1. Chiral symmetry breaking and structure of quark droplets
S. Yasui1)2), A. Hosaka2) and H. Toki2)
1)Yukawa Institute for Theoretical Physics (YITP), Kyoto Univ.
2)Research Center for Nuclear Physics (RCNP), Osaka Univ.
The title in my talk is chiral symmetry breaking and stability of strangelets.
Collaborators are Professor Hosaka and Professor Toki in RCNP in Osaka University.

2. contents 1. Stability of strangelets by strangeness flavor (Nf=3)
2. Chiral symmetry breaking and strangelets
“NJL+MIT bag” model
The contents in my talk is the followings.
In section 1, we review the past discussions about the stability of the strange matter from a view of increase of the degrees of freedom by strangeness in the quark matter.
In this talk, we focus on the stability of strangelet of finite size strange matter, so that we compare our results and experimental data.
In section 2, we discuss the effect of chiral symmetry breaking in the stability of strangelets, in order to consider the role of the fundamental properties of QCD in strangelets.
In section 3, we discuss our numerical results.
Finally, we come to a conclusion.
3. Results
strangelets, ud droplets and baryons
Conclusion

3. 1. Stability of strangelets
1. What’s quark matter ?
baryon
Quarks and gluons
appear explicitly.
Condensed matter physics
of quark matter.
meson
2. Where is quark matter ?
It has been discussed for a long periods that the nuclear matter at high density and temperature becomes quark matter.
There, quarks are not confined in hadrons.
The fundamental degrees of freedom are quark and gluon.
Such a phases is called Quark-Gluon Plasma or quark matter.
It is one of the most interesting subject in the quark and hadron physics to research the physics of the properties of the quark matter.
Such a matter is considered to be produced in the accelerators by relativistic heavy ion collisions.
We may observe the quark matter in the neutron stars, which have very high density.
Recently, there are several candidates of the quark stars, listed here.
It may be also possible to observe the QGP in the early universe.
Therefore, the physics of the quark matter is concerned with the astrophysics, too.
Heavy ion collisions
Compact stars
Early universe
RHIC, CERN, GSI
RX J1856,5-3754 4U
SAXJ

4. 1. Stability of strangelets
Bodmer (1971)
Witten(1983)
Farhi and Jaffe(1983)
Terazawa(1989)
Strange matter = uds quark matter
・Free quark Fermi gas model (mu=md=ms=0)
energy density
e ~ n m4
m; chemical potential
n; number of degeneracy
number density
n ~ n m3
・ Fix number density,
m ~ n-1/3
Energy per baryon number e/n ~ n-1/3
In this talk, especially, we focus on the strange matte.
The strange matter is a quark matter including the strange quarks in addition to the ud quarks.
Such a strange matter has been considered to be the most stable state in the quark matter.
We can see that stability by a simple calculation.
Here, we assume that the quark matter is massless quark Fermi gas.
This is an appropriate assumption in the limit of high density.
In this model, the energy density epsilon and the number density n is given by these relations.
Here, mu is a chemical potential and nu is the number of degeneracy.
We consider a fixed number density.
Then, we obtain the relationship between the mu and the nu.
Form these relations, we find the energy per baryon number epsilon over n is proportional to nu to the minus one third.
Now, let us compare the energy ratios in the ud quark matter and the strange matter.
The ud quark matter has mu equal to two, while the strange matter has nu equal to three.
Then we have the ratio of the energy 0.88 in the strange matter as compared with the ud quark matter.
Thus, we see that the strange matter is more stable than the ud quark matter.
This result is understood by a schematic figure, shown here.
The left one is the ud quark matter, and the right is the strange matter.
We see that the quark matter energy dscreases by dividing the energy to the strange quarks.
ud quark matter
n=2
strange matter
n=3 u d s
>
1/2
1/3
e/n(uds)
= (2/3)1/3 ~ 0.88
e/n(ud)
strange matter is more stable !!!

