# Question about the Mass, Volume and Shape of Elementary Particles

1. Jul 4, 2010

### Bararontok

These are some of the inconsistencies with physics that should be resolved on this forum.

Question #1. Relationship Between Mass and Gravity)

Gravitational fields exist in elementary particles but cannot be detected in laboratory experiments because they are too weak. However can indirect evidence of mass and the gravitational interaction of elementary particles come from gravitational lensing and the attraction of these particles to large masses like stars, galaxies and black holes?

From: http://en.wikipedia.org/wiki/Black_Hole

"A black hole, according to the general theory of relativity, is a region of space from which nothing, including light, can escape."

"according to the theory of relativity mass is just highly condensed energy (E = mc^2)."

From: http://en.wikipedia.org/wiki/Mass-energy_equivalence

"In physics, mass–energy equivalence is the concept that the mass of a body is a measure of its energy content. In this concept the total internal energy E of a body at rest is equal to the product of its rest mass m and a suitable conversion factor to transform from units of mass to units of energy."

These statements point that all forms of energy including radiation which is comprised of freely moving particles like photons and neutrons have mass because energy and mass are the same. So does that mean that anything that has mass must have a gravitational field by default and that the mass is directly proportional to its gravitational field?

Question #2. The Impenetrability Paradox of Point Charges)

Elementary particles are regarded as point charges with zero volume but this contradicts the non-violable law of matter called impenetrability and could mean that two particles can occupy the same space at the same time.

From: http://avstop.com/AC/apgeneral/matter.html

"h. Impenetrability - simply stated means that no two objects can occupy the same place at the same time. Thus, two portions of matter cannot at the same time occupy the same space."

Does this prove that even elementary particles must have some volume, negligible or not?

Question #3. The Relationship Between the Omnidirectional Exertion of Forces, The Fundamental Interactions, and Their Ability to Give Rise to The Shape and Properties of Elementary Particles)

Can an elementary particle's shape come from the force fields that give property to these particles which are omnidirectional because from observations, forces like gravity and electromganetism attract from all directions.

From: http://en.wikipedia.org/wiki/Force_field_(physics)

"A global Gravitational field consists of a spherical array of vectors pointing towards the center of gravity. It is a classical force field, in spherical coordinates,"

Take a look at this image and notice the omnidirectional spheroid shape of the black hole. Does this not mean that force fields are omnidirectional?

http://www.spacescan.org/images/bhppppp_69.jpg [Broken]

Does this prove that force fields are omnidirectional and spherical? And does this also mean that the particles that are the source of these fields will also assume a similar shape because that is what they are made of at the fundamental level? After all if not for these fundamental interactions, how can particles exist and have properties like mass, charge, and nuclear binding?

Last edited by a moderator: May 4, 2017
2. Jul 4, 2010

### the_house

One thing to remember is that the problem is not really that the effect of gravity on elementary particles is too weak to measure by itself, it's that gravitational effects are typically much, much, much, much weaker than the other forces acting on such a particle (e.g., electromagnetic forces, etc.). So the trick to measuring gravitational effects is to devise a system where all other forces cancel to such an amazing extent that you can pick out the gravitational signal from the background of other effects. This is no problem for measuring the gravitational effects on large neutral objects like planets and stars, but it gets more and more difficult for smaller systems.

I believe gravitational effects have been measured with neutron beams, although they are not exactly fundamental particles.

I don't really know much about the possibility of measuring gravitational lensing from galaxies, unless you count photons as "elementary particles", which as far as I know is how such lensing is usually measured.

Yes, in general anything with mass will cause an associated gravitational field. More specifically, the entire stress-energy tensor acts as a source of gravity---i.e., energy density, momentum density, pressure, shear stress, etc.

But be careful reading too much into the mass=energy statement. For example, a photon has energy (and this energy can contribute to a gravitational field), but an individual photon has no mass. Energy should have certain properties before you call it "mass" (typically it's some sort of well-localized lump of energy that has an associated rest frame.)

"Impenetrability" is not a non-violable law. For example, two identical bosons can occupy the exact same quantum state, including position. Even identical fermions can have the exact same position as long as they differ by some other quantum number (e.g., spin up and down electrons.) In fact, I am not aware of any notion of "impenetrability" built in to any current fundamental theory.

I'm not sure what you're getting at here, so I'll have to leave this one to someone else.

3. Jul 4, 2010

### arivero

Bararontok, your question is more related to history of physics and its concepts that to actual elementary particle theory. The discussion about volume has been a source of confusion since the Vth century BCE. Particularly, volume can be a property of particles, a property of space (thus separting particles). For each possiblility, it is possible to define an adequate concept of Impenetrability.

