# Geometric mean versus arithmetic mean

• I

## Main Question or Discussion Point

The Beer-Lambert law gives the intensity of monochromatic light as a function of depth $z$ in the form of an exponential attenuation:

$$I(z)=I_{0}e^{-\gamma z},$$

where $\gamma$ is the wavelength-dependent attenuation coefficient.

However, if two different wavelengths are present simultaneously, we will have two different coefficients to consider: $\gamma_{\lambda_{1}}$ and $\gamma_{\lambda_{2}}$. Would it be mathematically correct to combine the two attenuation coefficients into one using the weighted geometric mean?

I mean, if $p_1$ and $p_2$ are the fractions of the light beam which is of a given wavelength, using geometric mean (instead of the arithmetic formula) we would have:

$$I(z)=I_{o} e^{-\bar{\gamma}z},$$

where:

$$\bar{\gamma}=\left(\gamma_{\lambda_{1}} \times\gamma_{\lambda_{2}}\right)^{1 / (p_{1}+p_{2})}.$$

Is the use of the geometric mean valid here?

Related Classical Physics News on Phys.org
Homework Helper
Gold Member
Any kind of exponential behavior with multiple wavelengths with atmospheric attenuation is incorrect. What occurs in the infrared region with some atmospheric absorption bands is there are quite a number of very narrow absorption bands and absorption lines that remove a fair amount of the energy from those absorption bands and lines within about 50 feet, without appreciably affecting adjacent wavelengths. The result is the atmospheric transmission spectrum, especially at low to medium resolution, after about 100 feet looks nearly the same as it does after about 500 feet. Any kind of exponential averaging like you are proposing simply doesn't work. See also: https://en.wikipedia.org/wiki/Infrared_window

Last edited:
hilbert2
Gold Member
As Charles said, there's no way to write the sum of two exponentials $Ae^{-k_1 x} + Be^{-k_2 x}$ as a single exponential $Ce^{-k' x}$ if $k_1 \neq k_2$, no matter how you define the mean absorption coefficient $k'$.

Hi Charles,

I understand that Beer-Lambert law only applies to monochromatic light. But the definition of geometric mean does allow us to combine the two exponentials into one. I admit that I have never seen this done in literature before. So I wanted to know why this is mathematically incorrect (which is why I originally posted this under the mathematics sub-forum).

Note: my problem relates to the area of material processing by laser beams when two different wavelengths are simultaneously present in the beam (e.g. as a result of second harmonic generation). Suppose a laser operates at the wavelength $\lambda_1$. If $\lambda_2$ has a greater absorptance in a given material than $\lambda_1$, the overall absorptance of the laser energy increases by converting a fraction of the beam into $\lambda_2$. A convenient way to show this is to define an overall effective attenuation coefficient (if such a thing is possible).

As Charles said, there's no way to write the sum of two exponentials $Ae^{-k_1 x} + Be^{-k_2 x}$ as a single exponential $Ce^{-k' x}$ if $k_1 \neq k_2$, no matter how you define the mean absorption coefficient $k'$.
Why can you not use the weighted geometric mean?

It clearly works for exponentials and when you have geometric terms. From the definition given in Wikipedia we can write the combined attenuation coefficient as:

$$\bar{\gamma}=\left(\Pi_{i=1}^{n}\gamma_{\lambda_{i}}^{p_{i}}\right)^{1/\sum_{i=1}^{n}p_{i}},$$

where in this case $n=2$.

hilbert2
Gold Member
Let's see what happens if we assume that for some $k_1 \in \mathbb{R}$ and $k_2 \in \mathbb{R}$ with $k_1 \neq k_2$ and $k' , A, B, C \in \mathbb{R}$ we have

$Ae^{-k_1 x} + Be^{-k_2 x} = Ce^{-k' x}$, for any $x\in \mathbb{R}$.

Dividing both sides of the equation by $Ce^{-k' x}$, we obtain

$\frac{A}{C}e^{(k'-k_1 )x} + \frac{B}{C}e^{(k'-k_2 )x} = 1$.

But for the sum of exponentials on the LHS to be an x-independent constant, we would need to have

$k' - k_1 = 0$, and $k'-k_2 = 0$,

which is impossible when $k_1 \neq k_2$. Proved.

Edit: You also have to assume that $A,B$ and $C$ are greater than zero, but that's quite obvious...

Last edited:
Chestermiller
Mentor
What you did was incorrect mathematically. When I solve it I get: $$\bar{\gamma}=-\frac{\ln(p_1e^{-\gamma_1z}+p_2e^{-\gamma_2z})}{z}$$This does not reduce to the result you obtained.

Thank you Chestermiller. I can see what went wrong there.

So, we can't combine two absorption coefficients for two different wavelengths.

I guess absorption coefficients can only be combined for multiple absorbing species. For example, when two components are present in the sample you have the superposition:

$$\overline{\gamma} = \gamma_1 + \gamma_2 = \alpha_1 c_1 + \alpha_2 c_2,$$

where $c$ is the concentration of a component and $\alpha$ is its specific absorption coefficient.

mjc123