# Energy conditions for a “semi”-perfect fluid

Consider the follwing diagonal stress-energy tensor with different values for the pressures:

$$T_{MN}=\left( \begin{matrix} \rho & 0 & 0 & 0\\ 0 &p_1 &0 &0 \\ 0 &0 &p_2 &0 \\ 0 &0 &0 &p_3 \end{matrix} \right)$$

If all the pressures are equals ($p=p_1=p_2=p_3$), hence we have a perfect fluid where the energy conditions are imposed as:

The null energy condition: $\rho +p\geq 0.$ $\rho +p\geq 0.$

The weak energy condition: $\rho \geq 0,\;\;\rho +p\geq 0.$

The dominant energy condition: $\rho \geq |p|.$

The strong energy condition: $\rho +p\geq 0,\;\;\rho +3p\geq 0.$

However, if ($p_1 \neq p_2 \neq p_3$). There is some way to rewrite these energy conditions?

$$\rho + p_i \ge 0 \;\wedge \rho + \sum_{i=1}^{3}p_i \ge 0$$