The Friedmann equation is given by $$(\frac{\dot{a}}{a})^2 = \frac{8 \pi G \rho}{3} - \frac{kc^2}{a^2} + \frac{\lambda c^2}{3}.$$ Considering that the left hand side is analagous to the kinetic energy of expansion, and the term $\frac{8 \pi G \rho}{3}$ analagous to the G.P.E which resists expansion, does this imply that the cosmological constant represents an entity which opposes expansion?

Clearly this is not the case. Yet if it were not known that dark energy exerts a negative pressure, and thus represents a kind of 'anti-gravity', does the Friedmann equation not imply that the cosmological constant opposes expansion?


1 Answer 1


This whole equation, as you'll note, provides the basic rate of expansion. It's true that the $\frac{8\pi G\rho}{3}$ term decelerates expansion, however this is mostly because both $\rho$ and $\frac{1}{a^2}$ fall off as $a$ increases. Even with $\lambda$ positive and small, it remains constant, which means that as $\rho$ and $\frac{1}{a^2}$ drop to zero, the fraction $\frac{\dot a}{a}$ approaches a constant, which means the expansion becomes accelerated.

So the Friedmann equation implies that the cosmological constant eventually dominates and drives accelerated expansion.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.