# Feynman diagrams

Feynman diagrams are drawn on Space-time graphs. So is it possible for any particle path in that diagram to have a slope smaller than 45° or greater than 135°, i.e. can they travel faster than light? Does the slope or or velocity of the particles have any effect on the outcome of the diagram? Virtual particles (photons and the like) should have no limits for their properties, but what about incoming particles, say electrons?

• Also, for any increase in number of interactions in the diagram, the probability of that event decreases 100 times right? In these cases, do we also include cases where an electron emits and absorbs a (virtual photon), or when a photon splits into a particle-antiparticle pair? And does the actual location of the interactions matter? For simple electron scattering, what difference does it make if an electron emits and absorbs a photon while on its incoming path or on its outgoing one? – AP2261 Nov 1 '18 at 5:55
• If you want more accuracy, you have to include cases like you mention that have one or more loops. If you are computing to one-loop order, you need to include all topologically distinct one-loop diagrams, because they might all be comparable in size. So, for example, if you include a diagram with an emitted-and-absorbed photon on one of the incoming electrons, you also need to include the diagram with the photon on that outgoing electron, and the diagrams with the photon on the other incoming and outgoing electron, and the diagrams where a virtual photon travels “across each vertex”. – G. Smith Nov 1 '18 at 6:15

Incoming/outgoing particles with mass must travel at less than the speed of light. Incoming/outgoing massless particles must travel at the speed of light. The “virtual particles” inside the diagram can “travel” at any speed, but since they don’t really behave like particles this doesn’t matter.

Feynman diagrams aren’t pictures of what is “really happening”; they are just calculational tools. “Virtual particle” is an unfortunate misnomer, and the emphasis should definitely be on “virtual” rather than “particle”.

• I am a physics Ph.D. who recently started contributing to this forum. It would be very helpful to me if the person who downvoted my answer would explain what was objectionable about it. – G. Smith Nov 1 '18 at 6:55
• I find your answer fine. People who downvote and not support their disagreement may be carrying a private view of physics. Somtimes they downvote without comment because they do not consider the answer addresses the question. Remember this is an international site, and somebody might not understand/connect your "really happening" to the spacetime question. I still get downvoted without comments, and have gotten used to it :). – anna v Nov 1 '18 at 7:15
• @G.Smith Not me with the down vote. Unfortunately this is one small negative aspect of this otherwise excellent website which has not been addressed. When one is learning it is often not helpful if one gets a just a red cross and no explanation at the end of ones work. – Farcher Nov 1 '18 at 7:15
• @Farcher also they might confuse the downvote with "like" and "not like" of social media, not realizing that in a science site one should defend ones disagreement scientifically. – anna v Nov 1 '18 at 7:29

Feynman diagrams are not drawn on space time, that is a common misconception. It helps to think of them as being drawn on space time, but the reality is a bit different.

Feynman diagrams are nothing more than abstract diagrammatic pictures representing various terms in the expression for the transition amplitude between quantum states.

• So other than any vertices or interactions, the rest of the diagram has no effect on the result whatsoever? – AP2261 Nov 1 '18 at 5:47
• @user29774 The topology of the diagram has an effect. Not the positions of vertices (in fact, those positions don’t even exist in the physical sense) – Prof. Legolasov Nov 1 '18 at 5:48
• For each vertex, you get an extra factor of the fine structure constant, which is about 1/137. – G. Smith Nov 1 '18 at 5:58
• It’s not just the vertices that matter. For example, it matters which lines are for electrons, which for positrons, and which for photons. And the momenta and spins of the external particles matter. – G. Smith Nov 1 '18 at 6:01
• The spacetime positions of the vertices don’t matter because, according to the path integral point of view, you have to integrate over all possible interaction locations and times. – G. Smith Nov 1 '18 at 6:04

Feynman diagrams are multi-faceted, since their origin [1]. Feynman starts by transforming equation of motion (Schroedinger's or Dirac's) into an integral equation and draws diagrams according to the title, i.e. in spacetime. He names $$K_+$$ the kernel (later known as "Feynman's propagator").

Sec. 6 of the paper is entitled "Energy-momentum representation" and here you may find the form of Feynman diagrams in general use afterwards.

In path integral formulation the integral is over configuration space for particles, over field configuration space for fields. In both cases, not over spacetime.

[1] R.P. Feynman: "Space-Time Approach to Quantum Electrodynamics", Phys. Rev: 76 (1949), 769.