Wednesday, March 02, 2005 ... //

Maxwell's son knighted

Our students' former classmate and the son of Mary Maxwell - one of the famous women who gave their names to the Maxwell-Dworkin building at Harvard (is she related to James Clerk Maxwell?) was knighted by the British queen.

He can put KBE - Knight Commander of the British Empire - after his name. However, her majesty did not allow Bill to use "Sir" before his name because he's one of the rebel "Americans" who stole a colony from the British empire. Congratulations anyway! Also, Microsoft Windows may now be called Windows KBE, not to be confused with KDE. ;-)

Incidentally, Samsung Electronics Company's CEO Dr. Hwang is giving a talk in Maxwell-Dworkin today (or a nearby building) and all participants will receive a 256 MB USB flash memory or something like that. ;-) Oh no, it's not in Maxwell-Dworkin; Samsung just pays the shuttles between Maxwell-Dworkin and the Burden Auditorium (HBS).

... Hwang has made a good job. He showed ...

• how Samsung doubles the density of their memory every 20 months
• now they're working with tiny 16GB memory chips, and in 2007 he predicts a 200GB chip
• how their revenue equals the revenue of 3 next competitors combined
• their profit is however three times the profit of 3 competitors combined
• the annual revenue per an employee is almost 1 million dollars
• memory chips are more important than microprocessors, and therefore you should prefer Samsung over Intel
• the annual increase of the profit is roughly 60% while other companies sink

• investment and risky decisions were made even during the recession
• the competitors can't compete and most of them become Samsung's customers anyway
• one of the troubling statements by Hwang was that he believed that the size of the atom was 5 nanometers ;-) - this is what they use to calculate the limits of the semiconductor technologies (thanks to Quantoken for having corrected a typo)
• the memory chips will be a part of all gadgets, and all electronic devices will become a part of the universal "Smartphone"
Unfortunately, they heavily underestimated the number of people who would attend, and therefore they did not have enough flash drives for everyone. Their annual profit of order 20 billion dollars did not help, and therefore - despite the impressive high-tech video at the beginning - I recommend all the readers of my blog to sell their Samsung stock. ;-)

Incidentally, the audience (400 people) were mostly Asian people - a preview how the scientific and technological conferences and the developed world in general will look like in a few decades.

snail feedback (10) :

I think you mean Samsung Electronics CEO, Dr. Hwang. They did the same job, giving 256M to all participants at my school, KAIST, Korea. There, just before the talk, they showed the ad of Dr. Hwang talking at MIT with crowded audience. Now, I found that it's the power of 256M :) How's at Harvard?

ps. He's full of charisma and the talk is fun, anyway. But, the conclusion might be always the same, Samsung is the best and will keep that position.

Lubos, you mentioned "he believed that the size of the atom was 5 nanometers ;-) - this is what they use to calculate the limits of the superconducting technologies"

I guess that the reason is that if you go down in scale below that limit the thermal fluctuations will cause a lot of trouble. They will give rise to noise, non-ideal responses, and long-term memory loss. Better to have more atoms involved as then you will have average effects and not individual atoms dancing in the dark.

regards
Mike Ros

Dear Mike,

I tend to agree if you mean that 5nm may be the ultimate limit where one can go.

But he said that it was the size of an atom.

All the best
Lubos

Lubos said:
"one of the troubling statements by Hwang was that he believed that the size of the atom was 5 nanometers ;-) - this is what they use to calculate the limits of the superconducting technologies"

How big the atom is is very vaguely defined since electrons can be at different orbits. Highly excited electron orbits can be pretty far away from the nuclear. He is definitely not talking about Bohr radius of hydrogen atoms. He is not talking about the lattice constant either. He is probably talking about in the superconductor, how far the electron Cooper pairs can go before it is considered completely out of the influence of the original nuclear. That is also what material engineers are more interested in. In that sense 5nm is about right. Certainly it's quite possible that being a high level manager for years he probably forgot some of the most basic physics numbers already.

He must have meant 'point five' nanometer with little accent on 'point', I guess.

OK, I released some further note on how GUITAR gives the exactly correct He4 abundance in the universe, which is

P = (2/PI)^3 = 25.8%

My calculation of He abundance is based on the KNOWN scientific fact that thermal nuclear reactions in stars turn hydrogen into He4. Is any one going to call the H -> He4 process "crackpot". I dare you not since it's a known and undisputable fact that is what generate energy in stars!!!

I derived the He4 using the SAME g factor that was used to obtain the correct CMB temperature, as well as the correct solar constant. So you see it could not be numerology.

If you are uncomfortable with the g factor, you don't need to used it. You can just calculate from known CMB temperature. That way no new theory needs to be involved and you can still arive at the conclusion that thermal nuclear processes in stars provies 100% of the He4 abundance observed.

If 100% of observed He4 are generated in stars burning hydrogen, then that leaves no He4 to have been generated at the beginning of Big Bang. So that totally defeats the Big Bang model.

The CMB is merely star radiation energy. The 1/4 He4 abundance is simply the result of stars burning hydrogen into He4. So what's left for Big Bang to explain? Nothing.

