text:
## Shock Breakout

Here you can see the brilliant flash of a supernova as its core blasts through its surface. This is an animated cartoon made by NASA based on observations of a red supergiant star that exploded in 2011. It has been sped up by a factor of 240. You can see a graph of brightness showing the actual timescale at lower right.

When a star like this runs out of fuel for nuclear fusion, its core cools. That makes the pressure drop—so the core collapses under the force of gravity.

When the core of a supernova collapses, the infalling matter can reach almost a quarter the speed of light. So when it hits the center, this matter becomes very hot! Indeed, the temperature can reach 100 billion kelvin. That’s 6000 times the temperature of our Sun’s core!

For a supernova less than 25 solar masses, the collapse stops only when the core is compressed into a neutron star. As this happens, lots of electrons and protons become neutrons and neutrinos. Most of the resulting energy is instantly carried away by a ten-second burst of neutrinos. This burst can have an energy of 1046 joules.

It’s hard to comprehend this. It’s what you’d get if you suddenly converted the mass of 18,000 Earths into energy! Astronomers use a specially huge unit with such energies: the foe, which stands for ten to the fifty-one ergs.

That’s 1044 joules. So, a supernova can release 100 foe in neutrinos. By comparison, only 1 or 2 foe come out as light.

Why? Neutrinos can effortlessly breeze through matter. Light cannot! So it takes longer to actually see things happen at the star’s surface—especially since a red supergiant is large. This one was about 500 times the radius of our Sun.

So what happened? A shock wave rushed upward through the star. First it broke through the star’s surface in the form of finger-like plasma jets, which you can see in the animation.

20 minutes later, the full fury of the shock wave reached the surface—and the doomed star exploded in a blinding flash! This is called the shock breakout.

Then the star expanded as a blue-hot ball of plasma.

Here’s how the star’s luminosity changed with time, measured in multiples of the Sun’s luminosity:

Note that while the shock breakout seems very bright, it’s ultimately dwarfed by the luminosity of the expanding ball of plasma. So, KSN2011d was actually one of the first two supernovae for which the shock breakout was seen! For details, read this:

• P. M. Garnavich, B. E. Tucker, A. Rest, E. J. Shaya, R. P. Olling, D. Kasen and A. Villar, Shock breakout and early light curves of Type II-P supernovae observed with Kepler.

A Type II supernova is one that shows hydrogen in its spectral lines: these are commonly formed by the collapse of a star that has run out of fuel in its core, but retains hydrogen in its outer layers. A Type II-P is one that shows a plateau in its light curve: the P is for ‘plateau’. These are more common than the Type II-L, which show a more rapid (‘linear’) decay in their luminosity:

### 5 Responses to Shock Breakout

1. arch1 says:

Thanks, fascinating. Parts of that supernova animation reminded me of the next-to-last image here (though of course a supernova is ridiculously, outrageously more powerful than a mere nuclear bomb, as memorably described in this xkcd what if?)

2. Patrice Ayme says:

Science is not just a knowledge of facts, but a knowledge of beauty which would not have been otherwise revealed. Science is also the uncovering of logic of previously unsuspected subtlety, for all to see. Thus subjects as esoteric as how exactly supernovae explode can reveal how explanations can go about things.

Beauty itself, is partly a matter of logic: when god touches humanity with its finger in a famous painting, the beauty depicts a logic, and it is this uncovering of a logic what makes it, in part, beautiful… and certainly “interesting” as all great art is.

Knowing about supernovae is not just knowing how the chemistry which made Earth possible was created. It is not just about knowing the size of the universe, and how fast it is changing. It is also knowing about analogies, metaphors, logics and possibilities we never suspected, and also about our naivety, to never have suspected they were.

All of this to say it’s too bad the preceding post was not equipped with a “reblog” button. I had started an essay on the NASA cartoon above, but then this post was published, and it would have been easier for me to reblog it than to rehash known astrophysics. Rebloging gives full attribution, and, after a short extract, links back to the original post.

3. domenico says:

I am thinking that we see only a direction of supernova explosion, so that the theory cannot model the elementary interaction of the process; but if a similar process, like the inertial confinement fusion (in a lower energy without convective overturn), with different chemical elements and energies can be used to explore the explosion in many direction (with different spectroscopy and sensors, like a multiple particles collider), so that the correction to the Standard Model can be tested inside a controlled high density environment.

• John Baez says:

Nice! He estimates the distance from which the neutrinos released by a supernova would be fatal:

Karam calculates that the neutrino radiation dose at a distance of one parsec [6] would be around half a nanosievert, or 1/500th the dose from eating a banana. [7]

A fatal radiation dose is about 4 sieverts. Using the inverse-square law, we can calculate the radiation dose:

$\displaystyle{ 0.5 \textrm{ nanosieverts } \times \left(\frac{1 \textrm{ parsec }}{x}\right)^2 =5 \textrm{ sieverts }}$

$x=0.00001118 \textrm{ parsecs } = 2.3 \textrm{ AU }$

2.3 AU is a little more than the distance between the Sun and Mars.

The only mistake is to give two significant figures in the final answer; I’m sure it’s not known that accurately.

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text:
# Intro

My coding journey isn’t limited to these, I also have quite a few which aren’t (yet) publicly available for various reasons, mainly for being crappy or not yet in an acceptable state.

Those listed however, are the ones I am truly proud of.

# Wrap

Wrap is an open source command line tool that is able to convert Fountain and Wrap files into a correctly formatted screen- or stageplay as an HTML or a PDF.

Fountain is an awesome format to write screenplays in English, Wrap builds upon this to add support for different languages and stageplays whilst still fully supporting normal Fountain files.

All this in a FOSS package that you can use straight from the commandline.

# tldr

A tldr client in Go, which focusses on speed. It uses a database and, using Go, is able to run on all targets supported by Go.

# easyformat

easyformat is a LaTeX package that allows the use of _ (the underscore) to add italics and bold text and ^ (the circumflex) to add smallcaps.

## Usage:

Import the package with: \usepackage{easyformat}

Now you can do the following in textmode: _italics_, __bold__, ___bolditalic___ and ^smallcaps^.

The underscore’s (_) and circumflex’s (^) meaning in mathmode stay the same.

# cLC

cLC (commandline Lambda Calculator) is a commandline application to experiment with and learn about lambda calculus.

format: plain


text:
# Math Help - Choosing Random Integer

1. ## Choosing Random Integer

Each of four students chooses at random one integer from 1 to 10 inclusive. What is the probability that at least 2 of them choose the same integer?

Do you understand that there are $10^4$ ways to choose the integers?
That number is $10\cdot 9\cdot 8\cdot 7$. WHY?


text:
# How to Interpret the Euler Characteristic of Complex Algebraic Varieties

Let $X$ be a projective (or affine) variety over $\mathbb{C}$ defined by some homogenous ideal $I = (f_1,\ldots,f_n)$. How can we interpret the Euler characteristic of $X$ other than as just an invariant to distinguish non isomorphic objects? What I mean is what could it tell us if anything about either $X$ or the polynomials defining $I$ if $\chi(X) > 0$? How about if $\chi(X) = 0$ or $\chi(X) < 0$? How about in the case of $X$ being a hypersurface cut out by the single (homogeneous) polynomial $f$. What can we deduce about $f$ if $\chi(X) > 0, \chi(X) = 0,$ or $\chi(X) < 0$. I've been trying to relate conditions on the Euler characteristic to anything explicit about the polynomials defining $X$ but haven't had any success so anything that could be said would be interesting and helpful.

• The Euler characteristic is only a topological invariant, so it doesn't seem to tell too much of the scheme structure of $X.$ For a hypersurface, it only tells info about $f$ up to its degree: if the projective hypersurfaces defined by $f=0$ and $g=0$ are both smooth, and $\deg f=\deg g,$ then they are diffeomorphic and hence have the same Euler char. Even the degree of $f$ cannot in general be determined: think of a projective line and a conic in a plane. – shenghao Sep 8 '11 at 10:33
• While this doesn't answer your question about projective varieties, it is a notable fact that if a complex scheme $X$ is decomposed as a disjoint union of other schemes, then $\chi$ is additive. (Not true for arbitrary decompositions of topological spaces!) So you can think of $\chi$ as a characteristic zero analogue of "counting points over ${\mathbb F}_p$". – Allen Knutson Sep 8 '11 at 13:35

Hi Dori! I think that you will find Paolo Aluffi's paper "Computing characteristic classes of projective schemes" useful, if not for the exact formulas, at least for the algorithm.

