Staff Emeritus
Einstein's field equations, with cosmological constant, can be written as:

$G_{\mu \nu} + \Lambda g_{\mu \nu} = \kappa T_{\mu \nu}$

I understand that some physicists think that the cosmological constant, rather than being a free parameter, might instead be an effect of quantum field theory. Does that mean that mean that they are really moving the term to the other side of this equation:

$G_{\mu \nu} = - \Lambda g_{\mu \nu} + \kappa T_{\mu \nu} \equiv \kappa (T^{vac}_{\mu \nu} + T_{\mu \nu})$ where $T^{vac}$ is some vacuum energy?

PAllen
That is my understanding.

haushofer
Yes :)

ShayanJ
Gold Member
But how do they calculate it?

haushofer
They don't. They measure it, like the speed of light. Any attempt to explain this number in terms of QM failed.

ShayanJ
Gold Member
They don't. They measure it, like the speed of light. Any attempt to explain this number in terms of QM failed.
So what about the cosmological constant problem?

PAllen
ShayanJ
Gold Member
The referenced 'failure to explain the number' is the cosmological constant problem.
But the cosmological constant problem is usually stated as the huge(tens of orders of magnitude) discrepancy between the measured and calculated numbers for the cosmological constant. This suggest that there is a calculation!
e.g.
The quantum field theory prediction of the cosmological constant is 120 orders of magnitude higher than the observed value. This is known as the cosmological constant problem.
Source

George Jones
Staff Emeritus
Gold Member
For an interesting technical review, see "Everything You Always Wanted To Know About The Cosmological Constant Problem (But Were Afraid To Ask)" by Jerome Martin,

http://arxiv.org/abs/1205.3365

ShayanJ
PAllen
But the cosmological constant problem is usually stated as the huge(tens of orders of magnitude) discrepancy between the measured and calculated numbers for the cosmological constant. This suggest that there is a calculation!
e.g.

Source
That discrepancy is the failure to explain it. Amazingly, we have several back an forth posts where you and haushofer are in complete agreement! (I guess because he didn't use the exact words you are looking for??!!)

ShayanJ
Gold Member
That discrepancy is the failure to explain it. Amazingly, we have several back an forth posts where you and haushofer are in complete agreement! (I guess because he didn't use the exact words you are looking for??!!)
Yeah, I realized that!
I knew that whatever those calculations are, they don't work. But I wanted to know how those (unsuccessful) calculations are done!

haushofer
The usual number can be derived by considering a field as an (infinite) set of harmonic oscillators in their ground state. If you calculate the energy density of this ground state by integration over phase space using the Planck scale as cut-off, you get the huge discrepancy.

This is also one of the few examples (the only one I can think of) in which a low-energy effect (it involves cosmological scales) involves a UV-divergence.

ShayanJ
I think even Einstein himself considered introducing the cosmological constant "his greatest mistake" ...
But there is always hope that it will be expained.

haushofer
I have the feeling that it is strange that we try to "explain" the cosmological constant and not the speed of light. Our spacetime vacuum is characterized by two parameters: the speed of light c and the cosmological constant ##\Lambda##. In that sense we should describe special relativity and quantum field theory in a deSitter space. In QFT we never talk about quantum corrections/explanations of c, so why should we talk about the quantum nature of ##\Lambda## being an energy density? Somehow this feels like applying double standards.

jimmy1010100 and ShayanJ
I have the feeling that it is strange that we try to "explain" the cosmological constant and not the speed of light. Our spacetime vacuum is characterized by two parameters: the speed of light c and the cosmological constant
Well the speed of light is indeed the speed of light! (not just a vague constant c) But what does Λ represent? Which measurable quantity? (if any)
That's the real question, or at least one part of it.

