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Hypothesis about the formation of particles from fields


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1 hour ago, computer said:

I want to present my hypothesis for your consideration

ehveng.htm 76.17 kB · 1 download

No, you need to present a topic for discussion here, not require readers to click on links to take them elsewhere - and maybe pick up malware in the process. 

According to my limited understanding of QED, particles are modelled as disturbances in fields. Does your hypothesis relate to how these disturbances arise?

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2 hours ago, computer said:

I want to present my hypothesis for your consideration

ehveng.htm 76.17 kB · 1 download

I see someone has been foolhardy enough to download an dot_htm document, which could easily contain harmful code.

 

As @exchemist +1 says you you should post at least a sufficient summary here for discussion to take place.

additional documents and links to references are for links to supporting material to any claims you make here.

 

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3 hours ago, computer said:

I want to present my hypothesis for your consideration

!

Moderator Note

Then present it. Post the material here.

 
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Hypothesis about the formation of particles from fields

 

The hypothesis is an extension of field theory and an attempt to explain the internal structure of elementary particles.

 

Basic equations

 

Presumably, in three-dimensional space there is a field formed by vectors of electric intensity E = (Ex, Ey, Ez), magnetic intensity H = (Hx, Hy, Hz), and velocity V = (Vx, Vy, Vz). Also later in this article, the vectors of electrical induction in vacuum D = ε0 · E and magnetic induction in vacuum B = μ0 · H can be used.

 

E and H are "energy carriers" local density of energy is expressed as follows:

u = ε0/2 · E2 + μ0/2 · H2

where E2 = Ex2 + Ey2 + Ez2 and H2 = Hx2 + Hy2 + Hz2

 

Law of energy conservation: time derivative

u′ = - div W

where W = (Wx, Wy, Wz) is the energy flux vector.

In this case, W = [E × H] + ε0 · (E · V) · E

The scalar product EV = E · V = Ex · Vx + Ey · Vy + Ez · Vz

expresses the cosine of the angle between E and V.

In more detail,

Wx = Ey · Hz - Ez · Hy + ε0 · EV · Ex

Wy = Ez · Hx - Ex · Hz + ε0 · EV · Ey

Wz = Ex · Hy - Ey · Hx + ε0 · EV · Ez

Respectively,

div W = H · rot E - E · rot H + ε0 · E · grad EV + ε0 · EV · div E

 

Derivatives of the magnetic and electric field by time:

H = - 1/μ0 · rot E

E = 1/ε0 · rot H - grad EV - V · div E

In this case, div E is proportional to the local charge density q with a constant positive multiplier: q ~ div E, in the SI measurement system q = ε0 · div E.

 

Having performed the necessary transformations, we get:

u′ = ε0/2 · (2 · Ex · Ex + 2 · Ey · Ey + 2 · Ez · Ez)

+ μ0/2 · (2 · Hx · Hx + 2 · Hy · Hy + 2 · Hz · Hz)

= Ex · (∂Hz/∂y - ∂Hy/∂z - ε0 · ∂EV/∂x - ε0 · Vx · div E)

+ Ey · (∂Hx/∂z - ∂Hz/∂x - ε0 · ∂EV/∂y - ε0 · Vy · div E)

+ Ez · (∂Hy/∂x - ∂Hx/∂y - ε0 · ∂EV/∂z - ε0 · Vz · div E)

- Hx · (∂Ez/∂y - ∂Ey/∂z) - Hy · (∂Ex/∂z - ∂Ez/∂x) - Hz · (∂Ey/∂x - ∂Ex/∂y)

= E · rot H - H · rot E - ε0 · E · grad EV - ε0 · EV · div E = - div W

 

A term in the form of "grad EV" for E′ arises from the need to make an adequate expression of the energy conservation law, and although in the "natural" structures discussed below E is everywhere perpendicular to V, that is, EV = 0, it can play a role in maintaining the stability of field formations.