Among other things it proves the version of the formula in J.C. Ottem's answer for singular hypersurfaces. If $f\in \mathbb C[x_0,\dots,x_n]$ is a non-constant homogeneous polynomial, let $X=\mathcal Z(f)$ and $\Sigma = \mathcal Z_{\mathbb P^n}(\partial_0 f,\dots,\partial_n f)$. If we denote by $g_0,g_1,\dots,g_{n-1}$ the degrees of the gradient morphism $$\mathbb P^n\backslash \Sigma\to \mathbb P^n, x=(x_0,\dots,x_n)\to (\partial_0f(x),\dots,\partial_nf(x))$$ then we have $$\chi(X)=n+\sum g_i (-1)^{n+1-i}.$$

As far as restricting to graph hypersurfaces, I would say look at the recent papers of Aluffi and Marcolli but you know more than me about these matters so... :)

Added: A toy application of this is in the case of graphs $\Gamma_n$ which have two vertices and $n$ parallel edges, so called banana graphs. the formula above gives $$\chi(X_{\Gamma_n})=n+(-1)^n$$ so the Euler characteristic determines the graph. The motivic generalization is done in "Feynman motives of banana graphs". The motivic viewpoint also shows that one shouldn't expect a nice relation to hold for all graphs, since graph hypersurfaces are generators for the Grothendieck ring (details and actual statement are here).

• Thanks Gjergji! That's exactly the type of result I was looking for. I hadn't seen that paper or that result in any of their recent papers on graph hypersurfaces. – Dori Bejleri Sep 9 '11 at 7:57

This is unlikely unless $X$ is say, a hypersurface. The reason is that the Euler characteristic is invariant of the projective embedding (it is a topological invariant), while the equations defining a variety are certainly not. For example, it is known that any projective variety can be defined by determinental quadratic equations in some large projective space.

EDIT: Now, if $X\subset \mathbb{P}^n$ is a smooth hypersurface of degree $d$, then we have $$\chi(X)=((1-d)^{n+1}-1)d^{-1}+n+1.$$Hence when $d\ge 3$ or $n$ is odd, one finds that knowing the Euler characteristic would at least give you the degree of the polynomial (the ambiguous cases $d=1,2$ with $n$ even, are easy to distinguish). More generally, when $X$ is a complete intersection, I think there is also some explicit polynomial expressing $\chi(X)$ in terms of the degrees of the generators, but I doubt you can extract any useful information from this.

• Hi John, I think you forgot to divide by $d$ the first part of the formula for the Euler characteristic. – M P Sep 8 '11 at 10:33
• Also, note that you can reconstruct the degree of $X$ if you know its dimension, with the ambiguity that odd dimensional quadrics have the same cohomology groups (not ring!) as projective space. I think that, for fixed dimension, this is the only ambiguity. – M P Sep 8 '11 at 10:41
• Yes, you are right (and it is indeed the only ambiguous case). – J.C. Ottem Sep 8 '11 at 11:07
• This is quite well-known, and goes back to Mumfords "Varieties defined by quadratic equations", Questions on algebraic varieties, C.I.M.E. Varenna, 1969 , Cremonese (1970) pp. 29–100. To see why, just take any embedding $X\to \mathbb{P}^n$ and take sufficiently many veronese remembeddings of $\mathbb{P}^n$ into some large projective space $\mathbb{P}^N$. In this $\mathbb{P}^N$ the ideal defining $X$ will be defined by linear polynomials (coming from the original embedding) and quadratic binomials defining the veronese variety. – J.C. Ottem Sep 8 '11 at 17:30
• Thanks! Something like that formula is actually extremely useful for me. The hypersurfaces I'm working with are cut out by polynomials that have a very specific form, where each variable appears at most once in every monomial (specifically the Kerchoff polynomial of a graph) so knowing the degree from χ will still be very informative. Is there any generalization of this formula to singular hypersurfaces? Also @MP does your comment refer to using the expression for $\Chi$ that J. C. Ottem gave to deduce the degree or some other technique and if so what? – Dori Bejleri Sep 8 '11 at 18:09


text:
# A triangle has corners at (3 , 2 ), (6 ,7 ), and (5 ,8 ). What is the radius of the triangle's inscribed circle?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Feb 19, 2018

The radius of the circle is 0.581

#### Explanation:

We determine the coordinates of the center of inscribed circle by investigation of the intersection of the angular bisectors.

Let,$A \equiv \left(3 , 2\right) , B \equiv \left(6 , 7\right) , \mathmr{and} C \equiv \left(5 , 8\right)$represent the vertices of the triangle ABC

Angular bisector atA:
A-=(3,2) B-=(6,7) C-=(5,8)

Slope of line AB is ${m}_{A B} = \frac{7 - 2}{6 - 3} = \frac{5}{3}$

Slope of line AB is ${m}_{A C} = \frac{8 - 2}{5 - 3} = \frac{6}{2} = 3$

Let the slope of the bisector be m

$\frac{m - \frac{5}{3}}{1 + m \times \frac{5}{3}} = \frac{3 - m}{1 + 3 \times m}$

$= \frac{3 m - 5}{3 + 5 m} = \frac{3 - m}{1 + 3 m}$

Cross multiplying

$\left(1 + 3 m\right) \left(3 m - 5\right) = \left(3 - m\right) \left(3 + 5 m\right)$

$3 m - 5 + 9 {m}^{2} - 15 m = 9 - 3 m + 15 m - 5 {m}^{2}$

$9 {m}^{2} + 5 {m}^{2} + 3 m - 15 m + 3 m - 15 m - 5 - 9 = 0$

$14 {m}^{2} - 24 m - 14 = 0$

Dividing by 14

${m}^{2} - \frac{12}{7} m - 1 = 0$

$m = 2.17 , m = - 0.46$

Equation of the bisector through A(3,2) is

$\frac{y - 2}{x - 3} = 2.17$

Simplifying

$y - 2 = 2.17 \left(x - 3\right)$

$y - 2 = 2.17 x - 6.51$

$y = 2.17 x - 4.51$

Angular bisector atB:
A-=(3,2)
B-=(6,7)
C-=(5,8)

Slope of line AB is ${m}_{A B} = \frac{7 - 2}{6 - 3} = \frac{5}{3}$

Slope of line BC is ${m}_{B C} = \frac{8 - 7}{5 - 6} = \frac{1}{-} 1 = - 1$

Let the slope of the bisector be m

$\frac{m - \frac{5}{3}}{1 + m \times \frac{5}{3}} = \frac{- 1 - m}{1 + \left(- 1\right) \times m}$

$= \frac{3 m - 5}{3 + 5 m} = \frac{- 1 - m}{1 - m}$

Cross multiplying

$\left(1 - m\right) \left(3 m - 5\right) = \left(- 1 - m\right) \left(3 + 5 m\right)$

$3 m - 5 - 3 {m}^{2} + 5 m = - 3 - 5 m - 3 m - 5 {m}^{2}$

$5 {m}^{2} - 3 {m}^{2} + 5 m + 3 m + 5 m + 3 m + 3 - 5 = 0$

$2 {m}^{2} + 16 m - 2 = 0$

Dividing by 2

${m}^{2} + 8 m - 1 = 0$

$m = 0.12 , m = - 8.12$

Equation of the bisector through B(6,7) is

$\frac{y - 7}{x - 6} = 0.12$

Simplifying

$y - 7 = 0.12 \left(x - 6\right)$

$y - 7 = 0.12 x - 0.75$

$y = 0.12 x + 6.25$

The lines $y = 2.17 x - 4.51$ and $y = 0.12 x + 6.25$ intersect at the center of the incircle

Equating rhs

$2.17 x - 4.51 = 0.12 x + 6.25$

$2.17 x - 0.12 x = 6.25 + 4.51$

$2.05 x = 10.76$

$x = \frac{10.76}{2.05}$

$x = 5.25$

$y = 2.17 x - 4.51$

$y = 2.17 \times 5.25 - 4.51$

$y = 11.39 - 4.51$

$y = 6.88$

$y = 0.12 x + 6.25$

$y = 0.12 \times 5.25 + 6.25 = 0.63 + 6.25$

$y = 6.88$

Verified

The coordinates of the centre of incircle is

$O \equiv \left(5.25 , 6.88\right)$

Slope of line AB

${m}_{A B} = \frac{5}{3}$

Point A is $A \equiv \left(3 , 2\right)$

Equation of the line AB is

$\frac{y - 2}{x - 3} = \frac{5}{3}$

$3 \left(y - 2\right) = 5 \left(x - 3\right)$

$3 y - 6 = 5 x - 15$

$5 x - 3 y + 6 - 15 = 0$

$5 x - 3 y - 9 = 0$

Center is $O \equiv \left(5.25 , 6.88\right)$

Tangent is $5 x - 3 y - 9 = 0$

The distance from O to the line AB is

given by

$| \frac{5 \times 5.25 - 3 \times 6.88 - 9}{\sqrt{{5}^{2} + {\left(- 3\right)}^{2}}} | = | \frac{- 3.39}{5.83} | = 0.581$

The radius of the circle is 0.581

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text:
# How to ListContourPlot a functional eigenvalue spectrum?

