riemann &laquo; WordPress.com Tag Feed http://en.wordpress.com/tag/riemann/ Feed of posts on WordPress.com tagged "riemann" Fri, 27 Nov 2009 18:54:42 +0000 http://en.wordpress.com/tags/ en <![CDATA[A Formula for Primes]]> http://sumidiot.wordpress.com/2009/11/24/a-formula-for-primes/ Wed, 25 Nov 2009 04:55:08 +0000 sumidiot http://sumidiot.wordpress.com/2009/11/24/a-formula-for-primes/

I wanted to write this post yesterday, but I couldn’t work out some of the steps that I wanted to be able to work out, so I had to wait a day and try again. But I think I’ve got it now, let’s see…

We’ve been working with the function

\displaystyle J(x)=\sum_{p^n\leq x}\frac{1}{n}.

Last time, we arrived (skipping over some rather large gaps) at an analytic formula for J(x), namely:

\begin{array}{l}\displaystyle J(x)=Li(x)-\sum_{\text{Im }\rho>0}\left(Li(x^{\rho})+Li(x^{1-\rho})\right)\\ \displaystyle\qquad\qquad +\int_x^{\infty}\frac{dt}{t(t^2-1)\ln t}-\ln 2,\end{array}

where (mod issues)

\displaystyle Li(x)=\int_0^x \frac{dt}{\ln t}.

The goal today is to use this formula for J(x) to find a formula for primes. In particular, we will find a formula for \pi(x), which is the number of primes no larger than x. These two functions are related, as follows: recall that J(x) is a step function which jumps by 1/n at every prime power p^n. Pick an x_0. How many jumps of size 1 does J(x) make before getting to x_0? Well, it jumps by 1 at the primes, so it makes \pi(x_0) jumps of size 1. How many jumps of size 1/2? These are jumps at prime squares, p^2\leq x_0. This inequality is the same as p\leq x_0^{1/2}, so we make \pi(x_0^{1/2}) jumps of size 1/2. Similarly, we’ll make \pi(x_0^{1/n}) jumps of size 1/n to calculate J(x_0). Eventually (for some large enough N) x_0^{1/N} will be less than 2, and so \pi(x_0^{1/N})=0, and we can stop looking for jumps.

Translating the above into symbols, we have just decided that

J(x)=\pi(x)+\frac{1}{2}\pi(x^{1/2})+\frac{1}{3}\pi(x^{1/3})+\cdots+\frac{1}{n}\pi(x^{1/n})+\cdots.

I’ve written it as an infinite sum, but eventually (like I said) those \pi values will be 0, so we’ve really got a finite sum.

Now wait, our goal was a formula for \pi(x), in terms of J(x), and I’ve got a formula the other way, J(x) in terms of \pi(x).

This is the part I got stuck on yesterday. Edwards’ book says (or perhaps doesn’t, causing my problems) that basically you do Möbius inversion to the sum above, and obtain a formula for \pi(x) in terms of J(x). He says slightly more than that, but I wasn’t able to piece it together as easily as he seemed to indicate it should work. But anyway, what’s this Möbius inversion thing? I’ll take a pretty low-level account, leaving lots of room for things to be jazzed up (which I may return to and do sometime soonish).

An integer is said to be “squarefree” if it factors into the product of distinct primes or, equivalently, it is not divisible by a square. Define the following function, called the Möbius function, on positive integers (p_i is supposed to indicate a prime):

\displaystyle \mu(n) = \begin{cases}1& n=1 \\ 0 & n\text{ not squarefree} \\ (-1)^k & n=p_1\cdots p_k\text{ squarefree}\end{cases}

So, for example, \mu is -1 at any prime, \mu(6)=\mu(10)=1, and \mu(8)=\mu(12)=0.

Möbius inversion is then the following:

If f(n)=\sum_{d|n}g(d) then g(d)=\sum_{d|n}\mu(d)f(n/d), and conversely.

Well, I tried to figure out how my expression for J(x), in terms of \pi(x^{1/n})s was such an f of g. I don’t see it, as stated. Perhaps somebody can point me in the right direction in the comments.

The observation you are supposed to make comes when you consider something like \frac{1}{2}J(x^{1/2}). Using our formula above, we can write

\frac{1}{2}J(x^{1/2})=\frac{1}{2}\pi(x^{1/2})+\frac{1}{4}\pi(x^{1/4})+\frac{1}{6}\pi(x^{1/6})+\cdots

or, more generally, we have

\displaystyle \frac{1}{n}J(x^{1/n})=\sum_k \frac{1}{nk}\pi(x^{1/nk}).

If we subtract this equation (with n=2) from the equation for J(x) in terms of \frac{1}{n}\pi(x^{1/n})s, we can eliminate one term from the right-hand side. The idea is to iterate this process. I think a (reasonably, by my blog’s standards) clean way to write what is happening is as follows (it helps when you know the answer you are supposed to get :) ):

Consider the sum

\displaystyle \sum_n \frac{\mu(n)}{n}J(x^{1/n}),

which by our most recent formula is the same as

\displaystyle \sum_{n,k}\frac{\mu(n)}{nk}\pi(x^{1/nk}).

Now, fix a positive integer m. The coefficient of \pi(x^{1/m}) in this last double sum is

\displaystyle \sum_{nk=m}\frac{\mu(n)}{m}=\frac{1}{m}\sum_{n|m}\mu(n).

Almost there… what is \sum_{n|m}\mu(n)? Turns out, it is 1 if m=1 (which is clear), and 0 otherwise. Indeed, suppose m has prime factorization p_1^{e_1}\cdots p_k^{e_k}. Then all of the divisors, n in the sum we’re after, are of the form p_1^{f_1}\cdots p_k^{f_k} where each 0\leq f_i\leq e_i. The only divisors, n, where \mu(n)\neq 0 are those where the f_i are all either 0 or 1. So the worthwhile n could be 1, or some p_i, or a p_ip_j (i\neq j), or p_ip_jp_{\ell} (i\neq j, i\neq \ell, j\neq \ell), etc. So

\begin{array}{l}\displaystyle \sum_{n|m}\mu(n)=1+\sum_i \mu(p_i)+\sum_{i,j} \mu(p_ip_j)+\cdots+\mu(p_1\cdots p_k) \\ \qquad \qquad \displaystyle =1-k+\binom{k}{2}-\binom{k}{3}+\cdots+(-1)^k=(1-1)^k=0.\end{array}

(By the way, this is (almost verbatim) the proof in Hardy and Wright’s “An Introduction to the Theory of Numbers”, and I’ll count that as one of my references for anything I say about \mu).

So, let’s gather back up here. We’ve now shown that in the sum

\displaystyle \sum_n \frac{\mu(n)}{n}J(x^{1/n})=\sum_{n,k}\frac{\mu(n)}{nk}\pi(x^{1/nk})

the coefficient of \pi(x^{1/m}) is 0 unless m=1, when the coefficient is 1. Therefore, we have obtained out formula for \pi(x):

\begin{array}{l} \pi(x)=\displaystyle \sum_n \frac{\mu(n)}{n}J(x^{1/n})\\ \qquad \displaystyle=J(x)-\frac{1}{2}J(x^{1/2})-\frac{1}{3}J(x^{1/3})-\frac{1}{5}J(x^{1/5})+\frac{1}{6}J(x^{1/6})-\cdots\end{array}

I guess the point of Riemann’s paper was that the terrible integral formula he obtained for J(x) could then be used in this expression for \pi(x), to give an exact analytic expression (that is, one only an analyst would find pretty) for \pi(x). Pretty wild, in my mind.

It seems that if you get some feeling for the terms in J(x), you can decide that the “dominant” term is the Li(x) term. And then that still ends up as the dominant term in the expression above for \pi(x). And so you get to say

\pi(x)\sim Li(x).

So you’ve got that going for you, which is nice.

]]> <![CDATA[Another Formula for J(x)]]> http://sumidiot.wordpress.com/2009/11/22/another-formula-for-jx/ Mon, 23 Nov 2009 03:08:19 +0000 sumidiot http://sumidiot.wordpress.com/2009/11/22/another-formula-for-jx/

Yesterday, I related the logarithm of \zeta(s) to a piecewise linear function J(x). You may recall that J(x) was defined for positive reals by setting it equal to 0 at x=0, and then jumping by 1/n whenever x=p^n, for some prime p and integer n. At the end of the day, we got to

\displaystyle J(x)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\ln \zeta(s) x^s\frac{ds}{s}

where a=\text{Re}(s)>1. Today, we’ll analyze \ln \zeta(s) some more, and re-write the formula above.

When I introduced \zeta(s), I ended with the following formula:

\displaystyle \zeta(s)=\frac{\Gamma(1-s)}{2\pi i}\int_{+\infty}^{+\infty} \frac{(-x)^s}{e^x-1} \frac{dx}{x}

where the bounds on that integral are supposed to represent a curve that “starts” at the right-hand “end” of the real line, loops around 0, and then goes back out the positive axis to infinity. I’m not good enough at complex line integrals at this point to say any more about this. But apparently if you are good at these sorts of integrals, using Cauchy’s integral formula and things you can find the so-called “functional equation”

\zeta(s)=\Gamma(1-s)(2\pi)^{s-1}2\sin(s\pi/2)\zeta(1-s).

