I am trying to reproduce the results of this paper (https://arxiv.org/pdf/2003.03285.pdf). In particular Eq (25) and (26). I would like to convert the proton decay width (given in units of $\rm GeV$) to a lifetime in years. The formula is below:

$$\Gamma (p\to \pi^0 \mu^+)_{\rm flipped}= \frac{g_5^4m_p |V_{ud}|^2 |(U_l)_{21}|^2 }{32\pi M_X^4}(1-\frac{m_\pi^2}{m_p^2})^2 A_L^2 A_{S_1}^2 \left(\langle \pi^0\vert (ud)_Ru_L\vert p\rangle_{\mu}\right)^2.$$

Here $A_{L}=1.2$, $A_{S}=1.5$, $m_{\pi}=0.139$ GeV, $m_{p}=0.938$ GeV, $\left(\langle \pi^0\vert (ud)_Ru_L\vert p\rangle_{\mu}\right) = -0.118 \rm{GeV}^2$, $|V_{ud}|^2\approx 0.97$. The lifetime ($\tau=1/\Gamma$ in natural units) formula derive from the one above is

$$ \tau (p\to \pi^0 \mu^+)_{\rm flipped}\simeq 9.7 \times 10^{35} \times |(U_l)_{21}|^{-2} \biggl(\frac{M_X}{10^{16}~{\rm GeV}}\biggr)^4 \biggl(\frac{0.0378}{\alpha_5}\biggr)^2 ~{\rm yrs}~.$$

where $\alpha_{5}=\frac{g^2_{5}}{4\pi}$. I have tried plugging in these numbers and then converting from GeV to years ($1\rm{GeV}=4.79347\times10^{31} \rm{year}^{-1}$). Not sure what I am doing wrong as I get far too small a lifetime. Help is appreciated.

I am trying to reproduce the results of this paper¹. In particular Eqs. (25) and (26). I would like to convert the proton decay width (given in units of $\rm GeV$) to a lifetime in years. The formula is below:

$$\Gamma (p\to \pi^0 \mu^+)_{\rm flipped}= \frac{g_5^4m_p |V_{ud}|^2 |(U_l)_{21}|^2 }{32\pi M_X^4}(1-\frac{m_\pi^2}{m_p^2})^2 A_L^2 A_{S_1}^2 \left(\langle \pi^0\vert (ud)_Ru_L\vert p\rangle_{\mu}\right)^2.$$

Here $A_{L}=1.2$, $A_{S}=1.5$, $m_{\pi}=0.139$ GeV, $m_{p}=0.938$ GeV, $\left(\langle \pi^0\vert (ud)_Ru_L\vert p\rangle_{\mu}\right) = -0.118 \rm{GeV}^2$, $|V_{ud}|^2\approx 0.97$. The lifetime ($\tau=1/\Gamma$ in natural units) formula derived from the one above is

$$ \tau (p\to \pi^0 \mu^+)_{\rm flipped}\simeq 9.7 \times 10^{35} \times |(U_l)_{21}|^{-2} \biggl(\frac{M_X}{10^{16}~{\rm GeV}}\biggr)^4 \biggl(\frac{0.0378}{\alpha_5}\biggr)^2 ~{\rm yrs}~.$$

where $\alpha_{5}=\frac{g^2_{5}}{4\pi}$. I have tried plugging in these numbers and then converting from GeV to years $(1\rm{GeV}=4.79347\times10^{31} \rm{year}^{-1})$. Not sure what I am doing wrong as I get far too small a lifetime. Help is appreciated.

¹ Reference: J. Ellis, M. A. G. Garcia, N. Nagata, D. V. Nanopoulos & K. A. Olive, "Proton decay: flipped vs. unflipped $\rm SU(5)$", J. High Energ. Phys. 2020, 21 (2020).

I am trying to reproduce the results of a paper which looks at proton decay. I would like to convert the proton decay width (given in units of $\rm GeV$) to a lifetime in years. The formula is below:

$$\Gamma (p\to \pi^0 \mu^+)_{\rm flipped}= \frac{g_5^4m_p |V_{ud}|^2 |(U_l)_{21}|^2 }{32\pi M_X^4}(1-\frac{m_\pi^2}{m_p^2})^2 A_L^2 A_{S_1}^2 \left(\langle \pi^0\vert (ud)_Ru_L\vert p\rangle_{\mu}\right)^2.$$

Here $A_{L}=1.2$, $A_{S}=1.5$, $m_{\pi}=0.139$ GeV, $m_{p}=0.938$ GeV, $\left(\langle \pi^0\vert (ud)_Ru_L\vert p\rangle_{\mu}\right) = -0.118 \rm{GeV}^2$, $|V_{ud}|^2\approx 0.97$. The lifetime ($\tau=1/\Gamma$ in natural units) formula derive from the one above is

$$ \tau (p\to \pi^0 \mu^+)_{\rm flipped}\simeq 9.7 \times 10^{35} \times |(U_l)_{21}|^{-2} \biggl(\frac{M_X}{10^{16}~{\rm GeV}}\biggr)^4 \biggl(\frac{0.0378}{\alpha_5}\biggr)^2 ~{\rm yrs}~.$$

where $\alpha_{5}=\frac{g^2_{5}}{4\pi}$. I have tried plugging in these numbers and then converting from GeV to years ($1\rm{GeV}=4.79347\times10^{31} \rm{year}^{-1}$). Not sure what I am doing wrong as I get far too small a lifetime. Help is appreciated.

