The quadratic formula gives two algebraic possibilities. To select one consistent holomorphic square root, setwhere each factor is the principal square root. Both arguments lie in the upper half-plane, so neither crosses its branch cut. This choice satisfies $S(\zeta {)}^2={\zeta }^2-4$ and $S(\zeta )\sim \zeta$ at large $\zeta$ in the upper half-plane. The required inverse function isFor $\zeta =iR$, $R>0$, this gives $z=i(R+\sqrt{R^2+4})/2$, whose modulus exceeds one. As $\zeta$ varies continuously in the upper half-plane, this root cannot cross $|z|=1$, since that would make $\zeta$ real. Hence the selected root always lies in the fluid region. The other root is its reciprocal and lies inside the unit disk. Simply writing the principal square root of ${\zeta }^2-4$ would not specify this branch correctly throughout the upper half-plane.

For $z=x+iy$,Both factors are positive in the fluid region, so its image is contained in the upper half-plane. The boundary correspondence is also explicit: on the upper unit semicircle, $z=e^{i\theta }$ gives $g(z)=2\cos \theta \in [-2,2]$, while the real rays $x>1$ and $x<-1$ map onto the rays $\zeta >2$ and $\zeta <-2$.

To prove the image is the whole upper half-plane, fix any $\mathrm{Im}\zeta >0$. The equation $g(z)=\zeta$ isIts roots have product one. Neither root can have unit modulus, since that would make $\zeta =z+1/z$ real. Exactly one root therefore has modulus greater than one. For that root the formula for $\mathrm{Im}g$ forces $y>0$, so it belongs to the fluid region. Moreover, if $g(z_1)=g(z_2)$, thenTwo distinct roots in the fluid region would require $z_1{z}_2=1$, impossible when both moduli exceed one. Thus the Joukowski map of the upper exterior of a unit disk is a bijection from the fluid region onto the upper half-plane.

The Joukowski map is a holomorphic function for $z\ne 0$, withA holomorphic function is a local conformal map wherever its derivative is nonzero. Hence the two finite nonzero critical points areThey lie on the boundary of the fluid region. At $z=0$ the displayed formula has a pole and is undefined as a map to the finite complex plane; zero is outside the fluid region. Thus $g$ is locally conformal everywhere in the fluid.

Since $b>1$, the denominator inis positive. The positive root of $U{b}^2-b-U=0$ isThe other root is negative, so the horizontal force points in the positive $x$ direction for $1<b<b_{\ast }$, vanishes at $b=b_{\ast }$, and points in the negative $x$ direction for $b>b_{\ast }$. In particular, the nearby line source pulls the cylinder towards itself as $b\downarrow 1$, while at large $b$ the horizontal force is $F_x\sim -2\pi \rho U/b^2$.

Let $F_x,F_y$ denote the force per unit length exerted by the fluid on the cylinder, and let $\rho$ be the constant mass density. For a counterclockwise contour $C$ around the unit circle, the Blasius force theorem givesAn efficient use of the residue theorem finds the sum of the enclosed residues by subtracting the external pole at $b$ from the sum of all finite residues. At large $z$,The sum of all finite residues of $q^2$ is therefore $2U$. At $b$, write $q=(z-b{)}^{-1}+H(z)$, with $H$ a holomorphic function near $b$. ThenThusUsing $a=1/b$ and $1/(b-a)-1/b=1/[b(b^2-1)]$ givesThe vanishing vertical force also follows from reflection symmetry about the real axis. This force on a unit cylinder due to a uniform flow and an external source uses the counterclockwise orientation in the Blasius force theorem to fix the sign of the horizontal force.

Use $w=\varphi +i\psi$ for the complex potential of a two-dimensional potential flow, so its complex velocity is $q=w^{\prime }=u-iv$. The incident uniform flow and line source haveThe coefficient of the complex logarithm is one because a line source of strength $Q$ has complex potential of a two-dimensional potential flow $(Q/(2\pi ))\log (z-b)$, and here $Q=2\pi$. Applying the Milne-Thomson circle theorem to the unit circle givesSince $1/z-b=-b(z-1/b)/z$, locally consistent branches of the complex logarithm giveChanges of branch only alter the constant in a local expression; $w^{\prime }$ is single valued away from its singularities. The added terms represent a two-dimensional potential doublet at the origin, an image line source at $1/b$, and a compensating line sink at the origin. The image line source and line sink have equal and opposite strengths, so no net volume flux enters the cylinder.

