Static force fields are fields, such as a simple electric, magnetic or gravitational fields, that exist without excitations. The most common approximation method that physicists use for scattering calculations can be interpreted as static forces arising from the interactions between two bodies mediated by virtual particles, particles that exist for only a short time determined by the uncertainty principle.^[1] The virtual particles, also known as force carriers, are bosons, with different bosons associated with each force.^{[2]: 16–37}

The virtual-particle description of static forces is capable of identifying the spatial form of the forces, such as the inverse-square behavior in Newton's law of universal gravitation and in Coulomb's law. It is also able to predict whether the forces are attractive or repulsive for like bodies.

The path integral formulation is the natural language for describing force carriers. This article uses the path integral formulation to describe the force carriers for spin 0, 1, and 2 fields. Pions, photons, and gravitons fall into these respective categories.

There are limits to the validity of the virtual particle picture. The virtual-particle formulation is derived from a method known as perturbation theory which is an approximation assuming interactions are not too strong, and was intended for scattering problems, not bound states such as atoms. For the strong force binding quarks into nucleons at low energies, perturbation theory has never been shown to yield results in accord with experiments,^[3] thus, the validity of the "force-mediating particle" picture is questionable. Similarly, for bound states the method fails.^[4] In these cases, the physical interpretation must be re-examined. As an example, the calculations of atomic structure in atomic physics or of molecular structure in quantum chemistry could not easily be repeated, if at all, using the "force-mediating particle" picture.^{[citation needed]}

Use of the "force-mediating particle" picture (FMPP) is unnecessary in nonrelativistic quantum mechanics, and Coulomb's law is used as given in atomic physics and quantum chemistry to calculate both bound and scattering states. A non-perturbative relativistic quantum theory, in which Lorentz invariance is preserved, is achievable by evaluating Coulomb's law as a 4-space interaction using the 3-space position vector of a reference electron obeying Dirac's equation and the quantum trajectory of a second electron which depends only on the scaled time. The quantum trajectory of each electron in an ensemble is inferred from the Dirac current for each electron by setting it equal to a velocity field times a quantum density, calculating a position field from the time integral of the velocity field, and finally calculating a quantum trajectory from the expectation value of the position field. The quantum trajectories are of course spin dependent, and the theory can be validated by checking that Pauli's exclusion principle is obeyed for a collection of fermions.

Classical forces

The force exerted by one mass on another and the force exerted by one charge on another are strikingly similar. Both fall off as the square of the distance between the bodies. Both are proportional to the product of properties of the bodies, mass in the case of gravitation and charge in the case of electrostatics.

They also have a striking difference. Two masses attract each other, while two like charges repel each other.

In both cases, the bodies appear to act on each other over a distance. The concept of field was invented to mediate the interaction among bodies thus eliminating the need for action at a distance. The gravitational force is mediated by the gravitational field and the Coulomb force is mediated by the electromagnetic field.

Gravitational force

The gravitational force on a mass ${\displaystyle m}$ exerted by another mass ${\displaystyle M}$ is

$${\displaystyle \mathbf {F} =-G{\frac {mM}{r^{2))}\,{\hat {\mathbf {r} ))=m\mathbf {g} \left(\mathbf {r} \right),}$$

${\displaystyle {\hat {\mathbf {r} ))}$

${\displaystyle M}$

${\displaystyle m}$

The force can also be written

$${\displaystyle \mathbf {F} =m\mathbf {g} \left(\mathbf {r} \right),}$$

${\displaystyle \mathbf {g} \left(\mathbf {r} \right)}$

$${\displaystyle \nabla \cdot \mathbf {g} =-4\pi G\rho _{m},}$$

${\displaystyle \rho _{m))$

Coulomb force

The electrostatic Coulomb force on a charge ${\displaystyle q}$ exerted by a charge ${\displaystyle Q}$ is (SI units)

$${\displaystyle \mathbf {F} ={\frac {1}{4\pi \varepsilon _{0))}{\frac {qQ}{r^{2))}\mathbf {\hat {r)) ,}$$

${\displaystyle \varepsilon _{0))$

${\displaystyle r}$

${\displaystyle \mathbf {\hat {r)) }$

${\displaystyle Q}$

${\displaystyle q}$

The Coulomb force can also be written in terms of an electrostatic field:

$${\displaystyle \mathbf {F} =q\mathbf {E} \left(\mathbf {r} \right),}$$

$${\displaystyle \nabla \cdot \mathbf {E} ={\frac {\rho _{q)){\varepsilon _{0))};}$$

${\displaystyle \rho _{q))$

Virtual-particle exchange

In perturbation theory, forces are generated by the exchange of virtual particles. The mechanics of virtual-particle exchange is best described with the path integral formulation of quantum mechanics. There are insights that can be obtained, however, without going into the machinery of path integrals, such as why classical gravitational and electrostatic forces fall off as the inverse square of the distance between bodies.

Path-integral formulation of virtual-particle exchange

A virtual particle is created by a disturbance to the vacuum state, and the virtual particle is destroyed when it is absorbed back into the vacuum state by another disturbance. The disturbances are imagined to be due to bodies that interact with the virtual particle’s field.

