Shlomo (pdf)

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Shlomo (pdf)

freeze out surface. The validity of this model hinges on the assumption that the time scale involved in the expansion is

larger compared to the equilibration times in the expanding complex, which works when the flow velocity is smaller

compared to the average nucleonic velocity.

It was shown in Ref. [24] that the flow pressure P s

can be related to the kinetic energy of flow E

s

for the

fragments,

P = D(v , T) ⋅n ⋅ E .

(24)

s fs s s

In Eq. (24), v f

is the magnitude of the radial flow velocity, and D(v fs, T) ≈ 4.5. Experimental measurements [25]

indicate that the heavier fragments carry less flow energy per nucleon compared to the lighter ones. Thus E s

, in

Eq. (24), is assumed to take a simple parametric form

E = ε A α ,

(25)

s

where ε is the average flow energy carried by a single nucleon and the value of α is ≈ 0.95 . It can be easily seen from

Ref. [24] that if the flow effects are taken into account, the chemical potential is modified as

s

μ = μ N + μ Z −B − E () s + P / n ,

(26)

s n s p s s C s s

where Ps / ns = D⋅ ε A α

s

. For α = 1 , the fragment multiplicities remain unaltered with or without flow; only the chemical

potentials get renormalized. Using Eq. (17) the expression for the free neutron density is modified as

3/2

⎛ A ⎞ g

A

2

nn

= ⎜ ⎟ ⋅ ⋅ ⋅ exp ( ( B ( A , Z ) − B ( A+ 1, Z )) / T)

×

⎝ A+

1⎠

g λ

A+

1

3

T

α α

( EC

A Z EC

A Z T) ( Dε

A A T) R1

× exp −( ( , ) − ( + 1, )) / ⋅exp − (( + 1) − −1) / ⋅ , (27)

where R1 = Y( A+ 1, Z)/ Y( A, Z)

is the (single) ratio between the yields of two fragments differing by one neutron.

Also, Eq. (22) for the double ratio is modified as

R

Y( A′ , Z′

)/ Y( A, Z )

1 1 1 1

2

= =

Y( A′ 2, Z′

2)/ Y( A2, Z2)

3/2

⎛ A′ ⋅ A ⎞ ω( A′ , Z′

, T) ω( A , Z , T) exp / exp / exp / ,

( B T) ( E T) ( F T)

1 2 1 1 2 2

= ⎜ ⎟

⋅ Δ ⋅ −Δ

C

⋅ Δ

A1⋅

A′ 2

ω( A1, Z1, T) ω( A′ 2, Z′

2, T)



(28)

where

Δ F is given as

Δ F = Dε ( A′ − A + A − A′

) ,

(29)

α α α α

1 1 2 2

with A′

1

, A′

2

, etc. defined through Eq. (21).

Since (1 − α)

is very small, the last exponential in Eq. (27) can be very well approximated by

( D A α α

ε A T) ( Dε α A T)

exp − (( + 1) − −1) / ≈exp (1 − )(1 + ln ) / .

(30)

We thus have that the neutron density is increased by this factor, which is larger for smaller temperature, heavier

isotopes and larger flow energy. For example, from the yield ratio of, 4 He/ 3 He at T ≈ 4.0 MeV, α ≈ 0.95 ,

ε ≈ 10.0 MeV, one finds that the neutron density is increased by a factor of almost 4, and thus the freeze-out volume is

decreased by the same factor. With the inclusion of this correction (Eq. (30)), one finds that the freeze-out volume is

≈ 8V 0

which is closer to the values usually taken in macrocanonical calculations [14, 20] of nuclear multifragmentation.

We note that the extracted values of temperature can be modified, depending on the value of Δ F . Considering, for

example, the He-Li thermometer, one has from Eq. (29) that ΔF

≈ 0.11ε

, compared to the value of ΔB

≈ 13.32 MeV.

With ε ≈ 10 MeV, using Eq. (28), an increase in T of ≈ 10 % is obtained. Similar results are found for other

thermometers.

96

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