Static forces and virtual-particle exchange

Static force fields are fields, such as a simple electric, magnetic or gravitational fields, that exist without excitations that result in the carrying of information from place to place. The modern view of static forces is that the interactions between two bodies are mediated by virtual particles, particles that exist for only a short time determined by the uncertainty principle. The virtual particles, also known as force carriers, are bosons with a particular type of boson associated with each type of field.^[1]

The virtual-particle description of static forces is capable of identifying the spacial form of the forces, such as the inverse-square behavior in Newton's Universal Law of Gravitation and in Coulomb's Law. It is also able to predict whether the forces are attractive or repulsive for like bodies. Also, the description has been able to relate the mass and range of the force carriers for such forces as the strong force.

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.

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

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

where G is the gravitational constant, r is the distance between the masses, and ${\displaystyle \mathbf {\hat {r}} }$  is the unit vector from mass ${\displaystyle m}$  to mass ${\displaystyle M}$ .

The force can also be written

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

where ${\displaystyle \mathbf {g} \left(\mathbf {r} \right)}$  is the gravitational field described by the field equation

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

where ${\displaystyle \rho _{m}}$  is the mass density at each point in space.

Coulomb force

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

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

where ${\displaystyle \varepsilon _{0}}$  is the vacuum permittivity, ${\displaystyle r}$  is the separation of the two charges, and ${\displaystyle \mathbf {\hat {r}} }$  is a unit vector in the direction from charge ${\displaystyle q}$  to charge ${\displaystyle Q}$ .

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

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

where

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

where ${\displaystyle \rho _{q}}$  is the charge density at each point in space.

Virtual-particle exchange

In the modern picture, forces are generated by the exchange of virtual particles. A virtual particle acts as a carrier of force information between two bodies. 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.

Inverse square law

The virtual-particle picture can be used to provide a heuristic explanation of the inverse square law for gravitational and electrostatic forces. If we imagine that a body emits a virtual particle and that virtual particle is absorbed by another body a distance r away, then the uncertainty principle states that

${\displaystyle \Delta E\Delta t\approx \hbar }$ 

where ${\displaystyle \Delta t}$  is the time it takes the virtual particle to travel between bodies, and ${\displaystyle \Delta E}$  is the energy of the virtual particle. If we imagine that ${\displaystyle \Delta E}$  is a kind of potential energy and we assume the virtual particle travels at the speed of light, then

${\displaystyle \Delta t\approx {r \over c}}$ 

and

${\displaystyle \Delta E\approx {\hbar c \over r}}$ .

The force is the gradient of the potential energy, therefore

${\displaystyle F\propto {1 \over r^{2}}}$ ,

which yields the inverse square law seen in both electrostatic and gravitational forces.

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 field.

The probability amplitude

The probability amplitude for the creation, propagation, and destruction of a virtual particle is give, in the path integral formulation by

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

where ${\displaystyle {\hat {H}}}$  is the Hamiltonian operator, ${\displaystyle T}$  is elapsed time, ${\displaystyle E}$  is the energy change due to the disturbance, ${\displaystyle W=-ET}$  is the change in action due to the disturbance, ${\displaystyle \varphi }$  is the field of the virtual particle, the integral is over all paths, and the classical action is given by

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

where ${\displaystyle {\mathcal {L}}[\varphi (x)]}$  is the Lagrangian density. We are using natural units, ${\displaystyle \hbar =c=1}$ .

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

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

where ${\displaystyle {\hat {O}}}$  is a differential operator with ${\displaystyle \varphi }$  and ${\displaystyle J}$  functions of spacetime. The first term in the argument represents the free particle and the second term represents the disturbance to the field from an external source such as a charge or a mass.

The integral can be written (see Common integrals in quantum field theory)

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

where

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

is the change in the action due to the disturbances and the propagator ${\displaystyle D\left(x-y\right)}$  is the solution of

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

Static fields

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

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

${\displaystyle J_{1}=a_{1}\delta ^{3}\left({\vec {x}}-{\vec {x}}_{1}\right)}$ 

${\displaystyle J_{2}=a_{2}\delta ^{3}\left({\vec {x}}-{\vec {x}}_{2}\right)}$ 

where the delta functions are in space, the disturbances are located at ${\displaystyle {\vec {x}}_{1}}$  and ${\displaystyle {\vec {x}}_{2}}$ , and the coefficients ${\displaystyle {\vec {a}}_{1}}$  and ${\displaystyle {\vec {a}}_{2}}$  are the strengths of the disturbances.

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

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

which can be written

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

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

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

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

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

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

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

Consider the spin-0 Lagrangian density^[2]

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

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

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

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

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

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

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

The energy due to the static disturbances becomes (see Common integrals in quantum field theory)

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

with

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

which is attractive and has a range of

${\displaystyle {1 \over m}}$ .

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 force

Consider the spin-1 Proca Lagrangian with a disturbance^[3]

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

where

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

charge is conserved

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

and we choose the Lorenz gauge

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

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

${\displaystyle -{1 \over 4}\int d^{4}xF_{\mu \nu }F^{\mu \nu }=-{1 \over 4}\int d^{4}x\left(\partial _{\mu }A_{\nu }-\partial _{\nu }A_{\mu }\right)\left(\partial ^{\mu }A^{\nu }-\partial ^{\nu }A^{\mu }\right)}$ 

${\displaystyle ={1 \over 2}\int d^{4}x\;A_{\nu }\left(\partial ^{2}A^{\nu }-\partial ^{\nu }\partial _{\mu }A^{\mu }\right)={1 \over 2}\int d^{4}x\;A^{\mu }\left(\eta _{\mu \nu }\partial ^{2}\right)A^{\nu },}$ 

which implies

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

and

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

This yields

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

for the timelike propagator and

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

which has the opposite sign to the Yukawa case.

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

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

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.

Magnetostatics: The Darwin interaction

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

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

with a comparable expression for ${\displaystyle {\vec {J}}_{2}}$ .

The Fourier transform of this current is

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

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

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

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

${\displaystyle {\vec {k}}\cdot {\vec {J}}=-k_{0}J^{0}\rightarrow 0,}$ 

which results from charge conservation. Here ${\displaystyle k_{0}}$  vanishes because we are considering static forces.

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

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

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

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

which yields

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

which evaluates to (see Common integrals in quantum field theory)

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

which reduces to

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

in the limit of small m. The interaction energy is the negative of the interaction Lagrangian. For two like part particles traveling in the same direction, the interaction is attractive, which is the opposite of the Coulomb interaction.

Gravitation

The Lagrangian for the gravitational field, which we will not write down explicitly, is spin-2. The disturbance is generated by the stress-energy tensor ${\displaystyle T^{\mu \nu }}$ . If the disturbances are at rest, then the only component of the stress-energy tensor that survives 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

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

and

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

which is once again attractive rather than repulsive. The coefficients are proportional to the masses of the disturbances. In the limit of small graviton mass, we recover the inverse-square behavior of Newton's Law.^[4]

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.^[5]

References

1. A. Zee (2003). Quantum Field Theory in a Nutshell. Princeton University. ISBN 0-691-01019-6. pp. 16-37
2. Zee, pp. 21-29
3. Zee, pp. 30-31
4. Zee, pp. 32-37
5. Zee, p. 35