The performance audio and video were transmitted to **Earth** via SpaceX’s **Starlink** satellite network and were accompanied by performances of Earth-based musical ensembles. As space travel expands, musicians, artists, and poets will increasingly travel and live beyond Earth.

Sarah Gillis plays a Star Wars theme aboard the Polaris Dawn SpaceX Dragon spacecraft during orbital night.

Gillis with violin during orbital day. Polaris Dawn experiences a sunset and sunrise about every 106 minutes.

Polaris Dawn’s Commander **Jared Isaacman**, Pilot **Scott “Kidd” Poteet**, Mission Specialist **Sarah Gillis**, and Mission Specialist and Medical Officer **Anna Menon** had already broken the 1966 **Gemini XI** altitude record for Earth-orbital missions (excluding Apollo **cis-lunar **missions). This morning, the crew vented the **SpaceX** **Dragon** *Resilience*‘s cabin, exposing all four of them to the vacuum of space, realizing the first four-person spacewalk, and eclipsing the 1992 **STS-49** mission, which included history’s only three-person spacewalk.

At orbital sunset and near orbital apogee, Isaacman then floated outside to the Dragon’s **skywalker** mobility aid. For about 12 minutes, he tested the new SpaceX **extravehicular activity** (EVA) space suit in a carefully practiced choreography. Gillis followed next during orbital night and repeated the same tests, becoming the youngest person to perform a spacewalk, younger even than the very first spacewalker, **Alexei Leonov**. Later today, in yet another milestone, the crew transmitted images via high-bandwidth **laser** light beams to SpaceX’s **Starlink** satellite network.

Excitedly and inspirationally, Gillis and Menon are the first SpaceX engineers to fly in space. They have been intimately involved in human spaceflight development and training and will soon return to Earth to leverage their experience and help accelerate SpaceX’s progress toward making life and consciousness **interplanetary**.

Commander Jared Isaacman tests the new SpaceX EVA suit at orbital sunset near apogee. All four astronauts participated in the spacewalk by being exposed to the vacuum of space.

From left to right, the Polaris Dawn crew is Sarah Gillis, Kidd Poteet, Jared Isaacman, and Anna Menon. Gillis and Menon are the first SpaceX engineers to fly in space

Because I signed a **Non Disclosure Agreement**, I cannot discuss the project details. I can say the work included developing an app in the **Swift** programming language for the **iPhone** operating system **iOS**.

Towards an engaging and intuitive** Graphical User Interface**, I created virtual 3D buttons that cast virtual shadows, depending on the **iPhone**‘s physical orientation, due to a virtual light source directly above the phone. When touched, the buttons appear to depress (and click or vibrate the phone, depending on the user’s preferences). The hole in the cog-wheel button really works — if you touch the hole, the button will not depress — because a virtual hole in a virtual cog wheel is real!

Here, I outline a partial solution to the problem based on the work of **Fritz Rohrlich**. It’s a long calculation, which I checked using **Mathematica**. Round parentheses (\ ) group terms, square brackets [\ ] enclose function arguments, and boxes \square enclose matrix elements.

- Model static homogeneous gravity G with Rohrlich spacetime metric.
- Transform to free-fall coordinates F where G is in hyperbolic motion
- Find hyperbolic geodesics of constant proper acceleration g, as in Fig. 1..
- Find electromagnetic field of charge q at rest in G as observed by F.
- Transform to find electromagnetic field as observed by G.

Figure 1: Charge q (blue) at rest in static homogeneous gravitational field (left) and in hyperbolic motion in a freely falling reference frame (right). Gridlines are null (light) lines, which suggest a coordinate singularity at z_G = 1/g, and dashed lines are hyperbola asymptotes. Black lines outline future light cones.

If g is the non-relativistic gravitational field magnitude, relativistic units implies constant light speed c = 1 with time scale T = c/g_E \approx 1~\text{year}, and length scale L = c^2/g_E \approx 1~\text{light-year}. Restricting spatial motion to the vertical z direction implies cylindrical symmetry, so use **cylindrical** spacetime coordinates

The **line element**

gives the **proper time **d\tau between nearby events. Model a free-fall** **spacetime using the **Minkowski metric tensor** with matrix representation

and line element

\text d\tau^2 = \text dt^2-\text d\rho^2-\rho^2\text d\phi^2-\text dz^2 = \text dt^2-\text d\vec x^2,where

\frac{1}{\gamma^2} = \left( \frac{\text d\tau}{\text dt} \right)^2 = 1-\left( \frac{\text d\vec x}{\text dt} \right)^2.Simply model gravity using the **Rohrlich metric tensor** with matrix representation

where

u[z] = \text{sech}\left[\sqrt{(1-g z)^2-1}\right] = 1 + g z + O[z^2].The Rohrlich spacetime is **static** (independent of time t), **homogeneous** (independent of horizontal coordinates x = \rho\cos\phi, y=\rho\sin\phi), and depends only on height z and parameter g.

