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# Flat module

In homological algebra and algebraic geometry, a flat module over a ring R is an R-module M such that taking the tensor product over R with M preserves exact sequences. A module is faithfully flat if taking the tensor product with a sequence produces an exact sequence if and only if the original sequence is exact.

Vector spaces over a field are flat modules. Free modules, or more generally projective modules, are also flat, over any R. For finitely generated modules over a Noetherian ring, flatness and projectivity are equivalent. For finitely generated modules over local rings, flatness, projectivity and freeness are all equivalent.[1] The field of quotients of an integral domain, and, more generally, any localization of a commutative ring are flat modules. The product of the local rings of a commutative ring is a faithfully flat module.

Flatness was introduced by Serre (1956) in his paper Géometrie Algébrique et Géométrie Analytique. See also flat morphism.

## Definition

### Commutative rings

Let M be an R-module. The following conditions are all equivalent, so M is flat if it satisfies any (thus all) of them:

• The functor
${\displaystyle F_{M}:\operatorname {Mod} (R)\to \operatorname {Mod} (R),\quad N\mapsto M\otimes _{R}N}$
is exact, where ${\displaystyle \operatorname {Mod} (R)}$ is the category of ${\displaystyle R}$-modules.
• For all prime ideals ${\displaystyle {\mathfrak {p}}}$ of R, the module ${\displaystyle M_{\mathfrak {p}}}$ is a flat ${\displaystyle R_{\mathfrak {p}}}$-module.
• For all maximal ideals ${\displaystyle {\mathfrak {m}}}$ of R, the module ${\displaystyle M_{\mathfrak {m}}}$ is a flat ${\displaystyle R_{\mathfrak {m}}}$-module.
• For every injective morphism ${\displaystyle \phi :K\to L}$ of ${\displaystyle R}$-modules ${\displaystyle K}$ and ${\displaystyle L}$, the induced map
${\displaystyle F_{M}(\phi ):M\otimes _{R}K\to M\otimes _{R}L,}$
is injective.
• For every finitely generated ideal ${\displaystyle I\hookrightarrow R}$, the induced morphism ${\displaystyle I\otimes _{R}M\to R\otimes _{R}M\cong M}$ is injective.
• There exists a directed system of ${\displaystyle R}$-modules ${\displaystyle \{F_{\alpha }\}_{\alpha }}$ with the following properties:
1. For all ${\displaystyle \alpha }$, ${\displaystyle F_{\alpha }}$ is a finitely generated, free ${\displaystyle R}$-module.
2. The direct limit is ${\displaystyle M}$: ${\displaystyle \varinjlim _{\alpha }F_{\alpha }=M.}$
• [2] For every linear dependency in ${\displaystyle M,}$
${\displaystyle r^{T}x=\sum _{i=1}^{k}r_{i}x_{i}=0,}$
where ${\displaystyle r_{i}\in R,x_{i}\in M}$, there exists a matrix ${\displaystyle A\in R^{k\times j}}$ such that
1. ${\displaystyle Ay=x}$ has a solution for some ${\displaystyle y\in M^{j}.}$
2. ${\displaystyle r^{T}A=0.}$
• For every ${\displaystyle R}$-module ${\displaystyle N}$,
${\displaystyle \operatorname {Tor} _{1}^{R}(N,M)=0}$
• For every finitely generated ideal ${\displaystyle I\subset R,}$
${\displaystyle \operatorname {Tor} _{1}^{R}(R/I,M)=0.}$
• For every map ${\displaystyle f:F\to M}$, where ${\displaystyle F}$ is a finitely generated free ${\displaystyle R}$-module, and for every finitely generated ${\displaystyle R}$-submodule ${\displaystyle K\leq \ker f}$, ${\displaystyle f}$ factors through a map to a free ${\displaystyle R}$-module ${\displaystyle G}$ that kills ${\displaystyle K}$:

### General rings

When R isn't commutative one needs the more careful statement that, if M is a flat left R-module, the tensor product with M maps exact sequences of right R-modules to exact sequences of abelian groups.

Taking tensor products (over arbitrary rings) is always a right exact functor. Therefore, the R-module M is flat if and only if for any injective homomorphism KL of R-modules, the induced homomorphism K${\displaystyle \otimes }$ML${\displaystyle \otimes }$M is also injective.

## Examples

• For any multiplicatively closed subset ${\displaystyle S}$ of a commutative ring ${\displaystyle R}$, the localization ring ${\displaystyle S^{-1}R}$ is flat as an ${\displaystyle R}$-module. For example, ${\displaystyle \mathbb {Q} }$ is flat over ${\displaystyle \mathbb {Z} }$ (though not projective).
• Every free ${\displaystyle \mathbb {Z} }$-module (in fact any free module over a ring) is a flat module.
• For each integer ${\displaystyle n}$ > 1, ${\displaystyle \mathbb {Z} /n\mathbb {Z} }$ is not flat over ${\displaystyle \mathbb {Z} }$, because ${\displaystyle n:\mathbb {Z} \to \mathbb {Z} ,\,x\mapsto nx}$ is injective, but tensored with ${\displaystyle \mathbb {Z} /n\mathbb {Z} }$ it is not.
• Similarly, ${\displaystyle \mathbb {Q} /\mathbb {Z} }$ is not flat over ${\displaystyle \mathbb {Z} .}$
• Let ${\displaystyle R=k[t],k}$ a field, and ${\displaystyle S=R[x]/(tx-1)}$. Since ${\displaystyle S}$ is the same thing as the localization ${\displaystyle R[t^{-1}]}$, it is flat over ${\displaystyle R}$. On the other hand, ${\displaystyle R[x]/(tx-t)}$ is not flat over ${\displaystyle R}$ since ${\displaystyle t}$ is a torsion element on it (so it is not torsion-free).
• Let ${\displaystyle A}$ be a noetherian ring and ${\displaystyle I}$ an ideal. Then the completion ${\displaystyle A\to {\widehat {A}}}$ with respect to ${\displaystyle I}$ is flat.[3] It is faithfully flat if and only if ${\displaystyle I}$ is contained in the Jacobson radical of ${\displaystyle A}$.[4] (cf. Zariski ring.)
• The direct sum ${\displaystyle \bigoplus _{i\in I}M_{i}}$ is flat if and only if each ${\displaystyle M_{i}}$ is flat.
• Every product of flat ${\displaystyle A}$-modules is flat if and only if ${\displaystyle A}$ is a coherent ring.[5]
• (Kunz) A noetherian ring containing a field of characteristic ${\displaystyle p}$ is regular if and only if the Frobenius morphism ${\displaystyle R}$${\displaystyle R}$ is flat and ${\displaystyle R}$ is reduced.

## Case of commutative rings

When M is a finitely-generated R-module, being flat is the same as being locally free in the following sense: M is a flat R-module if and only if for every prime ideal (or even just for every maximal ideal) P of R, the localization ${\displaystyle M_{P}}$ is free as a module over the localization ${\displaystyle R_{P}.}$[1]

Let R be a local ring with nilpotent maximal ideal (e.g., an artinian local ring) and M a module over it. Then M flat implies M free.[1]

The local criterion for flatness states:[6]

Let R be a local noetherian ring, S a local noetherian R-algebra with ${\displaystyle {\mathfrak {m}}_{R}S\subset {\mathfrak {m}}_{S}}$, and M a finitely generated S-module. Then M is flat over R if and only if ${\displaystyle \operatorname {Tor} _{1}^{R}(M,R/{\mathfrak {m}}_{R})=0.}$

The significance of this is that S need not be finite over R and we only need to consider the maximal ideal of R instead of an arbitrary ideal of R.

The next criterion is also useful for testing flatness:[7]

Let R, S be as in the local criterion for flatness. Assume S is Cohen–Macaulay and R is regular. Then S is flat over R if and only if ${\displaystyle \dim S=\dim R+\dim S/{\mathfrak {m}}_{R}S.}$

If S is an R-algebra, i.e., we have a homomorphism ${\displaystyle f\colon R\to S}$, then S has the structure of an R-module, and hence it makes sense to ask if S is flat over R. In this setting we have:

If S is flat over R, then S is faithfully flat over R if and only if every prime ideal of R is the inverse image under f of a prime ideal in S. In other words, if and only if the induced map ${\displaystyle f^{*}\colon \operatorname {Spec} (S)\to \operatorname {Spec} (R)}$ is surjective.

