Yukawa potential

 ( Gauge Theories ) 

The electromagnetic wave equation reads $\backslash Box\; A^\backslash mu\; =\; \backslash text\{source\; terms\}$.

Where $A^\backslash mu$ is the electromagnetic four-potential and $\backslash Box\; =\; 1/c^2\; \backslash partial\_t^2\; -\; \backslash nabla^2$. The potential will go as $1/r$ for a point source.

This wave equation describes a photon. For the nuclear force, we hope to describe . The pion can be described by a scalar field, as opposed to a vector, and we modify the wave equation by adding a multiple of the field (which doesn't affect relativistic invariance):

$\backslash Box\; \backslash phi\; +\; \backslash mu^2\; \backslash phi\; =\; 0$

If $\backslash phi$ depends only on the radius $r$ in spherical coordinates and is time independent, then we have $\backslash nabla^2\; \backslash phi\; =\; \backslash mu^2\; \backslash phi$.

Using the expression for the Laplacian in spherical coordinates $\backslash nabla^2\; \backslash phi\; =\; \backslash frac\{1\}\{r\}\; \backslash partial\_r^2\; (r\; \backslash phi)$ we get

$\backslash partial\_r^2\; (r\; \backslash phi)\; =\; \backslash mu^2\; (r\; \backslash phi)$

which has solutions $r\; \backslash phi(r)\; =\; Ke^\{-\; \backslash mu\; r\}$ so that $\backslash phi(r)\; =\; K\; \backslash frac\{e^\{-\; \backslash mu\; r\}\}\{r\}$ which is the Yukawa potential.

For the photon, a wave solution will look like $\backslash exp(i\; (k\; \backslash dot\; r\; -\; \backslash omega\; t)\; )$. We can substitute this in to the homogenous electromagnetic wave equation to obtain $\backslash frac\{\backslash omega^2\}\{c^2\}\; =\; k^2$,

By using the Planck relation $E\; =\; \backslash hbar\; \backslash omega$ and the de Broglie relation $p\; =\; \backslash hbar\; k$, we get

$\backslash frac\{E^2\}\{c^2\}\; =\; p^2$.

The mass-energy equivalence tells us that $\backslash frac\{E^2\}\{c^2\}\; =\; p^2\; +\; m^2\; c^2$, so this tells us that the photon is massless.

If we repeat this analysis for the pion, we get $\backslash frac\{E^2\}\{c^2\}\; =\; k^2\; +\; \backslash mu^2\; \backslash hbar^2$, which tells us that the meson has a mass of $m\; =\; \backslash frac\{\backslash mu\; \backslash hbar\}\{c\}$.

$\backslash mu$ can be estimated from the observed range of the nuclear forces. The nuclear force has a range on the order of femtometers ($10^\{-15\}\; \backslash text\{m\}$), and therefore the pion mass should be about $170\; \backslash text\{MeV\}$, close to the actual value of around $140\; \backslash text\{MeV\}$.

$V(\backslash mathbf\{r\})\; =$

where the integral is performed over all possible values of the 3-vector momenta . In this form, and setting the scaling factor to one, $\backslash alpha\; =\; 1$, the fraction $\backslash frac\{4\; \backslash pi\}\{k^2\; +\; m^2\}$ is seen to be the propagator or Green's function of the Klein–Gordon equation.

$\backslash mathcal\{L\}\_\backslash mathrm\{int\}(x)\; =\; g~\backslash overline\{\backslash psi\}(x)~\backslash phi(x)~\backslash psi(x)~.$

The scattering amplitude for two fermions, one with initial momentum $p\_1$ and the other with momentum $p\_2$, exchanging a meson with momentum , is given by the Feynman diagram on the right.

The Feynman rules for each vertex associate a factor of $g$ with the amplitude; since this diagram has two vertices, the total amplitude will have a factor of $g^2$. The line in the middle, connecting the two fermion lines, represents the exchange of a meson. The Feynman rule for a particle exchange is to use the propagator; the propagator for a massive meson is $\backslash frac\{-4\backslash pi\}\{~k^2+m^2~\}$. Thus, we see that the Feynman amplitude for this graph is nothing more than

$V(\backslash mathbf\{k\})=-g^2\backslash frac\{4\backslash pi\}\{k^2+m^2\}~.$

From the previous section, this is seen to be the Fourier transform of the Yukawa potential.