5. 1. Stability of strangelets
Strangelets = finite volume strange matter
strangelets v.s. nuclei
MIT bag model
B1/4=150 MeV
mu0= md0= 0, ms0= 100 MeV
current mass
e=B e r
quark hadron
e=0
Strangelets are
stable for A>10.
56Fe
ud quark droplet
strangelet r R
quarks in a bag
quark
phase B
hadron
Baryon number A
(3A quarks)
Moreover, the strange matter is considered to be more stable than the nuclear matter.
Here, we consider an object of finite volume strange matter, which is called as strangelets.
In order to compare the stability of the strangelets and the nuclear matter, we consider the MIT bag model.
The MIT bag model is known as an effective theory of the baryon.
In this model, the energy difference between the inside and outside of the bag is taken into account by the bag constant.
Here, we consider that the bag constant is represents the energy difference between the quark matter and the hadronic vacuum for the quark matter.
First, we give a baryon number in the bag.
Second, The radius of the bag is determined by taking a variation of the energy.
In the right figure, we show the energy per baryon number E/A as a function of the baryon number A.
The black line is ud quark droplets, while the red line is the strangelets.
The energy per baryon number decreases as the baryon number increases, because the kinetic term decreases by large radius.
By taking parameter in the MIT bag model, we find the strangelets are more stable than the ud quark droplets.
The energy difference between the both of the strangelets and the ud quark droplets is almost 0.88 as we estimated in the previous page.
Moreover, the energy per baryon number of the strangelets is smller than that of iron nuclei, which are the most stable nuclei.
Thus, we can expect the true ground state of the nuclear matter may be the strangelets.
We mention here that only the current masses are considered in the MIT bag model.
Strangelets may be absolutely stable!!!

6. 1. Stability of strangelets
Strange matter hypothesis
strangelets ?
quark stars
(strange stars)
“strange nuclei”
“strange nuggets”
If the absolute stability of the strangelets is true, we can have a scenario of the strange matter hypothesis, which says there are many strange matter in the universe from the nuclei size a few fm to the strange stars about 10 km.
Such strangelets may be found in the universe as heavy cosmic rays.
Or the strange meteors may be an origin of dark matter.
〜fm
〜10km
scale

7. 1. Stability of strangelets
Chiral symmetry breaking
Tamagaki et al. (1992), Buballa(1996)
Dynamical quark mass (mu*= md*〜0.3 GeV, ms*〜0.5 GeV)
ud quark matter with nB0=0.17 fm-3
puF = 0.18 GeV
ud quark Fermi energy is
Can strangelets be stable state
for chiral symmetry breaking?
euF = puF 2 + mu* 2 = 0.35 GeV < ms*
euF < ms* u d
energy s
euF
ms*
・low density
・high density
In the vacuum and at the low density, chiral symmetry of quarks is dynamically broken, and the quarks acquire the dynamical masses.
There, it is not a trivial problem whether the strange matter is the ground state , or not.
Tamagaki discussed the strangelets by considering dynamical quark masses, and Buballa discussed that by the Nambu--Jona-Lasinio model.
For example, the ud quarks have about 0.3 GeV, and the s quarks have about 0.5 GeV.
If the quarks have dynamical masses in the quark matter, it is shown that the strange matter cannot be the ground state.
For example, at the density twice as the normal nuclear matter density, we have the ud quark Fermi momentum 0.23 GeV.
Then, we obtain the Fermi energy 0.37 GeV.
This is smaller than the mass of the strange quark.
Therefore, the strange quark cannot be generated in the ground state at the low density.
On the other hand, at high density, the ud quark Fermi energy increases and the strange quark mass may decrease by restoraction of chiral symmetry breaking.
There, we can expect that the strange matter becomes the ground state.
Thus, we discussed chiral symmetry breaking plays an important role in the stability of the strange matter.
Now, we have a question; the strangelets can be the stable state when we consider chiral symmetry breaking? u d s
euF ms
euF > ms
(restoration of
chiral symmetry)

8. 1. Stability of strangelets by strangeness flavor (Nf=3)
2. Chiral symmetry breaking and strangelets
“NJL+MIT bag” model
The contents in my talk is the followings.
In section 1, we review the past discussions about the stability of the strange matter from a view of increase of the degrees of freedom by strangeness in the quark matter.
In this talk, we focus on the stability of strangelet of finite size strange matter, so that we compare our results and experimental data.
In section 2, we discuss the effect of chiral symmetry breaking in the stability of strangelets, in order to consider the role of the fundamental properties of QCD in strangelets.
In section 3, we discuss our numerical results.
Finally, we come to a conclusion.
3. Results
strangelets, ud droplets and baryons
Conclusion

