Exchange force carrier between two protons

CAF123
Gold Member
Messages
2,918
Reaction score
87
I know that the virtual exchange particle between two electrons is the photon and this explains why they repel each other (ie virtual photon is spontaneously emitted from one electron A causing it to recoil. As the virtual photon hits electron B, it imparts momentum causing it to move away so that electron A and B move away in the opposite direction, as needed for conservation of momentum).

My question is can a Z particle also be the exchange carrier here because it too is neutral? I have the same question for the case of two protons.
Many thanks
 
Physics news on Phys.org
CAF123, You could regard the Z meson as a force carrier, like a photon, but there's a big difference -- the photon is massless, so the force field it produces (a Coulomb field) is long range, i.e. 1/r. The Z meson, on the other hand, is massive, very massive: 90 GeV. This means the force field it produces is short range (a Yukawa field), i.e. one that falls off exponentially, with a range equal to the Compton wavelength. The Compton wavelength of a 90 GeV particle is about .001 fermi, or 1/1000 the diameter of a proton.

Needless to say, at such a short range the interaction between two electrons (or two quarks) will be dominated by other effects, and so regarding the Z meson as having produced a force field is simply not a useful concept.
virtual photon is spontaneously emitted from one electron A causing it to recoil. As the virtual photon hits electron B, it imparts momentum causing it to move away so that electron A and B move away in the opposite direction
.
Might as well take this opportunity to point out that your attempt to visualize this interaction ignores its quantum nature. You're imagining that the momentum they exchange is in the same direction as the spatial separation between them. Therefore it appears to cause a repulsion. (And so, often we are asked next, then how can opposite charges possibly attract??) In fact, the interaction includes an integration over ALL momenta: forwards, backwards and sideways. The intensity of the interaction generates an effective potential, and the resulting force between the charges arises from the gradient of this potential.
 
I believe another reason is that the electromagnetic interaction at this scale is a lot stronger (order of 10^8 ) than that of the weak interaction
 
Nope, this is false. If you're talking about the interaction between the W or Z meson and other particles, it is about the same strength as electromagnetism.

The weak interactions only appear weak, due to the large masses of the W and Z. In a beta decay, for example, the W is not directly observed. It's virtual, an intermediate product of the decay. And since the decay energy is so much less than the W mass, the W is far off the mass shell, and this reduces the decay rate.

They're not intrinsically weak, just short range.
 
CAF123 said:
I know that the virtual exchange particle between two electrons is the photon and this explains why they repel each other (ie virtual photon is spontaneously emitted from one electron A causing it to recoil. As the virtual photon hits electron B, it imparts momentum causing it to move away so that electron A and B move away in the opposite direction, as needed for conservation of momentum).

My question is can a Z particle also be the exchange carrier here because it too is neutral? I have the same question for the case of two protons.
Many thanks

Protons are made up of quarks, which transmit forces via photons (electromagnetism) and gluons (strong force) themselves.

Z bosons can't mediate the interaction between two electrons or protons due to their mass. That is, they have mass so they can't travel as far as photons or gluons (both massless), which is required for the force particles of electromagnetism.

Additionally, the EM force has been observed repeatedly to travel at the speed of light, something impossible for the Z (which has mass), a direct consequence of special relativity.
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top