If you then use the relation I mentioned previously:

\Gamma(s)\Gamma(1-s)\sin(\pi s)=\pi

(well, you use this for s=s/2), and one I haven’t mentioned:

\Gamma(1-s)=2^{-s}\Gamma(1-\frac{s}{2})\Gamma(\frac{1-s}{2})\pi^{-1/2}

and move some symbols around, you arrive at a more symmetric equation:

\Gamma(\frac{s}{2})\pi^{-s/2}\zeta(s)=\Gamma(\frac{1-s}{2})\pi^{-(1-s)/2}\zeta(1-s).

Notice that if you plug 1-s in the formula on the left-hand side, you obtain the right-hand side.

This function on the left-hand side apparently has poles at 0 and 1, so if we define

\xi(s)=\frac{s(s-1)}{2}\Gamma(\frac{s}{2})\pi^{-s}{2}\zeta(s)

then we obtain an entire analytic function satisfying \xi(s)=\xi(1-s). Using the factorial relation for \Gamma, we can re-write \xi(s) as

\xi(s)=(s-1)\Gamma(\frac{s}{2}+1)\pi^{-s/2}\zeta(s).

I get the impression that if you know what you are doing, then the things above aren’t tooo hard to justify. Apparently the next part is a bit trickier. Apparently, you can write

\xi(s)=\xi(0)\prod_{\rho}(1-\frac{s}{\rho}),

where the product is indexed over the roots, \rho, of \xi (so \xi(\rho)=0).

If you’ve heard anything about the Riemann hypothesis, you know that the roots (the “non-trivial” ones, I didn’t talk about the trivial ones) of \zeta(s) are a big deal. Our second formula for \xi(s) shows that they are (basically) the same as the roots of \xi(s), and so they are the \rho that the sum above is indexed over. The symmetric equation from earlier has a little something to say about the zeroes, and it has been shown that all of the zeroes have real part bigger than 0 and less than 1 (this is called the “critical strip”). The hypothesis (whose truth won’t affect what we’re saying below) is that all of the zeroes have real part 1/2 (this is the “critical line”). Apparently Riemann didn’t need this hypothesis for the things in his paper that introduced it, so I don’t really have much more to say about it right now. Although, honestly, I still don’t see what all the fuss is about :) The formulas we’ll get below and tomorrow work even if the roots aren’t on the critical line (unless I’m missing something important. If I am, please comment).

Anyway, back to the topic at hand. Let me try to convince you that it isn’t horribly unreasonable to think about writing a function as a product over its roots, as I’ve done above. For the sake of example, let f(x)=3x^3+3x^2-30x+24 (or pick your own favorite polynomial). The usual way this would get factored, in all the classes I’ve ever taken or taught, is (up to permutation) f(x)=3(x+4)(x-1)(x-2), showing that the roots are x=1,2,-4. However, if you factor a 4 out of the x+4 term, and -1 and -2 out of the other terms, you can also write f(x)=24(1-\frac{x}{-4})(1-x)(1-\frac{x}{2}). You still see all the zeroes when you write the polynomial this way. You can also see that the coefficient in the front is f(0). So we’ve written f(x)=f(0)\prod_{\rho}(1-x/\rho), which is the same goal as what we’re doing with \xi above. Incidentally, the idea of writing a function this way was also used by Euler to establish \zeta(2)=\sum 1/n^2=\pi^2/6 (I’ve mentioned this briefly elsewhere).

We now have two formulas for \xi(s), so we can put them together to get

\xi(0)\prod_{\rho}(1-\frac{s}{\rho})=(s-1)\Gamma(\frac{s}{2}+1)\pi^{-s/2}\zeta(s).

Recalling that our formula for J(x), at the beginning, involved \ln\zeta(s), let’s take the log of the equation above and solve for the \zeta term:

\begin{array}{l} \ln \zeta(s)=\ln \xi(0)+\sum_{\rho}\ln(1-\frac{s}{\rho})\\ \qquad\qquad -\ln \Gamma(\frac{s}{2}+1)+\frac{s}{2}\ln \pi-\ln(s-1).\end{array}

The idea is now to plug this in the formula for J(x). Apparently if you do, though, you’ll have some issues with convergence. So actually try to do the integral in J(x), using integration by parts (hint: dv=x^s ds). The “uv” term goes to 0 and you obtain

\displaystyle J(x)=\frac{-1}{2\pi i}\cdot \frac{1}{\ln x}\int_{a-i\infty}^{a+i\infty}\frac{d}{dx}\left[\frac{\ln\zeta(s)}{s}\right]x^s\ ds,

where, as before a=\text{Re } s. Now plug in the 5 terms we’ve got above for \ln \zeta(s), and you get a formula for J(x). What happens to the terms? Can you actually work out any of the integrals?

Well, you might be able to. I’m not. Not right now anyway. But I can tell you about what others have figured out (rather like I’ve been doing all along, in fact)…

It’s clear that the \frac{s}{2}\ln \pi term drops out, because you divide by s and then take the derivative of a constant and just get 0. The term with \ln\xi(0) ends up just giving you \ln\xi(0), which is -\ln(2).

The term corresponding to the term with \Gamma in it can be rewritten as

\displaystyle \int_x^{\infty}\frac{dt}{t(t^2-1)\ln t}

(as if that were helpful).

The important terms seem to involve the function

\displaystyle Li(x)=\int_0^x \frac{dt}{\ln t}.

Of course, this integrand has a bit of an asymptote at 1, so really Li(x) (in Edwards’ book, anyway) is the “Cauchy principal value” of this integral, namely

\displaystyle Li(x)=\lim_{\epsilon\to 0^+}\int_0^{1-\epsilon}\frac{dt}{\ln t}+\int_{1+\epsilon}^x \frac{dt}{\ln t}.

This function is, rather famously, related to approximating the number of primes less than a given bound. In fact, tomorrow I plan on having more to say about this. But back to the terms in our integral for J(x).

The term corresponding to the sum over the roots ends up giving you

\displaystyle -\sum_{\text{Im }\rho>0}Li(x^{\rho})+Li(x^{1-\rho}).

But apparently the dominant term is the term corresponding to \ln (s-1). It actually gives you Li(x)

So, finally, we have written

\begin{array}{l}\displaystyle J(x)=Li(x)-\sum_{\text{Im }\rho>0}\left(Li(x^{\rho})+Li(x^{1-\rho})\right)\\ \displaystyle\qquad\qquad +\int_x^{\infty}\frac{dt}{t(t^2-1)\ln t}-\ln 2.\end{array}

Doesn’t that make you feel better? We started with the reasonably understandable

\displaystyle J(x)=\sum_{p^n\leq x}\frac{1}{n},

and created the monstrosity above. I guess this is why I’m not an analyst. To me, it seems worse to write J(x) as this terrible combination of lots of integrals. But apparently it’s useful in analysis to have such formulas. I guess we’ll see a use tomorrow…

]]> <![CDATA[Einstein speaks on General Relativity]]> http://adonis49.wordpress.com/2009/11/22/einstein-speaks-on-general-relativity/ Sun, 22 Nov 2009 09:53:56 +0000 adonis49 http://adonis49.wordpress.com/2009/11/22/einstein-speaks-on-general-relativity/

Einstein speaks on General Relativity; (Nov. 20, 2009)

 

I have already posted two articles in the series “Einstein speaks on…” This article describes Einstein’s theory of restricted relativity and then his concept for General Relativity. It is a theory meant to extend physics of fields (for example electrical and magnetic fields among others) to all natural phenomena, including gravity. Einstein declares that there was nothing speculative in his theory but it was adapted to observed facts.

The fundamentals are that the speed of light is constant in the void and that all systems of inertia are equally valid (each system of inertia has its own metric time). The experience of Michelson has demonstrated these fundamentals. The theory of restrained relativity adopts the continuum of space coordinates and time as absolute since they are measured by clocks and rigid bodies with a twist: the coordinates become relative because they depend on the movement of the selected system of inertia.

The theory of General Relativity is based on the verified numerical correspondence of inertia mass and weight. This discovery is obtained when coordinates posses relative accelerations with one another; thus each system of inertia has its own field of gravitation. Consequently, the movement of solid bodies does not correspond to the Euclidian geometry as well as the movement of clocks. The coordinates of space-time are no longer independent. This new kind of metrics existed mathematically thanks to the works of Gauss and Riemann.

Ernst Mach realized that classical mechanics movement is described without reference to the causes; thus, there are no movements but those in relation to other movements.  In this case, acceleration in classical mechanics can no longer be conceived with relative movement; Newton had to imagine a physical space where acceleration would exist and he logically announced an absolute space that did not satisfy Newton but that worked for two centuries. Mach tried to modify the equations so that they could be used in reference to a space represented by the other bodies under study.  Mach’s attempts failed in regard of the scientific knowledge of his time.

We know that space is influenced by the surrounding bodies and so far, I cannot think the general Relativity may surmount satisfactorily this difficulty except by considering space as a closed one assuming that the average density of maters in the universe has a finite value, however small it might be.

