Photon is massless why?

[swansont]

4 hours ago, martillo said:

In Physics by definition something has momentum p if it has mass and velocity and the momentum is defined as p = mv.

In classical mechanics, this would be the case. Physics is more than classical mechanics.

4 hours ago, martillo said:

The problem is with Relativity Theory. In Relativity Theory photons must be massless but there is that problem with the definitions of momentum and energy...

EM radiation possesses momentum. This was predicted by Maxwell (i.e. before relativity and QM) and experimentally confirmed in 1903 by Nichols and Hull.

[martillo]

2 hours ago, exchemist said:

Your "definition" of momentum is wrong...

17 minutes ago, swansont said:

In classical mechanics, this would be the case. Physics is more than classical mechanics.

The relativistic definition of momentum is p = mv for m = γm₀v where γ = (1 - v²/c²)^-1/2 and m₀ the rest mass.

 

Edited November 11, 2023 by martillo

[swansont]

2 minutes ago, martillo said:

The relativistic definition of momentum is p = mv for m = γm₀v where γ = (1 - v²/c²)^-1/2 and m₀ the rest mass.

 

For something with mass. 

[martillo]

18 minutes ago, swansont said:

21 minutes ago, martillo said:

The relativistic definition of momentum is p = mv for m = γm₀v where γ = (1 - v²/c²)^-1/2 and m₀ the rest mass.

 

For something with mass. 

Googling, that is the definition I find everywhere for relativistic momentum. Could you please give the definition for massless objects?

 

38 minutes ago, swansont said:

EM radiation possesses momentum. This was predicted by Maxwell (i.e. before relativity and QM) and experimentally confirmed in 1903 by Nichols and Hull.

I'm not questioning that electromagnetic waves have momentum. This is the wave-like behavior.

I'm questioning how the relativistic concept of momentum applies to a photon. This is the particles-like behavior.

 

2 hours ago, exchemist said:

...you are using the wrong formula. According to special relativity, E=mc² is not applicable to a photon. I went over this in post 11 of this thread. 

I know the relativistic formulation of energy in terms of momentum you mentioned. I have just said that applying De Broglie law in the classical approach would give E = mc².

Edited November 11, 2023 by martillo

[joigus]

4 minutes ago, martillo said:

Googling, that is the definition I find everywhere for relativistic momentum. Could you please give the definition for massless objects?

\[ \left|\boldsymbol{p}\right|=E/c \]

with,

\[ E=h\nu \]

with h Planck's constant, and \( |nu \) being the photon's frequency in Hertzs.

Direction of momentum given by,

\[ \frac{\boldsymbol{p}}{\left|\boldsymbol{p}\right|}=\boldsymbol{k} \]

[martillo]

10 minutes ago, joigus said:

 

|p|=E/c

 

with,

 

E=hν

 

with h Planck's constant, and |nu being the photon's frequency in Hertzs.

Direction of momentum given by,

 

p|p|=k

 

So, for massless objects you define p = E/c and define E = pc. Wouldn't that be a cyclic indetermination?

Edited November 11, 2023 by martillo

[joigus]

5 minutes ago, martillo said:

So you define p = E/c and define E = pc for massless objects. Wouldn't that be a cyclic indetermination?

No. It's similar as when you define mass as m=F/a, while you define F=ma. Things like F=-Gmm'/r² get you out of the tautology.

In the case of electromagnetism, you have many other relations that get you out of the tautology, like E=hx(frequency), or E(nu,k) is a solution of Maxwell's equations in free space. Etc. It all works out, and you never look like a dog chasing its own tail. Believe me.  

Edited November 11, 2023 by joigus
minor addition

[swansont]

1 hour ago, martillo said:

Googling, that is the definition I find everywhere for relativistic momentum. Could you please give the definition for massless objects?

 

I'm not questioning that electromagnetic waves have momentum. This is the wave-like behavior.

I'm questioning how the relativistic concept of momentum applies to a photon. This is the particles-like behavior.

