Do fundamental particles have density?

[gib65]

I was thinking about this question today and started wondering whether it even makes sense to talk about the density of a fundamental particle.

 

It makes sense to talk about the density of, say, a nucleus. You would define it as the amount of volume taken up by the protons and neutrons (as opposed to the space between them) over the amount of total volume enclosed by the nucleus as a whole. The density of protons and neutrons could in turn be likewise defined except in terms of quarks and the space between them. But a quark is (supposedly) fundamental. It has no internal parts.

 

So do we say that its density is 1 (occupied volume / total volume = 1), in which case all fundamental particles have a density of 1, or do we put it in terms of mass over volume, in which case some fundamental particles could be said to be more dense than others, or do we say that density simply doesn't apply to fundamental particles?

[ajb]

This is really a question about the size of a fundamental particle. This is quite subtle.

 

In classical theory a particle is just a point. It has no size. However, a particle has an effective size as determined by its interactions with fields, say the electromagnetic field. Look up scattering cross-section, which you should think of as the "target size" for scattering experiments. This produces a finite "size" of a particle, though it will depend on for example the energy of the photons scattered.

 

Another "size" is the Compton wavelength. You can think of this a quantum smearing of a particle. It is equivalent to the wavelength of a photon whose energy is the same as the rest-mass energy of the particle.

 

See what happens when you put the Compton wavelength of a particle equal to it's Schwartzchild radius.

[Horza2002]

You need to remember that particles are not particles....they are particles and also waves. so density on the qutanum scale loses its meaning compared to classical items. Even protons can be treated in a classical manor to a reasonable degree of accuracy.

 

Just a note, quarks are not considered fundamental particles because they are never observed on their own. They are always bound to at least one other quark or more commonly two others. Your right in that they dont appear to be amde of anything else but just thought I'd tell you that! Electrons are, however, considered fundamental since they can be observed on their own

[Bob_for_short]

I was thinking about this question today and started wondering whether it even makes sense to talk about the density of a fundamental particle.

The notion of density makes sense if your theory needs it. Normally the fundamental theories do not operate with particle densities so it does not make any sense - it is just not involved anywhere.

 

Another thing is that particles are always in interactions. Electron clouds determine the atomic sizes. We cannot say that the electron is smaller than that. Excited atoms are even larger. Ensembles of atoms (matter) can have mass and electron densities - they are involved in some calculations.

[Severian]

Just a note, quarks are not considered fundamental particles because they are never observed on their own.

 

Why would that stop them being fundamental? It is also not true - we have observed asymptotic freedom. At high energies quarks are not confined.

[Bob_for_short]

Why would that stop them being fundamental? It is also not true - we have observed asymptotic freedom. At high energies quarks are not confined.

 

There is some contradiction here between "fundamental" and free or independent. If quarks are always bound, i.e. come together with other quarks, who of them is more fundamental?

 

At high transferred energies nothing looks confined (electrons in atoms, nuclei, etc.).

[michel123456]

cancelling wrong comment of mine

Edited January 14, 2010 by michel123456
Consecutive posts merged.

[louis wu]

Well if reports of quark stars are true then colour superconducting quark matter is a real physical state of matter. From the volume and mass of candidate quark stars a density can be ascribed to quarks that matches the way density is calculated for other states of matter. Neutron degenerate matter has a calculated density ~5×10¹⁷kg/m3.

 

http://arxiv.org/PS_cache/arxiv/pdf/0709/0709.4635v2.pdf

This report uses some concepts that I am not very familiar with, so I will not endorse it unequivocally.

[occam]

In answer to the Exam question, Do Particles have density?

This is really a question about the size of a fundamental particle. This is quite subtle, It has no size.

Rubbish! The answer is yes, but it’s an elusive quantity.

There is no need to delve into the exotic, just consider a bar of iron, of a known mass. If you heat it, it expands, which means the constituent atoms occupy more space. Sub-Atomically a change in “Energy State” results in a change of radius.

Another "size" is the Compton wavelength. You can think of this a quantum smearing of a particle. It is equivalent to the wavelength of a photon whose energy is the same as the rest-mass energy of the particle

Tut! The “Compton wavelength” is a version of the “equivalence” which you will find in any Physics book. E=hf and f = c/l(lambda wavelength) so

Wavelength l= hc/E If now we substitute mc^2 for E we get l= h/mc. This is the “Compton” equation and applies to any mass.

For the Proton, this yields a value of 1.3214098446x10^-15 whereas all sources indicate that the (rms) radius of a Proton is 8.7460177430x10^-16. This is not a reliable way of calculating “size”. I cannot relate to “quantum smearing” as an excuse for this difference.

In classical theory a particle is just a point.

“classical”? A “point” is a useful mathematical technique for measuring for instance a fulcrum where forces interact. In calculating orbits Jupiter is envisaged as a “point”.