Newtonian free fall \vec x[t] satisfies \text d^2\vec x / \text dt^2 = \vec 0. More generally, **geodesic motion **x^\alpha[\tau] satisfies

with implied sums over the repeated indices, where the **connection coefficients**

encode the rates of basis vector component changes with coordinate changes. Here, the solutions

\boxed{\begin{array}{c} t_G \\ \rho_G \\ \phi_G \\ z_G \end{array}} = \boxed{\begin{array}{c} \text{arctanh}[g\tau]/g \\ 0 \\ 0 \\ (1-\sqrt{1+\text{arctanh}[g\tau]^2})/g \end{array}}imply **hyperbolic motion**

where

z_G = \frac{1-\sqrt{1+g^2 t_G^2}}{g} = -\frac{1}{2} g t_G^2 + O[t_G^4],as designed.

By the **equivalence principle**, a freely falling test particle defines a locally flat or inertial system. Equating the free and gravity line elements

implies the transformation

x^\mu_F = \boxed{\begin{array}{c} t_F \\ \rho_F \\ \phi_F \\ z_F \end{array}} = \boxed{\begin{array}{c} \sinh[g t_G] u[z_G]/g \\ \rho_G \\ \phi_G \\ \cosh[g t_G] u[z_G] / g \end{array}}with Jacobian

\underset{F\leftarrow G}{J^\mu_\nu} = \frac{\partial x^\mu_F}{\partial x^\nu_G} = \boxed{\begin{array}{cccc} \cosh[gt_G]u[z_G] & 0 & 0 & \sinh[gt_G]u^\prime[z_G]/g \\ 0 & 1 & 0 & 0\\0 & 0 & 1 & 0 \\ \sinh[gt_G]u[z_G] & 0 & 0 &\cosh[gt_G]u^\prime[z_G]/g \end{array}}Thus, the charge q at rest in G at x_G = y_G = z_G = 0 moves in F along the hyperbola

z_F^2\ -\ t_F^2 = u[z_G]^2/ g^2 = u[0]^2/ g^2 = 1 / g^2,where

z_F = \frac{\sqrt{1+g^2 t_F^2}}{g} = \frac{1}{g}+\frac{1}{2} g t_F^2 + O[t_F^4].Generally, the **spacetime acceleration** A^\mu = d^2 x^\mu / d^2\tau^2 implies the **Lorentz-invariant**

for \ddot z >0, where the over-dots indicate differentiation with respect to time t. Here, the **proper acceleration** of q fixed in G

is constant, as expected. (Furthermore, 1 < \gamma_F implies \ddot z_F < g, and \ddot z_F \rightarrow 0 as \dot z_F \rightarrow c, so F never observes q superluminal.)

Generally, for a charge q in arbitrary motion x^\mu_q[\tau], the **Coulomb potential** \varphi = q / r generalizes to the **Liénard-Wiechert potential**

where the **spacetime velocity** U^\mu = dx^\mu_q / d\tau, the **spacetime displacement** R^\mu = x^\mu-x^\mu_q, and because electromagnetic “news” travels at light speed, the right side is evaluated at the earlier, **retarded time** t_r defined implicitly by

Let the cylindrical coordinates of the field point at time t be \{\rho, \phi, z\}, and define

\begin{aligned}\delta^2 &= -t^2 + \rho^2 + z^2 + g^{-2}, \\ \Delta^2 &= +t^2 + \rho^2-z^2 + g^{-2}, \\ d^2 &= +t^2-\rho^2-z^2 + g^{-2}, \\ \xi &= \sqrt{4 g^{-2} \rho^2 +d^2}.\end{aligned}Here, for the charge’s hyperbolic motion in the free-fall frame, explicitly solving for the retarded time is possible, and

t_F-t_r = \sqrt{\rho_F^2 +(z_F-\sqrt{g^{-2}+t_r^2}\,)^2}implies

t_r = \frac{t_F\delta_F^2-z_F\xi_F}{2(z_F^2-t_F^2)}.Evaluating the Liénard-Wiechert potential at this time gives