Why on earth would the momentum change if you observed wave vs particle behavior?

p = E/c applies to both

 

[martillo]

1 hour ago, swansont said:

Why on earth would the momentum change if you observed wave vs particle behavior?

p = E/c applies to both

 

Everything seems to work fine if you define p = E/c for both massive or massless particles except that the concept of mass is then just omitted.

In the "waves-like" behavior mass does not play any role. Waves would not have mass. In the concept of "particles-like" behavior frequency has no meaning. There's no frequency associated to any particle in the Standard Model of particles.

I think Relativity Theory is about the "particles-like" behavior. When we begin to associate the frequency f for the energy E = hf we are entering in "Quantum Mechanics" (QM) and so we would be talking about "Relativistic Quantum Mechanics" (RQM), is that right?

Everything would work fine while not entering in the concept of mass.

By the way, with Relativity Theory we would need to talk about "longitudinal" and "transversal" masses, isn't it?

 

Edited November 11, 2023 by martillo

[joigus]

No, it's not \( E=\left|\boldsymbol{p}\right|c\) for massive particles. As @MigL said, it's,

On 11/9/2023 at 2:56 AM, MigL said:

The full equation is     E = root [(Mc²)² + (pc)²]

as Swansont has pointed out numerous times past.

The first tern   Mc² applies to massive stationary objects.
The second term   pc is added when that object is moving, and p denotes the momentum.

Obviously, for light, which is always moving at c and M is zero, the momentum term is the only one applicable.

for massive particles. So \( E=\left|\boldsymbol{p}\right|c\sqrt{1+\frac{m²c^{4}}{\left|\boldsymbol{p}\right|^{2}}} \). With this factorisation maybe it's clearer how and why it's not the same for massive particles?

[martillo]

2 hours ago, joigus said:

No, it's not E=|p|c for massive particles. As @MigL said, it's,

for massive particles. So E=|p|c1+m²c4|p|2−−−−−−−√ . With this factorisation maybe it's clearer how and why it's not the same for massive particles?

Right.

The general equation to be applied ever is E² = (Mc²)² + (pc)² and for massive particles p = mv = γMv while for massless ones, like the photon, the factor γ = (1 - v²/c²)^-1/2 is undefined for v = c and so must be defined p = E/c.

Electromagnetic waves would be massless so p = E/c would hold for them.

With analogy, just a question: Would the definition p = E/v apply to any other wave like sound and water waves?

Edited November 11, 2023 by martillo

[KJW]

In Minkowski spacetime, using natural units where c = 1, there is the four-dimensional energy-momentum vector. Energy is the time-component and momentum is the spatial components of this vector. The invariant mass is the magnitude of this vector. That is:

m² = E² – p²

Just as a tangent vector of a lightlike trajectory in spacetime is a null vector, so is the energy-momentum vector of a photon. That is, E² = p² and thus m = 0. As mass is an invariant, the mass of a photon is zero in every frame of reference.

[martillo]

51 minutes ago, martillo said:

Electromagnetic waves would be massless so p = E/c would hold for them.

With analogy, just a question: Would the definition p = E/v apply to any other wave like sound and water waves?

I have already found a good answer to my question at: https://as.nyu.edu/content/dam/nyu-as/as/documents/silverdialogues/SilverDialogues_Peskin.pdf

Seems the exception is on pure transversal waves which do not propagate. Would be the case of standing waves.

Edited November 11, 2023 by martillo

[joigus]

Sorry. A c⁴ should be c² inside the square root of what I wrote there.

1 hour ago, martillo said:

Right.

The general equation to be applied ever is E² = (Mc²)² + (pc)² and for massive particles p = mv = γMv while for massless ones, like the photon, the factor γ = (1 - v²/c²)^-1/2 is undefined for v = c and so must be defined p = E/c.

Electromagnetic waves would be massless so p = E/c would hold for them.

With analogy, just a question: Would the definition p = E/v apply to any other wave like sound and water waves?