ShayanJ
Gold Member
I have the feeling that it is strange that we try to "explain" the cosmological constant and not the speed of light. Our spacetime vacuum is characterized by two parameters: the speed of light c and the cosmological constant ##\Lambda##. In that sense we should describe special relativity and quantum field theory in a deSitter space. In QFT we never talk about quantum corrections/explanations of c, so why should we talk about the quantum nature of ##\Lambda## being an energy density? Somehow this feels like applying double standards.
I agree. If we say: "## \Lambda ## should be the same thing as the vacuum energy we know from QFT, and so the discrepancy is a problem", then this is really applying double standards. But if we say: "Hey guys, these two concepts seem similar. Maybe they are related. Lets see!" Then this is just pursuing a thread without any anticipation about where it leads. Its just that there is no such a thread about the speed of light! And of course this huge discrepancy suggests that the thread about ##\Lambda## may not lead anywhere but maybe there is actually a solution to this discrepancy which changes our worldview, just like Planck's solution to the UV catastrophe that led to QM.
But about describing SR and QFT in a deSitter spacetime, I'm definitely with you there!

Another example is planck's constant h (or h bar). When he introduced it, where did it come from? (other than giving a "correct" equation h = hf)
[Later, it is connected to certain observables or to the Uncertainty principle.]

Staff Emeritus
I don't see how one would calculate the vacuum energy from QFT, at all. In QFT without gravity, a constant background energy has no effect, so it would seem to me that you are free to modify your theory to have any background energy you like, and it makes no difference. So our current theories wouldn't (in this naive way of looking at it) constrain the background energy at all. So is there a simple explanation for how people hope to get a non-arbitrary number from QFT?

Haelfix
I have the feeling that it is strange that we try to "explain" the cosmological constant and not the speed of light. Our spacetime vacuum is characterized by two parameters: the speed of light c and the cosmological constant ##\Lambda##. In that sense we should describe special relativity and quantum field theory in a deSitter space. In QFT we never talk about quantum corrections/explanations of c, so why should we talk about the quantum nature of ##\Lambda## being an energy density? Somehow this feels like applying double standards.
This is not correct, and neither are most of the responses in this thread. The reason 'c' is not a problem for fundamental physics is that it is not a parameter that undergoes renormalization. The cosmological constant on the other hand is. It makes no difference if you put it on the right hand side of the equation like StevenDaryl does and interprets it as arising from matter contributions. Even if you have a mechanism or a reason to make that quantity zero, the LEFT hand side also undergoes renormalization and generates an effective cosmological constant. The only difference is we would interpret it as arising from some length or curvature scale in the geometry sector.

The reason physicists aren't used to seeing this, is b/c they are used to dealing with semiclassical gravity, where everything on the left hand side remains classical. However in quantum gravity done as an effective field theory, this is not the case, and those same dangerous renormalizations that haunt the matter sector also arise in the geometry sector unless there is a scale invariance symmetry within the fundamental quantum description of geometry.

This all relies on the premise that you can promote Einsteins field equations to operator valued equations.

Stavros Kiri
This all relies on the premise that you can promote Einsteins field equations to operator valued equations.
Exactly! You are assuming quantum gravity. Thus you could also be non correct (as of now). Gravity could be radically different from the other interactions in the fact that it is also the theory of space-time, thus providing the framework in which all phenomena and other interactions are studied. In other words, no one can guarrantee that applying Quantum Field Theory for Gravity is legitimate (although not impossible). Obviously something is missing here. That's why Quantum Gravity hasn't been succesfull yet.

Haelfix
I'm going to borrow parts of a recent post that I wrote in another thread to setup one aspect of the problem.
Consider for concreteness the theory of a single scalar field with potential $V\left ( \Phi \right )$. The action is given by
$$S= \int d^{4}x\sqrt{-g}\left ( \frac{1}{2}g^{uv}\partial _{u}\Phi \partial _{v}\Phi -V\left ( \Phi \right )\right )$$
The corresponding energy momentum tensor is computed as:
$$T_{uv}= \frac{1}{2}\partial _{u}\Phi \partial _{v}\Phi + \frac{1}{2}\left (g^{\alpha \beta }\partial _{\alpha }\Phi \partial _{\beta }\Phi \right ) g_{uv} - V\left(\Phi\right) g_{uv}$$
The lowest energy density configuration if it exists is obtained when both the kinetic and gradient terms vanishes, which gives us our definition for vacuum energy and implies for this particular theory that:
$$T_{uv}^{vac}\equiv -\rho _{vac} g_{uv} =-V\left(\Phi_{0}\right) g_{uv}$$
Where $\Phi _{0}$ is the value that minimizes the potential. Note that this is not necessarily zero. More generally you can argue by lorentz invariance that the form for vacuum energy is unique and fixed exactly as above.