 

Velocity derivative by time

 

From the point of the energy-flux view, the time derivative V′ can be any expression, but should not contain a common multiplier V or 1 - V2/c2, since when approaching zero or the speed of light, the vector would practically cease to change locally, which contradicts many experimental facts and theoretical studies. The most likely are the two-membered constituents for V′, where one part contains V as a multiplier in the scalar or vector product, the second does not.

 

For example, the pure field similarity of the Lorentz forces is of interest:

V′ ~ (D · V2 - [H × V]) · div E

where V2 = Vx2 + Vy2 + Vz2

 

The expressions D · V2 and H × V have the same dimension, A/s in SI, and after multiplying by the div E, it is still necessary to enter a coefficient to convert the resulting units into acceleration m/s2. The numerical value of the coefficient will probably have to be determined experimentally.

 

Although there are no strict restrictions on the absolute value of V, as we shall see later, for field formations common in nature, it is uncharacteristically |V| > c, and the speed of light is achieved at a mutually perpendicular arrangement of E, H, and V, when the local "E-energy" is equal to "H-energy", that is, E2 ~ 1/ε0, H2 ~ 1/μ0.

 

The exception is artificially created or simulated on the computer situations. Another hypothetical set of terms for the velocity derivative over time is V′ ~ W - u · V. In the models of particles discussed below, in this case, there is a "longitudinal" effect on the velocity vector, in contrast to the "transverse" one under the influence of an electric and magnetic field, with the mutual perpendicularity of all three vectors.

 

If indeed V′ ~ W - u · V, then although there is still no hard limit |V| ≤ c, the unlimited increase of the velocity in the absolute value is more explicitly limited by the member u · V with a negative sign. If the magnetic or electric field somehow disappears, the velocity will rush to zero, although the energy density may remain non-zero. Modulus of V reaches its maximum value (= c) when E and H are perpendicular and ε0/2 · E2 = μ0/2 · H2.

 

When the charged particle is in an external electric field, like created by another particle in the vicinity, due to the multiplier V2 in the expression V′ ~ (D · V2 - [H × V]) · div E

is independent of the sign of V, and the presence of significant velocities close to the speed of light inside the particle, the total acceleration acts in one direction (on average, although internal deformations may occur).

 

In an external magnetic field the velocity vector is involved in the first degree, in any projection about half of the currents are directed in one direction, and about half in the opposite direction, so only internal deformations occur. The shift of a particle as a whole is observed when it moves in an external magnetic field.

 

Let us consider the alleged structure of some elementary particles. To do this, we will use a cylindrical coordinate system (ρ,φ,z), where ρ2 = x2 + y2, φ is the angle counted from the positive direction of the x-axis counterclockwise if it is directed to the right, the y-axis upwards, and the z-axis is directed towards us (the right coordinate system). Also, for the particles under consideration, we will set the condition of cylindrical symmetry, that is, ∂/∂φ = 0 for any variables.

 

Neutral particles moving at the speed of light

 

Consider the motion of an "almost point" charge, spherically symmetrical at rest, along the z-axis with a constant velocity V = Vz. If the field structure moves as a whole unit, then

∂/∂t = - Vz · ∂/∂z for all variables.

As mentioned earlier, with cylindrical symmetry, ∂/∂φ = 0.

 

Let the scalar potential at rest to be

a = E0 / s, where s2 = R2 + ρ2 + K · z2

Here, the letter "a" is used to avoid confusion with the angle φ, the constant E0 expresses the amplitude of the field, the constant R characteristic dimensions (similarity of the wavelength or frequency of the wave), the constant K shows the likely deformation of the field along the axis of motion, as "compression" or "stretching" of the lines of force.

Note that ∂s/∂ρ = ρ / s, ∂s/∂z = K · z / s

 

The vector potential A = Az = E0 / s · Vz / c2 is directed along the z-axis,

Aρ = 0, Aφ = 0. H = 1/μ0 · rot A forms closed rings around z-axis,

Hρ = 1/μ0 · (- ∂Aφ/z) = 0

Hφ = 1/μ0 · (∂Aρ/∂z - ∂Az/∂ρ) = E00 · Vz / c2 · ρ / s3 = E0·ε0 · Vz · ρ / s3

Hz = 1/μ0 · (∂Aφ/∂ρ + Aφ / ρ) = 0,

since ε0 · μ0 = 1 / c2.

