The mass of light quanta

[decraig]

Given is a sufficiently flat spacetime.

 

Define the vector [math]\mu= (E/c^2, p/c) [/math], invariant under the Poisson group.

 

mu is the norm of the mass.

[math] < \mu, \mu> =m^2[/math]

Or [math] |m|=\mu [/math]

 

From this we can get [math]c^4m^2=E^2-p^2c^2[/math].

 

In the common understanding, [math]E=pc[/math]

so that the mass of light is zero.

 

However, for radiation emitted from an atom, or particle annihilation, in the center of momentum frame, the emission is radially symmetric.

 

This implies that the total momenta of an electromagnetic quanta is zero, giving nonzero mass;

[math]m=h\omega/c^2[/math].

[Enthalpy]

Why should radiation emitted by an atom or annihilation be symmetric?

Light is emitted in one direction, the atom recoils in the opposite direction.

An annihilation emits two photons at least. A reason is spin conservation.

Edited November 28, 2013 by Enthalpy

[swansont]

 

 

However, for radiation emitted from an atom, or particle annihilation, in the center of momentum frame, the emission is radially symmetric.

 

This implies that the total momenta of an electromagnetic quanta is zero, giving nonzero mass;

[math]m=h\omega/c^2[/math].

 

There is a large step (or steps) missing in the middle, where you show how you arrive at your conclusion.

An annihilation emits two photons at least. A reason is spin conservation.

 

Linear momentum conservation requires two photons. Spin could be conserved if both particles were spin up, spin 1/2 particles.

[timo]

The invariant mass of a multi-photon system, e.g. two photons, indeed is not necessarily zero. That is well known. For a radially-symmetry elementary excitation of the electromagnetic field, which in a sense you may call a photon, you'd have to define what mass is. Usually, the dispersion relation E(p) refers to cases with a definite momentum. A radially-symmetric "photon" is not a momentum eigenstate (obviously), so you'd have to pick a suitable generalization over the common usage. If you picked m^2 = <E>^2 - <p>^2 (with <.> being the expectation values for the respective measurements) then indeed that kind of photon would have a non-zero "generalized mass". The photons having zero mass refers to single particles in momentum eigenstates, since in every other case mass wouldn't be a property of the particle type.

 

Btw.: What's the point (in the sense of "intention") of your thread, anyways?

[decraig]

Why should radiation emitted by an atom or annihilation be symmetric?

Light is emitted in one direction, the atom recoils in the opposite direction.

An annihilation emits two photons at least. A reason is spin conservation.

The state of recoil is indeterminate with the same radial symmetry until it is measured or decohered.

[swansont]

The state of recoil is indeterminate with the same radial symmetry until it is measured or decohered.

 

I don't see how mass enters into that argument.

[decraig]

 

I don't see how mass enters into that argument.

 

What is the direction of recoil before the system decoheres? Name a direction.

 

The point is that the system is in a symmetric state.

Edited November 30, 2013 by decraig

[swansont]

 

What is the direction of recoil before the system decoheres? Name a direction.

 

The point is that the system is in a symmetric state.

 

How does that affect (or is affected by) mass?

[Enthalpy]

The state of recoil is indeterminate with the same radial symmetry until it is measured or decohered.

But as soon as you have a measure, which is necessary to give "mass" a meaning, the symmetry is lost.

You may also consider that the symmetry results from the superposition of all possible directions, and get the same result for each as for the single observed one.

[decraig]

Enthalpy, are you saying that a particle doesn't have mass until it's measured?

[decraig]

The invariant mass of a multi-photon system, e.g. two photons, indeed is not necessarily zero. That is well known. For a radially-symmetry elementary excitation of the electromagnetic field, which in a sense you may call a photon, you'd have to define what mass is. Usually, the dispersion relation E(p) refers to cases with a definite momentum. A radially-symmetric "photon" is not a momentum eigenstate (obviously), so you'd have to pick a suitable generalization over the common usage. If you picked m^2 = <E>^2 - <p>^2 (with <.> being the expectation values for the respective measurements) then indeed that kind of photon would have a non-zero "generalized mass". The photons having zero mass refers to single particles in momentum eigenstates, since in every other case mass wouldn't be a property of the particle type.

 

Btw.: What's the point (in the sense of "intention") of your thread, anyways?

 

Is special relativity is a good approximation of spacetime? Then so is invariant mass--an approximation, only.

