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- Thread starter jnorman
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I really don't know the details, but I think charge is not conserved in all cases because of violation of CP-symmetry which occurs in some rare cases.

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mathman

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CP violation does not effect charge. Basically the violation means that certain processes involving particles are not exactly the same (for example in decay route) as the equivalent process for antiparticles.

I really don't know the details, but I think charge is not conserved in all cases because of violation of CP-symmetry which occurs in some rare cases.

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K^2

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You are right to within flavor changes. A weak interaction can result in a flavor change, so you end up with a particle and anti-particle that aren't each-other's anti-particles. But charge must still be conserved in these.

For example, suppose you have a free neutron decaying into proton and electron. Electron isn't a part of neutron, and so it must be created. But if you create an electron and a positron, the net charge is not conserved if you also change neutron into proton. So instead, a neutrino and anti-neutrino are created. One of the down quarks in the neutron then weakly interacts with neutrino, exchanging electrical charge. Down quark becomes an up quark, and neutrino becomes an electron.

This is actually an example of a CP violation. Weak processes do not conserve CP.

There is also the bit where matter can change mass without particles or anti-particles being created, but then there isn't a need for charge to be created/destroyed either.

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I'm confused by this. Are you referring to the Dirac relativistic electron?If you do a "gauge transformation" on the Maxwell's equations (i.e. change A to A + del(f) and change V to V+df/dt where f is some function of space and time, and A and V are vector and scalar potentials) then Maxwell's equations stay the same and charge is conserved. Thats where momentum, energy, angular momentum and charge come from.

Classically, Maxwell's equations already contain conservation of charge in the form of the charge continuity equation.

Also, a regauging of (A,V) leaves F unchanged (and indirectly the charge and current). It is the regauging of F that directly leaves the charge unaltered. However, I don't know if there is such a real valued gauge.

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I should not act like an expert on this, because I'm not. I was not referring to Dirac's equation for the electron, but to electromagnetism in general - i.e. Maxwell's equations. Maxwell's equations express the conservation of charge, the way fluid mechanics equations express the conservation of momentum and energy. But with every conservation law, there is an associated transformation which leaves the equations (Maxwell's, fluid, etc) unchanged. Noether's theorem connects the two concepts, saying that every transformation which leaves the equations invariant corresponds to a conserved quantity. The fact that the fluid equations are invariant under translation corresponds to a conservation of momentum, for example.I'm confused by this. Are you referring to the Dirac relativistic electron?

Classically, Maxwell's equations already contain conservation of charge in the form of the charge continuity equation.

Also, a regauging of (A,V) leaves F unchanged (and indirectly the charge and current). It is the regauging of F that directly leaves the charge unaltered. However, I don't know if there is such a real valued gauge.

Regauging of (A,V) leaves F unchanged is, I think, a way of saying that Maxwell's equations are invariant under a gauge transformation. If regauging of (A,V) did not leave F unchanged, then, I'm guessing, a proof of charge conservation would fail.

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Yes, that's true--or at least we're not in disagreement. It's just that it's an indirect regauging. And without assuming the electric and magnetic fields are the derivatives of the more primative fieldsI should not act like an expert on this, because I'm not. I was not referring to Dirac's equation for the electron, but to electromagnetism in general - i.e. Maxwell's equations. Maxwell's equations express the conservation of charge, the way fluid mechanics equations express the conservation of momentum and energy. But with every conservation law, there is an associated transformation which leaves the equations (Maxwell's, fluid, etc) unchanged. Noether's theorem connects the two concepts, saying that every transformation which leaves the equations invariant corresponds to a conserved quantity. The fact that the fluid equations are invariant under translation corresponds to a conservation of momentum, for example.

Regauging of (A,V) leaves F unchanged is, I think, a way of saying that Maxwell's equations are invariant under a gauge transformation. If regauging of (A,V) did not leave F unchanged, then, I'm guessing, a proof of charge conservation would fail.

1) There may be a direct regauging, F <-- F' = F + dZ, where Z is the unknown gauge field, that leads to charge conservation.

2) Regauging A=(

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Hello, jnorman.

I think this is a very good question an I for one would be interested to read an adequate answer.

I am no expert in nuclear and (fundamental) particle reactions, but I do know that mass disappears in some and energy appears; what I am not aware of are any reactions where energy disappears and new particles appear instead.

However perhaps the experts in this field will lay out and discuss some of these?

I think this is a very good question an I for one would be interested to read an adequate answer.

I am no expert in nuclear and (fundamental) particle reactions, but I do know that mass disappears in some and energy appears; what I am not aware of are any reactions where energy disappears and new particles appear instead.

However perhaps the experts in this field will lay out and discuss some of these?

