Hypothesis about temperature (split from Physical mechanism how matter absorbs radiation.)

[martillo]

1 hour ago, swansont said:

They could have a next atom, but it would be some distance away. But the shouldn’t matter much to photons. And an atom doesn’t “know” where the photon it absorbs came from. 

The atom doesn't "know" that but momentum must be conserved so if the atom absorbs a photon coming in one direction it will "try" to re-emit it in the same direction. I mentioned "try" because the entire lattice will affect the emission direction and this way the path could curve. Conservation of momentum would play an essential role in the formation of the "linear" paths of conduction inside the metal. 

Other subject I must mention is that while an atom is not able to emit a photon it would be not able to absorb other photon and the same would happen with all the atoms in the entire path of absorption/re-emission I mentioned. This means that if in a conductor the conductance is broken at one of its ends it would not be able to absorb more photons in the other end and it would just reflect the incoming photons stopping the conduction of the heat from this side.

 

50 minutes ago, Mordred said:

truthfully put aside any concern on what the mediator particle is. It really doesn't matter as they can be used offshell being a mediator boson. Reduce the problem to 2 forms of energy. Potential energy and kinetic energy. Study the linear equations first via studying the mathematics of a classical spring. A graph of its momentum over time is dipolar just as the case with the EM field. This will further equate to the harmonic oscillations as well as the wave equations of QM/QFT. Make sure you have a good understanding of the energy/work/power/momentum relations involved. (all included in any decent book on thermodynamics thus far.).

 study each case separately conduction, convection and radiation. You should note the relationship between the potential energy terms and the kinetic energy terms an objects conductivity can be described via the spring and those energy terms.

After that for a field treatment for temperature use a scalar field this is also included in any decent book on the topic.

you will need to understand wave equation treatments if you plan on applying QM/QFT.

 

I appreciate your comment but I have another very ambitious plan. I pretend to to not enter in the "Wave Mechanics Theory" which is about half of the broad "Quantum Physics Theory" as far as I could.

I must mention the following:

50 minutes ago, Mordred said:

truthfully put aside any concern on what the mediator particle is. It really doesn't matter as they can be used offshell being a mediator boson.

The mediator particle is important to me. My intention is to work with known real particles and not any virtual particle which are for me a mathematical artifice to make the things work (they have never been proved to really exist). They have the value to maintain the thing within the particles-like' model and not invoke the wave-like model of them which is also something of my interest but I pretend a model without them.

Edited March 30 by martillo

[swansont]

2 hours ago, martillo said:

The atom doesn't "know" that but momentum must be conserved so if the atom absorbs a photon coming in one direction it will "try" to re-emit it in the same direction. I mentioned "try" because the entire lattice will affect the emission direction and this way the path could curve. Conservation of momentum would play an essential role in the formation of the "linear" paths of conduction inside the metal. 

No. Laser cooling (Nobel prize 1997) wouldn’t work if the photon was emitted in the same direction, since no net momentum would be imparted to the atom. But that’s not what happens. The emission is symmetric and not preferential, so momentum is imparted to the atom. (And depending on the specifics, could heat or cool the atoms. Cooling is usually more experimentally useful)

The momentum would be imparted regardless of separation distance. 

2 hours ago, martillo said:

Other subject I must mention is that while an atom is not able to emit a photon it would be not able to absorb other photon and the same would happen with all the atoms in the entire path of absorption/re-emission I mentioned. This means that if in a conductor the conductance is broken at one of its ends it would not be able to absorb more photons in the other end and it would just reflect the incoming photons stopping the conduction of the heat from this side.

This wouldn’t seem to explain the T vs T^4 difference we observe for conduction vs radiation.

[martillo]

1 hour ago, swansont said:

No. Laser cooling (Nobel prize 1997) wouldn’t work if the photon was emitted in the same direction, since no net momentum would be imparted to the atom. But that’s not what happens. The emission is symmetric and not preferential, so momentum is imparted to the atom. (And depending on the specifics, could heat or cool the atoms. Cooling is usually more experimentally useful)

I have read the subject at Wikipedia and there it is said that laser cooling works due to the Doppler effect on moving atoms only. It is said that static atoms are not affected by the laser beam and they pass through the atoms maintaining their original momentum. For me a photon cannot just pass through the nucleus of an atom but their photons are absorbed and re-emitted while the momentum is maintained. The atoms in a metal lattice are stationary and so they can absorb and re-emit photons with that Doppler effect which maintain their original momentum as I have sustained.

