• An academic in industry

    Recently, with members of NCSU’s Nonlinear Artificial Intelligence Lab, I completed a 3.5-year project as a subcontractor working on an industrial project. As an academic, this was a novel experience. Unlike most of my research, this work will not result in a journal article or conference presentation, although it might one day contribute to an industrial product.

    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!

    Animation of virtual 3D buttons casting virtual shadows as a real iPhone physically pitches and rolls. A virtual light source is always directly above the phone. (You may need to click to start the animation.)
  • Does a charge in gravity radiate?

    Caltech, Saturday night, grad student pizza. The conversation turns to a famous general relativity puzzle: does an electric charge at rest in a gravitational field radiate? According to Einstein’s equivalence principle, a static homogeneous gravitational field is indistinguishable from constant acceleration in empty space, and as is well known, accelerating charges radiate. Does that mean electrons in the table radiate as we eat? Sam says Kip says the electromagnetic field lines bend. We don’t resolve the paradox, and after dinner we return to our grad school homework.

    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.

    Outline

    • 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.

    Charge worldliness in the gravity and free fall frames.
    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.

    Model Gravity

    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

    x^\mu = \boxed{\begin{array}{c} x^0 \\ x^1 \\ x^2 \\ x^3 \\\end{array}} = \boxed{\begin{array}{c} ct \\ \rho \\ \phi \\ z \\\end{array}} = \boxed{\begin{array}{c} t \\ \rho \\ \phi \\ z \\\end{array}} = \boxed{\begin{array}{c} t \\ \vec x \\ \end{array}}.

    The line element

    \text d\tau^2 = \sum_{\mu=0}^3 \sum_{\nu=0}^3 g_{\mu\nu} \text dx^\mu \text dx^\nu = g_{\mu\nu} \text dx^\mu \text dx^\nu

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

    g^F_{\mu\nu} = \boxed{\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0\\0 & 0 & -\rho^2 & 0 \\0 & 0 & 0 & -1 \\ \end{array}} = \eta_{\mu\nu},

    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

    g^G_{\mu\nu} = \boxed{\begin{array}{cccc} u[z_G]^2 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\0 & 0 & -\rho^2 & 0 \\0 & 0 & 0 & -u'[z_G]^2/g^2 \\ \end{array}} = \boxed{g^{\mu\nu}_G}^{-1},

    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.

    Free fall

    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

    \frac{\text d^2 x^\alpha}{\text d \tau^2} = -\Gamma^\alpha_{\mu\nu} \frac{\text dx^\mu}{\text d \tau}\frac{\text d x^\nu}{\text d \tau},

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

    \Gamma^\alpha_{\beta\gamma} = \frac{1}{2}g^{\alpha\delta}\left(\frac{\partial g_{\delta\beta}}{\partial x^\gamma} + \frac{\partial g_{\gamma\delta}}{\partial x^\beta} – \frac{\partial g_{\beta\gamma}}{\partial x^\delta} \right)

    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

    (z_G-1/g)^2-t_G^2 = 1/g^2,

    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.

    Transform to free fall frame

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

    \begin{aligned}\text d\tau^2 &= \text dt_F^2-\text d\rho_F^2-\rho_F^2 \text d\phi_F^2-\text dz_F^2 \\ &= u[z_G]^2 \text dt_G^2-\text d\rho_G^2-\rho_G^2 \text d\phi_G^2-\frac{u^\prime[z_G]^2}{g^2} \text dz_G^2,\end{aligned}

    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

    \sqrt{-A^\mu A_\mu} = \sqrt{-g_{\mu\nu}A^\mu A^\nu} = \frac{\ddot{z}}{(1-\dot{z}^2)^{3/2}} = \gamma^3 \ddot z,

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

    \ddot z_G = \gamma_G^3 \ddot z_G^{} = \gamma_F^3 \ddot z_F = \frac{\ddot{z}_F}{(1-\dot{z}_F^2)^{3/2}} = 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.)