9. 2. Chiral symmetry breaking
“NJL+MIT bag”model
Strangelet with chiral symmetry breaking
Baryon number A
supplemented by quark confinement R r
quark
phase
hadron
U(3)LxU(3)R→SU(3)V
To incorporate this problem, we consider the NJL model, which is often used in the discussion of chiral symmety breaking.
In addition, we are now considering finite volume system of strangelets.
In order to confine the quarks in a strangelet, we impose a boundary condition of the MIT bag model.
Thus, we have a basis set of the quark wave functions in the strangelets by this boundary condition.
Our model may be called the “NJL + MIT bag” model.
Note the difference between our model and the MIT bag model.
In our model, we consider that quarks may acquire the dynamical masses in the strangelets , while, in the MIT bag model, only current masses are considered.
And, in our model, the energy difference between the inside and the outside of the bag is obtained automatically by the NJL interaction, while, in the MIT bag model, the energy difference was given by the bag constant as a parameter.
You may wonder in our model why chiral symmetry is explicitly broken at the bag surface.
In order to keep chiral symmetry, we need to introdues pions and kaons around the bag.
However, in this talk, we take the simplest equation by neglecting the meson clouds effects.
Let me summarize again, the feature of our model.
We assume that the quarks can acquire the dynamical mass by the NJL model.
And the finite size effect is taken into account by the boundary condition.
finite volume strange matter
O. Kiriyama and A. Hosaka PRD67, (2003)
S.Y., A. Hosaka and H. Toki, hep-ph/

10. 2. Chiral symmetry breaking
1. Chiral symmetry breaking in NJL int.
dynamical mass
constituent mass
equation of motion
2. Quark wave function in a bag
Spherical basis set
total angular momentum
Now, let me state how to discuss chiral symmetry breaking.
Concerning the NJL interaction, we consider only the scalar channel.
By a sum from a=0 to 8, we have the q¥bar{q} interaction.
In the flavor non- mixing term, we use a mean field approximation.
Here, we define the dynamcial mass m_{q}.
The quarks aqcuire the large mass due to the condensation of the q¥bar{q}.
We define the condensation part as contiuent mass ¥phi_{q}.
Then, we have a single particle equation for each flavor.
The dynamical mass is obtained by solving these equations self consistently.
parity
MIT bag boundary condition
for each j and P
Discrete energy levels

11. 2. Chiral symmetry breaking
Chiral symmetry breaking in bag
la are Gell-Mann matrices
Now, let me state how to discuss chiral symmetry breaking.
Concerning the NJL interaction, we consider only the scalar channel.
By a sum from a=0 to 8, we have the q¥bar{q} interaction.
In the flavor non- mixing term, we use a mean field approximation.
Here, we define the dynamcial mass m_{q}.
The quarks aqcuire the large mass due to the condensation of the q¥bar{q}.
We define the condensation part as contiuent mass ¥phi_{q}.
Then, we have a single particle equation for each flavor.
The dynamical mass is obtained by solving these equations self consistently.
constituent mass
dynamical mass
equation of motion

12. 2. Chiral symmetry breaking
Quark wave functions in a bag
Spherical basis set
total angular momentum
parity
MIT bag boundary condition
for each j and P
pR (M=0)
1/2-
3.81/R
One of the most interesting feature in our model is that the quark energy levels become discrete due to the boundary condition.
As a strangelet, we assume the spherical form.
There, good quantum number is the total angular momentum an parity.
Then, we prepare the spherical basis set for the quark wave functions.
Here, N is a normalization constant, j_{l} is a psherical Bessel function, p, E and momentum, energy and mass of a quark respectively.
Y is the spinor harmonics of eigen state of j.
By the boundary condition, we have discrete momentum for each j and parity, which is give by this equation.
We show a table of the eigen value obtained by this condition for M=0.
The lowest level is 1/2+.
The second is 3/2-, and the third is 1/2-.
Like in this figure, quarks occupies the energy levels by the Pauli principle.
3/2-
3.20/R
1/2+
2.04/R
vac