We have a (Hermitian) matrix of some finite dimension (n) whose elements are functions of real variables x and y. We wish to plot contours of the spectrum (level sets of the n real eigenvalues) at chosen values.

Here's a simple solution using ContourPlot:

matrix = Table[3 i x + j y, {i, 5}, {j, 5}];
ContourPlot[Sort@Re@Eigenvalues[matrix], {x, -2, 2}, {y, -2, 2},Contours->{0.3, 1.7}] /. _Polygon -> Sequence[]

To be honest, I am not quite sure why this does work--although it does--as the syntax does not appear to follow the documentation for ContourPlot precisely. Here, the argument of ContourPlot is an 5-dimensional list of eigenvalues, presumably sampled at some point in the x-y domain. Using PlotPoints seems to increase the grain size in the domain mesh.

Also, obviously in this example, the eigenvalues have analytic closed-form expressions as functions of x and y, but in my general problem, that is not possible. However, this ContourPlot method has worked in the past nonetheless.

However, I wish to discretize this process manually using ListContourPlot. I hope to gain the ability to manually control the domain and mesh size and also make the contour plotting significantly faster (the ContourPlot version above is quite slow in the real problem).

I am really struggling with the syntax of ListContourPlot, though. Note that the matrix we examine is given. I need to

1. Construct a domain (presumably a List of Lists like {0.234,-0.412}, etc., a list of {x,y} points in the domain).
2. Numerically evaluate the Eigenvalues of the matrix at every point in the domain, and store these results (n-dimensional lists) somehow.
3. Compile the many eigenvalue lists (one at every site in the discrete domain) into a complex array which fits the syntax for ListContourPlot
4. Perform the ListContourPlot and isolated chosen level sets using the Contours option.

The start of a weak working example

matrix = Table[3 i x + j y, {i, 5}, {j, 5}];
domain = Table[{x, y}, {x, -1, 1, 0.5}, {y, -1, 1, 0.5}];
II = Dimensions[domain][[1]];
JJ = Dimensions[domain][[2]];
array = domain;
Do[array[[i, j]] = {array[[i, j, 1]], array[[i, j, 2]],
Sort@Re@Eigenvalues[
ReplaceAll[
matrix, {x -> domain[[i, j, 1]],
y -> domain[[i, j, 2]]}]]}, {i, 1, II}, {j, 1, JJ}]

This is my best effort so far. However, I'm certain that someone can help me find much more elegant solutions to many of these steps.

1. I build a domain using Table, but this could definitely be improved. For example, suppose I wished to use all points (x,y) on a square lattice with some density which lie within a Polygon instead of a simple rectangle?

2. This point is at least accomplished in array, which contains a matrix of lists {x,y,{eigenvalues}}

3. However, this part is a huge failure. The syntax is not commensurable with ListContourPlot in the slightest, so we've really accomplished little.

• does a contour represent the locus where "any" eigenvalue is equal to some specified value? As noted your first example is not a documented usage and its not really clear what its doing. – george2079 Mar 28 '16 at 23:24
• Yes, the contours should represent level sets of any eigenvalue in the spectrum. So the contour at value 1.35 should represent the set (disjoint union of 1-dimensional closed curves) of all points in the x,y plane where there is an eigenvalue with value 1.35. – Steve Mar 29 '16 at 21:08

here is one approach: create a separate plot for each contour and combine.

matrix = Table[3 i x + j y, {i, 5}, {j, 5}];
domain = Table[{x, y}, {x, -2, 2, 0.05}, {y, -2, 2, 0.05}];
cp[lambda_, color_] := ListContourPlot[Flatten[Map[ Append[#,
Quiet[First@MinimalBy[
Cases[
Chop@Eigenvalues[
matrix /. { x -> #[[1]] , y -> #[[2]]  } ], _Real]
- lambda ,
Abs, 1]]] & , domain, {2}], 1], Contours -> {0},
ContourShading -> None, ContourStyle -> color]
contours = Range[-5, 5];
cf = ColorData["DarkRainbow"];
Show[Table[
cp[contours[[i]], cf[i/Length@contours]], {i, Length@contours}]]

note this is based on the real eigenvalues, not the real parts of all the eigenvalues. You can just stick a Re in there if that's what you really want.

for reference here is the ContourPlot version with the same contours. (this takes a long time with PlotPoints->100)

matrix = Table[3 i x + j y, {i, 5}, {j, 5}];
ContourPlot[
Sort@Cases[Chop@Eigenvalues[matrix], _Real], {x, -2, 2}, {y, -2, 2},
Contours -> Range[-5, 5], PlotPoints -> 100] /. _Polygon ->
Sequence[]

obviously similar. Not sure why my version has that extra set of bands.

• Since the OP claims his matrix is Hermitian, the eigenvalues ought to be always real. – J. M. will be back soon Mar 29 '16 at 14:20
• @J.M. not so for this example. Actually those spurious bands are in the region where there are complex eigenvalues. I guess not worth worrying about since its a toy example. – george2079 Mar 29 '16 at 14:58
• Unfortunately, your discrete version above gives errors (despite the Quiet) and does not work for me. Your continuous code below works, (but is quite slow, as you mentioned). edit: This is because MinimalBy was added in mma version 10, but I am using only version 9.0 – Steve Mar 29 '16 at 21:26
• In the real problem, all of the matrices I care about are indeed Hermitian. I should have given a Hermitian toy example, sorry. However, numerical machine error will often give eigenvalues with imaginary parts on the order of 10^(-17), etc., so I remove these confusions typically by just using a Re. – Steve Mar 29 '16 at 21:27


text:
When do MAP inference and full Bayesian Inference give the same solution and why?

I'm struggling to grasp some concepts regarding bayesian learning theory. As I understand it one can classify a dataset by finding a posterior distribution and simply selecting a $w$ in $f(x) = w^TX$ that maximises the density of the posterior distribution. I believe that I read somewhere that this gives the same solution as Full Bayesian inference in the case of Gaussian Process regression but not necessarily in general. Why is this the case? Why might this relation not hold true except in the case of GP regression? Thanks.

• What do you mean by "full Bayesian inference"? I guess you mean estimation the posterior distribution, but then MAP = mode of the posterior distribution, so mode never can't be the same as the distribution unless you have a degenerate distribution consisting of a single point...
– Tim
Jan 5 '18 at 21:48
• Hi, this quora answer explains what i mean: quora.com/… Jan 5 '18 at 22:13


text:
# Correlating Newtonian Model with Einstein's GR

Discussion in 'Alternative Theories' started by hansda, May 8, 2017.

1. ### NotEinsteinValued Senior Member

Messages:
1,431
So you refuse to give any here?
I guess I'll deal with them when I read your remaining two texts.

In other words, you are violating basic principes of Newtonian physics. Great, glad we got that cleared up.

So you agree with me that you are using a definition of the word "force" that is incompatible with its usage in physics. Great, glad we got that cleared up.

Perhaps you should make this more clear in your first two texts I read, because your definition isn't given there.

Yes, the universe appears to be very much non-Newtonian, and thus any Newtonian model will have great difficulty explaining it.

(I strongly suspect you have just introduced a preferred reference frame by claiming the universal expansion can be considered a "space-force", and that you are violating the light-speed limit. But I guess we'll go into that in more details later.)

Please explain how the title of a non-peer-reviewed paper you found has bearing on whether QFT is compatible with GR?

Except that the Newtonian Model is incompatible with the GR model, and thus at most only one of them can be true. Multiple solutions are fine (Newtonian mechanics and Lagrangian mechanics are a perfect example!), but incompatible solutions cannot coexist.

And that universe is not compatible with Newtonian physics, and thus your "Instantaneous Law of Inertia" can be rejected on that basis.