1 hour ago, swansont said:

This wouldn’t seem to explain the T vs T^4 difference we observe for conduction vs radiation.

Conduction involves two bodies with some temperature each. This heat transfer is governed by the Thermodynamics laws in which the energies are proportional to KT and the heat transfer is linearly proportional to the temperatures' difference. A metal conductor acts just as a perfect mediator between the bodies for the heat transfer and so its conductance present the same linear proportionality in T. Not perfect conductors would present a gradient of temperature inside and their conductance would not present a linear relation with T. Vacuum is one such type of conductors in which the heat transfer presents a relation to T⁴.

Edited March 31 by martillo

[Mordred]

4 hours ago, martillo said:

 

 

The mediator particle is important to me. My intention is to work with known real particles and not any virtual particle which are for me a mathematical artifice to make the things work (they have never been proved to really exist). They have the value to maintain the thing within the particles-like' model and not invoke the wave-like model of them which is also something of my interest but I pretend a model without them.

Well the more in depth you get into particle physics you will find the distinction between Real , quasi, and virtual particles becomes less and less important. Phonons however are not virtual particles they are quasi particles. They measurable properties of which they represent. Both real and quasi-particles are field excitations in so far as they have a minimal of a quanta of action. 

Individual virtual particles however do not hence they are oft treated as permutations in field theory as opposed to an excitation. You can never measure individual virtual particles its impossible. Now here is where the distinction becomes important. In Feymann path integrals virtual particles are represented by the internal lines as a type of field propagator.

 Quasi and real particles due to the same requirements of field excitation (localization of a wavefunction) fall on the external lines on path integrals. In QFT this doesn't really matter  as all particles are simply states where all particles are field excitations represented by that state. QFT is operator driven so it uses the creation/annihilation operator for thermodynamic treatments as well as for particle production.  Its variation of all these classical thermodynamic laws for conduction, convention and radiation apply those operators via the Euler Langrangian equations for its path integrals

for example Bose Einstein in classical radiation treatment which is used to calculate the number density of bosons at a given blackbody temperature looks like this

\[n_i = \frac {g_i} {e^{(\varepsilon_i-\mu)/kT} - 1}\]

PS at some point you will need this equation as well as Fermi-Dirac and Maxwell Boltzmann.

under QFT looks like this

Bose Einstein QFT format.

\[|\vec{k_1}\vec{k_2}\rangle\hat{a}^\dagger(\vec{k_1})\hat{a}^\dagger(\vec{k_2})|0\rangle\]

\[\Rightarrow |\vec{k_1}\vec{k_2}\rangle= |\vec{k_2}\vec{k_1}\rangle\]

note both versions use phase/momentum space. Practice your vector algebra lol you will need it. Anyways enough about which mediator to use. It isn't particularly important at this stage except to be aware of the distinction of its usage in anything you read on conduction vs radiation treatments. 

note those distinctions may or may not be compatible with photons with regards to phonons how they are used may mathematically differ.

for example an electron hole can behave the same as an electron but it isn't a particle but can be represented by a quasi particle

Edited March 31 by Mordred

[martillo]

49 minutes ago, Mordred said:

Well the more in depth you get into particle physics you will find the distinction between Real , quasi, and virtual particles becomes less and less important. Phonons however are not virtual particles they are quasi particles. They measurable properties of which they represent. Both real and quasi-particles are field excitations in so far as they have a minimal of a quanta of action. 

Individual virtual particles however do not hence they are oft treated as permutations in field theory as opposed to an excitation. You can never measure individual virtual particles its impossible. Now here is where the distinction becomes important. In Feymann path integrals virtual particles are represented by the internal lines as a type of field propagator.

 Quasi and real particles due to the same requirements of field excitation (localization of a wavefunction) fall on the external lines on path integrals. In QFT this doesn't really matter  as all particles are simply states where all particles are field excitations represented by that state. QFT is operator driven so it uses the creation/annihilation operator for thermodynamic treatments as well as for particle production.  Its variation of all these classical thermodynamic laws for conduction, convention and radiation apply those operators via the Euler Langrangian equations for its path integrals

 

 

Good comment about real, quasi and virtual particles.

I have now the intuitive concept that the quasi particles could come just because of a misunderstanding in the behavior of photons which could give the same cause for the considered behaviors on the interactions between real particles in QFT. I can imagine now that in the future all those quasi particles could be found as just particular instances of real photons. I have a better concept about them now.

