Using the derivation in previous work for the drag in three dimensions, it was found can be formulated as:

[math]\frac{1}{f}(\frac{mv^2}{n^2x^3} + \frac{mv^2}{n^2y^3} + \frac{mv^2}{n^2z^3}) = - \frac{A_b}{A_f} \cdot \rho v^2 = - \frac{A_b}{A_f} \cdot T^{00} [/math]

Solving for the stress energy alone we have

[math] \frac{1}{f}\frac{A_f}{A_b}(\frac{mv^2}{n^2x^3} + \frac{mv^2}{n^2y^3} + \frac{mv^2}{n^2z^3}) = - T^{00} [/math]

(concentrate on the sign, because this does not indicate a negative energy per se, we will see how the sign is connected specifically to the definition of the cosmological impetus [math]\Lambda[/math])

Einstein did think the cosmological constant as an independent parameter, but as we know today, it is inherently linked to the energy of the vacuum - in the mechanical explanation I have provided in my own work, it arises as a ''pressure difference'' and that slightly more momentum was owed to the expansion of the universe rather than it collapsing. Instead of writing that avenue out, all we need for now is to write the energy-momentum component as:

[math]T^{00} = - g^{00}\ \frac{\Lambda c^4}{8 \pi G}[/math]

And the density as

[math]\rho_{vac} = \frac{\Lambda c^2}{8 \pi G}[/math]

So the first equation can be formulated as

[math]- \frac{1}{f}\frac{A_f}{A_b}(\frac{mv^2}{n^2x^3} + \frac{mv^2}{n^2y^3} + \frac{mv^2}{n^2z^3}) = \frac{\Lambda c^2}{8 \pi G} = \rho_{vac} v^2 [/math]

Which relates the cosmological constant as a ''thrust.''

And the second replacement for the stress energy:

[math] - \frac{1}{f}\frac{A_f}{A_b}(\frac{mv^2}{n^2x^3} + \frac{mv^2}{n^2y^3} + \frac{mv^2}{n^2z^3}) = - g^{00}\ \frac{\Lambda c^4}{8 \pi G} = T^{00} [/math]

**Conclusions:**

The cosmological constant before was looked at as a mechanical explanation from fluid equations in which the universe had a little more momentum expanding than it did to collapse. The result of applying those drag equations in three dimensions to the stress energy relates the cosmological constant as a type of cosmological thrust.