1. Possible solution to the 7Li problem by the long lived stau
Toshifumi Jittoh (Saitama University)
Collaborators M. Koike, J. Sato, T. Shimomura, M. Yamanaka (Saitama University)
K. Kohri (Lancaster University)
arXiv: , To appear in PRD
I am Toshifumi Jittoh, belong to Saitama University.
The title of this talk is this.
In this talk, I solve a cosmological problem called Lithium 7 problem by taking into account the stau-nuclei interaction.
Collaborators of this work are Koike-san, Sato Joe-san, Yamanaka-kun, Shimomura-san, Kohri-san.
This work is appeared in this paper.
@KEKPH2007

2. Big Bang Nucleosynthesis (BBN） & ７Li problem
BBN　t = 1sec – O(100)sec
Prediction for light element abundances
parameter: baryon-to-photon ratio η
Predicted values and observational values are consistent
except for 7Li...
First, I explain what lithium 7 problem is.
It is a cosmological problem.
Light elements abundances synthesized at early universe are explained by Big Bang Nucleosynthesis called BBN.
Here, “early” means that the age of universe is between a few second and a few minutes.
This time region is sometimes called “The first three minutes”
BBN predicts light element abundances from a parameter, baryon to photon ratio.
This figure shows the predicted and observed abundances of the light element abundances.
In this figure the observed values are represented by these bands, and predicted values are represented by these lines.
At the observed value of baryon to photon ratio, observed values and predicted values are consistent, except for Lithium 7.
For lithium 7, predicted value is two or three times larger than observed value.
This inconsistency is called lithium 7 problem.
Since the predicted value exceed the observational value, we have to introduce a lithium 7 reducing process.
7Li problem
Bang Nucleosynthesis.ppt
Necessity to introduce 7Li reducing process

3. Solution to the 7Li problem
Reducing process naturally appears in the framework of the Minimal Supersymmetric Standard Model (MSSM)
Setting
Lightest Supersymmetric Particle = neutralino
Next Lightest Supersymmetric Particle = stau
Small mass difference between and
We found that such a reducing process naturally appears in the frame work of Minimal Supersymmetric Standard Model (called MSSM).
In our model the lightest supersymmetric particle (called LSP) is neutralino, next lightest supersymmetric particle (called NLSP) is stau.
As we will see, stau has a crucial role in this work.
Moreover, we suppose that the mass difference between LSP and NLSP is very small.
Such a small mass difference is cosmologically expected, to occur coannihilation process.
key point
･･･Cosmologically expected

4. Stau
TJ, J. Sato, T. Shimomura, M. Yamanaka PRD 73, (2006)
Diagram
Lifetime

5. Stau Diagram Lifetime Hadronic current → reducing 7Li
TJ, J. Sato, T. Shimomura, M. Yamanaka PRD 73, (2006)
Hadronic current → reducing 7Li
Diagram
Lifetime

6. Stau Diagram Lifetime Stau survive until the BBN era.
TJ, J. Sato, T. Shimomura, M. Yamanaka PRD 73, (2006)
Hadronic current → reducing 7Li
Diagram
Lifetime
First we see the nature of stau.
This is a diagram of stau decay process.
Staus decay into neutralino, nu tau, and hadronic current.
The hadronic current can interact with nuclei, and reduce the lithium 7.
And this figure shows the lifetime of stau.
From this figure, we see that stau survive until BBN era, if the mass difference is smaller than about one hundred MeV.
Because nucleosynthesis starts when the age of universe is 1 sec.
Stau survive until the BBN era.
Possibility to reduce the 7Li

7. Hadronic-current interaction with free stau
Hadronic current exchange
This diagram shows the interaction between free stau and nuclei.
Staus decay into these particles and hadronic currents, J plus minus.
The current interact with nuclei, and change it into other nuclei.
Here I note that, there is a cancellation processes;
J minus interact with lithium 7, and change it into beryllium 7.
J plus interact with beryllium 7, and change it into lithium 7.
Then these interactions can not reduce lithium 7 sufficiently.
Moreover the interaction is too weak to reduce lithium 7.
This figure shows the inverse interaction rate of such a process.
We see that the inverse interaction rate is between 10 to 30 sec and 10 to 24 sec.
This is much longer than BBN time scale.
So interaction between free stau and nuclei is ineffective.
1030 sec – 1024 sec
Γ-1 (sec)
ineffective

8. Formation of a bound state
If stau and nucleus form a bound state
･･stau
・・nucleus
Very close (∵ large )
However the situation is changed if stau and nucleus form a bound state.
Look at this picture.
Blue ball shows the stau, and red ball shows the nucleus.
The distance between stau and nucleus in the bound state is about nuclear radius, since stau is very heavy.
Then overlap of their wave functions becomes large, and the interaction rate also becomes large.
Moreover stau plus does no form a bound state, then cancellation process does not occur.
Then In the bound state, a exchange of hadronic current occurs effectively.
Overlap of wave functions: UP 　　 Interaction rate: UP
does not form a bound state no cancellation

