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?


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.

  • $\begingroup$ 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. $\endgroup$ – lux Oct 5 '19 at 19:50
  • $\begingroup$ @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. $\endgroup$ – anna v Oct 6 '19 at 5:18
  • $\begingroup$ 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 $\endgroup$ – lux Oct 6 '19 at 5:51
  • $\begingroup$ 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 $\endgroup$ – lux Oct 6 '19 at 5:54
  • $\begingroup$ I do not disagree on this, as I have not talked about which variables enter in the calculation of the diagrams. $\endgroup$ – anna v Oct 6 '19 at 5:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.