Again, please stop dodging questions. Demonstrate that your Law of Inertia can explain them. Show calculations, evidence. Just you saying "it can, trust me" isn't worth anything in science.

3. ### SchmelzerValued Senior Member

Messages:
4,067
There is. Essentially it is even quite easy - use a lattice regularization. The same what you do when you want to solve some equations on the computer - approximate the continuous field by a function on a lattice. The relativists are aware of this possibility, and use them even for computations. The only problem: They don't like it, because there is no such animal as Lorentz invariance on the lattice.
Of course, classical theory is an approximation of quantum theory, and if you want high enough accuracy, the classical approximation will fail. Less applicable, yes. But incomplete? No. The classical approximation also predicts something of the same type. Only not accurate enough.
Of course, there are. But they are quite useless.
They are. As explained, QG may be about gravity only, without anything about the SM, and TOE may be classical only.
Its simply nonsense.

5. ### NotEinsteinValued Senior Member

Messages:
1,431
Interesting! What's the name of this theory/framework, so I can look it up?

(With "incomplete" I was referring to the fact quantum effects would be absent.)
So a classical field theory that can only describe a subset of what Quantum Field Theory can, and probably with lower accuracy as well. So it cannot be a TOE, as it clearly is not all-encompassing.

(Note: I'm using Wikipedia's definition: "A theory of everything (ToE), final theory, ultimate theory, or master theory is a hypothetical single, all-encompassing, coherent theoretical framework of physics that fully explains and links together all physical aspects of the universe.")

So you need quantum theories to have a TOE.

(Note that there may be non-quantum explanations for the things we currently call "quantum effects", and those theories might be superior to quantum theories. In that case, I would be arguing in favor of those theories over quantum theories, obviously.)

Except it can't be, as you just said yourself: classical theories are not as good as quantum theories, and a TOE needs to be all-encompassing. If I have two possible-TOE's, and one is better than the other at modeling physical reality, the other is not a TOE.

7. ### SchmelzerValued Senior Member

Messages:
4,067
Simply lattice theory. Or lattice regularization. Problems which may appear in this approach you find with "chiral lattice theory" and "fermion doubling".
No, there is no such subset. In fact, the classical theory describes even more - namely, trajectories, instead of probability distributions only. So, what makes quantum theory better is only that it is more accurate.
A classical theory alone would not be viable because falsified by observations where quantum effects are important. But it would be a theory about exactly the same objects as the quantum TOE. So, it would be also a "single, all-encompassing, coherent theoretical framework of physics", and it would be also about "all physical aspects of the universe". It would also explain and link them all. Only incorrectly. But this could happen with a quantum TOE too.
Makes no sense. If General Relativity is better than Newtonian gravity, Newtonian gravity is no longer a theory of gravity? Same for theories of everything. If the two theories are about the same things, that means in this case about everything, they are TOEs, even if one is falsified and useful only as an approximation.

8. ### NotEinsteinValued Senior Member

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1,431
I can find mentions of the successful application of lattice field theory to QCD and gauge theories, but I can't find much information on whether lattice theory is as far as QFT in describing the entire SM.

And what I find funny is that you scoff at using some well-reasoned but seemingly arbitrary cut-off to create effective field theories, but arbitrarily deciding to do everything on some lattice is somehow not dubious.

So quantum effects are described by classical field theories?

I was under the impression that quantum field theories can do that too. Doesn't QFT describe an electron flying through empty space as having all kinds of quantum fluctuations around it, but when "zooming out" it reduces to the classical electron we all know?

If that's the only difference, it's more than enough to throw out classical theories, because obviously they are better.

Agreed.

But missing the quantum effects.

Except it missed the quantum effects, so it's not all-encompassing.

Except for the quantum aspect of it.

Except for quantum effects.

And thus it would make a terrible TOE.

Quantum: could. Classical: does. I prefer to stick with the one not (yet) proved to be incorrect.

It's no longer all-encompassing, because it doesn't have any relativistic effects. But more importantly: if it's wrong, it doesn't "explain all physical aspects of the universe", simply because it doesn't describe our universe.

If a theory is falsified, it cannot be describing our universe, and thus cannot be a TOE.

And before you claim that it's still a TOE because it describes a universe: that would make every theory a TOE, so that's a useless definition of the term TOE.

9. ### SchmelzerValued Senior Member

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4,067
You will find what people do to make computations. They don't consider a lattice as more than a tool for computations.
I don't. QFT considered as effective field theory is fine. You can do it using various regularizations, a lattice approximation is simply one method. I like it because it gives a well-defined theory with a finite number of degrees of freedom, thus, it gives a well-defined theory.
Quantum effects are not objects.
They describe not trajectories, but some wave packets or so. (Except realistic interpretations like dBB, which have trajectories.)
Terrible or not, TOE is TOE. Don't forget, I do not argue that one should use them, I simply clarify the meaning of the phrase TOE.
1.) A theory of gravity should not be all-encompassing, but encompassing gravity. This was not proposed not as a TOE, but as an example for the meaning of phrases like "theory of X". 2.) A theory of gravity should encompass the gravitational field, not whatever "effects".
It cannot describe it correctly but nonetheless describes it. Incorrectly. If one would follow your concept, we would never have a TOE. Because we can, in principle, never prove any physical theory. It has to be falsifiable, so falsification is always a theoretical possibility. So, we could never know if it is really a TOE.
No. Newton and Einstein had a theory of gravity, Maxwell had one of the electromagnetic field, both had no pretense to describe everything. The SM is also not a TOE, because it does not even pretend to describe gravity.

10. ### NotEinsteinValued Senior Member

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That doesn't answer the question. How is cutting off a bad integral to make the answers make sense not the same?

Ah OK, in that case I apologize for misinterpreting the intention of your statement about it in post #193.

Neither are trajectories.
But more importantly: do you really claim that quantum theories do not explain or describe quantum effects?

Edit: Here's where that question is coming from:
Post #203: Me: So a classical field theory that can only describe a subset of what Quantum Field Theory can, ...
Post #204: You: In fact, the classical theory describes even more - ...
Post #205: Me: So quantum effects are described by classical field theories?
Post #206: You: Quantum effects are not objects.

You clearly state that classical field theories can describe more than what QFT can, and insinuate that classical field theories cannot describe quantum effects. Ergo, quantum theories cannot either.

Yes, QFT replaces the (fundamental?) concept of trajectories with propagation. See it as the quantum-version of trajectories. Just like gravitational force disappears in GR. Doesn't mean Newtonian gravity describes something GR can't.

I technically agree with you here. The problem is that classical theories cannot describe quantum effects (by definition), so they can't be TOE's.

Agreed; I never claimed otherwise.

Please explain how "quantum effects" is not part of "everything".

Please demonstrate how Newtonian gravity explains black holes, frame dragging, Mercury's precession, etc. Those are all results of the gravitational field.

An incorrect model does not describe our universe. It at best attempts to describe it, and fails. It is clearly describing some universe incompatible with ours, some other universe. It's thus not a TOE.

Please don't put statements in my mouth. A theory that has not been proved incorrect (yet) is fully compatible with our understanding of the universe, and thus can be a TOE. As soon as some phenomena is found that cannot be described by the theory, it can no longer be a TOE, because it obviously isn't all-encompassing anymore.

I suggest you look up what the word "theory" means in a scientific context.

And I never claim otherwise. I'm just pointing out that if you are allowed to leave physical phenomena out of the "everything" in TOE, all those you mentioned are TOE's.

I fully agree: "all-encompassing".

Last edited: Nov 22, 2017
11. ### SchmelzerValued Senior Member

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I think it makes no sense to continue such word games. Language is in this case quite sloppy, and if one wants to play such word games, one will always find some interpretation of a phrase which is different, leading to yet another iteration. Is a given theory a "theory of X"? Does it describe X? These are not really scientific questions, but sloppy conventions.

I think I have described my position in a sufficiently clear way. A theory of something is such a theory even if it is false. So, NT is a theory of gravity even if it does not correctly describe Mercury precession. And a classical theory of some field is a theory of this field even if it is an incorrect one.

To summarize, let's collect some of the different meanings one can give to the phrase "theory of ...". The one I prefer is that if as the classical theory, as the quantum theory use the same configuration space Q. then they are simply different theories describing the same object, namely configurations $q \in Q$. The classical theory describes the behavior of these objects in a more detailed way, by a trajectory $q(t)$, while quantum theory gives only probability distributions $\rho(q)$. Thus, one can also say that the classical theory describes more. On the other hand, one can also talk about a theory describing "quantum effects". This is simply sloppy language, I see no precise way to distinguish what is a "quantum effect".

