The cosmological constant seems to be normally described as an energy (repulsive force, Dark Energy) of Space-Time. I was just wondering, if we were to interpret the cosmological constant as being geometrical (i.e. put it on the LHS of the equation), how would it be described. Would it correspond to an exponential increase in the 'stretch' of Space-Time from any given reference frame, or is it more subtle than that?
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2$\begingroup$ Einstein did originally see it as a modification to the LHS. If you think of the GR Lagrangian as a series in R, the cosmological constant is just the first term in the series $R^0$. The physics is the same if its put on the LHS or RHS. $\endgroup$– VirgoCommented May 29, 2015 at 0:39
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$\begingroup$ @physicsphile The mathematical effect is the same, but the physics of what's causing it is different. With dark energy the cosmological constant is a dynamical effect caused by a uniform filler, under Einstein's original interpretation it is purely kinematic. So the difference is similar to Lorentz's ether vs special relativity. Compared to standard GR one could say that in GR with CC space is 'internally stretchy'. $\endgroup$– ConifoldCommented May 29, 2015 at 3:38
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$\begingroup$ @Conifold I disagree with your ether analogy. The ether is disfavored observationally and theoretically compared to relativity. While in the CC case there is no strong experimental or physical argument that favors the term being on the LHS or RHS of the equation. $\endgroup$– VirgoCommented May 30, 2015 at 2:01
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$\begingroup$ I'm wondering if there is any the mathematics of the Cosmological Constant can be viewed as the dimensions of spacetime having a natural exponential stretch (i.e. if you lay out a set of rulers and clocks in a line in space, these rulers and clocks will be dilated such that the dilation increases exponentially with the distance from the observer). $\endgroup$– Chris LaforetCommented May 30, 2015 at 17:12
1 Answer
If we were to interpret the cosmological constant as being geometrical (i.e. put it on the LHS of the equation), how would it be described?
Dark energy is said to be responsible for the increasing expansion of the universe, so let set that aside. Let's focus on the cosmological constant, which is "the value of the energy density of the vacuum of space". Then let's think how the expanding universe is likened to an inflating balloon:
Image courtesy of the one-minute astronomer.
Imagine a balloon in a vacuum. The pressure of the air inside is balanced by the tension in the skin, and there's two ways to make it expand. One way is to blow more air into the balloon. However the dimensionality of energy is pressure x volume, so doing this is in breach of conservation of energy. But there is another way. Not by increasing the pressure, but by reducing the tension. You make the skin weaker. This sounds strange until you see the obvious: make it a bubble-gum balloon. As the balloon expands, the skin gets thinner and weaker, and less able to resist the expansion. So it expands further, so the skin gets weaker, and so on. The pressure drops, the volume increases, but energy is conserved.
Now take a look at page 5 of http://arxiv.org/abs/0912.2678 where Milgrom mentions the strength of space. Think in terms of the tensile strength of space, and it's something like our balloon analogy, but for a 3D bulk. Then go back to Einstein, who introduced the cosmological constant to stop his universe collapsing. That was akin to a pressure, but the cosmological constant is described as a negative pressure. And negative pressure is tension. So in my humble opinion the cosmological constant is described as a tension, rather ike the tension of the bag model. The tension is reducing, along with the energy density. And because the cosmological constant is "the value of the energy density of the vacuum of space", it isn't constant after all.
As to how it might end, there's something in this little article by Phil Plait that bothers me. Bubble-gum bubbles don't always end well.