Which side (of the Einstein equations) are you on?

As a cosmologist, I am often asked why it is that people talk about the cosmological constant as if it were some sort of vacuum energy or “dark energy“. I was explaining it again to a student today so I thought I’d jot something down here so I can use it for future reference. In a nutshell, it goes like this. The original form of Einstein’s equations for general relativity can be written

R_{ij}-\frac{1}{2}g_{ij}R = \frac{8\pi G}{c^4} T_{ij}.

The precise meaning of the terms on the left hand side doesn’t really matter, but basically they describe the curvature of space-time and are derived from the Ricci tensor R_{ij} and the metric tensor g_{ij}; this is how Einstein’s theory expresses the effect of gravity warping space. On the right hand side we have the energy-momentum tensor (sometimes called the stress tensor) T_{ij}, which describes the distribution of matter and its motion. Einstein’s equations can be summarised in John Archibald Wheeler’s pithy phrase: “Space tells matter how to move; matter tells space how to curve”.

In standard cosmology we usually assume that we can describe the matter-energy content of the Universe as a uniform perfect fluid, for which the energy-momentum tensor takes the simple form

T_{ij} = -pg_{ij} +\left(p+\rho c^2\right) U_i U_j,

in which p is the pressure and \rho the density; U_i is the fluid’s 4-velocity.

Einstein famously modified (or perhaps generalised) the original equations by adding a cosmological constant term \Lambda to the left hand side thus:

R_{ij}-\frac{1}{2}g_{ij}R -\Lambda g_{ij} = \frac{8\pi G}{c^4} T_{ij}.

Doing this essentially modifies the description of gravity, or appears to do so because it happens to be written on the left hand side of the equation. In fact one could equally well move the term involving \Lambda to the other side and absorb it into a redefined energy-momentum tensor, \tilde{T}_{ij}:

R_{ij}-\frac{1}{2}g_{ij}R = \frac{8\pi G}{c^4} \tilde{T}_{ij}.

The new energy-momentum tensor needed to make this work is of the form

\tilde{T}_{ij}=T_{ij}+ \left(\frac{\Lambda c^{4}}{8 \pi G} \right) g_{ij}= -\tilde{p} g_{ij} +\left(\tilde{p}+\tilde{\rho} c^2\right) U_i U_j


\tilde{p}=p-\frac{\Lambda c^4}{8\pi G}

\tilde{\rho}=\rho + \frac{\Lambda c^4}{8\pi G}

So the cosmological constant now looks like you didn’t modify gravity at all, but created an additional contribution to the pressure and density of the original fluid. In fact, considering the correction terms on their own it is clear that the cosmological constant acts exactly like an additional perfect fluid contribution with p=-\rho c^2.

This is just one simple example wherein a modification of the gravitational part of the theory can be made to look like the appearance of a peculiar form of matter. More complicated versions of this idea – most of them entirely speculative – abound in theoretical cosmology. That’s just what cosmologists are like.

Over the last few decades cosmology has suffered an invasion by been stimulated and enriched by particle physicists who would like to understand how such a mysterious form of energy might arise in their theories. That at least partly explains why, in one sense at least,  modern cosmologists prefer to dress to the right.

Incidentally, another interesting point is why people say such a fluid describes a cosmological “vacuum” energy. In the cosmological setting, i.e. assuming the fluid is distributed in  a homogeneous and isotropic fashion then the energy density of the expanding Universe varies with (cosmological proper) time according to

\dot{\rho}=-3\left(\frac{\dot{a}}{a}\right) \left(\rho + \frac{p}{c^2}\right)

so for our strange fluid, the second term in brackets vanishes and we have \dot{\rho}=0. As the universe expands, normal forms of matter and radiation get diluted, but the energy density of this stuff remains constant. It seems to me to be quite appropriate for a vacuum to something which, no matter how hard you try,  you can’t dilute!

I hope this clarifies the situation.


10 Responses to “Which side (of the Einstein equations) are you on?”

  1. Anton Garrett Says:

    Nice summary. It might be worth adding that the cosmological term can be included without wrecking the beautiful rationale underlying GR.

    How about a homeopathic universe model in which you CAN dilute the vacuum…

  2. Are you going to explain the various, persistently shaky, attempts at modifying the left side to get an effect like dark energy?

    • telescoper Says:

      That would take a while! But shaky modifications can actually be made on either side. One can modify the Einstein-Hilbert action to get a gravitational theory different to Einstein’s or dream up a peculiar form of matter that gives something with negative pressure (although more complicated than a pure Lambda) or both. It’s a popular game to play, but I find it all a bit, er, vacuous.

    • Anton Garrett Says:

      On the contrary Peter, it would be a gravitational theory different FROM Einstein’s.

      Yours pedantically

    • telescoper Says:

      Quite right. I grovel in mortification and will leave my error there as a penance.

  3. has suffered an invasion by been stimulated and enriched by particle physicists

    Your the ones who keep putting strange bounds on our constants 😉

  4. […] The particularly interesting case is w=-1 which corresponds to a cosmological constant term; see here for a technical discussion. However, we don’t know how to explain this dark energy from first […]

  5. […] When Einstein introduced the cosmological constant in 1915/6 he did it by modifying the left hand side of his field equations, essentially modifying the law of gravitation. This discussion shows that he could instead have modified the right hand side by introducing a vacuum energy with an equation of state p=-ρc2. A more detailed discussion of this can be found here. […]

  6. […] Finally, I’ll just mention another thing in the light of the Einstein (1917) paper. It is clear that Einstein thought of the cosmological as a modification of the left hand side of the field equations of general relativity, i.e. the part that expresses the effect of gravity through the curvature of space-time. Nowadays we tend to think of it instead as a peculiar form of energy (called dark energy) that has negative pressure. This sits on the right hand side of the field equations instead of the left so is not so much a modification of the law of gravity as an exotic form of energy. You can see the details in an older post here. […]

  7. […] if the dark energy is of the form of a cosmological constant (or vacuum energy). I explained why here. Non-relativistic matter (dominated by rest-mass energy) has w=0 while ultra-relativistic matter […]

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