Even a Photon is not a “point” unless someone has found a way to get red light through a blue sized hole.

[Klaynos]

I'll let someone else comment on your other points occam, but subwavelength hole emission for optical photons is a well documented phenomena.

[ajb]

There is no need to delve into the exotic, just consider a bar of iron, of a known mass.

 

Nothing I have said is exotic. Also a bar of iron is not a fundamental particle.

 

Tut! The “Compton wavelength” is a version of the “equivalence” which you will find in any Physics book. E=hf and f = c/l(lambda wavelength) so

Wavelength l= hc/E If now we substitute mc^2 for E we get l= h/mc. This is the “Compton” equation and applies to any mass.

For the Proton, this yields a value of 1.3214098446x10^-15 whereas all sources indicate that the (rms) radius of a Proton is 8.7460177430x10^-16. This is not a reliable way of calculating “size”. I cannot relate to “quantum smearing” as an excuse for this difference.

 

The Proton is again not a fundamental particle.

 

 

“classical”? A “point” is a useful mathematical technique for measuring for instance a fulcrum where forces interact. In calculating orbits Jupiter is envisaged as a “point”.

 

Really, when anyone talks of a particle we are thinking about a an idealised situation being described in the context of a mathematical model.

 

So, classically one thinks of fundamental particles as point-like. However a description that matches nature more closely includes quantum effects. One can think of these quantum effects as giving a size to the particles.

 

One method of defining some kind of extension without introducing further interactions is via the Compton wavelength. This, as you have noticed does not necessarily match up to the effective size or the classical radius.

 

You should thing of the Compton wavelength the minimum possible size of a "quantum particle", or better it is the limit of localising the position of the particle.

 

Another, maybe more useful interpretation of Compton wavelength is the scale at which quantum field theory comes into play.

 

So, one way we could define a density of a fundamental particle is directly from the Compton wavelength.

 

Let us assume the "particle" is spherical and that the radius is given by the Compoton wavelength. Then we can express the volume in terms of the wavelength as

 

[math] Vol = \frac{4}{3}\pi \left(\frac{h c}{mc^{2}} \right)^{3}[/math].

 

Now divide by the mass

 

[math]\frac{Vol}{Mass} = \frac{4}{3} \pi \frac{h^{3}c^{5}}{(mc^{2})^{4}} [/math].

 

Which looks like ("one over") a fundamental maximum density of a particle. However, as the derivation does not take onto account gravity in any way (nor in fact any other interaction) we cannot really associate this interpretation with it. I have no idea what this density really means, but we calculated something semiclassically.

Edited February 16, 2010 by ajb

[Mr Skeptic]

Well, I'd have to say that the minimum size for a fundamental particle would be its Schwarzschild radius, as smaller than that it would be a black hole. Then a density can be calculated, with the more massive particles being less dense. The Compton wavelength method results in more massive particles being more dense, instead.

 

How about if we use the quantum wavefunction of a particle, and multiply the probability of a particle being somewhere by it's volume, then divide by the mass? Not sure how useful it would be, but it would be a density of some kind.

 

In any case, the density is rather moot as a particle interacts as a whole, or not, with a certain probability.

[ajb]

Not sure how useful it would be, but it would be a density of some kind.

 

In any case, the density is rather moot as a particle interacts as a whole, or not, with a certain probability.

 

I have not come across any calculation in particle physics that requires any notion of the (mass) density of a fundamental particle.

[occam]

Nothing I have said is exotic. Also a bar of iron is not a fundamental particle.

The Proton is again not a fundamental particle.

 

It doesn’t matter, If I put my body weight into the Compton formula my “wavelength” is an absurd 2.25 x 10^-44 metres. Not surprising, more mass = more energy and more energy = shorter wavelength.

Quite simply Compton wavelength has absolutely nothing to do with particle size.

 

I'll let someone else comment on your other points occam, but subwavelength hole emission for optical photons is a well documented phenomena.

Yes I am aware of this, but as I understand it, this is due to the interaction between the incident photons and “Plasmons” in the material of the film which carries the sub-wavelength apertures. If I understand this correctly the transfer of energy is through the plasmons, and only occurs in an “array” of apertures, of precisely defined apertures and fringes, .and this is highly sensitive to the photon wavelength and incident angle. Curious things happen when photons interact with “atoms” As with “edge diffraction” if we take “white light” through a camera iris a spectrum appears at the edge of the image. If you use a single aperture is there is a progressive loss of wavelengths as the aperture is “stepped down”.

I stand by my assertion that a “photon” has to be a “volumetric” entity, with a diameter equal to its wavelength.

