Some remarks on Assignment 1¶

Relative difference¶

$$\Delta = \frac{|\mathrm{LHS}-\mathrm{RHS}|}{\max(|\mathrm{LHS}|,|\mathrm{RHS}|)}$$

This quantity is defined for each mass shell individually. If $\max$ was taken over the whole star, we would be comparing quantities in different regions, which limits the usefulness if the relative difference.

`max` vs `fmax`¶

In [2]:

A = array([1,5,1e10])
B = array([-17,24,16])

In [3]:

max(A) # overall maximum in a single array

Out[3]:

10000000000.0

In [4]:

fmax(A,B) # element-wise maximum, result is array of the same shape

Out[4]:

array([  1.00000000e+00,   2.40000000e+01,   1.00000000e+10])

In [5]:

m = mesa.profile('m-17.data')
rhs = - mesa.G * m.mass_grams / (4 * pi * m.radius_cm**4)
dmbar = (0.5 * (m.dm[:-1] + m.dm[1:]))
lhs = (m.pressure[:-1] - m.pressure[1:]) / dmbar
plot(m.mass[1:], abs(lhs - rhs[1:]) / np.fmax(np.abs(lhs),np.abs(rhs[1:])))
xlabel('$M/M_\odot$')
ylabel('$\Delta$')
yscale('log')

Some remarks on Python¶

Arrays¶

Python has lists of arbitrary objects marked by square brackets [...]. For computations these have to converted to an array (uniform datatype, supports mathematical operations on the whole array).

In [6]:

A = [1, 2, -1e-30] # list
B = array(A)

Selecting parts of an array¶

Indices in Python start at 0.

In [7]:

A = array([1, 2, 7.5, -1e-30])

In [8]:

A[1] # second element only

Out[8]:

2.0

In [9]:

A[-1] # last element

Out[9]:

-1.0000000000000001e-30

In [10]:

A[1:3] # elements 1 and 2 (last index not included)

Out[10]:

array([ 2. ,  7.5])

In [11]:

A[2:] # start at third (index 2) element to the end

Out[11]:

array([  7.50000000e+00,  -1.00000000e-30])

In [12]:

A.shape # shape of the whole array

Out[12]: