Attention conservation notice: This post requires a little general relativity to understand the early bits, and quite a bit of general relativity later on. Later in the post I’ve also used some of the munged LaTeX beloved of theorists collaborating by email; hopefully this won’t obscure the main point.

One of the most exciting developments in physics over the past twenty years was the discovery that the cosmological constant (“Einstein’s greatest mistake”) was not, in fact, a mistake, but appears to be very real.

In standard general relativity – the type where the cosmological constant is zero – it can be shown that test particles follow geodesics of spacetime, i.e., locally, they simply fall freely, with relative acceleration between different particles due to spacetime curvature.

(Note that this geodesic behaviour is sometimes presented as though it’s a fundamental assumption of Einstein’s theory. Actually, it’s not – it follows as a consequence of the Einstein field equations, as we’ll see below.)

I got to wondering recently what paths test particles take when the cosmological constant is non-zero. Do they still follow geodesics, or is their behaviour modified?

The answer, which I’m sure is well-known to relativists, is that they still follow geodesics. So even when the cosmological constant is non-zero, particles still fall freely. What does change is the geometry, since the modification of the field equations means that the same distribution of matter will give rise to a different geometry, and thus to different geodesics.

Here’s an outline of the argument deriving geodesic paths, which is a simple variation on the argument given in Dirac’s text on general relativity for the case where the cosmological constant \Lambda = 0.

Start with the field equations, G+ \Lambda g = 8 \pi T. Suppose we evaluate the divergence of both sides. In index notation we obtain:

G^{\mu \nu}_{; \nu} + \Lambda g^{\mu \nu}_{;\nu} = 8 \pi T^{\mu \nu}_{;\nu}

The first term on the left vanishes because of the Bianchi identies, while the second term vanishes because of the metric compatability condition imposed on the connection. This tells us that the divergence of the stress-energy tensor vanishes:

T^{\mu \nu}_{;\nu} = 0

Now, suppose the stress-energy tensor has the form T^{\mu \nu} = \rho v^{\mu} v^{\nu}, where \rho is matter density, and v^\mu is the four-velocity. Using the product rule to evaluate the divergence, and conservation of mass-energy (\rho v^{\mu})_{;\mu} = 0) we end up with the equation v^{\nu} v ^{\mu}_{;\,nu} = 0, which is the geodesic equation.

From → General

1. So what is the deal with the time-varying cosmological constant?

If the cosmological constant varies with time then the geodesics will change with time… but wait the geodesics are in the space AND time so it doesn’t make sense to talk about a time-varying cosmological constant.

There is a half-elegant solution to displaying math on the web using MathML. The solution uses a javascript that can be placed on your server and run at page display that automatically translates $g_{\mu\nu}$ to the proper MathML, which is displayed automatically with mozilla browsers. See

http://www1.chapman.edu/~jipsen/mathml/asciimath.html

and

http://pear.math.pitt.edu/mathzilla/itex2mml.html

2. I personally think it is absolutely mind-blowing that the geodesic postulate is not necessary. I mean, usually, you have a set of equations describing the interaction between charges and a field, another set for the field, and another set for the charges, but it is amazing that the Einstein field equations do all these jobs, at least for material particles.

It is also pretty nifty that in limited cases (where E \cdot B \neq 0), where E and B are the electric and magnetic fields, that you don’t need Maxwell’s equations either — just the Einstein field equations and the electromagnetic stress tensor.

3. Just another comment,

Your derivation of the geodesic equation is interesting, however, isn’t it the case that if you allow the mass of a particle to approach zero, you don’t have to make explicit use of 4-momentum conservation? It’s just that I tend to think of the Einstein equations as imposing the conservation of 4-momentum. (i.e. you have 10 field equations, giving you six things to set the geometry, and 4 to constrain 4-momentum, leaving 4 things to be arbitrarily determined, which would be arbitrary coordinate transformations).

4. Kaveh: Thanks for the MathML tip. I’ll have a look.

On the issue of a time varying cosmological constant, the divergence of the left hand side of the Einstein equations will no longer vanish, because the time derivative of Lamda will make a contribution to the divergence. As a result, we would no longer expect the geodesic equation to hold.

5. Whoops. I made a few errors.

1. Earlier, I said “I mean, usually, you have a set of equations describing the interaction between charges and a field, another set for the field, and another set for the charges”. Forget that — I meant to say you have one set of equations describing the how the charges affect the field, and another set of equations describing the effect of the field on the charges.

2. Now I see what you have done, Michael. I had forgotten some of the proof as given by Infeld and Schild. I was wondering how your derivation came to be so short and nice-looking, so I mistakingly thought you introduced energy conservation as an independent postulate. But I take that back. You used an explicit form for the stress energy tensor — that’s what got me. I’m used to seeing proofs of the geodesic postulate (for particles) that avoid using the stress tensor. I forgot about that earlier.

6. Hi Courtney: I’m not sure whether I’ve seen the Infeld-Schild proof, so I can’t really comment. This proof (which may be related) came out of section 25 of Dirac’s nice little book on GR. I certainly agree that it’s a lovely feature of GR that the dynamics comes out of the field equations.

7. 