[lightarrow]

Why do you need a photon for this? A light pulse wouldn't work?

[Pmb]

Why do you need a photon for this? A light pulse wouldn't work?

Who said a photon was needed? Einstein used a "burst of radiation". In any case it's sufficient, to use a photon, not neccesary. If you read the derivation I pointed to by Antippa below Eq. ( they you'd see that it uses an atom emitting a photon. You can't have an atom emitting light, pe se

[lightarrow]

Ok, anyway, if you want to use the term "photon" because it's emitted or absorbed by an atom, you can't say it can be localized in flight. Either is a photon, with its quantum properties, or is a classical pulse of light; you can't mix the two in a sort of "chimeric" beast.

[Pmb]

Ok, anyway, if you want to use the term "photon" because it's emitted or absorbed by an atom, you can't say it can be localized in flight. Either is a photon, with its quantum properties, or is a classical pulse of light; you can't mix the two in a sort of "chimeric" beast.

I'm afraid that you’ve made the same mistake here that you’ve made in your previous posts.  You say it can’t be done but give no proof. In this case you claim “you can't say it can be localized in flight” but don’t explain what that means or why you can’t say it and what it means not to be able to say something when in practice (i.e. in practical examples, math and all) it works just fine.

I feel like I’ll just be repeating myself here so I’m ending my contribution in this thread and am as such agreeing to disagree. That means I won’t respond to any further assertions you make even if I  know them to be wrong.

I rest easy in the knowledge that has been done successfully on numerous occasions by many physicists in many textbooks and at least some journal articles. Most notably its now being used in the new version of Exploring Black Holes. I’d like to note that one of the authors of that text, Edwin Taylor, is also co-author of the MIT Introductory Physics Series Quantum Mechanics text.

[lightarrow]

Ok, anyway, if you want to use the term "photon" because it's emitted or absorbed by an atom, you can't say it can be localized in flight. Either is a photon, with its quantum properties, or is a classical pulse of light; you can't mix the two in a sort of "chimeric" beast.

I'm afraid that you’ve made the same mistake here that you’ve made in your previous posts.  You say it can’t be done but give no proof. In this case you claim “you can't say it can be localized in flight” but don’t explain what that means or why you can’t say it and what it means not to be able to say something when in practice (i.e. in practical examples, math and all) it works just fine.

When I answered to the question of what is the "centre" of a system of two photons (question intended in the sense that the photons were emitted but not yet absorbed) I said that it's not possible because you can't say where a photon is exactly in flight, that is after emission and before detection.
The proof is very simple, it only needs the Young experiment with two slits and photons emitted one at a time: if you can say which slit the photon passed through, the interference pattern disappears...

[JP]

Lightarrow, is there any reason why you can't take a classical limit of the quantum theory to come up with classical photons like Pmb claims?  That approach is certainly valid for electrons (and explains why we have "classical" electrons" when we know they also behave like waves in the 2 slit experiment).

[AndroidNeox]

All matter and energy has mass. Even the kinetic energy of an object has mass. In fact, when you compress a spring, its mass increases... not detectably, but the potential energy added to the spring has its own contribution to the total mass.

[lightarrow]

Lightarrow, is there any reason why you can't take a classical limit of the quantum theory to come up with classical photons like Pmb claims?  That approach is certainly valid for electrons (and explains why we have "classical" electrons" when we know they also behave like waves in the 2 slit experiment).

When you take the classical limit for a photon (h --> 0) it gets zero energy, so it disappears.

[lightarrow]

All matter and energy has mass. Even the kinetic energy of an object has mass.

If you mean relativistic mass, ok. If you mean "mass", with this term it's usually intended "invariant" mass and then it's false, unless the object rotates around a fixed point.

In fact, when you compress a spring, its mass increases... not detectably, but the potential energy added to the spring has its own contribution to the total mass.

Correct.

[JP]

Lightarrow, is there any reason why you can't take a classical limit of the quantum theory to come up with classical photons like Pmb claims?  That approach is certainly valid for electrons (and explains why we have "classical" electrons" when we know they also behave like waves in the 2 slit experiment).

When you take the classical limit for a photon (h --> 0) it gets zero energy, so it disappears.