Virtual particles as fields propagators is something different. I think they come into place to substitute the classic concept of action at a distance of forces introducing also a delay in the action due to their finite velocity. I'm continuing the classical approach of instantaneous action a distance of the fields. I think I have made an important advance in this direction but this would be something off topic here in this thread.

49 minutes ago, Mordred said:

for example Bose Einstein in classical radiation treatment which is used to calculate the number density of bosons at a given blackbody temperature looks like this

 

ni=gie(εi−μ)/kT−1

 

PS at some point you will need this equation as well as Fermi-Dirac and Maxwell Boltzmann.

under QFT looks like this

Bose Einstein QFT format.

 

|k1→k2→〉a^†(k1→)a^†(k2→)|0〉

 

 

⇒|k1→k2→〉=|k2→k1→〉

 

note both versions use phase/momentum space. Practice your vector algebra lol you will need it.

I hope to not need to use the QFT format. I like algebra but not in the format of tensors. May be just because I didn't dedicate time to them, I admit.

Edited March 31 by martillo

[martillo]

Sorry. I was posting just another mistake here. Deleted.

Edited March 31 by martillo

[swansont]

11 hours ago, martillo said:

I have read the subject at Wikipedia and there it is said that laser cooling works due to the Doppler effect on moving atoms only. It is said that static atoms are not affected by the laser beam and they pass through the atoms maintaining their original momentum.

My PhD dissertation was based on laser cooling and trapping and I did projects based in it for 30 years. Your summary misses the point. Cooling happens on moving atoms (because the hot atoms are moving) but the photon absorption interaction is not dependent on that.

 

11 hours ago, martillo said:

For me a photon cannot just pass through the nucleus of an atom but their photons are absorbed and re-emitted while the momentum is maintained. The atoms in a metal lattice are stationary and so they can absorb and re-emit photons with that Doppler effect which maintain their original momentum as I have sustained.

“for me” isn’t how science works. If you don’t have experimental evidence for a notion, it’s worthless.

 

11 hours ago, martillo said:

Conduction involves two bodies with some temperature each. This heat transfer is governed by the Thermodynamics laws in which the energies are proportional to KT and the heat transfer is linearly proportional to the temperatures' difference. A metal conductor acts just as a perfect mediator between the bodies for the heat transfer and so its conductance present the same linear proportionality in T. Not perfect conductors would present a gradient of temperature inside and their conductance would not present a linear relation with T. Vacuum is one such type of conductors in which the heat transfer presents a relation to T⁴.

Then derive this relationship. Physics is based on models. 

[Mordred]

12 hours ago, martillo said:

Good comment about real, quasi and virtual particles.

I have now the intuitive concept that the quasi particles could come just because of a misunderstanding in the behavior of photons which could give the same cause for the considered behaviors on the interactions between real particles in QFT. I can imagine now that in the future all those quasi particles could be found as just particular instances of real photons. I have a better concept about them now.

Virtual particles as fields propagators is something different. I think they come into place to substitute the classic concept of action at a distance of forces introducing also a delay in the action due to their finite velocity. I'm continuing the classical approach of instantaneous action a distance of the fields. I think I have made an important advance in this direction but this would be something off topic here in this thread.

I hope to not need to use the QFT format. I like algebra but not in the format of tensors. May be just because I didn't dedicate time to them, I admit.

Like Swansont I too have done related experiments. For example you mentioned your feelings on virtual particles. However when you are conducting experiments that involve resonance in scatterings you will find that many of the exchanges involve the photon mediator. Great you have no issue with photons. However those photons are offshell. They do not have the same energy levels as real photons.

 The offshell mediator boson is typically offshell and as such falls under the virtual counterpart. 

Now this actually becomes unavoidable when you start applying Fermi-golden rule in resonance scatterings. A prime example is when electrons absorb a photon. In the lab you can detect this interaction however the energy level of mediator is far less than a real photon energy level.

Yet you cannot deny that an intermediate particle isn't involved. Repeatable experimentation clearly shows the intermediate interaction.

here is Fermi-Golden rule

\[\Gamma=\frac{2\pi}{\hbar}|V_{fi}|^2\frac{dN}{DE_f}\]

single phonon operator term

\[E_n=\hbar\omega_q(n_q+\frac{1}{2})\]

suffice it to say I've had a lot of dealings with detectable resonant scatterings that can one can only equate to virtual intermediate interactions. Good example is photons mediating charge of an EM field. Even though the photon itself has no charge it still mediates charge

A large part of the reason I mention this is that the resonance of a photon virtual ensemble or otherwise will differ from that of phonons they each have different resonant signatures.

the cross sections of photons/phonons vary depending on the interaction they are involved in. How they vary show distinctions as the two particles do have different properties.