13. 2. Chiral symmetry breaking
energy levels
Energy = valence + vacuum JP
Fermi surface
1/2-
valence
3/2-
Energy of each state jP
1/2+
with constituent mass
effective bag “constant”
In our model, we consider the vacuum structure in the strangelets explicitly.
Like in the valence quarks, the vacuum quarks have discrete energy levels.
Then, the total energy is given by a sum of the valence part and the vacuum part.
Here we mention that the quark mass is determined self-consistently.
We use constituent quark mass as an order parameter of chiral symmetry breaking.
Concerning the vacuum part, we take the zero point of the energy in the chirally broken vacuum outside of the bag, which is written by ¥omega_{bulk}[¥phi_{0}].
The energy difference between the vacuum quarks in the inside and the outside of the bag is defined as effective bag constant.
This bag constant represents the energy difference in the quark phase and the hadronic phase.
Truth to say, this is not a constant, but function depending on the radius of the bag.
However, let me use this word in analogy with that in the MIT bag model.
In the vacuum part, we introduce the Lorentz type regularization for momentum-cut off.
The parameters of the three dimensional momentum cut-off ¥Lambda and the diffuseness a are determined in the next page.
1/2-
3/2+
vacuum
Zero point: chirally broken vacuum
outside bag by the NJL model
1/2+
momentum cut-off (Lorentzian)
5/2-

14. 2. Chiral symmetry breaking
Parameter set
fp=93 MeV, mp=139 MeV, mu=378 MeV
GL2=4.6, L=600 MeV, a=20
m0u=m0d=5.9 MeV, m0s=119.5 MeV
We have three parameters of the coupling constant G in the NJL model, the momentum cut-off ¥Lambda, and the diffuseness a.
These values were determined by the pion decay constant, the pion mass, and the dynamical mass of the ud quark.
Here, we fixed the current mass of the ud and s quarks as appropriate values.
Please pay an attention the value of the momentum cut-off.
This gives us an interesting structure of the baryon, which is discussed soon.

15. 1. Stability of strangelets by strangeness flavor (Nf=3)
2. Chiral symmetry breaking and strangelets
“NJL+MIT bag” model
The contents in my talk is the followings.
In section 1, we review the past discussions about the stability of the strange matter from a view of increase of the degrees of freedom by strangeness in the quark matter.
In this talk, we focus on the stability of strangelet of finite size strange matter, so that we compare our results and experimental data.
In section 2, we discuss the effect of chiral symmetry breaking in the stability of strangelets, in order to consider the role of the fundamental properties of QCD in strangelets.
In section 3, we discuss our numerical results.
Finally, we come to a conclusion.
3. Results
strangelets, ud droplets and baryons
Conclusion

16. 3. Results A) Baryon number A=6 (Nq=18) for S=-6 ud s E/A=1.59 GeV
3/2-
2x3=6
2x2x3=12
1/2+
vac
2x2x3=12
2x3=6
1/2-
L=0.6GeV
3/2+
ud s
E/A=1.59 GeV
Constituent mass fu=0, fs=0
Bag constant B1/4=0.16 GeV
Number density nB=19nB0
S=-6 is the most stable state.

17. 3. Results A) Baryon number A=10 (Nq=30) for S=0 ud s E/A=1.79 GeV18
3/2-
2x2x3=12
1/2+
vac
2x2x3=12
1/2-
2x3=6
4x2x3=24
4x3=12
3/2+
L=0.6GeV
1/2+
Let us discuss the strangelets of the baryon number A=10, which include the 30 quarks.
Here we show the energy levels of the quarks in the bag.
Let me skip to state the values of the energy levels.
But, this is a sufficient to understand the stability of the strangelets.
When we put only the ud quarks in the state 1/2+ and 3/2-, such ud quark droplet is not the ground state.
In fact, we find an empty state of strange quark of 1/2+ state.
Therefore, it is better energetically to put strange quarks in this state instead of the ud quarks in the 3/2- state.
ud s
E/A=1.79 GeV
Constituent mass fu=0, fs=0
Bag constant B1/4=0.28 GeV
Number density nB=8.7nB0
S=-6 is the most stable state.
Cf. nB0=0.17fm-3

18. 3. Results A) Baryon number A=10 (Nq=30) for S=-6 ud s E/A=1.79 GeV12
3/2-
2x3=6
2x2x3=12
1/2+
vac
2x2x3=12
1/2-
2x3=6
4x2x3=24
4x3=12
3/2+
L=0.6GeV
1/2+
When we put six strange quarks in the 1/2+ state, the level of 1/2+ is closed.
If we want to add more strange quarks, the left seat is only in the 3/2-.
ud s
E/A=1.79 GeV
Constituent mass fu=0, fs=0
Bag constant B1/4=0.28 GeV
Number density nB=8.7nB0
S=-6 is the most stable state.
Cf. nB0=0.17fm-3