    Electromagnetic fields in free fall frame

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

    \mathcal{A}^\mu = q\frac{U^\mu}{R^\nu U_\nu} \bigg|_{t_r} = q\frac{U^\mu}{g_{\alpha\beta}R^\alpha U^\beta} \bigg|_{t_r},

    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

    t-t_r = \big| \vec x-\vec x_q[t_r] \big| / c.

    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

    \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}}_F = \mathcal{F}_F^{\mu\nu} = \frac{\partial \mathcal{A}_F^\nu}{\partial x_F^\mu}-\frac{\partial \mathcal{A}_F^\mu}{\partial x_F^\nu},

    generates the electric and magnetic field vectors

    \begin{aligned}\vec\mathcal{E}_F &= \frac{4 g^{-2} q}{\xi_F^3} \boxed{\begin{array}{c}2\rho_F z_F \\0\\-\Delta_F^2 \end{array}}, \\ \vec\mathcal{B}_F &= \frac{4 g^{-2} q}{\xi_F^3} \boxed{\begin{array}{c}0 \\2\rho_F t_F\\0 \end{array}}, \end{aligned}

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

    \vec\mathcal{S}_F = \vec\mathcal{E}_F \times \vec\mathcal{B}_F = \frac{32 g^{-4} q^2 t_F \rho_F}{\xi_F^6} \boxed{\begin{array}{c}\Delta_F^2 \\0 \\ 2\rho_F z_F \end{array}}.

    Electric and magnetic field in the free fall frame.
    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.

    Electromagnetic fields in gravity frame

    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}}.

    Electric and magnetic field in the gravity frame.
    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.

    Summary

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

  • Sabbatical trip to Europe – Part 3 (Otto Rössler)

    After the conference in Switzerland, I stopped in Tübingen, Germany to visit Otto Rössler. Nearly everyone who learned about nonlinear systems knows the nowadays named Rössler attractor and his work in chaos theory in the 1970s.
    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.

    Robert Timm (right) and John Cook (left).
    Otto Rössler and Niklas Manz on 20 July 2024 at Schwärzlocher Hof (farm and restaurant), which started as an abbey circa 1100.

    Robert Timm (right) and John Cook (left).
    Otto and Reimara Rössler on 20 July 2024 at Schwärzlocher – with the local peacock mom.

    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 &= \vphantom{}-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.

    Period-doubling route to chaos).
    Post-transient {x ,y, z} solutions to the Rössler system for 0.1 < a < 0.411, b = 2, c = 4, exhibiting a period-doubling route to chaos culminating in a 3-cycle window. (You may need to “click” on the figure to see the animation.)

    The animation was created by John Lindner.

  • Sabbatical trip to Europe – Part 2 (Switzerland)

    The second stop of my Europe trip was Switzerland. In Zürich, I visited places where Boris Belousov (the discoverer of the Belousov-Zhabotinsky reaction I am using in my lab) lived and studied during his time in exile in the early 20th century. But the theme of that part of my science history project keeps validating “Searching for a ghost”.

    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.

  • Sabbatical trip to Europe – Part 1 (Lviv, Ukraine)

    Since about 2018, I was interested in the work of Julian Hirniak, who published an article on periodic chemical systems in 1908 (and a follow-up in 1911), before Alfred Lotka’s famous theoretical 1910 paper and William Bray’s experimental work in 1921.
    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.

  • The Longest Flight

    As a kid pouring over the Guinness Book of World Records, I was astonished by the record longest flight, which lasted not just a few hours – as I would have guessed – but more than two months! Today, nearly 65 years later, that amazing achievement remains one of aviation’s most enduring records.

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

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

  • Where Are the Stars?

    When viewing space photography, such as Apollo or International Space Station photos, people often ask, “Where are the stars?” Typically such photos properly expose the bright lunar or space station surfaces and consequently underexpose the dim background stars, rendering space as featureless black.

