Static forces and virtual-particle exchange: Difference between revisions

Revision as of 18:16, 18 August 2010

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In the limit of zero [[photon]] mass, the Lagrangian reduces to the Lagrangian for [[electromagnetism]]. Therefore the energy reduces to the potential energy for the Coulomb force and the coefficients <math>a_1 </math> and <math>a_2 </math> 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 Lagrangian | Darwin interaction]]. To calculate this, consider the electrical currents in space generated by a moving charge

:<math>

\vec J_1\left( \vec x \right) = a_1 \vec v_1 \delta^3 \left( \vec x - \vec x_1 \right)

</math>

with a comparable expression for <math> \vec J_2 </math>.

The Fourier transform of this current is

:<math>

\vec J_1\left( \vec k \right) = a_1 \vec v_1 \exp\left( i \vec k \cdot \vec x_1 \right)

.</math>

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

:<math>

\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)

.</math>

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

:<math>

\vec k \cdot \vec J = -k_0 J^0 \rightarrow 0

,</math>

which results from charge conservation. Here <math>k_0 </math> vanishes because we are considering static forces.

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

:<math> E =

a_1 a_2\int {d^3k \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 )

</math>.

The propagator equation for the Proca Lagrangian is

:<math>

\eta_{\mu \nu} \left ( \partial^2 + m^2\right ) D\left ( x-y \right ) = \delta_{\mu \nu} \delta^4\left ( x-y \right )

.</math>.

The [[spacelike]] solution is

:<math>

D\left ( k \right )\mid_{k_0=0}\; = \;

-{1 \over \vec k^2 + m^2}

,</math>

which yields

:<math> E =

- a_1 a_2\int {d^3k \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 )

</math>

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

:<math> E =

- {1\over 2} {a_1 a_2 \over 4 \pi r } e^{ - m r } \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

</math>

which reduces to

:<math> 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

</math>

in the limit of small m. The interaction energy is the negative of the interaction Lagrangian.

==Gravitation==