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The Förster resonance energy transfer (FRET) spectral overlap integral looks like:

$$ J=\int_{0}^{\infty}\bar{F}_{D}(λ)ε_{A}(λ)λ^4 dλ $$ Where $λ$ is the wavelength, $\bar{F}_{D}(λ)$ is the normalized fluorescence emission spectrum of the donor(normalized to unity area), and $ε_{A}(λ)$ is the extinction coefficient ($M^{-1} \ cm^{-1}$).

I am not very familiar with resonance energy transfer but was just curious why would a $λ^4$ term be contained in this spectral overlap calculation. The intuitive implication would be that the longer the wavelength, the larger the spectral overlap?


See for example Eq. 12.1 on page 471 in FRET and FLIM Techniques (Theodorus W. J. Gadella, ed. 2008) or the corresponding $\omega^{-4}$ in the integrals of Eq. 14.30 of Sect. 14.5.1 in David L. Andrews' Chapter 14: Resonance Energy Transfer: Theoretical Foundations and Developing Applications

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  • $\begingroup$ I don't know why the ^4, but think about excitation energy with longer wavelengths... $\endgroup$
    – bob1
    Commented Mar 24, 2022 at 0:15
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    $\begingroup$ If you write the integral in terms of dipole strengths the $\omega^{-4}$ disappears[[1]](spiedigitallibrary.org/journals/journal-of-biomedical-optics/…). Perhaps, look at the relationship between emission spectra and dipole strength to see if that explains the $\omega^{-2}$ per factor. $\endgroup$
    – Retracted
    Commented Mar 24, 2022 at 6:11

1 Answer 1

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Why this is such a good question


After finishing teaching at 8pm EDT, it took me about 1 hour to find the answer and 1.5 hours writing this answer and confirming things. The Wikipedia article just gives that formula out of nowhere and cites [17] which is nothing but a table of contents (apparently not containing the string "rster" from Förster anywhere) for a book that I couldn't find online, [18] which is a Google Books preview of a book for which something related to the $\lambda^4$ does appear in Eq. 1.2 of pg 13 but explains nothing about where it came from apart from citations to {61}, which is missing from the Google Books preview, and 81-84 which are all books (unlikely to have a free PDF lying around), and [19] which gives the relevant formula in Eq. 14.1 but with no explanation and not even any citations to say where it came from.

Despite having worked in this field for a while, done my PhD in the general area, and even helped supervise a student whose Masters thesis gives a 5-page derivation of Foerster theory starting from page 30, I couldn't find where the $\lambda^4$ came from (the thesis presents the Foerster rate in a different way, which doesn't involve the $\lambda^4$, and I also looked at the reference in there to this paper on my friend Suggy Jang's generalization of Foerster-Dexter theory but it was done in the same spirit as my student's derivation, which unfortunately doesn't help us today with finding the source of the $\lambda^4$). I thought I was in luck, because my thorough studying of the field meant that I knew one of the best reviews on the topic (this PDF of a paper by Greg Scholes currently with 1238 citations on Google Scholar), but it turns out he literally just gives the integral on Pg. 62 with no explanation. The textbook by May and Kühn might have the desired explanation, but I couldn't find a copy online. Every single paper and/or article I found online with a title like "Introduction to Foerster Theory" or simply "Foerster Resonance Energy Transfer" gave me nothing (some examples of the better resources I found were this, this, some detailed discussions here, this very interesting article by someone who hung out with Foerster in real life entitled "Förster's resonance excitation transfer theory: not just a formula" but then disappointingly it did indeed give "just a formula" at Eq. 4.

The original 1948 paper by Foerster in German (currently with 9994 citations on Google Scholar) doesn't seem to have anything resembling anything close to "the" desired equation. The only research paper by Foerster mentioned in his Wikipedia article doesn't have the equation either, and the famous 1953 paper (currently with 9977 citations on Google Scholar) by Dexter that earned him a spot in the name "Foerster-Dexter theory" doesn't have $\lambda^4$ or $1/\nu^4$ or $1/\omega^4$ anywhere, although now that I have the answer and double-checked all these articles to make sure I wasn't discrediting anyone, I do see that there's a $1/E^4$ in Eq. 16 but it's not as close to your equation as what I'm about to show you.

The answer


This 1965 "report" by Foerster actually tells us how he got the $1/\nu^4 = (\lambda /c)^4$ right here:

enter image description here

Basically after 53 pages which largely set the reader up for it, he gets this expression for the transfer rate:

enter image description here

In that equation, you can plug in the following:

enter image description here

which will mean you'll get a factor of $c/\nu = \lambda$ in the numerator coming from the molar decadic excitation coefficient $\varepsilon(\nu)$ and a factor of $(c/\nu)^3 = \lambda^3$ coming from the normalized fluorescence quantum spectrum $f(\nu)$: the two combine together to form $\lambda^4$. Since Foerster tells us that these are "Einstein's well-known expressions" for $\varepsilon(\nu)$ and $f(\nu)$, I assume you don't need help knowing why those are proportional to $\nu$ and $\nu^3$ respectively...






... just kidding, explanations and more references about the Einstein coefficients can be found here and here and you'll see expressions proportional to $\nu$ and others proportional to $\nu^3$ in both of those resources.

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  • $\begingroup$ I even got out my QM text and skimmed through a few chapters... I didn't find it because physicists don't use those functions (the Einstein coefficients physicists are taught are these [en.wikipedia.org/wiki/Einstein_coefficients]) Great detective work! $\endgroup$
    – Retracted
    Commented Mar 26, 2022 at 17:37
  • $\begingroup$ This is cool! Thank you so much for such clear answer! $\endgroup$
    – Weiwei
    Commented Mar 27, 2022 at 2:45

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