The probability amplitude

Using natural units, ${\displaystyle \hbar =c=1}$, the probability amplitude for the creation, propagation, and destruction of a virtual particle is given, in the path integral formulation by

$${\displaystyle Z\equiv \langle 0|\exp \left(-i{\hat {H))T\right)|0\rangle =\exp \left(-iET\right)=\int D\varphi \;\exp \left(i{\mathcal {S))[\varphi ]\right)\;=\exp \left(iW\right)}$$

${\displaystyle {\hat {H))}$

${\displaystyle T}$

${\displaystyle E}$

${\displaystyle W=-ET}$

${\displaystyle \varphi }$

$${\displaystyle {\mathcal {S))[\varphi ]=\int \mathrm {d} ^{4}x\;((\mathcal {L))[\varphi (x)]\,))$$

${\displaystyle {\mathcal {L))[\varphi (x)]}$

Here, the spacetime metric is given by

$${\displaystyle \eta _{\mu \nu }={\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix)).}$$

The path integral often can be converted to the form

$${\displaystyle Z=\int \exp \left[i\int d^{4}x\left({\frac {1}{2))\varphi {\hat {O))\varphi +J\varphi \right)\right]D\varphi }$$

${\displaystyle {\hat {O))}$

${\displaystyle \varphi }$

${\displaystyle J}$

The integral can be written (see Common integrals in quantum field theory § Integrals with differential operators in the argument)

$${\displaystyle Z\propto \exp \left(iW\left(J\right)\right)}$$

$${\displaystyle W\left(J\right)=-{\frac {1}{2))\iint d^{4}x\;d^{4}y\;J\left(x\right)D\left(x-y\right)J\left(y\right)}$$

${\displaystyle D\left(x-y\right)}$

$${\displaystyle {\hat {O))D\left(x-y\right)=\delta ^{4}\left(x-y\right).}$$

Energy of interaction

We assume that there are two point disturbances representing two bodies and that the disturbances are motionless and constant in time. The disturbances can be written

$${\displaystyle J(x)=\left(J_{1}+J_{2},0,0,0\right)}$$

$${\displaystyle {\begin{aligned}J_{1}&=a_{1}\delta ^{3}\left(\mathbf {x} -\mathbf {x} _{1}\right)\\J_{2}&=a_{2}\delta ^{3}\left(\mathbf {x} -\mathbf {x} _{2}\right)\end{aligned))}$$

${\displaystyle \mathbf {x} _{1))$

${\displaystyle \mathbf {x} _{2))$

${\displaystyle a_{1))$

${\displaystyle a_{2))$

If we neglect self-interactions of the disturbances then W becomes

$${\displaystyle W\left(J\right)=-\iint d^{4}x\;d^{4}y\;J_{1}\left(x\right){\frac {1}{2))\left[D\left(x-y\right)+D\left(y-x\right)\right]J_{2}\left(y\right),}$$

which can be written

$${\displaystyle W\left(J\right)=-Ta_{1}a_{2}\int {\frac {d^{3}k}{(2\pi )^{3))}\;\;D\left(k\right)\mid _{k_{0}=0}\;\exp \left(i\mathbf {k} \cdot \left(\mathbf {x} _{1}-\mathbf {x} _{2}\right)\right).}$$

Here ${\displaystyle D\left(k\right)}$ is the Fourier transform of

$${\displaystyle {\frac {1}{2))\left[D\left(x-y\right)+D\left(y-x\right)\right].}$$

Finally, the change in energy due to the static disturbances of the vacuum is

$${\displaystyle E=-{\frac {W}{T))=a_{1}a_{2}\int {\frac {d^{3}k}{(2\pi )^{3))}\;\;D\left(k\right)\mid _{k_{0}=0}\;\exp \left(i\mathbf {k} \cdot \left(\mathbf {x} _{1}-\mathbf {x} _{2}\right)\right).}$$

If this quantity is negative, the force is attractive. If it is positive, the force is repulsive.

Examples of static, motionless, interacting currents are the Yukawa potential, the Coulomb potential in a vacuum, and the Coulomb potential in a simple plasma or electron gas.

The expression for the interaction energy can be generalized to the situation in which the point particles are moving, but the motion is slow compared with the speed of light. Examples are the Darwin interaction in a vacuum and in a plasma.

Finally, the expression for the interaction energy can be generalized to situations in which the disturbances are not point particles, but are possibly line charges, tubes of charges, or current vortices. Examples include: two line charges embedded in a plasma or electron gas, Coulomb potential between two current loops embedded in a magnetic field, and the magnetic interaction between current loops in a simple plasma or electron gas. As seen from the Coulomb interaction between tubes of charge example, shown below, these more complicated geometries can lead to such exotic phenomena as fractional quantum numbers.

Selected examples

The Yukawa potential: The force between two nucleons in an atomic nucleus

Consider the spin-0 Lagrangian density^{[2]: 21–29}

$${\displaystyle {\mathcal {L))[\varphi (x)]={\frac {1}{2))\left[\left(\partial \varphi \right)^{2}-m^{2}\varphi ^{2}\right].}$$

The equation of motion for this Lagrangian is the Klein–Gordon equation

$${\displaystyle \partial ^{2}\varphi +m^{2}\varphi =0.}$$

If we add a disturbance the probability amplitude becomes

$${\displaystyle Z=\int D\varphi \;\exp \left\{i\int d^{4}\mathbf {x} \;\left[{\frac {1}{2))\left(\left(\partial \varphi \right)^{2}-m^{2}\varphi ^{2}\right)+J\varphi \right]\right\}.}$$

If we integrate by parts and neglect boundary terms at infinity the probability amplitude becomes

$${\displaystyle Z=\int D\varphi \;\exp \left\{i\int d^{4}x\;\left[-{\frac {1}{2))\varphi \left(\partial ^{2}+m^{2}\right)\varphi +J\varphi \right]\right\}.}$$

With the amplitude in this form it can be seen that the propagator is the solution of

$${\displaystyle -\left(\partial ^{2}+m^{2}\right)D\left(x-y\right)=\delta ^{4}\left(x-y\right).}$$

From this it can be seen that

$${\displaystyle D\left(k\right)\mid _{k_{0}=0}\;=\;-{\frac {1}{k^{2}+m^{2))}.}$$

The energy due to the static disturbances becomes (see Common integrals in quantum field theory § Yukawa Potential: The Coulomb potential with mass)

$${\displaystyle E=-{\frac {a_{1}a_{2)){4\pi r))\exp \left(-mr\right)}$$

$${\displaystyle r^{2}=\left(\mathbf {x} _{1}-\mathbf {x} _{2}\right)^{2))$$

$${\displaystyle {\frac {1}{m)).}$$

Yukawa proposed that this field describes the force between two nucleons in an atomic nucleus. It allowed him to predict both the range and the mass of the particle, now known as the pion, associated with this field.

Electrostatics

The Coulomb potential in a vacuum

Consider the spin-1 Proca Lagrangian with a disturbance^{[2]: 30–31}

$${\displaystyle {\mathcal {L))[\varphi (x)]=-{\frac {1}{4))F_{\mu \nu }F^{\mu \nu }+{\frac {1}{2))m^{2}A_{\mu }A^{\mu }+A_{\mu }J^{\mu ))$$

$${\displaystyle F_{\mu \nu }=\partial _{\mu }A_{\nu }-\partial _{\nu }A_{\mu },}$$

$${\displaystyle \partial _{\mu }J^{\mu }=0,}$$

$${\displaystyle \partial _{\mu }A^{\mu }=0.}$$

Moreover, we assume that there is only a time-like component ${\displaystyle J^{0))$ to the disturbance. In ordinary language, this means that there is a charge at the points of disturbance, but there are no electric currents.

If we follow the same procedure as we did with the Yukawa potential we find that

$${\displaystyle {\begin{aligned}-{\frac {1}{4))\int d^{4}xF_{\mu \nu }F^{\mu \nu }&=-{\frac {1}{4))\int d^{4}x\left(\partial _{\mu }A_{\nu }-\partial _{\nu }A_{\mu }\right)\left(\partial ^{\mu }A^{\nu }-\partial ^{\nu }A^{\mu }\right)\\&={\frac {1}{2))\int d^{4}x\;A_{\nu }\left(\partial ^{2}A^{\nu }-\partial ^{\nu }\partial _{\mu }A^{\mu }\right)\\&={\frac {1}{2))\int d^{4}x\;A^{\mu }\left(\eta _{\mu \nu }\partial ^{2}\right)A^{\nu },\end{aligned))}$$

$${\displaystyle \eta _{\mu \alpha }\left(\partial ^{2}+m^{2}\right)D^{\alpha \nu }\left(x-y\right)=\delta _{\mu }^{\nu }\delta ^{4}\left(x-y\right)}$$

$${\displaystyle D_{\mu \nu }\left(k\right)\mid _{k_{0}=0}\;=\;\eta _{\mu \nu }{\frac {1}{-k^{2}+m^{2))}.}$$

This yields

$${\displaystyle D\left(k\right)\mid _{k_{0}=0}\;=\;{\frac {1}{\mathbf {k} ^{2}+m^{2))))$$

$${\displaystyle E=+{\frac {a_{1}a_{2)){4\pi r))\exp \left(-mr\right)}$$

In the limit of zero photon mass, the Lagrangian reduces to the Lagrangian for electromagnetism

$${\displaystyle E={\frac {a_{1}a_{2)){4\pi r)).}$$

Therefore the energy reduces to the potential energy for the Coulomb force and the coefficients ${\displaystyle a_{1))$ and ${\displaystyle a_{2))$ are proportional to the electric charge. Unlike the Yukawa case, like bodies, in this electrostatic case, repel each other.