EOD

12. ### NotEinsteinValued Senior Member

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It's indeed a theory of gravity, but not a theory of our gravity. You can always postulate a universe in which any (coherent, etc.) theory is the correct one.

Agreed.

Except that the classical theory doesn't allow for all configurations $q$; any configuration that needs quantum effect will be excluded. Take the photoelectric effect. Shoot a light beam with too low a frequency at a material. The configuration where no electron is freed (which is what happens in reality) is not allowed as a result by the classical theory.

I agree it's hard to define it precisely, but it's clear that they exist. There are certain results of experiments that cannot be explained through classical field theories.

Yes, that one wasn't particularly nice of me. I never made any claims about falsifiable (in fact, I agree with you), it's the word "everything" and its contextual usage that's important. Which is obvious from my post #207: "because it obviously isn't all-encompassing anymore." to which you were responding. I saw your (accidental?) straw-man, and decided to return the favor. I apologize; I should not have done that.

13. ### SchmelzerValued Senior Member

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What configurations give valid solutions and which don't give them is already the content of the particular theory. The classical theory allows a continuous spectrum of energy, quantum theory gives a discrete one, on the other hand, quantum theory gives some probability of observing a particle where classical theory does not allow it because of insufficient energy (tunneling). So there is not even a clear this theory allows more configurations than that. What is the same is the configuration space.
Obviously, else we would prefer the mathematically and conceptually much easier classical theories.

14. ### NotEinsteinValued Senior Member

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You are contradicting your own statement from post #204: "In fact, the classical theory describes even more - namely, trajectories, instead of probability distributions only."
And since reality seems to demand that configurations with quantum effects must be allowed (because they are measurably real), classical field theories cannot be TOE's, because they cannot cover all configurations (or possible transitions between them) that reality exhibits.

Indeed, obviously. So does that mean you are now withdrawing your above mentioned statement from post #204?

15. ### SchmelzerValued Senior Member

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I see no reason to withdraw anything, that classical theory describes trajectories but quantum theory only probability distributions is a fact. And that a trajectory contains more information than a probability distribution is also a fact. Of course, in nitpicking modus one can try to find some contradictions between the different specifications made. That's easy because all this talk about "more" is sloppy speech, nothing serious. The most rigorous thing one can say is that both have the same configuration space, and are, therefore, about the same objects.
Again, what reality demands is irrelevant once the question is what the theory is about. A false theory about gremlins is also a theory about gremlins.

16. ### NotEinsteinValued Senior Member

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And seeing that the probability distribution reduces to the trajectory predicted by the classical theory in the classical limit, quantum theory thus described trajectories too.

You've got that the wrong way around. I can make a probability distribution out of every single trajectory, but it cannot be done the other way around. Probability distributions are more general than trajectories, so a theory describing the former is (assuming it gets the numbers right) more powerful.

Except that, as I have pointed out numerous times now, classical theories cannot describe quantum effect by definition, while quantum theories can. Thus, quantum theories can describe more.

But if one has a theory about all gremlins, but it turns out it leaves out the green gremlins, it's not a theory of all gremlins.

17. ### SchmelzerValued Senior Member

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First of all, with quantum wave packets, you cannot construct trajectories -- the uncertainty relations prevent this. The second error is that you can easily construct probability distribution based on trajectories. This is done in the Liouville equation, which is the classical evolution equation for probability distributions. The third error is that more general, less specific information is less information, and less powerful.
Quantum theory also contains more Beauty, more Inspiration, and so on. In dependence of your feeling of Beauty, and Inspiration, of course. In other words, your "quantum effects" is sloppy speaking, much more sloppy that all the things I have given: The same configuration space, no preference if one considers values which are forbidden (each theory allows solutions with characteristics forbidden in the other), more information given by the exact trajectory in comparison with the probability distribution.
If it simply says "green gremlins do not exist", it is a theory of all gremlins.

18. ### NotEinsteinValued Senior Member

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Please provide evidence for this claim.

It is in fact trivial. Take a parametrization of the trajectory, and then construct a probability distribution where the probability is 1 if the coordinates conform to the parametrization, and zero elsewhere. Done!

Except that because a density distribution can contain all possible trajectories and more, it can contain more information, not less.

Why are you all talking metaphysics? This is irrelevant to the discussion at hand.

But even with the sloppy language, it's clear that quantum field theories can describe more than classical field theories. I've already given the example of the photoelectric effect.

Your terminology is horrible if you are suggesting that "there are no quantum effects" is a description of quantum effects.

But let's put your idea to the test. Show me (or link me to) any classical field theory that explains at least all aspects of the Standard Model of particle physics that QFT does.

Except that the theory doesn't describe green gremlins, and is thus not "all-encompassing".

19. ### SchmelzerValued Senior Member

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LOL, you are serious? What evidence do you need that information about a trajectory, with accuracy $\Delta p \Delta q = 0$, is more accurate, thus, contains more information than $\Delta p \Delta q \sim \hbar$?
No. You can get also, easily, a density distribution which contains all possible trajectories. This is very simple, start with an arbitrary density $\rho(p,q, t_0)$, and solving the Liouville equation gives you all the information how it changes in time, $\rho(p,q,t)$ for all times.

The funny idea that a probability distribution contains more information than the exact knowledge of the position and the momentum made my day.
Sorry, but this is not more what QT predicts, but simply something different. And in agreement with observation, while the classical solution, which gives much more information, is empirically falsified.
It is the language typical for mathematicians. They learn from the start that the most trivial examples, usually associated with numbers 0 or 1, have not to be excludes, but handled by exactly the same rules. This gives you, for example, zero-dimensional vector spaces (simply a single point). And a lot of other things. Almost every imaginable mathematical structure can be assigned to a single point. For example, make it a group, by defining the product $1 \cdot 1 = 1$ and so on. I will not change the usual terminology because you find it horrible.
Not interested, I see no reason why I should present you something which "explains" (a quite uncertain notion) all "aspects" (which is already completely arbitrary).

The classical version of the SM is very simple to obtain, take any textbook, copypaste the Lagrangian of the SM. This will be the Lagrangian of the classical SM. The configuration space of the classical SM is the same as that of the QFT. How it explains the beauty, neatness, vulgarity, naturalness and other "aspects" of the QFT I don't know and don't care.

Once they don't exist, there is no need to encompass them. But, ok, then no TOE is possible in principle if it does not describe green gremlins.

20. ### hansdaValued Senior Member

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Seems not very correct. https://en.wikipedia.org/wiki/Theory_of_everything . This wiki article essentially says that Quantum Gravity(QG) is TOE.

Why? Why you think so?

21. ### NotEinsteinValued Senior Member

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That's not an accuracy, that's the Heisenberg uncertainty. The accuracy of a description is how well it matches reality, not what the value of some uncertainty is.

Incorrect. The density distribution will contain more information, namely, the exact spread of the wavepacket. That is information that's not present in the classical case.

You say "no", but then go on to prove my point?

Except that apparently you don't understand what the Heisenberg uncertainty is. There is no "exact knowledge of the position and the momentum".

False, and demonstrated multiple times now.

How can a theory that is missing things have more information? Better yet: in the classical limit, quantum theories will reduce to the classical theories. So, quantum theories must at least contain as much information as classical theories. And because they describe more things, they contain strictly more information.

We are not talking mathematics here. Please stop generalizing specific mathematical terminology to general scientific terminology.

Yes, mathematics can describe points. But it doesn't not describe points, so your analog breaks down right out of the gate.

That is completely irrelevant to what we were talking about.

If you were using the "usual terminology" as used in science in general, not mathematics specifically, we wouldn't be having this discussion.

So you admit that your statements were baseless assertions with no evidence to back them up? OK.

Turns out that's kinda not how things work: https://physics.stackexchange.com/a/67937 "..., so the classical theory of muons is a theory with no muons in it." Oops?

Except you've removed everything to do with muons.

Why do you keep bringing those up? I am not talking about these; stop building straw-men.

Why are you talking about "aspects of QFT"? I am not talking about that.

But green gremlins (quantum effects) do exist, so they do need to be encompassed. That's been my entire point all this time. A theory is not a TOE if it does not encompass quantum effects.

22. ### hansdaValued Senior Member

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All these four unknown forces does not follow F=ma. The links are available. Anybody interested can read from there.