Good point.  After reading that and a bit more thinking, I believe that what's going on with these "classical photons" is two limits.  If we have a field made of photons, the classical limit does not correspond to taking h->0, but rather to taking many photons.  In this limit, you recover Maxwell's equations, but you've lost information about the behavior of individual photons.

Then you take a second limit corresponding to wavelength->0 which gets you ray optics.  So essentially, your "classical photon" is an arbitrary packet of energy assigned to propagate along a ray.  It's related to real photons only insofar as the sum over many photons gets you the classical field, which you then use to define rays.

[Pmb]

When you take the classical limit for a photon (h --> 0) it gets zero energy, so it disappears.

(sigh!) I'm clearly sorry that I asked.

That's wrong. If you were right then no classical particles exist in the classical limit. Don't forget what h physically means. I means that for every quantum mechanical particle that has inertial energy (defined as E = K + E₀ = Kinetic Energy + Rest Energy) has an associated frequency given by the relationship E = hf. What does it mean to take h -> 0 for an electron? It means that there is no associated wavelength.

Recall that in classical electrodynamics one can have a very small packet/burst of radiation (which can be described by a Fourier integral) which has enegy and momentum p. The relationship between them is a non-quantum mechanical relationship, i.e. E = pc. The shape of the light pulse can be selected such that the spatial extention is small enough for all practical purposes.

[Pmb]

If you mean relativistic mass, ok.

You know that its relativistic mass from the context in which he used it. Its not wise to assume that everyone who uses the description that he does needs to be reminded of these facts. To assume so is condescending to the poster because we're assuming that he doesn't know what he's talking about and we need to correct them.

I recommend that if someone says "mass depends on speed/energy etc." then we simply assume that by "mass" they mean relativistic mass. Having to remind people all the time is a waste of everybody's time. People can get irritated when someone has to comment on it every time they use the concept, don't you think

[JP]

Recall that in classical electrodynamics one can have a very small packet/burst of radiation (which can be described by a Fourier integral) which has enegy and momentum p. The relationship between them is a non-quantum mechanical relationship, i.e. E = pc. The shape of the light pulse can be selected such that the spatial extention is small enough for all practical purposes.

Yep, and a photon is usually defined similar to a Fourier component (monochromatic plane wave) of a pulse.  It has a well-defined frequency, momentum and polarization, but no simple position representation (much as a plane wave has a well-defined frequency, direction and polarization, but exists over all space).  Just like a pulse which has a confined position can be expressed as a superposition of many plane waves, a classical beam which has a limited area in space can be expressed as a state consisting of many photons.  Once you have that state, you can invoke geometrical optics or similar approximations to make it appear like a classical particle, but you've taken two approximations: many photons->classical pulse->geometrical/partical approximation to Maxwell's equations. 

I've no doubt that you can make these two approximations in that order because I know the math fairly well.  So if "classical photon" means (essentially) classical approximation to a solution of Maxwell's equations, which are themselves a many-photon approximation to a quantized field, then yes--"classical photons" are a thing.

What I'm less sure of is this: can you go directly from the mathematical description of a photon and, by expanding in powers of h, (presumably by expanding the action?) get to a classical particle description?  Or is there something about photons that prevents you from doing this directly?  I know, for example, that photon wave-functions are controversial and unlike those of massive particles (http://arxiv.org/pdf/quant-ph/0508202v1.pdf). 

[Pmb]

Yep, and a photon is usually defined similar to a Fourier component (monochromatic plane wave) of a pulse.

Huh? What is a "Fourier component (monochromatic plane wave) of a pulse."?

[JP]

A monochromatic plane wave is a solution to the wave equation that has both a definite direction and frequency.  You can write it as

where k is wave vector (2*pi/lambda*direction), x is position, omega is (angular) frequency and t is time.  Each of these solutions satisfies a wave equation, so any pulse built from these satisfies the wave equation.  Since they're exponential solutions, the integral is a Fourier integral that takes a weighting function in k and omega to x and t.

[lightarrow]

When you take the classical limit for a photon (h --> 0) it gets zero energy, so it disappears.

(sigh!) I'm clearly sorry that I asked.

