How Can Neutrons Emit Bosons Much Heavier Than Them

  • #1
The absorption or emission of W bosons changes neutrons to protons or protons to neutrons.Also,W bosons are almost 100 times as massive as the proton.The question is-How can a neutron emit a particle much more massive than itself and convert into a proton?
 

Answers and Replies

  • #2
Bill_K
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Because the W boson is virtual. Energy and momentum are conserved, but the virtual W in the intermediate state is not required to obey the mass-energy relation of a free particle.
 
  • #3
sorry,but I did not understand.What do you mean by 'virtual'?
 
  • #4
Drakkith
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sorry,but I did not understand.What do you mean by 'virtual'?
Virtual particles are a mathematical way to describe the interaction of forces. They aren't real particles.
 
  • #5
If they are a mathematical interpretation of interaction of forces,then how can they have mass?
 
  • #6
They mediate weak nuclear force, they are real particles, heavier then atoms of iron, they act as force carriers for weak nuclear force.
 
  • #8
Bill_K
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A freely propagating W boson has a mass of about 80 GeV, as you'll find it listed in the particle tables. A virtual particle exists for only a brief moment between the event when it is created and the event when it is absorbed. Whether you call it a 'real' particle or a 'mathematical fiction' is purely philosophical. Nevertheless it has all the properties of a real particle but one: its mass is not well defined. It's mass can be anything. The particle is therefore said to be off the mass shell. The probability associated with the virtual particle decreases the farther off the mass shell it is. The low probability for weak decays is what exactly led to the idea that they were mediated by a heavy virtual particle.
 
  • #9
649
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The mass of the W quantum field is defined in its Lagrangian, and therefore influences its behaviour, even for off-shell states. E.g. it appears in the expression for the propagator.

So I don't think it is appropriate to say that the mass of the virtual W is "not defined" or that "it can be anything". It is just not related through p^2 = m^2. p^2 can be anything, but its mass is fixed and can be read off from the Lagrangian, or by checking the singularities of the propagator.
 
  • #10
550
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The mass of the W quantum field is defined in its Lagrangian, and therefore influences its behaviour, even for off-shell states. E.g. it appears in the expression for the propagator.

So I don't think it is appropriate to say that the mass of the virtual W is "not defined" or that "it can be anything". It is just not related through p^2 = m^2. p^2 can be anything, but its mass is fixed and can be read off from the Lagrangian, or by checking the singularities of the propagator.
Since, [itex]p^2[/itex] is what determines both the inertial properties of a particle and the available energy for decays, it seems a little odd to insist that the pole mass of the field is really the particle's mass.
 
  • #11
649
2
Since, [itex]p^2[/itex] is what determines both the inertial properties of a particle and the available energy for decays, it seems a little odd to insist that the pole mass of the field is really the particle's mass.
Well, in the standard quantum field theory which governs particle physics as we know it, the mass of the quantum field is defined right there in the lagrangian, and can be obtained from the prefactor of the term square in the field operator.

I try to avoid defining single concepts in several different ways, and prefer the mass to be a constant value to minimize confusion.

I think it would cause unnecessary confusion to work with a mass of the quantum field (m), and a mass for each possible single-particle excitation of the field, depending on its value of p^2 which can have any value for internal states in loop perturbation expansions.
 
  • #12
550
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Well, in the standard quantum field theory which governs particle physics as we know it, the mass of the quantum field is defined right there in the lagrangian, and can be obtained from the prefactor of the term square in the field operator.

I try to avoid defining single concepts in several different ways, and prefer the mass to be a constant value to minimize confusion.

I think it would cause unnecessary confusion to work with a mass of the quantum field (m), and a mass for each possible single-particle excitation of the field, depending on its value of p^2 which can have any value for internal states in loop perturbation expansions.
If you're going to argue based on loops, it seems inconsistent to argue that the Lagrangian mass parameter is the mass of the field, since one-loop corrections require that this parameter actually be either 0 or infinite (in the limite of infinite cutoff scale) in order that the pole mass in the propagator be finite.

This, however, is really neither here nor there, since even real particles are not necessarily created right at the mass pole if they're unstable. The point I've been attempting to make here is that, while the pole mass of a field is extremely important for understanding its dynamics, calling it the mass, when talking about excitations of the field, disallows contact with the ideas we have about what mass means in pretty much any other area of physics. From the point of view of mass as a defining property of particle kinematics and dynamics, [itex]\sqrt{p^2}[/itex] is the correct quantity to call mass, not m.

As for the issue of particles in loops, there's nothing overtly wrong with those particles having really weird masses. After all, the only invariants that have any physical meaning are those that can be constructed entirely from external momenta. The difference, then, from the beta decay case that started this discussion is that, in that case, the mass of the W is one of the invariants that can be constructed from external momenta.
 
  • #13
649
2
If you're going to argue based on loops, it seems inconsistent to argue that the Lagrangian mass parameter is the mass of the field, since one-loop corrections require that this parameter actually be either 0 or infinite (in the limite of infinite cutoff scale) in order that the pole mass in the propagator be finite.
Since a quantum field theory should be accompanied by a choice of cut-off scale, I'd use the renormalized mass. At this choice of cut-off scale, the mass of an excitation of a given quantum field would be the same no matter what particle behaviour we are considering (on or off shell). And this mass would also appear in the Yukawa potential for interactions mediated by this field.

I would like to avoid getting into along discussion about terminology. I acknowledge that there are arguments for both viewpoints. To me, simplicity of terminology is important.
 
  • #14
88
2
The absorption or emission of W bosons changes neutrons to protons or protons to neutrons.Also,W bosons are almost 100 times as massive as the proton.The question is-How can a neutron emit a particle much more massive than itself and convert into a proton?
Let's see if I can kind of give you a simple, hand-waving explanation.

A virtual particle with a significantly large amount of mass can be created out of nothing - out of vacuum. The mass or energy is essentially "borrowed" from the vacuum - but you need to "put it back" extremely quickly before the universe notices that it's missing, and the more energy you "borrow", the less time it can be borrowed for. As long as these virtual particles only exist for a fleeting instant, normal ideas about conservation of mass or energy can be ostensibly ignored.

The more mass you create in a virtual particle, the shorter the lifetime of the virtual particle can be before it is destroyed. This can be expressed in the energy-time uncertainty principle.
 

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