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diff --git a/content/know/concept/euler-bernoulli-law/index.pdc b/content/know/concept/euler-bernoulli-law/index.pdc new file mode 100644 index 0000000..e9d3fdd --- /dev/null +++ b/content/know/concept/euler-bernoulli-law/index.pdc @@ -0,0 +1,314 @@ +--- +title: "Euler-Bernoulli law" +firstLetter: "E" +publishDate: 2021-06-03 +categories: +- Physics + +date: 2021-05-14T09:15:30+02:00 +draft: false +markup: pandoc +--- + +# Euler-Bernoulli law + +**Euler-Bernoulli beam theory** concerns itself with the bending of beams +(e.g. the metal beams used in large buildings), +subject to certain simplifying assumptions, +which are generally valid for beams that are narrow, +i.e. longitudinally much larger than transversely. + +Consider a beam of length $L$, placed upright +on the $z = 0$ plane, above the origin. +If we pull the top of this beam in the postive $y$-direction, +we assume that it bends uniformly, +i.e. with constant radius of [curvature](/know/concept/curvature/) $R$. +We also assume that the bending is **shear-free**: +if we treat the beam as a bundle of elastic strings, +then there is no friction between them. + +The central string has its length unchanged (i.e. still $L$), +while an arbitrary non-central string is extended or compressed to $L'$. +The [Cauchy strain tensor](/know/concept/cauchy-strain-tensor/) element $u_{zz}$ is then: + +$$\begin{aligned} + u_{zz} + = \frac{L' - L}{L} +\end{aligned}$$ + +Because the bending is uniform, the central string +is an arc with radius $R$ and central angle $\theta$, +where $L = \theta R$. +The non-central string has $L' = \theta R'$, +where $R'$ is geometrically shown to be $R' = R - y$, +with $y$ being the $y$-coordinate of that string at the beam's base. +So: + +$$\begin{aligned} + u_{zz} + = \frac{R' - R}{R} + = - \frac{y}{R} +\end{aligned}$$ + +By assumption, there are no shear stresses +and no forces acting on the beam's sides, +so the only nonzero component of the +[Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$ +is $\sigma_{zz}$, given by [Hooke's law](/know/concept/hookes-law/): + +$$\begin{aligned} + \sigma_{zz} = E u_{zz} +\end{aligned}$$ + +Where $E$ is the elastic modulus of the material. +By Hooke's inverse law, +the other nonzero strain components are as follows, +where $\nu$ is Poisson's ratio: + +$$\begin{aligned} + u_{xx} + = u_{yy} + = - \frac{\nu}{E} \sigma_{zz} + = - \nu u_{zz} + = \nu \frac{y}{R} +\end{aligned}$$ + +For completeness, we turn the strain tensor $\hat{u}$ +into a full displacement field $\va{u}$: + +$$\begin{aligned} + \boxed{ + u_x = \nu \frac{x y}{R} + \qquad + u_y = \frac{z^2}{2 R} + \nu \frac{y^2 - x^2}{2 R} + \qquad + u_z = - \frac{y z}{R} + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-field"/> +<label for="proof-field">Proof</label> +<div class="hidden"> +<label for="proof-field">Proof.