That's wrong. If you were right then no classical particles exist in the classical limit. Don't forget what h physically means. I means that for every quantum mechanical particle that has inertial energy (defined as E = K + E₀ = Kinetic Energy + Rest Energy) has an associated frequency given by the relationship E = hf. What does it mean to take h -> 0 for an electron? It means that there is no associated wavelength.

I have already replied you (in a previous post of this or another similar thread, don't remember) that a photon is different because has zero mass. For an electron, you can have non-zero momentum p even at very low speeds, because of its non-zero mass. Then De-Broglie relationship: = h/p tells that, in the limit h --> 0, = 0, that is, frequency should be infinite.
For a photon you can't do it, because its momentum p too would vanish, in the limit h --> 0.

[AndroidNeox]

All matter and energy has mass. Even the kinetic energy of an object has mass.

If you mean relativistic mass, ok. If you mean "mass", with this term it's usually intended "invariant" mass and then it's false

[/quote]

Naturally I'm referring to relativistic mass, since that's what the question is about. The rest mass doesn't change because it's never in motion and has no kinetic energy. I was specifically referring to kinetic energy having mass.

[lightarrow]

All matter and energy has mass. Even the kinetic energy of an object has mass.

If you mean relativistic mass, ok. If you mean "mass", with this term it's usually intended "invariant" mass and then it's false

Naturally I'm referring to relativistic mass, since that's what the question is about. The rest mass doesn't change because it's never in motion and has no kinetic energy. I was specifically referring to kinetic energy having mass.

Certainly. However we should be more precise when we discuss this subject because it's very easy to make confusion. For example, saying "All matter and energy has mass. Even the kinetic energy of an object has mass" is very confusing: in the first sentence, matter has invariant mass, "energy has mass" is incorrect, since energy is "a property" of a body, and a property cannot have mass (as if I would say that a number has a colour); we should say instead that "a body which has energy has mass", but in this case is not always invariant mass...
As you see, things are not so simple.

[AndroidNeox]

Certainly. However we should be more precise when we discuss this subject because it's very easy to make confusion. For example, saying "All matter and energy has mass. Even the kinetic energy of an object has mass" is very confusing: in the first sentence, matter has invariant mass, "energy has mass" is incorrect, since energy is "a property" of a body, and a property cannot have mass (as if I would say that a number has a colour); we should say instead that "a body which has energy has mass", but in this case is not always invariant mass...
As you see, things are not so simple.

Energy does have gravitational mass. Put a kilogram of matter and one of antimatter into an impregnable box, like a Schrödinger cat box, and the mass of the box (any category of mass you care to choose) will not change when the contents annihilate each other. Even if the box only contains light, the mass(es) will not change.

[lightarrow]

Certainly. However we should be more precise when we discuss this subject because it's very easy to make confusion. For example, saying "All matter and energy has mass. Even the kinetic energy of an object has mass" is very confusing: in the first sentence, matter has invariant mass, "energy has mass" is incorrect, since energy is "a property" of a body, and a property cannot have mass (as if I would say that a number has a colour); we should say instead that "a body which has energy has mass", but in this case is not always invariant mass...
As you see, things are not so simple.

Energy does have gravitational mass.

1. Energy *cannot* have mass, or charge, or lenght, colour, as a number cannot have mass, charge, lenght, or colour. I've already written it but you probably still haven't understood it. Energy is *a property* of a body, and *not a body itself*. What would you answer to someone who stated that his "age" has "weight"?
Please, answer this question, before stating another time that energy has mass.
2. Even when your statement is correctly written, that is: "a body which has energy
also have gravitational mass" is not exact because a photon, by itself, doesn't have gravitational mass (before you contest this, think to a physics book where this is written). Instead "a region of space which has electromagnetic energy" does have.

Put a kilogram of matter and one of antimatter into an impregnable box, like a Schrödinger cat box, and the mass of the box (any category of mass you care to choose) will not change when the contents annihilate each other. Even if the box only contains light, the mass(es) will not change.

Correct, but it doesn't confirm your statement.
By the way, there is no need of matter and antimatter and not even of light in a box,  two photons are enough, because such a system have a non-zero mass (I mean invariant mass, not relativistic mass), I have already showed it in a recent thread and in several others, during the years.