I am trying to reproduce the results of this paper (https://arxiv.org/pdf/2003.03285.pdf). In particular Eq (25) and (26). I would like to convert the proton decay width (given in units of $\rm GeV$) to a lifetime in years. The formula is below:

$$\Gamma (p\to \pi^0 \mu^+)_{\rm flipped}= \frac{g_5^4m_p |V_{ud}|^2 |(U_l)_{21}|^2 }{32\pi M_X^4}(1-\frac{m_\pi^2}{m_p^2})^2 A_L^2 A_{S_1}^2 \left(\langle \pi^0\vert (ud)_Ru_L\vert p\rangle_{\mu}\right)^2.$$

Here $A_{L}=1.2$, $A_{S}=1.5$, $m_{\pi}=0.139$ GeV, $m_{p}=0.938$ GeV, $\left(\langle \pi^0\vert (ud)_Ru_L\vert p\rangle_{\mu}\right) = -0.118 \rm{GeV}^2$, $|V_{ud}|^2\approx 0.97$. The lifetime ($\tau=1/\Gamma$ in natural units) formula derive from the one above is

$$ \tau (p\to \pi^0 \mu^+)_{\rm flipped}\simeq 9.7 \times 10^{35} \times |(U_l)_{21}|^{-2} \biggl(\frac{M_X}{10^{16}~{\rm GeV}}\biggr)^4 \biggl(\frac{0.0378}{\alpha_5}\biggr)^2 ~{\rm yrs}~.$$

where $\alpha_{5}=\frac{g^2_{5}}{4\pi}$. I have tried plugging in these numbers and then converting from GeV to years ($1\rm{GeV}=4.79347\times10^{31} \rm{year}^{-1}$). Not sure what I am doing wrong as I get far too small a lifetime. Help is appreciated.

I am trying to reproduce the results of a paper which looks at proton decay. I would like to convert the proton decay width (given in units of $\rm GeV$) to a lifetime in years. The formula is below:

$\Gamma (p\to \pi^0 \mu^+)_{\rm flipped}= \frac{g_5^4m_p |V_{ud}|^2 |(U_l)_{21}|^2 }{32\pi M_X^4}(1-\frac{m_\pi^2}{m_p^2})^2 A_L^2 A_{S_1}^2 \left(\langle \pi^0\vert (ud)_Ru_L\vert p\rangle_{\mu}\right)^2$

Here $A_{L}=1.2$, $A_{S}=1.5$, $m_{\pi}=0.139$ GeV, $m_{p}=0.938$ GeV, $\left(\langle \pi^0\vert (ud)_Ru_L\vert p\rangle_{\mu}\right) = -0.118 \rm{GeV}^2$, $|V_{ud}|^2\approx 0.97$. The lifetime ($\tau=1/\Gamma$ in natural units) formula derive from the one above is

$ \tau (p\to \pi^0 \mu^+)_{\rm flipped}\simeq 9.7 \times 10^{35} \times |(U_l)_{21}|^{-2} \biggl(\frac{M_X}{10^{16}~{\rm GeV}}\biggr)^4 \biggl(\frac{0.0378}{\alpha_5}\biggr)^2 ~{\rm yrs}~.$

where $\alpha_{5}=\frac{g^2_{5}}{4\pi}$. I have tried plugging in these numbers and then converting from GeV to years ($1\rm{GeV}=4.79347\times10^{31} \rm{year}^{-1}$). Not sure what I am doing wrong as I get far too small a lifetime. Help is appreciated.

I am trying to reproduce the results of a paper which looks at proton decay. I would like to convert the proton decay width (given in units of $\rm GeV$) to a lifetime in years. The formula is below:

$$\Gamma (p\to \pi^0 \mu^+)_{\rm flipped}= \frac{g_5^4m_p |V_{ud}|^2 |(U_l)_{21}|^2 }{32\pi M_X^4}(1-\frac{m_\pi^2}{m_p^2})^2 A_L^2 A_{S_1}^2 \left(\langle \pi^0\vert (ud)_Ru_L\vert p\rangle_{\mu}\right)^2.$$

Here $A_{L}=1.2$, $A_{S}=1.5$, $m_{\pi}=0.139$ GeV, $m_{p}=0.938$ GeV, $\left(\langle \pi^0\vert (ud)_Ru_L\vert p\rangle_{\mu}\right) = -0.118 \rm{GeV}^2$, $|V_{ud}|^2\approx 0.97$. The lifetime ($\tau=1/\Gamma$ in natural units) formula derive from the one above is

$$ \tau (p\to \pi^0 \mu^+)_{\rm flipped}\simeq 9.7 \times 10^{35} \times |(U_l)_{21}|^{-2} \biggl(\frac{M_X}{10^{16}~{\rm GeV}}\biggr)^4 \biggl(\frac{0.0378}{\alpha_5}\biggr)^2 ~{\rm yrs}~.$$

where $\alpha_{5}=\frac{g^2_{5}}{4\pi}$. I have tried plugging in these numbers and then converting from GeV to years ($1\rm{GeV}=4.79347\times10^{31} \rm{year}^{-1}$). Not sure what I am doing wrong as I get far too small a lifetime. Help is appreciated.