]]> <![CDATA[The Log of Zeta]]> http://sumidiot.wordpress.com/2009/11/21/the-log-of-zeta/ Sun, 22 Nov 2009 04:49:08 +0000 sumidiot http://sumidiot.wordpress.com/2009/11/21/the-log-of-zeta/

Last time (too long ago, sorry), I finally got around to talking about the function \zeta(s). Today, I think I’d like to talk about \log \zeta(s).

First, though, I should mention Euler’s product formula. The goal is to re-write the infinite series \zeta(s)=\sum_n n^{-s} as a product. For each prime p, we can tease out of the series above those terms where n is p to some power. That is, we can think about the sub-series \sum_{k\geq 0} (p^k)^{-s}. This is a geometric series that converges to (1-p^{-s})^{-1}. If we take two such series and multiply them together, which terms from our starting series for \zeta(s) will we recover?

Suppose p and q are two primes, and consider the product

\displaystyle\left(1+\frac{1}{p^s}+\frac{1}{p^{2s}}+\frac{1}{p^{3s}}+\cdots\right)\cdot \left(1+ \frac{1}{q^s}+\frac{1}{q^{2s}}+\frac{1}{q^{3s}}+\cdots\right).

The terms in \sum_n n^{-s} that we obtain from this product are those where n is a product of some power (possibly the 0 power) of p and some power (again, possibly 0) of q. We could then think about another prime, say r, and it’s geometric series, and multiply it by the above, and obtain all the terms in \sum_n n^{-s} where n=p^aq^br^c for some 0\leq a,b,c.

Continuing on, and recalling that every positive integer has a unique factorization as a product of primes, we obtain the magical formula (isn’t most of Euler’s work magical?):

\displaystyle \sum_n \frac{1}{n^s}=\prod_p (1-p^{-s})^{-1},

where the product is indexed by the primes. What’s nice about this, for us, at the moment, is that logarithms work well with products and powers, but not soo well for sums. Recalling the Taylor polynomial

\ln(1-x)=-1-x-x^2-x^3-\cdots=-\sum\limits_{n=0}^{\infty} x^n

we find

\displaystyle \ln \zeta(s)=\sum_p\sum_n \frac{1}{n}p^{-ns}.

That was fun. Incidentally, it’s not much of a jump from here to show that the series of prime reciprocals diverges. I mentioned it in a post a while ago.

Let’s switch gears for a little bit. I’m going to define a function J(x) on the positive reals. If you’re following Edwards’ book with me (as I jump all over the place), this will be almost exactly his J(x), differing only at the integers (for today anyway :) ). Let me start by setting J(0)=0. The function J(x) will be a step function, which I’ll define to mean piecewise linear with slope 0 on each piece. So to define J(x) I only need to tell you when to jump, and by how much. The rule is: when you get to an integer n, if n=p^k for some prime p, then you jump up by 1/k. So at 2,3,5,7,11,\ldots you jump by 1, at 4,9,25,49,\ldots you jump by 1/2, at 8, 27, \ldots you jump by 1/3, etc. Here’s a little picture of J(x), with a few values highlighted:

Slightly more precisely, we can write

\displaystyle J(x)=\sum_{p^n\leq x}\frac{1}{n}.

Now, let me see if I can convince you that there is some justification in writing

\displaystyle \ln \zeta(s)=s\int_0^{\infty} J(x)x^{-s-1}\ dx.

Let’s work with the right-hand side. Since J(x)=0 near x=0, I’ll actually start my integral at x=1. I think the way to go about it is as follows:

\begin{array}{rcl} \displaystyle s\int_1^{\infty}J(x)x^{-s-1}\ dx &=& \displaystyle s\sum_{m=1}^{\infty}\int_m^{m+1}J(x)x^{-s-1}\ dx \\ {} & = & \displaystyle \sum_{m=1}^{\infty}J(m)\left(\frac{1}{m^s}-\frac{1}{(m+1)^s}\right).\end{array}

Now suppose that J(m)=J(m+1) for some m. Then the terms corresponding to m and m+1 telescope as follows:

\begin{array}{c} \displaystyle J(m)\left(\frac{1}{m^s}-\frac{1}{(m+1)^s}\right)+J(m+1)\left(\frac{1}{(m+1)^s}-\frac{1}{(m+2)^s}\right)\\ =\displaystyle J(m)\left(\frac{1}{m^s}-\frac{1}{(m+2)^s}\right)\end{array}.

If, also, J(m)=J(m+2), this we can telescope another term into this one, and so on. So, really, the important m in this sum are those where J(x) jumps, which are the prime powers. Let m_i=p_i^{n_i} be the i-th prime power (i.e., the point where J(x) makes its i-th jump), starting with m_1=2. Then we can write

\begin{array}{rcl} \displaystyle s\int_0^{\infty}J(x)x^{-s-1}\ dx & = & \displaystyle \sum_i J(m_i)\left(\frac{1}{m_i^s}-\frac{1}{m_{i+1}^s}\right) \\ & = & \displaystyle \frac{1}{2}J(2)+\sum_{i=2}^{\infty} -J(m_{i-1})\frac{1}{m_i^s}+J(m_i)\frac{1}{m_i^s} \\ & = & \displaystyle \frac{1}{2}+\sum_{i=2}^{\infty} \frac{1}{m_i^s}\left(J(m_i)-J(m_{i-1})\right) \\ & = & \displaystyle \frac{1}{2}+\sum_{i=2}^{\infty} \frac{1}{m_i^s}\cdot \frac{1}{n_i} \\ & = & \displaystyle \sum_{p}\sum_n \frac{1}{n}p^{-ns} \\ & = & \ln \zeta(s).\end{array}

What good is writing \ln \zeta(s) this way? I guess if you know something about Fourier inversion (which I don’t) then you get to say that

\displaystyle J(x)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\log \zeta(s)x^s\frac{ds}{s},

for a=\text{Re}(s)>1. What good is that? I think I’ll have to read some more and tell you about it tomorrow, but it’ll turn out to be useful once we have yet another description of \ln \zeta(s), in terms of the 0s of \zeta(s) (finally getting to those 0s everybody cares so much about).

]]> <![CDATA[Einstein speaks of his mind processes on the origin of General Relativity]]> http://adonis49.wordpress.com/2009/11/21/einstein-speaks-of-his-mind-processes-on-the-origin-of-general-relativity/ Sat, 21 Nov 2009 08:52:16 +0000 adonis49 http://adonis49.wordpress.com/2009/11/21/einstein-speaks-of-his-mind-processes-on-the-origin-of-general-relativity/

Einstein speaks of his mind processes on the origin of General Relativity; (Nov. 21, 2009)

This article is how Einstein described his mind processes that lead to the theory of restricted relativity and then his concept for General Relativity. In 1905, restricted relativity discovered the equivalence of all systems of inertia for formulating physics equations. From a cinematic perspective there was no way to doubting relative movements; still there was the tendency to physically extend privileged significance to system of inertia.  The question was “if speed is relative do we have to consider acceleration as absolute?”

Ernest Mach considered that inertia did not resist acceleration except when related to the acceleration toward other masses. This idea impressed me greatly. First, I had to establish a law of gravitation field and suppress the concept of absolute simultaneity. Simplicity urged me to maintain Laplace’s scalar gravity potential and fine tune Poisson’s equation. Given the theorem of inertia of energy then inertia mass must be depended on gravitation potential but my research left me skeptical. In classical mechanics, vertical acceleration in a vertical field of gravity is independent of the horizontal component of velocity; it follows that vertical acceleration is exercised independently of the internal kinetic energy of the body in movement. I discovered that this independence did not exist in my draft theory; this evidence did not coincide with the affirmation that all bodies submit to the same acceleration in a gravitational field. Thus, the principle that there is equality between inertia mass and weight grew with striking significance. I was convinced of its validity though I had no knowledge of the results of experiments done by Eotvos.

Consequently, the principle of equality between inertia mass and weight would be explained as follow: in a homogenous gravitational field all movements are executed in relation to a system of coordinates accelerating uniformly as if in absence of gravity field. I conjectured that if this principle is applicable to any other events then it can be applied to system of coordinates not accelerating uniformly. These reflections occupied me from 1908 to 1911 and I figured that the principle of relativity needed to be extended (equations should retain their forms in non uniform accelerations of coordinates) in order to account for a rational theory of gravitation; the physical explanation of coordinates (measured by rules and clocks) has to go.

I reasoned that if in reality “a field of gravitation used in system of inertia” did not exist it could still be served in the Galilean expression that “a material point in a 4-dimentional space is represented by the shortest straight line”. Minkowski has demonstrated that this metric of the square of the distance of the line is a function of the squares of the differential coordinates.  If I introduced other coordinates by non linear transformation then the distance of the line stay homogeneous if coefficients dependent on coordinates are added to the metric (this is the Riemann metric in 4-dimension space not submitted to any gravity field). Thus, the coefficients describe the field of gravity in the selected system of coordinates; the physical significance is just related to the Riemannian metric. This resolved this dilemma in 1912.

Two other problems had to be resolved from 1912 to 1914 with the collaboration of Marcel Grossmann. The first problem is stated as follow: How can we transfer to a Riemannian metric a field law expressed in the language of restrained relativity?  I discovered that Ricci and Levi-Civia had answered it using infinitesimal differential calculus.  The second problem is: what are the differential laws that determine the coefficients of Riemann?  I needed to resolve invariant differential forms of the second order of Riemann’s coefficients. It turned out that Riemann had also answered the problem using curb tensors.