U^\mu_F = \frac{g}{2(z_F^2-t_F^2)}\boxed{\begin{array}{c} z_F \delta_F^2-t_F\xi_F \\ 0 \\ 0 \\ t_F \delta_F^2-z_F \xi_F \end{array}}and

R^\mu_F = \frac{1}{2(z_F^2-t_F^2)}\boxed{\begin{array}{c} z_F \xi_F-t_F\Delta_F^2 \\ 2\rho_F(z_F^2-t_F^2) \\ 0 \\ t_F \xi_F-z_F \Delta_F^2 \end{array}}and finally

\mathcal{A}^\mu_F = \frac{q}{\xi_F(z_F^2-t_F^2)}\boxed{\begin{array}{c} z_F \delta_F^2-t_F\xi_F \\ 0 \\ 0 \\ t_F \delta_F^2-z_F \xi_F^2 \end{array}}.Differentiating the **electromagnetic field tensor**

generates the **electric** and **magnetic field vectors**

for z_F> t_F, as in Fig. 2. The directional energy flux density or **Poynting vector**

Figure 2: Electric and (nonzero) magnetic fields in the free fall frame F for a charge q at rest in G and accelerating upward in F. Parameters are q=1, g = 0.2, and t_F=0.1.

The electromagnetic field transformed to the gravity frame is

\boxed{\begin{array}{cccc}0 & +\mathcal{E}_\rho & +\mathcal{E}_\phi & +\mathcal{E}_z \\-\mathcal{E}_\rho & 0 & -\mathcal{B}_z & +\mathcal{B}_\phi \\-\mathcal{E}_\phi & +\mathcal{B}_z & 0 & -\mathcal{B}_\rho \\-\mathcal{E}_z & -\mathcal{B}_\phi & +\mathcal{B}_\rho & 0 \end{array}}_G = \mathcal{F}_G^{\mu\nu} = J^{\mu}_\alpha J^{\nu}_\beta \mathcal{F}_F^{\alpha\beta},with implied sums over repeated indices. Hence,

\begin{aligned}\vec\mathcal{E}_G &= \frac{4 g^{-2}q}{\xi_G^3} \frac{\text{sech}\chi_G}{\chi_G} \boxed{\begin{array}{c}2\rho_G (z_G \cosh[g t_G]-t_G\sinh[g t_G])\chi_G \\ 0 \\ -\Delta_G^2 (1-g z_G)\text{sech}\chi_G \tanh\chi_G \end{array}}, \\ \vec\mathcal{B}_G &= \boxed{\begin{array}{c}0 \\ 0 \\ 0 \end{array}}, \end{aligned}where \chi = \sqrt{(1-g z)^2-1}, as in Fig. 3. As a check,

\lim_{g\rightarrow 0} \vec\mathcal{E}_G = \frac{q}{(z_G^2 + \rho_G^2)^{3/2}} \boxed{\begin{array}{c}\rho_G \\ 0 \\ z_G \end{array}} = \frac{q}{z_G^2 + \rho_G^2} \frac{\rho_G \hat\rho + z_G \hat z}{\sqrt{z_G^2 + \rho_G^2}} = \frac{q}{r_G^2} \hat r_G,as expected. The Poynting vector

\vec\mathcal{S}_G = \vec\mathcal{E}_G \times \vec\mathcal{B}_G = \boxed{\begin{array}{c} 0 \\ 0 \\ 0 \end{array}}.

Figure 3: Electric and (zero) magnetic fields in the gravity frame G for a charge q=1 at rest in G with g = 0.2.

Kip was right; the field lines bend (downward). But also, radiation is not a frame-invariant concept.

]]>For the last four years we were in contact for my science history project and this year, I finally visited him. We had a great day and I received MANY documents from conferences 40-50 years ago. He saved ‘everything’ in countless binders, sorted by years, and kept them in ceiling-heigh bookshelfs at his home.

In the 1970s, Otto Rössler designed a famous 3D flow that mimics the folding and bending of taffy in a taffy machine, which is now named the Rössler attractor. Described by the 3 coupled differential equations

\begin{aligned}\dot x &= -y - z, \\ \dot y &= x + a y,\\ \dot z &= b - c z + x z,\end{aligned}with just one nonlinear term, the x z in the third equation. The Rössler velocity field results from the interaction of two crossed vortices, one near the origin and the other far away pointing at the “fold” in the Rössler band. The parameter space \{a,b,c\} is large, but the system undergoes a period-doubling route to chaos in \{0<a<2, b=2, c=4\}, as in the animation below, which culminates in a period-3 window.

