String formatting and physical constants

P2.3.2String formatting and physical constants

The table below gives the names, symbols, values, uncertainties and units of some physical constants.

Name	Symbol	Value	Uncertainty	Units
Boltzmann constant	$k_\mathrm{B}$	$1.3806504\times10^{-23}$	$2.4\times 10^{-29}$	$\mathrm{J\,K^{-1}}$
Speed of light	$$c$$	$$299792458$$	(def)	$\mathrm{m\,s^{-1}}$
Planck constant	$$h$$	$6.62606896 \times 10^{-34}$	$3.3\times 10^{-41}$	$\mathrm{J\,s}$
Avogadro constant	$N_\mathrm{A}$	$\;\;\,6.02214179 \times 10^{23}$	$3\times 10^{16}$	$\mathrm{mol^{-1}}$
Electron magnetic moment	$\mu_e$	$-9.28476377 \times 10^{-24}$	$2.3\times 10^{-31}$	$\mathrm{J/T}$
Gravitational constant	$$G$$	$\;\;\,6.67428 \times 10^{-11}$	$6.7 \times 10^{-15}$	$\mathrm{N\,m^2\,kg^{-2}}$

Defining variables of the form:

kB = 1.3806504e-23  # J/K
kB_unc = 2.4e-29    # uncertainty
kB_units = 'J/K'

use the string object's format method to produce the following output:

(a)

kB = 1.381e-23 J/K

(b)

G = 0.0000000000667428 Nm^2/kg^2

kB   =  1.3807e-23 J/K
mu_e = -9.2848e-24 J/T
N_A  =  6.0221e+23 mol-1
c    =  2.9979e+08 m/s

(d) Again, using the same format specifier for each line,

===    G = +6.67E-11 [N m2 kg-2]    ===
===   μe = -9.28E-24 [      J/T]    ===

Hint: The Unicode codepoint for, μ, the lower case greek letter mu, is U+03BC.

(e) (Harder). Produce the output below, in which the uncertainty (one standard deviation) in the value of each constant is expressed as a number in parentheses relative the the preceding digits: that is, $6.62606896(33) \times 10^{-34}$ means $6.62606896 \times 10^{-34} \pm 3.3 \times 10^{-41}$ .

G = 6.67428(67)e-11 Nm2/kg2
mu_e = -9.28476377(23)e-24 J/T

Show solution

(a)

print('{:s} = {:.3e} {:s}'.format(kB_symb, kB, kB_units))

(b)

print('G = {:18.16f} {:s}'.format(G, G_units))

(c)

fmt = '{:4s} = {: 11.4e} {:s}\n'*4
print(fmt.format('kB', kB, kB_units, 'mu_e', mu_e, mu_e_units, 'N_A', N_A, N_A_units,
                 'c', c, c_units))

(d)

print('=== {:^31s} ==='.format('G = {0:+9.2E} [{1:>9s}]'.format(G,G_units)))
print('=== {:^31s} ==='.format('\u03BCe = {0:+9.2E [{1:>9s}]'
                               .format(mu_e,mu_e_units)))

(e)

# The number of digits in the uncertainty to output
nsd_digits = 2
# Separate our constant into mantissa * 10^exponent
exponent = int(math.floor(math.log10(abs(G))))
mantissa = G * 10**(-exponent)
# The uncertainty digits relative to the same exponent
sd = G_unc * 10**(-exponent)
# The number of decimal places to output
dp = int(-math.log(sd,10))+nsd_digits
# The uncertainty digits as an integer
sd_digits = int(round(sd*10**dp))
print('G = {mantissa:.{dp:d}f}({sd_digits:d})e{exponent:d} {units:s}'
            .format(mantissa=mantissa, dp=dp, sd_digits=sd_digits,
                    exponent=exponent, units=G_units))
6.67428(67)e-11

produces

G = 6.67428(67)e-11 Nm2/kg2

(and similarly for $\mu_e$ ).