See my BLOG:
http://quantoken.blogspot.comQuantoken

Dear Professor Quantoken,
The Hubble Space Telescope Administration regrets to inform you that our observations show conclusively that your prediction

He4 abundance in the universe, which is

P = (2/PI)^3 = 25.8%

is completely wrong. We would be grateful if you could abstain from further discussing your theory on Prof Motl's blog. Yours sincerely,
Fyodor Uckoff [Senior Manager, Hubble Project].

(1)A lot of the helium produced in stars during their main sequence phase subsequently gets burned to form carbon and nitrogen in the red giant phase. Type Ic supernovae have no helium lines since all the helium has been burned up.
(2) Only a tiny amount of the energy produced in the P-P cycle is carried off by neutrinos NOT half of it.
in their atmospheres and can never be producers of perfect black body radiation. The CMB radiation also has angular scale-invariant fluctuations, measured with great precision now, that star radiation can never have.

Mr Anonymous:

You are encouraged to go post your opinion on my BLOG and discuss there. I am following up here on Lubos's blog only because you post it here.

"(1)A lot of the helium produced in stars during their main sequence phase subsequently gets burned to form carbon and nitrogen in the red giant phase. Type Ic supernovae have no helium lines since all the helium has been burned up."

Evidently only an insignificant amount of He4 is converted this way. Astronomical observation shows a dorminate hydrogen at 3/4, and He4 at 1/4, and only a very tiny fraction of the baryon mass are other, heavier elements.

Do not read too much into the lack of He4 spectrum line in Supernova. At the kind of extreme temperature of supernova eruption, there is hardly any whole He4 atoms left. Everything is ionized so why do you expect to see any atomic emission spectrum lines? You would only be able to see absorption lines when the light pass through inter-galaxy gas clouds. So it only tells you the lack of He4 in those clouds.

"(2) Only a tiny amount of the energy produced in the P-P cycle is carried off by neutrinos NOT half of it."

Neutrino is still a big mistery so don't believe every data you are told. Basic physics instinctions will tell you that the energy is likely shared evenly by all degrees of freedoms. Therefore if neutrinos are massless particles like photons, I expect them to carry away the SAME amount of energy as photons.

Scientists could hardly produce REAL nuclear fusions in labs, the same condition like what goes on in the Sun, and in a very controled manner while doing accurate measurements. Controlable thermal nuclear fusion is still a few decades away.

Therefore, although the total energy released in solar proton processes is known, precisely, by looking at the difference of atomic weights. Exactly who carries how much portion of that energy is NEVER actually measured, but deduced from some models which may or may not be correct.

Doing an experimental measurement of the actual energy partition would require simulating the SAME temperature and density in the core of the Sun, not higher and not lower. And you would expect the reaction rate to be extremely low because it is extremely low on the Sun. No such experiment is feasible in the near future.

in their atmospheres and can never be producers of perfect black body radiation. The CMB radiation also has angular scale-invariant fluctuations, measured with great precision now, that star radiation can never have."

No one says the CMB photons come directly from the star. I only said the CMB energy originated from star radiations. The lights emitted from stars must have been mediated by some sort of intergalactic matters, before it is converted to CMB radiation. Such mediation would even it out to form perfect blackbody spectrum.

The universe is an incredibly huge shallow chamber, perfectly suited to create and maintain blackbody radiation. The mediating matter can be extremely thin (extremely low density), and does NOT need to fill out the whole space, and does not even need to be very absorptive or emittive at CMB wavelength range. As long as it has minimal interaction with CMB radiation at all, given long enough time the CMB will reach perfect blackbody equilibrium.

Remember how blackbody spectrum is derived in general physics class? The emissibility of the material forming the chamber does NOT matter. Once it reaches thermal equilibrium, it's perfect blackbody spectrum. That's because the energy will be perfectly evenly partitioned among all degrees of freddoms, upon thermal equilibrium.

As what you claim "The CMB radiation also has angular scale-invariant fluctuations, measured with great precision now" That's a complete lie. There is no great precision. The fluctuation is about the same order as instrumental errors. That's why just a few years ago when the instruments were slightly less precise they could not even detect such fluctuation at all. Now they just have some instrument slightly more sensitive and beginning to see the fluctuation, but it is in no may much bigger than the instrumental errors. The signal to noise ratio is still very low.

So stop lying about "great precision". There is none.

Do you realize "angular scale-invariant fluctuations" is just a fancier word describing what people normally would just call "white noise".

You detected some white noises. Big deal? Your theory predicted there would be some white noises. Big deal. Any physics system or any measurement contain white noises. White noises are every where you look.

Big Bang hasn't been able to predict the exact magnitude of that white noise fluctuation. All it ever predicted is the fluctuation will be "angular scale invariant", i.e., it should exihibit white noise characteristics, i.e., no recognizable pattern.

I would have taken them more seriously if they found some specific pattern that make the fluctuation different from white noise, and they were able to predict that characteristics before hand.

"Prediction" of existence of "angular scale-invariant" white noise is just a joke.

Quantoken