My Father-in-law

February 13, 2012

Jack King and baseball team in Papua, New Guinea

Jack King was my father-in-law, and my buddy. He served in New Guinea in WWII. Here he is with his baseball buddies!! He’s on the front row, second from left.

He was one of the best baseball players in Wythe County, Va (his home was Ivanhoe, Va), and was selected to play on a special team the Army created during the War.

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Big News in Number Theory

February 4, 2012

Terence Tao has posted a new paper on the arXiv giving a proof that all odd integers greater than 1 can be written as the sum of at most 5 primes. This was proven before, on the assumption of the Riemann Hypothesis, but he has removed the need this assumption. The key work in the title is ALL; previous work has shown that any LARGE ENOUGH integer can be written as the sum of at most three primes.

This is a pretty big step forward I think. For details, see Terence’s blog and the paper itself.

Happy 2012,

Tom

Shortwave Listening

June 4, 2011

I recently took up my old hobby of shortwave listening. It has changed a lot since the ’70s. In some ways it’s not as interesting, as budget cuts and the rise of the Internet have driven many broadcasters from the air.

In other ways, it’s got a lot to offer. I had always wanted to get a really nice receiver, of the traditional transistor variety. However, a type of radio called an SDR (software defined receiver) is popular these days, and I bought one, the Winradio Excalibur.

The ‘radio’ is just a receiver the size of an external hard drive that converts radio waves into a digital spectrum that can then be processed by a modern PC. This gives the user a ton of flexibility in defining filters, setting automatic gains, doing recording at specified times, etc. You can even record a chunk of spectrum and go back and analyze it later. You could get the entire AM band from 540 to 1700 kHz and look at it later for rare DX catches!

Here’s a look at the interface – there are three spectrum analyzers that can also be used as waterfall displays.

WinRadio Excalibur PC interface

I’ll have more to say about this in future posts, until then

’73s and good DX!

Tom

Calabi-Yau manifolds -a good book

January 18, 2011

I just finished reading the relatively new book by S. T. Yau (Field’s Medalist) and Steve Nadis “The Shape of Inner Space: String Theory and the Geometry of the Universe’s Hidden Dimensions” (see here).

I got my Ph. D in theoretical particle physics, back when string theory was just taking off, and have been trying to study it a bit since then, just for the sake of curiosity. Part of the basis of string theory is that there are tiny hidden dimensions that we can’t see, and that these have to be curled up in a particular way for the theory to work. The Calabi-Yau manifolds that the book explains are the best method of curling up these hidden dimensions.

Yau, the first author, did some work in the 70’s to confirm a conjecture of Eugenio Calabi, that manifolds with certain properties (that I won’t go into here) could actually be constructed, and it turned out that these were just the thing needed for string theory when its first revolution occurred in the mid 80’s. Since then there have been many developments in both math and physics centered around the geometry of Calabi-Yau’s. The book does an excellent job of telling this story, with enough human interest added to make for a very compelling story. In addition, the status of string theory as a barely, if at all, tested or testable theory is not overlooked. I particularly appreciated the even-handedness of the arguments.

The one thing I thought could be better was the end notes. As someone with a fair bit of physics and mathematics training, I would like to have seen more references to the original literature. (There are a few.) Anyway, with enough searching on the Internet I’ll probably be able to find the references I want.

This is one of the best books of the genre “make the complex math understandable to the rest of humanity.” Most of them are too dumbed-down; here, the authors went to a lot of trouble to give good pedagogical explanations.

On a side note, I knew a student of Yau’s when I was a grad student at Princeton. He will remain nameless, as all he ever did was play foosball!

Seven questions that keep physicists up at night

October 27, 2009

There’s an interesting post over at New Scientist describing a panel discussion held at the Perimeter Institute:

Seven questions that keep physicists up at night

On a more down-to-earth note, NASA plans to launch the Ares 1-X launch vehicle tomorrow morning, with a launch window beginning at 8 AM and extending until noon. The Ares will replace the shuttle as the vehicle for getting US astronauts into space. Tomorrow’s launch is just a short six-minute flight test.

