scalar field on space.
– Standard mechanics defines force as F = –∇U(x,y,z).
– This is a vector field, valid everywhere.
2. The energy of a particle following a path s(t).
– Along that trajectory one has E(t) = U(s(t)),
and the directional derivative is dE/ds = ∇U · t̂, a scalar.
Your "F = dE/ds" confuses the scalar directional derivative with the full gradient. At best it restates the work–energy theorem along a chosen path; it is not a universal law of forces.
Again this is first-year material. Please check in any standard reference:
– H. Goldstein, Classical Mechanics (Ch. 1–2)
– J.R. Taylor, Classical Mechanics (Ch. 4)
Beyond this, continuing the exchange is pointless.
Now, do whatever you want, think whatever you want. You’re only making yourself look ridiculous.
Igor Bayak ; Wolfgang Konle ;
If you really want to know what gravity is, the answer is already derived, not guessed.
https://www.researchgate.net/publication/395890083
Pure Time Theory -from Intuition to Genesis Chapter VI - One Spectrum, One Principle: Operational and Conceptual Unification of GR and QM
Gravity does emerge from a 4D geometry, but not the way you imagine. In Pure Time Theory it follows rigorously from the causal structure of time and its spectral relaxation, giving the exact weak–field projector and Einstein correspondence, with no free hypotheses
Essam Allou "Standard mechanics defines force as F = –∇U(x,y,z)."
"Now, do whatever you want, think whatever you want. You’re only making yourself look ridiculous."
Do you really think that you can enforce your nonsense with such a ridiculous announcement?
If you throw a ball, then you apply a force to that ball with your hand. There is no potential with a gradient to exert this force. Then the ball gains the kinetic energy along the path s where your hand provides the accelerating force. The kinetic energy is finally given by Ekin=∫Fds. Along that path we always have dEkin/ds=F.
Talking about a force caused by the gradient of a potential is fundamentally wrong in this "ball throwing" situation.
Standard mechanics does not generally define all forces as caused by a potential gradient. This is an unfounded claim. That a gravitational force in a conservative system can be expressed in a path independent way by a potential is correct, but the general law surely is F=dE/ds.
In reality a gradient is only applicable to a density. Pressure as an example is such an energy density. But a pressure gradient is not a force. Only the integration over a volume in which we have a pressure gradient leads to a force. We call such a force buoyancy.
But something real, which has a force as a gradient would be pure energy assigned to a position. But at a specific position there only is a point without a volume. We cannot assign energy to such a point. Only an energy density can be assigned to a point. Therefore your potential U is not something real.
Standard physics introduces forces as an example with a steel spring.
The force law is F=D*s with the spring constant D. But even this simple law is again F=dE/ds with the tension energy of the spring E=Ds²/2.
Wolfgang Konle ,
You talk nothing but nonsense. I shouldn’t even reply anymore, but you hid my last message with your nonsense. So I’ll take this chance to try to educate you one more time! And "bordel", open a textbook!
You are still mixing categories and confusing mathematical objects.
1) Conservative vs. non-conservative forces.
– In mechanics:
• If the force is conservative (gravity, spring, electrostatics):
F = –∇U(x,y,z).
• If the force is not conservative (hand pushing a ball, friction, etc.):
you use the work law W = ∫ F · ds and power P = F · v.
So saying “all forces are F = –∇U” would be too broad, but claiming “the general law is F = dE/ds” is simply false.
2) Scalar vs. vector.
– ∇E = (∂E/∂x, ∂E/∂y, ∂E/∂z) is a vector (the gradient).
– dE/ds = ∇E · t̂ is a scalar directional derivative along the tangent t̂.
Confusing these two is a category error. Force F is a vector, not a scalar.
3) On pressure and densities.
The correct continuum law is exactly f = –∇p (force density from a pressure gradient). So your statement “a gradient is not a force” is wrong: it is the standard relation in fluid mechanics.
4) Your spring example.
You write F = D s with energy E = (1/2) D s². That is exactly the conservative textbook case where F = –∇U and U(s) = (1/2) D s². Far from disproving the standard framework, your own example confirms it.
– In conservative cases: F = –∇U.
– In general: Work W = ∫ F · ds, Power P = F · v.
