Tuesday, January 21, 2014

A recursive evaluation of zeta of negative integers

arXiv: David Berenstein released a review of TeV strings that discusses stringy braneworlds where the excited strings may really be seen at the LHC. He offers some intro to strings, D-branes, effective actions, gauge groups, large extra dimensions, warped geometry, singularities, orbifolds, deformations, and constraints from LEP and proton decay and others and concludes with mostly negative concluding remarks that tend to disfavor this "excessively testable" class of stringy models.
The evaluation of the sum of positive integers may be viewed as a special example of the calculation of the Riemann zeta function\[

\zeta(x) = \sum_{n=1}^\infty \frac{1}{n^x}.

\] For \({\Re}(x)\gt 1\), the series above converges in the usual sense but the meromorphic function may be uniquely defined for all complex values of \(x\). The only single pole appears at \(x=1\) where we have \(\zeta(1)=\infty\).

Riemann zeta function in the complex plane as colors; black is zero, white is infinity, red is positive, other colors (hue) indicate other phases.

The values of the zeta function of other positive odd integers, like \(\zeta(3)\) or \(\zeta(5)\), are notoriously transcendental numbers, while the zeta function of positive even integers may be calculated and the result is a rational multiple of \(\pi^x\) in that case. For example,\[

\zeta(2) &= 1+\frac {1}{2^2} + \frac{1}{3^2}+\dots = \frac{\pi^2}{6}\\
\zeta(4) &= 1+\frac {1}{2^4} + \frac{1}{3^4}+\dots = \frac{\pi^4}{90}.

\] The last value is needed when you evaluate the total heat emitted by a black body (when you look for the constant in the Stefan-Boltzmann law by integrating the Planck curve).

These identities may be computed using Fourier series. You consider periodic functions \(f(t)\) with the periodicity \(2\pi\) that look like powers \(t^x\) in the interval \((-\pi,+\pi)\) and that may suffer from a discontinuity at \(t=\pm\pi\); it's actually useful to add a constant shift to \(f(t)\) so that the mean value is zero. The value of \(\int_{-\pi}^{+\pi} |f(t)|^2\) may be computed easily and directly – by those who know how to integrate polynomials (the integral yields factors including a high power of \(\pi\) from the powers of the limits) – but it may also be calculated, using a change of the basis on this Hilbert space of periodic functions, as the sum \(\sum_{n\in\ZZ} |c_n|^2\) of the squared Fourier coefficients of the function (with some appropriate normalization).

The Fourier coefficients \(c_n\) themselves are not hard to find as long as you know how to integrate things like \(t^m \cos(\omega t)\) over \(t\) and the most important term you will get in \(c_n\) will be proportional to \(n^x\).

The identity \(\int |f(t)|^2=\sum |c_n|^2\) then produces the required values of \(\zeta(2x)\) for positive integer values of \(x\). When we were college freshmen, about 21 years ago, a roommate at the student hostels (who studied teaching of maths and physics) learned this method from me and accepted the challenge. Who can compute \(\zeta(x)\) without any computer or calculator, just with a pen and paper, for a larger value of \(x\)?

He was very good so we obtained the results for \(x=2,4,6,8,10\) at about the same time. But then he gave up and I beat him with the result\[

\zeta(12) = \frac{691\pi^{12}}{638,512,875}.

\] Yes, this fraction is just slightly greater than one, of course. Twelve is the smallest integer for which \(\zeta(x)=\pi^x (p_x/q_x)\) so that the identity requires \(p\gt 1\); for smaller even values of \(x\), you just have \(\zeta(x) = \pi^x / q_x\) without any numerator.

We had to be crazy (maybe we have also calculated \(\zeta(14)\) together but without a calculator, I forgot the details). I am not sure whether I would overcome the laziness and find the motivation to calculate these stupid things today.

Zeta of non-positive integers

But we want to go in the opposite direction and calculate \(\zeta(x)\) for \(x\) which are negative integers or zero. The function is called the Riemann zeta function after Bernhard Riemann (1826-1866) but all these results had actually been obtained by Leonhard Euler (1707-1783) who first defined the Riemann zeta function and derived its most important properties and values.

Why do we call it after Riemann? Because Euler didn't really use complex analysis. If you think about it, this is a truly ironic reason not to credit Euler – a key co-founder of the complex number industry who made the complex numbers powerful by finding how they're exponentiated and how the exponentials are linked to the trigonometric functions.