Thanks. Earlier the better. It will be good for discussion.

Not violating but advancing Newtonian Physics.

This redefinition of force is based on my Instantaneous Law of Inertia.

I thought it will be understood.

Dont worry about difficulty. There is a possibility. http://hyperphysics.phy-astr.gsu.edu/hbase/Astro/expuni.html

I believe you are making this statement without reading my texts on space-force.

Let us not discuss about others paper. If there is fault, you observed; may be it is faulty.

You can read the OP. I made an attempt to correlate Newtonian Model with GR.
There is also a Newtonian Model for expanding universe. I gave the link earlier. So, you may be wrong here.

You read the four unknown forces. Compressive force can explain time dilation. This force can be tested also. The way time-dilation is tested, this compressive force also can be tested with a force-meter.

Last edited: Nov 24, 2017
23. ### NotEinsteinValued Senior Member

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1,431
Then I guess we'll get to them later.

In the mean time, feel free to address any of the issues I've already brought up.

Yes, as I said: you are violating some of the basic definitions of Newtonian mechanics. At most you can claim you've made a more generalized version.

Yes, in direct conflict with how Newtonian mechanics uses the term.

Clearly, it is not. If you use terms like "force" in a scientific context, but you are actually using a different definition than the standard one, it would be wise for you to explicitly mention that. Otherwise confusion ensues.

That link mentions that this is a first approximation, and cannot work as-is because of the acceleration of the universal expansion. In other words: the link itself states this model cannot explain our current observations. As I said: "Newtonian model will have great difficulty explaining it."

True, but can you promise me that your text either contains a space-force that doesn't need an absolute reference frame, or makes a strong case why an absolute reference frame exists?

Then stop bringing up others papers!

Right: "These two models can be correlated through my theory." I've already brought up all kinds of issues in your first two texts, issues that you refuse (cannot?) address. You have made an attempt, and looking at the current score, you've failed.

Turns out I wasn't wrong, and the link actually states that our universe is incompatible with the Newtonian model it discusses.

No calculations. No evidence. You are doding the question once again. I'll just repeat it verbatim: Demonstrate that your Law of Inertia can explain them. Show calculations, evidence. Just you saying "it can, trust me" isn't worth anything in science.


text:
# Coefficient of determination

Ordinary least squares regression of Okun's law. Since the regression line does not miss any of the points by very much, the R2 of the regression is relatively high.
Comparison of the Theil–Sen estimator (black) and simple linear regression (blue) for a set of points with outliers. Because of the many outliers, neither of the regression lines fits the data well, as measured by the fact that neither gives a very high R2.

In statistics, the coefficient of determination, denoted R2 or r2 and pronounced "R squared", is the proportion of the variance in the dependent variable that is predictable from the independent variable(s).

It is a statistic used in the context of statistical models whose main purpose is either the prediction of future outcomes or the testing of hypotheses, on the basis of other related information. It provides a measure of how well observed outcomes are replicated by the model, based on the proportion of total variation of outcomes explained by the model.[1][2][3]

There are several definitions of R2 that are only sometimes equivalent. One class of such cases includes that of simple linear regression where r2 is used instead of R2. When an intercept is included, then r2 is simply the square of the sample correlation coefficient (i.e., r) between the observed outcomes and the observed predictor values.[4] If additional regressors are included, R2 is the square of the coefficient of multiple correlation. In both such cases, the coefficient of determination ranges from 0 to 1.

There are cases where the computational definition of R2 can yield negative values, depending on the definition used. This can arise when the predictions that are being compared to the corresponding outcomes have not been derived from a model-fitting procedure using those data. Even if a model-fitting procedure has been used, R2 may still be negative, for example when linear regression is conducted without including an intercept, or when a non-linear function is used to fit the data.[5] In cases where negative values arise, the mean of the data provides a better fit to the outcomes than do the fitted function values, according to this particular criterion.[6]

## Definitions

${\displaystyle R^{2}=1-{\frac {\color {blue}{SS_{\text{res}}}}{\color {red}{SS_{\text{tot}}}}}}$
The better the linear regression (on the right) fits the data in comparison to the simple average (on the left graph), the closer the value of ${\displaystyle R^{2}}$ is to 1. The areas of the blue squares represent the squared residuals with respect to the linear regression. The areas of the red squares represent the squared residuals with respect to the average value.

A data set has n values marked y1,...,yn (collectively known as yi or as a vector y = [y1,...,yn]T), each associated with a predicted (or modeled) value f1,...,fn (known as fi, or sometimes ŷi, as a vector f).

Define the residuals as ei = yifi (forming a vector e).

If ${\displaystyle {\bar {y}}}$ is the mean of the observed data:

${\displaystyle {\bar {y}}={\frac {1}{n}}\sum _{i=1}^{n}y_{i}}$

then the variability of the data set can be measured using three sums of squares formulas:

${\displaystyle SS_{\text{tot}}=\sum _{i}(y_{i}-{\bar {y}})^{2},}$
${\displaystyle SS_{\text{reg}}=\sum _{i}(f_{i}-{\bar {y}})^{2},}$
${\displaystyle SS_{\text{res}}=\sum _{i}(y_{i}-f_{i})^{2}=\sum _{i}e_{i}^{2}\,}$

The most general definition of the coefficient of determination is

${\displaystyle R^{2}\equiv 1-{SS_{\rm {res}} \over SS_{\rm {tot}}}\,}$

### Relation to unexplained variance

In a general form, R2 can be seen to be related to the fraction of variance unexplained (FVU), since the second term compares the unexplained variance (variance of the model's errors) with the total variance (of the data):

{\displaystyle {\begin{aligned}R^{2}=1-{\text{FVU}}\end{aligned}}}

### As explained variance

Suppose R2 = 0.49. This implies that 49% of the variability of the dependent variable has been accounted for, and the remaining 51% of the variability is still unaccounted for. In some cases the total sum of squares equals the sum of the two other sums of squares defined above,

${\displaystyle SS_{\text{res}}+SS_{\text{reg}}=SS_{\text{tot}}.\,}$

See Partitioning in the general OLS model for a derivation of this result for one case where the relation holds. When this relation does hold, the above definition of R2 is equivalent to

${\displaystyle R^{2}={\frac {SS_{\text{reg}}}{SS_{\text{tot}}}}={\frac {SS_{\text{reg}}/n}{SS_{\text{tot}}/n}}}$

where n is the number of observations (cases) on the variables.

In this form R2 is expressed as the ratio of the explained variance (variance of the model's predictions, which is SSreg / n) to the total variance (sample variance of the dependent variable, which is SStot / n).

This partition of the sum of squares holds for instance when the model values ƒi have been obtained by linear regression. A milder sufficient condition reads as follows: The model has the form

${\displaystyle f_{i}={\hat {\alpha }}+{\hat {\beta }}q_{i}\,}$

where the qi are arbitrary values that may or may not depend on i or on other free parameters (the common choice qi = xi is just one special case), and the coefficient estimates ${\displaystyle {\hat {\alpha }}}$ and ${\displaystyle {\hat {\beta }}}$ are obtained by minimizing the residual sum of squares.

This set of conditions is an important one and it has a number of implications for the properties of the fitted residuals and the modelled values. In particular, under these conditions:

${\displaystyle {\bar {f}}={\bar {y}}.\,}$

### As squared correlation coefficient

In linear least squares multiple regression with an estimated intercept term, R2 equals the square of the Pearson correlation coefficient between the observed ${\displaystyle y}$ and modeled (predicted) ${\displaystyle f}$ data values of the dependent variable.

In a linear least squares regression with an intercept term and a single explanator, this is also equal to the squared Pearson correlation coefficient of the dependent variable ${\displaystyle y}$ and explanatory variable ${\displaystyle x.}$

It should not be confused with the correlation between two coefficient estimates, defined as

${\displaystyle \rho _{{\hat {\alpha }},{\hat {\beta }}}={\mathrm {cov} ({\hat {\alpha }},{\hat {\beta }}) \over \sigma _{\hat {\alpha }}\sigma _{\hat {\beta }}},}$

where the covariance between two coefficient estimates, as well as their standard deviations, are obtained from the covariance matrix of the coefficient estimates.