19. 3. Results A) Baryon number A=10 (Nq=30) for S=-10 ud s E/A=1.79 GeV4 8
3/2-
2x3=6
2x2x3=12
1/2+
vac
2x2x3=12
1/2-
2x3=6
4x2x3=24
4x3=12
3/2+
L=0.6GeV
1/2+
However, the energy level of 3/2- of strange quark is higher than that of the 3/2- of ud quark.
Therefore, S=-10 does not become the ground state.
In summary, we obtain the most stable strangelets with S=-6.
This result is also shown in the right figure, which is the energy per baryon number E/A as a function of the radius of the bag.
Each color line represents the strangeness in the strangelets.
The most stable radius is given by the local minimum by pressure condition.
As we have discussed, we obtain the ground state for S=-6.
In this state, we have restoration of chiral symmetry breaking, since the constituent mass is zero.
We have a effective bag constant B^{1/4}=0.28 GeV.
This is larger than that in the MIT bag model.
Therefore, we have a large mass as compared with the baryons.
Here, we mention the baryon number density is 8.7 times as the normal nuclear matter density nB0.
ud s
E/A=1.79 GeV
Constituent mass fu=0, fs=0
Bag constant B1/4=0.28 GeV
Number density nB=8.7nB0
S=-6 is the most stable state.
Cf. nB0=0.17fm-3

20. 3. Results B) Baryon number A=100 (Nq=300) eF,u ud(S=0) ms=m0s
=0.12GeV
uds(S=-100)
ud s
Next, we see the baryon number A=100.
For this baryon number, we do not show all the energy levels, since the number of state is too large to draw a picture.
Chiral symmetry is restored.
S=-100
Constituent mass fu=0, fs=0
Bag constant B1/4=0.26 GeV
Number density nB=7.7nB0
Cf.

21. 3. Results C) Baryon number A=5000 (Nq=15000) ms=0.50GeV eF,u
ud s
uds(S=-5000)
ud(S=0)
Chiral symmetry is broken.
S=0
Constituent mass fu=0.27, fs=0.38 GeV
Bag constant B1/4=0.21 GeV
Number density nB=3.9nB0
Cf.

22. 3. Results Bulk quark matter (A→∞) ground state: ud quark matter uds
Buballa(1996)
ground state: ud quark matter
uds
Energy e/nB=1.12GeV
Number density nB=2.59nB0
Constituent mass fu=0.047 GeV
(12.6 %)
fs=0.451 GeV
Bag constant B1/4=0.174GeV ud

23. 3. Results Comparison with liquid drop model
MRE (Multiple Reflection Expansion)
Balian and Bloch (1971)
Hansson and Jaffe (1983)
State density
R=1 fm
R=2 fm

24. 3. Results liquid drop model (MRE) and discrete level
MRE discrete level
A E/A [GeV] R [fm] E/A [GeV] R [fm] ‘ ‘
‘ ‘ ‘ ‘
‘ ‘ ‘
Liquid drop model is a good approximation for A>10.

25. 3. Results Energy E/A of strangelets with baryon number A
Strangelets for A < 1000
would be stable for weak decay
into ordinary (non-strange) nuclei. A=
In summary, we show the energy per baryon number from A = 1 to 5000 with several strangeness.
At A=1, we obtained the baryon mass.
Up to the 10, the energy per baryon number E/A increase, because the effective bag constant increases, too.
We obtain the maximum energy per baryon number at A=10.
There, the strangelets with strangeness S=-6 is the ground state.
From the A=10 to the infinity, the E/A decreases, because the effective bag constant decreases.
Up to the A=1000, chiral symmetry is restored in the strangelets.
Therefore, the strangelets become the ground state.
At A=1000, chiral symmetry is broken.
Then, the ground state is not strangelet, but ud quark droplet.
Therefore, the strangelets will be restricted for the baryon number from 10 to 1000.
It is interesting that we obtained the continuous discussion from baryon to the quark matter.
We can have a following scenario.
Assume that the initial state is the hot quark gluon plasma, including the ud and s quarks.
If the transition from the QGP to the normal nuclear matter is first order, we may have some quark droplets in this process.
If the baryon number of the quark droplet is smaller than 10, such droplets would decay into smaller droplets.
And finally, we have only baryons of A=1.
On the other hand, if the quark droplets have baryon number larger than 10, the fusion doe not occur spontaneouly.
Then, strangelets with some baryon number will remain.
Such heavy particles can make a fusion by collision of strangelets of ud quark droplets.
And finally, they may form quark stars.
How about decay into L(uds)? R= fm

