The concurrence for a state $\rho$ as defined here is \begin{equation} C(\rho) = {\rm max}\{0, \lambda_1-\lambda_2-\lambda_3-\lambda_4\}. \end{equation} Where $\lambda_i$ are the eigenvalues of matrix $ \sqrt{\sqrt{\rho} \tilde{\rho} \sqrt{\rho}}$, with $\tilde{\rho} = (\sigma_y \otimes \sigma_y) \rho^* (\sigma_y \otimes \sigma_y)$, $\sigma_y$ being the Pauli matrix and $\rho^*$ is the complex conjugate of state $\rho$.
In the case of Bell diagonal states of the form $\sum_{k=1}^{4} \lambda_k |\beta_k\rangle\langle\beta_k|$, where $|\beta_k\rangle$ is the Bell state. In this case, the concurrence, as defined here (Eq. B11), is given by
\begin{equation} C = 2~ {\rm max}[0, \Lambda-\frac{1}{2}] \end{equation} where $\Lambda = {\rm max}_k\{ \lambda_k\}$.
How are these two definitions connected?
Edit: Thanks for clearing the above doubts. Just found one more definition here (below Eq. 10). Here \begin{equation} C(\rho) = {\rm max}\{0, \lambda_1-\lambda_2-\lambda_3-\lambda_4\}. \end{equation} But this time $\lambda_k$ are $\textit{square root}$ o the eigenvalues of the matrix $\rho \tilde{\rho}$, rather than eigenvalue of the $\textit{square root}$ of matrix $ \sqrt{\sqrt{\rho} \tilde{\rho} \sqrt{\rho}}$. Even if we agree that $\sqrt{\rho} \tilde{\rho} \sqrt{\rho} = \rho$, in the Bell basis, then $\lambda_k$ should be $\textit{the eigenvalues of the square root of $\rho$, rather than square root of the eigenvalues of $\rho$}$.