When Einstein was formulating his General Theory of Relativity, he had to introduce a cosmological constant to negate his model universe’s expansion.
Later, astronomer Edwin Hubble discovered evidence that the universe was indeed expanding…a discovery which made Einstein declare the cosmological constant “the greatest blunder of his life”.
Lately, though, discoveries show that the universe’s expansion is not constant; apparently, over the past 14 billion years the expansion has speeded up and slowed down. In fact, the expansion rate is currently speeding up, thus renewing interest in the cosmological constant.
Does this imply the Hubble is a function of the Cosmological constant, or vice versa?
I’m not sure what your “Planck units” comment is supposed to be, if it’s not just a bump. But here’s the basic idea:
The relevant equations are called the Friedmann equations. These equations are what you get if you assume that the universe has homogeneous curvature and homogeneous, isotropic stress-energy tensor. In these equations, Lambda is the cosmological constant, a is the universal expansion factor, and rho and p are the density and pressure of matter (these are elements of the stress-energy tensor). H, in the first equation, is the Hubble “constant,” which clearly depends on Lambda.
Notice also that H need not be constant over cosmological time. H was originally an empirical “constant”: measurements showed that the distance and redshift of distant objects were, to the precision possible at the time, linearly related; H was just defined as that proportionality constant–hence the rather strange convention for units on H, of km/sec/Mpc, for something which is really an inverse-time. The modern definition of H in theoretical cosmology is H = (da/dt)/a, which no longer needs to be constant.
There are two different ways of interpreting the cosmological constant. The Einstein field equations (with cosmological term, and with G=c=1) look like G+8 Lambdag=8pi*T. The tensor G represents spacetime curvature (acceleration due to gravity, etc.), and the stress-energy tensor T represents all of the matter in spacetime. You can think of the Lambda term as a sort of intrinsic curvature to spacetime (i.e., adding to G) or as a strange sort of matter (adding to T).
As originally intended by Einstein, Lambda was indeed a constant (hence the name), just put in to make a static universe. But if you take this term to indicate some strange sort of matter, there’s no reason to assume it’s constant. Instead, this cosmological matter (or “vacuum energy” or “quintessence”; it’s been given lots of cute names, in different ideas for its origin and meaning) is just some extra stress-energy term, though one with behavior much different from that of ordinary matter. Different theories for the behavior of this stuff give different predictions for the evolution of spacetime.
This raises a huge question I’ve had sitting around. How can \Lambda not be constant? (warning: math ahead)
Basically, what sits on the right of the equation is the stress-energy tensor, which is basically the properly Lorentz-invariant notion of energy-momentum (see Noether’s theorem). This is automatically divergence-free, reflecting the (local) conservation of energy-momentum we must have from (local) spatiotemporal translation-invariance.
So, what sits on the right should also be divergence-free in and of itself, not as a consequence of the field equations (cf. Maxwell’s equations in the differential-geometric setting: charge is conserved <-> current 4-vector is divergence-free <-> d*F is identically divergence-free). So, what identically divergence-free tensors can be built from the metric and at most two derivatives? Basically two: the Einstein tensor G and the metric itself g. We get a linear combination of the two to set equal to the stress-energy tensor on the right, and the coefficient of g is the cosmological constant (basically).
So what happens if \Lambda isn’t constant? Well now the left hand side of the equation is not identically divergence-free, since it picks up divergence terms from the gradient of \Lambda. If this is still to be identified with stress-energy (like d*F is with the current 3-form in electromagnetism), then stress-energy is no longer divergence free and we lose conservation of energy-momentum.
Mathochist, I believe the two words you are looking for are “dark energy.”
Though I am by no means an expert here, what you have described is how a non-constant \Lambda is screwing up the divergence-free requirement of Einstein’s equation. The situation is clearly unacceptable so a first-order fix (and really, this is the scientific equivalent of putting duct tape on the Hindenberg) is to propose a new term that actually cancels out the \Lambda term.
The way you describe it, we’ll end up with Einstein’s equation being G + \Lambda g = k T, where k is some constant and T is the stress-energy tensor. We then require the dark energy stress tensor, say, T_dark, to fulfill kT_dark = -\Lambda g.
So then Einstein’s equation, when you add the dark energy to the total stress-energy, becomes G + \Lambda g = k(T + T_dark), which we can reduce appropriately to eliminate the problematic, non-constant \Lambda.
I doubt if this will actually assuage your doubts, but I’m pretty sure this is how it works. Apologies for the somewhat crude-looking equations.
Argh, I think I made a mistake in that last post (this is what happens when you run on memory alone).
You don’t actually want to cancel the lambda term. You want to subsume it into the dark energy term itself. That is, using the definition of T_dark above, Einstein’s equation becomes G = k(T + T_dark). The assumption is that the “new” stress-energy tensor, the total of the two, is divergence free.
Unfortunately I have to dash off somewhere, and I’m sorry I’m not 100% on the specifics here. I will be thinking more about this and try to come up with a more reasonable-sounding explanation, but given that the subject is dark energy I’m not optimistic.