26. 3. Results D) Baryon number A=1 (Nq=3) Structure of baryon for S=03
1/2+
vac
First, we discuss the case of the baryon number 1 or quark number 3, that is baryon.
In this figure, we see the structure of the baryon.
The upper part is the valence quarks, and the lower part is the vacuum quarks.
The left one is the ud quarks and the right one is the s quarks.
The energy levels in the ud and s quarks are not the same by difference of the current masses.
Therefore, the the ground state of the baryon is S=0, in which the three ud quarks occupy the lowest level 1/2+ and the strange quarks are no included.
Concerning the vacuum structure, we have several states of ud and s quaks.
For example, when the baryon has a radius fm, which will give by the energy minimization, the energy of the 1/2- in the vacuum is 0.54 GeV.
This is smaller than the momentum cut-off 0.6 GeV.
On the other hand, the energy level of 3/2- is GeV for the same radius.
Therefore, only the state of 1/2- is counted in the vacuum part in the bag.
The right figure is the energy of the baryon as function of the radius.
The black line is S=0, the blue one S=-1, the green one S=-2 and the red one S=-3.
The local minimum is give by pressure balance in the bag.
As we have seen in the left figure, the ground state is S=0.
Here we have restoration of chiral symmetry.
Therefore, the picture of the bag model is appropriate in our case.
Indeed, we have the effective bag constant B^{1/4}=0.11 GeV, which is consistent with the value in the MIT bag model.
Let us compare our results with the experimental data of the baryon mass listed here.
In the left colum, from S=0 to -3, we obtain these values.
For S=0, our result gives the mean value of the Nucleon and Delta.
The other values are also close to the experimental data.
ud s
S E [GeV]
Hadron S Mass [GeV] N D L S X
Constituent mass fu=0, fs=0
Bag constant B1/4=0.11 GeV

27. 3. Results D) Baryon number A=1 (Nq=3) Structure of baryon for S=03
1/2+
vac
2x2x3=12
2x3=6
1/2-
L=0.6GeV
First, we discuss the case of the baryon number 1 or quark number 3, that is baryon.
In this figure, we see the structure of the baryon.
The upper part is the valence quarks, and the lower part is the vacuum quarks.
The left one is the ud quarks and the right one is the s quarks.
The energy levels in the ud and s quarks are not the same by difference of the current masses.
Therefore, the the ground state of the baryon is S=0, in which the three ud quarks occupy the lowest level 1/2+ and the strange quarks are no included.
Concerning the vacuum structure, we have several states of ud and s quaks.
For example, when the baryon has a radius fm, which will give by the energy minimization, the energy of the 1/2- in the vacuum is 0.54 GeV.
This is smaller than the momentum cut-off 0.6 GeV.
On the other hand, the energy level of 3/2- is GeV for the same radius.
Therefore, only the state of 1/2- is counted in the vacuum part in the bag.
The right figure is the energy of the baryon as function of the radius.
The black line is S=0, the blue one S=-1, the green one S=-2 and the red one S=-3.
The local minimum is give by pressure balance in the bag.
As we have seen in the left figure, the ground state is S=0.
Here we have restoration of chiral symmetry.
Therefore, the picture of the bag model is appropriate in our case.
Indeed, we have the effective bag constant B^{1/4}=0.11 GeV, which is consistent with the value in the MIT bag model.
Let us compare our results with the experimental data of the baryon mass listed here.
In the left colum, from S=0 to -3, we obtain these values.
For S=0, our result gives the mean value of the Nucleon and Delta.
The other values are also close to the experimental data.
3/2+
ud s
S E [GeV]
Hadron S Mass [GeV] N D L S X
Constituent mass fu=0, fs=0
Bag constant B1/4=0.11 GeV

