# A Why isn't cosmological constant a mass of the graviton?

1. May 29, 2017

### Demystifier

The Lagrangian density for cosmological constant is
$${\cal L} = \sqrt{g}\Lambda$$
Let us write, schematically,
$$g=\eta+h$$
where $\eta$ is the flat Minkowski metric and $h$ is the spin-2 field. Expanding the square root for small $h$ we get something like $${\cal L} = \Lambda + h\Lambda + h^2\Lambda + h^3\Lambda + ...$$
In particular, the term quadratic in $h$, namely $h^2\Lambda$, looks like a mass term for spin-2 field, suggesting that mass$^2$ of the graviton is proportional to $\Lambda$.

I'm sure there is something wrong with this naive argument, but can someone tell me more precisely what exactly is wrong?

2. May 29, 2017

### stevendaryl

Staff Emeritus
What I've seen people claim is two things:
1. Gauge symmetry (which for GR means diffeomorphism invariance) implies that the "force carrier" must be massless.
2. Infinite-range forces imply massless force carriers (massive force carriers imply a short-range Yukawa type potential).
I don't know how these two claims square with your observation.

3. May 29, 2017

Staff Emeritus
Why do you think you can identify a term with units of mass with the mass of a graviton? You haven't put any QM into this, so ending up with a quantum field out would be surprising. Very surprising. Why not attribute it to something else with units of mass, like the Hubble radius in appropriate units?

Last edited: May 29, 2017
4. May 29, 2017

### RockyMarciano

There sems to be a clear gap between the perturbative approximation to Minkowskian background and the full nonlinear GR, even if people like Kip Thorne, Hawking, etc claim they are just equivalent sides of the same coin, but if they were I think your argument should work. The funny thing is that the claimed equivalence is based in mathematics that include the clear objections shown by stevendaryl.

5. May 29, 2017

### haushofer

Just a quickie: how is mass defined exactly in (A)dS spaces for massless spin-2 fields (Breitenlohner-Freedman bounds etc)?

Also,

https://arxiv.org/abs/1105.3735

might help.

6. May 29, 2017

### haushofer

7. May 30, 2017

### haushofer

I don't understand this. This is not the issue. The issue is how you define mass in a (A)dS background. As I stated, this differs from mass in Minkowski-backgrounds (the representations differ) and you see this back in the Lagrangian.

If you introduce a cosmological constant, the Minkowski metric is not a solution anymore and hence it makes no sense to write down a perturbation around it. Instead, Demystifier should expand around an (A)dS vacuum.

So, to answer the question in the OP: what's wrong is that your Minkowski-metric is not a solution to the field equations anymore.

8. May 30, 2017

### RockyMarciano

I was just ponting out that if one admits the mainstream view(routinely used in computations of GWs, BHs solution approximations, astroparticle physics, etc) that Minkowski metric plus perturbation is (perturbatively) equivalent to the full nonlinear GR FAPP, then that perturbation of the background Minkowski space(not just the Minkowski metric) should be valid as approximate solution in the presence of cosmological constant.
A different thing is how to interpret the mass term of the cosmological constant, the issues of GR with the mass concept are too large to comment here.

9. May 30, 2017

### Staff: Mentor

AFAIK all of these computations assume a zero cosmological constant. So, IIRC, do the theoretical treatments done by Feynman, Deser, and others in the 1960s and early 1970s, that showed that if you start with the "naive" QFT of a massless spin-2 field and do what's necessary to make it self-consistent, you end up with a QFT whose classical limit is GR (with zero cosmological constant).

Not if what I said above is correct (and I think it is).

10. May 30, 2017

### pervect

Staff Emeritus
I'm just not convinced that a non-zero mass of the graviton would have anything to do with the cosmological constant. The typical approach to the cosmological constant problem involves not the "mass of the graviton", which is assumed to be zero for reasons other posters have outlined, but the "vacuum energy density", which is the (effective?) stress-energy tensor of the vacuum. Then the theoretical basis for the relation of the vacuum energy density and the cosmological constant becomes clear, it is just Einstein's field equations.

It's rather well known that our estimates of the vacuum energy density are horrible. Wiki gives a reference, for instance, MP Hobson, GP Efstathiou & AN Lasenby (2006). General Relativity: An introduction for physicist, in which it's mentioned that it's off by 120 orders of magnitude, and is "the worst theoretical prediction in the history of physics". Hobson goes on to say that nobody knows how to make sense of this result.

11. May 30, 2017

### Staff: Mentor

One obvious question would be, if you're going to interpret the $h^2$ term as giving a mass to the graviton, the rest of the terms would need a physical interpretation as well. The $h^3$ and higher order terms are just interactions with 3 or more external lines (as the mass term is an "interaction" with just two external lines, i.e., a propagator). But the $h^0$ and $h^1$ terms don't seem to have any obvious interpretation like this; I'm not aware of any other QFT that even has such terms in its Lagrangian. So how would you explain the presence of these terms?