“Two years before the publication of my theory on General Relativity” said Einstein “I thought that my equations could not be confirmed by experiments. I was convinced that an invariant law of gravitation relative to any transformations of coordinates was not compatible with the causality principle. Astronomic experiments proved me right in 1915.”

Note:  I recall that during my last year in high school my physics teacher, an old Jesuit Brother, filled the blackboard with partial derivatives of Newton’s equation on the force applied to a mass; then he integrated and he got Einstein’s equation of energy equal mass by C square. At university, whenever I had problems to solve in classical mechanics on energy or momentum conservations I just applied the relativity equation for easy and quick results; pretty straightforward; not like the huge pain of describing or analyzing movements of an object in coordinate space.

]]> <![CDATA[Einem TI Voyage200 Integralrechnung beibringen...]]> http://pyjamaphysik.wordpress.com/2009/11/16/einem-ti-voyage200-integralrechnung-beibringen/ Mon, 16 Nov 2009 18:29:50 +0000 pyjamaphysik http://pyjamaphysik.wordpress.com/2009/11/16/einem-ti-voyage200-integralrechnung-beibringen/

Riemann Integral

Der TI-Voyage200, wie er an vielen Schulen in Deutschland eingesetzt wird, kann zwar sicherlich Integralrechnung, aber wir wollen ihm einmal ein bisschen mehr beibringen:

Wir fassen dazu das Integral abweichend von der Definition des Standardintegrals als Summe der Flächeninhalte von Rechtecken auf, die zwischen Funktion und Abszisse “gequetscht” werden. (Riemann-Integral)

Die Summe aller Rechtecksflächeninhalte ist, wenn man für die Höhe der Rechtecke den Funktionswert in der Mitte der Rechtecksbreite annimmt, gegeben durch:

\sum\limits_{i=0}^{n-1}\frac{b-a}{n}f\left(a+\frac{b-a}{2n}+\frac{b-a}{n}\,i\right)

B bezeichnet die Integrationsobergrenze, a die Integrationsuntergrenze und n die Anzahl der Rechtecke. Wir wollen dem Rechner nun eine Funktion beibringen, die den Flächeninhalt in Abhängigkeit von a,b und n angibt. Dazu bilden wir von dem gesamten Term den Grenzwert für n gegen p, damit wir für n sowohl jede reele Zahl als auch unendlich annehmen können:

\lim\limits_{n\rightarrow p}\sum\limits_{i=0}^{n-1}\frac{b-a}{n}f\left(a+\frac{b-a}{2n}+\frac{b-a}{n}\,i\right)

Nun können wir obigen Term unter i(a,b,p) abspeichern, wobei p nun die Anzahl der Rechtecke ist, für ein exaktes Ergebnis also unendlich betragen muss.

Rotationskörper

Nehmen wir nun als Körper unter der Abszisse keine Rechtecke, sondern Zylinder an, um einen Körper zu kreieren, der aussieht, wie die zwischen Graph und x-Achse entstandene Fläche, wenn sie um die x-Achse rotiert, so können wir den obigen Term mit der Formel für das Volumen eines Zylinders (V=h\cdot\pi r^2) abändern (bemerke h=\frac{b-a}{n} und r=f(x_i)):

\lim\limits_{n\rightarrow p}\sum\limits_{i=0}^{n-1}\frac{b-a}{n}\pi \left(f\left(a+\frac{b-a}{2n}+\frac{b-a}{n}\,i\right)\right)^2

Abgespeichert unter v(a,b,p) erhalten wir nun also eine allgemeine Funktion für die Berechnung des Volumens des Rotationskörpers.

]]> <![CDATA[Riemann's Zeta Function]]> http://sumidiot.wordpress.com/2009/11/13/riemanns-zeta-function/ Sat, 14 Nov 2009 03:28:33 +0000 sumidiot http://sumidiot.wordpress.com/2009/11/13/riemanns-zeta-function/

I guess it is about time to get to the zeta function side of this story, if we’re ever going to use Farey sequences to show how you could prove the Riemann hypothesis. I’ve been reading a bit of Edwards’ book, with the same title as this post, and thought I’d try to summarize the first chapter over the next few posts. I’m not sure that the content of the first chapter is hugely vital for the final goal of relating to the Farey sequences, but I wanted to try to learn some of it anyway.

I should mention, before talking about any of this, that I will not claim to know any complex analysis. The last complex analysis course I took was 5 years ago, I can’t be sure how much I paid attention at the time, and I haven’t used it since. I will be jumping over gaps of various sizes for quite a while in the upcoming posts. Perhaps sometimes I’ll mention that a gap is there. Mostly, though, what I’m after in this story is the outline, and how all of the parts fit together, as a big picture. Perhaps I’ll go back and fill in some gaps, as I understand more.

For today, let’s see if I can build up to an analytic expression of \zeta(s). Our starting point is the function

\zeta(s)=\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^s}

which is defined for real values of s larger than 1. The goal is to find a nice expression that is defined on more of the complex plane, but agrees with this definition on the reals larger than 1.

To get to that point, we’ll use the \Gamma function:

\Gamma(s)=\displaystyle\int_0^{\infty} e^{-x}x^{s-1}\ dx

This is an analytic function defined everywhere except at the negative integers and 0. If you are only interested in real values of s, this function is a continuous extension of the factorial function, which is only defined on the positive integers. Actually, there is a shift involved, so \Gamma(n)=(n-1)! (Edwards blames this shift on Legendre, whose reasons, he states, “are obscure”. Edwards uses \Pi(s)=\Gamma(s+1) throughout, so I might make some \pm 1 errors). In particular, s\Gamma(s)=\Gamma(s+1) when both sides make sense.

Another relation that we’ll use is that

\pi=\Gamma(s)\Gamma(1-s)\sin(\pi s)

If memory serves (from reading Artin’s lovely little book on the \Gamma function), you can obtain this by determining that the derivative of the expression on the right is 0. This tells you that \sin(\pi s)\Gamma(s)\Gamma(1-s) is a constant, and so you can pick your favorite value of s to obtain the value of that constant. Anyway, the identity I’ll use is a rearrangement of the one above, namely

\Gamma(s)\sin(\pi s)=\dfrac{\pi}{\Gamma(1-s)}.

Now, in the expression

\Gamma(s)=\displaystyle\int_0^{\infty}e^{-x}x^{s-1}\ dx,

substitute x=nu for some positive n. Messing about with that substitution for a few minutes (and then re-writing u=x by abuse of notation), you can arrive at

\dfrac{\Gamma(s)}{n^s}=\displaystyle\int_0^{\infty} e^{-nx}x^{s-1}\ dx.

That n^s in the denominator is useful for us, as that’s how it shows up in \zeta(s). In particular, we can obtain

\begin{array}{rcl}\Gamma(s)\zeta(s) &=& \displaystyle \Gamma(s)\sum_{n=1}^{\infty}\dfrac{1}{n^s}=\sum_{n=1}^{\infty} \frac{\Gamma(s)}{n^s} \\ &=& \displaystyle \sum_{n=1}^{\infty}\int_0^{\infty}e^{-nx}x^{s-1}\ dx \\ &=& \displaystyle \int_0^{\infty}x^{s-1}\sum_{n=1}^{\infty} e^{-nx}\ dx\\ &=& \displaystyle \int_0^{\infty} \dfrac{x^{s-1}}{e^x-1}\ dx\end{array}

That last transition coming about by summing the geometric series. There are probably some things analysts like to check in here, moving infinite sums and improper integrals past each other… I’ll let them.

Ok, we’re nearly there. Next up, you do some complex line integral and show that

\begin{array}{rcl} \displaystyle \int_{+\infty}^{+\infty} \dfrac{(-x)^s}{e^x-1}\ \dfrac{dx}{x} &=& \displaystyle (e^{i\pi s}-e^{-i\pi s})\int_0^{\infty}\dfrac{x^{s-1}}{e^x-1}\ dx \\ &=& 2i\sin(\pi s)\Gamma(s)\zeta(s)\end{array}.

That integral is a little weird, going “from +\infty to +\infty“. Really, we take a path that “starts at +\infty“, swings around 0, then goes back out to infinity. This is almost certainly one of the complex analysis things I should go back and learn more about. Since we are integrating (-x)^s=e^{s\log(- x)} along positive real values of x, we’re working with logs along the negative real axis. Perhaps I’ll return to this integral in a future post.

Using the identity about \sin(\pi s)\Gamma(s) from above, we obtain

\zeta(s)=\dfrac{\Gamma(1-s)}{2\pi i}\displaystyle \int_{+\infty}^{+\infty}\dfrac{(-x)^s}{e^x-1}\ \dfrac{dx}{x}.

We’re now at a point where we’ve (apparently) got an expression for \zeta(s) that is analytic in the complex plane except for a simple pole at s=1.

