Synthesis¶

Using Synthesis in Roland¶

roland has an Abund class for single abundance values and an Abundances class that holds the abundances for all elements. These are handy for modifying abundances when running synthesis.

Let’s import the Abundances class from the abundance module:

>>> from roland.abundance import Abundances

Internal Values and Hydrogen/Helium Abundances¶

Internally, an Abundances object stores the abundances values in linear N(X)/N(H) format.

The Hydrogen and Helium abundances are special and are held constant at N(H)/N(H)=1.0 and N(He)/N(H)=0.0851138.

Solar Abundance Values¶

The Abundances object stored internal values for solar abundances. The type of values can be selected at initialization time. The options are:

“asplund: Asplund, Grevesse and Sauval (2005)

“husser”: Asplund et al. 2009 chosen for the Husser et al. (2013) Phoenix model atmospheres

“kurucz”: abundances for the Kurucz model atmospheres (Grevesse+Sauval 1998)

“marcs”: abundances for the MARCS model atmospheres (Grevesse+2007 values with CNO abundances from Grevesse+Sauval 2008)

The “asplund” values are used by default.

Initializing¶

You can initialize an Abundances object using 1) a list of abundances, 2) a dictionary of abundance values, 3) a single metallicity or 4) just with the default solar values (no input).

This will use the default solar values:

>>> abu = Abundances()
>>> abu
Abundances([M/H]=0.00,solar=asplund)
Num Name   N(X)/N(H)   log(eps)   [X/H]
 1   H   1.000000e+00   12.000  [ ---- ]
 2   He  8.511380e-02   10.930  [ ---- ]
 3   Li  1.122018e-11    1.050  [ 0.000]
 4   Be  2.398833e-11    1.380  [ 0.000]
 5   B   5.011872e-10    2.700  [ 0.000]
 6   C   2.454709e-04    8.390  [ 0.000]
 ...
96   Cm  1.023293e-22   -9.990  [ 0.000]
97   Bk  1.023293e-22   -9.990  [ 0.000]
98   Cf  1.023293e-22   -9.990  [ 0.000]
99   Es  1.023293e-22   -9.990  [ 0.000]

Let’s start with [M/H]=-1.0 instead this time:

>>> abu = Abundances(-1.0)
>>> abu
Abundances([M/H]=-1.00,solar=asplund)
Num Name   N(X)/N(H)   log(eps)   [X/H]
 1   H   1.000000e+00   12.000  [ ---- ]
 2   He  8.511380e-02   10.930  [ ---- ]
 3   Li  1.122018e-12    0.050  [-1.000]
 4   Be  2.398833e-12    0.380  [-1.000]
 5   B   5.011872e-11    1.700  [-1.000]
 6   C   2.454709e-05    7.390  [-1.000]
 ...
96   Cm  1.023293e-23  -10.990  [-1.000]
97   Bk  1.023293e-23  -10.990  [-1.000]
98   Cf  1.023293e-23  -10.990  [-1.000]
99   Es  1.023293e-23  -10.990  [-1.000]

Now, let’s give it a dictionary of abundances values:

>>> abu = Abundances({"MG_H":-0.5,"SI_H":0.24})
>>> abu
Abundances([M/H]=0.00,solar=asplund)
Num Name   N(X)/N(H)   log(eps)   [X/H]
 H   1.000000e+00   12.000  [ ---- ]
 He  8.511380e-02   10.930  [ ---- ]
 Li  1.122018e-11    1.050  [ 0.000]
 Be  2.398833e-11    1.380  [ 0.000]
 B   5.011872e-10    2.700  [ 0.000]
 C   2.454709e-04    8.390  [ 0.000]
 N   6.025596e-05    7.780  [ 0.000]
 O   4.570882e-04    8.660  [ 0.000]
 F   3.630781e-08    4.560  [ 0.000]
 Ne  6.918310e-05    7.840  [ 0.000]
 Na  1.479108e-06    6.170  [ 0.000]
 Mg  1.071519e-05    7.030  [-0.500]
 Al  2.344229e-06    6.370  [ 0.000]
 Si  5.623413e-05    7.750  [ 0.240]
...

