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The Förster distance is the separation giving FRET efficiency $E=1/2$ for the specified pair and conditions. It is a property of the optical pair and its environment, rather than a universal length. In the usual pair model,where ${\kappa }^2$ is the FRET dipole orientation factor, $Q_D$ the FRET donor quantum yield, $J$ the FRET spectral overlap integral, and $n$ the medium's refractive index. The omitted proportionality constant depends on the unit convention for $J$.^[ref]

If FRET efficiency is known and those assumptions hold, inversion givesThe derivative explains why errors become troublesome near $E=0$ or $E=1$. An illustrative pair with $R_0=4$ nanometres and another with $R_0=6$ nanometres both give $E=1/2$ at their different Förster distances: knowing FRET efficiency alone does not determine $r$.

A heterogeneous population adds another ambiguity. In a synthetic example with equal weights per FRET donor excitation, efficiencies $0.1$ and $0.9$ average to $0.5$, but their separations are $9^{1/6}{R}_0$ and $9^{-1/6}{R}_0$. Their average separation is approximately $1.068R_0$, whereas inverting the average FRET efficiency gives $R_0$. A measured emission ratio has additional detection weights; it need not equal even this excitation-weighted average. The transferable lesson is that an average readout does not uniquely specify a distribution of molecular separations.

Förster resonance energy transfer is a nonradiative transfer of energy from an excited state of a FRET donor to a nearby FRET acceptor, through coupling of their electric dipoles. The transfer step does not emit a photon that the FRET acceptor subsequently absorbs. It competes with the FRET donor's own fluorescence and other routes out of its excited state. A fluorescent FRET acceptor can then emit fluorescence; an absorbing but nonfluorescent FRET acceptor can still receive the transferred energy.^[ref]

The striking use of FRET is to turn a change in molecular proximity into an optical readout. Proteins carrying a FRET donor and a FRET acceptor may change the readout when they associate, as in the chemotaxis FRET assay; two labels on one protein may report a change in its shape. Interpretation depends on the FRET spectral overlap, FRET dipole orientation factor and labeling geometry, as well as separation. A change in FRET is therefore evidence about proximity under specified conditions, rather than a direct measurement of an arbitrary distance.^[ref]

For a single pair with fixed Förster distance $R_0$ and label separation $r$, the FRET efficiency isAt $r=R_0$, half the FRET donor excitations transfer energy. At $r=R_0/2$, the model gives $E=64/65\approx 98.46\%$; at $r=2R_0$, it gives $E=1/65\approx 1.54\%$. The sixth power makes FRET especially useful near the Förster distance, with little sensitivity to further separation once transfer is already negligible. These are model calculations, not measurements of a particular pair.

The FRET dipole orientation factor describes the geometry of the two transition electric dipoles relative to their separation. Let ${\hat{\mathbf{\mu }}}_D$, ${\hat{\mathbf{\mu }}}_A$ and $\hat{\mathbf{r}}$ be their unit vectors. ThenFor a separation along $z$, parallel electric dipoles along $z$ give ${\kappa }^2=4$, parallel ones along $x$ give $1$, and one along $x$ with the other along $y$ gives $0$. Thus rotating a pair can change FRET efficiency without changing its separation.

The familiar value ${\kappa }^2=2/3$ requires suitable rapid, independent orientation averaging during the FRET donor lifetime. Random orientations across a population of slowly rotating fluorescent proteins do not establish that dynamic condition. ^[ref]

The FRET donor lifetime is the mean time spent in the excited state. For the constant-rate model of FRET efficiency,After a sufficiently brief excitation pulse, the FRET donor's fluorescence follows exponential decay, $I(t)=I(0)e^{-t/\tau }$. With the illustrative values ${\tau }_D=4$ nanoseconds and ${\tau }_{DA}=2$ nanoseconds, the inferred FRET efficiency is $1/2$. Multiplying $I(0)$ changes brightness without changing this decay time.

The lifetime comparison requires the same nontransfer decay rates in the reference and paired samples. Changes in the FRET donor's environment or a different nonradiative loss can also shorten the FRET donor lifetime. Heterogeneous FRET efficiencies or slowly varying FRET dipole orientation factors need an appropriate decay model; substituting any fitted average lifetime into the single-pair formula can mislead. ^[ref]

The FRET donor quantum yield is the emitted fluorescence photon fraction per absorbed excitation photon, measured without a FRET acceptor. In the constant-rate model used for FRET efficiency, $Q_D=k_{\mathrm{r}}/(k_{\mathrm{r}}+k_{\mathrm{nr}})$. Changing the FRET donor quantum yield changes the Förster distance, even if label positions remain fixed.

