The accepted answer to this question says

Since the electroweak interaction is mediated by spin 1 bosons, it is the case that "like (charge) repels like and opposites attract".

Another answer comments:

The Higgs is spin - 0 (scalar field) and the graviton is spin - 2; attractive.

From this other question

a force mediated by a spin-0 scalar is always attractive.

So what's going on here? Why does the spin of the boson determine whether the force is universally attractive or follows the opposites-attract-and-like-repels-like pattern?

What would it look like for a spin-3 or spin-4 boson, if such a particle existed?

  • $\begingroup$ "What would it look like for a spin-3 or spin-4 boson, if such a particle existed?" - See Q & A here: Why do we not have spin greater than 2? $\endgroup$
    – Hal Hollis
    Oct 10, 2018 at 16:51
  • $\begingroup$ @tparker this cannot be so, as the interaction between two same color quarks has an exchange of gluon , which has to carry two colors( color anticolor) , and I believe the result is always attractive, Do you have a link? ( this for the colors on gluon hyperphysics.phy-astr.gsu.edu/hbase/Particles/expar.html) $\endgroup$
    – anna v
    Apr 20, 2021 at 14:46

2 Answers 2


Here's a schematic approach that highlights where the sign of the force comes from, at least in the spin-0 and spin-1 cases. I don't know how to generalize it to higher-spin cases. The approach assumes that the Hamiltonian is quadratic in the force-mediating boson field, so it excludes QCD, whose higher-than-quadratic terms cannot be neglected. The approach uses a semi-classical approximation in which the force-mediating boson field is treated as a classical field, again excluding QCD. The one thing that the approach handles carefully is the signs.

Consider two models. One involves matter particles interacting with a spin 0 (scalar) field $\phi$. The other involves matter particles interacting with a spin 1 abelian gauge field $A_\mu$. The Lagrangians are $$ L = \frac{\big(\dot\phi(x)\big)^2 -\big(\nabla\phi(x)\big)^2}{2} - \phi(x) J(x) + \cdots \tag{1} $$ and $$ L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-A_\mu J^\mu+\cdots \tag{2} $$ with $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$. The operators $J$ and $J^\mu$ are constructed from the matter field, which I'll be more explicit about later. The dots denote the kinetic terms for the matter field. The signs of the coupling terms $\phi J$ and $A_\mu J^\mu$ are not important: they can be chosen as desired by choosing the sign-convention for the boson fields. However, the signs of the kinetic terms for $\phi$ and $A_\mu$ are important: they must be chosen so that the energy (the Hamiltonian) has a finite lower bound. This positive-energy requirement is the key to explaining the connection between the spin of the boson and the sign of the force that it mediates.

Now consider situations in which the motion of the matter and the time-dependence of the boson fields are both negligible. This simplifies things in a few ways: (1) we can neglect all terms involving time-derivatives, (2) the temporal component $J^0$ is the only significant component of the current, and (3) with the help of a gauge-fixing choice, the temporal component $A_0$ can be regarded as the only significant component of the gauge field (as in electrostatics). After these simplifications, the Hamiltonians reduce to $$ H = \int d^3x\ \left(\frac{ \big(\nabla\phi(x)\big)^2}{2} + \phi(x) J(x) + \cdots\right) \tag{3} $$ and $$ H = \int d^3x\ \left(\frac{ \big(\nabla A_0(x)\big)^2}{2} + A_0(x) J^0(x) + \cdots\right), \tag{4} $$ and the equations of motion for the boson fields reduce to $$ \nabla^2\phi(x)=J(x) \hskip2cm -\nabla^2 A_0(x) = J^0(x). \tag{5} $$ Notice the opposite signs here. Using the same Green function $G(x-y)$ in both cases, we can express the boson fields in terms of $J$ and $J^0$ like this: $$ \phi(x) = \int dy\ G(x-y)J(y) \hskip2cm A_0(x) = -\int dy\ G(x-y)J^0(y). \tag{6} $$ Substitute these back into the Hamiltonian to see that the energy in the spin 0 case is $$ H = \frac{1}{2}\int d^3x\ J(x)G(x-y)J(y) + \cdots \tag{7} $$ and that the energy in the spin 1 case is $$ H = \frac{-1}{2}\int d^3x\ J^0(x)G(x-y)J^0(y) + \cdots \tag{8} $$ Again, notice the opposite signs. Given an explicit expression for the Green function, we can determine the sign of the force between particles of matter by looking at how the energy varies with the distance $|x-y|$. Or, since we already know that like charges repel each other in electrodynamics, we can just compare the spin-0 and spin-1 results to conclude that like charges must attract each other in the spin-0 case.

