`irfpy.util.physics`¶

Collection of physical formulae

Code author: Yoshifumi Futaana

A simple command line calculator for physical formulae.

See also temperature_2_pressure().

>>> p = 1e5 * 1e9      # 1 bar as nPa
>>> loschmidt = 2.6868e19   # in /cm3
>>> temperature = pressure_2_temperature(p, loschmidt)
>>> print('{:.3f}'.format(temperature))   # Should be around 0degC
269.576

Unitful version also exists

>>> p = 1 * u.bar      # 1 bar
>>> loschmidt = 2.6868e19 / u.cm **3
>>> temperature = pressure_2_temperature_u(p, loschmidt)
>>> temperature = float(temperature.rescale(u.K))
>>> print('{:.3f}'.format(temperature))
269.576

irfpy.util.physics.pressure_2_temperature_u(*args)¶: Unitful version of temperature_2_pressure()

irfpy.util.physics.dflux_2_vdf(j, mass, energy)[source]¶

Differnetial flux to velocity distribution function

This is MKSA version. For plasma-unit version (but still unitless), you may refer to dflux_2_vdf_pl(). For unitq version, you may refere to “math:dflux_2_vdf_u.

The relation is

\[f = \frac{m^2}{2E} j\]

Let us consider the differential flux of j=1e-5 /cm2 eV sr s. The particle is proton (for simplicity mass is assumed to be 1 amu), and the energy is 500 eV.

>>> j = 1e+5   # /cm2 eV sr s
>>> m = 1      # amu
>>> e = 500    # eV
>>> print('%.2e' % dflux_2_vdf_pl(j, m, e))
1.07e-10

Now, we get the velocity distribution function of 1.1e-10 s^3/m^6.

Obviously, we also call via MKSA unit version.

>>> jmksa = j * 1e4 / k.qe
>>> print('%.2e' % jmksa)   # in /m2 J sr s
6.24e+27

>>> mmksa = m * k.amu   # kg
>>> emksa = e * k.qe   # J

>>> print('%.2e' % dflux_2_vdf(jmksa, mmksa, emksa))
1.07e-10

For interactive use, unitq module will be a significant help.

>>> j_u = 1e5 / u.cm**2 / u.eV / u.sr / u.s
>>> m_u = 1 * u.amu
>>> e_u = 500 * u.eV
>>> vdf_u = dflux_2_vdf_u(j_u, m_u, e_u)
>>> u.pprint(vdf_u, fmt='%.2e')
1.07e-10 s**3/m**6

irfpy.util.physics.j2f(j, mass, energy)¶: Alias to dflux_2_vdf()

irfpy.util.physics.dflux_2_vdf_pl(j, mass, energy)[source]¶

Differnetial flux with plasma-frequently-used units

This provides the alias but different unit system to dflux_2_vdf().

The input unit is j as [/cm2 eV sr s], mass for [amu], and energy for [eV].

The returned vdf is [s**3 / m**6].

irfpy.util.physics.j2f_pl(j, mass, energy)¶: Alias to dflux_2_vdf_pl()

irfpy.util.physics.vdf_2_dflux(f, mass, velocity)[source]¶

J from f. (MKSA)

\[J = \frac{v^2}{m} f\]

>>> f = 1.07e-10    # [s^3 / m^6]
>>> j = vdf_2_dflux(f, 1.66e-27, 300e3)
>>> print('%.2e' % j)    # [/m^2 J sr s]
5.80e+27

The dflux should be converted back to the origical vdf.

>>> f2 = dflux_2_vdf(j, 1.66e-27, kinetic_energy(1.66e-27, 300e3))
>>> print('%.2e' % f2)
1.07e-10

The unitful version:

>>> f_u = 1.07e-10 * u.s ** 3 / u.m ** 6
>>> j_u = vdf_2_dflux_u(f_u, 1 * u.amu, 300 * u.km/u.s)
>>> u.pprint(j_u, fmt='%.2e')
5.80e+27 1/(m**2*s*sr*J)

>>> j_u.units = u.cu_diffflux
>>> u.pprint(j_u, fmt='%.2e')
9.29e+04 (1/cm**2 /sr /eV / s)

irfpy.util.physics.f2j(f, mass, velocity)¶

J from f. (MKSA)

\[J = \frac{v^2}{m} f\]

>>> f = 1.07e-10    # [s^3 / m^6]
>>> j = vdf_2_dflux(f, 1.66e-27, 300e3)
>>> print('%.2e' % j)    # [/m^2 J sr s]
5.80e+27

The dflux should be converted back to the origical vdf.

