Z boson emission by a neutrino in de Sitter expanding universe

The problem of electro-weak interactions in a de Sitter space-time by using perturbative methods was studied only recently in [1, 19, 20]. In [1, 19] the general formalism for studying the neutral current interactions intermediated by the Z boson was constructed in a curved space-time. This allows us to explore processes of interaction that generate production of massive bosons and fermions in an expanding de Sitter universe by adapting the electro-weak perturbation theory [5, 6, 7, 8, 9, 10, 11, 13, 12] to a curved space-time. It is well known that the massive bosons were produced in early universe [8], and it is important to explore all possible processes that could produce them, including the first order perturbative processes that are forbidden in Minkowski theory [13] by energy and momentum conservation. In a non-stationary space-time the translational invariance with respect to time is lost and the amplitudes and probabilities corresponding to processes of spontaneous particle production are non-vanishing [16, 17].

The idea of exploring the problem of particle generation at fields interactions was first proposed in [14], and the main results of this paper are related to the fact that the perturbative calculations can be translated in the number of particles. Another result established in [14] is related to the conditions in which the perturbative particle production becomes dominant in rapport with the cosmological particle production. In the present paper we want to study for the first time the process of Z boson emission by a neutrino in a de Sitter metric. We will use in our study the solutions of the Dirac equation and Proca equation in a de Sitter geometry [3, 4], which have a defined momentum and helicity. We will use the perturnative formalism employed in the field theory including renormalization procedures for extracting finite results from our computations. Our approach allows us to explore the interesting limit cases when the expansion parameter is vanishing and the case when the expansion is large comparatively with the Z boson mass.

The paper begins in section two with the computations of the transition amplitude and probability for the process of Z boson emission by a neutrino. In section three we compute the total probability of the process and in section four the transition rate is obtained. Density number of Z bosons obtained in emission processes by neutrinos are analysed in section five and in Appendices we present the free fields solutions in de Sitter geometry and the integrals that help us to establish the analytical results. We use natural units with $\hbar=1,c=1$.

For analyse the emission of Z bosons by neutrinos in early universe we start with the de Sitter metric [2]:

$$ds^{2}=dt^{2}-e^{2\omega t}d\vec{x}\,^{2}=\frac{1}{(\omega t_{c})^{2}}(dt_{c}^{2}-d\vec{x}\,^{2}),$$

(1)

where the conformal time is given in terms of proper time by $t_{c}=\frac{-e^{-\omega t}}{\omega}$, and $\omega$ is the expansion factor ($\omega>0$). The first order transition amplitude in electro-weak theory on curved space-time for the interaction between Z bosons and the neutrino-antineutrino field was obtained in [1]. For the process of Z emission by a single neutrino $\nu\rightarrow\nu+Z$, the transition amplitude is:

$\displaystyle\mathcal{A}_{\nu\rightarrow Z\nu}=-\int d^{4}x\sqrt{-g}\left(\frac{e_{0}}{\sin(2\theta_{W})}\right)(\overline{U}_{p^{\prime},\sigma^{\prime}})_{\nu}(x)\gamma^{\hat{\mu}}e_{\,\hat{\mu}}^{\alpha}\left(\frac{1-\gamma^{5}}{2}\right)(U_{p\sigma})_{\nu}(x)f_{\alpha\mathcal{P},\lambda,Z}^{*}(x),$

(2)

where $e_{0}$ is the electric charge, $\theta_{W}$ is the Weinberg angle and, $(U_{p\sigma})_{\nu}(x)$ is the solution of zero mass for the Dirac equation in de Sitter space-time [3], while $A_{\alpha}(Z)$ designates the Z boson field. The solutions for the free fields equations in momentum-helicity basis are presented in Appendix A. We also mention that we use point independent Dirac matrices $\gamma^{\hat{\mu}}$ and the tetrad fields $e_{\hat{\mu}}^{\alpha}$. For the line element (1), in the Cartesian gauge, the tetrad components are:

$$e^{0}_{\widehat{0}}=-\omega t_{c};\,\,\,e^{i}_{\widehat{j}}=-\delta^{i}_{\widehat{j}}\,\omega t_{c}.$$

(3)

Our computations are done in the chart with conformal time $t_{c}\in(-\infty,0)$, which covers the expanding portion of de Sitter space.

