# I Question about cosmological constant

1. May 29, 2016

### stevendaryl

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?

2. May 29, 2016

### PAllen

That is my understanding.

3. May 29, 2016

### haushofer

Yes :)

4. May 29, 2016

### ShayanJ

But how do they calculate it?

5. May 29, 2016

### haushofer

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

6. May 29, 2016

### ShayanJ

So what about the cosmological constant problem?

7. May 29, 2016

### PAllen

8. May 29, 2016

### ShayanJ

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

9. May 29, 2016

### George Jones

Staff Emeritus
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

10. May 29, 2016

### PAllen

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??!!)

11. May 29, 2016

### ShayanJ

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!

12. May 30, 2016

### 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.

13. May 30, 2016

### Stavros Kiri

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

14. May 30, 2016

### 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.

15. May 30, 2016

### Stavros Kiri

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.

16. May 30, 2016

### ShayanJ

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!

17. May 30, 2016

### Stavros Kiri

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.]

18. May 30, 2016

### stevendaryl

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?

19. May 30, 2016

### Haelfix

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.

20. May 30, 2016

### Stavros Kiri

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.