Source code for astropy.cosmology.funcs

# Licensed under a 3-clause BSD style license - see LICENSE.rst
Convenience functions for `astropy.cosmology`.

import warnings
import numpy as np

from .core import CosmologyError
from astropy.units import Quantity

__all__ = ['z_at_value']

__doctest_requires__ = {'*': ['scipy']}

[docs]def z_at_value(func, fval, zmin=1e-8, zmax=1000, ztol=1e-8, maxfun=500): """ Find the redshift ``z`` at which ``func(z) = fval``. This finds the redshift at which one of the cosmology functions or methods (for example Planck13.distmod) is equal to a known value. .. warning:: Make sure you understand the behavior of the function that you are trying to invert! Depending on the cosmology, there may not be a unique solution. For example, in the standard Lambda CDM cosmology, there are two redshifts which give an angular diameter distance of 1500 Mpc, z ~ 0.7 and z ~ 3.8. To force ``z_at_value`` to find the solution you are interested in, use the ``zmin`` and ``zmax`` keywords to limit the search range (see the example below). Parameters ---------- func : function or method A function that takes a redshift as input. fval : astropy.Quantity instance The value of ``func(z)``. zmin : float, optional The lower search limit for ``z``. Beware of divergences in some cosmological functions, such as distance moduli, at z=0 (default 1e-8). zmax : float, optional The upper search limit for ``z`` (default 1000). ztol : float, optional The relative error in ``z`` acceptable for convergence. maxfun : int, optional The maximum number of function evaluations allowed in the optimization routine (default 500). Returns ------- z : float The redshift ``z`` satisfying ``zmin < z < zmax`` and ``func(z) = fval`` within ``ztol``. Notes ----- This works for any arbitrary input cosmology, but is inefficient if you want to invert a large number of values for the same cosmology. In this case, it is faster to instead generate an array of values at many closely-spaced redshifts that cover the relevant redshift range, and then use interpolation to find the redshift at each value you're interested in. For example, to efficiently find the redshifts corresponding to 10^6 values of the distance modulus in a Planck13 cosmology, you could do the following: >>> import astropy.units as u >>> from astropy.cosmology import Planck13, z_at_value Generate 10^6 distance moduli between 24 and 43 for which we want to find the corresponding redshifts: >>> Dvals = (24 + np.random.rand(1000000) * 20) * u.mag Make a grid of distance moduli covering the redshift range we need using 50 equally log-spaced values between zmin and zmax. We use log spacing to adequately sample the steep part of the curve at low distance moduli: >>> zmin = z_at_value(Planck13.distmod, Dvals.min()) >>> zmax = z_at_value(Planck13.distmod, Dvals.max()) >>> zgrid = np.logspace(np.log10(zmin), np.log10(zmax), 50) >>> Dgrid = Planck13.distmod(zgrid) Finally interpolate to find the redshift at each distance modulus: >>> zvals = np.interp(Dvals.value, Dgrid.value, zgrid) Examples -------- >>> import astropy.units as u >>> from astropy.cosmology import Planck13, z_at_value The age and lookback time are monotonic with redshift, and so a unique solution can be found: >>> z_at_value(Planck13.age, 2 * u.Gyr) # doctest: +FLOAT_CMP 3.19812268 The angular diameter is not monotonic however, and there are two redshifts that give a value of 1500 Mpc. Use the zmin and zmax keywords to find the one you're interested in: >>> z_at_value(Planck13.angular_diameter_distance, ... 1500 * u.Mpc, zmax=1.5) # doctest: +FLOAT_CMP 0.6812769577 >>> z_at_value(Planck13.angular_diameter_distance, ... 1500 * u.Mpc, zmin=2.5) # doctest: +FLOAT_CMP 3.7914913242 Also note that the luminosity distance and distance modulus (two other commonly inverted quantities) are monotonic in flat and open universes, but not in closed universes. """ from scipy.optimize import fminbound fval_zmin = func(zmin) fval_zmax = func(zmax) if np.sign(fval - fval_zmin) != np.sign(fval_zmax - fval): warnings.warn("""\ fval is not bracketed by func(zmin) and func(zmax). This means either there is no solution, or that there is more than one solution between zmin and zmax satisfying fval = func(z).""") if isinstance(fval_zmin, Quantity): val = fval.to_value(fval_zmin.unit) f = lambda z: abs(func(z).value - val) else: f = lambda z: abs(func(z) - fval) zbest, resval, ierr, ncall = fminbound(f, zmin, zmax, maxfun=maxfun, full_output=1, xtol=ztol) if ierr != 0: warnings.warn('Maximum number of function calls ({}) reached'.format( ncall)) if np.allclose(zbest, zmax): raise CosmologyError("Best guess z is very close the upper z limit.\n" "Try re-running with a different zmax.") elif np.allclose(zbest, zmin): raise CosmologyError("Best guess z is very close the lower z limit.\n" "Try re-running with a different zmin.") return zbest