In this section we will analyse the situation when the expansion parameter $\omega$ is much larger than the Z boson mass. We will analyse the case of transversal polarization with $\lambda=\pm 1$. This can be done by observing that in equation (II.1) the algebraic argument of the Bessel K functions become very small when $t\rightarrow\infty$ and we can use the formula [15]:

$$K_{\nu}(z)\simeq\frac{\Gamma(\nu)}{2}\left(\frac{2}{z}\right)^{\nu},\,z\rightarrow 0,$$

(14)

where the real part of the index must meet the condition $Re(\nu)>0$. In order to study the case where the expansion parameter is larger than the mass of the Z boson, we will consider the situation where $M_{Z}/\omega<<1/2$. We mention that both the probabilities and amplitudes have a good behaviour in terms of parameter of the $M_{Z}/\omega$ (see Figs. (1),(2)). We compute the total probability of transition in the case of large expansion, when $M_{Z}/\omega\rightarrow 0$ and the index of Bessel K functions becomes $-ik\rightarrow\frac{1}{2}$. The total probability is obtained by solving the integrals after the final momenta. These kind of integrals are in general divergent, and we will apply a method for regularization.

$\displaystyle P_{tot}$

$\displaystyle=$

$\displaystyle\int d^{3}P\int d^{3}p^{\prime}\,P_{i\rightarrow f}.$

(15)

First we obtain the transition amplitude in the limit $M_{Z}/\omega\rightarrow 0$, by using equations (II.1) and (II.1) :

	$\displaystyle A_{\nu\rightarrow Z\nu}(\lambda=\pm 1)$	$\displaystyle=$	$\displaystyle\frac{e_{0}}{\sin(2\theta_{W})}\frac{\delta^{3}(\vec{P}+\vec{p}\,^{\prime}-\vec{p})}{(2\pi)^{3/2}}\Big{(}\frac{1}{2}-\sigma\Big{)}\Big{(}\frac{1}{2}-\sigma^{\prime}\Big{)}\frac{1}{\sqrt{2iP}(P+p^{\prime}-p)}$		(16)
			$\displaystyle\times\xi^{+}_{\sigma\,^{\prime}}(\vec{p}\,^{\prime})\vec{\sigma}\cdot\vec{\epsilon}^{*}\xi_{\sigma}(\vec{p}).$

Then the total probability expression of Z boson emission by e neutrino in the limit of large expansion is

$$P_{tot}=\frac{1}{2}\sum_{\lambda}\int d^{3}P\int d^{3}p^{\prime}\,\frac{e^{2}\delta^{3}(\vec{P}+\vec{p}\,^{\prime}-\vec{p})}{2(2\pi)^{3}\sin^{2}(2\theta_{W})P(P+p^{\prime}-p)^{2}}\Big{(}\frac{1}{2}-\sigma\Big{)}^{2}\Big{(}\frac{1}{2}-\sigma^{\prime}\Big{)}^{2}\big{|}\xi^{+}_{\sigma^{\prime}}(\vec{p}\,^{\prime})\vec{\sigma}\cdot\vec{\epsilon}^{*}\xi_{\sigma}(\vec{p}\,)\big{|}^{2}$$

(17)

To facilitate our calculations we need to consider the case where the particles momenta are on the same direction, that is the $z$ axis, in such a way that the angle between the momenta vectors is $0$ i.e $\vec{p}=p\,\vec{e}_{3}\,,\,\vec{p}\,^{\prime}=p^{\prime}\,\vec{e}_{3}$. In this case the bispinor summation is reduced to a number. The momentum of the Z boson is also considered on the third axis such that $\vec{P}=P\,\vec{e}_{3}$. Then the momenta integrals in total probability expression are:

$$I=\frac{1}{(2\pi)^{3}}\int d^{3}P\int d^{3}p^{\prime}\,\frac{\delta^{3}(\vec{P}+\vec{p}\,^{\prime}-\vec{p})}{P(P+p^{\prime}-p)}=\frac{1}{(2\pi)^{3}}\int d^{3}P\,\frac{1}{2P(p-P)^{2}},$$

(18)

where we perform the delta Dirac integration and finally we obtain an integral that is logarithmical divergent. To obtain a finite result we will apply the dimensional regularization proposed in [26, 27, 28]. Then we will replace our integral with a $D$ dimensional momenta integral as follows:

$$I(D)=\frac{1}{2}\int\frac{d^{D}P}{(2\pi)^{D}}\frac{1}{P(p-P)^{2}}=\frac{2\pi^{D/2}}{2(2\pi)^{D}\Gamma\big{(}\frac{D}{2}\big{)}}\int_{0}^{\infty}dP\,\frac{P^{D-1}}{P(p-P)^{2}}.$$