The animation was created by John Lindner.

]]>None of the buildings mentioned in publications still exist. And searching for original document is two archived did not provide any new information. Therefore, two impressions from Zürich: the beautiful Zürich train station and the street sign of the street Belousov supposedly lived in.

From Zürich, I traveled to Les Diablerets in the southwest of Switzerland to attend the Gordon Research Conference on *Oscillations and Dynamic Instabilities in Chemical Systems*. This is a beautiful location at 1200 m to interact with scientists from all over the world and to spend the free afternoons. One afternoon, I ‘ran’ up the switchback road to the Col de la Croix and took several hiking trails back to the village – to be back in time for a poster session.

After the conference, I walked towards the Lake Geneva to take the ‘train’ at the last possible ‘train station’ before the track winds into the next valley. It was a beautiful hike to relax the brain.

The article had been published in a journal of the Shevchenko Society in Lviv (Austria-Hungarian empire at that time) in, as I read everywhere, in Ruthenian language. All those years, I could not get a scan of the article, or the journal, not even from the Shevchenko Society archive in New York City.

In January, I contacted a physicist at the Ukrainian National Academy of Science, who recently published about the Shevchenko Society and asked him about Julian Hirniak and the journal. He immediately responded and shortly after, he sent me pictures of the article in question. He translated that article into English, I translated Hirniak’s German 1911 article into English and, together with another co-author, we submitted a manuscript in June.

When I mentioned that I will be in Europe this summer, he invited me to visit him in Lviv, Ukraine. It became the first trip during my nearly 6-week time in Europe.

Impression of Lviv, located close to the Polish border in Western Ukraine: surreal!

Despite the war in the Eastern/Southern part of the country, life is going on. Electricity outages are compensated by power generators outside every shop/restaurant. On Saturday evening, we saw Mozart’s Don Giovanni in Lviv’s Opera Theater.

We also visited the archive of the Shevchenko society – on a Saturday morning. The director Kostiantyn Kurylyshyn (Head of the Department at the Vasyl Stefanyk National Scientific Library of Ukraine in Lviv) invited us because he is, as he said, nevertheless there and could show us around. Finally, I visited the place, build in 1912, where ‘every’ Ukrainian publication can be found. This is similar to the Library of Congress in Washington DC. And I could finally see the original 1908 publication – and take my own picture.

For over **64 days** in 1958-1959 (!), **Robert Timm** and **John Cook** flew a modified **Cessna 172** above and around **Las Vegas**. Modifications included an extra fuel tank, a mattress, a small steel sink, and a camping toilet. The duo took turns piloting, and they refueled and resupplied every 12 hours by flying low and slow above a speeding truck.

Robert Timm (right) and John Cook (left) flying their modified Cessna 172 near Las Vegas, Nevada, 1958-1959. (Howard W. Cannon Aviation Museum)

Twice a day Timm and Cook refueled and resupplied from a fast truck. (Howard W. Cannon Aviation Museum)

Current ISS astronaut **Matthew Dominick** has been experimenting with photography, and his photo below, of a docked **SpaceX Dragon **taken from a docked **Boeing Starliner**, just after orbital sunset and just as **Earth’s moon **rises, *does* show stars from our **Milky Way **galaxy, with the spacecraft dimly illuminated by moonlight. Note the face in the Dragon window.

Erdős was one of the most prolific twentieth century mathematicians, publishing about 1500 articles with more than 500 coauthors. (Indeed, my **Erdős number**, or collaboration distance, is 5.) Reportedly, Erdős liked to talk about **The Book**, in which **God** maintains the perfect proofs for mathematical theorems. Like Aigner & Ziegler’s presentation in their **Proofs from The Book**, my “illustrated” version is a modest attempt at such a proof, a deep result proved by elementary means: bounding the **central binomial coefficient **(2n)! / (n!n!)** **above and below exposes the necessity of primes p for all n < p \le 2n. Enjoy!

