# Luboš Motl

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bio website motls.blogspot.com location Czech Republic age 40 member for 2 years, 11 months seen Mar 12 at 7:48 profile views 2,479

Hi, I am a string theorist and a publicist.

 Mar6 comment Why does $1+2+3+\dots = {-1\over 12}$? Sure, it's a completely analogous sum. Just like $1+2+3+\dots$ quantifies the ground state energy in 1+1 dimensions (of a string), $1+1+1+\dots$ quantifies the ground state's charge (of a system of free fermions, for example). It's really the reason why the two degenerate states $|0\rangle$ and $c_0|0\rangle$ have ghost numbers $\pm 1/2$, for example. Feb10 comment Why does $1+2+3+\dots = {-1\over 12}$? It might be but it is a wrong conclusion. The key property of this $\infty$, which may be subtracted by a counterterm, is that it is independent of all independent variables, so it is formally a scale-invariant constant, and $0$ is the number to be associated with it. The fact that the sum of integers equals $-1/12$ is perhaps the most important fact about mathematics among the laymen - and among mathematicians without breadth and depth. Jan24 comment $1 + 1 + 1 +\cdots = -\frac{1}{2}$ Dear @Pedro, I am saying that $\zeta(2k+1)$ are transcendental numbers because I am ready to bet all my wealth on the validity of the claim. After all, even $\pi^n$ are transcendental numbers for rational nonzero values of $n$ so even zeta of even positive integers are (provably) transcendental. But the zetas of odd positive integers are (I claim without a rigorous proof) much more transcendental than the powers of $\pi$. I never claimed that I had a rigorous proof of the transcendentality but if you want to claim that I said something incorrect, you should have a proof which you don't have. Jan24 comment $1 + 1 + 1 +\cdots = -\frac{1}{2}$ Nope, it's just believed by me and almost everyone that they're transcendental. The much "easier" proof that zeta(3) is irrational came just recently, in 1978, see en.wikipedia.org/wiki/Ap%C3%A9ry's_theorem Jan23 comment $1 + 1 + 1 +\cdots = -\frac{1}{2}$ Nice, +1. An alternative calculation of $\zeta(0)=-1/2$ appears early in this article: motls.blogspot.com/2014/01/… - See my comment under the OP's question for 4 more articles about the topic if you wish... Jan23 comment $1 + 1 + 1 +\cdots = -\frac{1}{2}$ Five more articles with 8 different ways to compute the sum of integers (most of the methods extend to the "sum of ones", too) etc.: motls.blogspot.com/2007/09/… motls.blogspot.com/2011/07/… motls.blogspot.com/2014/01/… motls.blogspot.com/2014/01/… motls.blogspot.com/2014/01/… Jan11 comment Why does $1+2+3+\dots = {-1\over 12}$? "Incorrect", "not right", and "opposite of right" are the same thing, also known as "synonyma". Sep20 comment Showing Dirac Delta Behavior Well, I think you would also have to discuss that the negative regions of $(\sin x)/x$ don't combine to other distributions like $\delta'$ etc. But otherwise OK, I upvote it. Jun25 comment Fourier transform of a signal sequence? I would be sort of surprised if there is an analytic formula for the Fourier transform of this contrived function. May24 comment Grover Algorithm Orthogonal vectors Sorry, I made a mistake. I meant that $\omega$ and $s'$ are orthogonal. My answer is hopefully fixed. I don't understand the question "how I will be able to define $\sqrt{N}|\omega\rangle$". It's just multiplication, right? One should know how to multiply vectors by numbers. Dec9 comment Rotation matrix in terms of axis of rotation It's just a conjugation of the simple matrix for a rotation around the $z$-axis, which is effectively just 2 x 2 matrix, by another rotational matrix that rotates the North pole to the point $V$, which is a product of rotation in the theta-direction and the phi-direction to get where you need to get. Conjugation by $U$ is $URU^{-1}$ where the product is matrix product. It's possible ineffective to write these things without matrices so if you don't know matrices, this is a reason to learn them. At any rate, it's not really physics, it's linear algebra and geometry and a basic one. Dec2 comment Kummer's Equation Dear Alex, the equation for $L_n$ is nothing else than a coefficient in front of $t^n$ in the Taylor expansion of the equation for $g$ with respect to $t$. You just compare the terms term-by-term. Please just expand my equation for $g$ as a Taylor expansion in $t$ and don't ask any more questions. May23 comment A general pattern to find the roots of the classical lie algebras Sorry, your updated question got very confusing. I was answering your original question. The new question talks about "levels" etc. There aren't levels in ordinary Lie algebras. The roots are very simple objects and be sure I could easily enumerate all of them for all 7 classes of the Lie algebras, A,B,C,D,E,F,G. They're just the non-negative integral combinations of the simple roots that have the right length (either the same as simple roots or, in non-simply-laced groups, sqrt2 or sqrt3 times longer). ... If you have an algebra, the roots are just defined as the eigenvalues under the Cartan. Nov5 comment How do I determine the measure for a volume integral? Sorry, this was the integral of $r^2=x^2+y^2+z^2$. If I calculate the correct $r_z^2$, the squared distance from the $z$-axis, it's only $x^2+y^2$ so I only get two terms out of three. Due to the spherical symmetry, it's $2/3$ of the result above, so that the moment of inertia is $(2/5)MR^2$ for a solid ball. Nov5 comment How do I determine the measure for a volume integral? Let me calculate the moment of inertia of a ball of radius $R$ here. I choose spherical coordinatets so $dV=r^2\,dr\,d\Omega$. The integral over $d\Omega$ gives $4\pi$, the integral of $r^4dr$ (there was $r^2$ from the moment) from $0$ to $R$ gives $R^5/5$: I can factorize them here. So the result is $4\pi \rho R^5/5=3/5 MR^2$. Less symmetric shapes are more complicated but it's pure maths. Nov5 comment How do I determine the measure for a volume integral? Also, your question as well as your comments below my answer make it spectacularly self-evident that you don't understand what either of these symbols mean, in a striking contradiction with your assertion. Nov5 comment How do I determine the measure for a volume integral? Dear @jnm2, the first equation in your comment above isn't a "magic transformation": it's a totally trivial identity. $\int$ is the integral sign but it's still a sum of infinitesimal pieces. The infinitesimal pieces are proportional to $dV$ and one may choose cubic shapes of $dV$ to make it clear that it may be written as $dx\,dy\,dz$. I don't understand what's your problem. If you can't understand this simple thing about the volume integral, you should give up studying physics quantitatively. Also, I wrote you how you can calculate the triple integral for any shape you mentioned. Jul9 comment True, false, or meaningless? I see, so $\forall i\in \{\}: T(i)$ holds for any $T(i)$ because there is no counterexample in an empty set for which $T(i)$ would fail - it holds for everyone (all zero of them). On the other hand, $\exists i\in\{\}: T(i)$ is always untrue because there doesn't exist any $i$ in an empty set that has a property - whatever property - because there's nothing in an empty set even without adjectives. Jun27 comment Validity of $\sum_{i=1}^n(a_i^2+b_i^2+c_i^2+d_i^2)\lambda_i\geq\lambda_1+\lambda_2+\lambda_3+\lambda_4$? Oh, I see, you're right. Jun27 comment Puzzle: $(\Box @)+(\Box @) = (\Box\bigstar\Box$) Thanks for your care and implicit compliments, @JTL and @Jyrki, but I don't think it's a big issue. Let's be generous, people have the right to vote the way they want.