Under more general modeling conditions, where the predicted values might be generated from a model different from linear least squares regression, an R2 value can be calculated as the square of the correlation coefficient between the original ${\displaystyle y}$ and modeled ${\displaystyle f}$ data values. In this case, the value is not directly a measure of how good the modeled values are, but rather a measure of how good a predictor might be constructed from the modeled values (by creating a revised predictor of the form α + βƒi).[citation needed] According to Everitt (p. 78),[7] this usage is specifically the definition of the term "coefficient of determination": the square of the correlation between two (general) variables.

## Interpretation

R2 is a statistic that will give some information about the goodness of fit of a model. In regression, the R2 coefficient of determination is a statistical measure of how well the regression predictions approximate the real data points. An R2 of 1 indicates that the regression predictions perfectly fit the data.

Values of R2 outside the range 0 to 1 can occur when the model fits the data worse than a horizontal hyperplane. This would occur when the wrong model was chosen, or nonsensical constraints were applied by mistake. If equation 1 of Kvalseth[8] is used (this is the equation used most often), R2 can be less than zero. If equation 2 of Kvalseth is used, R2 can be greater than one.

In all instances where R2 is used, the predictors are calculated by ordinary least-squares regression: that is, by minimizing SSres. In this case R2 increases as we increase the number of variables in the model (R2 is monotone increasing with the number of variables included—i.e., it will never decrease). This illustrates a drawback to one possible use of R2, where one might keep adding variables (Kitchen sink regression) to increase the R2 value. For example, if one is trying to predict the sales of a model of car from the car's gas mileage, price, and engine power, one can include such irrelevant factors as the first letter of the model's name or the height of the lead engineer designing the car because the R2 will never decrease as variables are added and will probably experience an increase due to chance alone.

This leads to the alternative approach of looking at the adjusted R2. The explanation of this statistic is almost the same as R2 but it penalizes the statistic as extra variables are included in the model. For cases other than fitting by ordinary least squares, the R2 statistic can be calculated as above and may still be a useful measure. If fitting is by weighted least squares or generalized least squares, alternative versions of R2 can be calculated appropriate to those statistical frameworks, while the "raw" R2 may still be useful if it is more easily interpreted. Values for R2 can be calculated for any type of predictive model, which need not have a statistical basis.

### In a non-simple linear model

Consider a linear model with more than a single explanatory variable, of the form

${\displaystyle Y_{i}=\beta _{0}+\sum _{j=1}^{p}\beta _{j}X_{i,j}+\varepsilon _{i},}$

where, for the ith case, ${\displaystyle {Y_{i}}}$ is the response variable, ${\displaystyle X_{i,1},\dots ,X_{i,p}}$ are p regressors, and ${\displaystyle \varepsilon _{i}}$ is a mean zero error term. The quantities ${\displaystyle \beta _{0},\dots ,\beta _{p}}$ are unknown coefficients, whose values are estimated by least squares. The coefficient of determination R2 is a measure of the global fit of the model. Specifically, R2 is an element of [0, 1] and represents the proportion of variability in Yi that may be attributed to some linear combination of the regressors (explanatory variables) in X.[9]

R2 is often interpreted as the proportion of response variation "explained" by the regressors in the model. Thus, R2 = 1 indicates that the fitted model explains all variability in ${\displaystyle y}$, while R2 = 0 indicates no 'linear' relationship (for straight line regression, this means that the straight line model is a constant line (slope = 0, intercept = ${\displaystyle {\bar {y}}}$) between the response variable and regressors). An interior value such as R2 = 0.7 may be interpreted as follows: "Seventy percent of the variance in the response variable can be explained by the explanatory variables. The remaining thirty percent can be attributed to unknown, lurking variables or inherent variability."

A caution that applies to R2, as to other statistical descriptions of correlation and association is that "correlation does not imply causation." In other words, while correlations may sometimes provide valuable clues in uncovering causal relationships among variables, a non-zero estimated correlation between two variables is not, on its own, evidence that changing the value of one variable would result in changes in the values of other variables. For example, the practice of carrying matches (or a lighter) is correlated with incidence of lung cancer, but carrying matches does not cause cancer (in the standard sense of "cause").

In case of a single regressor, fitted by least squares, R2 is the square of the Pearson product-moment correlation coefficient relating the regressor and the response variable. More generally, R2 is the square of the correlation between the constructed predictor and the response variable. With more than one regressor, the R2 can be referred to as the coefficient of multiple determination.

### Inflation of R2

In least squares regression, R2 is weakly increasing with increases in the number of regressors in the model. Because increases in the number of regressors increase the value of R2, R2 alone cannot be used as a meaningful comparison of models with very different numbers of independent variables. For a meaningful comparison between two models, an F-test can be performed on the residual sum of squares, similar to the F-tests in Granger causality, though this is not always appropriate. As a reminder of this, some authors denote R2 by Rq2, where q is the number of columns in X (the number of explanators including the constant).

To demonstrate this property, first recall that the objective of least squares linear regression is

${\displaystyle \min _{b}SS_{\text{res}}(b)\Rightarrow \min _{b}\sum _{i}(y_{i}-X_{i}b)^{2}\,}$

where Xi is a row vector of values of explanatory variables for case i and b is a column vector of coefficients of the respective elements of Xi.

The optimal value of the objective is weakly smaller as more explanatory variables are added and hence additional columns of ${\displaystyle X}$ (the explanatory data matrix whose ith row is Xi) are added, by the fact that less constrained minimization leads to an optimal cost which is weakly smaller than more constrained minimization does. Given the previous conclusion and noting that ${\displaystyle SS_{tot}}$ depends only on y, the non-decreasing property of R2 follows directly from the definition above.

The intuitive reason that using an additional explanatory variable cannot lower the R2 is this: Minimizing ${\displaystyle SS_{\text{res}}}$ is equivalent to maximizing R2. When the extra variable is included, the data always have the option of giving it an estimated coefficient of zero, leaving the predicted values and the R2 unchanged. The only way that the optimization problem will give a non-zero coefficient is if doing so improves the R2.

### Caveats

R2 does not indicate whether:

• the independent variables are a cause of the changes in the dependent variable;
• omitted-variable bias exists;
• the correct regression was used;
• the most appropriate set of independent variables has been chosen;
• there is collinearity present in the data on the explanatory variables;
• the model might be improved by using transformed versions of the existing set of independent variables;
• there are enough data points to make a solid conclusion.

## Extensions

The use of an adjusted R2 (one common notation is ${\displaystyle {\bar {R}}^{2}}$, pronounced "R bar squared"; another is ${\displaystyle R_{\text{adj}}^{2}}$) is an attempt to take account of the phenomenon of the R2 automatically and spuriously increasing when extra explanatory variables are added to the model. It is a modification due to Henri Theil of R2 that adjusts for the number of explanatory terms in a model relative to the number of data points.[10] The adjusted R2 can be negative, and its value will always be less than or equal to that of R2. Unlike R2, the adjusted R2 increases only when the increase in R2 (due to the inclusion of a new explanatory variable) is more than one would expect to see by chance. If a set of explanatory variables with a predetermined hierarchy of importance are introduced into a regression one at a time, with the adjusted R2 computed each time, the level at which adjusted R2 reaches a maximum, and decreases afterward, would be the regression with the ideal combination of having the best fit without excess/unnecessary terms. The adjusted R2 is defined as

${\displaystyle {\bar {R}}^{2}={1-(1-R^{2}){n-1 \over n-p-1}}}$

where p is the total number of explanatory variables in the model (not including the constant term), and n is the sample size.

Adjusted R2 can also be written as

${\displaystyle {\bar {R}}^{2}={1-{SS_{\text{res}}/{\text{df}}_{e} \over SS_{\text{tot}}/{\text{df}}_{t}}}}$

where dft is the degrees of freedom n– 1 of the estimate of the population variance of the dependent variable, and dfe is the degrees of freedom np – 1 of the estimate of the underlying population error variance.

The principle behind the adjusted R2 statistic can be seen by rewriting the ordinary R2 as

${\displaystyle R^{2}={1-{{\textit {VAR}}_{\text{res}} \over {\textit {VAR}}_{\text{tot}}}}}$

where ${\displaystyle {\text{VAR}}_{\text{res}}=SS_{\text{res}}/n}$ and ${\displaystyle {\text{VAR}}_{\text{tot}}=SS_{\text{tot}}/n}$ are the sample variances of the estimated residuals and the dependent variable respectively, which can be seen as biased estimates of the population variances of the errors and of the dependent variable. These estimates are replaced by statistically unbiased versions: ${\displaystyle {\text{VAR}}_{\text{res}}=SS_{\text{res}}/(n-p-1)}$ and ${\displaystyle {\text{VAR}}_{\text{tot}}=SS_{\text{tot}}/(n-1)}$.

