I cannot understand how can the force carrier of electromagnetic force be the same particle that makes up light. Electrically charged objects emit photons (by the way, why don't we see them?), and the exchange of these photons make unlike charges attract, like charges repel. But then why can't one move charged objects by lighting them?

Edit: it seems many people don't/misunderstand my question. I'm not really asking why light can't move objects, because it can, via transferring momentum, but not because photons mediated electromagnetic force between its emitter and receiver. You can push objects with photons (but it will accelerate very very slowly), but you can't pull objects with photons is real life, unlike in quantum-mechanics. Now pick a charged object, and light it. It won't move. Why won't these photons mediate force to the object? Or will it accelerate, but at a very slow rate that is far too small to be observable?


closed as unclear what you're asking by Emilio Pisanty, ZeroTheHero, Yashas, Wolpertinger, Jon Custer Apr 3 '17 at 22:29

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    $\begingroup$ Who says that you can't? $\endgroup$ – Emilio Pisanty Apr 2 '17 at 16:05
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    $\begingroup$ @EmilioPisanty ok, but that is because photons have momentum. It has nothing to do with electric charge. $\endgroup$ – AstroRP Apr 2 '17 at 17:05
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    $\begingroup$ I don't understand your comment - radiation pressure can only work through interactions with the electric charges inside materials. You probably have a sharper question in mind than is currently written down, but we can't guess at what you're thinking unless you clarify the post. $\endgroup$ – Emilio Pisanty Apr 2 '17 at 17:09
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    $\begingroup$ As I understand (correct me if I'm wrong) photons have momentum (given by E/c), and because of conservation of momentum, if light hits an object, that object has to start moving in the direction light was moving (this is independent of the charge of the object). Imagine that an electron attracts a positron. The electron emits a photon, and when it reaches the positron, it (the positron) will start to move in the direction light came from. So this is different from radiation pressure. $\endgroup$ – AstroRP Apr 2 '17 at 17:18
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    $\begingroup$ @AstroRP, when you said "light hits an object", you're talking about an interaction where the light is absorbed by the object. This only happens if the object contains charged particles. $\endgroup$ – The Photon Apr 2 '17 at 17:36

I want to clear this up:

Electrically charged objects emit photons (by the way, why don't we see them?), and the exchange of these photons make unlike charges attract, like charges repel.

Electrically macroscopic objects belong to the classical regime. Once one introduces photons, one is in the quantum mechanical regime, and needs quantum mechanical equations to describe interactions.

These photons are virtual, i.e they are a mathematical construct that is necessary to write down in quantum mechanical terms the interaction of like and/or unlike charges.


This is what happens between electrons of two separated charged objects ( supposed they are negatively charged)

The photons that are exchanged have all the quantum numbers of real photons, except the mass, they are off mass shell. Innumerable such exchanges build up the classical electric field between the two separated macroscopic objects. but these virtual particles are not something that can be seen. only the outside lines, the electrons are real, and their motion can be seen.

So as carriers of force , photons are always virtual, because they have to be mediators of interactions. As free particles on mass shell they are called real and and an enormous number of them build up the classical electromagnetic wave, we call light. This might help in intuition of how photons build up classical waves.

In general, light, and its constituent photons have momentum and can move neutral or charged particles, at the quantum level it will be diagrams of photon electron (or ion) scattering. As real photons which are building up light will interact with the appropriate diagrams with a charged object, and also with a neutral object,( because photons as point particles see the atoms with their electron clouds) , transferring momentum. This is seen macroscopically as the radiation pressure of light.

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    $\begingroup$ I would just like to comment, that I find it nicer to say, that the virtual photons don't follow the dispersion relation, instead of saying they are off mass shell. Otherwise, very nice answer. $\endgroup$ – jure Apr 2 '17 at 22:56
  • $\begingroup$ Are virtual photons still a wave phenomena? Does it make sense to ask what frequency virtual photons oscillate at? $\endgroup$ – Michael Apr 3 '17 at 15:58
  • $\begingroup$ @m They are under an integral, where the energy is one of the variables involved in the integration, so there is no unique energy so a variable frequency $\endgroup$ – anna v Apr 3 '17 at 17:13
  • $\begingroup$ Okay, but why can't virtual photons be seen? Does vision somehow have to do with the mass of a photon? (As from your answer that seems to be the only difference between real and virtual photons.) The answer as it stands now comes of a little tautological: "You don't see force carrying photons because they're not something that can be seen." $\endgroup$ – R.M. Apr 3 '17 at 17:31
  • $\begingroup$ to see a photon, it has to interact with your eye, transfer energy to the cones in the retina. the virtual photon is a mathematical construct under an integral between two charged particles . In the diagram above, even the electrons do not see an individual virtual photon they see a continuum from the minimum to the maximum value allowed by the integration over the momentum transfer variables. The integral will give the probability of scattering, i.e. an angular distribution for repeated scatters. $\endgroup$ – anna v Apr 3 '17 at 19:38

Why can't light move electrically charged object[?] ... But then why can't one move charged objects by lighting them?