I suggest to start with the very old physics framework, from lucretius etc. You have atoms (things) and space (no-thing), in a dual relationship: atoms are separated by space, space is divided by atoms (think about a line, and points in the line). Then you have idols, or "eidola", "phantoms", the images sent by the objects (by the atoms) and perceived in your detectors (eyes, etc). Idols do not have the property of impenetrability.

For atoms, impenetrability implies some need of ordering, to stack them in a consistent way (for instance, to proof Gauss or Stokes theorem). This was called by the ancients "diatigué", "contact-touch", and "trope", "position/orientation". There was a third property, "rythmos" or frequency, which we could consider akin to modern mass-density or Energy-momentum density. To define a solid object, you need to consider not only the atoms, the "thing", mut also the vacuum, the "no-thing".

To this framework, favoured in modern calculus theory and in field theory, you can oppose other approaches where "space" is a property of objects. This is the case of string theory, but it has happened in the past.

Moreover, you have different approaches to build macroscopic objects from all this theory.

4. Jul 5, 2010

### Bararontok

According to mass-energy equivalence, photons in motion can actually acquire mass which comes in the form of a corresponding relative gravity charge, and will only lose the gravity charge once it has been absorbed by a charged particle such as an electron or proton and converted into another form of energy such as heat. This means that even a freely moving photon is subject to interaction with gravity but this is only observable under extreme conditions.

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The statement that matter cannot occupy the same space at the same time might actually be true because fermions are the class of particles that give rise to matter which in turn gives rise to the force carrier particles. The force carrier particles can occupy the same space at the same time because it would be impossible for these forces to interact if their fields do not come in contact with each other and overlap. Proof of this is due to the fact that objects can stay within the gravitational fields of large masses such as the earth. If the force particles are impenetrable then no objects can fall into gravitational fields.

According to this statement:

"This "antisymmetric wavefunction" behavior implies that fermions are subject to the Pauli exclusion principle — no two fermions can occupy the same quantum state at the same time. This results in "rigidity" or "stiffness" of states which include fermions (atomic nuclei, atoms, molecules, etc.), so fermions are sometimes said to be the constituents of matter, while bosons are said to be the particles that transmit interactions (force carriers), or the constituents of radiation."

Source: http://en.wikipedia.org/wiki/Fermion

Could we then use the quantum state as the new definition of volume and use this property to calculate volume coupled with the omnidirectional exertion of the force carriers of the matter particle to achieve a definitive volumetric quantity for the fermions?

5. Jul 5, 2010

### the_house

In the standard nomenclature, a photon has exactly zero mass. As I mentioned, this does not mean that photons do not interact with gravity, since gravity couples to the entire energy-momentum tensor. Perhaps you are confused by the concept of "relativistic mass" that seems to pop up on this forum from time to time, but this is not standard nomenclature and (in my opinion) it is confusing and not useful.

As I mentioned, fermions do obey the Pauli exclusion principle. This indeed means that two identical fermions cannot be in the same quantum state. They can, however, have the same exact position, as long as they differ in any other quantum number. Moreover, this says nothing about two fermions that are not identical, or bosons, for neither of which is there any notion of impenetrability in the Standard Model. The Pauli exclusion principle does have important ramifications for the behavior of matter, but you appear to have things a bit confused.

6. Jul 5, 2010

### ZapperZ

Staff Emeritus
You're missing the point. There is ANOTHER class of matter, which are bosons. They don't follow the exclusion principle nor the Fermi statistics. And that in itself indicates that your general classification of matter is incomplete.

Furthermore, as has already been stated, there's nothing about the exclusion principle that prevents two fermions from being located at the same place if the spins are antisymmetric.

Until you correct these basic QM misunderstandings, you shouldn't be making extrapolations based on them.

Zz.

7. Jul 5, 2010

### Bararontok

So that means if these conditions are met then it would be possible to fit billions of fermions into a single space?

I have another question, what if we could use the range of the fermion's force fields which must be finite to define their volume assuming the quantum spins are symmetric?

8. Jul 6, 2010

### inflector

Ask yourself: How can more than two fermions have anti-symmetric spin at the same time?

9. Jul 6, 2010

### Bararontok

Good point so could we just take the finitely ranged fields of the two fermions to calculate the volume of the overlapping particles and make that our definition of volume after all once the space is occupied by two fermions at once then no more overlap can occur right? So does this impose a degree of impenetrability? After all an antisymmetric elementary particle pair can be collided with a similar pair in a particle accelerator and they would just scatter from one another but if they were completely dimensionless, they would just slip through each other as if nothing happened.