28. 3. Results Mechanism for baryon mass effective bag “constant”
Cf.
effective bag “constant”
1) R < 2.04/L=0.67fm
vac L
-2.04/R
1/2-
In the baryon structure by our model, the vacuum contribution in the bag plays an interesting role.
Let us see the details of the calculation of the effective bag constant.
As we have shown, the effective bag constant is defined by the difference of the vacuum energy inside and outside of the bag.
Let see the case 1.
When the radius of the bag is too small, there is no contribution from the vacuum quark in the bag.
Indeed, as we have seen in the previous page, for radius smaller than 0.67 fm, the highest energy level 1/2- is not counted.
Thus, the effective bag constant has only the term of bulk vacuum energy outside of the bag.
Note that ¥omega_{bulk}[¥phi_{0}] is a negative value.
Therefore, the effective bag energy has a positive value.
This is shown in the right figure, which we show the effective bag constant as function of the radius.
For the radius smaller than 0.67 fm, we have a constant value given by -¥omega_{bilk}[¥phi_{0}].
Next, we see the case 2.
When the radius is between the 0.67 fm and 1.05 fm, the latter is given by the energy level of 3/2+,only the 1/2- state is counted in the vacuum energy.
Then, the contribution from 1/2- is added to the effective bag constant.
Here, we have written 18 by considering the ud and s quarks are almost degenerate for convenience.
Of course, in the final result, we consider the difference of the ud and s quarks exactly.
Anyway, in this region of the radius, the effective bag constant is an increasing function with respect to the radius R.
Therefore, we find a local minimum around R=0.67 fm as shown in the right figure.
This radius is very close to that obtained in the previous page, R=0.745 fm.
The value fm is obtained by considering kinetic term, since the kinetic term pushes the radius outside.
In the right figure, we also show the effective bag constant of the A=6, 10, 100.
In the next some pages, we come to the discussion of the strangelets with these baryon numbers.
2) 0.67fm < R < 3.20/L=1.05fm
vac
-2.04/R
1/2- L
-3.20/R
3/2+
Cf. L=0.6GeV

29. 3. Results Energy E/A of strangelets with baryon number A
In summary, we show the energy per baryon number from A = 1 to 5000 with several strangeness.
At A=1, we obtained the baryon mass.
Up to the 10, the energy per baryon number E/A increase, because the effective bag constant increases, too.
We obtain the maximum energy per baryon number at A=10.
There, the strangelets with strangeness S=-6 is the ground state.
From the A=10 to the infinity, the E/A decreases, because the effective bag constant decreases.
Up to the A=1000, chiral symmetry is restored in the strangelets.
Therefore, the strangelets become the ground state.
At A=1000, chiral symmetry is broken.
Then, the ground state is not strangelet, but ud quark droplet.
Therefore, the strangelets will be restricted for the baryon number from 10 to 1000.
It is interesting that we obtained the continuous discussion from baryon to the quark matter.
We can have a following scenario.
Assume that the initial state is the hot quark gluon plasma, including the ud and s quarks.
If the transition from the QGP to the normal nuclear matter is first order, we may have some quark droplets in this process.
If the baryon number of the quark droplet is smaller than 10, such droplets would decay into smaller droplets.
And finally, we have only baryons of A=1.
On the other hand, if the quark droplets have baryon number larger than 10, the fusion doe not occur spontaneouly.
Then, strangelets with some baryon number will remain.
Such heavy particles can make a fusion by collision of strangelets of ud quark droplets.
And finally, they may form quark stars.

30. 3. Results Energy E/A of strangelets with baryon number A
In summary, we show the energy per baryon number from A = 1 to 5000 with several strangeness.
At A=1, we obtained the baryon mass.
Up to the 10, the energy per baryon number E/A increase, because the effective bag constant increases, too.
We obtain the maximum energy per baryon number at A=10.
There, the strangelets with strangeness S=-6 is the ground state.
From the A=10 to the infinity, the E/A decreases, because the effective bag constant decreases.
Up to the A=1000, chiral symmetry is restored in the strangelets.
Therefore, the strangelets become the ground state.
At A=1000, chiral symmetry is broken.
Then, the ground state is not strangelet, but ud quark droplet.
Therefore, the strangelets will be restricted for the baryon number from 10 to 1000.
It is interesting that we obtained the continuous discussion from baryon to the quark matter.
We can have a following scenario.
Assume that the initial state is the hot quark gluon plasma, including the ud and s quarks.
If the transition from the QGP to the normal nuclear matter is first order, we may have some quark droplets in this process.
If the baryon number of the quark droplet is smaller than 10, such droplets would decay into smaller droplets.
And finally, we have only baryons of A=1.
On the other hand, if the quark droplets have baryon number larger than 10, the fusion doe not occur spontaneouly.
Then, strangelets with some baryon number will remain.
Such heavy particles can make a fusion by collision of strangelets of ud quark droplets.
And finally, they may form quark stars.