]]> <![CDATA[book: Riemann hypothesis]]> http://ocmcatalog.wordpress.com/2009/11/11/book-riemann-hypothesis/ Wed, 11 Nov 2009 20:47:53 +0000 ocmpoma http://ocmcatalog.wordpress.com/2009/11/11/book-riemann-hypothesis/

The Riemann hypothesis: The greatest unsolved problem in mathematics
QA246 .S23
511

]]> <![CDATA[book: Bernhard Riemann]]> http://ocmcatalog.wordpress.com/2009/11/11/book-bernhard-riemann/ Wed, 11 Nov 2009 20:44:25 +0000 ocmpoma http://ocmcatalog.wordpress.com/2009/11/11/book-bernhard-riemann/

Prime obsession: Bernhard Riemann and the greatest unsolved problem in mathematics
QA246 .D47
512.73

]]> <![CDATA[Comic #39 - Wer wird Millionär?]]> http://24lightyears.wordpress.com/2009/11/09/comic-39-wer-wird-millionar/ Mon, 09 Nov 2009 10:19:57 +0000 Cauti http://24lightyears.wordpress.com/2009/11/09/comic-39-wer-wird-millionar/

39<- Mehr von Cautis Comix ->

]]> <![CDATA[MaBloWriMo]]> http://sumidiot.wordpress.com/2009/11/03/mablowrimo/ Tue, 03 Nov 2009 01:42:03 +0000 sumidiot http://sumidiot.wordpress.com/2009/11/03/mablowrimo/

I’m at least a day behind, but I’m thinking I might try this “Math Blog Writing Month” suggested by Charles Siegel over at Rigorous Trivialities.

A few months ago I was reading Hardy and Wright’s “An Introduction to the Theory of Numbers” with some friends. While that reading seems to have ended (we made it further than originally expected, to be honest), there was a topic I read about that I decided I wanted to learn more about. When H&W covered Farey sequences, I checked out the Wikipedia page and was fairly delighted to find that there is a statement about Farey sequences that is equivalent to the Riemann hypothesis (RH).

I know nearly nothing about the RH. You take a function defined by an infinite series, extend to the complex plane, notice that the zeroes have to lie in a particular region, and then the claim is that you can narrow down that region even more. Great. Why do I care about the zeroes of this function? Oh, right, because they are related to the distribution of the primes. And how is that again? I have no idea. Probably I should be embarrassed about this, being a math graduate student and not knowing the math behind the most famous open problem.

Well, now seems like as good a time as any to learn more. Farey sequences look fun, and close to some pretty pictures (Ford circles and the ? function). And so my plan is to learn enough about all of that, and the RH, to see if I can understand the relationship between the two. Not so that I can try to prove RH, mind you (I seriously have no delusions about this). It just seems like a fun thing to learn about. Plus maybe it’ll give me an excuse to give a talk in my department’s “graduate seminar”, which I always think is a fun thing to do.

I’ve got a few sources that I plan on looking at (and, indeed, have already been looking at). I found Rademacher’s “Higher Mathematics from an Elementary Point of View” to be quite delightful, and it contains information about the Farey and Ford part of the story. Of course, H&W might also help out here. I’ve got some papers tucked away somewhere I may dig up as well, and when I do, I’ll mention them. And, on the Riemann side of the story, I think I’ll look at Edwards’ “Riemann’s Zeta Function”, which has a section on the connection with Farey sequences.

I figure I’ll try to read at least a little bit every day, and share what I come across here. Perhaps in a month I’ll have some vague understanding (goal: less vague than it is now) about what’s going with the topic. Or perhaps I’ll crap out after a week (or less). But for now, let’s say I’m trying. Starting off a day late (and with an essentially non-mathematical post) means I probably should try to find a day or two and write more than one post. We’ll see. Also, I don’t expect I’ll get close to the 1000 word mark very often. We’ll see.

]]> <![CDATA[Grothendieck, Sistemas Dinâmicos, Dualidade de Langlands e Higgs Bundles…]]> http://arsphysica.wordpress.com/2009/10/31/grothendieck-sistemas-dinamicos-dualidade-de-langlands-e-higgs-bundles/ Sun, 01 Nov 2009 02:41:38 +0000 Daniel http://arsphysica.wordpress.com/2009/10/31/grothendieck-sistemas-dinamicos-dualidade-de-langlands-e-higgs-bundles/

Por falta de coisa melhor pra fazer num sábado de Halloween, à noite, …, eu estou aqui, lendo o artigo

e pensando na conexão dele com Local Dynamical Systems e Dinâmica Topológica (juntamente com Topological entropy), principalmente no contexto de Higgs bundles and local systems ou The Self-Duality Equations on a Riemann Surface — isso pra não falar em A mad day’s work: from Grothendieck to Connes and Kontsevich The evolution of concepts of space and symmetry, ou, melhor ainda, na continuação analítica de TFTs.

Taí o que me tira o sono… :evil:

(Atualizado (2009-Nov-01 @ 11:30h): Pra “apimentar” ainda mais essa mistura toda, pensar em termos de Discrete gauge theory: from lattices to TQFT também corrói…, afinal de contas, o T. Tao, acima, descreve tudo de modo simplicial, combinatório — e o que é Lattice QFT senão uma versão discretizada, simplicial, de QFTs contínuas? :razz: E já existem técnicas robustas pra se discretizar via ‘lattices’ que não são apenas o ingênuo ‘lattice’ cúbico — muito dessa tecnologia foi desenvolvida pra simulações numéricas de GR; aliás, mais ainda, hoje em dia, já é possível até se construir ‘lattices’ supersimétricos. Resumindo: no final das contas, é possível até se pensar em simulações numéricas dessas construções! :shock: )

Mas, agora é hora de fazer uma boquinha… :cool:

P.S.: Claro, o relógio abaixo também me tira o sono… :twisted:

Suunto Elementum Terra with the negative display and the stainless steel bracelet

Suunto Elementum Terra with the negative display and the stainless steel bracelet

]]> <![CDATA[Sobre el número de números primos menores a una cantidad dada]]> http://seriesdivergentes.wordpress.com/2009/10/19/sobre-el-numero-de-numeros-primos-menores-a-una-cantidad-dada/ Tue, 20 Oct 2009 04:42:23 +0000 ricardos http://seriesdivergentes.wordpress.com/2009/10/19/sobre-el-numero-de-numeros-primos-menores-a-una-cantidad-dada/

Creo que la mejor manera de agradecer el honor que la Academia me ha conferido, al admitirme como uno de sus miembros correspondientes, es si rápidamente hago uso de la licencia recibida para comunicar una investigación sobre la acumulación de los números primos; tal vez este tema no es del todo indigno de tal comunicación, dado el interés que mostraron Gauss y Dirichlet en él por mucho tiempo.

Así inicia el artículo que presentó Bernhard Riemann ante la Academia de Ciencias de Berlín el 19 de octubre de 1859, hace 150 años, a los pocos meses de ser admitido. En el artículo (Ueber die Anzahl der Primzahlen unter einer gegebenen Grösse) discute la relación de la llamada por él mismo función zeta, \zeta(s), y el número de primos menores o iguales a un número dado.

En el transcurso del trabajo conjetura que los ceros (no triviales) de \zeta(s), extendida en todo el plano complejo, se encuentran en la recta s=\dfrac{1}{2} + it. Hasta ahora, la conjetura ha permanecido como uno de los problemas abiertos más importantes de las matemáticas.

]]> <![CDATA[He-Man Models !!!!]]> http://janstephens.wordpress.com/2009/09/25/he-man-models/ Fri, 25 Sep 2009 21:59:40 +0000 janstephens http://janstephens.wordpress.com/2009/09/25/he-man-models/

German artist Adrian Riemann has recreated classic He-Man characters as hip fashion models.

View HERE.

]]> <![CDATA[Basics of the Kahler manifolds]]> http://andyoctavian.com/2009/09/22/basics-of-the-kahler-manifolds/ Mon, 21 Sep 2009 22:54:16 +0000 Octavian http://andyoctavian.com/2009/09/22/basics-of-the-kahler-manifolds/

Suppose we have a complex manifold {M}&fg=000000 endowed with the Riemannian metric {g}&fg=000000. We know that this metric is positive-definite simmetric bilinear tensor. {g}&fg=000000 is called Hermitian metric if

\displaystyle \begin{array}{rcl} g(X,Y) = g(JX,JY) \end{array} &fg=000000

for every {X, Y \in \chi (M)}&fg=000000, and {J}&fg=000000 is a complex structure of {M}&fg=000000. Componentwise, we can state the Hermitian metric {g}&fg=000000 as

\displaystyle g = g_{\mu \bar{\nu}} dz^\mu \otimes d\bar{z}^\nu + g_{\bar{\mu} \nu} d\bar{z}^\mu \otimes dz^\nu \ \ \ \ \ (1)&fg=000000

Define a Kahler form of {g}&fg=000000 as

\displaystyle \begin{array}{rcl} \omega(X,Y) = g(JX,Y) \end{array} &fg=000000

for every {X, Y \in \chi (M)}&fg=000000, and {J}&fg=000000 is a complex structure of {M}&fg=000000. Hence we can write the componentwise version of Kahler form as

\displaystyle \omega = i g_{\mu \bar{\nu}} dz^\mu \wedge d\bar{z}^\nu \ \ \ \ \ (2)&fg=000000

where {g_{\mu \bar{\nu}}}&fg=000000 in equation (2) above is Hermitian. The complex manifold endowed with a Hermitian metric {g}&fg=000000 is called Hermitian manifold.