</label> +By integrating the above strains $u_{ii} = \pdv*{u_i}{i}$, +we get the components of $\va{u}$: + +$$\begin{aligned} + u_x + = \nu \frac{x y}{R} + f_x(y, z) + \qquad + u_y + = \nu \frac{y^2}{2 R} + f_y(x, z) + \qquad + u_z + = - \frac{y z}{R} + f_z(x, y) +\end{aligned}$$ + +Where $f_x$, $f_y$ and $f_z$ are integration constants, +which we find by demanding that the off-diagonal strains $u_{ij}$ are zero. +Starting with $u_{xz} = 0$: + +$$\begin{aligned} + 0 + = u_{xz} + = \frac{1}{2} \Big( \pdv{u_x}{z} + \pdv{u_z}{x} \Big) + = \frac{1}{2} \Big( \pdv{f_x}{z} + \pdv{f_z}{x} \Big) +\end{aligned}$$ + +Here, only $f_x$ may depend on $z$, +and only $f_z$ may depend on $x$. +This equation thus tell us: + +$$\begin{aligned} + f_x(y, z) + = z \: g(y) + \qquad \quad + f_z(x, y) + = - x \: g(y) +\end{aligned}$$ + +Where $g(y)$ is an unknown integration constant. +Moving on to $u_{xy} = 0$: + +$$\begin{aligned} + 0 + = \frac{1}{2} \Big( \pdv{u_x}{y} + \pdv{u_y}{x} \Big) + = \frac{1}{2} \Big( \nu \frac{x}{R} + \pdv{f_x}{y} + \pdv{f_y}{x} \Big) +\end{aligned}$$ + +Only $f_x$ may contain $y$, +so its $y$-derivative must be a constant, +so $g(y) = C y$. Therefore: + +$$\begin{aligned} + f_x(y, z) + = C y z + \qquad + f_y(x, z) + = - \nu \frac{x^2}{2 R} - C x z + h(z) + \qquad + f_z(x, y) + = - C x y +\end{aligned}$$ + +Where $h(z)$ is an unknown integration constant. +Finally, we put everything in $u_{yz} = 0$: + +$$\begin{aligned} + 0 + = \frac{1}{2} \Big( \pdv{u_y}{z} + \pdv{u_z}{y} \Big) + = \frac{1}{2} \Big( \pdv{f_y}{z} - \frac{z}{R} + \pdv{f_z}{y} \Big) + = \frac{1}{2} \Big( \!-\! 2 C x + \dv{h}{z} - \frac{z}{R} \Big) +\end{aligned}$$ + +Only the first term contains $x$, so to satisfy this equation, we must set $C = 0$. +The remaining terms then tell us that $h(z) = z^2 / (2 R)$. +Therefore: + +$$\begin{aligned} + f_x = 0 + \qquad + f_y = - \nu \frac{x^2}{2 R} + \frac{z^2}{2 R} + \qquad + f_z = 0 +\end{aligned}$$ + +Inserting this into the components $u_x$, $u_y$ and $u_z$ +then yields the full displacement field. +</div> +</div> + +In any case, the beam experiences a bending torque with an $x$-component $T_x$ given by: + +$$\begin{aligned} + T_x + = - \int_A y \sigma_{zz} \dd{A} + = - \frac{E}{R} \int_A y^2 \dd{A} +\end{aligned}$$ + +Where $A$ is the cross-section. +Th above integral is known as the **area moment**, +and is typically abbreviated by $I$. +This brings us to the **Euler-Bernoulli law**: + +$$\begin{aligned} + \boxed{ + T_x + = - \frac{E I}{R} + } + \qquad \quad + I + \equiv \int_A y^2 \dd{A} +\end{aligned}$$ + +The product $E I$ is called the **flexural rigidity**, +i.e. the beam's "stiffness". +For a small deformation, i.e. a large radius of curvature $R$, +the law can be approximated by: + +$$\begin{aligned} + T_x + \approx - E I \dv[2]{y}{z} +\end{aligned}$$ + + + +## Slender rods + +A beam that is very thin in the transverse directions ($x$ and $y$ in this case), +can be approximated as a single string or rod $y(z)$. +Each infinitesimal piece $(\dd{y}, \dd{z})$ of the rod +exerts forces $F_y$ and $F_z$ on the next piece, +and is feels external forces-per-length $K_y$ and $K_z$, e.g. gravity. +In order to have equilibrium, the total force must be zero: + +$$\begin{aligned} + 0 + &= F_y(z + \dd{z}) - F_y(z) + K_y(z) \dd{z} + \\ + 0 + &= F_z(z + \dd{z}) - F_z(z) + K_z(z) \dd{z} +\end{aligned}$$ + +Rearranging these relations yields these equations for the internal forces $F_y$ and $F_z$: + +$$\begin{aligned} + \boxed{ + \dv{F_y}{z} + = - K_y + } + \qquad \quad + \boxed{ + \dv{F_z}{z} + = - K_z + } +\end{aligned}$$ + +Meanwhile, the rod also feels a torque with $x$-component $T_x$, +where equilibrium entails: + +$$\begin{aligned} + 