Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I really need help understanding this equation ,i am new to quantum mechanics and i cant understand the math, so i need every single symbol to be explained or given a value if it is a constant , ( lets say X is Hydrogen) enter image description here

share|cite|improve this question
What is the formula supposed to describe? What are the quantities involved? I can't make any sense about it without any additional information. – Ondřej Černotík Jan 14 '13 at 17:13
up vote 2 down vote accepted

I don't really think you need to understand a lot of quantum mechanics to make sense of this. This is what I make of it:

$\phi_\infty(\lambda)$: The light intensity as a function of wavelength $\lambda$. Probably refers to the solar light before reaching earth's atmosphere.

$\exp\left[-\sum_m\sigma_m^{a}(\lambda)\int_z^\infty n_m(s)ds)\right]$: Factor representing the transmission of light throgh the atmosphere depending on wavelength. This is basically the Beer-Lambert Law with a summation over all absorbing species in the atmosphere and an integral along the light path.

$\sigma_m^{a}(\lambda)$: The absorption crossection of species $m$ at wavelength $\lambda$.

$n_m(s)$ : The number density (i.e. molecules per unit of volume) of species $m$ at point $s$ along the light path.

$\sigma_X^{(i)}(\lambda)$: Probably the photoioniztion crossection for $X$ at wavelength $\lambda$.

$n_n(X)$: Probably the number density of $X$.

The only thing that i find strange is that there is a sum over wavelengths. I would have expected an integral instead. But then I'm not used to notation and conventions for describing photoioniztion.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.