Now, when you introduce quantum corrections, you can take my word for it, or you can perform the calculation yourself and you will get the first corrections that look like:
$$T_{uv}^{quantvac}\equiv -\rho _{quantvac} g_{uv} =-V\left(\Phi_{0}\right) g_{uv} + \left [ \sum_{k} n_{k}w_{k} \right ] g_{uv}$$

The first term on the right is the classical part of the cosmological constant, the latter term on the right is the first oscillator correction to the scalar field's classical value. The sum of these two numbers + the contribution from anything you want on the left hand side (classical value, as well as quantum corrections) must be equal to the effective cosmological constant that is observed in nature. In practise people just group everything that is naively classical to one part, and everything that is quantum to the other part.

A few notes: The quantum part of this diverges as k -- > infinity, so it must be regulated. If you are completely naive and perform a hard cutoff at some length scale L, you will see that you will get an answer that depends on the fourth power of this length scale. This is part of the problem. Why should this number (presumably derived from the microphysics of fundamental physics) match a classical part to an incredible 10^120 decimal places of accuracy (the classical value is in principle sensitive to cosmological contributions, like domain walls and things of that nature). This incredible conspiracy of hierarchies is why physicists believe there is a problem with the logic of this setup, but it is very difficult to sort of undo this, without violating some other known part of physics. For instance if you try to force the quantum part to take the value zero, you also spoil things like inflation.

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haushofer, Stavros Kiri and ShayanJ
haushofer
This is not correct, and neither are most of the responses in this thread. The reason 'c' is not a problem for fundamental physics is that it is not a parameter that undergoes renormalization.
I understand. But if you would write down a QFT in deSitter space, is ##\Lambda## then not just part of the causal structure of spacetime, like c? To put it differently, does ##\Lambda## then undergo renormalization? Or do we still need a symmetry in order to 'protect' ##\Lambda## for undergoing renormalization? (edit: I now see your last post, so the answer is 'yes' apparently)

I'm not familiar with QFT's based on "deSitter special relativity", so I could say silly things here.

haushofer
I guess my question boils down to: So why is the speed of light c not renormalized in QFT, then?

Is there a relationship between the cosmological constant and the net effective mass of quantum foam? If QF particles consist of positive and negative mass, a deSitter concept, and there is a slight asymmetry in the annihilation it could account for dark energy and a non zero cosmological constant.

Stavros Kiri
Correction of a typo about Planck's constant in a previous comment of mine (#17):
Another example is planck's constant h (or h bar). When he introduced it, where did it come from? (other than giving a "correct" equation h = hf)
[Later, it is connected to certain observables or to the Uncertainty principle.]
I meant E = hf (for the energy quanta [introduced by Planck], ... and later by Einstein, energy of photons) ...

Of course there is no problem whatsoever with Planck's constant, similar (or any other) to the problem of the cosmological constant. So this may not be a good example either. But mainly, earlier, I wanted to make a point in the discussion about Λ and c. A good answer may have probably been given above by:
The reason 'c' is not a problem for fundamental physics is that it is not a parameter that undergoes renormalization.The cosmological constant on the other hand is.
but it assumes quantum gravity, with problems still. Unless the following gives a way out:
However in quantum gravity done as an effective field theory, this is not the case, and those same dangerous renormalizations that haunt the matter sector also arise in the geometry sector unless there is a scale invariance symmetry within the fundamental quantum description of geometry.
I am not a strong expert in the field of Quantum Gravity, but I think that either the above gives a way out or the solution has to be looked for totally elsewhere (perhaps outside of Quantum Gravity). I will have to do some more studying (including Haelfix's other post) to securely offer an opinion, but anyone else can.