Now the point of this is even though you may or may not get the math to work out using photons for conductance. You may very well find the cross sections of the scatterings involved will match phonons and not photons. I didn't bother posting details on the photon as they are readily available. Its more tricky to find the needed details for phonons. There is also no point posting cross section formulas as each scattering will have its own. They all apply the Breit Wigner cross section formula

\[\sigma(E)=\frac{2J+1}{2s_1+1)(2S_2+1)}\frac{4\pi}{k^2}[\frac{\Gamma^2/4}{(E-E_0)^2+\Gamma/4)}]B_{in}B_{out}\]

Edited March 31 by Mordred

[Mordred]

here is one paper that applies the phonon cross sections in its research via resonant scatterings

https://arxiv.org/pdf/2109.12117.pdf

Edited March 31 by Mordred

[martillo]

3 hours ago, swansont said:

My PhD dissertation was based on laser cooling and trapping and I did projects based in it for 30 years. Your summary misses the point. Cooling happens on moving atoms (because the hot atoms are moving) but the photon absorption interaction is not dependent on that.

Well, I need to precisely define certain concepts here.

Photons absorption involves energy absorption by the atom and a force on the atom due to the momentum conservation, right? Then if I consider some little time after the atom emit other photon of the same kind -I mean same energy and same momentum absolute value (this what I call re-emission of a photon)- and also in the same direction then the energy and momentum are both conserved. This is what I really meant while saying that the photons could "pass through" the atoms maintaining their original momentum. Now as a photon is emitted with same momentum it has exerted a force back opposite to that force in the absorption. The net momentum change on the atom is zero. This atom had no net effect as needed when cooling an atom by laser as you say. This does not mean laser cooling would not work. This mean that this atom in particular was not cooled. As you say laser cooling does actually imply a non zero net change in the momentum of the atom. It is just that this is not what happens in the atoms in the heat conducting process.

Now you would ask for some experimental evidence about this process I described as you commented:

3 hours ago, swansont said:

“for me” isn’t how science works. If you don’t have experimental evidence for a notion, it’s worthless.

As an evidence on that the photons' "passing through" the nucleus of atoms actually happens by the absorption with posterior emission of a photon of the same energy and momentum process I would mention that certain time delay is needed to this process take place. If not, if a photon would really pass through an atom practically unaltered this time delay would be negligible since the dimension of the nucleus would be negligible. Now, that time delay effect in each atom of the sequence of atoms involved in the path of the linear transmission of the heat I consider, is easily observed in practice as the heat transmission happens in a much slower velocity than the velocity of the photons. The heat transferring happens with the accumulation of all the delays of all the atoms in the sequence of the path and so the observable delay in the heat transmission in the total path. 

3 hours ago, swansont said:

Then derive this relationship. Physics is based on models. 

Strong challenge to me I must say. I will make my try, please give me some time.

 

52 minutes ago, Mordred said:

here is one paper that applies the phonon cross sections in its research via resonant scatterings

https://arxiv.org/pdf/2109.12117.pdf

I'm reading your two recent posts now. Please tell me what is meant by "cross section" when describing "cross sections of the photons/phonons scatterings".

Edited March 31 by martillo

[Mordred]

the cross section is the waveform you get when you have multiple particles interact. It gives us valuable information such as determining say the mass of the Higgs Boson. The cross sections of how that boson interacts with neutrinos, leptons etc. Allows us to determine its mass value as one example. Cross sections for the standard model for all particles are also applied in the CKMS and PMNS mass mixing matrix using the Weinberg angles (requires the cross section).

 Here is some cross sections I had in my own thread dealing with Early universe processes as an example you can see every particle interaction generates its own cross section. Those cross sections vary depending on the properties such as spin, charge, mass, flavor etc. 

