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Is a proton on Earth attracts an electron on the moon? According to F=kx1x1/r^2, the answer is yes?
Possibly that’s what you meant by: “…the field is too small to be defined at that distance.”
In quantum field theory, fields are things that permeate all space and time. The "electromagnetic field generated by an electron" is described by the electron coupling to this background electromagnetic field and changing it. The electron doesn't create the field, but rather creates a disturbance in the field that we call "the electromagnetic field of the electron."You can think of it loosely like boats on an ocean: the ocean is the background field and would exist even in the absence of boats. The boats are like particles, which interact with the ocean and can disturb it. The boats don't generate the ocean, but they do generate particular disturbances on the ocean.
Quote from: Bill SPossibly that’s what you meant by: “…the field is too small to be defined at that distance.”Yes. Exactly. Also there are uncertainty relations with regard to the components of the fields that have to be taken into account too. Think about how you'd measure such a field. In classical electrodynamics you choose what's called a test charge which is a charge you use to determine the value of the field at a given point and you make much much smaller than the charge which whose field you're measuring. You can't do that with a proton since the electron has the same charge. Then you'd have to precisely measure the location of the source charge and the test charge which can't be done due to the uncertaintly principle, especially since you want to measure the force on the charge and force. And force is not an observable in quantum mechanics so all you can do is find the expectation value of it and <f> = d<p>/dt. Since there's some uncertainty in p this means that there is some uncertainty in <f> and therefore the proton's electric field. So the field can be so small as to be measured to be zero if the uncertainty is too large. But this is too complicated for me to determine myself. Perhaps JP can do the calculations? What do you say JP?