Finally, we can give an entire array or list of abundances values. You have give the type of abundance values you are giving in the second parameter. The options are linear, log, logeps, or x_h:

>>> abu = Abundances([12.  , 10.93,  1.05,  1.38,  2.7 ,  8.39,  7.78,
                      8.66,  4.56,  7.84,  6.17,  7.03,  6.37,  7.75,
                      5.36,  7.14,  5.5 ,  6.18],'logeps')
>>> abu
Abundances([M/H]=0.00,solar=asplund)
Num Name   N(X)/N(H)   log(eps)   [X/H]
 1   H   1.000000e+00   12.000  [ ---- ]
 2   He  8.511380e-02   10.930  [ ---- ]
 3   Li  1.122018e-11    1.050  [ 0.000]
 4   Be  2.398833e-11    1.380  [ 0.000]
 5   B   5.011872e-10    2.700  [ 0.000]
 6   C   2.454709e-04    8.390  [ 0.000]
...
97   Bk  1.023293e-22   -9.990  [ 0.000]
98   Cf  1.023293e-22   -9.990  [ 0.000]
99   Es  1.023293e-22   -9.990  [ 0.000]

Modifying an Abundances Object¶

You can always modify an Abundances object in place:

>>> abu['O_H'] = -0.5
Abundances([M/H]=-0.17,solar=asplund)
Num Name   N(X)/N(H)   log(eps)   [X/H]
 H   1.000000e+00   12.000  [ ---- ]
 He  8.511380e-02   10.930  [ ---- ]
 Li  1.122018e-11    1.050  [ 0.000]
 Be  2.398833e-11    1.380  [ 0.000]
 B   5.011872e-10    2.700  [ 0.000]
 C   2.454709e-04    8.390  [ 0.000]
 N   6.025596e-05    7.780  [ 0.000]
 O   1.445440e-04    8.160  [-0.500]
 F   3.630781e-08    4.560  [ 0.000]
...
 Cm  1.023293e-22   -9.990  [ 0.000]
 Bk  1.023293e-22   -9.990  [ 0.000]
 Cf  1.023293e-22   -9.990  [ 0.000]
 Es  1.023293e-22   -9.990  [ 0.000]

Or change the metallicity:

>>> abu['M_H'] = -0.5
Abundances([M/H]=-0.50,solar=asplund)
Num Name   N(X)/N(H)   log(eps)   [X/H]
 H   1.000000e+00   12.000  [ ---- ]
 He  8.511380e-02   10.930  [ ---- ]
 Li  3.548134e-12    0.550  [-0.500]
 Be  7.585776e-12    0.880  [-0.500]
 B   1.584893e-10    2.200  [-0.500]
 C   7.762471e-05    7.890  [-0.500]
 N   1.905461e-05    7.280  [-0.500]
 O   1.445440e-04    8.160  [-0.500]
...
 Am  3.235937e-23  -10.490  [-0.500]
 Cm  3.235937e-23  -10.490  [-0.500]
 Bk  3.235937e-23  -10.490  [-0.500]
 Cf  3.235937e-23  -10.490  [-0.500]
 Es  3.235937e-23  -10.490  [-0.500]

You can also change the entire metallicity by an increment amount:

>>> abu += 0.5
Abundances([M/H]=0.50,solar=asplund)
Num Name   N(X)/N(H)   log(eps)   [X/H]
 H   1.000000e+00   12.000  [ ---- ]
 He  8.511380e-02   10.930  [ ---- ]
 Li  3.548134e-11    1.550  [ 0.500]
 Be  7.585776e-11    1.880  [ 0.500]
 B   1.584893e-09    3.200  [ 0.500]
 C   7.762471e-04    8.890  [ 0.500]
 N   1.905461e-04    8.280  [ 0.500]
...
 Cm  3.235937e-22   -9.490  [ 0.500]
 Bk  3.235937e-22   -9.490  [ 0.500]
 Cf  3.235937e-22   -9.490  [ 0.500]
 Es  3.235937e-22   -9.490  [ 0.500]

Or the alpha abundances:

>>> abu['alpha'] -= 0.5
Abundances([M/H]=-0.25,solar=asplund)
Num Name   N(X)/N(H)   log(eps)   [X/H]
 H   1.000000e+00   12.000  [ ---- ]
 He  8.511380e-02   10.930  [ ---- ]
 Li  1.122018e-11    1.050  [ 0.000]
 Be  2.398833e-11    1.380  [ 0.000]
 B   5.011872e-10    2.700  [ 0.000]
 C   2.454709e-04    8.390  [ 0.000]
 N   6.025596e-05    7.780  [ 0.000]
 O   1.445440e-04    8.160  [-0.500]
 F   3.630781e-08    4.560  [ 0.000]
 Ne  2.187762e-05    7.340  [-0.500]
 Na  1.479108e-06    6.170  [ 0.000]
 Mg  1.071519e-05    7.030  [-0.500]
 Al  2.344229e-06    6.370  [ 0.000]
 Si  1.023293e-05    7.010  [-0.500]
 P   2.290868e-07    5.360  [ 0.000]
 S   4.365158e-06    6.640  [-0.500]
 Cl  3.162278e-07    5.500  [ 0.000]
 Ar  4.786301e-07    5.680  [-0.500]
 K   1.202264e-07    5.080  [ 0.000]
 Ca  6.456542e-07    5.810  [-0.500]
 Sc  1.122018e-09    3.050  [ 0.000]
 Ti  2.511886e-08    4.400  [-0.500]
 V   1.000000e-08    4.000  [ 0.000]
...
 Cm  1.023293e-22   -9.990  [ 0.000]
 Bk  1.023293e-22   -9.990  [ 0.000]
 Cf  1.023293e-22   -9.990  [ 0.000]
 Es  1.023293e-22   -9.990  [ 0.000]

Creating a New, Modified Abundances Object¶

You can also call the object and create a new, modified object.

Create a new Abundances object with a metallicity of -1.5:

>>> abu2 = abu(-1.5)
>>> abu2
Abundances([M/H]=-1.50,solar=asplund)
Num Name   N(X)/N(H)   log(eps)   [X/H]
 1   H   1.000000e+00   12.000  [ ---- ]
 2   He  8.511380e-02   10.930  [ ---- ]
 3   Li  3.548134e-13   -0.450  [-1.500]
 4   Be  7.585776e-13   -0.120  [-1.500]
 5   B   1.584893e-11    1.200  [-1.500]
 6   C   7.762471e-06    6.890  [-1.500]
...
96   Cm  3.235937e-24  -11.490  [-1.500]
97   Bk  3.235937e-24  -11.490  [-1.500]
98   Cf  3.235937e-24  -11.490  [-1.500]
99   Es  3.235937e-24  -11.490  [-1.500]

You can also input a dictionary of abundances values:

>>> abu2 = abu({"c_h":-1.5})
>>> abu2
Abundances([M/H]=-0.12,solar=asplund)
Num Name   N(X)/N(H)   log(eps)   [X/H]
 1   H   1.000000e+00   12.000  [ ---- ]
 2   He  8.511380e-02   10.930  [ ---- ]
 3   Li  1.122018e-11    1.050  [ 0.000]
 4   Be  2.398833e-11    1.380  [ 0.000]
 5   B   5.011872e-10    2.700  [ 0.000]
 6   C   7.762471e-06    6.890  [-1.500]
 7   N   6.025596e-05    7.780  [ 0.000]
...
96   Cm  1.023293e-22   -9.990  [ 0.000]
97   Bk  1.023293e-22   -9.990  [ 0.000]
98   Cf  1.023293e-22   -9.990  [ 0.000]
99   Es  1.023293e-22   -9.990  [ 0.000]

Abundances Output¶

The Abundances class can output the information in several ways.

If you select a single element (by element name or index), an Abund object will be returned.:

>>> abu['Ca']
Abund(20 Ca N(Ca)/N(H)=2.042e-06 log(eps)=6.310)

>>> abu[10]
Abund(11 Na N(Na)/N(H)=1.479e-06 log(eps)=6.170)

Selecting values in bracket notation will only return the value. Abundance versus H:

>>> abu['Ca_H']
0.0

Abundance versus M:

>>> abu['Ca_M']
-0.0003221051142099841

There are several useful properties that will print out all of the abundances.

Print the linear or N(X)/N(H) values with linear.:

>>> abu.linear
array([1.00000000e+00, 8.51138038e-02, 1.12201845e-11, 2.39883292e-11,
   5.01187234e-10, 2.45470892e-04, 6.02559586e-05, 4.57088190e-04,
   3.63078055e-08, 6.91830971e-05, 1.47910839e-06, 1.07151931e-05,
   ...
   1.02329299e-22, 1.14815362e-12, 1.02329299e-22, 3.01995172e-13,
   1.02329299e-22, 1.02329299e-22, 1.02329299e-22, 1.02329299e-22,
   1.02329299e-22, 1.02329299e-22, 1.02329299e-22])

Or you can also use log, logeps, xh, or xm. The log abundances::