Early Universe Cross section list

Breit Wigner cross section

\[\sigma(E)=\frac{2J+1}{2s_1+1)(2S_2+1)}\frac{4\pi}{k^2}[\frac{\Gamma^2/4}{(E-E_0)^2+\Gamma/4)}]B_{in}B_{out}\]

E=c.m energy, J is spin of resonance, (2S_1+1)(2s_2+1) is the #of polarization states of the two incident particles, the c.m., initial momentum k E_0 is the energy c.m. at resonance, \Gamma is full width at half max amplitude, B_[in} B_{out] are the initial and final state for narrow resonance the [] can be replaced by

\[\pi\Gamma\delta(E-E_0)^2/2\]

The production of point-like, spin-1/2 fermions in e+e− annihilation through a virtual photon at c.m.

\[e^+,e^-\longrightarrow\gamma^\ast\longrightarrow f\bar{f}\]

\[\frac{d\sigma}{d\Omega}=N_c{\alpha^2}{4S}\beta[1+\cos^2\theta+(1-\beta^2)\sin^2\theta]Q^2_f\]

where

\[\beta=v/c\]

c/m frame scattering angle

\[\theta\] 

fermion charge

\[Q_f\]

if factor [N_c=1=charged leptons if N_c=3 for quarks.

if v=c then (ultrarelativistic particles)

\[\sigma=N_cQ^2_f\frac{4\pi\alpha^2}{3s}=N_cQ^2_f\frac{86.8 nb}{s (GeV^2)}\]

2 pair quark to 2 pair quark

\[\frac{d\sigma}{d\Omega}(q\bar{q}\rightarrow \acute{q}\acute{\bar{q}})=\frac{\alpha^2_s}{9s}\frac{t^2+u^2}{s^2}\]

cross pair symmetry gives

\[\frac{d\sigma}{d\Omega}(q\bar{q}\rightarrow \acute{q}\acute{\bar{q}})=\frac{\alpha^2_s}{9s}\frac{t^2+u^2}{t^2}\]

here is a sample of phonon cross sections.

https://e-learning.pan-training.eu/wiki/index.php/The_scattering_cross_section_for_phonons

One detail you should note is that the cross sections and subsequent mean lifetime for accoustic phonons will differ from optical phonons.

however using phonons vs photons for conduction has far more involved than merely some personal choice of which to use. Which one is chosen is determined via direct experimentation and not some personal or group choice. There is in fact distinctive and measurable differences between phonons (acoustic and optical) and photons or any other gauge boson virtual or otherwise. That last link for example mentions the effects of a particles polarizations on cross sections. Longitudinal polarizations have different effects than transverse polarizations.

Edited March 31 by Mordred

[swansont]

23 minutes ago, martillo said:

Well, I need to precisely define certain concepts here.

Photons absorption involves energy absorption by the atom and a force on the atom due to the momentum conservation, right? Then if I consider some little time after the atom emit other photon of the same kind -I mean same energy and same momentum absolute value (this what I call re-emission of a photon)- and also in the same direction then the energy and momentum are both conserved.

The emission is not preferential in that direction. The atom doesn’t “remember” the direction a photon came from. It’s simply in an excited state, and the subsequent emission probability is symmetric. It’s just as likely to emit in the direction the photon came from as in the opposite direction. 

 

23 minutes ago, martillo said:

 

This is what I really meant while saying that the photons could "pass through" the atoms maintaining their original momentum. Now as a photon is emitted with same momentum it has exerted a force back opposite to that force in the absorption. The net momentum change on the atom is zero. This atom had no net effect as needed when cooling an atom by laser as you say. This does not mean laser cooling would not work. This mean that this atom in particular was not cooled. As you say laser cooling does actually imply a non zero net change in the momentum of the atom. It is just that this is not what happens in the atoms in the heat conducting process.

How does an atom “know” the difference between these photons?

23 minutes ago, martillo said:

Now you would ask for some experimental evidence about this process I described as you commented:

As an evidence on that the photons' "passing through" the nucleus of atoms actually happens by the absorption with posterior emission of a photon of the same energy and momentum process I would mention that certain time delay is needed to this process take place. If not, if a photon would really pass through an atom practically unaltered this time delay would be negligible since the dimension of the nucleus would be negligible. Now, that time delay effect in each atom of the sequence of atoms involved in the path of the linear transmission of the heat I consider, is easily observed in practice as the heat transmission happens in a much slower velocity than the velocity of the photons. The heat transferring happens with the accumulation of all the delays of all the atoms in the sequence of the path and so the observable delay in the heat transmission in the total path. 

Strong challenge to me I must say. I will make my try, please give me some time.

We can measure how long it takes for light to pass through various materials. Polyethylene, for example, has an index of refraction of around 1.5 for infrared light. So light goes at around 2/3 c through it.

[Mordred]

Ordinarily I avoid posting whiteboard lesson notes however this one has some valuable images to better help understand cross sections than an article that simply has the relevant math.

https://www.precision.hep.phy.cam.ac.uk/wp-content/people/mitov/lectures/ParticlePhysics-2018/H01_Introduction.pdf

[martillo]

59 minutes ago, swansont said:

The emission is not preferential in that direction. The atom doesn’t “remember” the direction a photon came from. It’s simply in an excited state, and the subsequent emission probability is symmetric. It’s just as likely to emit in the direction the photon came from as in the opposite direction. 

Isolated atoms don't "remember" the direction of the incoming photon, the probability is symmetric I agree. That's why I preciselly mentioned: "Then if I consider some little time after the atom emit other photon of the same kind -I mean same energy and same momentum absolute value (this what I call re-emission of a photon)- ". I after added the condition that: "and also in the same direction then the energy and momentum are both conserved." So I reasoned at the inverse: If the absorption and posterior emission of a photon happens in the same direction then energy and, particularly for us now, the momentum is conserved. We are analyzing the conduction of heat through a conductor between two bodies, right? We are analyzing how the heat flows in the direction of one body to the other one. When applying the Thermodynamics laws it is considered this flow between the two bodies and assuming there is no loss of heat so, in this case, there is a preferred direction of heat conduction. I'm explaining then, with my approach, how this directional flow happens with the mediating photons proposed.

59 minutes ago, swansont said:

How does an atom “know” the difference between these photons?

As I said in principle the atom doesn't "know" the difference. It is the inverse reasoning: if photons are emitted in the same direction as the absorbed one then there is no net effect in the momentum of the atom and so it "transmit" the energy in that direction and "laser cooling" effect does not take place. Now if photons are not emitted in the same direction then is right to say that then some "laser cooling" effect could take place.

59 minutes ago, swansont said:

We can measure how long it takes for light to pass through various materials. Polyethylene, for example, has an index of refraction of around 1.5 for infrared light. So light goes at around 2/3 c through it.

And how do you explain that slower velocity? I explain it through the absorption with posterior delayed emission of photons. 

Edited March 31 by martillo

[swansont]

1 minute ago, martillo said:

Isolated atoms don't "remember" the direction of the incoming photon, the probability is symmetric I agree. That's why I preciselly mentioned: "Then if I consider some little time after the atom emit other photon of the same kind -I mean same energy and same momentum absolute value (this what I call re-emission of a photon)- ". I after added the condition that: "and also in the same direction then the energy and momentum are both conserved." So I reasoned at the inverse: If the absorption and posterior emission of a photon happens in the same direction then energy and, particularly for us now, the momentum is conserved.

Energy and momentum will be conserved regardless of the direction the photon is emitted. The difference is whether any energy and momentum is imparted to the atom.

 

1 minute ago, martillo said:

We are analyzing the conduction of heat through a conductor between two bodies, right? We are analyzing how the heat flows in the direction of one body to the other one. When applying the Thermodynamics laws it is considered this flow between the two bodies and assuming there is no loss of heat so, in this case, there is a preferred direction of heat conduction. I'm explaining then, with my approach, how this directional flow happens with the mediating photons proposed.

Your proposed mechanism is not observed to happen. i.e. your prediction fails. 

1 minute ago, martillo said:

 

As I said in principle the atom doesn't "know" the difference. It is the inverse reasoning: if photons are emitted in the same direction as the absorbed one then there is no net effect in the momentum of the atom and so it "transmit" the energy in that direction and "laser cooling" effect does not take place. Now if photons are not emitted in the same direction then is right to say that then some "laser cooling" effect could take place.

And since the photons can be emitted in other directions, this happens only rarely.

 

1 minute ago, martillo said:

And how do you explain that slower velocity? I explain it through the absorption with posterior delayed emission of photons. Is there other explanation? If so please tell me because I don't know any other one.

You did more than that. You predicted a speed. Does thermal conduction happen at the speed that your idea predicts?

[martillo]

It's a bit difficult to discuss with you when you instantaneously think in a possible objection and just post a few words about it without much care. I dedicate time and effort to understand what you are saying trying to get the real key point and answer appropriately and as precisely as possible.

24 minutes ago, swansont said:

Energy and momentum will be conserved regardless of the direction the photon is emitted. The difference is whether any energy and momentum is imparted to the atom.

I was talking about the conservation of momentum in the photons with no net change in the momentum of the atom.

24 minutes ago, swansont said:

Your proposed mechanism is not observed to happen. i.e. your prediction fails. 

The photons cannot be observed. I'm giving an explanation compatible with the observational evidence that is known about heat transfer and so my explanation is factible. If not give an observational evidence against it.