OK, what are the values of the Riemann zeta function for non-positive integers \(x\)? Well, you can prove and use the "left-right symmetry" functional equation obeyed by the Riemann zeta function\[

\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s) \!,

\] and reduce the problem of negative values of \(x\) (or zero) to the problem of the positive ones. But it would take some time to prove the identity above and the evaluation of \(\zeta(x)\) for positive integer values of \(x\) isn't "trivial", either, as we discussed above. So is there some direct way to calculate \(\zeta(x)\) for non-positive integers \(x\)?

The functional equation above or a method of mine that I use below may be used to prove that\[

\zeta(-2)=\zeta(-4)=\zeta(-6)=\dots = \zeta(2q)=0.

\] The zeta function vanishes for all negative even integers! These are the "trivial" zeroes of the function; the other ones are "nontrivial" and they sit at seemingly random points of the vertical axis \(+1/2+it\), \(t\in\RR\).

We also want to calculate the interesting values\[

\zeta(0)=-\frac 12,\,\,\zeta(-1)&=-\frac{1}{12},\\

\] or at least the first ones. Yes, the signs are alternating here (except for \(\zeta(0)\) at the beginning). It's a bit surprising that the complexity of these rational numbers doesn't seem to increase much, does it? Note that the first one result above is\[

\zeta(0) = 1+1+1+\dots = -\frac{1}{2}.

\] One must be a bit careful about this particular sum because we must remember from "what values of \(n\)" the terms equal to \(1\) arise. If we're summing over \(n\) from one to infinity, we get \(-1/2\). If we were summing from zero to infinity, we would get \(+1/2\), of course, even though the sum would "look" like \(1+1+1+\dots\) as well. (The sum of \(1\) over all integers vanishes, \(-1/2+1-1/2\), a fact you may justify in various other ways.) But just to be sure, if you get this "sum of ones" in field theory or string theory, the theory tells you what the value of \(n\) is for the individual terms so there is no ambiguity left!

Let's ask again: How do we calculate these values of the zeta function? Euler designed a clever method that allows you to calculate the values on several lines and in general, the values of the zeta function may be linked to the Bernoulli numbers. You may study it but the Bernoulli numbers may be hard.
GmailTeX: all math-literate Gmail users must have it!

Gordon gave me a nice link to Fields medalist Terry Tao's presentation of these divergent sums (recommended) and I also learned about MathJax and CodeCogs-based Gmail\(\rm \TeX\), see the image above. It adds entries in the sidebar that may reformat an e-mail message you are just writing or you have received to full rich maths (single-dollar or double-dollar delimiters must be right, also F8 shortcut) or simple maths (basic signs, subscripts, superscripts, Greek letters using "normal fonts", doesn't require the correct dollar delimiters, guesses what is pseudo-maths, also F9 shortcut). Badly recommended. The full rich maths expressions are embedded as images so the recipient doesn't have to have the Gmail\(\rm \TeX\) Chrome extension. Install Gmail\(\rm \TeX\) for Chrome (extension), Firefox (Greasemonkey needed first), Safari, and Opera now.
Instead, let me offer you a seemingly inequivalent method by your humble correspondent reported in a student journal at Charles University's Math-Phys Faculty, The Pictures of Yellow Roses, in 1994. The journal had 110 copies (in Czech) so it's a bit unlikely that you managed to have grabbed one. ;-)

The article begins with some teasing of mathematicians detached from the reality and the reasons why the divergent geometric series may still be assigned the value \(1/(1-q)\). But we want to calculate the zeta function for non-positive integral arguments. Let's generalize the zeta function to\[

\zeta_s(x)=\sum_{n=1}^\infty (n+s)^{-x}

\] The old zeta function – and one that we will try to use at almost all times – is obtained if the subscript is \(s=0\). For example, the sum of positive integers is \(\zeta_0(-1)\). The extra subscript \(s\) has the effect of shifting the integers that are exponentiated to the \((-x)\)-th power. A cool identity you should notice is\[

\zeta_1(x) = \zeta_0(x) - 1.

\] If you shift the base \(n\) by one, the sum will go from two to infinity, so the effect is that you just skip the first term \(n=1\) in the sum, and the term is one. The identity above demonstrably holds whenever both sides are convergent. But the philosophy behind these assignments of finite values says that if something is equal to something else in a whole region of the complex plane, you may extrapolate the identity to the whole complex plane. So there exist functions \(\zeta_1(x)\) and \(\zeta_0(x)\) that are given by the series above at least when the series converge and that differ by one. Sometimes, functions of complex variables may have branch cuts, numerous sheets, and ambiguities about the values follow from that. However, the zeta functions are actually single-valued.

