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Most often we think of photons mediating a force between charged particles (e.g. electrons and positrons), and it's a standard exercise in QFT to do the computation and find that, to leading order in $\alpha$, there is a $q_1 q_2/r^2$ repulsive force between particles (which of course becomes attractive if $q_1q_2 < 0$).

But in the absence of other matter, otherwise freely propagating photons feel forces from each other, mediated by electrons and positrons (and additional contributions from all other charged particles). The point where this becomes relevant is the Schwinger limit, I suppose.

Question: what kind of form does this interaction take? My QFT is too shaky to try to do the algebra myself. Some of my first questions are "is it attractive or repulsive", but that feels like a lousy question to be asking given the fact that photons don't really localize to a point the way that electrons do.

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    $\begingroup$ photons don't really localize to a point the way that electrons do The point of the photoelectric effect is that they do. $\endgroup$
    – G. Smith
    Commented Dec 8, 2020 at 22:02
  • $\begingroup$ The comment by @G.Smith is valid: photons can indeed be roughly localized, and that's sufficient for the question. But the last sentence in the question is also correct: a photon can't be strictly localized at a point: What's the physical meaning of the statement that “photons don't have positions”?. Neither can any other particle in relativistic QFT, including electrons, except in the nonrelativistic approximation. So the comment and the question's assertion are both correct, depending on how strictly we interpret "localize to a point". $\endgroup$
    – Chiral Anomaly
    Commented Dec 9, 2020 at 1:36

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Photon-photon scattering was first calculated by Euler and Kockel in 1935. At low energy the differential cross section in the zero-momentum frame, in natural units, is

$$\frac{d\sigma}{d\Omega}=\frac{139\alpha^4}{(180\pi)^2}\frac{\omega^6}{m^8}(3+\cos^2\theta)^2$$

where $\alpha$ is the fine-structure constant, $m$ the electron mass, $\omega$ the energy of each of the photons, and $\theta$ the scattering angle. See this paper for more details.

I am not aware of any way to say whether the photons are attracting or repelling each other. Their fields are interacting and the photons are scattering as a result.

In classical physics, two particles interacting electrostatically can have the same differential cross section for scattering regardless of whether they attract or repel. For example, the differential cross section for Rutherford scattering stays the same if you change the sign of $Z_1$ or $Z_2$. The hyperbolic trajectories are similar at infinity; all that differs is what happens when the particles are close. (Does the hyperbola of one particle go in front of the other particle or behind it?) But this is meaningless in a quantum theory because there are no classical trajectories.

Since photons cannot be at rest, there cannot be a QFT calculation for photons that is analogous to the derivation of the Coulomb potential for charged particles at rest.

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  • $\begingroup$ Thanks for the thoughtful answer. I suppose I'm thinking analogous to how photons are deflected in the presence of black holes: with the photon's wavelength much smaller than the distance between the photon and the black hole, it's meaningful to say that the photon is attracted by the black hole. (Granted, this is a weird one, since the gravity does not present itself as a force, exactly, so maybe it doesn't carry over.) $\endgroup$
    – Alex Meiburg
    Commented Dec 8, 2020 at 23:32
  • $\begingroup$ So I suppose that the scenario in my head might be: two photonic wave packets are traveling along skew lines. They will pass some distance from each other, which is much greater than the size of the wave packet. Then what does the modification to their trajectory look like? For this question to be well-formed, we need the wave packets to be localized (wavelength << packet size) and the packets to be separated (distance between >> packet size). Maybe when those are both true, the scattering disappears too quickly to attribute to anything like a force. $\endgroup$
    – Alex Meiburg
    Commented Dec 8, 2020 at 23:34
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    $\begingroup$ Regarding your first comment... Yes, but that is a classical calculation. It looks at null geodesics, which are the well-defined classical trajectories of photons. $\endgroup$
    – G. Smith
    Commented Dec 8, 2020 at 23:35
  • $\begingroup$ Mm. Maybe you just hit the nail on the head when you said: there is no analogous computation because electrons can be at rest. Electrons, at any speed, will exchange photons more rapidly than they are moving and so can feel a force. Photons will necessarily travel faster than the electrons they want to signal each other with, so we can't talk about any kind of "slow-varying potential", or force, at all. Thanks for your help. $\endgroup$
    – Alex Meiburg
    Commented Dec 8, 2020 at 23:37
  • $\begingroup$ Regarding your second comment... This seems like an interesting idea. I don’t know whether anyone has tried doing this calculation. $\endgroup$
    – G. Smith
    Commented Dec 8, 2020 at 23:37

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