Flat modules over commutative rings are always torsion-free (that is, the multiplication by any regular element of the ring in injective in the module). Projective modules (and thus free modules) are always flat. For certain common classes of rings, these statements can be reversed (for example, every torsion-free module over a Dedekind ring is automatically flat and flat modules over perfect rings are always projective), as is subsumed in the following diagram of module properties:

An integral domain is called a Prüfer domain if every torsion-free module over it is flat.

## Categorical colimits

In general, arbitrary direct sums and direct limits of flat modules are flat, a consequence of the fact that the tensor product commutes with direct sums and direct limits (in fact with all colimits), and that both direct sums and direct limits are exact functors. Submodules and factor modules of flat modules need not be flat in general (e.g. Z/nZ is not a flat Z-module for n>1). However we have the following result: the homomorphic image of a flat module M is flat if and only if the kernel is a pure submodule of M.

Daniel Lazard proved in 1969 that a module M is flat if and only if it is a direct limit of finitely-generated free modules.[8] As a consequence, one can deduce that every finitely-presented flat module is projective.

An abelian group is flat (viewed as a Z-module) if and only if it is torsion-free.

## Homological algebra

Flatness may also be expressed using the Tor functors, the left derived functors of the tensor product. A left R-module M is flat if and only if TornR(–, M) = 0 for all ${\displaystyle n\geq 1}$ (i.e., if and only if TornR(X, M) = 0 for all ${\displaystyle n\geq 1}$ and all right R-modules X). Similarly, a right R-module M is flat if and only if TornR(M, X) = 0 for all ${\displaystyle n\geq 1}$ and all left R-modules X. Using the Tor functor's long exact sequences, one can then easily prove facts about a short exact sequence

${\displaystyle 0\;\rightarrow \;A\;{\overset {f}{\longrightarrow }}\;B\;{\overset {g}{\longrightarrow }}\;C\;\rightarrow \;0}$
• If A and C are flat, then so is B
• If B and C are flat, then so is A

If A and B are flat, C need not be flat in general. However, it can be shown that

• If A is pure in B and B is flat, then A and C are flat.

## Flat resolutions

A flat resolution of a module M is a resolution of the form

... → F2F1F0M → 0

where the Fi are all flat modules. Any free or projective resolution is necessarily a flat resolution. Flat resolutions can be used to compute the Tor functor.

The length of a finite flat resolution is the first subscript n such that Fn is nonzero and Fi = 0 for i greater than n. If a module M admits a finite flat resolution, the minimal length among all finite flat resolutions of M is called its flat dimension[9] and denoted fd(M). If M does not admit a finite flat resolution, then by convention the flat dimension is said to be infinite. As an example, consider a module M such that fd(M) = 0. In this situation, the exactness of the sequence 0 → F0M → 0 indicates that the arrow in the center is an isomorphism, and hence M itself is flat.[10]

In some areas of module theory, a flat resolution must satisfy the additional requirement that each map is a flat pre-cover of the kernel of the map to the right. For projective resolutions, this condition is almost invisible: a projective pre-cover is simply an epimorphism from a projective module. These ideas are inspired from Auslander's work in approximations. These ideas are also familiar from the more common notion of minimal projective resolutions, where each map is required to be a projective cover of the kernel of the map to the right. However, projective covers need not exist in general, so minimal projective resolutions are only of limited use over rings like the integers.

## Flat covers

While projective covers for modules do not always exist, it was speculated that for general rings, every module would have a flat cover, that is, every module M would be the epimorphic image of a flat module F such that every map from a flat module onto M factors through F, and any endomorphism of F over M is an automoprhism. This flat cover conjecture was explicitly first stated in (Enochs 1981, p 196). The conjecture turned out to be true, resolved positively and proved simultaneously by L. Bican, R. El Bashir and E. Enochs.[11] This was preceded by important contributions by P. Eklof, J. Trlifaj and J. Xu.

Since flat covers exist for all modules over all rings, minimal flat resolutions can take the place of minimal projective resolutions in many circumstances. The measurement of the departure of flat resolutions from projective resolutions is called relative homological algebra, and is covered in classics such as (MacLane 1963) and in more recent works focussing on flat resolutions such as (Enochs & Jenda 2000).

## In constructive mathematics

Flat modules have increased importance in constructive mathematics, where projective modules are less useful. For example, that all free modules are projective is equivalent to the full axiom of choice, so theorems about projective modules, even if proved constructively, do not necessarily apply to free modules. In contrast, no choice is needed to prove that free modules are flat, so theorems about flat modules can still apply.[12]

## References

1. ^ a b c Matsumura 1970, Proposition 3.G
2. ^ Bourbaki, Ch. I, § 2. Proposition 13, Corollay 1.
3. ^ Matsumura 1970, Corollary 1 of Theorem 55, p. 170
4. ^ Matsumura 1970, Theorem 56
5. ^ http://mathoverflow.net/questions/120403/flatness-of-power-series-rings/
6. ^ Eisenbud 1994, Theorem 6.8
7. ^ Eisenbud 1994, Theorem 18.16
8. ^ Lazard, D. (1969), "Autour de la platitude", Bulletin de la Société Mathématique de France, 97: 81–128
9. ^ Lam 1999, p. 183.
10. ^ A module isomorphic to an flat module is of course flat.
11. ^
12. ^


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# Weak-* sequential compactness and separability

Let $X$ be a Banach space, and let $B$ be the closed unit ball of $X^*$, equipped with the weak-* topology. Alaoglu's theorem says that $B$ is compact. If $X$ is separable, then $B$ is metrizable, and in particular it is sequentially compact.

What about the converse? If $B$ is sequentially compact, must $X$ be separable?

This question was inspired by this one.

-
Wouldn't $B$ be weak* sequentially compact in any reflexive space by Eberlein's theorem? –  David Mitra Jul 9 '12 at 17:42
@DavidMitra: You are quite right. I had forgotten about Eberlein. Want to post this as an answer? –  Nate Eldredge Jul 9 '12 at 18:00

Also, and more generally, it follows from Rosenthal's $\ell_1$ Theorem (a Banach space $X$ does not contain $\ell_1$ isomorphically if and only if every every bounded sequence in $X$ has a weakly Cauchy subsequence) that if $X^*$ does not contain $\ell_1$ isomorphically, then the unit ball of $X^*$ is weak* sequentially compact.
For an explicit counterexample: let $H$ be a non-separable Hilbert space, and suppose we have a sequence of vectors $x_n$ in the unit ball of $H$. Let $H_0$ be the closed linear span of the $x_n$, so that $H_0$ is a separable Hilbert space. Then the $x_n$ have a subsequence converging weakly in $H_0$. But because of the orthogonal decomposition $H = H_0 \oplus H_0^\perp$, it is easy to see that this subsequence also converges weakly in $H$.


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EXCHANGE RINGS SATISFYING STABLE RANGE CONDITIONS

Title & Authors
EXCHANGE RINGS SATISFYING STABLE RANGE CONDITIONS
Chen, Huanyin; Chen, Miaosen;

Abstract
In this paper, we establish necessary and sufficient conditions for an exchange ring R to satisfy the n-stable range condition. It is shown that an exchange ring R satisfies the n-stable range condition if and only if for any regular a $\small{\in}$ R$\small{^n}$, there exists a unimodular u $\small{\in}$$\small{^n}$ R such that au $\small{\in}$ R is a group member, and if and only if whenever a$\small{\simeq}$$\small{_n}$b with a $\small{\in}$ R, b $\small{\in}$ M$\small{_n}$(R), there exist u $\small{\in}$ R$\small{^n}$, v $\small{\in}$$\small{^n}$ R such that a
Keywords
exchange ring;stable range condition;n-pseudo similarity;
Language
English
Cited by
1.
A CRASH COURSE ON STABLE RANGE, CANCELLATION, SUBSTITUTION AND EXCHANGE, Journal of Algebra and Its Applications, 2004, 03, 03, 301
2.
A general vectorial Ekeland’s variational principle with a P-distance, Acta Mathematica Sinica, English Series, 2013, 29, 9, 1655
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H. Chen, Generalized stable exchange rings, SEAMS Bull. Math. 24 (2000), 19-23.