0 + = T_x(z + \dd{z}) - T_x(z) + F_z(z) \dd{y} - F_y(z) \dd{z} +\end{aligned}$$ + +This can be rearranged to get a differential equation for $T_x$, namely: + +$$\begin{aligned} + \boxed{ + \dv{T_x}{z} + = F_y - F_z \dv{y}{z} + } +\end{aligned}$$ + +If $F_z$ and $\dv*{y}{z}$ are small, the last term can be dropped. +These equations are widely applicable, +but there is one especially important application, +so much so that it is usually what is meant by "Euler-Bernoulli law": +the shape of a laterally loaded rod. + +Consider a beam along the $z$-axis, carrying a lateral load $K_y$, +e.g. its own weight $A g \rho$ or more. +Assuming there is no other load $K_z = 0$ +and $F_y \ll F_z$, the above equations become: + +$$\begin{aligned} + T_x + = - E I \dv[2]{y}{z} + \qquad + \dv{T_x}{z} + = F_y + \qquad + \dv{F_y}{z} + = - K_y +\end{aligned}$$ + +Which we can simply substitute into each other, +eventually leading to: + +$$\begin{aligned} + \boxed{ + K_y + = \dv[2]{z} \Big( E I \dv[2]{y}{z} \Big) + } +\end{aligned}$$ + +This is often referred to as the **Euler-Bernoulli law** as well. +Typically the flexural rigidity $EI$ is a constant in $z$, +in which case we can reduce the equation to: + +$$\begin{aligned} + K_y + = E I \dv[4]{y}{z} +\end{aligned}$$ + +Which is clearly solved by a fourth-order polynomial, +given some boundary conditions. + + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. diff --git a/content/know/concept/impulse-response/index.pdc b/content/know/concept/impulse-response/index.pdc index b055fe7..fa921fa 100644 --- a/content/know/concept/impulse-response/index.pdc +++ b/content/know/concept/impulse-response/index.pdc @@ -35,9 +35,14 @@ $$\begin{aligned} } \end{aligned}$$ -*__Proof.__ Starting from the definition of $u_p(t)$, +<div class="accordion"> +<input type="checkbox" id="proof-main"/> +<label for="proof-main">Proof</label> +<div class="hidden"> +<label for="proof-main">Proof.</label> +Starting from the definition of $u_p(t)$, we shift the argument by some constant $\tau$, -and multiply both sides by the constant $f(\tau)$:* +and multiply both sides by the constant $f(\tau)$: $$\begin{aligned} \hat{L} \{ u_p(t - \tau) \} &= \delta(t - \tau) @@ -45,8 +50,8 @@ $$\begin{aligned} \hat{L} \{ f(\tau) \: u_p(t - \tau) \} &= f(\tau) \: \delta(t - \tau) \end{aligned}$$ -*Where $f(\tau)$ can be moved inside using the -linearity of $\hat{L}$. Integrating over $\tau$ then gives us:* +Where $f(\tau)$ can be moved inside using the +linearity of $\hat{L}$. Integrating over $\tau$ then gives us: $$\begin{aligned} \int_0^\infty \hat{L} \{ f(\tau) \: u_p(t - \tau) \} \dd{\tau} @@ -54,14 +59,14 @@ $$\begin{aligned} = f(t) \end{aligned}$$ -*The integral and $\hat{L}$ are operators of different variables, so we reorder them:* +The integral and $\hat{L}$ are operators of different variables, so we reorder them: $$\begin{aligned} \hat{L} \int_0^\infty f(\tau) \: u_p(t - \tau) \dd{\tau} &= (f * u_p)(t) = \hat{L}\{ u(t) \} = f(t) \end{aligned}$$ - -*__Q.E.D.__* +</div> +</div> diff --git a/content/know/concept/lubrication-theory/index.pdc b/content/know/concept/lubrication-theory/index.pdc new file mode 100644 index 0000000..7641440 --- /dev/null +++ b/content/know/concept/lubrication-theory/index.pdc @@ -0,0 +1,223 @@ +--- +title: "Lubrication theory" +firstLetter: "L" +publishDate: 2021-06-03 +categories: +- Physics +- Fluid mechanics +- Fluid dynamics + +date: 2021-05-11T14:34:07+02:00 +draft: false +markup: pandoc +--- + +# Lubrication theory + +**Lubricants** are widely used +to reduce friction between two moving surfaces. +In fluid mechanics, **lubrication theory** +is the study of fluids that are tightly constrained in one dimension, +especially those in small gaps between moving surfaces. + +For simplicity, we limit ourselves to 2D +by assuming that everything is constant along the $z$-axis. +Consider a gap of width $d$ (along $y$) and length $L$ (along $x$), +where $d \ll L$, containing the fluid. +Outside the gap, the lubricant has a +[Reynolds number](/know/concept/reynolds-number/) $\mathrm{Re} \approx U L / \nu$. + +Inside the gap, the Reynolds number $\mathrm{Re}_\mathrm{gap}$ is different. +This is because advection will dominate along the $x$-axis (gap length), +and viscosity along the $y$-axis (gap width). +Therefore: + +$$\begin{aligned} + \mathrm{Re}_\mathrm{gap} + \approx \frac{|(\va{v} \cdot \nabla) \va{v}|}{|\nu \nabla^2 \va{v}|} + \approx \frac{U^2 / L}{\nu U / d^2} + \approx \frac{d^2}{L^2} \mathrm{Re} +\end{aligned}$$ + +If $d$ is small enough compared to $L$, +then $\mathrm{Re}_\mathrm{gap} \ll 1$. +More formally, we need $d \ll L / \sqrt{\mathrm{Re}}$, +so we are inside the boundary layer, +in the realm of the [Prandtl equations](/know/concept/prandtl-equations/). + +Let $\mathrm{Re}_\mathrm{gap} \ll 1$. +We are thus dealing with *Stokes flow*, in which case +the [Navier-Stokes equations](/know/concept/navier-stokes/equations/) +can be reduced to the following *Stokes equations*: + +$$\begin{aligned} + \pdv{p}{x} + = \eta \: \Big( \pdv[2]{v_x}{x} + \pdv[2]{v_x}{y} \Big) + \qquad \quad + \pdv{p}{y} + = \eta \: \Big( \pdv[2]{v_y}{x} + \pdv[2]{v_y}{y} \Big) +\end{aligned}$$ + +Let the $y = 0$ plane be an infinite flat surface, +sliding in the positive $x$-direction at a constant velocity $U$. +On the other side of the gap, +an arbitrary surface is described by $h(x)$. + +Since the gap is so narrow, +and the surfaces' movements cause large shear stresses inside, +$v_y$ is negligible compared to $v_x$. +Furthermore, because the gap is so long, +we assume that $\pdv*{v_x}{x}$ is negligible compared to $\pdv*{v_x}{y}$. +This reduces the Stokes equations to: + +$$\begin{aligned} + \pdv{p}{x} + = \eta \pdv[2]{v_x}{y} + \qquad \quad + \pdv{p}{y} + = 0 +\end{aligned}$$ + +This result could also be derived from the Prandtl equations. +In any case, it tells us that $p$ only depends on $x$, +allowing us to integrate the former equation: + +$$\begin{aligned} + v_x + = \frac{p'}{2 \eta} y^2 + C_1 y + C_2 +\end{aligned}$$ + +Where $C_1$ and $C_2$ are integration constants. +At $y = 0$, the viscous *no-slip* condition demands that $v_x = U$, so $C_2 = U$. +Likewise, at $y = h(x)$, we need $v_x = 0$, leading us to: + +$$\begin{aligned} + v_x + = \frac{p'}{2 \eta} y^2 - \Big( \frac{p'}{2 \eta} h + \frac{U}{h} \Big) y + U +\end{aligned}$$ + +The moving bottom surface drags fluid in the $x$-direction +at a volumetric rate $Q$, given by: + +$$\begin{aligned} + Q + = \int_0^{h(x)} v_x(x, y) \dd{y} + = \bigg[ \frac{p'}{6 \eta} y^3 - \frac{p'}{4 \eta} h y^2 - \frac{U}{2 h} y^2 + U y \bigg]_0^{h} + = - \frac{p'}{12 \eta} h^3 + \frac{U}{2} h +\end{aligned}$$ + +Assuming that the lubricant is incompressible, +meaning that the same volume of fluid must be leaving a point as is entering it. +In other words, $Q$ is independent of $x$, +which allows us to write $p'(x)$ in terms of +measurable constants and the known function $h(x)$: + +$$\begin{aligned} + \boxed{ + p' + = 6 \eta \: \Big( \frac{U}{h^2} - \frac{2 Q}{h^3} \Big) + } +\end{aligned}$$ + +Then we insert this into our earlier expression for $v_x$, yielding: + +$$\begin{aligned} + v_x + &= 3 y (y - h) \Big( \frac{U}{h^2} - \frac{2 Q}{h^3} \Big) - \frac{U h}{h^2} y + \frac{U h^2}{h^2} + %\\ + %&= U \frac{3 y (y - h) - h y + h^2}{h^2} - Q \frac{6 y (y - h)}{h^3} +\end{aligned}$$ + +Which, after some rearranging, can be written in the following form: + +$$\begin{aligned} + \boxed{ + v_x + = U \frac{(3 y - h) (y - h)}{h^2} - Q \frac{6 y (y - h)}{h^3} + } +\end{aligned}$$ + +With this, we can find $v_y$ by exploiting incompressibility, +i.e. the continuity equation states: + +$$\begin{aligned} + \pdv{v_y}{y} + = - \pdv{v_x}{x} + = - 2 h' \frac{U h - 3 Q}{h^4} \big( 2 h y - 3 y^2 \big) +\end{aligned}$$ + +Integrating with respect to $y$ thus leads to the following transverse velocity $v_y$: + +$$\begin{aligned} + \boxed{ + v_y + = - 2 h' \frac{U h - 3 Q}{h^4} y^2 (h - y) + } +\end{aligned}$$ + +Typically, the lubricant is not in a preexisting pressure differential, +i.e it is not getting pumped through the system. +Although the pressure gradient $p'$ need not be zero, +we therefore expect that its integral vanishes: + +$$\begin{aligned} + 0 + = \int_L p'(x) \dd{x} + = 6 \eta U \int_L \frac{1}{h(x)^2} \dd{x} - 12 \eta Q \int_L \frac{1}{h(x)^3} \dd{x} +\end{aligned}$$ + +Isolating this for $Q$, and defining $q$ as below, yields a simple equation: + +$$\begin{aligned} + Q + = \frac{1}{2} U q + \qquad \quad + q + \equiv \frac{\int_L h^{-2} \dd{x}}{\int_L h^{-3} \dd{x}} +\end{aligned}$$ + +We substitute this into $v_x$ and rearrange to get an interesting expression: + +$$\begin{aligned} + v_x + &= U \frac{3 y^2 - h y - 3 h y + h^2}{h^2} - U q \frac{3 y^2 - 3 h y}{h^3} + \\ + &= U \Big( 1 - \frac{y}{h} \Big) \Big( 1 - \frac{3 y (h - q)}{h^2} \Big) +\end{aligned}$$ + +The first factor is always positive, +but the second can be negative, +if for some $y$-values: + +$$\begin{aligned} + h^2 < 3 y (h - q) + \quad \implies \quad + y > \frac{h^2}{3 (h - q)} +\end{aligned}$$ + +Since $h > y$, such $y$-values will only exist +if $h$ is larger than some threshold: + +$$\begin{aligned} + 3 (h - q) > h + \quad \implies \quad + h > \frac{3}{2} q +\end{aligned}$$ + +If this condition is satisfied, +there will be some flow reversal: +rather than just getting dragged by the shearing motion, +the lubricant instead "rolls" inside the gap. +This is confirmed by $v_y$: + +$$\begin{aligned} + v_y + = - U h' \frac{2 h - 3 q}{h^4} y^2 (h - y) +\end{aligned}$$ + + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. diff --git a/content/know/concept/selection-rules/index.pdc b/content/know/concept/selection-rules/index.pdc index 5da97e7..22dfd64 100644 --- a/content/know/concept/selection-rules/index.pdc +++ b/content/know/concept/selection-rules/index.pdc @@ -20,7 +20,7 @@ the total angular momentum and its $z$-component: $$\begin{aligned} \matrixel{f}{\hat{O}}{i} - = \matrixel{n_f \ell_f m_f}{\hat{O}}{n_i \ell_i m_i} + = \matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} \end{aligned}$$ Where $\hat{O}$ is an operator, $\ket{i}$ is an initial state, and @@ -28,8 +28,6 @@ $\ket{f}$ is a final state (usually at least; $\ket{i}$ and $\ket{f}$ can be any states). **Selection rules** are requirements on the relations between $\ell_i$, $\ell_f$, $m_i$ and $m_f$, which, if not met, guarantee that the above matrix element is zero. -Note that $n_f$ and $n_i$ typically do not matter in this context, -so they will be omitted from now on. ## Parity rules @@ -478,11 +476,180 @@ $$\begin{aligned} </div> +## Rotational rules + +Given a general (pseudo)scalar operator $\hat{s}$, +which, by nature, must satisfy the +following relations with the angular momentum operators: + +$$\begin{aligned} + \comm*{\hat{L}^2}{\hat{s}} = 0 + \qquad + \comm*{\hat{L}_z}{\hat{s}} = 0 + \qquad + \comm*{\hat{L}_{\pm}}{\hat{s}} = 0 +\end{aligned}$$ + +Where $\hat{L}_\pm \equiv \hat{L}_x \pm i \hat{L}_y$. +The inner product of any such $\hat{s}$ must obey these selection rules: + +$$\begin{aligned} + \boxed{ + \Delta \ell = 0 + } + \qquad \quad + \boxed{ + \Delta m = 0 + } +\end{aligned}$$ + +It is common to write this in the following more complete way, where +$\matrixel{\ell_f}{|\hat{s}|}{\ell_i}$ is the **reduced matrix element**, +which is identical to $\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}$, but +with a different notation to say that it does not depend on $m_f$ or $m_i$: + +$$\begin{aligned} + \boxed{ + \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} + = \delta_{\ell_f \ell_i} \delta_{m_f m_i} \matrixel{\ell_f}{|\hat{s}|}{\ell_i} + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-rot-scalar"/> +<label for="proof-rot-scalar">Proof</label> +<div class="hidden"> +<label for="proof-rot-scalar">Proof.</label> +Firstly, we look at the commutator of $\hat{s}$ with the $z$-component $\hat{L}_z$: +$$\begin{aligned} + 0 + = \matrixel{\ell_f m_f}{\comm*{\hat{L}_z}{\hat{s}}}{\ell_i m_i} + &= \matrixel{\ell_f m_f}{\hat{L}_z \hat{s}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{s} \hat{L}_z}{\ell_i m_i} + \\ + &= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} +\end{aligned}$$ + +Which can only be true if $m_f \!-\! m_i = 0$, unless, +of course, $\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} = 0$ by itself. + +Secondly, we look at the commutator of $\hat{s}$ with the total angular momentum $\hat{L}^2$: + +$$\begin{aligned} + 0 + = \matrixel{\ell_f m_f}{\comm*{\hat{L}^2}{\hat{s}}}{\ell_i m_i} + &= \matrixel{\ell_f m_f}{\hat{L}^2 \hat{s}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{s} \hat{L}^2}{\ell_i m_i} + \\ + &= \hbar^2 \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big) \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} +\end{aligned}$$ + +Assuming $\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} \neq 0$, +this can only be satisfied if the following holds: + +$$\begin{aligned} + 0 + = \ell_f^2 + \ell_f - \ell_i^2 - \ell_i + = (\ell_f + \ell_i) (\ell_f - \ell_i) + (\ell_f - \ell_i) +\end{aligned}$$ + +If $\ell_f = \ell_i = 0$ this equation is trivially satisfied. +Otherwise, the only option is $\ell_f \!