Coulomb potential in a simple plasma or electron gas

Plasma waves

The dispersion relation for plasma waves is^{[5]: 75–82}

$${\displaystyle \omega ^{2}=\omega _{p}^{2}+\gamma \left(\omega \right){\frac {T_{e)){m))\mathbf {k} ^{2}.}$$

${\displaystyle \omega }$

$${\displaystyle \omega _{p}^{2}={\frac {4\pi ne^{2)){m))}$$

${\displaystyle e}$

${\displaystyle m}$

${\displaystyle T_{e))$

${\displaystyle \gamma \left(\omega \right)}$

${\displaystyle \gamma \left(\omega \right)}$

${\displaystyle \gamma \left(\omega \right)}$

For low frequencies, the dispersion relation becomes

$${\displaystyle \mathbf {k} ^{2}+\mathbf {k} _{D}^{2}=0}$$

$${\displaystyle k_{D}^{2}={\frac {4\pi ne^{2)){T_{e))))$$

$${\displaystyle D\left(k\right)\mid _{k_{0}=0}\;=\;{\frac {1}{k^{2}+k_{D}^{2))}.}$$

In fact, if the retardation effects are not neglected, then the dispersion relation is

$${\displaystyle -k_{0}^{2}+k^{2}+k_{D}^{2}-{\frac {m}{T_{e))}k_{0}^{2}=0,}$$

$${\displaystyle E={\frac {a_{1}a_{2)){4\pi r))\exp \left(-k_{D}r\right).}$$

Plasmons

In a quantum electron gas, plasma waves are known as plasmons. Debye screening is replaced with Thomas–Fermi screening to yield^[6]

$${\displaystyle E={\frac {a_{1}a_{2)){4\pi r))\exp \left(-k_{s}r\right)}$$

$${\displaystyle k_{s}^{2}={\frac {6\pi ne^{2)){\varepsilon _{F))))$$

${\displaystyle \varepsilon _{F))$

${\textstyle \varepsilon _{F}={\frac {\hbar ^{2)){2m))\left({3\pi ^{2}n}\right)^{2/3}.}$

This expression can be derived from the chemical potential for an electron gas and from Poisson's equation. The chemical potential for an electron gas near equilibrium is constant and given by

$${\displaystyle \mu =-e\varphi +\varepsilon _{F))$$

${\displaystyle \varphi }$

Two line charges embedded in a plasma or electron gas

We consider a line of charge with axis in the z direction embedded in an electron gas

$${\displaystyle J_{1}\left(x\right)={\frac {a_{1)){L_{B))}{\frac {1}{2\pi r))\delta ^{2}\left(r\right)}$$

${\displaystyle r}$

${\displaystyle L_{B))$

$${\displaystyle D\left(k\right)\mid _{k_{0}=0}\;=\;{\frac {1}{\mathbf {k} ^{2}+k_{Ds}^{2))))$$

${\displaystyle k_{Ds))$

The interaction energy is

$${\displaystyle E=\left({\frac {a_{1}\,a_{2)){2\pi L_{B))}\right)\int _{0}^{\infty }{\frac {k\,dk}{k^{2}+k_{Ds}^{2))}{\mathcal {J))_{0}(kr_{12})=\left({\frac {a_{1}\,a_{2)){2\pi L_{B))}\right)K_{0}\left(k_{Ds}r_{12}\right)}$$

${\displaystyle {\mathcal {J))_{n}(x)}$

${\displaystyle K_{0}(x)}$

${\displaystyle r_{12))$

$${\displaystyle \int _{0}^{2\pi }{\frac {d\varphi }{2\pi ))\exp \left(ip\cos \left(\varphi \right)\right)={\mathcal {J))_{0}(p)}$$

$${\displaystyle \int _{0}^{\infty }{\frac {k\,dk}{k^{2}+m^{2))}{\mathcal {J))_{0}(kr)=K_{0}(mr).}$$

For ${\displaystyle k_{Ds}r_{12}\ll 1}$, we have

$${\displaystyle K_{0}\left(k_{Ds}r_{12}\right)\to -\ln \left({\frac {k_{Ds}r_{12)){2))\right)+0.5772.}$$

Coulomb potential between two current loops embedded in a magnetic field

Interaction energy for vortices

We consider a charge density in tube with axis along a magnetic field embedded in an electron gas

$${\displaystyle J_{1}\left(x\right)={\frac {a_{1)){L_{b))}{\frac {1}{2\pi r))\delta ^{2}{\left(r-r_{B1}\right)))$$

${\displaystyle r}$

${\displaystyle L_{B))$

$${\displaystyle r_{B1}={\frac ((\sqrt {4\pi ))m_{1}v_{1)){a_{1}B))={\sqrt {\frac {2\hbar }{m_{1}\omega _{c))))}$$

$${\displaystyle \omega _{c}={\frac {a_{1}B}((\sqrt {4\pi ))m_{1}c))}$$

$${\displaystyle v_{1}={\sqrt {\frac {2\hbar \omega _{c)){m_{1))))}$$

In this geometry, the interaction energy can be written

$${\displaystyle E=\left({\frac {a_{1}\,a_{2)){2\pi L_{B))}\right)\int _{0}^{\infty }{k\;dk\;}D\left(k\right)\mid _{k_{0}=k_{B}=0}{\mathcal {J))_{0}\left(kr_{B1}\right){\mathcal {J))_{0}\left(kr_{B2}\right){\mathcal {J))_{0}\left(kr_{12}\right)}$$