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14.
H. Chen and F. Li, Exchange rings having ideal-stable range one, Science in China (Series A) 44 (2001), 580-586.

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E. Pardo, Comparability, separativity, and exchange rings, Comm. Algebra 24 (1996), 2915-2929.

18.
T. Wu and Y. Xu, On the stable range condition of exchange rings, Comm. Algebra 25 (1997), 2355-2363.


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# Beyond Basic: Advanced simulations in EDEM using the Application Programming and Coupling Interfaces

Image credit: EDEM

### Abstract

EDEM simulations can be customised and extended using the EDEM Application Programming Interface (API). THE API allows users to create custom contact model, particle body force models and particle factories. There are also various coupling options that allow CFD and rigid body dynamics to be included.

Date
May 4, 2016 12:25 PM — 1:00 PM
Location
Alexander Graham Bell Building, School of Engineering
Thomas Bayes rd., The King's Buildings, Edinburgh, Scotland, EH9 3JL
Click on the Slides button above to view the built-in slides feature.

Slides can be added in a few ways:

• Create slides using Wowchemy’s Slides feature and link using slides parameter in the front matter of the talk file
• Upload an existing slide deck to static/ and link using url_slides parameter in the front matter of the talk file


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Question

# To find the first 5 multiples of 12

Factors and multiples
To find the first 5 multiples of 12

2021-08-08
$$\displaystyle{12}\times{1}={12}$$
$$\displaystyle{12}\times{2}={24}$$
$$\displaystyle{12}\times{3}={36}$$
$$\displaystyle{12}\times{4}={48}$$
$$\displaystyle{12}\times{5}={60}$$
Step 3
Hence, the first 5 multiples of 12 are 12, 24, 36, 48 and 60


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I am a mathematician at Duke University, where I am a first-year PhD student. I completed my undergraduate at the University of Utah.

When I'm not doing math, I enjoy rock climbing, cooking, reading, and playing video/board games.

Pronouns: he/him, they/them
Email:
Office: Phytotron 110

## Research

My early research interests lean toward algebraic number theory.

## Teaching

The best way to learn mathematics is to do mathematics. As such, I don't just present concepts to students; I also work to build each student's confidence in their personal problem-solving ability. That self-confidence inspires them to continue working on difficult exercises as well as independently seek out math which interests them. Similarly, students are more successful when they have role models with whom they can identify. For this reason, I try to highlight the accomplishments of diverse mathematicians, use language which makes no assumptions about the look or level of a mathematician, and think critically about how my identity informs my mathematical life. Courses whose instruction I have participated in are tabulated below.

Organization My Role Course Term Comments
Duke University Grader Math 404: Mathematical Cryptography Spring 2021
Duke University Grader Math 305S: Number Theory Seminar Fall 2020
University of Utah Teaching Assistant Math 1220: Calculus II Spring 2020 Course website
University of Utah Help Session Tutor Math 3210/3220: Foundations of Analysis Fall 2018 — Spring 2020

I aspire that my teaching is consistent with Federico Ardila's axioms in Todos Cuentan: Cultivating Diversity in Combinatorics, which together constitute a "pressing call to action" for math educators:

1. Mathematical talent is distributed equally among different groups, irrespective of geographic, demographic, and economic boundaries.

2. Everyone can have joyful, meaningful, and empowering mathematical experiences.

3. Mathematics is a powerful, malleable tool that can be shaped and used differently by various communities to serve their needs.

4. Every student deserves to be treated with dignity and respect.

## Expository Writing

Below are expository notes I wrote to organize my understanding at the time. I collect them here so others may find them useful.

1. $$\ell$$-adic Galois Representations

In this paper, we introduce objects which are central to modern number theory: $$\ell$$-adic Galois representations. These special representations correspond with both highly-symmetric meromorphic functions on the upper half of the complex plane — so-called "automorphic forms" — and with the rational zero-sets of polynomials — so-called "algebraic varieties" over $$\mathbb{Q}$$. In section 1, we motivate our study by describing how this correspondence proved Fermat's Last Theorem, one of the great triumphs of twentieth century mathematics. In section 2, we lay down a foundation of essential number theory. In section 3, we precisely define $$\ell$$-adic Galois representations and study some examples. And finally in section 4, we return to the big picture and discuss the vast frontier of current research in this area.
2. Modular Forms, Elliptic Curves, and their Connection to Fermat's Last Theorem
Fermat's Last Theorem (FLT) states that for an integer $$n > 2$$, the equation $$x^n + y^n = z^n$$ has no integer solutions with $$xyz \neq 0$$. This incredible statement eluded proof for over three-hundred years: in that time, mathematicians developed numerous tools which finally proved FLT in 1995. In this paper, we introduce some of the essential objects which enter the proof — especially modular forms, elliptic curves, and Galois representations — with an emphasis on precisely stating the Shimura-Taniyama Conjecture and explaining how its proof finally settled FLT. We offer proofs whenever they clarify a definition or elucidate an idea, but generally prefer examples and exposition which make concrete a truly beautiful body of mathematical theory.


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# Polynomials: Adding, Subtracting, Dividing & Multiplying

••• Littlewitz/iStock/GettyImages
Print

All math students and many science students encounter polynomials at some stage during their studies, but thankfully they’re easy to deal with once you learn the basics. The main operations you’ll need to do with polynomial expressions are adding, subtracting, multiplying and dividing, and while division can be complex, most of the time you’ll be able to handle the basics with ease.

## Polynomials: Definition and Examples

Polynomial describes an algebraic expression with one or more terms involving a variable (or more than one), with exponents and possibly constants. They can’t include division by a variable, can’t have negative or fractional exponents and must have a finite number of terms.

This example shows a polynomial:

x^3 + 2 x^ 2 - 9 x - 4

And this shows another one:

xy^2 - 3 x + y

There are many ways of classifying polynomials, including by degree (the sum of the exponents on the highest power term, e.g. 3 in the first example) and by the number of terms they contain, such as monomials (one term), binomials (two terms) and trinomials (three terms).

Adding and subtracting polynomials depends on combining “like” terms. A like term is one with the same variables and exponents as another, but the number they’re multiplied by (the coefficient) can be different. For example, x2 and 4 x2 are like terms because they have the same variable and exponent, and 2 xy4 and 6 xy4 are like terms as well. However, x2, x3, x2y2 and y2 are not like terms, because each one contains different combinations of variables and exponents.

Add polynomials by combining like terms in the same way you would with other algebraic terms. For example, look at the problem:

(x^3 + 3 x ) + (9 x^3 + 2 x + y)

Collect the like terms to get:

(x^3 + 9 x^3) + (3 x + 2 x ) + y

And then evaluate by simply adding together the coefficients and combining into a single term:

10 x^3 + 5 x + y

Note that you can’t do anything with y because it has no like term.

Subtraction works in the same way:

(4 x^4 + 3 y^2 + 6 y ) - (2 x^4 + 2 y^2 + y)

First, note that all the terms in the right hand bracket are subtracted from those in the left hand bracket, so write it as:

4 x^4 + 3 y^2 + 6 y - 2 x^4 - 2 y^2- y

Combine like terms and evaluate to get:

(4 x^4 - 2 x^4) + (3 y^2 - 2 y^2) + (6 y - y) = 2 x^4 + y^2 + 5 y

For a problem like this:

(4 xy + x^2) - (6 xy - 3 x^2)

Note that the minus sign is applied to the whole expression in the right bracket, so the two negative signs before 3x2 become an addition sign:

(4 xy + x^2) - (6 xy - 3 x^2) = 4 xy + x^2 - 6 xy + 3 x^2

Then calculate as before.