In analogy with the **factorial function**

as the product of all positive integers m not greater than n, define the **primorial function**

as the product of all *primes* p not greater than n. Recall the **binomial** **function**

and the **floor function**

The primorial function is upper bounded by the exponential

x\# = \prod_{p\le x}p \le 4^{x-1} < 4^x.Proving this for the largest prime q < x is sufficient, as the substitution unchanges the left side and lowers the right side. For x = 2, the bound 2 < 4 is correct. For induction, assume it’s true for primes x < 2m and for x = 2m+1 split the product

(2m+1)\# = \prod_{p\le 2m+1} \hspace{-0.7em}p = \prod_{p\le m+1} \hspace{-0.5em}p\ \prod_{m+1 < p\le 2m+1} \hspace{-1.7em}p\hspace{1em}.The first factor is bounded by the induction hypothesis. For the second factor, consider the binomial expansion

2^{2m+1}=(1+1)^{2m+1}=\sum_{k=1}^{2m+1}\binom{2m+1}{k} =\cdots + \binom{2m+1}{m} + \binom{2m+1}{m+1} + \cdots \ge \binom{2m+1}{m} + \binom{2m+1}{m+1} = 2\binom{2m+1}{m},where the *two* central binomial coefficients are equal (as in Pascal’s triangle). But the integer

where the integer M>1, as the bounded primes p divide the numerator but not the denominator. Combine these results to get

(2m+1)\# \le4^m \cdot 2^{2m} = 4^{2m}as desired.

Consider the *one* central binomial coefficient

Since \lfloor n/p \rfloor factors of n! are divisible by p, and \lfloor n/p^2 \rfloor factors of n! are divisible by p^2, and so on, n! contains the prime p exactly \sum_{k\ge1} \lfloor n/p^k\rfloor times. Thus,

n!=\prod_p p^{\sum_k \lfloor n/p^k\rfloor}and

\binom{2n}{n}=\prod_p p^{\sum_k \left( \lfloor 2n/p^k\rfloor-2\lfloor n/p^k\rfloor \right) }.Since

x-1<\lfloor x \rfloor \le x,the integer summands difference

\left\lfloor \frac{2n}{p^k} \right\rfloor -2\left\lfloor \frac{n}{p^k}\right\rfloor < \frac{2n}{p^k}-2\left(\frac{n}{p^k}-1\right) = 2and thus must be either 0 or 1.

If p^k > 2n, \lfloor 2n / p^k \rfloor = 0 and \lfloor n / p^k \rfloor = 0 and no power of p divides (2n)!/(n!)^2. If p^k \le 2n, then the divisor’s highest power k \le \log 2n / \log p, but if p > \sqrt{2n}, then \log 2n / \log p < 2, and the power must be 0 or 1.

For n \ge 3, if 2n/3 < p \le n, then p \le n < 2p \le 2n < 3 p, which implies that (2n)! contains p and 2p and not 3p while n! contains p and not 2p, so the powers of p in (2n)!/(n!)^2 cancel.

Split the central binomial coefficient into products of successive ranges of primes and generously bound the factors from above by the previous results to get

\binom{2n}{n}=\prod_{\smash{p}} p^{k_p} =\prod_{\smash{p \le \sqrt{2n}}} \hspace{-0.5em} p^{k_p} \prod_{\smash{\sqrt{2n} < p \le 2n/3}} \hspace{-1.5em} p^{k_p}\ \prod_{\smash{2n/3 < p \le n}} \hspace{-1.1em} p^{k_p} \prod_{\smash{n < p \le 2n}} \hspace{-0.7em}p^{k_p} \le\prod_{\smash{p \le \sqrt{2n}}} \hspace{-0.5em}2n \prod_{\smash{p \le 2n/3}} \hspace{-0.4em}p \prod_{\smash{2n/3 < p \le n}} \hspace{-0.9em}p^{0} \prod_{\smash{n < p \le 2n}} \hspace{-0.6em}p < (2n)^{\sqrt{2n}} \cdot 4^{2n/3} \cdot 1 \cdot (2n)^N,where N is the number of primes between n and 2n, if any.

Because the central binomial coefficient is the largest,

2^{2n}=(1+1)^{2n}=\sum_{m=0}^{2n}\binom{2n}{m} = 2 + \sum_{m=1}^{2n-1}\binom{2n}{m} < 2n\binom{2n}{n},and so

\frac{4^n}{2n} < \binom{2n}{n}.Combine the central binomial coefficient upper and lower bounds to get

\frac{4^n}{2n} < \binom{2n}{n} < (2n)^{\sqrt{2n}} \cdot 4^{2n/3}(2n)^N,which simplifies to

4^{n/3} < (2n)^{\sqrt{2n}+N},and so the number of primes in n < p \le 2n is

N > \frac{2n}{3 \log_2(2n)}-\sqrt{2n}-1.Evaluate the right side to find N>1 for all n > 507. For n \le 507, the sequence of primes

2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 521,where each is smaller than twice its predecessor, then suffices to prove Bertrand’s postulate for all n \ge 1.