We may Taylor-expand the generalized zeta function with respect to the subscript \(s\) and move from \(s=0\) to \(s=1\). Taylor's identity reads\[

\zeta_1(x) &= \zeta_0(x) + \frac{\partial}{\partial s} \zeta_s(x)|_{s=0} +\\
&+ \frac{1}{2!}\frac{\partial^2}{\partial s^2} \zeta_s(x)|_{s=0} + \frac{1}{3!}\frac{\partial^3}{\partial s^3} \zeta_s(x)|_{s=0}+\dots

\] It's a helpful identity because the \(s\)-derivatives of the generalized zeta function are easily calculable. Because of the simple rule for the derivative of a power, the first derivative equals\[

\frac{\partial}{\partial s} \zeta_s(x) = \sum_{n=1}^\infty (n+s)^{(-x-1)} (-x) = (-x)\zeta_s(x+1)

\] and similarly the \(m\)-th derivative is equal to\[

\frac{\partial^m}{\partial s^m} \zeta_s(x) = \underbrace{(-x)(-x-1)\cdots (-x-m+1)}_{m\text{ factors}}\zeta_s(x+m)

\] If we substitute \(\zeta_1(x) \to \zeta_0(x) - 1\) and then subtract \(\zeta_0(x)\) from both sides of the Taylor identity above and if we substitute the formula for the \(m\)-th derivative we just calculated, we conclude that\[

-1&= +(-x)\zeta_0(x+1)+\frac{(-x)(-x-1)}{2!}\zeta_0(x+2)-\\

\] Let's substitute \(x\to 0\) to this Taylor identity. Because \(\zeta(x)\) is finite for \({\rm Re}(x)\gt 1\) and these finite numbers are being multiplied by factors that include \((-x)\) which therefore go to zero, only the first term on the right hand side survives and the identity reduces to\[

\lim_{x\to 0} x\zeta_0(x+1) = 1.

\] This tells us that \(\zeta(x)\) indeed has a pole for \(x\to 1\) i.e. around the pole, it behaves like \(\zeta(1+\varepsilon)=1/\varepsilon+{\rm finite}\). Similarly, we may substitute \(x\to -1\) to the identity. The latter simplifies to\[

-1 = \zeta_0(0)+\frac{1}{2!}(-x-1)\zeta(x+2).

\] But \(\lim_{x\to (-1)} (-x-1)\zeta_0(x+2)=-1\) as have calculated in the previous displayed equation (where \(x\) was just shifted by one relatively to what we need here), and therefore the last displayed formula implies \(\zeta_0(0)=-1+1/2=-1/2\). That's one of the sums we wanted: \(1+1+1+\dots = -1/2\).

Finally, I want to explicitly calculate \(\zeta(-1)=1+2+3+4+5+\dots\). We substitute \(x\to -2\) to the identity and the identity dumbs down to\[

-\!1 = 2\zeta_0(-\!1) + \frac{1}{2!} 2\cdot 1 \zeta_0(0)+ \frac{1}{3!}2\cdot 1(-x\!-\!2)\zeta_0(x+3)

\] which boils down, after a trivial simplification, to\[

-1 = 2\zeta_0(-1)-\frac 12-\frac 13 \quad \Rightarrow \quad \zeta_0(-1) = -\frac{1}{12}.

\] We were able to calculate the limit of \(\zeta(x)\) for \(x\to -1\) and because the limit didn't depend on the direction from which we approach the point (the function has no essential singularities etc.), we may deduce that this is the value of \(\zeta(-1)\) itself. The moral justification of all the steps – that ultimately allowed us to prove that a seemingly divergent sum is equal to \(-1/12\) – was the fact that each identity was valid at least in a region (neighborhood) of the complex plane and identities obeying this condition may be "formally" (according to these rules of analytic continuation) extrapolated to the whole complex plane. In other words, the business is to carefully derive identities for complex functions and their sums and integrals and simply avoid the kind of complaint that "I don't like this identity for a sum because it diverges for certain values of the variables".

Zeta of negative even integers

I invite you to substitute \(x\to -3\) and then \(x\to -4\) etc. and recursively compute the zeta function of even more negative integers. The values of \(\zeta(x)\) for odd negative integers will be nonzero rational numbers with an alternating sign. But you will also find out that \[

\zeta_0(-2)=\zeta_0(-4)=\zeta_0(-6)=\dots = 0.