The Riemann curvature tensor of a Hermitian manifold is defined as

\displaystyle R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z \ \ \ \ \ (3)&fg=000000

for {X,Y,Z \in \chi(M)}&fg=000000. From Riemann curvature tensor we can get the Ricci form defined as

\displaystyle \mathcal{R} = i R_{\mu \bar{\nu}} dz^\mu \wedge d\bar{z}^\nu \ \ \ \ \ (4)&fg=000000

where {R_{\mu \bar{\nu}}}&fg=000000 is the contraction of Riemann tensor:

\displaystyle R_{\mu \bar{\nu}} = {R^\kappa}_{\kappa \mu \bar{\nu}} = - \partial_\mu \partial_{\bar{\nu}} \log{G} \ \ \ \ \ (5)&fg=000000

where {G = \det{(g_{\mu \bar{\nu}})}}&fg=000000.

The metric {g}&fg=000000 of a Hermitian manifold {M}&fg=000000 is called Kahler metric if the Kahler form is a closed {(1,1)}&fg=000000-form, i.e.

\displaystyle d \omega = 0 \ \ \ \ \ (6)&fg=000000

And from equation (6) above we can get

\displaystyle \frac{\partial g_{\mu \bar{\nu}}}{\partial z^\lambda} = \frac{\partial g_{\lambda \bar{\nu}}}{\partial z^\mu} \ \ \ \ \ (7)&fg=000000

\displaystyle \frac{\partial g_{\mu \bar{\nu}}}{\partial \bar{z}^\lambda} = \frac{\partial g_{\mu \bar{\lambda}}}{\partial \bar{z}^\nu} \ \ \ \ \ (8)&fg=000000

which from the Kahler conditions (7) and (8) above we can find an explicit component of {g}&fg=000000, i.e.

\displaystyle g_{\mu \bar{\nu}} = \partial_\mu \partial_{\bar{\nu}} K \ \ \ \ \ (9)&fg=000000

{K}&fg=000000 is called the Kahler potential. For if the metric {g}&fg=000000 is Kahler, then we can get the expression for Kahler form as

\displaystyle \omega = i \partial \bar{\partial} K \ \ \ \ \ (10)&fg=000000

The Hermitian manifold endowed with a Kahler metric is called Kahler manifold.

Kahler manifold is torsion free. The Ricci form of Kahler manifold is defined as before,

\displaystyle Ric = -i \partial_\mu \partial_{\bar{\nu}} \log{G} dz^\mu \wedge d\bar{z}^\nu \ \ \ \ \ (11)&fg=000000

where we have used the notation {Ric}&fg=000000 instead of {\mathcal{R}}&fg=000000 since in Kahler manifold, the component of Ricci form, {R_{\mu \bar{\nu}}}&fg=000000, is same as Ricci curvature {Ric_{\mu \bar{\nu}} = {R^\kappa}_{\mu \kappa \bar{\nu}}}&fg=000000, due to the additional simmetry of the components of Riemann tensor results from the Kahler condition.

]]> <![CDATA[Proving the tensorial property of some certain object]]> http://andyoctavian.com/2009/08/13/proving-the-tensorial-property-of-some-certain-object/ Thu, 13 Aug 2009 06:34:52 +0000 Octavian http://andyoctavian.com/2009/08/13/proving-the-tensorial-property-of-some-certain-object/

If you have a tensor-like object, what you usually do to prove whether this is a tensor or not? Maybe you are familiar with the concept that the tensor is something that behaves certainly under coordinate transformation. For instance, if you are given an object with three indexes like this one

{T^{\alpha}}_{\beta \gamma \delta}

and also you are given a fact that in a different chart (and hence in a different coordinate system), this object transforms in this way

{\bar{T}^{\mu}}_{\nu \omega \kappa} = \frac{\partial y^{\mu}}{\partial x^{\alpha}} \frac{\partial x^{\beta}}{\partial y^{\nu}} \frac{\partial x^{\gamma}}{\partial y^{\omega}} \frac{\partial x^{\delta}}{\partial y^{\kappa}} {T^{\alpha}}_{\beta \gamma \delta}

then you can say that this guy is a tensor, surely. But it’s not the only way. To show the alternatives, I think it’s better to give out the example. This morning in a bedroom, I did this calculation, of which the main goal is to show that Riemann curvature tensor is really a tensor, as its name already does.

Riemann curvature tensor is defined as

R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z

where X, Y, Z are the vectors on a manifold, \nabla is affine connection, and [\quad , \quad] is a Lie bracket. To prove that this is a tensor, then you traditionally can check the relation between {\bar{R}^{\mu}}_{\nu \omega \kappa} in chart with coordinate system \{ y^\eta \} with {R^{\alpha}}_{\beta \gamma \delta} in an another chart with coordinate system \{ x^{\epsilon} \}. Or, you can also prove that this relation holds

R(fX,gY)hZ = fgh R(X,Y)Z

for any smooth functions f, g and h, defined on a manifold. This is really a simple calculation.

R(fX,gY)hZ = f \nabla_X (g \nabla_Y (hZ)) - g \nabla_Y (f \nabla_X (hZ)) - \nabla_{[fX,gY]}(hZ)

R(fX,gY)hZ = f \nabla_X (g \nabla_Y (hZ)) - g \nabla_Y (f \nabla_X (hZ)) - fg \nabla_{[X,Y]} (hZ) - f X[g] \nabla_Y (hZ)

+ g Y[f] \nabla_X (hZ)

R(fX,gY)hZ = f \nabla_X (g Y[h] Z + gh \nabla_Y Z) - g \nabla_Y (f X[h] Z + fh \nabla_X Z) - fg [X,Y][h]Z

- fgh \nabla_{[X,Y]} Z - f X[g] Y[h] Z - fh X[g] \nabla_Y Z

+ g Y[f] X[h] Z + gh Y[f] \nabla_X Z

R(fX,gY)hZ = f X[g] Y[h] Z + fg X[Y[h]] Z + fg Y[h] \nabla_X Z

+ fh X[g] \nabla_Y Z + fg X[h] \nabla_Y Z + fgh \nabla_X \nabla_Y Z

- g Y[f] X[h] Z - fg Y[X[h]] Z - fg X[h] \nabla_Y Z

- gh Y[f] \nabla_X Z - fg Y[h] \nabla_X Z - fgh \nabla_Y \nabla_X Z

- fg [X,Y][h] Z - fgh \nabla_{[X,Y]} Z - f X[g] Y[h] Z

- fh X[g] \nabla_Y Z + g Y[f] X[h] Z + gh Y[f] \nabla_X Z

R(fX,gY)hZ = fg X[Y[h]] Z + fgh \nabla_X\nabla_Y Z - fg Y[X[h]] Z

- fgh \nabla_Y \nabla_X Z - fg [X,Y][h] Z - fgh \nabla_{[X,Y]} Z

R(fX,gY)hZ = fgh (\nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z)

R(fX,gY)hZ = fgh R(X,Y)Z

From the last relation, we can get more concrete relation that verifies the tensorial property of Riemann curvature tensor, i.e.

R(X,Y)Z = R(X^\beta e_\beta, Y^\gamma e_\gamma) Z^\delta e_\delta = X^\beta Y^\gamma Z^\delta R(e_\beta, e_\gamma) e_\delta

]]> <![CDATA[Ancora Problemi di Hilbert]]> http://seipernove42.wordpress.com/2009/08/10/ancora-problemi-di-hilbert/ Mon, 10 Aug 2009 17:21:18 +0000 scardax http://seipernove42.wordpress.com/2009/08/10/ancora-problemi-di-hilbert/

Ho aggiunto una pagina ai problemi di Hilbert dedicata alla prima metà dell’ottavo problema, l’Ipotesi di Riemann:

http://seipernove42.wordpress.com/problemi-di-hilbert/lipotesi-di-riemann/

Per un po’ smettero’ con questa sezione, e cerchero’ di passare ad altri argomenti.

]]> <![CDATA[Is P=NP an Ill Posed Problem?]]> http://rjlipton.wordpress.com/2009/07/03/is-pnp-an-ill-posed-problem/ Fri, 03 Jul 2009 13:48:23 +0000 rjlipton http://rjlipton.wordpress.com/2009/07/03/is-pnp-an-ill-posed-problem/


P=NP has an infinite number of possible outcomes

images

Russell Impagliazzo is one of the most original thinkers in complexity theory, and is famous for many wonderful things. One of his papers is on many worlds. The worlds are: Algorithmica, Heuristica, Pessiland, Minicrypt, and Cryptomania. See his paper for the definitions and see this for a recent conference on this topics.

Today I want to talk about a different take on P=NP and possible worlds that we live in. I think that the situation is much more complex than Russell’s view. As usual, I have a different view of things, which I hope you will find interesting.

Some of you may have been at Russell’s invited talk at STOC 2006. It was a great talk, as all his presentations are. But, the coolest part–for me–was that he arrived to the conference with transparencies, overheads. The room for the invited talk was very wide and not very deep. So it was critical that his slides appear on two projectors, otherwise part of the audience would be unable to see his slides.