## Multiplying Polynomial Expressions

Multiply polynomial expressions by using the distributive property of multiplication. In short, multiply every term in the first polynomial by every term in the second one. Look at this simple example:

4 x × (2 x^2 + y)

You solve this using the distributive property, so:

\begin{aligned} 4 x × (2 x^2 + y) &= (4 x × 2 x^2) + (4 x × y) \\ &= 8 x^3 + 4 xy \end{aligned}

Tackle more complicated problems in the same way:

\begin{aligned} (2 y^3 + 3 x ) × &(5 x^2 + 2 x ) \\ &= (2 y^3 × (5 x^2 + 2 x )) + (3 x × (5 x^2 + 2 x )) \\ &= (2 y^3 × 5 x^2) + (2 y^3 × 2 x ) + (3 x × 5 x^2) + (3 x × 2 x ) \\ &= 10 y^3x^2 + 4 y^3x + 15 x^3 + 6 x^2 \end{aligned}

These problems can get complicated for bigger groupings, but the basic process is still the same.

## Dividing Polynomial Expressions

Dividing polynomial expressions takes longer but you can tackle it in steps. Look at the expression:

\frac{x^2 - 3 x - 10}{x + 2}

First, write the expression like a long division, with the divisor on the left and the dividend on the right:

x + 2 )\overline{x^2 - 3 x - 10}

Divide the first term in the dividend by the first term in the divisor, and put the result on the line above the division. In this case, x2 ÷ x = x, so:

\begin{aligned} &x \\ x + 2 )&\overline{x^2 - 3 x - 10} \end{aligned}

Multiply this result by the whole divisor, so in this case, (x + 2) × x = x2 + 2 x . Put this result below the division:

\begin{aligned} &x \\ x + 2 )&\overline{x^2 - 3 x - 10} \\ &x^2 + 2 x \end{aligned}

Subtract the result on the new line from the terms directly above it (note that technically you change the sign, so if you had a negative result you’d add it instead), and put this on a line below it. Move the final term from the original dividend down too.

\begin{aligned} &x \\ x + 2 )&\overline{x^2 - 3 x - 10} \\ &x^2 + 2 x \\ &0 - 5 x - 10 \end{aligned}

Now repeat the process with the divisor and the new polynomial on the bottom line. So divide the first term of the divisor (x) by the first term of the dividend (−5 x ) and put this above:

\begin{aligned} &x -5\\ x + 2 )&\overline{x^2 - 3 x - 10} \\ &x^2 + 2 x \\ &0 - 5 x - 10 \end{aligned}

Multiply this result (−5 x ÷ x = −5) by the original divisor (so (x + 2) × −5 = −5 x −10) and put the result on a new bottom line:

\begin{aligned} &x -5\\ x + 2 )&\overline{x^2 - 3 x - 10} \\ &x^2 + 2 x \\ &0 - 5 x - 10 \\ & -5 x - 10 \end{aligned}

Then subtract the bottom line from the next one up (so in this case change the sign and add), and put the result on a new bottom line:

\begin{aligned} &x -5\\ x + 2 )&\overline{x^2 - 3 x - 10} \\ &x^2 + 2 x \\ &0 - 5 x - 10 \\ &-5 x - 10 \\ & 0 \quad 0 \end{aligned}

Since there is now a row of zeros at the bottom, the process is finished. If there were non-zero terms remaining, you would repeat the process again. The result is on the top line, so:

\frac{x^2 - 3 x - 10}{x + 2} = x - 5

This division and some others can be solved more simply if you can factor the polynomial in the dividend.


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1:52 AM
3

If I'm in outer space, initially at rest, and every single particle in my body accelerates at the same rate in the same direction, will I feel that? My brain is fried thinking about this. There are two possibilities: initially at rest and then accelerating with const acceleration w.r.t. some ine...

17 hours later…
6:46 PM
1

In beta decay neutrons convert into proton, in that process they'll release an electron (called beta particle) but mass still remains same $$_zX^A \longrightarrow\; _{z+1}Y^{A}+ e^-$$ The question is where did that beta particle came out from no where like matter can't be created or destroyed?...

2 hours later…
8:27 PM
1

The name itself suggests that we can hear acoustic branch of phonon modes (and not the optical branch), but other than that I do not see any reason whatsoever why we are able to hear them. In order to be completely clear, I define the acoustic phonons as the ones with the property that \$\omega \p...

3 hours later…
11:25 PM
1

A body travels from A to B at 40 m/s and from B to A at 60 m/s. Calculate the average speed. Can't we just add the given speeds and divide by 2 to get the average speed? This way answer would be 50 m/s In my book, however, they calculated total time taken and distance covered separately and the...


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# Excercise 5.3 Data Handling- NCERT Solution Class 8

Go back to  'Data Handling'

## Chapter 5 Ex.5.3 Question 1

List the outcomes you can see in these experiments.

(a) Spinning a wheel

(b) Tossing two coins together

### Solution

What is known?

Two different condition.

What is unknown?

Outcomes of the given conditions.

Reasoning:

All the possibilities of an event are known as its outcomes.

Steps:

(a) There are four letters $$A, B, C$$ and $$D$$ in a spinning wheel. So, there are $$4$$ outcomes.

(b) When two coins are tossed together there are four possible outcomes

$$HH, HT, TH, TT.$$

(Here $$HT$$ means head on first coin and tail on second coin and so on.)

## Chapter 5 Ex.5.3 Question 2

When a die is thrown, list the outcomes of an event of getting:

(i)  (a) a prime number

(b) not a prime number

(ii) (a) a number greater than $$5$$

(b) a number not greater than $$5$$

### Solution

What is known?

A situation in which a die is thrown.

When a dice is thrown,the possible outcomes are  $$1,2,3,4,5$$ and $$6.$$

What is unknown?

Outcomes of the situation under given conditions.

Reasoning:

All the possibilities of an event are known as its outcomes.

Steps:

(i)

(a) Outcomes of event of getting a prime number are $$2, 3$$ and $$5.$$

(b) Outcomes of event of not getting a prime number are $$1, 4$$ and $$6.$$

(ii)

(a) Outcomes of event of getting a number greater than $$5$$ is $$6.$$

(b) Outcomes of event of not getting a number greater than $$5$$ are $$1, 2, 3, 4$$ and $$5.$$

## Chapter 5 Ex.5.3 Question 3

Find the:

(a) Probability of the pointer stopping on $$D$$ in (Question 1 (a))?

(b) Probability of getting an ace from a well shuffled deck of $$52$$ playing cards?

(c) Probability of getting a red apple? (See figure alongside)

### Solution

What is known?

(i)  Spinning wheel image

(ii) Deck of $$52$$ playing cards

(iii) Image depicting Red and Green apples

What is unknown?

(i) Probability of the pointer stopping on $$D.$$

(ii) Probability of getting an ace from a well shuffled deck of $$52$$ playing cards.

(iii) Probability of getting a red apple

Reasoning:

Probability

\begin{align}= \! \frac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\end{align}

Steps:

(a) In a spinning wheel, there are five pointers $$A, A, B, C, D.$$ So, there are five outcomes. Pointer stops at $$D$$ which is $$1$$ outcome. So, the probability of the pointer stopping on \begin{align}D = \frac{1}{5}\end{align}

(b) There are $$4$$ aces in a deck of $$52$$ playing cards. So, there are $$4$$ events of getting an ace.

So, probability of getting an ace \begin{align} = \frac{4}{{52}} = \frac{1}{{13}}\end{align}

(c) Total number of apples $$= 7$$

Number of red apples $$= 4$$

So, probability of getting red apple \begin{align} = \frac{4}{7}\end{align}

## Chapter 5 Ex.5.3 Question 4

Numbers $$1$$ to $$10$$ are written on ten separate slips (one number on one slip), kept in a box and mixed well. One slip is chosen from the box without looking into it.

What is the probability of:

(i)  getting a number $$6?$$

(ii)  getting a number less than $$6?$$

(iii) getting a number greater than $$6?$$

(iv)  getting a number less than $$6?$$

(v) getting a $$1-$$digit number?

### Solution

What is known?

Numbers $$1$$ to $$10$$ are written on ten separate slips (one number on one slip), kept in a box and mixed well

What is unknown?