    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.

    Spacecraft and stars
    Dragon spacecraft with Milky Way stars illuminated faintly by moonlight, 2024 June 29. A 1s, f1.4, ISO 5000, 28mm photo by NASA astronaut Matthew Dominick. Click for a larger version.

  • Bertrand’s Postulate

    When searching for prime numbers, the next prime number is no larger than twice the current number. Postulated by Joseph Bertrand, first proved by Pafnuty Chebyshev, I present an elementary proof based on one by the teenage Paul Erdős.

    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!

    Some Notation

    In analogy with the factorial function

    n! = \prod_{m\le n} m

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

    n\# = \prod_{p\le n} p

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

    \binom{n}{m} = \frac{n!}{m!(n-m)!}

    and the floor function

    \lfloor x\rfloor =\max\{n\in \mathbb {Z} \mid n\leq x\}.

    Primorial Function Upper Bound

    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

    \binom{2m+1}{m} = \frac{(2m+1)!}{m!(m+1)!} = M \prod_{m+1 < p\le 2m+1} \hspace{-1.6em}p \hspace{1em}\ge \prod_{m+1 < p\le 2m+1} \hspace{-1.6em}p\hspace{1.2em},

    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.

    primorial
    Primorial function x\# (blue) and its upper bound (red).

    Central Binomial Prime Factors

    Consider the one central binomial coefficient

    \binom{2n}{n} = \frac{(2n)!}{n!n!} = \frac{(2n)!}{(n!)^2}.

    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) = 2

    and 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.

    Largest prime powers dividing the central binomial coefficient.
    Largest prime powers dividing the central binomial coefficient. Only the “smallest” prime powers can divide the binomial multiple times.

    Central Binomial Upper Bound

    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.

    Central Binomial Lower Bound

    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}.

    Central Binomial Squeeze

    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.

    Number of primes
    Number of primes N in n < p \le 2n (blue) and its generous lower bound (red).

  • Aero thermo dynamics

    Up early this morning to watch the spectacular fourth integrated flight test of SpaceX’s Superheavy Starship, the largest rocket ever built. Each IFT has greatly improved on the previous one, and the fourth was no exception. For the first time, both the booster and the ship softly splashed down in the ocean!

    Especially impressive was watching live onboard views from the ship as it reentered the atmosphere from orbital speed. Adiabatic heating (not friction) ionized the surrounding air. The resulting plasma sheath would have caused a customary communication blackout, but starship’s large size and SpaceX’s low-Earth-orbit Starlink satellite internet constellation (operating “up” instead of “down”) enabled nearly continuous live video of the descent.

    One camera was pointed toward a forward control flap, which did suffer some heating damage, but continued to function, controlling the descent attitude and enabling the final flip-and-burn deceleration maneuver. Forward flaps on near-future versions of starship may be moved leeward to improve reliability and ease manufacturing.

    Starship 29 re-enters Earth's atmosphere.
    Starship 29 re-enters Earth’s atmosphere, 2024 June 6. Adiabatic heated and ionized gas wraps around the ship, like a meteoroid creating a meteor in its wake. S29 splashed down softly in the ocean, the largest object ever to survive reentry intact. (SpaceX)

  • Stegosaurus Tiling

    John Chase, the head of the Walter Johnson High School Math Department, in Maryland, near Washington DC, liked my Stegosaurus variation of the Spectre monotile so much that he had his students paint it on the wall of their math office! Attached are a couple of photos he shared.

    SmithMyersKaplan, and Goodman-Strauss recently discovered an infinite continuum of aperiodic monotiles, of which the stegosaurus is a specially simple equilateral example with only ±60° and ±90° turns (and a single 0° turn).

    Stegosaurus mural.
    Stegosaurus mural in the Walter Johnson High School Math Department office. (John Chase)

    Mural detail.
    Mural detail with a few upright stegosauruses near the middle. (John Chase)

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