## Understanding the entropy generation rate

In my previous post, I computed the entropy generation rate during my constant-entropy r-process calculation. The calculation showed substantial entropy generation at certain points during the calculation. In this post, I will seek to understand that entropy generation.

I will begin by considering the entropy generation near T9 = 5, which is a local peak in the entropy generation. I note that I can infer that this entropy generation is not due to weak decays from the figure in the previous post. From the file props.txt I created, I see that the step number corresponding to T9 = 5 is around step 28. From the file sdot_all.txt, I see that the peak in this temperature range is at step 30. I thus compute the entropy generation rate by reactions by first compiling the relevant codes

cd nucnet-tools-code/examples/thermo

./examples_make

Again, if you have trouble with clang, you can simply type

make compute_sdot_by_reactions

Once the code is compiled, I can compute the reactions that contribute to the entropy generation rate at step 30 by typing

./compute_sdot_by_reaction ../../my_examples/network/my_output1.xml --zone_xpath "[position() = 30]"

The zone XPath selects step number 30. The output from my calculation gives a long list headed by

``` time(s) = 0.0844315  t9 = 4.77324 rho(g/cc) = 135551 sdot = 8.10825

he4 + ar46 -> n + ca49  7.993122e-01
n + mg25 -> mg26 + gamma  5.583167e-01
he4 + fe62 -> n + ni65  3.390110e-01
he4 + ar44 -> n + ca47  3.368543e-01
```

From this, I see that the reaction 4He + 46Ar → n + 49Ca is the leading contributor to the entropy generation rate. I will seek to understand this.

First, I note that the change in the entropy in a system is given by

Here T is the temperature, ds is the change in the entropy per nucleon, dq is the heat gained or lost by the system per nucleon, μi is the chemical potential of species i, and dYi is the change in the abundance of species i per nucleon. From this, I can compute the time rate of change of the entropy per nucleon of the system in terms of Boltzmann's constant k as

It is convenient to consider the chemical potential of the nuclear species in terms of an offset from the value in nuclear statistical equilibrium. I thus can write

The other species in the system are the electrons and positrons and neutrinos. The net electron-to-nucleon ratio is Ye and is given by

where Ye- is the abundance per nucleon of electrons and Ye+ is the abundance per nucleon of positrons. The second equality results from charge neutrality and the sum is over nuclear species only. The net electron-type neutrino abundance per nucleon is given by

where νe- refers to neutrinos associated with electrons (referred to as "electron-type neutrinos") and νe+ refers to neutrinos associated with positrons ("electron-type anti-neutrinos"). With these equations, the entropy generation rate is

I now consider the reaction 4He + 46Ar → n + 49Ca. This reaction is not a weak reaction, so the net electron-to-nucleon ratio and the net electron-type neutrino abundance do not change. The rate of change of the abundance of 46Ar due to this reaction is

NA is Avogadro's number, ⟨ σ v ⟩ is the Maxwellian averaged cross section for the reaction, ρ is the mass density, and the subscripts 1 and 2 refer to the forward and reverse reactions, respectively. If we assume no heat is entering or leaving the system (dq = 0), and we note that dY(46Ar) = dY(4He) = -dY(49Ca) = -dY(n), the entropy generation rate from this reaction is then

I now compute the relevant terms. I change into the analysis directory and type

./examples_make

I then compute the chemical potential offsets from the NSE values for step 30 by typing

cd ../analysis

./compute_zone_mu ../../my_examples/network/my_output1.xml "[position() = 30]" > mu_30.txt

The output file mu_30.txt contains a header and then columns giving the species, Z, A, the abundance, and the chemical potential offset divided by kT. From the data, I find the Δ values for 4He, 46Ar, and 49Ca to be 2.31590e-03, -4.84017e+01, and -6.18981e+01, respectively. The total chemical potential offset is then 2.31590e-03 - 4.84017e+01 - (-6.18981e+01) = 13.498718. I next compute the flow, which is the part in between the square brackets in the above equation for the entropy generation for the reaction. I type

./compute_flows ../../my_examples/network/my_output1.xml "[position() = 30]" > flows_30.txt

I search the output file flows_30.txt for the reaction and find the line

```he4 + ar46 -> n + ca49                                    5.921e-02   8.128e-08   5.921e-02
```

The forward flow is 5.921e-2. The reverse flow is 8.128e-8. The net flow is thus 5.921e-2. The entropy generation rate per nucleon from this reaction is therefore 13.498718 x 5.921e-2 = 0.7993, in agreement with our calculation above.

Entropy is being generated at this point in the calculation because the large equilibrium present is breaking down into smaller clusters. The species 46Ar and 49Ca do not belong to the same cluster; thus, since these species are not in equilibrium with each other, flow between them generates entropy.

Posted by 2016-04-27

Anonymous