24 minutes ago, swansont said:

And since the photons can be emitted in other directions, this happens only rarely.

As I said in the case we are analyzing of heat transferring of two bodies there is a preferred direction of flow and so that is what happens while quite no "laser cooling" effect is present.

24 minutes ago, swansont said:

You did more than that. You predicted a speed. Does thermal conduction happen at the speed that your idea predicts?

Not at all. I didn't predict any velocity. I just explained the known refractive velocity you provided to be less than c.

Edited March 31 by martillo

[swansont]

1 hour ago, martillo said:

It's a bit difficult to discuss with you when you instantaneously think in a possible objection and just post a few words about it without much care. I dedicate time and effort to understand what you are saying trying to get the real key point and answer appropriately and as precisely as possible.

That’s the process - see if you can poke holes in an idea. Falsifiability is a key component of science.

1 hour ago, martillo said:

I was talking about the conservation of momentum in the photons with no net change in the momentum of the atom.

And I’m telling you it’s a rare occurrence. The photon gan go in many directions, and you require a specific one, over and over again.

1 hour ago, martillo said:

The photons cannot be observed. I'm giving an explanation compatible with the observational evidence that is known about heat transfer and so my explanation is factible. If not give an observational evidence against it.

If they can’t be observed how can they be responsible for heat transfer? Isn’t heating something up an observable process?

1 hour ago, martillo said:

As I said in the case we are analyzing of heat transferring of two bodies there is a preferred direction of flow and so that is what happens while quite no "laser cooling" effect is present.

There’s no evidence it works this way, since the evidence we have says it doesn’t.

Wishing does not make it so.

1 hour ago, martillo said:

Not at all. I didn't predict any velocity. I just explained the known refractive velocity you provided to be less than c.

Yo predicted a time for heat transfer, which depends on this speed. Why does the light for heat transfer behave differently than other light? (without resorting to magic or special pleading, please)

[martillo]

1 hour ago, swansont said:

And I’m telling you it’s a rare occurrence. The photon gan go in many directions, and you require a specific one, over and over again.

Rare? What is actually rare is the "laser cooling" effect. 

1 hour ago, swansont said:

If they can’t be observed how can they be responsible for heat transfer? Isn’t heating something up an observable process?

I was referring to no direct observation. 

1 hour ago, swansont said:

There’s no evidence it works this way, since the evidence we have says it doesn’t.

Heat transfer is mediated by photons and of course the photons responsible for the transfer must travel in the direction of flow at any place.

1 hour ago, swansont said:

Yo predicted a time for heat transfer, which depends on this speed. Why does the light for heat transfer behave differently than other light?

I only predicted a time delay when an atom absorb a photon and a posteriori emit a photon. Valid for "light", "heat" whatever energy flow through a material substance composed by atoms.

Edited March 31 by martillo

[swansont]

5 minutes ago, martillo said:

Rare? What is actually rare is the "laser cooling" effect. 

Is this a view from your extensive experience in atomic physics?

Absorption followed by de-excitation in a dipole pattern is the normal process in an electric dipole transition.

5 minutes ago, martillo said:

I was referring to no direct observation. 

With no explanation as to why this should magically be unobservable.

5 minutes ago, martillo said:

Heat transfer is mediated by photons and of course the photons responsible for the transfer must travel in the direction of flow at any place.

This is your assertion. To conclude this is circular logic.

5 minutes ago, martillo said:

I only predicted a time delay when an atom absorb a photon and a posteriori emit a photon. Valid for "light", "heat" whatever energy flow through a material substance composed by atoms.

But you said this time delay is responsible for the delay in heat transmission through the material. Why can light that isn’t heat travel through the material at 2/3 c, while the heat transmission delay - which you say is due to light - is much, much longer?

[martillo]

23 minutes ago, swansont said:

But you said this time delay is responsible for the delay in heat transmission through the material. Why can light that isn’t heat travel through the material at 2/3 c, while the heat transmission delay - which you say is due to light - is much, much longer?

In both cases the velocity is slowed down because of the time delay of the same absorption/emission interaction of photons with the atoms. The velocity is different just depending on how much atoms the "light"/"heat" had interacted with before reaching the end.

Edited March 31 by martillo

[swansont]

1 hour ago, martillo said:

In both cases the velocity is slowed down because of the time delay of the same absorption/emission interaction of photons with the atoms. The velocity is different just depending on how much atoms the "light"/"heat" had interacted with before reaching the end.