(i) Probability of getting a number $$6$$

(ii) Probability of getting a number less than $$6$$

(iii) Probability of getting a number greater than $$6$$

(iv) Probability of getting a number less than $$6$$

(v) Probability of getting a $$1-$$digit number

Reasoning:

Probability\begin{align} = \frac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\end{align}

Steps:

(i) Outcome of getting a number $$6$$ from ten separate slips is $$1$$. Therefore, probability of getting a number \begin{align}6 = \frac{1}{{10}}\end{align}

(ii) Numbers less than $$6$$ are $$1, 2, 3, 4$$ and $$5$$. So, there are $$5$$ outcomes. Therefore, probability of getting a number less than

\begin{align} 6 = \frac{5}{{10}} = \frac{1}{2} \end{align}

(iii) Number greater than 6 are $$7, 8, 9, 10.$$ So there are $$4$$ possible outcomes. Therefore, probability of getting a number greater than

\begin{align} 6 = \frac{4}{{10}} = \frac{2}{5} \end{align}

(iv) One-digit numbers are

$$1, 2, 3, 4, 5, 6, 7, 8, 9$$ out of ten.

Therefore, probability of getting a $$1$$-digit number

\begin{align} = \frac{9}{{10}} \end{align}

## Chapter 5 Ex.5.3 Question 5

If you have a spinning wheel with $$3$$ green sectors, $$1$$ blue sector and $$1$$ red sector, what is the probability of getting a green sector?

What is the probability of getting a non-blue sector?

### Solution

What is known?

Spinning wheel with $$3$$ green sectors, $$1$$ blue sector and $$1$$ red sector

What is unknown?

i)  probability of getting a green sector

ii) probability of getting a non-blue sector

Reasoning:

Probability

\begin{align}= \frac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\end{align}

Steps:

Total number of sectors $$= 5$$

Total number of green sectors $$= 3$$

Therefore, probability of getting a green sector\begin{align} = \frac{3}{5} \end{align}

Total number of blue sectors $$= 1$$

Total number of non-blue sectors $$= 5 – 1 = 4$$

Therefore, probability of getting a non-blue sector \begin{align} = \frac{4}{5}\end{align}

## Chapter 5 Ex.5.3 Question 6

Find the probability of the events given in Question $$2.$$

### Solution

What is known?

A situation in which a die is thrown.

What is unknown?

Probability of the events

Reasoning:

Probability

\begin{align} =\frac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}} \end{align}

Steps:

When a die is thrown, there are total six outcomes, i.e., $$1, 2, 3, 4, 5$$ and $$6.$$

(i) (a) $$2, 3, 5$$ are prime numbers. So, there are $$3$$ outcomes out of $$6.$$

Therefore,probability of getting a prime number

\begin{align} = \frac{3}{6} = \frac{1}{2} \end{align}

(b) $$1, 4, 6$$ are not the prime numbers. So, there are $$3$$ outcomes out of $$6.$$

Therefore, probability of getting a prime number

\begin{align} = \frac{3}{6} = \frac{1}{2} \end{align}

(ii) (a) Only $$6$$ is greater than $$5.$$ So, there is one outcome out of $$6.$$

Therefore, probability of getting a number greater than

\begin{align} 5 = \frac{1}{6} \end{align}

(b) Numbers not greater than $$5$$ are $$1, 2, 3, 4$$ and $$5.$$ So there are $$5$$ outcomes out of $$6.$$

Therefore, probability of not getting a number greater than

\begin{align} 5 = \frac{1}{6} \end{align}


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# వికీపీడియా:Alternative text for images

Every image with more than decorative value, including math-mode equations, should specify alternative text (alt text). Alternative text is used as a replacement for an image, whenever the image cannot be seen. This can happen, for example, when someone:

• uses a screen reader (e.g. a visually impaired person)
• uses a text-only browser (e.g. browsing from a mobile phone)
• uses a graphical browser with images turned off
• browses results from a Web search
• copies an extract from a Web page into a word processor.

Alternative text is also used for other purposes. For example, Google's image search uses it to help return appropriate images. Finally, good choice of alternative text and captions makes life easier for people who are viewing the source of an article, either when editing it, or in a diff, or in Wikipedia's internal search.

To add alt text to an image, specify an "alt=" parameter. Here is an example image specified with alt text, next to what its output looks like:

National flag of France.
`[[Image:Flag of France.svg|thumb|alt=Vertical tricolor flag (blue, white, red).|National flag of France.]]`

In equations formatted in Wikipedia's math mode, the alt text is specified within the initial math tag. For example, this:

`<math alt="3/5 = 0.6">\tfrac{3}{5} = 0.6</math>`

generates this output:

${\displaystyle {\tfrac {3}{5}}=0.6}$

Unlike images, alt text in the math tag takes quotation marks.

Crafting appropriate alternative text can be difficult since it must try to meet several criteria. Above all, it should convey a correct description to visually impaired Wikipedians using screen readers such as JAWS. However, alt text should also make sense in a graphical browser with images turned off, and it should fit with the surrounding text when viewed with a text-only browser. Trying to accommodate all these criteria can be daunting! This article aims to make the whole process easier.

Alt text is not the same as a caption. The alt text is meant for those who cannot see the image, whereas the caption is intended for all readers. In general, the alt text summarizes the image, whereas the caption helps all readers to interpret the image, to focus on its most essential elements and to connect it with the article text.

## Captions and alt text serve different functions

A 3rd century BC coin depicts the co-rulers of Ptolemaic Egypt: Ptolemy II Philadelphus (left), patron and ex-pupil of Philitas; and Philadelphus' sister and wife Arsinoe II, possibly also an ex-pupil.

The alt text description should not duplicate information already present in the caption or the main text of the article. However, material can be added to help the listener identify the subject of the image. Here is an example adapted from the article Philitas of Cos:

``` [[Image:Oktadrachmon Ptolemaios II Arsinoe II.jpg|thumb|alt=Gold coin showing heads and shoulders of a well-fed king and queen in ancient Greek clothing; the king is more prominent.|A 3rd century BC coin depicts the co-rulers of [[Ptolemaic Egypt]]: [[Ptolemy II Philadelphus]] (left), patron and ex-pupil of Philitas; and Philadelphus' sister and wife [[Arsinoe II of Egypt|Arsinoe II]], possibly also an ex-pupil.]] ```

The alt text describes the coin's appearance, whereas the caption focuses on its explanation. Again, the basic rule is that alt text is written for those who can't see the image, whereas captions are written for everyone.

## Choosing good alternative text

Alternative text exists as a substitute for the image, for those who cannot see it. It should not interpret the image, or suggest its meaning, since that role belongs to the caption instead. Original research should be scrupulously avoided when writing alt text. In general, a sighted reader should be able to confirm by eye that the alt text describes the image. On the other hand, the alt text need not summarize irrelevant details of the image.

The alt text should make sense when read aloud and in the absence of the image itself. It should be composed of complete sentences and punctuated correctly, with screen readers in mind. It should sit comfortably in the flow of the article.

A helpful way to think about alternative text is to imagine that the Web page text, including all alternative text, is the script for an audio cassette. Your listeners aren't necessarily blind, so they may be interested in hearing about what something looks like. But they can't possibly see any images on the cassette, so referring to an image itself will sound silly. If you write your alternative text with this in mind, it should work well in all the situations listed above.

## Alternative text is always necessary on Wikipedia

Because the Wikipedia images are also links, the alternative text must always be specified. Otherwise, a screen reader would read the file name by default.[WCAG-TECH 1]

### Alternative text for icons

Since icons does not carry important information, only a short and general alt text is needed. For example, stub icon should have the following alt text: `alt=Stub icon`.

Icons usually have no `thumb` parameter. In this case, and if there is a caption, MediaWiki will copy the caption in the alt parameter. For example, the result of [[Image:P Eiffel.png|48x24px|Eiffel tower icon]] will be:

•

When the link on the image is not relevant (for example, link on icons are not relevant), you may decide to suppress it. To do so, you have two options:

1. When you add an empty `link=` parameter, Mediawiki produces an empty alt text (`alt=""`). Since there is no link the image becomes accessible.[WCAG-TECH 2]
• For example, the result of [[Image:P Eiffel.png|48x24px|link=]] is:
2. Use css to have the icon in background. Here is an example:

## Describe the image, not its file

Alternative text describes the image, not its file or format. Thus, you shouldn't mention

• the filename
• the file format
• the size of the file
• the dimensions of the file
• instructions on how to view a bigger version of the image.
• the fact that it is a substitute for an image:
• Good: "The straight chain form consists of four C H O H groups..."
• Bad: "This image shows that the straight chain form consists of four C H O H groups..."

## Example

### Wikipedia code

``` [[Image:Antikythera philosopher.JPG|thumb|left|upright|alt=Bronze head of bearded man.|The ''Philosopher'' ({{circa|250–200 BC}}) from the [[Antikythera wreck]] illustrates the style used by Hecataeus in his bronze of Philitas.]] ```

The phrase "Bronze head of bearded man." is the text that will be rendered in place of the image.