9. Internal conversion 1 sec – 10-4 sec J 7Be → 7Li Γ-1 (sec)
C. Bird, K. Koopmans, M. Pospelov, hep-ph/
7Be → 7Li J
Γ-1 (sec)
This is the diagram of hadronic current exchange process in the bound state.
The shaded circle means bound state.
Here we consider not only lithium 7 reducing process, but also beryllium 7.
This is because, beryllium 7 naturally decays into lithium 7, therefore to reduce the lithium 7 we have to reduce also beryllium 7.
The inverse interaction rate is shown in this figure.
From this figure, the value of the inverse interaction rate is between 1 sec to 10 to minus 4 sec.
This is shorter than the time scale of BBN.
This means that once stau makes a bound state, the stau immediately decays and destruct the nucleus.
1 sec – 10-4 sec

10. New decay chain New decay chain reducing the 7Li Internal conversion
Scattering with background particles
This is a new decay chain which reduce the lithium 7.
By the internal conversion, beryllium 7 change into lithium 7.
The lithium 7 change into helium 7, by also internal conversion.
Helium 7 is very unstable.
In 10 to –20 sec, helium 7 emit a neutron, and change into helium 6.
Helium 6 is also unstable.
However the lifetime is about 1 sec.
Then it interact with background particles, like proton, and becomes helium3, helium4, deuteron, or some light element.
Someone may think that, these created elements make some now problems.
But such a new problem does not appear.
Because, the number of beryllium 7 and lithium 7 is too small to change the predicted values of these elements effectively.
n-emission

11. Number density of the bound state
Evaluation the number density Saha equation
: number density of the bound state
: number density of the nucleus
: binding energy
: temperature of the universe
We have used Saha equation to evaluate the number density of the bound state.
In the frame of Saha equation, the number density is determined by these values.

12. Numerical result 7Li problem is solved Forbidden areas
：number density of
：entropy density
Forbidden areas
　　　　 ： 7Li/H
　　　　 ： 6Li/7Li
　　　 　： 4He
　　　 　： D/H
We calculate nucleosynthesis process including our new decay chain.
This is the result.
The horizontal axis is the mass difference between stau and neutralino, and the longitudinal axis is yield value of stau, defined as this;
Number density of stau over entropy density of the universe.
Roughly, Y stau means the number of stau.
While these colored regions are forbidden by these observational values, in this white region lithium 7 problem is actually solved.
The thick-dotted line represents a yield value of stau whose daughter particle, neutralino, accounts for all the dark matter component.
7Li problem is solved

13. No Internal conversion
Physical reasons
7Li: down
6Li/7Li: up
Forbidden by constraint
Stau decays before the formation of the bound state
No Internal conversion
Why the allowed region has such a strange shape?
This is because of three reasons.
First, the right side of the vertical area.
In this parameter region, stau decays before the formation of the bound state.
Therefore, the internal conversion does not occur, and the lithium 7 is not reduced.
Second, the left side.
In this region, lithium 7 is reduced too much.
therefore the ratio of lithium 6 to lithium 7 becomes too large,
and exceed the observed value.
Third, the lower region.
In this region, the number of stau is smaller than the number of lithium 7.
Therefore stau can not reduce lithium 7 enough.
In this region, Not enough reducing

14. Summary What we done Future work (most important) upward
We investigated a possible solution for the 7Li problem by internal conversion of a stau-nucleus bound state
There is a parameter region in which 7Li problem is solved
Future work (most important)
To use the Boltzmann equation instead of Saha equation
Now I summarize this talk .
What we done.
We investigated a possible solution for the lithium 7 by internal conversion.
There exist a parameter region in which lithium 7 problem is solved.
There are some improvable point in this work.
I consider that the most important point is to use the Boltzmann equation instead of Saha equation.
Saha equation is valid only when effect of the expansion of the universe is negligible.
However, in this case, the effect of the expansion may be non-negligible.
So, using Saha equation is not good assumption, and we should use Boltzmann equation.
If we use the Boltzman equation, the allowed region will be shifted upward,
and close to the likely abundance of dark matter.
By a numerical calculation, now we consider that they will be very close.
But we have not get the precise result.
Then we will show the results another time.
Allowed region DM abundance
upward
close (consistent ?)

15. Boltzmann equation Very close Allowed region DM abundance
This figure represents the ratio between the number of bound lithium 7 and free lithium 7, as a function of the temperature of the universe.
These are calculated by using Boltzmann equation.
Number of the bound lithium 7 is almost equal to the number of reduced lithium7.
Remember that to solve the lithium 7 problem, we have to reduce the half or two third of lithium 7.
Then the ratio should be about a few ten percent.
And Y stau should be 10 to minus 12.
In other hand, the expected number density is also 10 to minus 12.
So, they are very close.
And this model is promising.