No. Each wave has a different dispersion relation. E=pc gives you a photon D.R. Matter waves are a different matter. 😬

[martillo]

2 minutes ago, joigus said:

Sorry. A c⁴ should be c² inside the square root of what I wrote there.

No problem for me.

 

[KJW]

As for why a photon is massless, (IIRC) this has something to do with the Higgs field and the symmetry-breaking of the four electroweak bosons.
 

[martillo]

12 minutes ago, KJW said:

As for why a photon is massless, (IIRC) this has something to do with the Higgs field and the symmetry-breaking of the four electroweak bosons.
 

Nothing simple...

25 minutes ago, joigus said:

No. Each wave has a different dispersion relation. E=pc gives you a photon D.R. Matter waves are a different matter. 😬

In the link I posted after, quite all the waves relate their energy and momentum in the form p = E/v excepting for transversal waves only.

Edited November 11, 2023 by martillo

[swansont]

4 hours ago, martillo said:

Everything seems to work fine if you define p = E/c for both massive or massless particles except that the concept of mass is then just omitted.

A particle at rest has energy, so I don’t see how that would work. A particle at rest has a momentum of mc?

 

4 hours ago, martillo said:

In the "waves-like" behavior mass does not play any role. Waves would not have mass. In the concept of "particles-like" behavior frequency has no meaning. There's no frequency associated to any particle in the Standard Model of particles.

DeBroglie wavelength of an electron and proton moving at the same speed is not the same. Mass definitely play a role.

 

 

[martillo]

On 11/11/2023 at 6:28 PM, swansont said:

On 11/11/2023 at 1:55 PM, martillo said:

Everything seems to work fine if you define p = E/c for both massive or massless particles except that the concept of mass is then just omitted.

A particle at rest has energy, so I don’t see how that would work. A particle at rest has a momentum of mc?

I was wrong: p = E/c works for massless particles only like the photon.

 

On 11/11/2023 at 6:28 PM, swansont said:

DeBroglie wavelength of an electron and proton moving at the same speed is not the same. Mass definitely play a role.

You are right. Mass plays a role in the "matter waves".

[swansont]

!

Moderator Note

Matter wave discussion has been split

https://www.scienceforums.net/topic/132885-matter-waves-split-from-photon-is-massless-why/

 

 

[Muhammad Owais Isaac]

But wave is produces by the fluctuating energy 

[swansont]

12 minutes ago, Muhammad Owais Isaac said:

But wave is produces by the fluctuating energy 

What energy is allegedly fluctuating?

[Alysdexic]

Immassive motes (particles) don’t really exist but are convenient mathematic models for the action between massive motes. Motes must interact in a field that extends indefinitely far so each action must acquire a mass by the Goldstone effect; thus gravitòns, fòtòns, and gluòns don’t exist but pressuròns, polaritòns, and piòns do. Whichever set exists the state of one of these quanta is simply the initial state of the massive bodies that emit and absorb it subtracted from the final state. Bodies—substantives, nouns, objects, subjects, targets, arguments—keep distinct to haps—actions, verbs, processes, emissions, absorptions, events, histories. Bodies are what elementary interaction laws like Newton's, Coulomb's, and Yucawa's laws and their temporal derivatives Lense-Thirring, Ampère's, and Lund-Schifman(?) laws posit. Haps are posited by propagation laws of the formers' temporal and spatial derivatives like Heaviside's, Compton's, de Broglie's, Schrödinger's, Lorentz's, and Rindler's laws. Concepts combine the two like Dirac's wave equation. Quanta of bodies are motes and quanta of haps are waves. Motes wave; in so doing motes are behind the waves. Scientists mistakenly call wave quanta motes, especially when they're immassive. The immassive free-range excitations-relaxations of mass, charge, and "color" are the gravitòn, fòtòn, and gluòn. However they onely exist in a matter-free field, which doesn't exist; thus, none of them really exist. When bodies interact they bind to each other and transfer mass under the Goldstone effect; thus onely the pressuròn, polaritòn, and mesòn carriers of radiation exist (proper radiation, not the radioactive debris of convection).