31. 3. Results Energy E/A of strangelets with baryon number A
In summary, we show the energy per baryon number from A = 1 to 5000 with several strangeness.
At A=1, we obtained the baryon mass.
Up to the 10, the energy per baryon number E/A increase, because the effective bag constant increases, too.
We obtain the maximum energy per baryon number at A=10.
There, the strangelets with strangeness S=-6 is the ground state.
From the A=10 to the infinity, the E/A decreases, because the effective bag constant decreases.
Up to the A=1000, chiral symmetry is restored in the strangelets.
Therefore, the strangelets become the ground state.
At A=1000, chiral symmetry is broken.
Then, the ground state is not strangelet, but ud quark droplet.
Therefore, the strangelets will be restricted for the baryon number from 10 to 1000.
It is interesting that we obtained the continuous discussion from baryon to the quark matter.
We can have a following scenario.
Assume that the initial state is the hot quark gluon plasma, including the ud and s quarks.
If the transition from the QGP to the normal nuclear matter is first order, we may have some quark droplets in this process.
If the baryon number of the quark droplet is smaller than 10, such droplets would decay into smaller droplets.
And finally, we have only baryons of A=1.
On the other hand, if the quark droplets have baryon number larger than 10, the fusion doe not occur spontaneouly.
Then, strangelets with some baryon number will remain.
Such heavy particles can make a fusion by collision of strangelets of ud quark droplets.
And finally, they may form quark stars.

32. 3. Results Decay mode of strangelets 2 x A=500 1. Fission 22 GeV A=500
forbidden
A=500
2. Hyperon emission
We discuss the decay mode of strangelets.
As we have mentioned, the fission is not possible in the strangelets.
For example, When we consider that the strangelets with baryon number 1000 decay into two strangelets with baryon number 500, the energy becomes very high.
Therefore, this fission is forbidden energetically.
However, hyperon emission is possible.
Indeed, when the strangelet with baryon number 1000 decay to the lambda and strangelets of A=999, the energy 0.27 GeV is released.
To discuss this process in detail, we have to consider hadronization process, but this would be a future works.
A=1000
L(uds)
L + A=999
0.27 GeV
A=1000
allowed
A=999

33. 3. Results
Electric charge Q = (2/3)Nu - (1/3)Nd - (1/3)Ns
1) strange matter; Q= for Nu= Nd= Ns
2) ud quark matter; Q/A=0.5 for Nu= Nd and Ns=0
Our results
heavy particles in cosmic rays
A S Q/A
1000 ‘
A Q/A
[Kasuya,1993]
[Capdevielle,1996]
[Ichimura,1993]
[Price,1978]
In this last page, we discuss observables of strangelets.
As such a quantity, we consider electric charge.
Because, the electric charge of ud and s quarks is 2/3, -1/3 and -1/3 respectively.
There are several data of heavy particles in the cosmic rays.
Our results are consistent
with experimental data.
→ AMS-02, ECCO on space ship (2005)

34. Conclusion We discussed stability of strangelets by considering
chiral symmetry breaking.
2. We formulated “NJL+MIT bag” model for strangelets.
3. Our model derives baryon mass and radius for A=1.
4. Strangelets and ud quark droplets are not absolutely stable.

35. 3. Results Vacuum structure for one flavor R=2.04/L (0.67fm) R=3.20/L
1/2-
2x3=6
1/2-
4x3=12
1/2-
3/2+
1/2-
3/2+
L=0.6 GeV
3/2+
(R=0.75 fm)
Cf.

36. 3. Results
Effective bag constant
R=2.04/L
(0.67fm)
R=3.20/L
(1.05fm)

37. 2. Chiral symmetry breaking
Thermodynamical potential for baryon
kinetic energy + effective bag “constant”
w[f] is obtained as a sum by flavors