On $|z|=1$, the Milne-Thomson circle theorem makes $w=f(z)+\overline{f(z)}$ up to a constant. Consequently the streamfunction is constant there, establishing the no-penetration boundary condition. All line source strengths are real, and the two-dimensional potential doublet has zero circulation, so the circulation around the cylinder is zero. There is no additional line vortex term.

Adopt the streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$. Integrating the first component of the velocity field in $y$ gives $\psi =-xy+\frac{1}{2}\Omega e^{-2t}{y}^2+f(x,t)$. The second component then requires $f_x=\Omega e^{2t}x$. Thus, omitting an arbitrary function of time,At a fixed instant, the streamlines are the connected curves $\psi =K$. To classify this quadratic form, introduce the invertible scaled and rotated coordinatesThenFor $\Omega >1$, both coefficients are positive. Every $K>0$ therefore gives an ellipse, and $K=0$ is the stagnation point at the origin. For $0<\Omega <1$, the quadratic form is an indefinite quadratic form: each $K\ne 0$ gives a hyperbola. The level $K=0$ consists of the straight separatriceswhose rays approach or leave the origin; the origin itself is a stagnation point. At the transition $\Omega =1$, $\psi =d^2$, so the nonstationary streamlines are parallel straight lines and $d=0$ is a line of stagnation points. These are instantaneous streamlines; the time dependence means that they need not coincide with pathlines.

For a planar incompressible flow, the divergence of the velocity field must vanish. HereEquivalently, the Jacobian determinant of the deformation map isso every material area is preserved. However, the out-of-plane vorticity isSince $\Omega >0$, this vorticity never vanishes. The motion is an incompressible flow but is not an irrotational flow; its local spin is clockwise.

The velocity of a fluid element is the time derivative of its position with its Lagrangian coordinates $X,Y$ held fixed. Direct differentiation givesReplacing the Lagrangian coordinates by the current Eulerian coordinates therefore gives the velocity fieldThe distinction between following a fluid element and differentiating at fixed Eulerian coordinates is what connects the given deformation map to this velocity field.

The orientation-dependent results in the official notes show that the measured spring constant is not a Lorentz scalar. A Lorentz transformation changes both the simultaneous extension and the measured three-force; their ratio depends on the orientation. This does not make the spring intrinsically anisotropic in its own rest frame, or require a preferred inertial frame. It is a quasistatic effective stiffness comparison between frames; an extended real spring also has finite internal signal speeds.

Covered by the official notes: spring constant orthogonal to the boost (PDF p. 13; handwritten p. 8).^[ref]

Covered by the official notes: spring constant parallel to the boost (PDF p. 13; handwritten p. 8).^[ref]

Covered by the official notes: transverse three-force transformation (PDF p. 13; handwritten p. 8).^[ref]

Covered by the official notes: longitudinal three-force transformation (PDF p. 13; handwritten p. 8).^[ref]

Covered by the official notes: definition of a pure force via constant rest mass (PDF p. 13; handwritten p. 8).^[ref]

The page labels $d{p}^{\mu }/d\tau$ as a proper force; that is the four-force. The requested proper force is the three-force measured in the instantaneous rest frame of the object. For a pure force its magnitude is $m_0$ times the proper acceleration, and in that frame the four-force has components $(0,{\mathbf{f}}_0)$.

Covered by the official notes: transverse-emission neutrino energy and laboratory angle (PDF p. 12; handwritten p. 7).^[ref]

Covered by the official notes: forward neutrino energy for the specified kaon beam (PDF p. 12; handwritten p. 7).^[ref]

Covered by the official notes: transverse-emission laboratory energy and laboratory angle (PDF p. 11; handwritten p. 6).^[ref]

Covered by the official notes: forward collinear laboratory energy (PDF p. 11; handwritten p. 6).^[ref]

Covered by the official notes: launch-pad speed as a function of launch-pad time (PDF pp. 8; handwritten pp. 3).^[ref]