 

Let be d = 1/μ0 · div A = 1/μ0 · ∂Az/∂z

= - E00 · Vz / c2 · K · z / s3 = - E0·ε0 · Vz · K · z / s3

Obviously a = - 1/ε0 · d = - c2 · div A,

since a = - Vz · ∂a/∂z = E0 · Vz · K · z / s3

that corresponds to the classical field equations.

 

E we will find from the field equation A = - E - G

where G = grad a

Gρ = ∂a/∂ρ = - E0 · ρ / s3

Gφ = 0

Gz = ∂a/∂z = - E0 · K · z / s3

Eρ = - Aρ - Gρ = Vz · ∂Aρ/∂z - Gρ = E0 · ρ / s3

Eφ = 0

Ez = - Az - Gz = Vz · ∂Az/∂z - Gz = E0 · (1 - Vz2 / c2) · K · z / s3

As can be seen, when Vz raises to the speed of light, Ez "disappears" and only the non-zero radial Eρ, perpendicular to the z-axis, remains.

 

The time derivatives G′, d′, H′ are computed trivially, and the time derivative E′ requires special attention.

 

Let be J = rot H,

Jρ = - ∂Hφ/∂z = E0·ε0 · 3 · Vz · K · ρ · z / s5

Jφ = 0

Jz = ∂Hφ/∂ρ + Hφ / ρ = E0·ε0 · Vz · (2 / s3 - 3 · ρ2 / s5)

 

div E = ∂Eρ/∂ρ + Eρ / ρ + ∂Ez/∂z

= E0· {2 / s3 - 3 · ρ2 / s5 + (1 - Vz 2 / c2) · K · (1 / s3 - 3 · K · z2 / s5)}

 

Must be observed equality

Eρ =  - Vz · ∂Eρ/∂z = E0 · 3 · Vz · K · ρ · z / s5

According to a hypothetical equation, and since E is perpendicular to V (E · V = 0), and Vρ = 0:

Eρ = 1/ε0 · Jρ - ∂EV/∂ρ - Vρ · div E

= 1/ε0 · E0·ε0 · 3 · Vz · K · ρ · z / s5 = E0 · 3 · Vz · K · ρ · z / s5

 

Also must be observed equality

Ez =  - Vz · ∂Ez/∂z = - E0 · Vz · (1 - Vz2 / c2) · K · (1 / s3 - 3 · K · z2 / s5)

According to the hypothetical equation, and since E is perpendicular to V (EV = 0):

Ez = 1/ε0 · Jz - ∂EV/z - Vz · div E

= 1/ε0 · E0·ε0 · Vz · (2 / s3 - 3 · ρ2 / s5)

- Vz · E0 · {2 / s3 - 3 · ρ2 / s5 + (1 - Vz 2 / c2) · K · (1 / s3 - 3 · K · z2 / s5)}

= - E0 · Vz · (1 - Vz 2 / c2) · K · (1 / s3 - 3 · K · z2 / s5)

 

The equations for E′ are true for any Vz, E0, R and K, but if Vz = c and Ez = 0, the all-space integral of div E multiplied by volume unit is zero:

 

-∞+∞02·π0 (2 / s3 - 3 · ρ2 / s5) · ρ ∂ρ ∂φ ∂z = 0

 

That is, when accelerating to the speed of light, the whole field formation will be charged neutrally, although locally the charge density changes.

 

Probably a similar structure, relatively simple, have neutrinos, and we will also use the given example of the direct motion of a particle to check the adequacy of expressions for V′.

 

If V′ ~ (D · V2 - [H × V]) · div E

then Vρ′ ~ 0 · Eρ · Vz2 - Hφ · Vz) · div E

= (ε0 · E0 · ρ / s3 · Vz2 - E0·ε0 · Vz · ρ / s3 · Vz) · div E = 0

and the radial velocity remains zero. If D · div E would be used in the "similarity of the Lorentz forces" without multiplication by V2, like the classical effect of an electric field on charge, equality would not be observed.