– Your slogan “F = dE/ds” is not general and not even dimensionally correct: dE/ds is a scalar, while F is a vector.
You are only re-stating special conservative cases while denying the general formalism of mechanics.
If you really want to know what gravity is, the answer is already derived, not guessed.
https://www.researchgate.net/publication/395890083
Pure Time Theory -from Intuition to Genesis Chapter VI - One Spectrum, One Principle: Operational and Conceptual Unification of GR and QM
Gravity does emerge from a 4D geometry, but not the way you imagine. In Pure Time Theory it follows rigorously from the causal structure of time and its spectral relaxation, giving the exact weak–field projector and Einstein correspondence, with no free hypotheses
Essam Allou "dE/ds is a scalar, while F is a vector."
I had explained it often enough that generally dE/ds is a vector, a directional derivative.
dE/ds=(∂E/∂sx, ∂E/∂sy, ∂E/∂sz), the inversion of dE=F*ds=
Fx*dsx+Fy*dsy+Fz*dsz
If you always repeat that dE/ds is a scalar though ds is a vector, your claim does not become true.
F=dE/ds is a concrete force applied to an object. This force is generated by an energy E which depends on the position s of the object. This situation does not correspond to a gradient situation.
You can repeat your false claim as often as you want but this does not make your claim true.
To everyone who follows this thread.
This is a general statement on the possible source of gravity.
If F=∇U(x,y,z) were a possible source of the gravitational force F on every object on the Earth's surface, then the potential U(x,y,z) would be a physical amount of energy located at the location (x,y,z).
But how could such an amount of energy equally and at the same time provide exactly the force needed for each individual object? The idea that such a universal physical energy U(x,y,z) could exist is absolutely insane.
But it's also crazy that any object could harbor such an amount of energy, or that a potential would cause each object to have just the right amount of potential energy inside.
The problem would be that the conversion of part of this stored amount of energy into kinetic energy also requires an impulse. And there is no possibility at all that an object could have a supply of momentum at its disposal in addition to a reservoir of energy.
Therefore, we can definitively rule out F=∇U(x,y,z) as a possible source of gravitational force.
The actual source of the gravitational force is described by the law F=dE/ds.
The energy E is contained in the combined field g of the earth and the object. We have E=∫constant*(g(object)+g(earth))²dV.
This energy depends only on the position s of the object, and dE/ds is not a gradient. It is a directed derivative defined as F=(∂E/∂sx, ∂E/∂sy, ∂E/∂sz).
I apologize for my endless and fruitless discussion with Essam Allou on the topic raised in this RG thread.
If you really want to know what gravity is, the answer is already derived, not guessed.
https://www.researchgate.net/publication/395890083
Pure Time Theory -from Intuition to Genesis Chapter VI - One Spectrum, One Principle: Operational and Conceptual Unification of GR and QM
Gravity does emerge from a 4D geometry, but not the way you imagine. In Pure Time Theory it follows rigorously from the causal structure of time and its spectral relaxation, giving the exact weak–field projector and Einstein correspondence, with no free hypotheses
Essam Allou "Gravity does emerge from a 4D geometry, but not the way you imagine."
Sorry, but this story is extremely far fetched and not comprehensible at all.
I had intensively tried to follow and to understand your lengthy and complex descriptions, full of unjustified assumptions, but finally I had to surrender.
My general problem with your assumptions is that you assess them all as fully justified, and for me they fall from the sky.
But the cause for my problem in understanding your arguments obviously are our differences in considering basic physical facts, as we have seen in our recent discussion.
I hope that you consider our dispute as entirely factual without any personal aspect. I appreciate it that now no artificial intelligence seems to contribute to the discussion in an essential way.
Wolfgang Konle ,
Enough confusion. Let’s set the record straight, one last time.
1) On “dE/ds”
– By definition: ∇E = (∂E/∂x, ∂E/∂y, ∂E/∂z) is a vector field.
– Along a path s(t) with tangent t̂ = ds/d|s|, the *directional derivative* is dE/ds = ∇E · t̂, a scalar.
– Force F is a vector. Equating F with dE/ds is a category mistake: you are confusing the scalar directional derivative with the vector gradient.