 

My intent: I have learned a great deal from reading the questions of neophites. Certainly more than the accompanying answers of those who might call themselves experts. However, that is not to say, I have not learned a great deal from these as well.

 

How is the house of physics today? Is it well constructed?

 

It's in disarray. But this doesn't stop the carpenters from constructing new additions of the same decayed material. This will not persist. This is offers opportunity, unrequenchable.

 

I make some simple calculations, and viola, light has non-zero mass. Why should this be when another calculation says it is exactly zero?

 

My intent is to provoke, in the hope to learn.

 

Relativity theory is fairly concrete. I can make a statement about relativity and be either right or wrong within the bounds of theory. I can always feel confident in my response by appeal the work of respected men. They can argue with Einstein, etc. if they don't like it.

 

Quantum theory, by contrast, is anarchy. Without an interpretation, even the question of mass is ambiguous, though relativity also has its allotment of peculiar issues.

 

If I were hell bent on expressing superior knowledge, I would be expounding in homework help. I am not so much. Though I am hardly sophomoric, I am a student hoping to learn.

Edited December 1, 2013 by decraig

[timo]

If your intent is to provoke, then forgive me for not being keen on discussing with you. If your intent is to learn, I already gave a reply. You are correct that a naive and sloppy interpretation of pop-sci books and Wikipedia runs into problems at some point, like in case of "light has zero mass". That's sloppy because I am pretty sure that even pop-sci will say "photons have zero mass". And it is naive to assume that from understanding the words of a sentence one can directly understand the sentence.

Contrary to popular misconception statements in physics are strongly context-based. And especially at higher levels of physics knowledge, understanding these contexts and their limitations, and therefore understanding the validity limits of the statements, becomes increasingly important. In your example, if you knew particle physics you'd know how fields are commonly quantized and how particle physics interactions are being calculated: in the basis of momentum eigenstates. Also, even if you knew particle physics you'd have to understand why and how a calculation in terms of momentum eigenstates (the Feynman diagrams you've undoubtedly seen) relates to the real world. The statement about photons being massless has to be understood in this context.

On top of that, many terms in physics are not defined very sharply. In your example, I do not know if there is a universal rule whether a superposition of two or more different momentum eigenstates would be called a photon or if only the eigenstate to a momentum is called a photon. I'd call it a photon, maybe Swansont wouldn't. Same goes for a definition of mass, which I only need when I actually need mass (why would I care about a mass of light in your example?). This fuzzyness about terms usually does not become a problem in interaction between scientists, since they (a) are trained to properly define what they mean where necessary situationally, and (b) usually have interactions with the intent to understand each other, not to correct each other.

 

Just because Prof. Hawking doesn't put his statements into context in his pop-sci books (which he is proud of that they contain no math at all) that does not mean that Prof. Hawking does not know about their context. Your asking about things that seem incorrect is an important step towards understanding contexts. Putting forward such questions with the intention to shed light on the rotten building of physics rather than the humble attempt to increase one's understanding is understandable (I've been a young physics student myself at some point, and in hindsight I really don't envy by tutors) but somewhat silly. In fact, picturing a "constructed house of physics" is questionable in the first place.

[swansont]

 

I make some simple calculations, and viola, light has non-zero mass. Why should this be when another calculation says it is exactly zero?

 

 

Then share them with us, as you have been asked. What you've shown here is insufficient.

[decraig]

 

Then share them with us, as you have been asked. What you've shown here is insufficient.

I really have no idea what you are asking for. Could you be more specific?

[swansont]

You claim to have calculations that show that light has nonzero mass, but the calculations you've given don't show that. All you've done is assert that radial symmetry implies a nonzero mass. So how about showing the calculations that show what you claim?

[decraig]

If your intent is to provoke, then forgive me for not being keen on discussing with you. If your intent is to learn, I already gave a reply. You are correct that a naive and sloppy interpretation of pop-sci books and Wikipedia runs into problems at some point, like in case of "light has zero mass". That's sloppy because I am pretty sure that even pop-sci will say "photons have zero mass". And it is naive to assume that from understanding the words of a sentence one can directly understand the sentence.

Contrary to popular misconception statements in physics are strongly context-based. And especially at higher levels of physics knowledge, understanding these contexts and their limitations, and therefore understanding the validity limits of the statements, becomes increasingly important. In your example, if you knew particle physics you'd know how fields are commonly quantized and how particle physics interactions are being calculated: in the basis of momentum eigenstates. Also, even if you knew particle physics you'd have to understand why and how a calculation in terms of momentum eigenstates (the Feynman diagrams you've undoubtedly seen) relates to the real world. The statement about photons being massless has to be understood in this context.