\] In fact, it is not hard to prove all these identities using the Taylor expansion, either. The novelty will be that we will be Taylor-extrapolating the zeta function from \(s=0\) to \(s=-1\) instead of \(s=+1\) which we used above. For \(x\gt 1\), we may see that\[

\lim_{s\to (-1)} \zav{ \zeta_s(x)-\frac{1}{ (s+1)^x } } = \zeta_0(x)

\] i.e. the zeta function at \(s=-1\) only contains one additional term relatively to the zeta function at \(s=0\); this is analogous to our claim that the first term "one" was missing for \(s=1\). However, for \(x\lt 0\), the limit \(\lim_{s\to(-1)} 1/ (s+1)^x=0\); this identity wouldn't be kosher if extrapolated to non-negative \(x\) because for a non-negative \(x\), the singularity \(1/(s+1)^x\) for \(s\to -1\) is "genuine" because its values for \(s=-1+\varepsilon\) for \(\varepsilon\) tiny yet small are particular large numbers whose limit is infinite. So for \(x\lt 0\), the last displayed equation implies\[

\zeta_{-1}(x) = \zeta_0(x)

\] The Taylor expansion extrapolating the zeta function from \(s=0\) to \(s=-1\) produces the following identity analogous to one that we encountered for \(s=+1\) above:\[

0&= -(-x)\zeta_0(x+1)+\frac{(-x)(-x-1)}{2!}\zeta_0(x+2)-\\

\] which you're invited to check by substituting \(x\to -1,-2,-3\). To prove the vanishing of the zeta function for even negative arguments, we just add the last displayed equation and the previous one that we obtained from \(s=1\) and that differed by not having the alternating signs and by having \(-1\) rather than zero as the left hand side. One-half of the sum of these two equations (we could derive the same conclusions if we considered the difference instead of the sum) reads\[

-\frac 12 &= \frac{1}{2!} (-x)(-x-1) \zeta_0(x+2)+\\
&+ \frac{1}{4!} (-x)(-x-1)(-x-2)(-x-3)\zeta_0(x+4)+\dots

\] If we substitute values \(-4,-6,-8\) etc. for \(x\) into this equation, the last nonzero term on the right hand side will be a multiple of \(\zeta_0(0)\) and this term will be equal to \(\zeta_0(0)=-1/2\) itself (because the product of \((x+{\rm something})\) will cancel against the factorial) which may be compensated against the same term \(-1/2\) on the left hand side.

The remaining terms in the equations will be a combination of \(\zeta_0(-2)\), \(\zeta_0(-4)\), \(\zeta_0(-6)\), and other even negative integers, and such an equation may be used to recursively prove that e.g. if \(\zeta_0(-2)\), \(\zeta_0(-4)\) vanish, then so does \(\zeta_0(-6)\). By mathematical induction, this proves that the zeta function vanishes for all even negative values of the arguments (these are the trivial zeroes of the function).

My 1994 article in The Pictures of the Yellow Roses contained an extra section that calculated the critical dimension \(D=26\) out of the sum of integers (in the light cone gauge, by demanding that the first excited level had to be massless because there would have to be \((D-1)\) and not just \((D-2)\) polarizations for a massive vector) but I have done similar things in various previous blog posts so let me omit it. If you can't find it, I may add the argument here.

I obviously became keen on the zeta-function tricks in 1994 (twenty years ago) because the first preprint I ever sent to the arXiv was about them.

The fact that the sum of integers should be assigned the value \(-1/12\) is the reason for the omnipresence of factors like \(1/12\) or \(1/24\) etc. in perturbative string theory.


  1. Climatology is the Carbon Tax on Everything. Geoengineering sulfuric acid mist stratospheric haze mirroring sunlight back into space requires leaving sulfur in jet fuel, reducing its price, with free delivery. MADNESS!

    Enviro-process massively transports pure sulfur dioxide into the stratosphere at astounding cost. 1) Air is average MW = 29, SO2 is 64, 2.2 times heavier, containing only 50 wt-% sulfur. Power innovative new transporters with innovative new Enviro-biofuels. 2) Not just any sulfur, bio-sulfur! Burn Brussels sprouts, cabbage, and cauliflower. The UK and Russia starve. Strength through Sacrifice.

    Transplant polar bears to Antarctica. You can't save a species without breaking a few penguins.