$\backslash psi(\backslash vec\{r\})\; \backslash approx\; A\; \backslash left(e^\{ipr\})$

where $\backslash vec\{p\}\; =\; p\; \backslash hat\{z\}$ is the particle's incoming momentum. The function $f(\backslash theta)$ is given by:

$f(\backslash theta)\; =\; \backslash frac\{-2\backslash mu\}\{\backslash hbar^2\; \backslash left|\backslash vec\{p\}-\backslash vec\{p\}\text{'}\; \backslash right|\}\; \backslash ,\; \backslash int\_\{0\}^\{\backslash infty\}\; r\; \backslash ,\; V(r)\; \backslash ,\; \backslash sin\backslash left(\backslash left|\backslash vec\{p\}-\backslash vec\{p\}\text{'}\backslash right|\; r\; \backslash right)\; ~\; \backslash mathrm\{d\}r$

where $\backslash vec\; p\text{'}\; =\; p\; \backslash hat\; r$ is the particle's outgoing scattered momentum and $\backslash mu$ is the incoming particles' mass (not to be confused with $m,$ the pion's mass). We calculate $f(\backslash theta)$ by plugging in $V\_\backslash text\{Yukawa\}$:

$f(\backslash theta)\; =\; \backslash frac\{2\backslash mu\}\{\backslash hbar^2\; \backslash left|\backslash vec\{p\}-\backslash vec\{p\}\text{'}\backslash right|\}\; \backslash ,\; g^2\; \backslash int\_\{0\}^\{\backslash infty\}\; e^\{-\backslash alpha\; m\; r\}\; \backslash ,\; \backslash sin\; \backslash left(\backslash left|\backslash vec\{p\}\; -\; \backslash vec\{p\}\text{'}\backslash right|\; \backslash ,\; r\; \backslash right)\; \backslash ,\; \backslash mathrm\{d\}r$

Evaluating the integral gives

$f(\backslash theta)\; =\; \backslash frac\{2\; \backslash mu\; g^2\}\{\backslash hbar^2\backslash ,\backslash left(\backslash alpha\}$

Energy conservation implies

$\backslash bigl|\backslash vec\; p\backslash bigr|\; =\; \backslash bigl|\backslash vec\; p\text{'}\backslash bigr|\; =\; p~$

so that

$\backslash left|\backslash vec\; p\; -\; \backslash vec\; p\text{'}\backslash right|\; =\; 2\backslash ,p\backslash ,\backslash sin\backslash left(\backslash tfrac\{1\}\{2\}\backslash theta\backslash right)~$

Plugging in, we get:

$f(\backslash theta)\; =\; \backslash frac\{2\; \backslash mu\; g^2\}\{\backslash hbar^2\; \backslash left(\backslash alpha\}$

We thus get a differential cross section of:

$\backslash frac\{\backslash mathrm\; d\; \backslash sigma\}\{\backslash mathrm\; d\; \backslash Omega\}\; =\; \backslash left|f(\backslash theta)\backslash right|^2\; =\; \backslash frac\{4\; \backslash mu^2\; g^4\}\{\backslash hbar^4\backslash \; \backslash left^2\}$

Integrating, the total cross section is:

$\backslash sigma\; =\; \backslash int\; \backslash frac\{\backslash mathrm\{d\}\; \backslash sigma\}\{\backslash mathrm\{d\}\; \backslash Omega\}\; \backslash mathrm\{d\}\backslash Omega$

$V(r\; >\; R)\; =\; G\; g\; \backslash frac\{e^\{-\backslash alpha\; m\; r\}\}\{r\}\; \backslash frac\{\backslash sinh\; \backslash alpha\; mR\}\{\backslash alpha\; mR\}.$