-\! \ell_i = 0$, +which is another part of the selection rule. + +Thirdly, we look at the commutator of $\hat{s}$ with the ladder operators $\hat{L}_\pm$: + +$$\begin{aligned} + 0 + = \matrixel{\ell_f m_f}{\comm*{\hat{L}_\pm}{\hat{s}}}{\ell_i m_i} + &= \matrixel{\ell_f m_f}{\hat{L}_\pm \hat{s}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{s} \hat{L}_\pm}{\ell_i m_i} + \\ + &= C_f \matrixel{\ell_f (m_f\!\mp\!1)}{\hat{s}}{\ell_i m_i} - C_i \matrixel{\ell_f m_f}{\hat{s}}{\ell_i (m_i\!\pm\!1)} +\end{aligned}$$ + +Where $C_f$ and $C_i$ are constants given below. +We already know that $\Delta \ell = 0$ and $\Delta m = 0$, +so the above matrix elements are only nonzero if $m_f = m_i \pm 1$. +Therefore: + +$$\begin{aligned} + C_i + &= \hbar \sqrt{\ell_i (\ell_i + 1) - m_i (m_i \pm 1)} + \\ + C_f + &= \hbar \sqrt{\ell_f (\ell_f \!+\! 1) - m_f (m_f \!\mp\! 1)} + \\ + &= \hbar \sqrt{\ell_f (\ell_f \!+\! 1) - (m_i \!\pm\! 1) (m_i \!\pm\! 1 \!\mp\! 1)} + \\ + &= \hbar \sqrt{\ell_f (\ell_f \!+\! 1) - m_i (m_i \!\pm\! 1)} +\end{aligned}$$ + +In other words, $C_f = C_i$. The above equation therefore reduces to: + +$$\begin{aligned} + \matrixel{\ell_f m_i}{\hat{s}}{\ell_i m_i} + &= \matrixel{\ell_f (m_i \!\pm\! 1)}{\hat{s}}{\ell_i (m_i\!\pm\!1)} +\end{aligned}$$ + +Which means that the value of the matrix element +does not depend on $m_i$ (or $m_f$) at all. +</div> +</div> + +Similarly, given a general (pseudo)vector operator $\vu{V}$, +which, by nature, must satisfy the following commutation relations, +where $\hat{V}_\pm \equiv \hat{V}_x \pm i \hat{V}_y$: + +$$\begin{gathered} + \comm*{\hat{L}_z}{\hat{V}_z} = 0 + \qquad + \comm*{\hat{L}_z}{\hat{V}_{\pm}} = \pm \hbar \hat{V}_{\pm} + \qquad + \comm*{\hat{L}_{\pm}}{\hat{V}_z} = \mp \hbar \hat{V}_{\pm} + \\ + \comm*{\hat{L}_{\pm}}{\hat{V}_{\pm}} = 0 + \qquad + \comm*{\hat{L}_{\pm}}{\hat{V}_{\mp}} = \pm 2 \hbar \hat{V}_z +\end{gathered}$$ + +The inner product of any such $\vu{V}$ must obey the following selection rules: + +$$\begin{aligned} + \boxed{ + \Delta \ell + = 0 \:\:\mathrm{or}\: \pm 1 + } + \qquad + \boxed{ + \Delta m + = 0 \:\:\mathrm{or}\: \pm 1 + } +\end{aligned}$$ + +In fact, the complete result involves the Clebsch-Gordan coefficients (from spin addition): + +$$\begin{gathered} + \boxed{ + \matrixel{\ell_f m_f}{\hat{V}_{z}}{\ell_i m_i} + = C^{\ell_i \: 1 \: \ell_f}_{m_i \: 0 \:m_f} \matrixel{\ell_f}{|\hat{V}|}{\ell_i} + } + \\ + \boxed{ + \matrixel{\ell_f m_f}{\hat{V}_{+}}{\ell_i m_i} + = - \sqrt{2} C^{\ell_i \: 1 \: \ell_f}_{m_i \: 1 \:m_f} \matrixel{\ell_f}{|\hat{V}|}{\ell_i} + } + \\ + \boxed{ + \matrixel{\ell_f m_f}{\hat{V}_{-}}{\ell_i m_i} + = \sqrt{2} C^{\ell_i \: 1 \: \ell_f}_{m_i \: -1 \:m_f} \matrixel{\ell_f}{|\hat{V}}{|\ell_i} + } +\end{gathered}$$ + + ## Superselection rule -Selection rules need not always be about atomic electron transitions. +Selection rules are not always about atomic electron transitions, or angular momenta even. + According to the **principle of indistinguishability**, -permutating identical particles never leads to an observable difference. +permuting identical particles never leads to an observable difference. In other words, the particles are fundamentally indistinguishable, so for any observable $\hat{O}$ and multi-particle state $\ket{\Psi}$, we can say: |