${\displaystyle r_{12))$

${\displaystyle {\mathcal {J))_{n}(x)}$

$${\displaystyle \int _{0}^{2\pi }{\frac {d\varphi }{2\pi ))\exp \left(ip\cos(\varphi )\right)={\mathcal {J))_{0}(p).}$$

Electric field due to a density perturbation

The chemical potential near equilibrium, is given by

$${\displaystyle \mu =-e\varphi +N\hbar \omega _{c}=N_{0}\hbar \omega _{c))$$

${\displaystyle -e\varphi }$

${\displaystyle N_{0))$

${\displaystyle N}$

The density fluctuation is then

$${\displaystyle \delta n={\frac {e\varphi }{\hbar \omega _{c}A_{M}L_{B))))$$

${\displaystyle A_{M))$

Poisson's equation yields

$${\displaystyle \left(k^{2}+k_{B}^{2}\right)\varphi =0}$$

$${\displaystyle k_{B}^{2}={\frac {4\pi e^{2)){\hbar \omega _{c}A_{M}L_{B))}.}$$

The propagator is then

$${\displaystyle D\left(k\right)\mid _{k_{0}=k_{B}=0}={\frac {1}{k^{2}+k_{B}^{2))))$$

$${\displaystyle E=\left({\frac {a_{1}\,a_{2)){2\pi L_{B))}\right)\int _{0}^{\infty }{\frac {k\;dk\;}{k^{2}+k_{B}^{2))}{\mathcal {J))_{0}\left(kr_{B1}\right){\mathcal {J))_{0}\left(kr_{B2}\right){\mathcal {J))_{0}\left(kr_{12}\right)=\left({\frac {2e^{2)){L_{B))}\right)\int _{0}^{\infty }{\frac {k\;dk\;}{k^{2}+k_{B}^{2}r_{B}^{2))}{\mathcal {J))_{0}^{2}\left(k\right){\mathcal {J))_{0}\left(k{\frac {r_{12)){r_{B))}\right)}$$

In analogy with plasmons, the force carrier is the quantum version of the upper hybrid oscillation which is a longitudinal plasma wave that propagates perpendicular to the magnetic field.

Currents with angular momentum

Delta function currents

Unlike classical currents, quantum current loops can have various values of the Larmor radius for a given energy.^{[7]: 187–190} Landau levels, the energy states of a charged particle in the presence of a magnetic field, are multiply degenerate. The current loops correspond to angular momentum states of the charged particle that may have the same energy. Specifically, the charge density is peaked around radii of

$${\displaystyle r_{\ell }={\sqrt {\ell ))\;r_{B}\;\;\;\ell =0,1,2,\ldots }$$

${\displaystyle \ell }$

${\displaystyle \ell =1}$

${\displaystyle \ell >0}$

${\displaystyle \ell '\geq \ell }$

${\displaystyle r_{\ell ))$

$${\displaystyle E=\left({\frac {2e^{2)){L_{B))}\right)\int _{0}^{\infty }{\frac {k\;dk\;}{k^{2}+k_{B}^{2}r_{\ell }^{2))}\;{\mathcal {J))_{0}\left(k\right)\;{\mathcal {J))_{0}\left({\sqrt {\frac {\ell '}{\ell ))}\;k\right)\;{\mathcal {J))_{0}\left(k{\frac {r_{12)){r_{\ell ))}\right).}$$

The interaction energy for ${\displaystyle \ell =\ell '}$ is given in Figure 1 for various values of ${\displaystyle k_{B}r_{\ell ))$. The energy for two different values is given in Figure 2.

Quasiparticles

For large values of angular momentum, the energy can have local minima at distances other than zero and infinity. It can be numerically verified that the minima occur at

$${\displaystyle r_{12}=r_{\ell \ell '}={\sqrt {\ell +\ell '))\;r_{B}.}$$

This suggests that the pair of particles that are bound and separated by a distance ${\displaystyle r_{\ell \ell '))$ act as a single quasiparticle with angular momentum ${\displaystyle \ell +\ell '}$.

If we scale the lengths as ${\displaystyle r_{\ell \ell '))$, then the interaction energy becomes

$${\displaystyle E={\frac {2e^{2)){L_{B))}\int _{0}^{\infty }{\frac {k\,dk}{k^{2}+k_{B}^{2}r_{\ell \ell '}^{2))}\;{\mathcal {J))_{0}\left(\cos \theta \,k\right)\;{\mathcal {J))_{0}(\sin \theta \,k)\;{\mathcal {J))_{0}{\left(k{\frac {r_{12)){r_{\ell \ell '))}\right)))$$

$${\displaystyle \tan \theta ={\sqrt {\frac {\ell }{\ell '))}.}$$

The value of the ${\displaystyle r_{12))$ at which the energy is minimum, ${\displaystyle r_{12}=r_{\ell \ell '))$, is independent of the ratio ${\textstyle \tan \theta ={\sqrt ((\ell }/{\ell '))))$. However the value of the energy at the minimum depends on the ratio. The lowest energy minimum occurs when