This is not my invention, it is standard mechanics. Check Goldstein, *Classical Mechanics* §1.5–1.7; Taylor, *Classical Mechanics* §4.2. End of discussion.
2) Conservative vs non-conservative forces
– If the force is conservative (gravity, spring, electrostatics): F = –∇U(x,y,z), U a scalar potential.
– If the force is non-conservative (a hand push, friction): you use work W = ∫ F·ds and power P = F·v.
So no, “F = dE/ds” is not a universal law. It collapses conservative and non-conservative cases into nonsense.
3) On “unjustified assumptions” PTT
This is simply false. The entire Pure Time Theory rests on four axioms A0–A3, clearly stated and justified:
– A0: pure causal time and relaxed spectral time (positive-affine gauge).
– A1: rank drop from 4D cone to 3D base (self-duality preserved).
– A2: cadence operator on annuli (unitary recurrence).
– A3: visibility/ergodicity (positive lower density of detection times).
From these, everything else is derived : weak-field projector, clock law, intrinsic constants, Schrödinger limit, Einstein–Hilbert action. That is the opposite of “falling from the sky.”
So let’s be clear:
– Claiming dE/ds is a vector is a textbook-level error.
– Dismissing a logically derived results as “unjustified assumptions” only shows you never engaged with the actual axioms or you grasp nothing. Bu i ask you anything but shut up if you don't grasp anything. and lecture others !!
You can repeat your private slogans forever, but it won’t change standard mechanics, and it won’t erase the fact that PTT is built on a rigorously minimal and explicit axiom set.
Essam Allou "Enough confusion. Let’s set the record straight, one last time."
Yes, let us stop the unfortunate confusion you are causing.
The force F in the force law F=dE/ds is defined as a directional derivative. If you also want to define another force in another force law F=∇E(x,y,z) as something meaningful, you can do that and explain its meaning as whatever it may mean for you. But you cannot declare your private new definition as something which replaces or denies another definition. Even declaring your definition as an already existing standard does not allow to deny or replace another definition.
In physics and mathematics there is one fix and basic rule. Definitions are untouchable. Only theorems can be challenged.
That you don't accept the rule of the untouchable definition you are just showing with your PTT theory which even tries to redefine time.
But the rule of the untouchable definitions has its reasons. Allowing modifications in definitions would have uncalculable side effects because nobody can know every proof and every theorem in science in which the considered definition has been essential.
Essam Allou "Time will soon tell us. Don't worry"
Switching to philosophy is a good choice for you. Much better than challenging basic definitions in physics and math.
Essam Allou "You can repeat your private slogans forever, but it won’t change standard mechanics, and it won’t erase the fact that PTT is built on a rigorously minimal and explicit axiom set."
With the force law F=dE/ds I am not presenting anything private. This law is basic mechanics and has nothing to do with your PTT.
Yes PTT is based on an independent, rigourously, minimal, and explicit axiom set in parallel to the axiom set of standard mechanics.
But even with this unique and surely remarkable set of axioms, you cannot replace or deny any of the definitions in standard mechanics. This includes all the various indirect definitions of time which implicitly are present in standard mechanics.
With PTT you are trying to establish a new world in parallel to the existing world of physics. But the problem with such a new world is that it is new from scratch and this means you cannot use anything from the old world to compare or explain something in your new world. You must explicitly define everything from scratch in your new world.
But consequently this is not an affordable task.
Wolfgang Konle ,
Normally, your last reply is enough for anyone to see your weak level in physics, modern or even classical. Claiming that “definitions are untouchable” is exactly the opposite of how science has ever progressed. Newton redefined motion, Einstein redefined simultaneity and gravity, quantum mechanics redefined determinism. If physicists had treated definitions as dogma, we would still be stuck with Aristotelian physics.
You clearly do not understand that in physics, definitions are only tools, never sacred truths. They are refined when deeper structures appear. GR did not “deny” Newton; it showed Newtonian laws as approximations in the right limit. Likewise, PTT recovers Newton, GR, QM in their appropriate regimes while reformulating them from a more fundamental causal structure. That is exactly how physics evolves.
So no, your “untouchable definitions” mantra is not wisdom, it only proves you don’t grasp the history or the logic of physics. Real progress comes from challenging and replacing inadequate definitions when a more coherent framework is found.