 

Also Vz ~ 0 · Ez · Vz2 + Hφ · Vρ) · div E

= {ε0 · E0 · (1 - Vz2 / c2) · K · z / s3 · Vz2} · div E

Vz = 0 and the velocity Vz remains constant only at Vz = c.

Apparently, this is related to the fact that experimentally detected neutrino-like particles move at the speed of light, while the movement of massive particles at low speeds is a much more complex process at the field level.

 

When Vz = c and the electrical energy density uE = ε0/2 · E2 is equal to the magnetic energy density uH = μ0/2 · H2, the equality W = u · V is also fulfilled, which is an argument in favor of considering the hypothesis V′ ~ W - u · V. The same fact occurs in field structures with zero electric field divergence (electromagnetic waves and presumably photons).

 

Stable charged particles with cylindrical symmetry

 

Probably, the basis of electrons and other leptons is the E, H, V field in the form of closed rings of energy flow and velocity vector. The structure of the particle is not similar to the classical "infinitely thin" circuit with an electric current, where on the elements of the ring the electric and magnetic field differ from zero, due to the fields created by other parts of the circular current.

 

On the line of "main circuit", the electric and magnetic field is zero, whereas the charge density (~ div E) is close to the local maximum.

 

In computer modeling in a cylindrical coordinate system, the following field values can be a good initial approximation (with s2 = R2 + ρ2 + z2😞

vector potential A = Aφ ~ ρ3 / s5

Hρ ~ - ∂Aφ/∂z ~ 5 · ρ3 · z / s7

Hφ = 0

Hz ~ ∂Aφ/∂ρ + Aφ / ρ ~ 4 · ρ 2 / s5 - 5 · ρ 4 / s7

Jρ = 0

Jφ = ∂Hρ/∂z - ∂Hz/∂ρ ~ - 8 · ρ / s5 + 10 · ρ3 / s7 + 35 · ρ3 · R2 / s9

Jz = 0

scalar potential a ~ ρ 4 / s5

Eρ ~ - ∂a/∂ρ ~ - 4 · ρ 3 / s5 + 5 · ρ 5 / s7

Eφ = 0

Ez ~ - ∂a/∂z ~ 5 · ρ 4 · z / s7

div E = ∂Eρ/∂ρ + Eρ / ρ + ∂Ez/∂z

~ - 16 · ρ2 / s5 + 20 · ρ4 / s7 + 35 · ρ4 · R2 / s9

 

Note that Eρ · Hρ + Ez · Hz = 0, E is perpendicular to H everywhere.

 

Near the center of the particle is a region where the values div E and rot H are opposite in sign to those found in the rest of space. Meanwhile, V has the same sign everywhere. If the conditional magnetic dipole is directed along the z-axis, with a positive multiplier for A and H, and the total charge of the particle is positive, then near the center there will be a region with a negative divergence E and a negative rotor H, but at the great distance these values are positive.

 

The velocity is positive everywhere, that is, it is directed counterclockwise with the direction of the z-axis towards us and the x-axis (the start of the counting φ) to the right. div E and rot H must change the sign synchronously so that equality is observed:

E = 1/ε0 · rot H - grad EV - V · div E = 0

since in a more or less stable particle all fields derivatives in time are zero.

EV = 0, V is perpendicular to E and H, that is, we are talking about a mutually perpendicular triple of vectors in any combination.

 

A model with a field arrangement closer to the z-axis, for example:

A = Aφ ~ ρ / s3

Hρ ~ 3 · ρ · z / s5

Hφ = 0

Hz ~ 2 / s3 - 3 · ρ2 / s5

a ~ ρ2 / s3

Eρ ~ - 2 · ρ / s3 + 3 · ρ3 / s5

Eφ = 0

Ez ~ 3 · ρ 2 · z / s5

where on the z-axis there is a pronounced maximum of Hz

and J = Jφ = ∂Hρ/∂z - ∂Hz/∂ρ ~ 15 · R2 · ρ / s7

does not fit, because rot H is positive everywhere, and div E changes sign in the central part. Models with a spherically symmetric scalar potential and electric field are even more inadequate:

a ~ 1 / s

Eρ ~ ρ / s3

Eφ = 0

Ez ~ z / s3

div E ~ 3 · R2 / s5

there E is not even perpendicular to H, if we take A and H from the previous model.