FRET efficiency is the probability that an excitation of the FRET donor ends in transfer to the FRET acceptor. Write $k_D=k_{\mathrm{r}}+k_{\mathrm{nr}}$ for the total decay rate without a FRET acceptor, including radiative fluorescence and other decay, and $k_T$ for the transfer rate. For constant competing rates,The integral counts transfer events among all excitations in a population with exponential decay. Substitution gives the separation law above. This short rate argument assumes one FRET donor, one FRET acceptor, weak coupling, an available FRET acceptor, and fixed geometry during the decay; it is not a model for arbitrary interacting labels.^[ref]

FRET optical crosstalk includes FRET donor emission leaking into the FRET acceptor detection band and direct excitation of the FRET acceptor by the nominal FRET donor excitation light. Measure donor-only, acceptor-only and unlabeled preparations with the same fluorescence microscope settings to separate these contributions and background.^[ref]

An explicit linear measurement model makes the controls' purpose clear. Let $D$ be background-subtracted donor-band counts under donor excitation, $A$ acceptor-band counts under acceptor excitation, and $F$ acceptor-band counts under donor excitation. Donor-only and acceptor-only preparations determine leakage coefficients $a$ and $b$. Thenis a corrected FRET sensitized emission signal under the assumed linear, fixed optical response. In a synthetic example $D=100$, $A=80$, $a=0.10$, $b=0.05$ and $F=34$ give $S=20$. Even a nontransferring sample would give $F=14$ from these two contaminants. Correcting that signal still does not convert $S$ into FRET efficiency without a calibration of excitation, yields and detection.

A biological comparison also needs a pairing-negative preparation and a known transferring positive preparation. Match label abundance and localization where possible; changing concentration can change encounter frequency as well as brightness. Check that attaching the labels preserves the protein function being investigated. Otherwise the measurement can faithfully report a perturbation introduced by the labels themselves.

FRET sensitized emission is FRET acceptor fluorescence caused by transfer from an excited FRET donor. Excite predominantly the FRET donor and collect both emission bands. Reduced FRET donor emission and increased FRET acceptor emission are useful together, but an acceptor-to-donor emission ratio is not automatically the FRET efficiency: brightness, label abundance and channel collection also enter.^[ref]

For n identical independent available FRET acceptors, transfer rates add. With single-pair FRET efficiency E, the donor exit probability is $E_n=nE/[1+(n-1)E]$. It is not nE; competition with the donor's other decay routes must be retained.

For FRET efficiency $E(R)=1/[1+(R/R_0{)}^6]$ and distinct positive reporter distances A and B, the absolute contrast is maximized at the geometric mean Förster distance $R_0=\sqrt{AB}$. Substitution of $x=R_0^6$ makes the derivative have the sign of $A^6{B}^6-x^2$. The maximum is $|B^3-A^3|/(B^3+A^3)$ and the two efficiencies then sum to one. If $A=B$, every choice gives zero contrast and there is no unique optimum.

Let $k_D=k_r+k_{nr}$, so the lifetime without transfer is ${\tau }_D=1/k_D$. For a fixed isolated donor–acceptor pair, the specified transfer rate is $k_T=k_D(R_0/R{)}^6$. Inserting this rate into the definition of FRET efficiency givesThe donor population in the presence of the acceptor obeys ${\dot{N}}_D=-(k_D+k_T)N_D$, so its fluorescence decays exponentially with lifetime ${\tau }_{DA}=1/(k_D+k_T)$. ThereforeBoth relations assume the donor's nontransfer rates are unchanged by adding the acceptor, a single approximately fixed separation during each excited-state lifetime, and an available acceptor. The transfer law also presumes the usual weak-coupling dipole approximation and the spectral and orientation conditions contained in $R_0$. If the reference donor is differently quenched, the lifetime difference cannot all be attributed to transfer. Heterogeneous mixtures or a fluctuating separation can produce multiexponential decay, in which case blindly substituting an arbitrary fitted average lifetime into the single-pair formula need not give the ensemble's mean efficiency.

Let $k_D$ be the donor's total decay rate without acceptors, including radiative and nonradiative decay, and let $k_T$ be its transfer rate to one acceptor. In weak excitation, independent equivalent acceptors supply additive transfer rates, so FRET efficiency isEliminating $k_T/k_D=E/(1-E)$ givesThis assumes the same Förster distance, equivalent donor–acceptor distances and unchanged donor decay, with no acceptor saturation or other transfer pathways affecting donor de-excitation.

For Förster resonance energy transfer, $k_T/k_D=(R_0/R{)}^6$, henceHere the single-pair FRET efficiency would be $E=0.60$, which indeed produces $E_2=0.75$.