(19)

The new variable $P=-py$ is introduced to obtain the integral of the Beta Euler function [15], as given in Appendix B and the final result is:

	$\displaystyle I(D)$	$\displaystyle=$	$\displaystyle\frac{2\pi^{D/2}p^{D-3}}{2(2\pi)^{D}\Gamma\big{(}\frac{D}{2}\big{)}}\int_{0}^{\infty}\frac{dy\cdot y^{D-2}}{(1+y)^{2}}$		(20)
		$\displaystyle=$	$\displaystyle\frac{2\pi^{D/2}p^{D-3}\Gamma(D-1)\Gamma(3-D)}{2(2\pi)^{D}\Gamma\big{(}\frac{D}{2}\big{)}}.$

Because the result is still divergent for $D=3$ we will use the relation between gamma Euler functions $z\Gamma(z)=\Gamma(1+z)$ to obtain $\Gamma(3-D)=\frac{\Gamma(4-D)}{(3-D)}$ ,that allows to rewrite $I(D)$ as:

$$I(D)=\frac{2\pi^{D/2}p^{D-3}\Gamma(D-1)\Gamma(4-D)}{2(2\pi)^{D}\Gamma\big{(}\frac{D}{2}\big{)}(3-D)}.$$

(21)

We observe that the above function is divergent in $D=3$, and to remove the divergence we use the method of minimal substraction proposed in [28]. The pole of $I(D)$ in $D=3$ have the residue:

$$ResI(D)=\lim_{D\rightarrow 3}(D-3)I(D)=-\frac{1}{4\pi^{2}}.$$

(22)

The method of minimal substraction allows one to choose a counter-term dependent on a mass parameter $\mu$ with the general form $\frac{\mu^{s}R}{D-3}$, where $s$ is taken such the dimension of $I(D)$ remains the same. Then the renormalized integral is:

	$\displaystyle I(D)_{r}$	$\displaystyle=$	$\displaystyle I(D)+\frac{\mu^{D-3}}{4\pi^{2}(D-3)}$		(23)
		$\displaystyle=$	$\displaystyle\frac{1}{D-3}\bigg{(}\frac{2\pi^{D/2}p^{D-3}\Gamma(D-1)\Gamma(4-D)}{2(2\pi)^{D}\Gamma\big{(}\frac{D}{2}\big{)}}-\frac{\mu^{D-3}}{4\pi^{2}}\bigg{)}.$		(24)

The parenthesis from the above equation can be expanded around $D=3$, and the result is:

$$\frac{2\pi^{D/2}p^{D-3}\Gamma(D-1)\Gamma(4-D)}{2(2\pi)^{D}\Gamma\big{(}\frac{D}{2}\big{)}}-\frac{\mu^{D-3}}{4\pi^{2}}=\frac{(D-3)}{8\pi^{2}}\bigg{[}2-\ln\Big{(}\frac{4\pi\mu^{2}}{p^{2}}\Big{)}-\psi\Big{(}\frac{3}{2}\Big{)}\bigg{]}+\mathcal{O}((D-3)^{2}),$$

(25)

where $\Psi(3/2)$ is the digamma Euler function. Using this method we successfully cancel the divergent term $D-3$, and the final result of the renormalised integral is finite:

$$I(D)_{r}=\frac{1}{8\pi^{2}}\bigg{[}2-\ln\Big{(}\frac{4\pi\mu^{2}}{p^{2}}\Big{)}-\psi\Big{(}\frac{3}{2}\Big{)}\bigg{]}.$$

(26)

The final expression for the total probability for the process of Z boson emission by a neutrino in the early universe depends on the fundamental constants of the electro-weak theory

$$P_{tot}=\frac{e^{2}}{16\pi^{2}\sin^{2}(2\theta_{W})}\bigg{[}2-\ln\Big{(}\frac{4\pi\mu^{2}}{p^{2}}\Big{)}-\psi\Big{(}\frac{3}{2}\Big{)}\bigg{]}=\frac{\sqrt{2}M_{W}^{2}G_{F}\sin^{2}\theta_{W}}{4\pi^{2}\sin^{2}(2\theta_{W})}\bigg{[}2-\ln\Big{(}\frac{4\pi\mu^{2}}{p^{2}}\Big{)}-\psi\Big{(}\frac{3}{2}\Big{)}\bigg{]}$$

(27)

where in the second equality we introduce the Fermi constant $G_{F}$ i.e. $e^{2}=4\sqrt{2}M_{W}^{2}G_{F}\sin^{2}\theta_{W}$, with $M_{W}$ the mass of the W boson.