1. The time is massively different. There are only so many atoms with which they can interact.

2. How do the atoms know which of these identical photons to interact with? Not interacting with light from a laser, that passes through at 2/3 c, but interacting with thermal radiation, which you claim goes much, much slower?

For a small block of material the light might take 0.1 nanosecond (2 cm of travel) but the heat transfer takes perhaps a second? 

What is the physics that differentiates these two cases? Not some fairy tale, which is not science. Where’s the science?

[martillo]

7 hours ago, swansont said:

1. The time is massively different. There are only so many atoms with which they can interact.

2. How do the atoms know which of these identical photons to interact with? Not interacting with light from a laser, that passes through at 2/3 c, but interacting with thermal radiation, which you claim goes much, much slower?

For a small block of material the light might take 0.1 nanosecond (2 cm of travel) but the heat transfer takes perhaps a second? 

What is the physics that differentiates these two cases? Not some fairy tale, which is not science. Where’s the science?

I have given a wrong answer. The same material can present different velocities in transmitting "light" or "heat". Glass for instance is an excellent light transmitter, it is transparent to light while a very bad heat transmitter, it is an insulator to heat. The difference is in the frequency of the photons involved. "Light" frequency is in the optical range while "heat" is mainly in the infrared range. Glass is an excellent transmitter in the optical range while a very bad transmitter in the infrared range. I must answer now why this happens with photons? It is because the glass' atoms cannot absorb the relative high energetic photons and they pass through the lattice or are reflected while they can well absorb and emit quite all the less energetic photons of heat. Now, how is that? Well, the absorption of photons by atoms depend on the available levels of energy in the atoms according to their particular characteristic energy spectrum with their characteristic levels of energies. I must also explain how the velocity of light in the glass is anyway less than in vacuum and this is because the light's photons pass through the lattice but have another interaction with atoms rather than absorption and emission. The interaction would be elastic collisions in which the photons are slowed down. The photons have an EM field which interacts with the EM field of the atoms with an EM force between them. The force act braking a bit the photons at first while accelerating them again after but a delay of time is involved in this process. 

Edited April 1 by martillo

[Mordred]

40 minutes ago, martillo said:

 

The question is: I can understand the need of a virtual particle as field propagator to explain the action at a distance of forces at far distances if the action is not instantaneous. I don't get why the need of a virtual particle in very close interactions. If the particles have an EM field they can directly interact through an EM force that would exist between them so, why the need of a mediating extra particle to represent this EM force? 

One of the first lessons in physics is that energy is a property of some state, object etc. Energy doesn't exist on its own, hence the use of mediator particles. Energy as a property of an  objects/state etc ability to perform work. Mass is another property that doesn't exist on its own. It too is a property. Mass being the objects/state etc resistance to inertia change or shortly described acceleration.

Those definitions apply regardless of physics theory. Including string theory as properties they must have some state etc to be applied to.

Edited April 1 by Mordred

[martillo]

On 3/30/2024 at 10:42 PM, Mordred said:

Now here is where the distinction becomes important. In Feymann path integrals virtual particles are represented by the internal lines as a type of field propagator.

The question is: I could understand the need of a virtual particle as field propagator to explain the action at a distance of forces at far distances where the fields seem to vanish. I don't get why the need of a virtual particle in very close interactions. If the particles have an EM field they can directly interact through an EM force that would exist between them so, why the need of a mediating extra particle to represent this EM force? 

Sorry, I was editing the post and now it appears after your answer. I don't know why.

Edited April 1 by martillo

[Mordred]

1 minute ago, martillo said:

The question is: I could understand the need of a virtual particle as field propagator to explain the action at a distance of forces at far distances where the fields seem to vanish. I don't get why the need of a virtual particle in very close interactions. If the particles have an EM field they can directly interact through an EM force that would exist between them so, why the need of a mediating extra particle to represent this EM force? 

You don't particularly have to worry about it you can simply use the field treatments without specifying the mediator by using operator states.  All mediator bosons are massless all you need from there is sufficient energy/momentum for mediation.

 However you don't need to learn the relations via the quantum world most statistical mechanics textbooks barely touch on the quantum regime. However at the quantum level then even at short range you will need a mediator. 

 

Let's put it this way you can mathematically describe everything you need using nothing more than classical mechanics. When you get to the quantum regime the number density of your mediator can then be determined by whatever the energy level of field or state your measuring by simply applying the Bose-Einstein statistics for bosons.So it's something you can worry about later provided your not breaking any laws such as the speed limit.