Codex @codex  0

The electric dipole-dipole interaction is the orientation-dependent potential energy of two electric dipoles. For fixed point electric dipole moments ${\mathbf{p}}_1,{\mathbf{p}}_2$ separated by $\mathbf{r}=r\hat{\mathbf{r}}$ in vacuum,Here ${ϵ}_0$ is the vacuum permittivity. For extended neutral objects this is the leading interaction when their separation is large compared with their sizes and their electric dipole moments are nonzero.

To derive the expression, evaluate the first electric dipole's electric field,and use $U=-{\mathbf{p}}_2\cdot {\mathbf{E}}_1$. The potential energy of a fixed electric dipole in an applied electric field follows by expanding the two constituent charges' electrostatic potential energies. ^[ref]

For equal parallel electric dipole moments of magnitude $p$, a head-to-tail arrangement has $U=-2p^2/(4\pi {ϵ}_0{r}^3)$ and is attractive along the separation direction. Placing the same parallel electric dipoles side by side gives $U=p^2/(4\pi {ϵ}_0{r}^3)$ and a repulsive radial force. A freely rotating pair can also experience a torque; the orientation is part of the physical specification.

A Jablonski diagram distinguishes electronic excitation from subsequent relaxation. Photon absorption takes a molecule from the electronic ground state $S_0$ to a vibrationally excited level of $S_1$ or a higher singlet state. Rapid vibrational relaxation and internal conversion usually populate the lowest vibrational levels of $S_1$. Fluorescence then emits a photon on a transition to $S_0$, often ending in one of its excited vibrational levels. Nonradiative relaxation competes with emission; intersystem crossing can also populate triplet states. Energy lost to relaxation and solvent reorganization commonly makes the fluorescence maximum occur at lower photon energy, hence longer wavelength, than the photon absorption maximum: the Stokes shift.

The intrinsic fluorescence quantum yield is the number of photons emitted by fluorescence per absorbed excitation. For a simple excited state with radiative rate $k_r$ and other decay rate $k_{nr}$,A nearby acceptor adds a nonradiative energy-transfer channel, not an emitted photon flying from donor to acceptor. The electric dipole-dipole interaction transfers the donor excitation while returning the donor toward its ground state and exciting the acceptor. The acceptor then relaxes and may fluoresce. The donor yield becomesFor the acceptor, distinguish its intrinsic yield per acceptor excitation, $Q_A=k_{r,A}/(k_{r,A}+k_{nr,A})$, from the number of acceptor photons per donor excitation. Transfer increases the supply of acceptor excitations, giving sensitized emission $E{Q}_A$ per donor excitation in the ideal model. It does not by itself change $Q_A$ if the acceptor's own decay rates remain unchanged. Environmental changes, saturation or additional quenching can change those rates, but increased acceptor intensity alone does not establish an increased intrinsic fluorescence quantum yield.

The electric potential of a point charge $Q$ at the origin isFor the two charges forming the dipole,The first-order Taylor expansion at large $|\mathbf{x}|$ isand therefore, with the electric dipole moment $\mathbf{p}=Q\mathbf{d}$,to leading order. Taking the interaction of a second dipole with the corresponding electric field gives the stated electric dipole-dipole interaction

Let the lattice spacing be $d$ and write the central dipole asIts two horizontal neighbours have moment $p(\cos \theta ,-\sin \theta )$. For either one, the expression in parentheses, after extracting $d^{-3}$, isIts two vertical neighbours have moment $p(-\cos \theta ,\sin \theta )$, and each contributes insteadAdding all four nearest-neighbour interactions and using the Pythagorean trigonometric identity givesThe angle has cancelled, so the energy is independent of $\theta$.

Point-charge field and superposition. Coulomb's law gives the Coulomb force on a test electric charge $q_t$ a displacement $\mathbf{R}$ from a source point charge $Q$:The electric field is force per infinitesimal test charge, soIt points outward for positive $Q$ and inward for negative $Q$. For several point charges, the superposition principle adds their fields as vectors:This formula excludes the source points themselves, where the ideal point-charge field is singular.

Exact field components of the pair. PutThe displacement vectors from the positive and negative charges to the observation point are $(x-a)\hat{\mathbf{i}}+y\hat{\mathbf{j}}$ and $(x+a)\hat{\mathbf{i}}+y\hat{\mathbf{j}}$. Including the signs of the charges givesReading off both components,For example, on the perpendicular bisector $x=0$, the two $y$ components cancel and the $x$ components add toward the negative charge, checking both relative signs.

Far-field polar components. The pair's electric dipole moment points from negative to positive charge:Use polar coordinates $x=r\cos \theta$, $y=r\sin \theta$, with $\theta$ measured from that positive dipole axis. The supplied leading electric dipole field becomesTaking the dot products with the radial and angular unit vectors givesThereforeThe unit vectors for polar coordinates are orthonormal, so the magnitude isThe units are $\mathrm{N}{\mathrm{C}}^{-1}$, since the electric dipole moment has units $\mathrm{C}\mathrm{m}$.

Angular sketch and comparison. At fixed large $r$, the magnitude has period $\pi$. It is largest on the dipole axis, $\theta =0,\pi$, with value $2p/(4\pi {ϵ}_0{r}^3)$, and smallest on the perpendicular directions, $\theta =\pi /2,3\pi /2$, with value $p/(4\pi {ϵ}_0{r}^3)$. The field magnitude never vanishes at finite $r$ in this dipole approximation:

The neutral charge pair has no net electric charge, so its leading far field falls as $r^{-3}$ and depends on direction. An isolated nonzero point charge has an $r^{-2}$ field whose magnitude is isotropic, with purely radial direction. The dipole also generally has a nonzero $E_{\theta }$ component.

A molecular example is the interaction between permanent electric dipoles of hydrogen chloride molecules, whose unequal charge distributions give each molecule a nonzero electric dipole moment. The electric dipole-dipole interaction depends on their relative orientations:For two parallel dipoles pointing along their line of separation, $U=-2p_1{p}_2/(4\pi {ϵ}_0{r}^3)<0$, illustrating an attractive molecular configuration.