On top of that, many terms in physics are not defined very sharply. In your example, I do not know if there is a universal rule whether a superposition of two or more different momentum eigenstates would be called a photon or if only the eigenstate to a momentum is called a photon. I'd call it a photon, maybe Swansont wouldn't. Same goes for a definition of mass, which I only need when I actually need mass (why would I care about a mass of light in your example?). This fuzzyness about terms usually does not become a problem in interaction between scientists, since they (a) are trained to properly define what they mean where necessary situationally, and (b) usually have interactions with the intent to understand each other, not to correct each other.

 

Just because Prof. Hawking doesn't put his statements into context in his pop-sci books (which he is proud of that they contain no math at all) that does not mean that Prof. Hawking does not know about their context. Your asking about things that seem incorrect is an important step towards understanding contexts. Putting forward such questions with the intention to shed light on the rotten building of physics rather than the humble attempt to increase one's understanding is understandable (I've been a young physics student myself at some point, and in hindsight I really don't envy by tutors) but somewhat silly. In fact, picturing a "constructed house of physics" is questionable in the first place.

Thanks for all that. I especially enjoyed you're multiple comments on 'context'.

 

Perhaps you can help me resolve a contextual issue. Refer to Weinberg, QTF, volume 1, section 5.9. The discussion begins expressing the need for gauge invariance.

 

By eq. 5.9.30 the need seems to have been dropped, where the temporal component of the vector potential must vanish in all inertial frames.

 

However, if we begin with a Maxwell's equations, then reconstruct using a complex (or real) gauge invariant vector potential on a pseudo Riemann manifold, we can eliminate all 8 equations, and replace them with physical correspondences. This reduces to the dynamical constraint dJ/dA=0 upon demanding extremal action.

 

The contrast is large where, 5.9.30 assumes the Lorentz group. Weinburg's context seems to be only flat space time. We seem to have at least two points of difference.

 

I'm sure Weinberg will correct the context of quantum field theory in volume 2 upon a more realistic manifold, though I don't posess volume 2. Do you?

Edited December 3, 2013 by decraig

[decraig]

You claim to have calculations that show that light has nonzero mass, but the calculations you've given don't show that. All you've done is assert that radial symmetry implies a nonzero mass. So how about showing the calculations that show what you claim?

Ok, I see.

 

[math]c^4m^2=E^2-p^2c^2[/math] and radial symmetry implies [math]p=0[/math], implies [math] m = +/- E/c^2[/math]

[swansont]

Ok, I see.

 

[math]c^4m^2=E^2-p^2c^2[/math] and radial symmetry implies [math]p=0[/math], implies [math] m = +/- E/c^2[/math]

 

But that's not what radial symmetry implies. Saying p = 0 owing to radial symmetry would be true if averaged over many events, and you were looking at the vector sum. Or, equivalently, if you look at the probability. When you actually measure the momentum for any individual event, it's not zero. The photon has momentum and emitting particle recoils. The effect is exploited in laser cooling.

[decraig]

 

But that's not what radial symmetry implies. Saying p = 0 owing to radial symmetry would be true if averaged over many events, and you were looking at the vector sum. Or, equivalently, if you look at the probability. When you actually measure the momentum for any individual event, it's not zero. The photon has momentum and emitting particle recoils. The effect is exploited in laser cooling.

Yes it does. But we're quibbling n getting no where.

 

For a single quanta radial symmetry is broken upon the measurement you invoke. However you think trajectory has objective existence before measurement.

 

OK, but now you can't explain interference objectively without invoking Bohmian pilot waves. Quantum mechanics is a mess of senselessness.

 

We can just leave quantum mechanics as a theory in search of "making sense". Nobody else really cares who wins this butting contest between you and me. I was just looking for ideas. Truce? Unless you happen to have some special insight outside the mainstream philosophizing....

Edited December 17, 2013 by decraig

[Enthalpy]

You implicitly suppose, in spherical symmetry, that a particle (or worse, two particles; not a "single quantum" here) is described by one single wave. This is not the case, for instance with entangled photons, who can be of linear or circular polarization; a single wave does not explain the correlation both in linear and in circular detectors.

 

If you properly describe the spherical symmetry as a spherical distribution of possible waves, then these individual waves have the proper spin, polarization and so on.