Of course if he had used Powerpoint this would have been automatic. Since Russell arrived with plastic slides to the conference at the last minute there was a slightly-panicked run to Kinko’s. There they made identical copies of his colorful slides. Later, as he spoke and turned his slides the slides on the other projector also flipped–magically. Of course someone had the job of trying to keep this all in synch. It was a great talk, in any event.

Now let’s turn first to another famous open problem the Riemann Hypothesis.

Riemann Hypothesis

The prime number theorem states that {\pi(x)}&fg=000000, the number of primes less than {x}&fg=000000, is well approximated by the logarithmic integral:

\displaystyle \pi(x) = \text{Li}(x) + E(x)&fg=000000

where

\displaystyle \text{Li}(x) = \int_{0}^{x} \frac{dt}{\log (t)}&fg=000000

and {E(x) = o(x)}&fg=000000 is the error term.

The Riemann Hypothesis (RH) implies that the error term, {E(x)}&fg=000000, is order {x^{\frac{1}{2}}}&fg=000000–ignoring logarithmic terms. But, the RH is “binary” in the sense that even one zero that is not on the critical line destroys this bound. More precisely, if there is a zero at {\sigma + it}&fg=000000 with {\sigma > 1/2}&fg=000000, then the error term in the prime number theorem is at least {x^\sigma}&fg=000000–again ignoring some logarithmic factors. I thank Andy Odlyzko for pointing out this tight relationship between a non-critical zero and the error term.

This is a relatively simple behavior, one bad apple and the whole distribution of the prime numbers changes. Famously, Enrico Bombieri has said that “The failure of the Riemann hypothesis would create havoc in the distribution of prime numbers.”

P=NP is Different

I claim that our problem P=NP is different than the RH, since there is no “one bad apple” behavior. Consider first what a proof that P=NP could look like.

{\bullet}&fg=000000 There might be an algorithm for SAT that runs in a reasonable time bound. This would be an immense result, and would change the world. Crypto-systems would all be destroyed, whole parts of theory would have to radically change. This is why many people believe that P=NP is impossible–at least that’s my impression–they say it would be too good to be true. I agree that it would be an amazing situation, but it is only one of a multitude of possibilities.

{\bullet}&fg=000000 There might be an algorithm for SAT that is polynomial but with a very high exponent, for example, {O(n^{10})}&fg=000000. This would win the million Clay dollars, but would have a more philosophical impact. Crypto-researchers would worry that eventually the algorithm could be made practical, but it would not have any immediate practical impact. Research papers would have to do a song and dance: “we study an approximation to the following NP-hard problem, that is in P of course thanks to the famous result of X, but we feel our results are still interesting because {\dots}&fg=000000”

{\bullet}&fg=000000 There might be an algorithm that is polynomial in a very weak sense. What if its running time is

\displaystyle n^{2^{2^{2^{2^{100}}}}}.&fg=000000

That would be great, I think you still get the money, but crypto people could still sleep at night, without any sleep aids.

{\bullet}&fg=000000 There might exist an algorithm that runs in {n^{c}}&fg=000000 time for some constant {c}&fg=000000 that is not even known. Results like this are quite possible, if the proof of the algorithm’s correctness used some indirect argument. Again you still get the cash–I believe.

Now consider what a proof that P is not equal to NP could look like.

{\bullet}&fg=000000 SAT might require exponential time. The best algorithm might run in time {1.2^{n}}&fg=000000, for example. This type of behavior is what many believe. There even is a “formal” conjecture that the best algorithm for 3-SAT will have this type of running time. Note, even here one needs to be careful: what if the running time is {1.0000001^{n}}&fg=000000? Recall Alan Perlis said:

For every polynomial algorithm you have, I have an exponential algorithm that I would rather run.

{\bullet}&fg=000000 SAT might have wild complexity. Perhaps for some lengths {n}&fg=000000 the best running is exponential, but infinitely often the best time is polynomial in {n}&fg=000000. This could happen.

{\bullet}&fg=000000 SAT might be barely super polynomial: what if the lower bound is

\displaystyle n^{\log \log \log n}.&fg=000000

Does this really mean anything? Yet it would technically resolve the P=NP question, would be an immense achievement, would win the Clay Prize, and probably many other prizes too. But does it really imply that SAT is “intractable?”

{\bullet}&fg=000000 SAT might have no polynomial time algorithm, but there could be random algorithms {\dots}&fg=000000

The Riemann Hypothesis Again

You might be thinking what if the RH is true, could it have the same complex behavior that P=NP can? In particular, could the RH be true, but with a crazy error term:

\displaystyle 2^{2^{2^{100}}}x^{\frac{1}{2}}\log(x)?&fg=000000

Isn’t this the same issue that I raised above? The answer is that this cannot happen for the RH: The following is a theorem:

Theorem: If RH, then for all {x \ge 2657}&bg=e8e8e8&fg=000000,

\displaystyle | \pi(x) - \text{Li}(x)| < \frac{1}{8\pi}\sqrt x \log(x). &bg=e8e8e8&fg=000000

This is due to Lowell Schoenfeld, who actually proved that RH is equivalent to his inequality.

Is P=NP Ill Posed?

I love the P=NP question, but it occurs to me that in a sense the question is “ill posed.” David Hilbert said once:

“If I were to awaken after having slept for a thousand years, my first question would be: Has the Riemann hypothesis been proven?”

After sleeping for a thousand years knowing if P has been proved to be not equal to NP, would not be nearly as meaningful as knowing that RH has been proved.

To make my point consider three players: Paula, Ted, and George. Paula is a practical minded computer scientist, Ted is a theory minded computer scientist, and George is the best gossip in the department.

There are two scenes. In the first P=NP and in the second P{\neq}&fg=000000NP.

George barges into Paula’s office without knocking. She is about to complain that she is in a meeting with Ted, when George says, “Ted did you hear that P=NP has been proved! I just got off the phone with Lance and the proof is correct.”

Paula sees Ted jump up at this and he says, “are you kidding?” George is out of breathe, but smiling. It’s no joke. “Unbelievable, unbelievable”, is all that Ted can manage.

Paula has never seen George this excited, “no it’s really true. The FOCS program committee accepted the paper, even though she did not submit it. I guess that’s a first, she just missed the deadline or something.”

Paula sits behind a large desk that is covered in books on circuit design and VLSI, says “this is great. Finally, you theory guys have done something that I can use. I can’t wait to get my team implementing the algorithm.” She sees Ted sit down, he is turning pale and mutters to himself, “Dammit, this wipes out Fred’s thesis. I told that idiot to finish it last spring.”

Paula ignores him, “this will change VLSI design and testing and–”

George interrupts her, “well not quite Paula. I hear that the running time of her algorithm is bad.”

“What? How bad is bad?”

“I think its running time is {n^{2^{2^{100}}}}&fg=000000, although I forget if there is a logarithmic factor too. But, P=NP!”’

George barges into Paula’s office without knocking. She is about to complain that she is in a meeting with Ted, when George says, “Ted did you hear that P{\neq}&fg=000000NP has been proved! I just got off the phone with Lance and the proof is correct.”

Paula sees Ted jump up at this and he says, “are you kidding?” George is out of breathe, but smiling. It’s no joke. “Unbelievable, unbelievable”, is all that Ted can manage.

Paula has never seen George this excited, “no it’s really true. The STOC program committee accepted the paper, even though she did not submit it. I guess that’s a first, she just missed the deadline or something.”

Paula who is listening from behind a large desk that is covered in books on circuit design and VLSI, idly checks her Seiko. It’s almost twenty to 12, she needs to leave for her lunch meeting soon.

Ted says, “I knew it, that crazy blogger–what’s his name?–always talking about P could equal NP. Well he was wrong. What an idiot.”

George says, “yeah.”

“So how does the proof go?”

“It’s a circuit lower bound, she proves SAT requires at least {n^{\log \log \log n}}&fg=000000 size circuits. Avoids the usual roadblocks by using that new combinatorial lemma of Terry. Isn’t this wonderful.”

Paula who a moment ago was hardly listening says, “Did I hear you right? That’s not a very big bound. Let’s see if {n}&fg=000000 is a million, let me see, that’s not even a quadratic lower bound. Were those log’s base two?”

“Of course they’re base two, log’s are always base two. But it’s a super polynomial lower bound, P{\neq}&fg=000000NP!”

My apologies to all fiction writers that have ever lived.

Open Problems

My point is simple: There is a huge variance in the potential answers to the question: Does P equal NP?

It’s not simple. Equal, not equal, Equal, not equal. There are an infinite number of possibilities. I think that this is what makes me think that we need to work hard on this problem.

A class of open problems is to prove something like Schoenfeld’s theorem for P=NP. Can we show that some of the above pathological behaviors cannot occur? Or can we show that one pathological behavior implies another? Or precludes another?

]]> <![CDATA[Stalking the Riemann Hypothesis]]> http://bookskeepers.wordpress.com/2009/06/25/stalking-the-riemann-hypothesis/ Thu, 25 Jun 2009 14:20:39 +0000 j t http://bookskeepers.wordpress.com/2009/06/25/stalking-the-riemann-hypothesis/

History on discovery of the natural and prime numbers, cartographers Legendre and Gauss’s reestimation of primes, Dirichlet and Euler’s revelation of analytical techniques on series, the connection between Hamiltonian matrices and the hypothesis, Riemann’s legacy in physics and the embodiment of primes in Riemann’s hypothesis. A technical book providing much insights into modern day progress by matheticians on the illusive hypothesis.