The Philosopher (కాలం. 250–200 BC) from the Antikythera wreck illustrates the style used by Hecataeus in his bronze of Philitas.

The above might be displayed in a graphical web browser that is capable of showing images. In this case, the image serves to illustrate the description of a (different) bronze statue.

### What people see with images turned off

The Philosopher (కాలం. 250–200 BC) from the Antikythera wreck illustrates the style used by Hecataeus in his bronze of Philitas.

This example shows what might be displayed in a web browser with images turned off. Many modern browsers allow users to turn off images (for example, if they are using a low-bandwidth connection). The contents of the `alt` text are often rendered in place of the image. Here, the phrase serves as a useful replacement for the image.

Caveat: Though this is not technically considered an appropriate usage for the `alt` text contents, many browsers nevertheless display it in this manner. If possible, choose `alt` text which makes sense in this format.

## History

Until 2008, the alt text of an image was automatically the same as its caption, an approach with two drawbacks. First, this meant that the image caption was read twice by the screen reader to the listener, once for the alt text and once for the caption itself. Second, captions do not always describe their image, but rather describe its meaning for the reader.

A new system was introduced in October 2008. It allows Wikipedians to specify the image alt text independently of its caption. The caption is always the last parameter in the Image link, whereas the alt text is specified by an "alt=" parameter. In this new system, the alt text serves to describe the image to those who cannot see it, whereas the caption conveys essential information beyond what can be seen.


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# 위성통신용 고이득 광대역 마이크로스트립 배열 안테나의 설계연구

• 윤현보 (동아대학교 전자공학과) ;
• 임계재 (동아대학교 전자공학과)
• Published : 1991.09.01

#### Abstract

A modified transmission line model is proposed for input impedance analysis of a square microstip antenna with single-circularly-polarized. Alog-periodic arrangement with microstrip antenna (LPMA) which consists of 3 resonant element ($3\times1$) is designed for broadband operation ranging from 11.2 GHz to 12.4 GHz, based on this pro- posed model from transmission line model. Aplanar array ($3\times4$array), in which 4sets of the above LPMA are in the same plane, is fabricated for high gain X-band satellite communication antennas. The optimum spacing between each LPMA is determined using computer simulation for the minimum bariation of far field pattern and null. This $3\times4$array has a measured values of gain greater than 8dBi, VSWR les then 2.3, and bandwidth greater than 7% respectively.


text:
## Monday, February 18, 2013 ... /////

### Nima at STScl: don't modify gravity, understand it

...because it is a moral issue...

It's not the first time when Nima Arkani-Hamed gave a "totally negative" talk on a similar issue but it's a fun time.

Twelve days ago he came to The Space Telescope Science Institute (STScI) in Baltimore, Maryland – the terrestrial headquarters for Hubble (past) and James Webb Telescope (future) and a loose part of Johns Hopkins University – and had the following things to say.

Video:

Don't Modify Gravity—Understand It! (video, 85 minutes)
He is introduced by a host who enumerates some prizes Nima has received, including a nice one from Milner, and mentions that the only person over there who would understand the talk is Mario Livio who is just recovering from an illness (fortunately for his health, unfortunately for the readiness of the audience).

Nima claims to have a special moral right to give an entirely negative talk about the sinfulness of modifications of gravity because he's been a dirty sinner himself. ;-) So that's why he has the credentials to crawl from his hole, confess, and repent. Of course, he means it's wrong to modify gravity at long distances. Of course, GR isn't adequate at super short distances, an issue he later (around 15:00) talks about, too. Gravity just becomes an intrinsically strong force over there (at the Planck scale).

Nima debunks the usual vague philosophical attitudes about general relativity's being justified by the beauty, sexiness of geometry, and similar stuff. Instead, GR works because it's inevitable; it's dictated by consistency on top of special relativity and quantum mechanics. This provable inevitability shows how deep our knowledge of physics has become. He sketches his plans to discuss the cosmological constant problem and learn from the epic failure of every single attempt to solve the problem by modifying gravity at long distances. He divides the theories to the deeply flawed ones that he would spend his time with, anyway; and complete idiocies that are not worth talking about at all, despite a thousand of deluded papers about them.

The experience has numerously shown that physicists have to be radically conservative instead of conservatively radical. Well, in this case, he claims that the conservatism leads to the landscape. OK, his more technical comments start with the proof that gauge theories and gravity are inevitable.

Nima attacks the claims in the popular books that "we don't know how to combine gravity and quantum mechanics". Obviously, neutron interferometry may be done and measured (and used to show that Erik's Verlinde entropic gravity is crap). Obviously, we may talk about gravity and quantum mechanics in the same sentence – otherwise we couldn't have discovered quantum mechanics on planet Earth (where we have gravity) at all. In fact, we may even calculated the leading-corrected Newton's law:$F = \frac{G m_1 m_2}{r^2} \zav{ 1 - \frac{27}{2\pi^2}\frac{G\hbar}{r^2c^2} +\dots}$ We can questions like that! The effect is minuscule but we may still calculate it. Quantum mechanics is a robust framework that can deal with anything. The real difficulty – sad properly – is that we don't know what happens with various terms in gravity at short distances which is why GR needs a replacement. I couldn't agree with him more. Note that Nima attacks not only really crappy popular books such as Lee Smolin's ones but even the really mainstream ones such as Brian Greene's books. Needless to say, I think that Brian's confusing presentation of the conflict is biased towards his completely flawed tendency to modify quantum mechanics whenever he sees a vague opportunity or excuse for that. ;-)

Fine. Nima returns to long distances. Photons would produce a wrong negative-norm polarization. Gauge invariance ("redundancy", using his more favorite words) is needed. The same goes for the diffeomorphism redundancy and gravity and spin-2 fields. These gauge redundancies aren't real symmetries – they're artifacts of our imagination. He continues with Weinberg's arguments from the 1960s that the interactions with added photons – regardless of the nature of the theory – have to depend on the polarization vectors for these interactions to remain nonzero at long distances. Now, pure gauge polarizations have to decouple (interaction amplitude going to zero) which implies that the total charge is zero and therefore, there has to be the "symmetry".

The same with spin-2. You don't have to know anything about the "culture" of gravity or the word. The spin-2 needs to decouple the bad polarizations. This forces the conservation laws for the momentum in this case. As I have explained many times, the allowed spins of massless/light particles are $0,1/2,1,3/2,2$ and the last three choices inevitably arrive with increasingly constrained gauge redundancies.

Nima says many things I repeat many times. Someone comes to you and tells you that he may have a completely new spiritual viewpoint on gravity, based on some torsion, wakalixes, or any other word. We just don't need to care about these vague words and fog! We may study what the theory predicts for the interactions and whatever it is, it will either agree with the "gauge descriptions above" or violate the general postulates of relativity or quantum mechanics. That's it, we're finished, "anything goes" is just wrong.

At 35:40, he begins to talk about the cosmological constant problem, the most frequent excuse for modifications of gravity at long distances. Everything is Planckian in the fundamental theory, so should be the density or curvature of empty space, but it is empirically 123 orders of magnitude smaller. There are contributions to the C.C. that are Planckian, after all, and they must cancel with the huge relative accuracy (classical terms against quantum corrections etc.). The cancellation isn't an inconsistency, however ludicrous it may seem. People used to believe in the fantasy that there was a deep so far unknown reasons why the C.C. ultimately exactly cancels. Its nonzero value made this scenario far less plausible: it almost cancels but there must still be a tiny leftover. Oops.