 

If V′ ~ (D · V2 - [H × V]) · div E

then the multiplier V2 prevents the destruction of the particle due to electrical repulsion near the z-axis, although exactly onto it can be also D = 0. With V = Vφ, Vρ = 0 and Vz = 0, there must inevitably be a zone with zero velocity near the z-axis, since the values of the existing quantities in the physical world are finite and only smooth functions with continuous derivatives are permissible.

 

Precise solutions in degrees of s for real lepton-like particles seem impossible, numerical simulations are required. In the first time after setting the initial approximation, occurs a rapid adaptation of the fields to more accurate values, further stability depends on the adequacy of the model and the accumulation of numerical errors.

 

Let's try to estimate the magnitudes of what order are the fields at a considerable distance from the z-axis in the perpendicular plane (z = 0, s ≈ r, r2 = ρ2 + z2). From experimental data and works on classical physics it is known that E = Eρ ~ 1 / r2, H = Hz ~ - 1 / r3, u ≈ E2 ~ 1 / r4

D · V2 - [H × V] will tend to zero if V = Vφ ~ 1 / r

The same order of Vφ ~ 1 / r follows from the equation

W - u · V = 0, given that W = Wφ = - Eρ · Hz ~ 1 / r5

 

Further directions of research

 

All the above mentioned equations are linear in E and H. That is, when these vectors are both multiplied by the same number, they remain true. Since the particles observed in the physical world have strictly defined charges and masses (internal energies), it is logical to assume that the expression V′ may contain nonlinear terms with relatively small factors. Although due to statistical factors, greater resistance to random disturbances, some field formations may be more stable than others.

Excuse me, I thought the nice HTML-file here would be turned into plain text, so attached it.

The main differences from the more classical approach are:

1) Velocity vector V had introduced as an "independent" and essential part of field,

as physical reality along with E and H

2) The expression for the energy flux W was changed, with an additional member ε0 · (E · V) · E

and I hope someone will conduct experiments to confirm or reverse that assumption

3) There are non-trivial suggestions, what "forces" may affect the velocity derivative over time

4) Assumptions about the internal structure of electron-like particles, that inside the main part with a predominant charge

is a zone with a charge of the opposite sign, and due to what factors it arises

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OK you are serious.

We are all very glad to learn this.

Welcome, my apologies for doubting, but unfortunately there are too many nasty people about and this site encounters its fair share of them.

 

Looking quickly at the development of your discussion, a couple of things stand out.

 

1) I see no sinks or sources in your treatment.  These are very important in the real world. Are you assuming that these three fields were and are always present everywhere and everywhen ?

2) I see no discussion of the clash between the continuous Maxwell field theory (which is compatible with both special and general relativity) and the Plank relationship E =hv which defines Quantum theory.

I look forward to your answers

 

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studiot, if you mean "infinitely small" points-particles by sources of field, it's a simplified abstraction from electrodynamics.
Of course, in the macroscopic, and even in the microscopic world sources are very important.
But I talk about "pure" field theory, where is explained the very internal structure of particles.
Here is more accurate level.

These three vectors CAN exist everywhere, but at the time of observation all or part of them may be zero at a certain point.
As an example: near the pole of a permanent magnet only the magnetic field is non-zero, E and V are equal to zero.
 

Тhe experimental ratio E = hv is true for photons, and there is a need to find out how photons are arranged.
What is "v" in this case. If we consider a monochromatic wave, it is infinite in space and carries infinite energy.
Obviously, there are no such formations in the real world. If a photon has restricted dimensions and form-factors,
like amplitude E0 and positive constant R in my article, it is necessary to explain, why photons are the way they are.
In the real world most photons are emitted during electron transitions between energy levels,
and they can "inherit" the amplitude E0 of electrons, and differ only by R, that experimenters conventionally call
frequency or wavelength. Maybe I was wrong and the neutral particles I described are actually photons, not neutrinos.
 