]]> <![CDATA[Evolution of geometric quantities of a manifold]]> http://andyoctavian.com/2009/06/14/evolution-of-geometric-quantities-of-a-manifold/ Sun, 14 Jun 2009 16:58:48 +0000 Octavian http://andyoctavian.com/2009/06/14/evolution-of-geometric-quantities-of-a-manifold/

Lemma

For a Riemannian manifold M with a time-dependent Riemannian metric g(t) which satisfies

\frac{\partial}{\partial t} g_{ij}(t) = h_{ij}(t)

then these will follow:

  1. Metric inverse
    \frac{\partial}{\partial t} g^{ij} = -h^{ij} = -g^{ik} g_{jl} h_{kl}
  2. Christoffel symbol
    \frac{\partial}{\partial t} \Gamma_{ij}^k = \frac{1}{2} g^{kl} \big( \nabla_i h_{jl} + \nabla_j h_{il} - \nabla_l h_{ij} \big)
  3. Riemann curvature tensor
    \frac{\partial}{\partial t} R_{ijk}^l = \frac{1}{2} g^{lp} \big( \nabla_i \nabla_j h_{kp} + \nabla_i \nabla_k h_{jp} - \nabla_i \nabla_p h_{jk} - \nabla_j \nabla_i h_{kp} - \nabla_j \nabla_k h_{ip} + \nabla_j \nabla_p h_{ik} \big)
  4. Ricci tensor
    \frac{\partial}{\partial t} R_{ij} = \frac{1}{2} g^{pq} \big( \nabla_q \nabla_i h_{jp} + \nabla_q \nabla_j h_{ip} - \nabla_q \nabla_p h_{ij} - \nabla_i \nabla_j h_{qp} \big)
  5. Scalar curvature
    \frac{\partial}{\partial t} R = -\Delta H + \nabla^p \nabla^q h_{pq} - h^{pq} R_{pq}, where H = g^{ij} h_{ij}
  6. Volume element
    \frac{\partial}{\partial t} d \mu = \frac{H}{2} d \mu
  7. Volume of manifold M
    \frac{\partial}{\partial t} \int_M d \mu = \int_M \frac{H}{2} d \mu
  8. Total scalar curvature on a closed manifold M
    \frac{\partial}{\partial t} \int_M R d \mu = \int_M (\frac{1}{2} RH - h^{ij} R_{ij}) d \mu
]]> <![CDATA[La sorgente della vita eterna]]> http://2senxcosx.wordpress.com/2009/06/06/la-sorgente-della-vita-eterna/ Sat, 06 Jun 2009 09:45:29 +0000 ippaso http://2senxcosx.wordpress.com/2009/06/06/la-sorgente-della-vita-eterna/

L’Ipotesi di Riemann (RH) è una congettura che se dimostrata metterebbe ordine nel caos dei numeri primi, ma ancora nessuno è mai riuscito a dimostrare tale ipotesi. Molti matematici ormai la danno per buona, mentre altri continuano a essere scettici.
Certo è che questo dilemma resiste da ben un secolo e mezzo, da quando cioè un brillante matematico ex-allievo di Gauss, Riemann, non pubblicò il proprio risultato in un articoletto. Siamo sicuri che chi dovesse dimostrarlo riceverà fama eterna. In aggiunta riceverà anche dei cospicui premi in denaro (la RH era uno dei problemi di Hilbert per il XX secolo, ed pra è uno dei problemi da un milione di dollari dei Clay Prize).
Ma i matematici meno “scientifici” hanno azzardato teorie molto particolari sulla RH.

Jacques Hadamard e Charles de la Vallée Poussin han dato un contributo all’avvicinamento alla soluzione di questo problema dimostrando il Teorema dei Numeri Primi e son stati ricompensati sfiorando entrambi i 100 anni di vita. Atle Selberg e Paul Erdos hanno migliorato la dimostrazione di Hadamard-Poussin ed anche loro sono stati molto molto longevi.

Inoltre dimostrare la RH vorrebbe dire trovare la legge che regola la disposizione dei numeri dei primi, ovvero degli atomi dell’aritmetia, delle entità più elementari e più sfuggevoli dell’Universo.

Mettendo assieme tutti questi indizi alcuni matematici si sono convinti che chi dimostrasse l’Ipotesi otterrebbe sicuramente la Vita Eterna. Il problema è che non si ha la più pallida idea di come procedere nella dimostrazione. Ultimamente abbiamo assistito a tentativi anche attraverso la Geometria Algebrica e attraverso la Meccanica Quantistica, ma niente da fare. Alcuni esponenti autorevoli della comunità scientifica hanno iniziato allora a convincersi che la RH sia falsa.

E qui si è sparsa un’altra voce: probabilmente qualcuno è già riuscito a dimostrare che l’Ipotesi di Riemann è falsa, ma è morto sul colpo.

]]> <![CDATA[(06.03.09) Student Meeting]]> http://texaslarouchemovement.wordpress.com/2009/06/04/06-03-09-student-meeting/ Thu, 04 Jun 2009 15:54:55 +0000 texaslarouche http://texaslarouchemovement.wordpress.com/2009/06/04/06-03-09-student-meeting/

CeresGauss

Houston, TX –Wednesday (June 3rd) @ 7:00 pm

Strategic Political Briefing:   Harley Schlanger

Audio:  060309_harley_briefing.mp3

Class:  Phil Rubinstein, on “The Narrow Path from Kepler through Leibniz, Gauss, to Riemann & Einstein”, or on “Humanity”

Audio:  060309_phil_tensors.mp3

(Click here to watch the Basement’s exciting new animation described in Phil’s class.)

]]> <![CDATA[1 milione di Euro e la gloria imperitura]]> http://andreamacco.wordpress.com/2009/05/22/1-milione-di-euro-e-la-gloria-imperitura/ Thu, 21 May 2009 23:24:08 +0000 Andrea "feynman82" http://andreamacco.wordpress.com/2009/05/22/1-milione-di-euro-e-la-gloria-imperitura/

Il mio docente di Geometria dell’Università, il prof. Arezzo, era solito entusiasmarci con lezioni corredate di aneddoti e/o piccole curiosità.
È uno stile che ho sempre apprezzato, non solo in lui, ma anche in altri (pochi e rari) professori. Senza un minimo di passione la materia perde quel quid che dà un colpo d’ala in più e permette di non far affievolire l’entusiasmo e accresce la voglia di studiare e di applicarsi nella materia.

Durante le lezioni di questi giorni affrontavo alcune questioni di matematica che sono state risolte dai matematici di fine ‘800 e inizio ‘900. E mi è venuto spontaneo citare la grande sfida lanciata da Hilbert al grande Congresso di Parigi del 1900 quando dimostrò la sua genialità nel riuscire a cogliere i 10 grandi problemi aperti della Matematica che la Comunità Scientifica avrebbe dovuto affrontare nei decenni a seguire.

David Hilbert (1862-1943)
David Hilbert (1862-1943)

In realtà i problemi proposti da Hilbert furono da lui stessi integrati con altri quesiti e in tutto si contano 23 problemi. Di questi 2 restano tutt’oggi aperti, e 4 presentano una soluzione solo parziale.
(Su Wikipedia a questo link una formulazione riassuntiva ma un po’ stringata, su questo documento pdf qualche notazione in più sui 23 problemi di Hilbert)

Che centra tutto questo con i soldi e la ricchezza? Fin’ora, al massimo, si è parlato di gloria imperitura.
Come forse alcuni sanno, nel 2000 sono stati formulati dall’ Istituto Matematico Clay (CMI) 7 Problemi del Millennio. Per chi li risolverà è previsto un premio di 1 milione di dollari per problema risolto e, ovviamente, la fama eterna.
I sette problemi sono considerati dal CMI i “più importanti problemi classici che hanno resistito ai tentativi di soluzione nel corso degli anni”.
Qui – sito ufficiale potete leggere di che si tratta. Pure su wikipedia in italiano si trova una sommaria presentazione (cliccare qui). Avvertenza: Alcuni problemi richiedono, solo per la loro comprensione, di conoscenze di matematica avanzata; altri, riguardanti anche l’ambito della Fisica, sono di comprensione più immediata.

ATTENZIONE: uno dei problemi del Millennio è stato recentemente risolto da Grigory Perelman, l’eccentrico matematico russo che si è dedicato alla congettura di Poincaré. La comunità scientifica sta inoltre passando al vaglio la soluzione presentata recentemente da Louis de Branges de Bourcia della Purdue University (lo stesso che ha risolto la congettura di Bieberbach) di un’altro problema del Millennio: l’Ipotesi di Riemann, che compariva già tra i problemi enunciati da Hilbert nel 1900!

DUNQUE: amanti delle sfide e matematici che non avete ancora sviluppato a pieno le potenzialità della vostra mente: datevi da fare!
Magari – dicevo oggi ai miei studenti – un giorno potrò dire di avere preparato o di avere stretto la mano ad un matematico milionario passato alla storia …

Andrea Macco

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