Thanks for this stupid message, you expose your level and your limit all by yourself!
Essam Allou "Newton redefined motion, Einstein redefined simultaneity and gravity, quantum mechanics redefined determinism."
No, Newton did not redefine motion. You have a problem distinguishing between definitions and theorems.
As an example, velocity is defined as the path traveled per time. Acceleration is defined as the change in velocity per time. But Newton's law F=m*a is a theorem. This theorem is based on a valid definition of F, m, and a. However, sometimes theorems are used in definitions. However, this is only permissible if all entities occurring in it are clearly defined and if the theorem is proven. The network of theorems and definitions must have a valid basis, the axioms. The axioms can be theorems and definitions at the same time. But then any further definition must be derived from the base and can then be added to the base. Only valid definitions from the base can be used in theorems. If a theorem has been proven, it can also be added to the base and used in further definitions.
But no one can remove or deny definitions from the common base. However, it is possible to add new definitions. And it is also possible that old definitions will become obsolete if the defined terms are no longer used.
However, this must first be proven before outdated definitions can actually be removed.
But for someone to enter the system and claim "I have found the absolute truth, we just need to replace a set of definitions as I suggest" is out of the question.
Your way of doing this would immediately lead to inconsistent chaos in science.
Wolfgang Konle ,
Clearly you are really determined to keep making a fool of yourself I reply here only for the sake of science and for young students who might read this thread, because your level is really not what one expects from someone with a doctorate. Please, stop doing physics, for your own sake!
Let’s strip down your long pseudo-logic:
- You claim definitions are untouchable.
- Theorems come afterwards.
- You can add definitions, but never remove or replace them.
- Therefore, if someone (like me with PTT) “redefines” anything, it creates chaos.
The problem
1. You confuse science with a dictionary.
In physics, a “definition” (e.g. absolute time for Newton) is not carved in stone. Einstein *redefined* it with relativity. That’s what progress in science means.
2. You deny the actual history of physics.
– Aristotle: motion requires continuous force.
– Newton: redefined motion with inertia and F = ma.
– Einstein: redefined simultaneity and gravit, what was “force” became spacetime curvature.
– Quantum mechanics: redefined determinism, dynamics became probabilistic.
So your dogma “we cannot touch definitions” is flatly contradicted by 400 years of physics.
3. Your rhetoric is circular.
You pretend to build a logical hierarchy of “axioms and theorems,” but you cite none. You only defend your conceptual comfort zone by shouting “forbidden to redefine!” That is fear of change, not an argument.
In summary:
– You have provided nothing scientific.
– You deny that science evolves by reformulating its core definitions.
– Your position is ahistorical and absurd, since every revolution in physics came from *new definitions*.
PTT simply continues this tradition: it recovers Newton, GR, and QM in their proper limits while reformulating them more deeply. That is what real progress looks like.
So yes, keep repeating your private dogma if you want, but readers here will see clearly: science moves forward by daring to redefine. History proves it.
Essam Allou "In physics, a “definition” (e.g. absolute time for Newton) is not carved in stone. Einstein *redefined* it with relativity. That’s what progress in science means."
You always confuse definitions and theorems. A theorem relates physical entities. But first these entities must be defined because relating undefined items does not make any sense.
Based on well defined items, theorems can be proven or rebutted.
But if later definitions are changed the whole logical network of proven theorems and consistent definitions gets confused.
"Einstein: redefined simultaneity and gravit, what was “force” became spacetime curvature."
Einstein did not at all redefine force or anything else. Relativity is something which extends physics and does not replace anything.
All definitions of physical entities like, mass, energy, time, force, distance, velocity, charge, ... remain exactly valid as they have been originally specified.
But Einstein has introduced new definitions like spacetime, relativistic mass, the Lorentz-factor, the energy-momentum tensor, ...
Wolfgang Konle ,
You say: "All definitions of physical entities like, mass, energy, time, force, distance, velocity, charge, ... remain exactly valid as they have been originally specified."
Yes, but only within their own regime of validity, not at all scales. That is exactly what I am trying to make you understand, but you keep missing the nuance: the notion of a regime.