Last edited: Jul 6, 2010
10. Aug 13, 2010

### Bararontok

There is a question that has been boggling my mind lately. Since photons are the particles carrying the electromagnetic force and the charged particles carrying photons have finite amounts of energy stored in their electromagnetic fields, that must mean that the charged particles only carry a finite number of photons. So when a charged particle is energized to the radiative excitation level so that it can emit radiation, is it possible for a charged particle to run out of photons or does it continuously replenish its supply as long as an energy source delivers energy to the charged particle?

And since Bosons are not restricted by the Pauli Exclusion Principle, is it possible to compress infinite amounts of photons in a stream of radiation to have a signal that has an infinite radiant flux and frequency?

11. Aug 13, 2010

### the_house

Your questions seem a bit jumbled and confused, but I'll just try to state a bunch of true things and see if any of them help.

It seems you may have a classical picture in your head of a charged particle "carrying" some fixed number of photons along with it that it can then send out to interact with other particles. This picture will definitely do you no good. There are not necessarily any actual photons present near a charged particle. Even if it is interacting with another charged particle and you hear people talking about "exchanging virtual photons" (which probably only causes confusion if you're not familiar with quantum field theory), there's no use trying to imagine some set number of actual photons anywhere in the problem.

The notion of conservation of energy is a good one, though, so if some system actually radiates energy (in the form of electromagnetic energy or any other way) it is true that it can only radiate as much energy as it initially had available unless it absorbs energy from some other system.

On the other hand, a photon can have an energy that is arbitrarily close to zero. So even if you have a system of just photons with a finite amount of total energy, there is no limit to the number of photons there can be in that system.

If some composite system (like an atom) is excited, it will only emit energy until it is back into its ground state. Again, the actual number of photons is not the issue, just the total energy that they carry. Yes, if an energy source delivers energy to the system, that gives it more energy to radiate.

If you're thinking of a fundamental particle, it can radiate if it is being accelerated. This can go on indefinitely as long as whatever force is accelerating it continues to transfer energy to it.

In principle, one can have an arbitrary number of identical non-interacting bosons with the same wavefunction (e.g., in the same place). To a first approximation photons do not interact with each other, so this would mean you can have a stream of an arbitrary number of photons. However, the frequency has nothing to do with the number of photons. The intensity of the beam would be arbitrarily large, though (and it would of course take an arbitrarily large amount of energy to create such a beam).

Of course, in the real world, at some extremely large density, this picture will break down. Photon self-interaction will become important, or some other physics will come into play. You can't necessarily put an infinite number of bosons in the same place if they are interacting (e.g., if they repel each other strongly).

I hope this was at least somewhat helpful.

12. Aug 14, 2010

### Bararontok

So this means that an indefinite number of photons can be released to carry away excess energy so long as the system is not allowed to rest.

And talking about the limit imposed by photon self interaction, may this be used to calculate a finite but very small charge radius for the photon? Because the charge radius of the electron has already been determined by Penning Trap experiments to be 10^-22m.

13. Aug 15, 2010

### the_house

I don't know what you mean by " so long as the system is not allowed to rest," and I again am not sure what you're getting at.

I don't follow your logic, but no, there is probably no charge radius for the photon. Also, the charge radius of the electron has not been determined to be finite. It is consistent with zero, with an upper bound (presumably that is the 10^-22m number that you quote).

14. Aug 15, 2010

### Bararontok

What I mean to say is that so long as the system is energized so that it can emit charged particles. If no more energy is delivered to the system it will certainly stop generating photons.

The definition of charge radius that I am proposing is based on the notion of self-interaction and the fact that an infinite number of bosons cannot be compressed into a single space as you have stated earlier and not the conventional notion of a charge radius.

15. Aug 15, 2010

Staff Emeritus
Except that you can't just redefine something. There's already a definition of the charge radius. I am guessing that it is something like this:

$$r \equiv \sqrt{\frac{\int{r^2 \rho(r) d^3r}}{\int{\rho(r) d^3r}}$$

I may have some factors of 2 or pi missing, but this is the idea. Note that the denominator can be simplified:

$$r \equiv \sqrt{\int{r^2 \frac{\rho(r)}{q} d^3r}}$$

16. Aug 15, 2010

### Bararontok

Then it will not be called a charge radius. Based on the limitation of compression, the new term could be called a compressive limitation radius.