[Klaynos]

Yes I am aware of this, but as I understand it, this is due to the interaction between the incident photons and “Plasmons” in the material of the film which carries the sub-wavelength apertures. If I understand this correctly the transfer of energy is through the plasmons, and only occurs in an “array” of apertures, of precisely defined apertures and fringes, .and this is highly sensitive to the photon wavelength and incident angle. Curious things happen when photons interact with “atoms” As with “edge diffraction” if we take “white light” through a camera iris a spectrum appears at the edge of the image. If you use a single aperture is there is a progressive loss of wavelengths as the aperture is “stepped down”.

I stand by my assertion that a “photon” has to be a “volumetric” entity, with a diameter equal to its wavelength.

 

Your understanding is OK, ish, but only for optical frequencies. Metals at microwave frequencies cannot suppor surface plasmon polaritons, and therefore that kind of understanding cannot be applied, yet you can easily create systems where you have propagation through subwavelength apertures.

[ajb]

You can think of the Compton wavelength as being a fundamental limit on the localisation of a point particle taking into account special relativity and quantum mechanics.

 

In this sense, it gives a "size" to a point particle.

 

This size is not the same as the effective size as defined via interactions and the scattering cross section. Which is a quasi-classical notion. You think of point particles coming in and scattering off a tiny solid ball. The problem is the cross section depends on the momentum of the scattering particles, I don't see that it gives a "fundamental size".

 

And here I think is the problem. It is difficult to apply classical notions to fundamental particles.

 

So Occum, how would you define the size?

[occam]

Your understanding is OK, ish, but only for optical frequencies. Metals at microwave frequencies cannot suppor surface plasmon polaritons, and therefore that kind of understanding cannot be applied, yet you can easily create systems where you have propagation through subwavelength apertures.

I think discussing photon/atom interactioss here will take us a long way off the topic of sub atomic particles. I will send you a message suggesting another thread.

Occam.

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Merged post follows:

Consecutive posts merged

And here I think is the problem. It is difficult to apply classical notions to fundamental particles.

But you can’t just ignore it. If you have an object which you can (just about) measure, and you postulate that this is made of objects you are unable to measure, then these must also have some properties which lead to the properties of the object you can measure.

I’m not a fan of the “pawnbroker sign” concept of the three quarks in a photon, but the available data shows “Up” quarks at 5 MeV, and “Down” quarks a 6Mev, but two ups and a down at 16Mev is only 1.7% of the mass pf a proton. So what’s the rest of it made of?

 

By implication when you use the words “point” and “fundamental” you claim that there is nothing more to be found.

I find that rather sad.

[ajb]

I’m not a fan of the “pawnbroker sign” concept of the three quarks in a photon...

 

 

You mean proton and similar.

 

The three particle construct is nieve when you learn a bit more about quantum field theory.

 

but the available data shows “Up” quarks at 5 MeV, and “Down” quarks a 6Mev, but two ups and a down at 16Mev is only 1.7% of the mass pf a proton. So what’s the rest of it made of?

 

This is because it us a bound system. Look up mass defect.

 

By implication when you use the words “point” and “fundamental” you claim that there is nothing more to be found.

 

Look up fundamental particle.

 

By fundamental we mean no known substructure.

 

I find that rather sad.

 

I don't understand why you are sad.

[occam]

By fundamental we mean no known substructure.

OK I accept the semantic correction, but the question is not about the sub structure. The quarks have two known properties, mass and spin, so they have some “structure”. I find the concept of a point with angular momentum bizarre. What I find sad is that you do not seem to accept the possibility that these particles may also have “displacement”, whilst their assembly certainly does.

 

This is because it us a bound system. Look up mass defect.

Actually, I have done some work on this:-.Simply if you take the published masses of the Proton and Neutron the smallest combination (deuterium) falls a bit short of P+N. This Atomic mass unit is normalised at Carbon = 12. What becomes interesting is that if the data is arranged by “average nucleon mass” the progression of atoms becomes a catenary, with the lowest average mass around Iron. Analysing further, taking only the fully stable isotopes, this resolves into three distinct “curves” corresponding to “fermions”= particles with odd numbers of nucleons with fractional spins, Even numbered nucleons (with zero or unitary spins) also divide into those which can be divided by four, and the other even numbers. Each of these “stable” isotopes sits in a “well” of beta decay isotopes. This analysis, plus the data file of the 3000+ known isotopes, is much too large for an attachment in this forum but if anyone would like a copy, please contact me.

 

Returning to the exam question: The “bound” system is where the net mass is less than the components, the “Binding Energy” being that required to unbind the constituents to a “free” state. In the case of the Quark to Proton, the requirement is for a 98.3% increase in mass. So how does “quantum field” theory suggest where this additional mass come from?

[ajb]

What I find sad is that you do not seem to accept the possibility that these particles may also have “displacement”, whilst their assembly certainly does.

 

"Displacement"? What do you mean?

 

Do you mean "extent"? If so we are just going round in circles.

 

So how does “quantum field” theory suggest where this additional mass come from?

