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This question is based on this blog post 'Is Dark Matter Lurking in Neutron Decays?' and a following comment I re-read recently. It has something I have seen often in popular writing, which I believe is confusing the on-shell and off-shell descriptions.

For example, the post says

The objection would be correct if the emitted W boson were a real final-state particle in the reaction. But the W is “virtual” here: it exists for a time so short that the energy budget does not get affected by its sudden appearance and disappearance. It is as if somebody stole a trillion dollars from the bank, and then put the sum back in place before any accountant could realize that the balance does not make sense.

But then

The W boson likes to be on-mass-shell: you can rather easily materialize a W boson if you invest 80.4 GeV or more of energy in your reaction, but if you ask the W to content itself with a smaller mass you will make a deal much less often. The poor down quark, having a single MeV or so of energy to invest, has to wait an eternity to pull it off: 15 minutes are a mindbogglingly long time for a quark!

My question is, are the two views contradictory? In the first, it is argued that the energy of 80.4 GeV is “borrowed” for a short time and then put back without noticing (on-mass-shell view where relativity is temporarily violated, i.e. the equation E^2 = p^2 + m^4). In the second, there is the different view (off-mass-shell view) where relativity and energy conservation are obeyed but the particle does not have the “real” mass of 80.4 GeV.

In my understanding these are alternative viewpoints but cannot both be true (while the off-mass shell view is the one which is covariant so may be preferred). Is this correct?

user50229
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1 Answers1

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The first view is mostly wrong, a handwaving popularization that leads to wrong conclusions. Virtual particles exist only in the Feynman diagrams of any reaction one can think of, and they are always internal lines in the diagram.

External lines represent a particle with a given identity, and a four vector on the mass shell of the particle. The identity of a particle depends on its quantum numbers, and its mass.

Here is a simple Feynman diagram that clearly shows the difference of internal and external lines:

enter image description here

The internal line represents in the Feynman integral the propagator of a photon, which has the mass of the photon (zero) and its quantum numbers. The four vector of the represented photon does not have a mass of zero as its length, it is off mass shell and varies within the interval of integration. It is called virtual to distinguish it from the on mass shell particle and it is just a line with quantum numbers so as to keep quantum number conservation at the vertices.

The same is true for all internal lines. W because of its mass when offshell in ordinary resonance decays, of the rho or the phi for example, will be in higher order diagrams with very much smaller contribution to the total cross section of the interaction of the incoming particles, ( calculating cross sections and decays is what the whole fuss of Feynman diagrams is about).

Off shell particles never can become on shell without adding or taking away a four vector. Take Compton scattering:

enter image description here

Pictorially an electron is hit by a photon and becomes "virtual" and then another photon leaves and the electron becomes "real". It has to get on mass shell to exist as a particle, the infinity under the integration is not seen in the pictorial mnemonic, but the mathematics is there and inexorable.

The second has little meaning either, except as hand waving and talking of particles as conscious entities. As far as the mainstream physics goes, elementary particles have no consciousness, and obey well the quantum field theory of the standard model, i.e Feynman diagram calculation using the standard model are validated by the data.