  2. I do think that most of these geoengineering schemes are "insane" but not quite for the reasons given by Gore. Most of them are "insane" in the conventional sense that they are massively expensive harebrained boondoggles that will accomplish nothing positive-assuming they accomplish anything at all, which is unlikely.

    The worst case scenario would be them working "too well" and jump starting the next ice age, in which case it would seem Messrs. Niven, Pournelle, and Flynn will have proven quite prescient in their novel Fallen Angels, albeit not quite for the right reasons.

  3. We do geoengineering all the time on various scales and there is nothing inherently wrong with doing it on a global scale. Gore is wrong even if all the ideas presented so far are useless and/or harmful.
    It is of minor consequence but I think that it would be better to put all the investments of public money in wind and solar power into the technology of natural gas recovery. At least the money would not be completely wasted.

  4. thejollygreenmanJan 21, 2014, 7:01:00 PM

    They share one common trait, their aversion to soap and a shower.

  5. I support geoengineering research. Maybe we can find something which works and the idea of having control over a planet's thermostat seems rather attractive.

  6. Yes, leave him in peace to make millions from the scam.

  7. If we were to have 2nd thoughts about the sulphur after releasing it, how would we get rid of that shit?

    Deflecting from space seems better in this regard. Like a fresnel lens, or something, at L1. Then you could just push it out of the way if it's deemed unwanted; no permanent harm done(except the ridiculous amount of money wasted on that thing...)

    Is something like that technologically feasible?

  8. It has become very clear that the mechanism of climate alarmism has had its day. Are any of you so enamored with the lure of geoengineering you would cede control of Earth’s climate to real mad scientists? Not everything that is possible should be attempted.

    Climate change is an enemy murdering our grandchildren? Must we mount a war effort second to none to attack and defeat this venal killing
    machine? We’re fighting for our very existence and we must win at any cost?

    I don’t think so………

  9. What Gore is doing IMO, is covering his own back.
    He sees the game is up and he's probably extorted all of the squillions he can out of CAGW, so now he has to make sure he doesn't attract the blame for the damage to the earth and us all that desperate geo-engineering measures his alarmism [ aka carpetbagging] has precipitated, could wreak.
    Probably scared someone---some country even---is going to sue the socks off him.

  10. Sulfur dioxide's lífetime in the troposphere is about one day. Once it reaches the stratosphere it's several days to several weeks. But the key to successful engineering is to research it thoroughly before moving on to a pilot phase. I wouldn't be backing a particular solution...yet.

  11. I think Al Gore is against geoengineering because he has no investments positioned to profit from the concept. This could in fact detract from the returns on his green energy BS investments.

  12. Hi,

    Your obviously thinking at a level magnitudes above my capability. When I say man made climate change deniers are science deniers please explain the flaw in my logic.

    , its true is that if you increase the concentration of CO2 in an atmosphere it will retain more heat. No one questions we have an atmosphere, a large complex totally organic HVAC element providing for a thriving human existence. No one denies the concentration of CO2 in the atmosphere has been rising dramatically over the last one hundred years. We can debate about the source, if you said the majority was natural you would be wrong. Remember 2010 eruptions of Eyjafjallajökull shut down air travel in Europe. Well it cut carbon emission from European airplanes by about two thirds. Yes the volcano put out a third of the gas that planes do during the same period. Maybe we could activate more volcanoes resulting in an end to carbon based airtravel and save the planet, or maybe not, but what if that where true?

    I am Canadian. There is a huge debate over development of what the Oil lobby calls the Oil sands. If this label was correct they would be lakes of oil, not solids banks of tar. The extraction of oil precursors from this single geologic formation creates CO2 emissions equal to the output of half the cars in Canada. Consider that the product then need energy intensive refining to make fuel, and then further energy intensive transport over many thousands of kilometers to market. Therefore just from the standpoint of economic efficiency does the whole thing make sense. It reminds me of the Soviet Union where the KGB sold natural resources for pennies on the dollar to the west. Would we not be better off creating hydrogen with this energy?

    In summary, CO2 traps heat in a sealed environment. The CO2 released by Man is large enough to have some effect. We can measure the amount of CO2 in the atmosphere going back hundreds of thousands of years. It now equals the an amount similar to the highest levels seen. How can anyone deny man made climate change?

  13. "its true is that if you increase the concentration of CO2 in an atmosphere it will retain more heat".
    Only if you assume no other changes from increasing the CO2 concentration, and even then only if you assume no changes from the assumed increase in heat. There would be change, as long as everything stays the same. Ignore changes in flora, clouds, and the rest of the planet.
    Even if we make those unreasonable assumptions, and that there would be a net increase in heat retained, it does not mean the amount of heat would be significant. You are speaking qualitatively, not quantitatively. My cat walking around my apartment heats in more than when he sleeps, yet this is of no significance. But, yes, it is a scientific fact that my apartment is warmer from this feline activity.