$${\displaystyle {\frac {\ell }{\ell '))=1.}$$

When the ratio differs from 1, then the energy minimum is higher (Figure 3). Therefore, for even values of total momentum, the lowest energy occurs when (Figure 4)

$${\displaystyle \ell =\ell '=1}$$

$${\displaystyle {\frac {\ell }{\ell ^{*))}={\frac {1}{2))}$$

$${\displaystyle \ell ^{*}=\ell +\ell '.}$$

When the total angular momentum is odd, the minima cannot occur for ${\displaystyle \ell =\ell '.}$ The lowest energy states for odd total angular momentum occur when

$${\displaystyle {\frac {\ell }{\ell ^{*))}=\;{\frac {\ell ^{*}\pm 1}{2\ell ^{*))))$$

$${\displaystyle {\frac {\ell }{\ell ^{*))}={\frac {1}{3)),{\frac {2}{5)),{\frac {3}{7)),{\text{etc.,))}$$

$${\displaystyle {\frac {\ell }{\ell ^{*))}={\frac {2}{3)),{\frac {3}{5)),{\frac {4}{7)),{\text{etc.,))}$$

Charge density spread over a wave function

The charge density is not actually concentrated in a delta function. The charge is spread over a wave function. In that case the electron density is^[7]: 189

$${\displaystyle {\frac {1}{\pi r_{B}^{2}L_{B))}{\frac {1}{n!))\left({\frac {r}{r_{B))}\right)^{2l}\exp \left(-{\frac {r^{2)){r_{B}^{2))}\right).}$$

The interaction energy becomes

$${\displaystyle E=\left({\frac {2e^{2)){L_{B))}\right)\int _{0}^{\infty }{\frac {k\;dk\;}{k^{2}+k_{B}^{2}r_{B}^{2))}\;M{\left(\ell +1,1,-{\frac {k^{2)){4))\right)}\;M{\left(\ell '+1,1,-{\frac {k^{2)){4))\right)}\;{\mathcal {J))_{0}{\left(k{\frac {r_{12)){r_{B))}\right)))$$

${\displaystyle M}$

$${\displaystyle {\frac {2}{n!))\int _{0}^{\infty }dr\;r^{2n+1}e^{-r^{2))J_{0}(kr)=M\left(n+1,1,-{\frac {k^{2)){4))\right).}$$

As with delta function charges, the value of ${\displaystyle r_{12))$ in which the energy is a local minimum only depends on the total angular momentum, not on the angular momenta of the individual currents. Also, as with the delta function charges, the energy at the minimum increases as the ratio of angular momenta varies from one. Therefore, the series

$${\displaystyle {\frac {\ell }{\ell ^{*))}={\frac {1}{3)),{\frac {2}{5)),{\frac {3}{7)),{\text{etc.,))}$$

$${\displaystyle {\frac {\ell }{\ell ^{*))}={\frac {2}{3)),{\frac {3}{5)),{\frac {4}{7)),{\text{etc.,))}$$

appear as well in the case of charges spread by the wave function.

The Laughlin wavefunction is an ansatz for the quasiparticle wavefunction. If the expectation value of the interaction energy is taken over a Laughlin wavefunction, these series are also preserved.

Magnetostatics

Darwin interaction in a vacuum

A charged moving particle can generate a magnetic field that affects the motion of another charged particle. The static version of this effect is called the Darwin interaction. To calculate this, consider the electrical currents in space generated by a moving charge

$${\displaystyle \mathbf {J} _{1}{\left(\mathbf {x} \right)}=a_{1}\mathbf {v} _{1}\delta ^{3}{\left(\mathbf {x} -\mathbf {x} _{1}\right)))$$

${\displaystyle \mathbf {J} _{2))$

The Fourier transform of this current is

$${\displaystyle \mathbf {J} _{1}{\left(\mathbf {k} \right)}=a_{1}\mathbf {v} _{1}\exp \left(i\mathbf {k} \cdot \mathbf {x} _{1}\right).}$$

The current can be decomposed into a transverse and a longitudinal part (see Helmholtz decomposition).

$${\displaystyle \mathbf {J} _{1}{\left(\mathbf {k} \right)}=a_{1}\left[1-{\hat {\mathbf {k} )){\hat {\mathbf {k} ))\right]\cdot \mathbf {v} _{1}\exp \left(i\mathbf {k} \cdot \mathbf {x} _{1}\right)+a_{1}\left[{\hat {\mathbf {k} )){\hat {\mathbf {k} ))\right]\cdot \mathbf {v} _{1}\exp \left(i\mathbf {k} \cdot \mathbf {x} _{1}\right).}$$

The hat indicates a unit vector. The last term disappears because

$${\displaystyle \mathbf {k} \cdot \mathbf {J} =-k_{0}J^{0}\to 0,}$$

${\displaystyle k_{0))$

With the current in this form the energy of interaction can be written

$${\displaystyle E=a_{1}a_{2}\int {\frac {d^{3}\mathbf {k} }{(2\pi )^{3))}\;\;D\left(k\right)\mid _{k_{0}=0}\;\mathbf {v} _{1}\cdot \left[1-{\hat {\mathbf {k} )){\hat {\mathbf {k} ))\right]\cdot \mathbf {v} _{2}\;\exp \left(i\mathbf {k} \cdot \left(\mathbf {x} _{1}-\mathbf {x} _{2}\right)\right).}$$