The polarity of the electromagnetic wave associated with visible light changes hundreds of trillions of times per second. This is fast enough that a charged macroscopic object won't accelerate noticeably before the polarity changes and it starts accelerating back in the other direction.

Electrically charged objects emit photons (by the way, why don't we see them?)

If you're asking about the virtual photons that are involved with Cuolomb forces, please see Anna's answer.

If you're asking about true photons emitted from objects either because of blackbody radiation or because they hold oscillating charges, the chemicals in our eyes that respond to light only respond to a narrow band of frequencies (or if you prefer, a narrow band of photon energies). Therefore we only see light in this band, which we call "visible light".

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    $\begingroup$ I think that your answer to the second part is misleading, you should add the comment you have on the OP , to clarify . You are talking of black body radiation I guess in this answer, but the question is about how fields build up by photons, as you correctly clarify in your comment $\endgroup$ – anna v Apr 3 '17 at 3:31
  • $\begingroup$ The first sentence should be cleaned up. In it’s current form, it says: “The polarity … changes polarity …”. $\endgroup$ – Holger Apr 3 '17 at 14:41

Just thought I'd add some interesting examples to illustrate that light can, and does, in fact, move charged objects:

  1. The Photoelectric effect. For example: UV light shining on an aluminum plate. The UV photons carry so much energy that they "kick" electrons off" from the aluminum atoms. In a general scale, this is essentially called ionization. You can read some more about it here and here.

  2. Laser. Generally, (or in layman's terms), incident photons are absorbed by the outer electrons on the medium, which "jump" to a higher energy state (but without leaving their atoms; this is called excitation as opposed to ionization). See also Stimulated emission and Laser pumping.

  3. A microwave oven heats food by shooting RF photons at it, which causes water molecules to rotate very fast in order to align with the oscillating EM field. This happens because water molecules are polar, and is called Dielectric heating.

  4. Solar sails, as mentioned by user Emilio Pisanty in OP's comments. He says, "radiation pressure can only work through interactions with the electric charges inside materials".
    User The photon also adds, "when you said "light hits an object", you're talking about an interaction where the light is absorbed by the object. This only happens if the object contains charged particles".

  5. Optical tweezers. I haven't personally read a lot about this just yet but it's certainly cool.

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    $\begingroup$ 3. Such molecules (water, organics) have net neutral charges. If you are referring to radio photons interacting with the charged subatomic particles within such molecules, I don't see how that would produce rotational molecular motion. $\endgroup$ – electronpusher Apr 3 '17 at 6:11
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    $\begingroup$ @electronpusher it's because H2O is a dipole. Added this to my text. $\endgroup$ – Marc.2377 Apr 3 '17 at 13:23

I think it's clear from the above answers that real photons (with the direction of the momentum always equal to the direction of the velocity) can move objects, like the space ship with a big sail, propelled by photons wich are absorbed, regardless if the charges which absorb them are positive or negative (the propulsion would be even more effective if the photons are reflected).

Now if we consider the interaction between an electron and a positron, the exchanged photon is virtual, which is to say that it can have energies and momenta that are independent of each other. Imagine the electron and positron exchanging a virtual photon. The end result is that both particles move towards each other. At each real electron-real positron-virtual photon vertex the photon must have a four-momentum to conserve the four-momentum of the real electron and positron. If you consider the positron and electron at rest first, then at the positron vertex the virtual photon has an energy equal to the positron's energy and a momentum that's opposite to the momentum of the outgoing positron (which sounds counter-intuitive, because you expect the positron to have the same momentum as the momentum of the virtual photon). Likewise, at the electron vertex, the virtual photon has the same energy as the outgoing electron and a momentum in the opposite direction as the electron. So at each vertex the momenta of the virtual photons have the same energy but their momenta are opposite. For a virtual photon it does matter if the charge at the interaction vertex is positive or negative.

You can ask yourself how an electron "knows" that it's interacting with a positron or an electron (or how a positron knows that it's interacting with an electron or a positron). After all, a virtual photon absorbed by an electron is the same as a virtual photon absorbed by a positron. I'll leave it to you to find out.

You asked why we can't see photons. Well, seeing an object is realized by receiving photons on our retina, which in the brain is transformed into an image. So if you want to see a photon, it has to emit many photons to see an image of the photon. If a single photon (with a frequency in the range of different frequencies so that our retina can react with it) hits the retina, you'll maybe see a little flash of light (though I don't know if we actually see something after a single photon hits the retina). Of course, as many photons hit the retina we see an image of something but never an image of the photon itself.

  • $\begingroup$ By seeing a photon I meant that why aren't charged objects shining bright if they emit a lot of photons. $\endgroup$ – AstroRP Apr 5 '17 at 15:52
  • $\begingroup$ Ah, I see what you mean. This is because charged objects have a stationary electric field around it, built up out of virtual photons (though I don't "see" how this happens) . Only if you shake the charged object with a high-frequency, electromagnetic waves are produced. These consist of large numbers of real photons which can excite the chemicals in our eyes (dependent on their frequency). $\endgroup$ – descheleschilder Apr 5 '17 at 17:57

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