  14. True our atmosphere is complex, but can we not agree pumping billions of tons of CO2 into it will have some effect?
    could we not calculate the total size of the atmosphere, and the total amount of man made CO2 and draw reasonable conclusions?

  15. Increasing CO2 has some effects. Most importantly, a 1% of CO2 increase raises the productivity of plants by something like 0.5%. That's good for all life on Earth.

    So the raising CO2 has a rather clear and direct positive effect due to CO2's being the main plant food. It's still silly to try to raise CO2 artificially because it would be too expensive.

    The indirect effects of increasing CO2 concentrations, like the effect on the temperatures through the greenhouse effect, are tiny and virtually negligible (but in principle nonzero and beneficial).

    Because the effects of increasing CO2 on other things are negligible, a rational person recognizes the increasing CO2 as what it actually is - a negligible side effect of some processes and activities that are vastly more important than the side effect.

  16. What evidence do you have that they are negligible? I believe the trend is upward. If air temperatures as I am told have stablized, I think there is a connection to risiing sea temperatures, That is the oceans are becoming a heat sink.

    Using fuel more efficiently thereby reducing emission has far more benifts than burning coal to get cheap electricity.

  17. There are no net benefits in reducing energy production and consumption. It's enough to look in poor Africa or North Korea what life without enough energy looks like.

    There are about 1,000 articles on this blog explaining why the indirect effects of CO2 are negligible. The claim that is really being proven is that the scientically conceivable changes caused by changing CO2 are orders of magnitude smaller than other contributions that matter.

    The fact that a higher temperature is beneficial (tiny temperature increase is a tiny benefit) may be seen by comparing the interglacials with ice ages and dividing the result by 10.

  18. Lubos your close to absolute statement about there being no net benefits certainly needs some qualification,

    First the source of energy is very important to make that statement true. Lets assume your correct and the earth can easily handle an infinite amount of human production of CO2.

    Nuclear Energy has been proven unsafe. No running of cost benifts will come up with a profit for Japan and Ukraine for generating energy this way..

    Coal has billions of health care costs and degradation of the quality of life on its tally.

    Natural Gas seems a pretty good compromise. However the wind, sun, water and earth are the winning team.

    I say that from the point of view of energy consumed to provide the input of production. Efficiency means cheaper goods and more prosperity. Less pollution means more valuable land because you do not have the NIMBY effect distorting decisions about power plant placement.

    Africa and North Korea do not have enough energy but that has nothing to do with a choice to have less energy. They have made serious political mistakes.

    As we can see today in China, just producing energy has serious unintended consequences.

    I still think its just common sense to adopt as sustainable a profile as possible.

    Look at all the crazy stuff we do today. Like developing tar sands when the energy consumed to make benzine exceeds the energy produced by the end product, not logical.

    Worldwide about 30% of the energy in oilfields is wasted because the natural gas is burned off instead of harvested. Do not get me started on helium,

  19. Dear Steve, you are fighting extraterrestrial aliens if you talk about "infinite CO2" and similar stuff.

    We can't reach "infinite CO2 concentrations". Even if we burn all the classical and alternative fossil fuels, the concentration will stay below 2,000 ppm which is totally fine from all points of view. I would have trouble if the Earth were to go to 5,000 ppm - people may start to feel uncomfortable - but it can't happen. At any rate, even 1,000 is science-fiction today because we will probably reach at most 600 ppm by 2100.

    Wind, solar etc. are financial catastrophes and even the greenest nations are canceling these things - simply by stopping the subsidies on which these sources relied upon.

    Nuclear energy is safe when done right, coal has been environmentally fine for decades since filters were added to the plants.

    Africa hasn't made any political mistake. It just hasn't developed yet. That's why most African economies look like ours many centuries ago or worse.

    North Korea has made political mistakes and is doing them all the time, but my point is that you want the West to do some pretty analogous mistakes.

    There may be inefficiencies in the production of oil and natural gas etc. but be sure that those who own the resources are trying hard not to lose their money. If you have a way to easily improve the efficiency by 30%, you may earn quite some money.


  20. You are completely correct, Lubos. Only a fool would doubt it. But Africa will advance much more quickly than we did.