The propagator equation for the Proca Lagrangian is

$${\displaystyle \eta _{\mu \alpha }\left(\partial ^{2}+m^{2}\right)D^{\alpha \nu }\left(x-y\right)=\delta _{\mu }^{\nu }\delta ^{4}\left(x-y\right).}$$

The spacelike solution is

$${\displaystyle D\left(k\right)\mid _{k_{0}=0}\;=\;-{\frac {1}{k^{2}+m^{2))},}$$

$${\displaystyle E=-a_{1}a_{2}\int {\frac {d^{3}\mathbf {k} }{(2\pi )^{3))}\;\;{\frac {\mathbf {v} _{1}\cdot \left[1-{\hat {\mathbf {k} )){\hat {\mathbf {k} ))\right]\cdot \mathbf {v} _{2)){k^{2}+m^{2))}\;\exp \left(i\mathbf {k} \cdot \left(\mathbf {x} _{1}-\mathbf {x} _{2}\right)\right),}$$

${\textstyle k=|\mathbf {k} |}$

$${\displaystyle E=-{\frac {1}{2)){\frac {a_{1}a_{2)){4\pi r))e^{-mr}\left\((\frac {2}{\left(mr\right)^{2))}\left(e^{mr}-1\right)-{\frac {2}{mr))\right\}\mathbf {v} _{1}\cdot \left[1+{\hat {\mathbf {r} )){\hat {\mathbf {r} ))\right]\cdot \mathbf {v} _{2))$$

$${\displaystyle E=-{\frac {1}{2)){\frac {a_{1}a_{2)){4\pi r))\mathbf {v} _{1}\cdot \left[1+{\hat {\mathbf {r} )){\hat {\mathbf {r} ))\right]\cdot \mathbf {v} _{2))$$

Darwin interaction in a plasma

In a plasma, the dispersion relation for an electromagnetic wave is^{[5]: 100–103} (${\displaystyle c=1}$)

$${\displaystyle k_{0}^{2}=\omega _{p}^{2}+k^{2},}$$

$${\displaystyle D\left(k\right)\mid _{k_{0}=0}\;=\;-{\frac {1}{k^{2}+\omega _{p}^{2))}.}$$

Here ${\displaystyle \omega _{p))$ is the plasma frequency. The interaction energy is therefore

$${\displaystyle E=-{\frac {1}{2)){\frac {a_{1}a_{2)){4\pi r))\mathbf {v} _{1}\cdot \left[1+{\hat {\mathbf {r} )){\hat {\mathbf {r} ))\right]\cdot \mathbf {v} _{2}\;e^{-\omega _{p}r}\left\((\frac {2}{\left(\omega _{p}r\right)^{2))}\left(e^{\omega _{p}r}-1\right)-{\frac {2}{\omega _{p}r))\right\}.}$$

Magnetic interaction between current loops in a simple plasma or electron gas

The interaction energy

Consider a tube of current rotating in a magnetic field embedded in a simple plasma or electron gas. The current, which lies in the plane perpendicular to the magnetic field, is defined as

$${\displaystyle \mathbf {J} _{1}(\mathbf {x} )=a_{1}v_{1}{\frac {1}{2\pi rL_{B))}\;\delta ^{2}{\left(r-r_{B1}\right)}\left({\hat {\mathbf {b} ))\times {\hat {\mathbf {r} ))\right)}$$

$${\displaystyle r_{B1}={\frac ((\sqrt {4\pi ))m_{1}v_{1)){a_{1}B))}$$

${\displaystyle {\hat {\mathbf {b} ))}$

${\displaystyle L_{B))$

The energy of interaction is

$${\displaystyle E=\left({\frac {a_{1}\,a_{2)){2\pi L_{B))}\right)v_{1}\,v_{2}\,\int _{0}^{\infty }{k\;dk\;}D\left(k\right)\mid _{k_{0}=k_{B}=0}{\mathcal {J))_{1}{\left(kr_{B1}\right)}{\mathcal {J))_{1}{\left(kr_{B2}\right)}{\mathcal {J))_{0}{\left(kr_{12}\right)))$$

${\displaystyle r_{12))$

${\displaystyle {\mathcal {J))_{n}(x)}$

$${\displaystyle \int _{0}^{2\pi }{\frac {d\varphi }{2\pi ))\exp \left(ip\cos \left(\varphi \right)\right)={\mathcal {J))_{0}\left(p\right)}$$

$${\displaystyle \int _{0}^{2\pi }{\frac {d\varphi }{2\pi ))\cos \left(\varphi \right)\exp \left(ip\cos \left(\varphi \right)\right)=i{\mathcal {J))_{1}\left(p\right).}$$

See Common integrals in quantum field theory § Angular integration in cylindrical coordinates.