What about opposite charges? If the matter field is a spin-1/2 fermion field, then we have $$ J\propto \overline{\psi}\psi = \psi^\dagger\gamma^0\psi \hskip2cm J^0\propto \overline{\psi}\gamma^0\psi = \psi^\dagger\psi. \tag{9} $$ If we define $$ \psi_\pm = \frac{1\pm\gamma^0}{2}\psi, \tag{10} $$ then $$ J\propto\psi^\dagger_+\psi_+ - \psi^\dagger_-\psi_- \hskip2cm J^0\propto\psi^\dagger_+\psi_+ + \psi^\dagger_-\psi_-. \tag{11} $$ Use the fact that fermion fields are anticommutative to get $$ J\propto\psi^\dagger_+\psi_+ + \psi_-\psi^\dagger_- \hskip2cm J^0\propto\psi^\dagger_+\psi_+ -\psi_- \psi^\dagger_-. \tag{12} $$ I didn't write the kinetic terms for the matter field, but thanks to the positive-energy requirement, we can (schematically!) identify the first term in each of these expressions as the charge density of the particle, and the second term as the charge density of the corresponding antiparticle. Now, substitute these expressions for $J$ and $J^0$ back into the previous expressions for the energy. The fact that both terms in $J$ have the same sign means that in the spin 0 case, the sign of the force is the same for particle-particle, particle-antiparticle, and antiparticle-antiparticle. The fact that the two terms in $J^0$ have the opposite signs means that in the spin 1 case, the sign of the force for particle-antiparticle is opposite the sign of the force for particle-particle or antiparticle-antiparticle.

Granted this was all very schematic and specific to the spin-0 and spin-1 cases, but maybe it at least captures part of the answer to your question.

  • $\begingroup$ Thank you for the interesting explanation. Finally it seems that it is the difference of the sign of the antiparticles in $J$ and $J^0$ that matters. But then, I don't understand what importance does the sign have in the equations of motion $ \nabla^2\phi(x)=J(x)$ and $-\nabla^2 A_0(x) = J^0(x)$. Could you clarify this point? $\endgroup$
    – Davius
    Jul 1, 2022 at 20:31
  • 1
    $\begingroup$ @Davius Say we already know that the spin 1 mediated force between same-species particles is repulsive. Then the signs in eqs (11) and (12) relate that to the direction of the spin 1 force between the other combinations of particle/antiparticle. The signs in eqs (7) and (8), which come from eq (5), are used to compare the directions of the spin 1 and spin 0 forces. By combining the signs in eqs (11) and (12) with the signs in eqs (7) and (8), we can infer the force-directions in all 6 combinations: both spins of the mediating field, and all 3 combos of particle/antiparticle. $\endgroup$ Jul 2, 2022 at 14:40
  • $\begingroup$ Another example, interactions between nucleons (neutrons and protons) appear to be mediated primarily by $\pi$ mesons that have spin 0. However, the force seems to be essentially attractive, not repulsive. Isn't this partially in tension with the argument of the answer? $\endgroup$
    – Davius
    Jul 4, 2022 at 20:44
  • 1
    $\begingroup$ @Davius That's a good example, and I think this answer is consistent with it. Start with the fact that the Coulomb (spin 1) interaction between two protons is repulsive. Then argument indicates that an interaction between protons mediated by a spin 0 field, like the pion field, should be attractive: that's the sign difference between (7) and (8). [part 1 of 2] $\endgroup$ Jul 4, 2022 at 22:57
  • 1
    $\begingroup$ @Davius [part 2 of 2] Then the text below equation (12) says that in the spin 0 case, the particle-antiparticle and antiparticle-antiparticle forces have the same sign as the particle-particle one -- so the pion-mediated interaction between all combinations of protons and antiprotons is predicted to be attractive, in agreement with experiment at least for long distances, where the nucleon-nucleon interaction is mostly due to the pion-mediated part. $\endgroup$ Jul 4, 2022 at 22:57

For bi-linear interactions it is simply a fact that the algebra of Dirac $\gamma$ matrices leads to uniform attraction for spin 0 boson exchange and attraction/repulsion for unlike/like charges for spin 1 boson exchanges. The nonlinear structure of QCD may alter this relationship for gluon exchanges as pointed out in a comment by @annav. I don't remember the details of the derivations from my graduate school days 50 years ago, but I do remember that the final result was clear and convincing.


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