>>> abu.log
array([  0.  ,  -1.07, -10.95, -10.62,  -9.3 ,  -3.61,  -4.22,  -3.34,
        -7.44,  -4.16,  -5.83,  -4.97,  -5.63,  -4.25,  -6.64,  -4.86,
        -6.5 ,  -5.82,  -6.92,  -5.69,  -8.95,  -7.1 ,  -8.  ,  -6.36,
       ...
       -11.1 , -10.  , -11.35, -21.99, -21.99, -21.99, -21.99, -21.99,
       -21.99, -11.94, -21.99, -12.52, -21.99, -21.99, -21.99, -21.99,
       -21.99, -21.99, -21.99])

Abundances in log(eps) notation, or log(N(X)/N(H))+12.0:

>>> abu.logeps
array([12.  , 10.93,  1.05,  1.38,  2.7 ,  8.39,  7.78,  8.66,  4.56,
    7.84,  6.17,  7.03,  6.37,  7.75,  5.36,  7.14,  5.5 ,  6.18,
    5.08,  6.31,  3.05,  4.9 ,  4.  ,  5.64,  5.39,  7.45,  4.92,
    ...
   -0.17,  1.11,  0.23,  1.45,  1.38,  1.64,  1.01,  1.13,  0.9 ,
    2.  ,  0.65, -9.99, -9.99, -9.99, -9.99, -9.99, -9.99,  0.06,
   -9.99, -0.52, -9.99, -9.99, -9.99, -9.99, -9.99, -9.99, -9.99])

Bracket notation, relative to H:

>>> abu.xh
array([ 0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,
        0.  ,  0.  , -0.5 ,  0.  ,  0.24,  0.  ,  0.  ,  0.  ,  0.  ,
        0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,
        ...
        0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,
        0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,
        0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ])

Bracket notation, relative to M:

>>> abu.xm
array([-3.22105114e-04, -3.22105114e-04, -3.22105114e-04, -3.22105114e-04,
       -3.22105114e-04, -3.22105114e-04, -3.22105114e-04, -3.22105114e-04,
       -3.22105114e-04, -3.22105114e-04, -3.22105114e-04, -5.00322105e-01,
       ...
       -3.22105114e-04, -3.22105114e-04, -3.22105114e-04, -3.22105114e-04,
       -3.22105114e-04, -3.22105114e-04, -3.22105114e-04, -3.22105114e-04,
       -3.22105114e-04, -3.22105114e-04, -3.22105114e-04])

You can also return the abundances in formats that are useful for model atmospheres.

Return abundance values in the format for Kurucz model atmospheres:

>>> abu.to_kurucz()
array([  0.92075543,   0.07836899, -10.98585571, -10.65585571,
        -9.33585571,  -3.64585571,  -4.25585571,  -3.37585571,
        -7.47585571,  -4.19585571,  -5.86585571,  -4.50585571,
        -6.64585571,  -4.58585571,  -7.11585571,  -5.80585571,
        ...
        -20.        , -11.97585571, -20.        , -12.55585571,
        -20.        , -20.        , -20.        , -20.        ,
        -20.        , -20.        , -20.        ]

Return abundance values in the format for MARCS model atmospheres:

>>> abu.to_marcs()
array([12.  , 10.93,  1.05,  1.38,  2.7 ,  8.39,  7.78,  8.66,  4.56,
       7.84,  6.17,  7.53,  6.37,  7.51,  5.36,  7.14,  5.5 ,  6.18,
       5.08,  6.31,  3.05,  4.9 ,  4.  ,  5.64,  5.39,  7.45,  4.92,
       ...
      -0.17,  1.11,  0.23,  1.45,  1.38,  1.64,  1.01,  1.13,  0.9 ,
       2.  ,  0.65, -9.99, -9.99, -9.99, -9.99, -9.99, -9.99,  0.06,
      -9.99, -0.52, -9.99, -9.99, -9.99, -9.99, -9.99, -9.99, -9.99])

Other useful properties and methods:

# Return the metallicity as [M/H]
>>> abu.metallicity
0.0

# Return the metallicity as Sum(N(X)/N(H) over all metals
>>> abu.metals
0.0009509329459494126

# Return all of the element symbols
>>> abu.symbol
['H','He','Li','Be','B','C','N','O','F',
...
 'U','Np','Pu','Am','Cm','Bk','Cf','Es']

# Return all of the element mass values (in amu).
>>> abu.mass
[1.00794, 4.0026, 6.941, 9.01218, 10.811, 12.0107,
...
 244.0, 243.0, 247.0, 247.0, 251.0, 252.0]

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