Nima compares this situation to the fantasies in the 1930s when the first divergent loop corrections to QED were seen. It was believed that those infinities were fantasies that would ultimately see a redefinition of QED that makes all of them zero. Well, they were finite but they were definitely not zero. Loop corrections at every order matter and are nonzero, although much smaller than the naive (infinite) value. As a stumbling block, the case of the C.C. is more serious; as a stimulation of the progress, the story of the loop corrections to the magnetic moment was far more dramatic.

When written as an energy density, the C.C. is the fourth power of the inverse millimeter. So some new physics could be appearing at the millimeter scale. Except that we have looked and there seems to be nothing over there. (We think that the analogous solution does explain the hierarchy problem, the unbearable lightness of the Higgs' being, however.)

Now, at the level of linguistics, a great idea to explain why the C.C. is so small is to "degravitate" the vacuum energy, all the modes of the fields or those whose wavelengths are shorter than the millimeter (this refinement is needed to preserve the tested behavior of gravity at distances longer than 1 mm). Everything else gravitates but this form of energy just happens to exert no gravitational influence, the proposal says. However, physics only starts once you begin to convert the word-level ideas to equations and 99.9% of the word-level ideas simply fail at that point.

A slightly clever mechanism to realize similar ideas was "dumping curvature in the bulk" of a higher-dimensional space. Of course it's easy to "solve" the C.C. problem if you're allowed to modify the behavior of gravity, e.g. by changing the exponent in Newton's power law. The discussion turns to examples in lower-dimensional gravity (3D gravity with deficit angles), too.

A modification of the idea is based on Randall-Sundrum who managed to "trap" gravity on the brane. At the beginning, it looked like the tower of apparently massless Kaluza-Klein modes of the gravitons may provide you with a loophole. However, one may ultimately see that the RS theories end up being effectively equivalent to normal four-dimensional theories, GR with massless matter. I know that Lisa doesn't like these "RS is equivalent to..." comments but I am afraid (or happy to see) that Nima is right. It's not just the full-fledged AdS/CFT methods that argue in this direction.

More exotic ideas (53:00) came from DGP, Dvali-Gabadadze-Porrati (it's a messed up GDP; I really disliked those things from the beginning and without an interruption because they seemed so contrary to all the lessons I learned from string theory). The Einstein-Hilbert term is weighted by a coefficient that only makes gravity matter on our brane; it's extra 4D gravity added on top of the normal 5D gravity. Your humble correspondent thinks it's inconsistent at the quantum level (swampland). Surely in string theory, it looks like gravity only affects "all of space" and the "$D$-dimensional space with all the dimensions one can find". At any rate, Nima shows that if the crossover state obeys the inequality needed to dump the curvature etc., gravity on the brane will be modified at millimeter-like distances, too.

Deficit angle didn't work because it can't go over $2\pi$ or so. With an excess angle, this limitation seems to be lifted. Things look fine. However, if you're diligent, you derive the propagator in your theory and alas, it is modified, too. Some $2\pi$ becomes $2\pi+|T|$, in some units; the tension is added. For a large tension, the tension term dominates. You recalculate at what distances gravity starts to behave properly and you find out that these distances must be larger than the Hubble scale (the size of the Universe). Too bad. ;-)

A model after model after model fails like that. Why all of them fail? It's useful to hit your head with a hammer 10 or 20 times, then you may start to get the message. ;-) The reason behind the failure of all these things is relativity. Relativity relates space and time. If you modify physics at some distances, you modify it at some time scale, too – or break relativity or causality.

More technically, your stress energy tensor $T^{\mu\nu}={\rm diag}(\rho,p,p,p)$. The different components have different sources – the C.C., visible matter, radiation – and there's simply no relativistically invariant way to "isolate" the piece that you would like to make non-gravitating without spoiling the proper behavior of matter and radiation at the same moment! The only way to avoid the conclusion would be to modify gravity by "knowing in advance" what the impact would be, and that would violate causality.

At 1:00:00, he switches to the question whether one may modify gravity at all – a purely theoretical question. Can one make gravity a bit massive etc.? Every modification of gravity has to add a new degree of freedom, typically a scalar field that couples to $T_\mu^\mu$. That's morally nothing else than boring extension of GR with scalars such as Brans-Dicke etc., something that's been around since the 1940s. If you add the scalars, you modify gravity in a boring way – e.g. you modify the bending of light so that you are in conflict with the observations at the same time unless you boringly choose the coefficients of the extra terms to be much smaller than one (obnoxious imitators/deformations of GR that can't solve its fundamental problems such as the C.C. problem).

Exciting possibilities are those where the scalar fields are more exotic, e.g. massive gravity or Galileons or ghosts or fields with specific non-linear self-interactions. Lots of phenomenology but as I always said, these theories lead to deep conflicts with locality and/or thermodynamics. At any rate, Nima promotes the Galileons. Scalar fields are equipped not just by symmetry $\phi\to\phi+c$ but also $\phi\to\phi+v_\mu x^\mu$, a "Galilean symmetry" but acting on fields rather than $\vec x$. Lunar phenomenology, fun, but to make the story short, some radial modes propagate superluminally. Despite formal Lorentz invariance, the "nice structure" is inseparable from the lethal acausality.

Similarly, "higgsed gravity" allow you to produce "black hole hair" which is enough to create a perpetual motion machine of second kind by making a cycle between two distinct event horizons of a black hole valid for two different particle species/fields. Too bad. Nima spent 8 years of his life with similar things, he says. Gravity is much more constraining than other types of dynamics. He declares that the eternal inflation plus the multiverse are the only possible way to the C.C. problem given the failures discussed above. I disagree with that. I don't have to modify gravity at all but the selection that makes the C.C. tiny may still reject the existence of any multiverse.

Of course, I agree with Nima that one has to be radically conservative instead of a left-wing as*hole because the declaration that the previous knowledge was "just wrong and may be scratched" has never been successful. Even quantum mechanics overthrew classical physics in a very respectful way – when it explained some surprising features of classical physics that should have been asked even by classical physicists, e.g. "Why can the equations of motion be derived from a Hamiltonian as well as the principle of least action?" No trashing here.

Radical conservatism – the right attitude – means that one is prepared to push the tested principles as far as they can go and only when one becomes absolutely sure that there is a problem, he may think about modifications. Nima repeated his slogan, Don't modify gravity, just understand it. And the host proposed not to modify Nima's talk by stupid questions, instead, let's drink some beer which is what they are conservatively good at.

#### snail feedback (18) :

speaking of milner prize, new ones for 2013 are out

http://www.fundamentalphysicsprize.org/news/news3

Just to be clear, when Nima says QM and relativity by themselves imply all these other things he is referring to special relativity? Or general relativity?

[disregard previous comment, let me rephrase]

So Nima (and Lubos) are saying QM and relativity (ie, special relativity) by themselves are enough to prove the laws of gravity, the way it warps space and time, etc.?

Right, only you are 2+ months late:

Ha ha, this looks like a lot of fun and exciting, the title seems to contain a typical cool Nima comment :-D

Too bad that I can only watch the video at the weekend, since I am too busy for a talk I want to give @work on Friday :-/

So do not write to many similar nice things that would pile up from the floor to the ceiling until the weekend :-D

Cheers

It has taken me 30 years to get the experimental results that is shown in the picture posted below. With with other embodiments of my invention which I call Reactionless
Thermal Drive (Pat Pending) I have gotten two 99% loss of weight and another 90% loss of weight as well smaller loss of weight. It my hope that these results and my heat-based gravity theory which has motivated these experiments will one day put and end to this domination of mathemagicians like Lubos, Nima Arkani-Hamid and assorted String theorists who think it is a smart idea that General Relativity should be united with the other forces. But I suspect you will ignore my post and it will not be until the general population are using my invention for a "better way to go from A to B" will your crowd will come to realize that what is says in the introductory physics textbook is not in every word the "veritable gospel truth".

