I know they don't interact with matter nor any electromagnetic forces, but I know they have mass, very little mass but they have. When something has mass it must be attracted by gravity and other objects (electromagnetic) right?
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1$\begingroup$ Possible duplicate of: physics.stackexchange.com/q/22734 $\endgroup$– UKHCommented May 31, 2016 at 0:40
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First, their being influenced by gravity is something completely different than their electromagnetic interaction.
All objects and fields that have a nonzero mass, energy, or momentum interact gravitationally, and so do neutrinos – although they're very light and hard to produce so the gravitational force from any neutrinos we know is undetectable at this time.
Neutrinos also have negligible but nonzero interactions with the electromagnetic field. They're uncharged and have no "tree level" interaction with the photons. However, a neutrino may split to a virtual electron and a virtual W-boson, those may do some thing, attach to (influence) other particles, and then the virtual particles become a neutrino again. These higher-order Feynman diagrams predict some interactions that are undetectably tiny.
In practice, neutrinos only interact via the virtual W-bosons, the weak nuclear force, and this interaction always looks like some kind of a beta-decay (or its reverse). Neutrinos also oscillate which is given by the mixing in their mass matrix – and this behavior isn't considered an interaction because it involves no other particles.
answered May 31, 2016 at 4:57
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$\begingroup$ Am I correct in believing that every particle species in the Standard Model must logically interact with every other species via some high-order Feynman diagrams (possibly involving several species of virtual particles)? Because if we made a graph with the particle species as nodes and the tree-level interactions as edges, then the graph must be connected - otherwise whichever component we lived in, we would be completely unable to detect any of the particles in any other component. $\endgroup$– tparkerCommented May 31, 2016 at 5:20
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1$\begingroup$ Yes, of course, it's always possible to construct a connected higher-order Feynman diagram that has the predetermined external lines (i.e. some weak interaction involving the particles) as long as the conservation laws hold. The conservation laws include the charge and spin (basically just whether the sum of spins is integer or not) but in perturbative Standard Model also (the total) lepton and baryon numbers. Those are conserved "accidentally", accidental symmetries. There may exist nonperturbative processes that violate L and B, too. In particular black hole formation+evaporation does. $\endgroup$ Commented May 31, 2016 at 5:29