anna v
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  • Whilst it is true that the photon will in general be off shell at tree level, there is no integration to be done over its momentum because momentum conservation at the vertex completely fixes it as the sum of the ingoing momenta of the electrons. – nox Oct 05 '19 at 19:50
  • @lux the integration is usually done over $Q^2$ limits, and that is why the mass is off shell to keep momentum and energy conservation at the vertices. – anna v Oct 06 '19 at 05:18
  • In the diagrams you have referred to there are no momentum integrals. The momentum of the internal photon is fixed by the momenta of the external, on shell particles. The photon is off shell because the sum of the external momenta is not necessarily a null vector – nox Oct 06 '19 at 05:51
  • At one loop order it is different - there the internal loop momentum is not uniquely fixed by the external particle momenta and one integrates over one internal momentum – nox Oct 06 '19 at 05:54
  • I do not disagree on this, as I have not talked about which variables enter in the calculation of the diagrams. – anna v Oct 06 '19 at 05:56
  • Your answer refers to integrals over the photon momenta as the reason for them being off shell. At tree level this is misleading: there are no such integrals over photon momenta – nox Oct 06 '19 at 07:15
  • @lux I find no particular variable for the integration, can you count the line? The word "momentum" is not in my answer. You mean the wikipedia lnk? take it up with the wikipedia. (anyway momentum is not in their final limit, it is in the method of finding it.) – anna v Oct 06 '19 at 10:54
  • I think the p in wikipedia is the total momentum of the interaction. Generally crossections are a function of energy, and that is what the calculations should give you, for compton effect or e-e- scattering they give it as a function of a total variable p – anna v Oct 06 '19 at 11:05
  • "and varies within the interval of integration" in your comment you also mention integration over Q^2 – nox Oct 06 '19 at 16:05
  • Of course the four vector varies, the length of the four vector is the mass and the mass of the particle is off shell for virtual particles. Momentum is not a four vector but a three vector. Which variable one uses for the integration depends on the method used and what is being calculated. – anna v Oct 06 '19 at 16:14
  • If one fixes the momenta (of course momenta in this relativistic context means four vector and most certainly the momenta in Feynman diagrams are four vectors) $p_1$ and $p_2$ of the incoming external particles then the internal photon momentum (again, obviously the four momentum) is fixed by $p = p_1 + p_2$ and there is no integration over $p$. P will not on general be on shell. If you vary the incoming momentum then $p$ will change, but for a single experiment at fixed energy there is absolutely no "interval of integration" for the internal photon 4-momentum – nox Oct 06 '19 at 16:18
  • I do agree that to find the total cross section one must integrate over the phase space of the incoming and outgoing external, on shell particles, but this is an additional step after one has calculated the amplitude for the process from the feynam diagram – nox Oct 06 '19 at 16:21
  • Look at this for example https://arxiv.org/pdf/1412.2296.pdf .. page 4. k, the integration , is different than p. Lets stop here, I am not a theorist so can only depend on theoretical publications. I have written the way I understand that things work mathematically imo – anna v Oct 06 '19 at 16:28
  • The integral on page four is at one loop - the box involves one undetermined momentum that is not fixed by momentum conservation at the vertices. This is unlike the diagrams in your answer. I am enjoying this discussion but in terms of providing an accurate answer to op's question it it is important to clarify that internal particles can be off shell even without an integration over undetermined momentum. In the diagrams of your answer it is a consequence of the fact that the sum of two four vectors is not always null. – nox Oct 06 '19 at 16:34
  • @lux look at the integrations discussed here https://arxiv.org/pdf/1602.04182.pdf , page 21 for example. where in page 20 the lorenz nvariant phase space is defined – anna v Oct 06 '19 at 18:21
  • Υes, that's fine - notice these are three momenta integrals over the three momenta of the final states. This integration is completely different to the integration over the four momenta of internal lines in the case that they are undetermined. I repeat, in your answer in both diagrams the photon has a single, fixed, four momentum (for fixed incoming momenta) that is never integrated over; you are right that is is virtual because it is off shell (sum of incoming four momenta is not null) but wrong to say "it varies within the interval of integration" since there is no such integration. – nox Oct 07 '19 at 13:59
  • Your answer misleadingly suggests the integration is the reason for it being off shell but this is not the case. The integration over phase space is because fixing the incoming particles' momenta fixes the sum of the outgoing momenta but not their individual momenta and so one integrates essentially over all possible angles at which the outgoing particles could... "go out" - for all of these angles the internal photon has the same four momentum. – nox Oct 07 '19 at 14:00
  • I have been taught that Feynman diagrams are a pictorial way of finding the integrals for the scattering amplitude in the perturbative expansion., so I do not understand you. Thats ok, lets agree to disagree. – anna v Oct 07 '19 at 16:04
  • In a sense yes - the diagram leads to a mathematica expression that involves integrals over undetermined momenta. In your tree level example, there are no undetermined momenta - momentum conservation at the vertex fixes the internal photon momentum once and for all. Hence no integral in this special case. It's not about agreeing to disagree - there is a correct prescription that the op must apply and we should help them to do it. – nox Oct 08 '19 at 05:15
  • For example, the electron electron scattering involves a photon propagator $\frac {\delta^{\mu \nu}}{ q^2 }$ where $q = p_1+p_2$ is the sum of momenta flowing into the vertex that is not to be integrated over. (propagator in Feynman gauge) – nox Oct 08 '19 at 05:22