>>> f2 = dflux_2_vdf(j, 1.66e-27, kinetic_energy(1.66e-27, 300e3))
>>> print('%.2e' % f2)
1.07e-10

The unitful version:

>>> f_u = 1.07e-10 * u.s ** 3 / u.m ** 6
>>> j_u = vdf_2_dflux_u(f_u, 1 * u.amu, 300 * u.km/u.s)
>>> u.pprint(j_u, fmt='%.2e')
5.80e+27 1/(m**2*s*sr*J)

>>> j_u.units = u.cu_diffflux
>>> u.pprint(j_u, fmt='%.2e')
9.29e+04 (1/cm**2 /sr /eV / s)

irfpy.util.physics.kinetic_energy(mass, velocity)[source]¶

Return the kinetic energy.

\[KE = \frac{1}{2}mv^2\]

>>> print(kinetic_energy(4, 15))   # Kinetic energy for 4kg mass point with 15m/s speed. 450J.
450.0
>>> print(kinetic_energy(u.nit('4 kg'), u.nit('15 m/s')).rescale(u.J))
450.0 J

irfpy.util.physics.kinetic_energy_u(*args)¶: Unitful version of kinetic_energy()

irfpy.util.physics.kinetic_energy_proton(velocity_in_km_s)[source]¶

Return the proton energy.

Parameters:: velocity_in_m_s – Velocity in km/s.
Returns:: The kinetic energy in eV.

Simple formulation: 1 eV energy proton is about 13.8 km/s

>>> print('{:.3f}'.format(kinetic_energy_proton(13.8)))
0.994

Solar wind (4e5 m/s) proton has energy of 835 eV. The call below is the unitful version of the formulation.

>>> energy = kinetic_energy_proton_u(400000 * u.m / u.s)
>>> print('{:.3f}'.format(float(energy.nr(u.keV))))
0.835

irfpy.util.physics.kinetic_energy_electron(velocity_in_km_s)[source]¶

Return the electron energy

Parameters:: velocity_in_km_s – Velocity in km/s
Returns:: The kinetic energy in eV

Simple formulation: 1 eV energy electron is about 600 km/s

>>> print('{:.3f}'.format(kinetic_energy_electron(600)))
1.023

or one can use unitful version.

>>> energy = kinetic_energy_electron_u(600000 * u.m / u.s)
>>> print('{:.3e}'.format(float(energy.nr(u.J))))
1.640e-19

irfpy.util.physics.energy_2_velocity(mass, energy)[source]¶

Return the velocity from the kinetic energy

Return the velocity from the kinetic energy. This is the inversion function of kinetic_energy().

>>> print(energy_2_velocity(4, 450.))   # 4kg mass with 450 J => v=15 m/s
15.0

You may also use unitful version.

>>> print(energy_2_velocity_u(u.nit('4 kg'), u.nit('450 J')))
15.0 m/s

irfpy.util.physics.energy_2_velocity_proton(energy_eV)[source]¶

Return the velocity (in km/s) of the proton with the given energy (eV)

Parameters:: energy_eV – Energy of proton in eV.
Returns:: Speed of proton in km/s.

>>> print('{:.2f}'.format(energy_2_velocity_proton(1)))    # 1 eV proton has 13.8 km/s
13.84

or use unit-version.

>>> speed = energy_2_velocity_proton_u(1 * u.eV)
>>> u.pprint(speed)
13.841 km/s

irfpy.util.physics.energy_2_velocity_electron(energy_eV)[source]¶

Return the velocity (in km/s) of the electron with the given energy (eV)

Parameters:: energy_eV – Energy of electron in eV.
Returns:: Speed of electron in km/s.

>>> print('{:.2f}'.format(energy_2_velocity_electron(1)))    # 1 eV proton has 593 km/s
593.10

or use unit-version.

>>> speed = energy_2_velocity_electron_u(1 * u.eV)
>>> u.pprint(speed)
593.097 km/s

irfpy.util.physics.temperature_2_thermal_velocity(mass, temperature)[source]¶

Return the thermal velocity from temperature

\[v_{th} = \sqrt{\frac{kT}{m}}\]

Thermal velocity is calculated from the temperature.

Roughly the proton with 1eV temperature has 9.8 km/s thermal velocity.

>>> print('%.3f' % temperature_2_thermal_velocity(1.67e-27, 11600))
9792.924

Unitful version also supported.

>>> print('%.3f' % temperature_2_thermal_velocity_u(k.mp_u,
...        1 * u.eVT).nr(u.nit('km/s')))
9.787

irfpy.util.physics.temperature_2_thermal_velocity_u(*args)¶: Unitful version of temperature_2_thermal_velocity().

irfpy.util.physics.vth(mass, temperature)¶

Return the thermal velocity from temperature

\[v_{th} = \sqrt{\frac{kT}{m}}\]

Thermal velocity is calculated from the temperature.

Roughly the proton with 1eV temperature has 9.8 km/s thermal velocity.

>>> print('%.3f' % temperature_2_thermal_velocity(1.67e-27, 11600))
9792.924

Unitful version also supported.

>>> print('%.3f' % temperature_2_thermal_velocity_u(k.mp_u,
...        1 * u.eVT).nr(u.nit('km/s')))
9.787

irfpy.util.physics.gyrofrequency(mass, charge, magnetic_flux_density)[source]¶