In the previous section we computed the total probability for the process of Z boson emission by a neutrino in de Sitter geometry, but a fundamental problem is related to the computation of measurable quantities for such a process. For that we will compute the rate of transition in this geometry, following the results obtained in [18] for the Milne universe. Consider the transition amplitude between the initial and final state in a de Sitter space-time of the form:

$$A_{if}=\delta^{3}(\vec{p}_{f}-\vec{p}_{i})M_{if}I_{if},$$

(28)

where the delta functions assure the momentum conservation in the process. In equation (28) the usual delta function of energy $\delta(E_{f}-E_{i})$, is missing in de Sitter amplitudes. This problem comes from the temporal integrals denoted by $I_{if}$:

$$I_{if}=\int_{0}^{\infty}dt\mathcal{K}_{if},$$

(29)

whose results do not give the usual $\delta(E_{f}-E_{i})$ as in Minkowski spacetime, but instead have a rather complex dependence on hypergeometric Gauss functions as in eqs.(II.1),(II.1), (II.1). The rate of transition is defined in Minkowski space-time by using the fact that the four delta Dirac functions $\delta^{4}(p)$ when squared give the usual $\delta(0)\delta^{3}(0)=\frac{1}{(2\pi)^{4}}VT$, where $V$ is the volume and $T$ is the interaction time. Then the rate is obtained dividing the probability by $VT$, or in other words the rate is the probability derivative with respect to time. Since in de Sitter geometry this is no longer valid, we adopt the definition given in [18], where the derivative with respect to time is applied on the integrals $I_{if}$. These integrals are written in terms of conformal time $t_{c}$ and the transition rate will be defined in conformal chart $\{t_{c},\vec{x}\}$ where we expect that the results are similar with those from Minkowski metric:

$$R_{if}=\lim_{t_{c}\rightarrow 0}\frac{1}{2V}\frac{d}{dt_{c}}|A_{if}|^{2}=\lim_{t\rightarrow\infty}\frac{e^{\omega t}}{2V}\frac{d}{dt}|A_{if}|^{2}.$$

(30)

The above result for the rate of transition can be written now as:

$$R_{if}=\frac{1}{(2\pi)^{3}}\,\delta^{3}(\vec{p}_{f}-\vec{p}_{i})|M_{if}|^{2}|I_{if}|\lim_{t\rightarrow\infty}|e^{\omega t}\mathcal{K}_{if}|,$$

(31)

where $\mathcal{K}_{if}$ are the integrands from the temporal integrals given in equations (II.1). Since in the present paper the transition between states in continuum spectrum are discussed, then to obtain the rate we must integrate after the final momenta in equation (31). Then the rate definition for the process of Z emission by a neutrino is:

$\displaystyle R_{\nu\rightarrow Z\nu}=\frac{1}{(2\pi)^{3}}\int\frac{{d^{3}p^{\prime}}}{(2\pi)^{3}}\int\frac{d^{3}\mathcal{P}}{(2\pi)^{3}}\,\delta^{3}(\vec{\mathcal{P}}+\vec{p}\,^{\prime}-\vec{p}\,)\sum_{\lambda}|M_{\nu\rightarrow Z\nu}(\lambda)|^{2}|I_{\nu\rightarrow Z\nu}(\lambda)|\lim_{t\rightarrow\infty}|e^{\omega t}\mathcal{K}_{\nu\rightarrow Z\nu}(\lambda)|,$

where $I_{\nu\rightarrow Z\nu}$ are the temporal integrals given in equation (II.1) with the results given in equations (II.1), (II.1). The quantity $M_{\nu\rightarrow Z\nu}(\lambda)$, is proportional to the coupling constants and the polarization terms $\xi^{+}_{-\frac{1}{2}}(\vec{p}\,^{\prime}\,)\vec{\sigma}\cdot\vec{\epsilon}\,^{*}(\vec{n}_{\mathcal{P}},\lambda)\xi_{-\frac{1}{2}}(\vec{p}\,)$. The function $\mathcal{K}_{\nu\rightarrow Z\nu}(\lambda)$ represents the integrands from equation (II.1). All these quantities depend on $\lambda$. Knowing that the neutrino polarizations are fixed $\sigma=\sigma^{\prime}=-\frac{1}{2}$, we need to sum only after the Z boson polarizations $\lambda$.

The transition rate can be evaluated in the limit where the expansion parameter is larger than the mass of the Z boson, by using the amplitude equation in this limit:

	$\displaystyle A_{\nu\rightarrow Z\nu}(\lambda=\pm 1)$	$\displaystyle=$	$\displaystyle\frac{ie_{0}}{\sin(2\theta_{W})}\frac{\delta^{3}(\vec{P}+\vec{p}\,^{\prime}-\vec{p}\,)}{(2\pi)^{3/2}}\Big{(}\frac{1}{2}-\sigma\Big{)}\Big{(}\frac{1}{2}-\sigma^{\prime}\Big{)}$		(33)
			$\displaystyle\times\int_{0}^{\infty}dz\sqrt{z}\,e^{-i(p^{\prime}-p)z}K_{1/2}(i\mathcal{P}z)\xi^{+}_{\sigma\,^{\prime}}(\vec{p}\,^{\prime})\vec{\sigma}\cdot\vec{\epsilon}^{*}\xi_{\sigma}(\vec{p}\,).$

In the present case of large expansion $I_{\nu\rightarrow Z\nu}$ has the expression:

$$I_{\nu\rightarrow Z\nu}=i\int_{0}^{\infty}dz\sqrt{z}\,e^{-i(p^{\prime}-p)z}K_{1/2}(i\mathcal{P}z)=\sqrt{\frac{\pi}{2i\mathcal{P}}}\frac{1}{p^{\prime}+\mathcal{P}-p},$$

(34)

while $M_{if}$ is given by:

$$M_{if}=\frac{e_{0}}{(2\pi)^{3/2}\sin(2\theta_{W})}\,\xi^{+}_{\sigma\,^{\prime}}(\vec{p}\,^{\prime})\vec{\sigma}\cdot\vec{\epsilon}^{*}\xi_{\sigma}(\vec{p}\,).$$

(35)

The limit from equation (31) can be obtained by using the integrand in equation (34):

$$\lim_{t\to\infty}\Big{|}e^{\omega t}\mathcal{K}_{\nu\rightarrow Z\nu}\Big{(}\frac{M_{Z}}{\omega}=0\Big{)}\Big{|}=\frac{\sqrt{\pi}}{\sqrt{2P}}.$$

(36)

Putting all the above quantities together and summing after the polarization we obtain the transition rate:

$$R_{i\rightarrow f}=\frac{1}{2}\sum_{\lambda}\frac{e_{0}^{2}\,\delta^{3}(\vec{P}+\vec{p}\,^{\prime}-\vec{p})\big{|}\xi^{+}_{\sigma^{\prime}}(\vec{p}\,^{\prime})\vec{\sigma}\cdot\vec{\epsilon}^{*}\xi_{\sigma}(\vec{p}\,)\big{|}^{2}}{(2\pi)^{6}\sin^{2}(2\theta_{W})2P(P+p^{\prime}-p)},$$

(37)

which can be simplified by choosing the momenta of particles on the same direction:

$$R_{i\rightarrow f}=\frac{e_{0}^{2}}{\sin^{2}(2\theta_{W})}\frac{\delta^{3}(\vec{P}+\vec{p}\,^{\prime}-\vec{p}\,)}{(2\pi)^{6}4P(P+p^{\prime}-p)}$$

(38)

The total transition rate is obtained by integration after the final momenta:

$$R_{tot}=\int d^{3}P\int d^{3}p^{\prime}\,R_{i\rightarrow f}=\frac{1}{(2\pi)^{3}}\int d^{3}p^{\prime}\int d^{3}P\frac{\delta^{3}(\vec{P}+\vec{p}\,^{\prime}-\vec{p}\,)}{P(P+p^{\prime}-p)}.$$

(39)

The momenta integral that needs to be computed is:

$$I=\frac{1}{2(2\pi)^{3}}\int d^{3}P\,\frac{1}{P(p-P)}.$$

(40)

In this case we also use the dimensional regularization [26, 27, 28] and the $D$ dimensional integral reads:

	$\displaystyle I(D)$	$\displaystyle=$	$\displaystyle\frac{1}{2(2\pi)^{D}}\int\frac{d^{D}P}{P(p-P)}=\frac{1}{2(2\pi)^{D}}\frac{2\pi^{D/2}}{\Gamma\big{(}\frac{D}{2}\big{)}}\int_{0}^{\infty}dP\,\frac{P^{D-1}}{P(p-P)}$		(41)
		$\displaystyle=$	$\displaystyle\frac{-2\pi^{D/2}p^{D-2}}{2(2\pi)^{D}\Gamma\big{(}\frac{D}{2}\big{)}}\int_{0}^{\infty}\frac{dy\cdot y^{D-2}}{(1+y)},$

where the last equality in the above equation is obtained by making the variable change $P=-py$. The final result of the integral is obtained by using the definition of Beta Euler function as:

$$I(D)=-\frac{2\pi^{D/2}p^{D-2}}{2(2\pi)^{D}\Gamma\big{(}\frac{D}{2}\big{)}}\frac{\Gamma(D-1)\Gamma(2-D)}{\Gamma(1)},$$

(42)

which is divergent for $D=3$. To remove the divergence we apply the method of minimal substraction [28]. First we rewrite the divergent gamma function as:

$$\Gamma(2-D)=\frac{\Gamma(3-D)}{(2-D)}=\frac{\Gamma(4-D)}{(2-D)(3-D)}$$

(43)

Then $I(D)$ becomes

$$I(D)=-\frac{2\pi^{D/2}p^{D-2}\Gamma(D-1)}{2(2\pi)^{D}\Gamma\big{(}\frac{D}{2}\big{)}}\frac{\Gamma(4-D)}{(2-D)(3-D)}$$

(44)

The residue in $D=3$ is then computed

$$ResI(D)=\lim_{D\rightarrow 3}(D-3)I(D)=-\frac{p}{4\pi^{2}}.$$

(45)

Then the regularized integral is obtained by choosing a counter-term with the same dimension [28]

$\displaystyle I(D)_{r}$

$\displaystyle=$

$\displaystyle\frac{1}{(D-3)}\Bigg{[}\frac{2\pi^{D/3}p^{D-2}\Gamma(D-1)\Gamma(4-D)}{(2\pi)^{D}\Gamma\big{(}\frac{D}{2}\big{)}(2-D)}+\frac{p\mu^{D-3}}{4\pi^{2}}\Bigg{]},$

(46)

where $\mu$ is a bosonic mass parameter. The parenthesis expansion around $D=3$ then gives:

			$\displaystyle\Bigg{[}\frac{2\pi^{D/3}p^{D-2}\Gamma(D-1)\Gamma(4-D)}{(2\pi)^{D}\Gamma\big{(}\frac{D}{2}\big{)}(2-D)}+\frac{p\mu^{D-3}}{4\pi^{2}}\Bigg{]}$		(47)
		$\displaystyle\simeq$	$\displaystyle\bigg{(}\frac{p}{8\pi^{2}}\bigg{)}\bigg{[}\psi\Big{(}\frac{3}{2}\Big{)}+\ln\Big{(}\frac{4\pi\mu^{2}}{p^{2}}\Big{)}\bigg{]}+\mathcal{O}\big{(}(D-3)^{2}\big{)}.$

In this way one obtain the regularized result for the momenta integral:

$$I(D)_{r}=\frac{p}{8\pi^{2}}\bigg{[}\psi\Big{(}\frac{3}{2}\Big{)}+\ln\Big{(}\frac{4\pi\mu^{2}}{p^{2}}\Big{)}\bigg{]}.$$

(48)

The total transition rate is then given by a finite expression:

$$R_{tot}=\frac{e^{2}}{(2\pi)^{3}\sin^{2}(2\theta_{W})}\frac{p}{32\pi^{2}}\bigg{[}\psi\Big{(}\frac{3}{2}\Big{)}+\ln\Big{(}\frac{4\pi\mu^{2}}{p^{2}}\Big{)}\bigg{]}=\frac{\sqrt{2}M_{W}^{2}G_{F}\sin^{2}\theta_{W}}{(2\pi)^{3}\sin^{2}(2\theta_{W})}\frac{p}{8\pi^{2}}\bigg{[}\psi\Big{(}\frac{3}{2}\Big{)}+\ln\Big{(}\frac{4\pi\mu^{2}}{p^{2}}\Big{)}\bigg{]}$$

(49)