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On the misconception: mass converted to energy

http://en.wikipedia.org/wiki/Mass#Mass_and_energy_in_special_relativity"

http://en.wikipedia.org/wiki/Mass#Mass_and_energy_in_special_relativity"

In as much as energy is conserved in closed systems in relativity, the mass of a system is also a quantity which is conserved: this means it does not change over time, even as some types of particles are converted to others. For any given observer, the mass of any system is separately conserved and cannot change over time, just as energy is separately conserved and cannot change over time. The incorrect popular idea that mass may be converted to (massless) energy in relativity is because some matter particles may in some cases be converted to types of energy which are not matter (such as light, kinetic energy, and the potential energy in magnetic, electric, and other fields). However, this confuses "matter" (a non-conserved and ill-defined thing) with mass (which is well-defined and is conserved). Even if not considered "matter," all types of energy still continue to exhibit mass in relativity. Thus, mass and energy do not change into one another in relativity; rather, both are names for the same thing, and neither mass nor energy appear without the other. "Matter" particles may not be conserved in reactions in relativity, but closed-system mass always is.

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K^2

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Don't just complain; commit yourself. Why don't you explain how you think this

[tex]E=mc^2[/tex]

really means

[tex]E + mc^2 = constant[/tex]

or whatever it is you have in mind.

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K^2

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Because by convention, m means rest mass. It is true that

[tex]E = m_{rel}c^2[/tex]

But that's not the same thing. The equation for rest mass is different.

[tex]E^2 = p^2c^2 + m^2c^4[/tex]

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I think by convention, m means the relativistic mass, and rest mass m0

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OK, I see where my use of [itex]E = mc^2[/itex] was misunderstood.

Because by convention, m means rest mass. It is true that

[tex]E = m_{rel}c^2[/tex]

But that's not the same thing. The equation for rest mass is different.

[tex]E^2 = p^2c^2 + m^2c^4[/tex]

Forget about relativistic mass. It just confuses things.

Mass is the magnitude of the Lorentz convariant 4-vector of momentum, [itex]p^\mu[/itex], from which [itex]m^2c^4 = E^2 - p^2c^2 [/itex] immediately follows.

This continues to hold for a system of particles where the domain is special relativity but not general relativity so it's not really true anyways except in a narrow sense. In the center of mass frame,

[tex]E =mc^2 \ .[/tex]

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That is easily debatable if not wrong. Google "rest mass" and you will not find a reference which uses m for rest mass, all use m sub zero. I doubt you can find any reference which introduces m as the rest mass without explicitly stating such in order to eliminate confusion with the conventional m sub zero. Regardless, Einstein's statement [TEX]E=mc^2[/TEX] is the signature equation of special relativity, seen every where, and to ingenuously imply that it is an error or incomplete is misleading.That is a very old convention. Nobody uses it. In any modern article, m means rest mass because it is a Lorentz invariant quantity.

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K^2

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Open ANY particle physics paper from the past 50 years. You will never see a single mThat is easily debatable if not wrong. Google "rest mass" and you will not find a reference which uses m for rest mass, all use m sub zero. I doubt you can find any reference which introduces m as the rest mass without explicitly stating such in order to eliminate confusion with the conventional m sub zero. Regardless, Einstein's statement [TEX]E=mc^2[/TEX] is the signature equation of special relativity, seen every where, and to ingenuously imply that it is an error or incomplete is misleading.

Then your statement is plainly wrong. CoM frame mass of an atom that has emitted a photon changes without matter creation/annihilation process. (The photon has zero rest mass.)Phrak said:Forget about relativistic mass. It just confuses things.

Relativistic mass

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I thought this thread was all about nuclear and particle reactions such as power our nuclear power stations.

Thanks

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No expert here, but I think:Hello, jnorman.

I think this is a very good question an I for one would be interested to read an adequate answer.

I am no expert in nuclear and (fundamental) particle reactions, but I do know that mass disappears in some and energy appears; what I am not aware of are any reactions where energy disappears and new particles appear instead.

However perhaps the experts in this field will lay out and discuss some of these?

1. Gamma ray turning into an electron/positron pair

2. I think firing a fast electron into a nucleon may produce an extra neutron from the energy of the electron (ie. you end up with more mass then you started with and less KE then you started with)

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This is an extraordinary claim. Your emitted photons are interesting objects with definite momentum. Isn't the uncertainty in momentum in the order their energy? What is the expectation value, <p>?Then your statement is plainly wrong. CoM frame mass of an atom that has emitted a photon changes without matter creation/annihilation process. (The photon has zero rest mass.)

This is an extraordinary claim. Your emitted photons are interesting objects with definite momentum.

Relativistic massisconserved. Invariant mass is not. If you don't want to talk about relativistic mass, you are just wrong.

Do you have a citation for this theory or is it your own?

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Thank you, zincshow and welcome.

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