What A Joke!

October 10, 2009

You may already know to what I’m referring – President Obama winning the Nobel Peace Prize! For what??

I believe this was the Nobel committee’s chance to rub dirt on Pres. Bush, who I doubt if the Norwegians liked very much. He of course did far more than Obama, who has barely started his Presidency, to promote world peace. (I’m not talking about Iraq or Afghanistan!)

It will be interesting in the future to see if we get any behind-the-scenes details about the selection process, which by the way happened 11 days after Obama was inaugurated!! What a first 11 days…..

Only A Test

March 21, 2009

This is simply a test of the Latex to WordPress converter available here.

Start with

\displaystyle  -\sum_{n=1}^\infty \frac{\log n}{n} \times \left\{-1,-1,\cdots,k-1,-1,-1,\cdots,k-1,\cdots \right\}

where the term {k-1} appears at {n=0 \bmod k}.

We can rewrite this as

\displaystyle  -\frac{\partial}{\partial \nu} \sum_{m=1}^\infty \left[ \sum_{n=1}^{k-1} \frac{1}{(n +k(m-1))^\nu} + (1-k)\frac{1}{(m k)^\nu} \right] \vert_{\nu=1}

\displaystyle  = -\frac{\partial}{\partial \nu} \left[ \frac{1-k}{k^\nu}\zeta(\nu) + \sum_{n=1}^{k-1} \sum_{m=1}^\infty \frac{1}{ k^\nu} \frac{1}{(m-1+\frac{n}{k})^\nu} \right] \vert_{\nu=1}

\displaystyle  = -\frac{\partial}{\partial \nu} \left[ \frac{1-k}{k^\nu}\zeta(\nu) + \sum_{n=1}^{k-1} \frac{1}{ k^\nu} \sum_{m=0}^\infty \frac{1}{(m+\frac{n}{k})^\nu} \right] \vert_{\nu=1}

\displaystyle  = -\frac{\partial}{\partial \nu} \left[ \frac{1-k}{k^\nu}\zeta(\nu) + \sum_{n=1}^{k-1} \frac{1}{ k^\nu} \hat{\zeta} \left(\nu,\frac{n}{k}\right) \right] \vert_{\nu=1}

\displaystyle  = -\frac{\partial}{\partial \nu} \left[ \frac{1-k}{k^\nu}\zeta(\nu) + \sum_{n=1}^{k} \frac{1}{ k^\nu} \hat{\zeta} \left(\nu,\frac{n}{k}\right) - \hat{\zeta}(\nu,1) \right] \vert_{\nu=1}

\displaystyle  = -\frac{\partial}{\partial \nu} \left[ \frac{1-k}{k^\nu}\zeta(\nu) + \zeta(\nu) - \hat{\zeta}(\nu,1) \right] \vert_{\nu=1}

\displaystyle  = -k^{-\nu} \left[ k \log k \zeta(\nu) + (k^\nu - k) \zeta'(\nu) \right] \vert_{\nu=1}

Now use the series, for {z} near 1,

\displaystyle  \zeta(z) \approx \frac{1}{z-1} + \gamma -\gamma_1 (z-1)

and put {\nu = 1 + \epsilon} to obtain

\displaystyle  = -\frac{1}{k} \left[ k \log k \zeta(1+\epsilon) + k(\epsilon \log k + \frac{1}{2} \epsilon^2 \log^2 k ) \zeta'(1+\epsilon) \right]_{\epsilon=0}

\displaystyle  = -\gamma \log k + \frac{1}{2} \log^2 k

as desired.

My Cats

March 19, 2009

floyd_zeke_smaller

Well, the tan one has passed on 😦

Hello world!

March 19, 2009

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