12. May 30, 2017

### RockyMarciano

It is correct that they assumed zero lambda because it appeared like an unnecessary addition in the absence of physical justification, but adding a cosmological constant is always possible for the EFE, they admit the addition of this term by construction, and in that sense one may choose its value to be zero or nonzero. By contrast a solution like deSitter's is not possible for vanishing lambda.

13. May 30, 2017

### Staff: Mentor

No, that's not why they assumed zero $\Lambda$. They assumed it because without it their model would not work. (More precisely, they didn't assume it, they derived it; see below.)

Yes, but the QFT models I'm talking about did not start with the EFE. They started by assuming a fixed background Minkowski spacetime and constructing the QFT of a massless, spin-2 field on it. The EFE is derived in this approach, as the field equation that the massless, spin-2 field satisfies once you have made all of the necessary corrections to the initial "naive" QFT you get by just writing down the massless, spin-2 field Lagrangian. But this EFE, derived with Minkowski spacetime as the fixed background, does not contain $\Lambda$.

Only after all of the above has been done is it observed that the massless, spin-2 field has a geometric interpretation, i.e., that solutions to the QFT can be interpreted, not as a field on a fixed background Minkowski spacetime, but as the geometry of a curved spacetime, with metric $g_{ab} = \eta_{ab} + h_{ab}$, where $h_{ab}$ is the spin-2 field. However, AFAIK one cannot obtain spacetimes with a nonzero $\Lambda$ by this approach; to do that, one would have to go back and redo the entire procedure with de Sitter spacetime (for positive $\Lambda$) or anti-de Sitter spacetime (for negative $\Lambda$) as the fixed background. But if you do that, then the spin-2 field $h_{ab}$ does not contain $\Lambda$ (because it's part of the fixed background) and you can't write the $\sqrt{-g} \Lambda$ term in the Lagrangian the way the OP writes it.

14. May 30, 2017

### RockyMarciano

Why not? It is not a question of what's derived from what, I'm assuming as starting point(for the sake of the argument in this thread) the equivalence of the two approaches, if this is so then it doen't matter how they derived one from the other or viceversa, the perturbative linearized EFE should admit a lambda term just like the full EFE.
You could instead argue that there is no such equivalence.

15. May 30, 2017

### dextercioby

If you start with the free spin-2 field and its Lagrangian density (known as Pauli-Fierz field), then the full nonlinear theory of Einstein and Hilbert is reached perturbatively in the presence of a non-zero cosmological constant. This was shown by the ULB school behind Marc Henneaux https://doi.org/10.1016/S0550-3213(00)00718-5
It is one of their most significant theoretical achievements. In this article they show more, that there's no Yang-Mills-like theory of gravity (cross-interactions between families of gravitons). My former school (the Craiova one behind Constantin Bizdadea) took this result further and showed there is no Yang-Mills-like theory of gravity coupled to Dirac fields http://stacks.iop.org/1126-6708/2005/i=02/a=016 and no Yang-Mills theory of gravity in the presence of massive gravitinos. https://link.springer.com/article/10.1140/epjc/s2006-02620-9.

16. May 30, 2017

### Staff: Mentor

Hm, I was not aware of this. Interesting.

The arxiv preprint of the paper, for those who, like me, can't get behind paywalls, is here:

https://arxiv.org/abs/hep-th/0007220

17. May 30, 2017

### Staff: Mentor

My belief when I made that previous post was that the equivalence only holds for the EFE with zero $\Lambda$. However, I now need to look at the paper @dextercioby posted, which seems to indicate that it holds in the presence of nonzero $\Lambda$ as well.

18. May 30, 2017

### Staff: Mentor

On a quick skim of the paper, I'm not entirely sure which of the two following possibilities it is claiming to demonstrate:

(1) Start with flat Minkowski spacetime as the fixed background. Construct the perturbative QFT of a massless, spin-2 field on this background. Find that this perturbative QFT leads to an effective Lagrangian in which part of the perturbation $h$ "becomes" the scalar curvature $R$ and part of it "becomes" the cosmological constant $\Lambda$ (the latter part, if I'm reading correctly, is the trace $h^{\mu}{}_{\mu}$ of the perturbation).

(2) Start with some spacetime that already has a nonzero $\Lambda$ as the background. Construct the perturbative QFT of a massless, spin-2 field on this background. Find that this perturbative QFT is self-consistent, in that the effective Lagrangian still contains a nonzero $\Lambda$ term as well as the scalar curvature $R$.

I think it's #1, but I'm not sure since I'm not very familiar with the mathematical tools being used in the paper.

19. May 30, 2017

### dextercioby

It is #1, because the assumption on the Lagrangian density has no Lambda in it. The existence of Lambda at a perturbative level is proved in the same way as in the EH Lagrangian. A mathematical necessity.

20. May 30, 2017

### Staff: Mentor

Hm, ok, but that then leads to my next question: the $\Lambda$ term comes from the trace $h^{\mu}{}_{\mu}$, correct? But I can always make the tensor $h_{\mu \nu}$ traceless by a gauge transformation, so it seems like the trace should not have any physical content. Doesn't that mean I can't actually say this theory shows the existence of $\Lambda$?