Vacuum energy, cosmological constant problem and the worst theoretical prediction ever

Transcription

1 Vacuum energy, cosmological constant problem and the worst theoretical prediction ever Juraj Tekel Student Perspectives in Physics, Graduate Center CUNY 3/25/2011

2 Ground state of quantum harmonic oscillator has energy E = 1 2 ω Quantum field is equivalent to collection of harmonic oscillators with ω k = k 2 + m 2 So the ground state has energy density ρ = d 3 k (2π) ω k

3 After regularization, this is proportional to with ε the cutoff distance. ρ ε 4 And with some nice story this is usually discarded... however...

4 The stories do not work if we work in curved space-time. Here the action is S = d 4 x gl(φ, φ) and constant shift in energy is no longer a constant term in the action. It effects equations of motion for the metric itself.

5 For the Einstein-Hilbert action S = d 4 x [ ] 1 g R + L(φ, φ) 16πG we get equations of motion for the space-time metric R µν 1 2 g µνr + 8πG ρ g µν = 8πGT µν So the vacuum energy is equivalent to a cosmological constant.

6 Indeed the most general action S = d 4 x [ ] 1 g 16πG (R 2Λ 0) + L(φ, φ) Gives equations R µν 1 2 g µνr + Λg µν = 8πG With an effective cosmological constant Λ = Λ 0 + 8πGρ

7 So where is the problem?

8 Cosmological constant matter has some very peculiar properties energy-momentum tensor and thus negative pressure T µν = Λ 8πG g µν the energy density is constant repulsive gravity Plus all observations were compatible with the assumption Λ = 0, so nobody really cared... however...

9 We now believe that the universe has positive cosmological constant, because the expansion of universe is accelerating (distant supernovae) repulsive gravity of Λ the universe is flat (CMB anisotropies) but baryons and dark matter make only 30% of critical density

10 So where is the problem? Vacuum energy TRULY IS the cosmological constant. Problem solved... however...

11 To predict the value, we take the cutoff to be Plank length and get the corresponding energy density ρ GeV 4 Value of the cosmological constant energy density consistent with the measurements is ρ GeV 4

12 The difference is 118 ORDERS OF MAGNITUDE!!!

13 And nobody really knows what to do now. Supersymmetry? Supergravity and Superstrings? Antropic principle? Different source of dark energy? etc. All the solutions to the problem require enormous fine tuning. As of now, no physical principle is know that requires Λ to be so small.

14 What do we have to say about this?

15 We do not offer a solution. We rather claim that the problem is ill defined from the very beginning.

16 Vacuum fluctuations in curved space can produce configurations, that create black holes. Such fluctuations would make the vacuum unstable, showing that we can not combine quantum fields and general relativity just like that. We try to compute rate, at which black holes occur in vacuum.

17 In vacuum 0 the probability distribution for any observable A is P(α)dα = 0 δ (α A) 0 dα This yields αp(α)dα = ᾱ = 0 A 0 α 2 P(α)dα = ᾱ2 = 0 A 2 0

18 Probability, that we will find energy in interval E, E + de in volume V as a result of fluctuation in vacuum is therefor ( ) P(E, V )de = 0 δ E T 00 (x)d 3 x 0 de If the energy inside volume V is greater than 3 V, this region will collapse into a black hole. We convert the probability into rate by assuming, that the sample time is proportional to the light travel time across the volume V. V

19 Seems rather random, but this choice is well motivated the only meaningful time in the problem information about rearrangement cross-section formula for black hole formation and particles colliding in V

20 For the probability, we get Gaussian distribution P(E, V )de = 1 2πσ 2 (E µ)2 e 2σ 2 de with µ = σ 2 = V V d 3 x 0 T 00 (x) 0 d 3 xd 3 y 0 T 00 (x)t 00 (y) 0 Note that the first integral reproduces the formula for the vacuum energy from the first slide. Indeed, up to the first order P(E, V )de = δ(e µ)de

21 Some technicalities We work with a simple theory of a massless scalar field. We regulate the momentum integrals by introducing d 3 k d 3 ke ε k and later setting ε to be the Plank length. We also take the volume V to be a sphere.

22 The final formula for the total rate is R = 6 1/6 5 1/2 π 5/6 ε 2 ε de 1 5π 4/3 (3E 2 π 2 ε 4 ) 2 e 6 1/3 ε 4 E 7/3 The problem is that this gives rates much higher than expected. So far seems to give results around black holes every second in very cm 3.

23 Well, we still work on that. Thanks for your attention!