Nicholas Kang Posted August 2, 2014 Posted August 2, 2014 A vacuum isn`t empty, there are virtual particles. If light/EM wave can travel through vacuum, can it be affected by the virtual particles? Or light, which is a wave, travel through vacuum and needs a medium, can travel through vacuum is caused by these particles?
derek w Posted August 3, 2014 Posted August 3, 2014 Virtual particles are the force carrying particles between charged particles.
ajb Posted August 3, 2014 Posted August 3, 2014 Virtual particles are the force carrying particles between charged particles. Virtual means that they do not obey the classical equations of motion. You are right that in perturbative quantum field theory "force carrying" particles are virtual (they correspond to internal lines of the Feynman diagrams). But note that all speices of particle can be realised as virtual particles. I think the answer to the opening question is not very simple. In QFT we have the notion of bare particles and dressed particles. The properties of the particles we measure are that of the dressed particles, the cloud of virtual particles around the bare particle affects the particle's properties. This is where the notion of renormalisation comes in. It is hard to spell out details here, but in essence you have to make sense of how the contributions to a particle's properties due to all these virtual particles yeilds a finite measurable property, that is why it is not "measured" to be infinite.
Enthalpy Posted August 3, 2014 Posted August 3, 2014 Do I properly imagine that, when a gamma ray produces an electron-positron pair near a nucleus, the photon interacts with a virtual pair to give it energy and make it real, and then the nucleus' electric field separates the new particles? And that pair creation by photons without a nucleus, which has been hard to observe, is rare because the electric field of photons uses to be much weaker than that of nuclei?
swansont Posted August 3, 2014 Posted August 3, 2014 One reason pair production is hard to observe without a nucleus present is that the reaction gamma —> e+ + e- does not conserve momentum. The reverse gives you more than one gamma (usually two)
Enthalpy Posted August 4, 2014 Posted August 4, 2014 Do you mean the momentum E/c? The electrons moving apart to the sides, and slightly forward (versus the gama direction) would conserve this momentum, wouldn't they? Or do you mean an other momentum? The gamma's spin=1 seems to combine nicely with two spin=1/2, doesn't it?
Nicholas Kang Posted August 5, 2014 Author Posted August 5, 2014 So, does light, or photons are being affected by these virtual particles when in empty vacuum space? Now, a new question. What is the name of the particle that travels in a magnetic field, photons? electrons? I mean what is the name of the particle that make up the magnetic field?
ajb Posted August 5, 2014 Posted August 5, 2014 I mean what is the name of the particle that make up the magnetic field? Photons. Be aware that really we have the electromagnetic field.
beefpatty Posted August 8, 2014 Posted August 8, 2014 (edited) Do you mean the momentum E/c? The electrons moving apart to the sides, and slightly forward (versus the gama direction) would conserve this momentum, wouldn't they? Or do you mean an other momentum? The gamma's spin=1 seems to combine nicely with two spin=1/2, doesn't it? You can always find a reference frame in which the two particles are moving directly away from each other. That is, in your case where they separate both with an "x" and "y" velocity, there is a reference frame in which they only have an "x" or "y" velocity. Since physics must be valid in all reference frames, the fact that it does not conserve momentum in the "directly away" reference frame forbids it in all other frames as well. I posted a couple pictures to help illustrate. What you are talking about would be this frame: However, if I moved "up" with the electron and positron, it would look like this: which obviously does not conserve momentum. Edited August 8, 2014 by beefpatty 1
Nicholas Kang Posted August 13, 2014 Author Posted August 13, 2014 Photons. Be aware that really we have the electromagnetic field. Photons make up magnetic field, then what makes up electric field and what makes up electromagnetic field?
ajb Posted August 13, 2014 Posted August 13, 2014 Photons make up magnetic field, then what makes up electric field and what makes up electromagnetic field? Okay, so a better statement is "the quanta of the electromagnetic field are photons". That is if we apply the rules of quantum mechanics to the electromagnetic field the quanta we get are photons.
Nicholas Kang Posted August 13, 2014 Author Posted August 13, 2014 Is it the same for electric field and magnetic field?
beefpatty Posted August 13, 2014 Posted August 13, 2014 Electric fields and magnetic fields are not independent phenomena; you can only fully describe them by combining them into what is known as the electromagnetic field. A moving electric field induces a magnetic field, and vice versa. Even if you have just a static electric field, this is merely the special case where you are at rest with respect to the electric field. You could easily "boost" into a different frame where the E-field is moving with respect to you, and thus it would also have a magnetic field. Thus, there is no quanta for electric fields or magnetic fields. There are only quanta for electromagnetic fields.
Nicholas Kang Posted August 13, 2014 Author Posted August 13, 2014 So, you mean quanta only exist in moving EM field but not static electric and magnetic field?
Enthalpy Posted August 14, 2014 Posted August 14, 2014 Static electric and magnetic fields don't propagate and stay local, decreasing quickly over distance. In that case people introduce "virtual" photons that have a negative energy, so that their wavevector is imaginary and lets the amplitude decrease over distance instead of oscillate. Though virtual particles fit nicely to make theories more homogenous, I have seen no case in electromagnetism where the notion is fertile. It brings often no result at all, or when it does, it's the same as a computation of the fields but less direct.
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