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4 hours ago, computer said:

studiot, if you mean "infinitely small" points-particles by sources of field, it's a simplified abstraction from electrodynamics.
Of course, in the macroscopic, and even in the microscopic world sources are very important.
But I talk about "pure" field theory, where is explained the very internal structure of particles.
Here is more accurate level.

No I don't mean particles of any description or size by a source or a sink.

Sources and sinks are an inseparable part of Field Theory in general and Maxwell's field theory in particular.
If you are developing a Field theory, you should understand this and provide for them.

You magnetic pole example is such a source, the other pole is then a sink.

 

5 hours ago, computer said:

Тhe experimental ratio E = hv is true for photons, and there is a need to find out how photons are arranged.
What is "v" in this case. If we consider a monochromatic wave, it is infinite in space and carries infinite energy.
Obviously, there are no such formations in the real world. If a photon has restricted dimensions and form-factors,
like amplitude E0 and positive constant R in my article, it is necessary to explain, why photons are the way they are.
In the real world most photons are emitted during electron transitions between energy levels,
and they can "inherit" the amplitude E0 of electrons, and differ only by R, that experimenters conventionally call
frequency or wavelength. Maybe I was wrong and the neutral particles I described are actually photons, not neutrinos.

 

Again you missed my point.

Maxwell's equations lead directly to a (continuous) wave theory, which is incompatible with photons.

See in particular 8.7 here

Quote

Lipton and Lipton  Optical Physics  Cambridge University Press

 

photons1.thumb.jpg.942bbb35086e7e37336b9928b08a8908.jpg

 

 

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19 hours ago, studiot said:

magnetic pole example is such a source, the other pole is then a sink

Magnetic poles exist in "large" objects, where many electrons are oriented in one direction. At the pure field level, there are only closed magnetic field lines, as in the example I gave:

Hρ = 3 · ρ · z / s5

Hφ = 0

Hz = 2 / s3 - 3 · ρ2 / s5

It is in a some sense mathematically simplest dipole, with minimal integer powers of (s2 = R2 + ρ2 + K · z2).

Тhe magnetic field lines are always closed, since it is formed only by the rotor (curl) of E. Even if in the early stages of the formation of the Universe were not closed, they quickly became so, since natural fields are in motion and pass into each other.

Regarding the pages of the excerpt from the book you cited, I want to note:
It's not clear what mass means. If it is a measure of inertia, the ability to change direction of motion under the influence of forces, but light is deflected under the influence of huge attractions in cosmic galaxies, and photon mass is not zero. If mass has a close relationship with internal energy, it would be logical to call mass density simply a quantity u / c2

The difference between particles and waves is more technical than philosophical. It is only necessary to find mechanisms for "self-assembly" of fields into more stable formations, with sufficient for this internal energies.

Although there are few truly stable particles: electrons (positrons), protons (+anti), photons and neutrinos. Muons and neutrons still have quite a long lifetime. The rest it seems to be very short-term field spikes.
 

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7 hours ago, computer said:

but light is deflected under the influence of huge attractions in cosmic galaxies

No light is not 'deflected' by gravity.  Spacetime is warped by gravity and a photon travels in a straight line through the warped spacetime.

7 hours ago, computer said:

and photon mass is not zero

Do you have citation for this claim.

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I don't want to be nit picky.

Especially as I sometimes use the analogy of eddies in a stream to discuss the idea of a semi permanent structural feature in a stream to represent what is meant by the quantum idea of particles being just such structures in quantum fields.

But here we are discussing classicla EM fields as far as I know.

 

So  a few comments on the text.

 

15 hours ago, computer said:

Тhe magnetic field lines are always closed, since it is formed only by the rotor (curl) of E.

The electric field E has zero curl, it is the magnetic field that has non zero curl.

epsilon and mu are not scalar constants , except in special circumstances, actually they are second order tensors that interact with the E and D   and   B and H fields which may result in these not pointing in common directions.