Simple example:
Newton’s law of force F = ma and universal gravitation F = GMm/r^2 are perfectly valid in the regime of low velocities and weak gravitational fields. But as soon as we approach the speed of light or the horizon of a black hole, this framework breaks down, and General Relativity takes over (spacetime curvature, energy-momentum tensor, etc.).
Newton is not “abolished”; it is an approximation valid in a regime of the more general theory. That is what scientific progress means: understanding the hierarchy of regimes, not confusing local validity with universal truth.
Essam Allou "All definitions of physical entities like, mass, energy, time, force, distance, velocity, charge, ... remain exactly valid as they have been originally specified.
Yes, but only within their own regime of validity, not at all scales."
Are you serious? If yes, you should know that the definition of entities does not have a regime of validity and it has no scales. The definition of an entity only exactly defines/specifies what the entity is. The various properties of entities are subject of theorems.
"Newton’s law of force F = ma and universal gravitation F = GMm/r^2 are perfectly valid in the regime of low velocities and weak gravitational fields. But as soon as we approach the speed of light or the horizon of a black hole, this framework breaks down, and General Relativity takes over (spacetime curvature, energy-momentum tensor, etc.)."
Yes, Newton's theorems are valid in the domain of weak gravitation and of relative velocities much smaller than the speed of light.
But the definition of force, mass, velocity, charge, ..., and of whatever entity is valid in all domains. In the relativistic domain F=m*a must be replaced by F=dP/dt. (P is the momentum)
You really have no plan to differentiate between definitions and theorems. For you, everything is a theorem and can be challenged.
Wolfgang Konle ,
You have now reached the peak of absurdity.
You claim:
– Definitions are “valid in all domains.”
– Only theorems have regimes of validity.
This is simply wrong and historically ridiculous.
1) Time.
– Newton: absolute, universal, ticking independently.
– Einstein: redefined time as relative to observers, linked to spacetime structure.
→ The definition itself changed.
2) Mass.
– Newton: invariant scalar “amount of matter.”
– Relativity: rest mass vs. relativistic mass, energy–momentum tensor.
→ Again, the definition changed.
3) Force.
– Newton: defined by F = ma.
– Relativity: replaced by F = dP/dt with P = γmv.
– In GR: “force” is not even fundamental, free fall is geodesic motion.
→ Your claim that the “definition” of force never changes is simply false.
4) Charge.
– Classical: fixed quantity.
– Quantum field theory: renormalized, scale-dependent, running with energy.
→ Even charge has a regime of validity.
Definitions in physics are not eternal dogmas. They are models of reality, refined and sometimes redefined when experiments and theory demand it. That is how science progresses.
Repeating “definitions are untouchable” only shows you never understood how physics actually works. It is philosophy-by-dogma, not science.
Newton → Einstein → Quantum mechanics → Modern QFT → each time, definitions themselves evolved.
That is not chaos. That is progress.
So please stop embarrassing yourself further. You are arguing against the history of physics itself.
Essam Allou "You claim: Definitions are “valid in all domains.” Only theorems have regimes of validity. This is simply wrong and historically ridiculous."
You still don't understand what definitions are. A definition has not been proven. It has simply come into being and has established itself as a convention.
It just doesn't make sense to question a definition.
In physics, the relationship between a definition and a theorem is sometimes not as strict as in mathematics. Nevertheless, when a definition is changed, everything related to that definition becomes invalid and, this is the main problem, no one knows that the theorems proven with the now modified definition are no longer valid.
If physics were an unsorted pile of unrelated claims, then everything would be modifiable in your sense.
But physics has a strict structure based on mechanics. Emmy Noether has brought a lot of basic knowledge to this structure. Symmetry rules and the associated conservation laws form the backbone of this structure. But this structure is based on the definitions of energy, momentum, charge, mass, force, time, length, ... and on many more basic entities.
Questioning these basic definitions shakes the entire structure of physics. But it is precisely this structure that makes the difference between physics and all esoteric and mystical pseudoscience.
Do you even see what you are questioning and threatening to cause with your unfortunate claims? Do you really want to turn physics into a disjointed collection of vague assumptions, like all the myriad pseudo sciences that are spread on the internet?
But the problem is that you have to know the underlying structure of physics in order to recognize what a valuable asset this structure actually is.
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