 

You need to think about bound states in QCD. This is difficult, mainly due to the nonperturbative behaviour of QCD at large distances. You can get a handle on this using lattice gauge theory. Masses decay constants and similar can be calculated.

 

I don't know much about this. I am sure a quick google will produce lots of papers.

[occam]

"Displacement"? What do you mean?

Do you mean "extent"? If so we are just going round in circles.

 

No “displacement” in the sense of Volume, which in turn implies “density”.

But you are quite correct; the current theories do not include this parameter.

But the answer does lie in the Standard Model and QCD, where Hadrons, such as the Proton is made of Quarks held together by “gluons” which are the gauge bosons carrying the strong nuclear force. These were experimentally detected in the PETRA experiments in 1979. According to the literature the Gluon has spin 1 and mass zero. (Or in some references “too small to measure”). If this “gluon” is to be “fundamental”, and the Quantum theory is to hold then this has to have a minimum value greater than zero. For Energy this minimum value is Planck’s constant, but this is not a unit of “Energy”. Its dimensional definition is Joule-seconds, which implies that Time is also a factor in calculating “Energy”. If mc^2 is to hold, its equivalent for mass is h/c^2, which dimensionally is Kilogramme-seconds with a numerical value of 7.3725 x 10^-51.

I will come clean. Using this basic value, and including the properties of “space” and “time”, I have managed to construct a numerical and spatial model of the Proton.

Now I must turn to the moderators. The paper is in this forum, Swansont and Klaynos know where it is. It is up to them whether the link appears in this thread.

 

Occam

[ajb]

If the gauge symmetry of QCD is not broken then the mass of the gluons must be identically zero. The experimental bounds put the mass to be less than a few MeV.

[pioneer]

If fundamental particles don't have density, where does the density come from when they combine to form classical particles? If we took a 1Kg weight of iron and compare that to all its fundamental particles, the first has density while the second does not. What changes? Does the particle-wave duality shift more toward the particle side of equilibrium for density and more toward the wave side of equilibrium to not form density?

[occam]

If the gauge symmetry of QCD is not broken then the mass of the gluons must be identically zero. The experimental bounds put the mass to be less than a few MeV.

Note my definition - Kilogram-seconds.

Mass is a time related multiple.

Occam

[occam]

It’s gone quiet. There are two things missing from the Standard Model and QCD: Space-time, and Gravity.

The key to this question is Density. Consider this table of numbers:

It is in .csv format and you may find it useful to put these into a spreadsheet and draw a line graph of the four columns..

-1,-4,-30,-88

-2,-7,-27,-82

-3,-10,-24,-76

-4,-13,-21,-70

-5,-16,-18,-64

-6,-19,-15,-58

-7,-22,-12,-52

-8,-25,-9,-46

-9,-28,-6,-40

-10,-31,-3,-34

-11,-34,0,-28

-12,-37,3,-22

-13,-40,6,-16

-14,-43,9,-10

-15,-46,12,-4

-16,-49,15,2

-17,-52,18,8

-18,-55,21,14

-19,-58,24,20

-20,-61,27,26

-21,-64,30,32

-22,-67,33,38

-23,-70,36,44

-24,-73,39,50

-25,-76,42,56

-26,-79,45,62

-27,-82,48,68

-28,-85,51,74

-29,-88,54,80

-30,-91,57,86

-31,-94,60,92

-32,-97,63,98

-33,-100,66,104

-34,-103,69,110

-35,-106,72,116

 

The figures are exponentials. The first column is “wavelength” -2 is centimetre waves, -6 is optical wavelengths, up to the Planck limit~-35.

Column 2 makes the assumption wave length = wave width = diameter and gives the notional spherical volume of the boson.

The Numerical value of the constants is in SI units, and to obtain the conversion values these must be considered against a time base of one second. Therefore the crucial calculation is how many “Boson volumes” will fit into a volume 1 light second in diameter. This gives “space-time – Volume per second”

Using the first constant h we can calculate the “Energy Density” and thus the coulomb force exerted by the volume. This is column 3.

Now in Column 4 we can use the second constant h/c^2 to calculate the “Mass density” and thus the gravitational force.

From the data you can see that for each order of magnitude decrease in wavelength the Coulomb force increases by 3 orders of magnitude, but the Gravitational force increases by 6 orders of magnitude.

Most significant is that these curves cross. At a wavelength of 6.8967 x10-21 metres the forces are equal, and at shorter wavelengths, or greater “density” the gravitational force is dominant.

 

The consequence is that at a lesser density the Boson momentum is linear, and is observed as a Photon. At higher densities the momentum is a closed angular momentum, and the Boson is observed as a “gluon” or more properly is the Graviton.

The full explanation can be found in the thread;

http://www.scienceforums.net/forum/showthread.php?t=42325&highlight=elephant