The energy density is one half E cross D and one half B cross H for the electric and magnetic fields respectively.

 

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15 hours ago, Bufofrog said:

Spacetime is warped by gravity and a photon travels in a straight line through the warped spacetime.

Thanks for clarifying. It seems that due to this there is also an attraction of massive bodies, they slide into a common gravitational pit.

Although inertia when a moving body is exposed to an electric or magnetic field may have a different origin.

6 hours ago, studiot said:

quantum idea of particles being just such structures in quantum fields

I had not seen detailed descriptions of the internal structure of particles or formulas suitable for computer simulation within the framework of quantum theories.

Explanations, suitable for implementation, end with the Schrödinger equation or some of its modifications, which is really useful in the modeling of atoms and molecules.

6 hours ago, studiot said:

The electric field E has zero curl, it is the magnetic field that has non zero curl

The curl of the electric field is zero in some stable formations, like electrons, where the field can be expressed as a gradient of the scalar potential. But in a radio wave the curl is not zero.

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3 hours ago, computer said:

I had not seen detailed descriptions of the internal structure of particles

That’s because elementary particles - as opposed to composite particles - do not have any internal structure; that’s why they’re termed ‘elementary’.

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12 minutes ago, Markus Hanke said:

That’s because elementary particles - as opposed to composite particles - do not have any internal structure; that’s why they’re termed ‘elementary’.

They must have an internal structure. Another thing is that it is unknown to us, so this internal structure is a "black box" for us.

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8 hours ago, SergUpstart said:

They must have an internal structure. Another thing is that it is unknown to us, so this internal structure is a "black box" for us.

What is the evidence that they have an internal structure? We've found evidence for all the particles we've discovered that have an internal structure - that we can cause excitations, and that we can break them up in some way.

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3 hours ago, swansont said:

What is the evidence that they have an internal structure?

Here is the question in this regard? The particle has energy E=mc^2. Only a system of objects can possess energy. So what is the nature of this energy E=mc^2? There are two possible answers.

1. This is the interaction energy of the component parts of the particle.
2. This is the energy of interaction of a particle with the entire universe, based on C^2=-Phi, E=-mPhi

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1 hour ago, SergUpstart said:

Here is the question in this regard? The particle has energy E=mc^2. Only a system of objects can possess energy.

What’s your evidence of the bolded statement?

1 hour ago, SergUpstart said:

So what is the nature of this energy E=mc^2? There are two possible answers.

1. This is the interaction energy of the component parts of the particle.
2. This is the energy of interaction of a particle with the entire universe, based on C^2=-Phi, E=-mPhi

Or it’s the energy something has owing to is mass.

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6 hours ago, swansont said:
7 hours ago, SergUpstart said:

Only a system of objects can possess energy.

What’s your evidence of the bolded statement?

Potential energy is always the energy of a system of interacting bodies. Spring - the atoms in the spring interact, the potential energy of the suspended load W = mgH, - the load interacts with the gravity source, the the charged capacitor - charges on its plates interact in it....

The kinetic energy mv^2/2 is not related to the interaction, but a second object is still needed, with which the reference frame is connected, in which the velocity equal to v.

E=mc^2 is the potential energy.

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21 hours ago, Markus Hanke said:

do not have any internal structure

It seems that the authors of such statements are referring to a "very simple" structure. For example, an electric field of the form

Eρ ~ ρ / s, Eφ = 0 , Ez ~ z / s3

and magnetic

Hρ ~ 3 · ρ · z / s5 , Hφ = 0 , Hz ~ 2 / s3 - 3 · ρ2 / s5

with some effective size parameter R, s2 = R2 + r2 because we can not divide by zero at the center if we write ρ / ror z / r3

But it is impossible to find detailed descriptions of experiments that would give rise to unambiguous conclusions about a very simple structure.

The same goes for the "size" of electron. Electron microscopy resolution is about 10-9 m, even with advanced software processing. Estimation of 10-13 or 10-15 m arises from theoretical reasoning, as if all the mass of electron was in the squares of electric and magnetic field.