A current in a plasma confined to the plane perpendicular to the magnetic field generates an extraordinary wave.^{[5]: 110–112} This wave generates Hall currents that interact and modify the electromagnetic field. The dispersion relation for extraordinary waves is^[5]: 112

$${\displaystyle -k_{0}^{2}+k^{2}+\omega _{p}^{2}{\frac {k_{0}^{2}-\omega _{p}^{2)){k_{0}^{2}-\omega _{H}^{2))}=0,}$$

$${\displaystyle D\left(k\right)\mid _{k_{0}=k_{B}=0}\;=\;-\left({\frac {1}{k^{2}+k_{X}^{2))}\right)}$$

$${\displaystyle k_{X}\equiv {\frac {\omega _{p}^{2)){\omega _{H))))$$

$${\displaystyle \omega _{H}^{2}=\omega _{p}^{2}+\omega _{c}^{2},}$$

$${\displaystyle \omega _{c}={\frac {eB}{mc)),}$$

$${\displaystyle \omega _{p}^{2}={\frac {4\pi ne^{2)){m)).}$$

Here n is the electron density, e is the magnitude of the electron charge, and m is the electron mass.

The interaction energy becomes, for like currents,

$${\displaystyle E=-\left({\frac {a^{2)){2\pi L_{B))}\right)v^{2}\,\int _{0}^{\infty }{\frac {k\;dk}{k^{2}+k_{X}^{2))}{\mathcal {J))_{1}^{2}\left(kr_{B}\right){\mathcal {J))_{0}\left(kr_{12}\right)}$$

Limit of small distance between current loops

In the limit that the distance between current loops is small,

$${\displaystyle E=-E_{0}\;I_{1}{\left(\mu \right)}K_{1}{\left(\mu \right)))$$

$${\displaystyle E_{0}=\left({\frac {a^{2)){2\pi L_{B))}\right)v^{2))$$

$${\displaystyle \mu ={\frac {\omega _{p}^{2}r_{B)){\omega _{H))}=k_{X}\;r_{B))$$

We have made use of the integral (see Common integrals in quantum field theory § Integration of the cylindrical propagator with mass)

$${\displaystyle \int _{o}^{\infty }{\frac {k\;dk}{k^{2}+m^{2))}{\mathcal {J))_{1}^{2}\left(kr\right)=I_{1}\left(mr\right)K_{1}\left(mr\right).}$$

For small mr the integral becomes

$${\displaystyle I_{1}{\left(mr\right)}K_{1}{\left(mr\right)}\to {\frac {1}{2))\left[1-{\frac {1}{8))\left(mr\right)^{2}\right].}$$

For large mr the integral becomes

$${\displaystyle I_{1}\left(mr\right)K_{1}\left(mr\right)\rightarrow {\frac {1}{2))\;\left({\frac {1}{mr))\right).}$$

Relation to the quantum Hall effect

The screening wavenumber can be written (Gaussian units)

$${\displaystyle \mu ={\frac {\omega _{p}^{2}r_{B)){\omega _{H}c))=\left({\frac {2e^{2}r_{B)){L_{B}\hbar c))\right){\frac {\nu }{\sqrt {1+{\frac {\omega _{p}^{2)){\omega _{c}^{2))))))=2\alpha \left({\frac {r_{B)){L_{B))}\right)\left({\frac {1}{\sqrt {1+{\frac {\omega _{p}^{2)){\omega _{c}^{2))))))\right)\nu }$$

${\displaystyle \alpha }$

$${\displaystyle \nu ={\frac {2\pi N\hbar c}{eBA))}$$

For cases of interest in the quantum Hall effect, ${\displaystyle \mu }$ is small. In that case the interaction energy is

$${\displaystyle E=-{\frac {E_{0)){2))\left[1-{\frac {1}{8))\mu ^{2}\right]}$$

$${\displaystyle E_{0}={4\pi }{\frac {e^{2)){L_{B))}{\frac {v^{2)){c^{2))}={8\pi }{\frac {e^{2)){L_{B))}\left({\frac {\hbar \omega _{c)){mc^{2))}\right)}$$

$${\displaystyle {\frac {1}{2))mv^{2}=\hbar \omega _{c}.}$$

Gravitation

A gravitational disturbance is generated by the stress–energy tensor ${\displaystyle T^{\mu \nu ))$; consequently, the Lagrangian for the gravitational field is spin-2. If the disturbances are at rest, then the only component of the stress–energy tensor that persists is the ${\displaystyle 00}$ component. If we use the same trick of giving the graviton some mass and then taking the mass to zero at the end of the calculation the propagator becomes

$${\displaystyle D\left(k\right)\mid _{k_{0}=0}\;=\;-{\frac {4}{3)){\frac {1}{k^{2}+m^{2))))$$

$${\displaystyle E=-{\frac {4}{3)){\frac {a_{1}a_{2)){4\pi r))\exp \left(-mr\right),}$$

Unlike the electrostatic case, however, taking the small-mass limit of the boson does not yield the correct result. A more rigorous treatment yields a factor of one in the energy rather than 4/3.^[2]: 35