Is there any quantum explanation of Newtonian Reaction other than photon pressure? I.e. when I put my foot on the ground is it supported by slightly above ambient temperature infrared radiation?

  • $\begingroup$ Does this answer your question? Forces other than the fundamental interactions, e.g. friction $\endgroup$
    – Paul T.
    Mar 22, 2021 at 15:44
  • $\begingroup$ No. It seems to me all forces are produced by photon exchange. We are told they are force carrying bosons. Friction produces forces just like gravity etc- matter is distorted away from equilibrium and photons are produced. $\endgroup$
    – Nick Green
    Mar 22, 2021 at 21:06

2 Answers 2


I think when you put your foot on the ground, it is supported by the repulsion between the electrons in the quantum shells of 1. the compounds on the sole of you shoe and 2. the floor material

  • 1
    $\begingroup$ Yes but when a quantum shell is perturbed by a load it emits a photon surely? So it's all photons... $\endgroup$
    – Nick Green
    Mar 22, 2021 at 21:00

It seems to me that you may be confusing the classical phenomenon of photon pressure or radiation pressure with the quantum field theory interpretation of virtual photons mediating electromagnetic interactions.

photon pressure

Photon pressure is a classical phenomenon that can be explained in terms of electromagnetic waves or photons. It is a consequence of conservation of momentum.

A single photon's momentum is determined by its frequency through the energy-momentum relation in special relativity. When a photon collides with an object it imparts momentum. If the photon is absorbed by the object, the stuck together object-photon system will have the same momentum as the photon did originally. If the photon reflects back, the object will gain up to twice the photon's original momentum (depending on the angles). In the maximal case: $$ p_i = p_f $$ $$ +p_\mathrm{photon} = -p_\mathrm{photon} + p_\mathrm{object}$$ $$ \implies p_\mathrm{obj} = 2p_\mathrm{ph}.$$

From Newton's second law we see the object's change in momentum as a force, $\vec{F} = \frac{d\vec{p}}{dt}$. In order to average the effect of many photons spread out over a macroscopic object it can be helpful to define a pressure, $P = F/A$, from this force.

The same momentum argument would apply if we thought about an EM wave colliding with an object. In that case we would calculate the momentum of the wave using classical electrodynamics rather than S.R. When we are talking about large numbers of photons that make up a classical EM wave, we'd get the same answer.

The photons in this description are real, energy carrying photons that physically collide with the object. In a classical description the contact force of the ground pushing up on you is not from photons. In order to be supported by photon pressure you would need an incredible source of light shining up on the small area of your feet.

contact forces

You may have heard that the contact force of the ground pushing up on your feet is actually an electromagnetic force. It is. The electrons in the atoms that make up your feet (or shoes) are repelled by the electrons in the atoms that make up the ground. The atoms don't touch at all. There is a non-contact electrostatic force. This force is very different from photon pressure. The ground doesn't make light that shines upward, the electric fields of the ground and your feet interact, resulting in a force that obeys Coulomb's law.

Answers to the question I linked in a comment discuss how all contact forces, like friction and normal forces, arise from the four fundamental forces like the electromagnetic force.

quantum field theory

In the quantum field theory (QFT) description of the universe, everything is just a bunch of quantum fields. Interactions between particles or wavefunctions are mediated by gauge bosons: photons for the electromagnetic force, gluons for the strong nuclear force, and W and Z particles for the weak nuclear force. Like the regular quantum wave-particle duality we can think of things like photons and electrons as particles or wave-packet like excitations of the underlying quantum fields.

In the wave way of thinking about QFT the two interacting electrons aren't particles at all. We might think of them as wave packets on the field that don't have well defined spatial extent. We can interpret their interaction as resulting from the overlap of their wavefunctions into a common wavefunction, like the interference of classical waves. Within the same paradigm it might even make more sense to say that their isn't a defined number of electrons, there is only single wavefunction and we could calculate the expectation value for observing two electrons given the field configuration, but there may be some non-zero probability we would observe one or three.

In this framework the electromagnetic interaction is wrapped up in the evolution of the wavefunction and there is no call to photons, real or virtual.

In a particle focused description we could say that the electrons are two separate point particles and the electromagnetic interaction between them is mediated by an exchange of virtual particles. One electron loses momentum by emitting a virtual photon, and the other gains momentum by absorbing it. This interaction is notably about virtual photons not real ones.

This framework is helpful when keeping track of the perturbation series to calculate the outcome of the interaction. The sum over paths formalism says that this interaction isn't just from the exchange of one virtual photon. There is a superposition of an exchanges: an exchange of one photon, an exchange of two photons, an exchange of four ... In this interaction series there will be terms that involve the exchange of virtual electrons and positrons too. There are even terms in the interaction series with exchanges of virtual gluons and quarks from the strong nuclear force and virtual W and Z particles and neutrinos from the weak nuclear force.

It is important to note that these are not real particles. In no "path" for the interaction where an emitted photon misses the target electron, so it could be observed separately. These virtual particles don't necessarily exist. They are a tool used to keep track of the series approximation to the full calculation.

but what's really happening?

I think confusion like this stems from the fact that there is no coherent description of "what is really happening" in quantum mechanics. Adherents of the Copenhagen interpretation would say that it doesn't matter what's really happening. We calculate observables, who cares what happens under the hood. Many of us try to make sense of what is really happening by using mental models and analogies to classical mechanics. But there is no definitive statement about what "really" happens when two electrons interact.

One thing to remember is that terms like photon pressure and force carrier particles have very specific definitions based in particular formulations of theory. I think everyone can agree that:

  • The contact force of the ground pushing on your feet is not photon pressure.

  • There is a formulation of quantum field theory where the force of the ground pushing on your feet is mediated by the exchange of virtual particles, including photons.

  • $\begingroup$ Paul T.Thank you very much for such a detailed response. $\endgroup$
    – Nick Green
    Mar 25, 2021 at 18:14
  • $\begingroup$ Paul T.Thank you very much for such a detailed response. You go to the Wave Equation then invoke unobservable Virtual Photons (the usual view) to produce a fictional rigidity in a structure agreed to be oscillating. But this doesn't escape the need for a rate of momentum transfer process sufficient to produce equilibrium of the load. Every electron going round an atom produces a current of very roughly 16 amps- we're not short of options when it comes to real observable fields. Have we considered them all? Recall van der Waals forces exerting one atmosphere of pressure at 100 atoms distance. $\endgroup$
    – Nick Green
    Mar 25, 2021 at 18:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.