9 hours ago, SergUpstart said:

Only a system of objects can possess energy

A system of objects may have "excess" energy. If two electrons are forcibly approached, energy will be stored in the grown square of the electric field within and between them. Then released on repulsion and converted into excess of magnetic field square, if electrons are released.

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On 8/4/2022 at 5:39 PM, SergUpstart said:

They must have an internal structure.

Elementary particles don’t have an internal structure, because they are local excitations of quantum fields. Such fields don’t have structures. In tech speak, elementary particles are irreducible representations of symmetry groups.

There is no physical principle requiring all particles to have internal structure.

11 hours ago, SergUpstart said:

Only a system of objects can possess energy.

E=mc^2 has nothing to do with potential energy, which is what you must be referring to in this statement, or else it doesn’t make sense.

11 hours ago, SergUpstart said:

So what is the nature of this energy E=mc^2?

It’s the energy equivalent of the particle’s rest mass, and this relationship is true only in a massive particle’s rest frame. This has nothing to do with any potentials or internal structures.

4 hours ago, SergUpstart said:

Potential energy is always the energy of a system of interacting bodies.

E=mc^2 has nothing to do with potential energy, nor internal structure. Only with rest mass.

4 hours ago, SergUpstart said:

E=mc^2 is the potential energy.

No, see above.

3 hours ago, computer said:

It seems that the authors of such statements are referring to a "very simple" structure.

No, I am saying that there is no internal structure, according to current understanding of the Standard Model. Elementary particles are irreducible - and that’s true for all of them, irrespective of whether they carry electric charge, colour charge, flavour, isospin, or mass.

To experimentally verify the elementary-ness of such particles, you use a technique called deep inelastic scattering. This is, however, limited by the available energy of the accelerator.

Proposing internal structure for these particles means you need to introduce new physics.

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1 hour ago, Markus Hanke said:

E=mc^2 has nothing to do with potential energy, nor internal structure. Only with rest mass.

Isn't the explosion of a thermonuclear bomb an act of releasing part of the potential energy contained in the mass of a nuclear charge?

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8 hours ago, SergUpstart said:

Potential energy is always the energy of a system of interacting bodies.

But there is no interaction. It’s just a particle.

8 hours ago, SergUpstart said:

Spring - the atoms in the spring interact, the potential energy of the suspended load W = mgH, - the load interacts with the gravity source, the the charged capacitor - charges on its plates interact in it....

The kinetic energy mv^2/2 is not related to the interaction, but a second object is still needed, with which the reference frame is connected, in which the velocity equal to v.

E=mc^2 is the potential energy.

 

Mass energy is mass energy. Interactions can affect it, but they are not the source of energy for fundamental particles. 

 

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22 hours ago, SergUpstart said:

Isn't the explosion of a thermonuclear bomb an act of releasing part of the potential energy contained in the mass of a nuclear charge?

It’s a bit more complicated than that - a fission bomb fuses hydrogen atoms into heavier elements. Initially you need to supply energy to do this, in order to overcome the electromagnetic repulsion between protons (among other things). However, once the protons are close enough for the residual strong force to kick in, they “fall” into its respective potential well, forming a stable nucleus - which is structured such that you end up with a net surplus energy at the end of the interaction. 

There’s actually a lot more going on, but that’s the gist of it.

I reiterate again that E=mc^2 works only for massive particles at rest - it’s a special case of the energy-momentum relation, which is in turn just the magnitude of the energy-momentum 4-vector. 

Edited by Markus Hanke
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2 hours ago, Markus Hanke said:

I reiterate again that E=mc^2 works only for massive particles at rest - it’s a special case of the energy-momentum relation, which is in turn just the magnitude of the energy-momentum 4-vector. 

But the interpretation of the energy E is scalar through the stress-energy-momentum tensor.

It's just for confirmation.

Thanks.

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9 hours ago, Markus Hanke said:

a fission bomb fuses hydrogen atoms into heavier elements

Fusion, though fission is used to create the high temperature and pressure.

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