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If photons are the carriers for the electromagnetic force, then why can we not see electromagnetic fields, given that photons are involved in this interaction?

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  • $\begingroup$ It's not clear how the part after 'then' here is supposed to follow from the bit after 'if'. $\endgroup$
    – ACuriousMind
    Oct 8 '15 at 22:46
  • $\begingroup$ ouch! 2 downs. In feel bad. $\endgroup$
    – Justin
    Oct 8 '15 at 22:55
  • $\begingroup$ But really. How about. "Why cant we see virtual photons?" $\endgroup$
    – Justin
    Oct 8 '15 at 22:57
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    $\begingroup$ We cannot see electric field either, so it's not "just" the magnetic field. $\endgroup$
    – Kyle Kanos
    Oct 8 '15 at 23:44
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    $\begingroup$ I'm voting to close this question as off-topic because it belongs in biology (how a retina works) $\endgroup$ Oct 9 '15 at 12:16
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Virtual photons and generally virtual particles are mathematical constructs coming from the iconal representation of scattering amplitudes by Feynman diagrams. To make the point clear here is a virtual particle with an enormous mass with respect to the incoming and outgoing real particles. A Feynman diagram is a prescription for writing the mathematical formulae that have to be integrated over in order to have predictive numbers.

feyndiagr

A free neutron will decay by emitting a W-, which produces an electron and an antineutrino.

The W- has a mass of order 80GeV. The neutron has a mass of order 1 GeV. Since energy conservation is a fundamental law, it is evident that this so called W is not a real particle but represents a mathematical function, a propagator, that has to be integrated over when one wants to turn the Feynman diagram into the calculation of the lifetime.

Why is it labeled W-? because it carries all the quantum numbers of W and the mass in the propagator is the mass of the W. It cannot be seen or measured and it is there for mathematical consistency and conservation of quantum numbers.

In a diagram where a virtual photon is involved, one also cannot see it because it is a mathematical functions to be integrated over, not a real free particle.

If photons are the carriers for the electromagnetic force, then why can we not see electromagnetic fields, given that photons are involved in this interaction?

Because photons when functioning as carriers of the electromagnetic force are virtual, are mathematical constructs, (like the W above which is the carrier of the weak force). They only exist in a mathematical formula which has to be integrated in order to get real numbers predicting a crossection for a scattering event, or a lifetime for a decay.

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  • $\begingroup$ Interesting Anna, your comment about virtual particles only existing in Maths certainly explains why we cannot observe them but this approach seems to contradict Hawking radiation idea that a virtual photo pair can be separated so i still don't quite understand this answer. $\endgroup$
    – Justin
    Oct 12 '15 at 0:01
  • $\begingroup$ The Hawking radiation can be seen as a Feynman diagram: the pair is created through a graviton or photon exchange with the whole hole field, and the kinematics of the created pair keeps one of the two back into the hole. The separation is a separation in the quantum numbers of the pair. It is interesting to hear this Feynman story 3quarksdaily.com/3quarksdaily/2014/09/… . The discussion is about photons , but the mathematics will be similar. $\endgroup$
    – anna v
    Oct 12 '15 at 3:43
  • $\begingroup$ To think of how to construct the Feynman diagram look at this en.wikipedia.org/wiki/QED_vacuum . All one needs is to attach three of the four virtual photon outgoing legs to an interaction with the field of the hole and the fourth has a QM probability of getting free on mass and away $\endgroup$
    – anna v
    Oct 12 '15 at 4:02
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If you had ask a slightly different question "Why can't we interact directly with virtual photons" I could say that, in the formulation of Feynman series as scattering off-shell segments of virtual photons and virtual electrically charged matter-antimatter pairs, any virtual particle is by definition not part of the asymptotic states

What is an asymptotic state? When we study a Feynman propagator, we usually think of them to behave as a numerical machine that takes some input physical state at time $t$, and return some output physical state at time $t+T$. In the case of a scattering interaction (the canonical case of study in QED), both input and output states are assumed to be states in the irreducible representations of the Poincare algebra. Usually, these are described as asymptotic states

In other words, if a 'virtual photon' interacted with the detector, we wouldn't call it virtual to begin with, but we still would account it as a real photon in the perturbation expansion

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  • $\begingroup$ So how can see if we cannot interact directly with photons? did you mean "Why can't we interact directly with virtual photons" $\endgroup$
    – Justin
    Oct 9 '15 at 1:18
  • $\begingroup$ I still thinks the question deserves more than -1 points :-) $\endgroup$
    – Justin
    Oct 9 '15 at 2:48
  • $\begingroup$ don't pay attention to downvoters. Those are mostly a few regular haters in this site that wish only physicists made questions in here $\endgroup$
    – lurscher
    Oct 9 '15 at 3:00
  • $\begingroup$ @lurscher please don't post your personal anger here (last comment you made). Most of us downvote questions which are either not competently worded or are not physics questions. $\endgroup$ Oct 9 '15 at 12:17
  • $\begingroup$ Personal anger? you are confused. Actually, the site policy is that a downvote should be explained, something which is rarely done by downvoters $\endgroup$
    – lurscher
    Oct 9 '15 at 14:43
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Hawking radiation starts with the creation of a pair of virtual photons, I believe. One is absorbed by the Black Hole and the other could be observed if it were not so faint.

My understanding is that in the general case a virtual pair re-combines (annihilates) within a time on the order of a Planck time and cannot be observed.

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