[decraig]

You implicitly suppose, in spherical symmetry, that a particle (or worse, two particles; not a "single quantum" here) is described by one single wave. This is not the case, for instance with entangled photons, who can be of linear or circular polarization; a single wave does not explain the correlation both in linear and in circular detectors.

 

If you properly describe the spherical symmetry as a spherical distribution of possible waves, then these individual waves have the proper spin, polarization and so on.

Not spherical symmetry, but radial symmetry.

 

A particle is described by a single non-particulate field, so I'm not sure why you seem to be arguing differently. Again, I get the argument about detectors or measurements.

 

Detection breaks symmetry. You are changing the scenario. Once measured, or detected the mass/energy of a quanta has no issue. The issue I present is before measurement.

 

I'm perplexed. It seems to be a standard way of thinking among amatures that fundamental particles act like little bee-bees. But no theory beyond bohmian mechanics supports propagation of particle-like matter. And bohemian mechanics has it's own issues requiring field propagation as well as particulate matter, and has other problems.

 

I think the confusion in particle physics stems from the naming. It shouldn't be called 'particle physics' at all. 'Wave mechanics' would be closer.

 

I suggest watching the Feynman lectures he gave in New Zeeland. http://www.youtube.com/playlist?list=PL5DB4C82BDD7375E4

Edited December 21, 2013 by decraig

[Enthalpy]

I do argue that a single wave (including a weighed sum of base waves) does not suffice to describe some emitted photons. With two polarization-entangled photons, if you write the wave as linear polarized, you can't explain to correlation between circularly polarized detectors. If you write the wave as circularly polarized, you can't explain the correlation between linear detectors. You have to say "it can be this wave, it can be an other".

 

Then, accepting that a single wave does not suffice, you can write a spherical sum of waves that do have a polarization - just the usual EM fields for photons.

[Enthalpy]

Changed my mind about a single wavefunction describing an (or several) emitted photon. If using the electric field as a wavefunction for the photon, sometimes one function doesn't suffice. If using the ubiquitous Psi function, it does. More thoughts there

http://www.scienceforums.net/topic/80780-spin-2-what-wavefunction/#entry783582

after the second ----------.

[Extracted from the original question]

[math]c^4m^2=E^2-p^2c^2[/math].

 

However, for radiation emitted from an atom, or particle annihilation, in the center of momentum frame, the emission is radially symmetric.

This implies that the total momenta of an electromagnetic quanta is zero, giving nonzero mass. [...]

 

I would use

E² = m₀²c⁴ + p²c²

only for one particle, or for a group with identical speed vectors. Not for the summed vector p of several particles.

 

Then each photon can keep its zero mass.

-----------------------------------

 

There is more in the original question.

 

If we take a more intuitive system of several objects, say, a lead atom, or a satellite with a spinning momentum wheel, instead of two photons:

 

The electrons it the atom, or the momentum wheel in the satellite, have kinetic energy. If we decide to accelerate the atom or satellite, this kinetic energy adds inertia to what the rest mass of the electrons or wheel would bring. It also creates more gravity than the rest mass alone.

 

As seen globally, we can ignore (for not knowing the details, or not wanting to know them) the kinetic energy of the electrons or the momentum wheel, and just take an atom or satellite mass bigger than the constituents' rest mass. This bigger mass serves both to accelerate the global object and to compute the gravity it creates.

 

An interesting case are the particles of unclear composition, as are still the proton and the neutron: no consensus about how much kinetic energy, rest mass of constituents (if this makes sense) and so on. But we know a rest mass for the global particle, perfectly useable. Just compute with its center of inertia and observed mass.

 

When computing just the movements of the composite object's center of inertia, what is kinetic energy and interaction energy of its constituents becomes rest mass of the composite. This means also that we can't tell by that way if a particle is fundamental.

 

Now, for your two photons:

 

If someone decided to consider the system of two photons, with their vector sum p=0, then he would attribute a rest mass to the system. But not to each photon: the rest mass of the system results from the kinetic energy p*c of the constituents instead.

 

This would be useful if one can act on a system of two photons: "accelerate" both, measure the gravity field they create... Nothing easy, and computing with separate photons is more natural since the system is free, not bound.

 

In that sense,

E² = m₀²c⁴ + p²c²

still applies to several objects with different speed vectors, but then you must take a system rest mass m₀ which is not the sum of the constituents' rest mass. m₀ must include the kinetic energies of the constituents and their interaction energy.

Edited December 25, 2013 by Enthalpy