Thank you for this interesting post Lubos - I have not yet listened to Nima's talk at the Space Telescope Science Institute, but surely will before the end of the week. I've always been interested by the cosmological constant problem, it is one of the "big elephants in the room" (or in this instance a very tiny elephant, much smaller than one expects).

In your post, you say that "a great idea to explain why the C.C. is so small is to "degravitate" the vacuum energy", and then goes on to explain why this and other models fail.

I used to slightly favor this idea, partly for aesthetic reasons. In the absence of gravity the vacuum energy is not observable, being swept under the carpet by the procedure of normal ordering. We can only measure changes in energy. For example, in the Casimir effect one measures the force between two plates as their separation increases or decreases. The force is minus the derivative of the electrodynamic vacuum energy between the two plates and if you shift the vacuum energy by an arbitrary constant that won't affect the measurement. Now this picture changes brutally when gravity is turned on; suddenly the quantum zero point fluctuations yield a huge vacuum energy with observable consequences and Minkowski space is not stable. This "discontinuity" (for lack of a better term) as gravity is turned on/off used to bother me a little bit. Of course that's just a vague aesthetic statement and also it's not valid when supersymmetry is not broken.

But another reason why I like the above idea is the following. If you calculate the bare vacuum energy density in a FLRW spacetime, with a cut-off k_c in momentum space, then you can show by dimensional analysis that it has the expansion \rho_{vac, 0} = k_c^4 + k_c^2 R + R^2 + ... (we need LaTeX in the comments 8) ), where R is related to the curvature and is roughly on the order of H_0^2 with H_0 the value of the Hubble constant today. The principle of stability of empty Minkowski space allows you to discard the first term, while the third term is negligible. So you end up with \rho_vac = (k_c H_0)^2 and if k_c is close to the Planck scale then the result is of the same order of magnitude as the observed value of the cosmological constant. This model was recently proposed by Denis Bernard and Andre LeClair. I find it nice and simple. If it's true (we don't know and perhaps will never know), then the cosmological constant is analogous to a Casimir effect in an expanding spacetime.

Hi, Lubos. I would like to know your opinion regarding fields vs. particles: Specifically, some (e.g., Matt Strassler, and I think Frank Wilczek) believe fields are 'real'; but others (e.g., Nima Arkani Hamed) say that fields are a useful fiction, and that only particles exist. Thank you for your time. Kevin

Dear Kevin, it depends how you define "real". You would have to give an operational definition. I think it's fair to say that fields are as real or as fundamental as particles - fields and particles only differ by the choice of the basis in the Hilbert space. I think that Nima's assertion that they're not real only means that one may build everything in terms of particles only so that fields are just an auxiliary concept that may be useful and I agree with that. However, fields may also be measured - E_z(x,y,z,t) is an observable etc. so I would still prefer to say that they're "real".

It's nice you like those ideas ;-) but - assuming that you have already listened (if my blog entries haven't made the job), have you actually tried to look at their lethal flaws and/or whether a fix may be found?

Always the same message: Nima is right and the rest are complete idiots. It's hard to listen to this neurotic more than a few minutes ....

Maybe some people have trouble to listen to deep and sensible talks like that because they belong to the "rest" as you described it? ;-)

;-)

Regarding the leading edge correction to Newton's law of gravity, this depends on your choice of coordinate system for the (I assume non rotating) black hole. What coordinate system gives a 27/pi^2 coefficient? The complete (exact) sequence for Einstein's black hole gravitation in terms of F=ma is given in my paper that won an honorable mention at the annual gravitation contest here:
http://arxiv.org/abs/0907.0660

Come on, Carl, you're not going to argue by references to an "honorable mention" by a crackpot foundation, are you?

At the accuracy indicated, there's nothing coordinate-dependent about the first subleading term. It's a C/r^2 correction to the gravitational potential. To separate it from the usual 1/r term, one just needs not to change the definition of "r" by more than O(1 Planck length) at infinity which is easy to do and a natural convention, too.

Lubos, the argument doesn't use quantum calculations and the truth of the argument isn't based on the gravity research foundation (which Hawking won several times, some crackpot). The argument is based on quite simple calculus which you can easily verify. Try to avoid using the argument techniques that the CAGW believers have taught.

The standard coordinates for black holes is Schwarzschild. The GR results are not the order you've given. If you know what coordinates the result you gave are for, say so. Or better, give a link to the paper.

Furthermore, the "r" used in GP coordinates is identical to the "r" used in Schwarzschild. To convert (r,t) from one to the other all you change is t. Schwarzschild metric is given by equation (3) and the conversion to GP is given in equation (5) of http://arxiv.org/abs/gr-qc/0411060

So from the expansion I gave you can see that there are huge differences in the form of the GP and Schwarzschild accelerations despite their sharing the same r (the changes show up as powers of the speed of the test particle and do not appear in the acceleration of a "stationary" test particle).


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If $a$ is an integer, $b$ is an integer and $c$ is irrational, which of the following are rational numbers?

1. $\dfrac{5}{6}$
2. $\dfrac{a}{3}$
3. $\dfrac{-2}{b}$
4. $\dfrac{1}{c}$
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1 Answer

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1. rational

2. rational

3. rational

4. irrational
by Diamond (39,701 points)

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The Operator Algebra of Bounded Linear Operators on X

# The Operator Algebra of Bounded Linear Operators on X

 Definition: Let $X$ be a Banach space and let $\mathcal B(X) = \mathcal B(X, X)$ denote the space of all bounded linear operators on $X$. The Operator Algebra of Bounded Linear Operators on $X$ is the space $\mathcal B(X)$ with the operations of operator addition, scalar multiplication, and composition as operator multiplication, specified for all $S, T \in \mathcal B(X)$ and for all $\alpha \in \mathbb{C}$ by: 1) $(S + T)(x) := S(x) + T(x)$ for all $x \in X$. 2) $(\alpha S)(x) := \alpha S(x)$ for all $x \in X$. 3) $(S \circ T)(x) :=S(T(x))$ for all $x \in X$. and with norm on $B(X)$ as the usual operator norm, $\displaystyle{\| S \| := \sup_{\| x \| = 1} \| S(x) \|}$.

Note that the operator norm on $\mathcal B(X)$ is indeed an algebra norm, since for all $S, T \in \mathcal B(X)$ we have that:

(1)
\begin{align} \quad \| S \circ T \| = \sup_{\| x \| = 1} \| S(T(x)) \| \leq \sup_{\| x \| = 1} \| S \| \| T(x) \| = \| S \| \sup_{\| x \| = 1} \| T(x) \| = \| S \| \| T \| \end{align}

Moreover, recall from the Criterion for B(X, Y) to be a Banach Space page, that if $Y$ is a Banach space then $\mathcal B(X, Y)$ is also a Banach space. In particular, if $X$ is Banach space, then $\mathcal B(X)$ is a Banach space and thus $\mathcal B(X)$ is a Banach algebra.

Note that in general, composition of two functions (when defined) is not commutative. The same applies to compositions of two bounded linear operators, so $\mathcal B(X)$ is a noncommutative Banach algebra.


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# Integral of a numerical second derivative

I need to find the integral $$M(r)$$ of a second derivative of equation $$f(r)$$ where this equation has only a numerical solution sol = NDSolve[f[r],...].

M[r]=Integrate[the second derivative of f(r),{r,0,10}]

Because lack of example code I show the principal way to solve the problem:

Create NDSolve solution f[r]

f = NDSolveValue[{F'[r] + F[r] == 0, F[0] == 1}, F, {r, 0, 10}]

Integrate NDSolve result f''[r]

NIntegrate[f''[r], {r, 0, 10}]
(*0.999955*)

That's it!

Let's say the original call has a form something like the following:

sol = NDSolve[f[r], u, {r, 0, 10}] (* OP says f[r] is an 'equation' but does
not indicate the dependent variable
or boundary conditions              *)

By the Fundamental Theorem of Calculus, the answer should be given by

u'[10] - u'[0] /. sol

This will return the value(s) for the integral, one for each solution returned by NDSolve. (A nonlinear f[r] may have more than one solution.)