Adjusted R2 can be interpreted as an unbiased (or less biased) estimator of the population R2, whereas the observed sample R2 is a positively biased estimate of the population value.[11] Adjusted R2 is more appropriate when evaluating model fit (the variance in the dependent variable accounted for by the independent variables) and in comparing alternative models in the feature selection stage of model building.[11]

### Coefficient of partial determination

The coefficient of partial determination can be defined as the proportion of variation that cannot be explained in a reduced model, but can be explained by the predictors specified in a full(er) model.[12][13][14] This coefficient is used to provide insight into whether or not one or more additional predictors may be useful in a more fully specified regression model.

The calculation for the partial R2 is relatively straightforward after estimating two models and generating the ANOVA tables for them. The calculation for the partial R2 is

${\displaystyle {\frac {SS_{\text{ res, reduced}}-SS_{\text{ res, full}}}{SS_{\text{ res, reduced}}}},}$

which is analogous to the usual coefficient of determination:

${\displaystyle {\frac {SS_{\text{tot}}-SS_{\text{res}}}{SS_{\text{tot}}}}.}$

### Generalized R2

In the case of logistic regression, usually fit by maximum likelihood, there are several choices of pseudo-R2.

One is the generalized R2 originally proposed by Cox & Snell,[15] and independently by Magee:[16]

${\displaystyle R^{2}=1-\left({{\mathcal {L}}(0) \over {\mathcal {L}}({\hat {\theta }})}\right)^{2/n}}$

where ${\displaystyle {\mathcal {L}}(0)}$ is the likelihood of the model with only the intercept, ${\displaystyle {{\mathcal {L}}({\hat {\theta }})}}$ is the likelihood of the estimated model (i.e., the model with a given set of parameter estimates) and n is the sample size. It is easily rewritten to:

${\displaystyle R^{2}=1-e^{{\frac {2}{n}}(ln({\mathcal {L}}(0))-ln({\mathcal {L}}({\hat {\theta }}))}=1-e^{-{\frac {D}{n}}}}$

where D is the test statistic of the likelihood ratio test.

Nagelkerke[17] noted that it had the following properties:

1. It is consistent with the classical coefficient of determination when both can be computed;
2. Its value is maximised by the maximum likelihood estimation of a model;
3. It is asymptotically independent of the sample size;
4. The interpretation is the proportion of the variation explained by the model;
5. The values are between 0 and 1, with 0 denoting that model does not explain any variation and 1 denoting that it perfectly explains the observed variation;
6. It does not have any unit.

However, in the case of a logistic model, where ${\displaystyle {\mathcal {L}}({\hat {\theta }})}$ cannot be greater than 1, R2 is between 0 and ${\displaystyle R_{\max }^{2}=1-({\mathcal {L}}(0))^{2/n}}$: thus, Nagelkerke suggested the possibility to define a scaled R2 as R²/R2max.[18]

## Comparison with norm of residuals

Occasionally, the norm of residuals is used for indicating goodness of fit. This term is calculated as the square-root of the sum of squared residuals:

${\displaystyle {\text{norm of residuals}}={\sqrt {SS_{\text{res}}}}=\|e\|.}$

Both R2 and the norm of residuals have their relative merits. For least squares analysis R2 varies between 0 and 1, with larger numbers indicating better fits and 1 representing a perfect fit. The norm of residuals varies from 0 to infinity with smaller numbers indicating better fits and zero indicating a perfect fit. One advantage and disadvantage of R2 is the ${\displaystyle SS_{\text{tot}}}$ term acts to normalize the value. If the yi values are all multiplied by a constant, the norm of residuals will also change by that constant but R2 will stay the same. As a basic example, for the linear least squares fit to the set of data:

${\displaystyle x=1,\ 2,\ 3,\ 4,\ 5}$
${\displaystyle y=1.9,\ 3.7,\ 5.8,\ 8.0,\ 9.6}$

R2 = 0.998, and norm of residuals = 0.302. If all values of y are multiplied by 1000 (for example, in an SI prefix change), then R2 remains the same, but norm of residuals = 302.

Other single parameter indicators of fit include the standard deviation of the residuals, or the RMSE of the residuals. These would have values of 0.151 and 0.174 respectively for the above example given that the fit was linear with an unforced intercept.[19]

## History

The creation of the coefficient of determination has been attributed to the geneticist Sewall Wright and was first published in 1921.[20]

## Notes

1. ^ Steel, R. G. D.; Torrie, J. H. (1960). Principles and Procedures of Statistics with Special Reference to the Biological Sciences. McGraw Hill.
2. ^ Glantz, Stanton A.; Slinker, B. K. (1990). Primer of Applied Regression and Analysis of Variance. McGraw-Hill. ISBN 978-0-07-023407-9.
3. ^ Draper, N. R.; Smith, H. (1998). Applied Regression Analysis. Wiley-Interscience. ISBN 978-0-471-17082-2.
4. ^ Devore, Jay L. (2011). Probability and Statistics for Engineering and the Sciences (8th ed.). Boston, MA: Cengage Learning. pp. 508–510. ISBN 978-0-538-73352-6.
5. ^ Colin Cameron, A.; Windmeijer, Frank A.G. (1997). "An R-squared measure of goodness of fit for some common nonlinear regression models". Journal of Econometrics. 77 (2): 1790–2. doi:10.1016/S0304-4076(96)01818-0.
7. ^ Everitt, B. S. (2002). Cambridge Dictionary of Statistics (2nd ed.). CUP. ISBN 978-0-521-81099-9.
8. ^ Kvalseth, Tarald O. (1985). "Cautionary Note about R2". The American Statistician. 39 (4): 279–285. doi:10.2307/2683704. JSTOR 2683704.
9. ^ Computing Adjusted R2 for Polynomial Regressions
10. ^ Theil, Henri (1961). Economic Forecasts and Policy. Holland, Amsterdam: North.[page needed]
11. ^ a b Shieh, Gwowen (2008-04-01). "Improved shrinkage estimation of squared multiple correlation coefficient and squared cross-validity coefficient". Organizational Research Methods. 11 (2): 387–407. doi:10.1177/1094428106292901. ISSN 1094-4281.
12. ^ Richard Anderson-Sprecher, "Model Comparisons and R2", The American Statistician, Volume 48, Issue 2, 1994, pp. 113–117.
13. ^ (generalized to Maximum Likelihood) N. J. D. Nagelkerke, "A Note on a General Definition of the Coefficient of Determination", Biometrika, Vol. 78, No. 3. (Sep., 1991), pp. 691–692.
14. ^
15. ^ Cox, D. D.; Snell, E. J. (1989). The Analysis of Binary Data (2nd ed.). Chapman and Hall.
16. ^ Magee, L. (1990). "R2 measures based on Wald and likelihood ratio joint significance tests". The American Statistician. 44. pp. 250–3. doi:10.1080/00031305.1990.10475731.
17. ^ Nagelkerke, Nico J. D. (1992). Maximum Likelihood Estimation of Functional Relationships, Pays-Bas. Lecture Notes in Statistics. 69. ISBN 978-0-387-97721-8.
18. ^ Nagelkerke, N. J. D. (1991). "A Note on a General Definition of the Coefficient of Determination". Biometrika. 78 (3): 691–2. doi:10.1093/biomet/78.3.691. JSTOR 2337038.
19. ^ OriginLab webpage, http://www.originlab.com/doc/Origin-Help/LR-Algorithm. Retrieved February 9, 2016.
20. ^ Wright, Sewell (January 1921). "Correlation and causation". Journal of Agricultural Research. 20: 557–585.


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# Talk:Cofinality

## Incoherent sentence

There is this statement: "Moreover, any cofinal subset of B whose cardinality is equal to the cofinality of B is well-ordered and order isomorphic to its own cardinality." Is this correct? I don't think so... The last part "order isomorphic to its own cardinality." would imply that B is a cardinal?! Anyone? 82.157.131.133 21:21, 30 Mar 2005 (UTC)

Yup, that sentence didn't make sense. I think I have it stated correctly now.--Luke Gustafson 11:05, 22 December 2005 (UTC)

## Countable choice needed?

Countable union of countable sets is countable - this requires at least the Countable Axiom of Choice, as far as I know. Correct?

Correct, I added a mention of that.--Luke Gustafson 10:36, 22 December 2005 (UTC)

## Why is cf(card(R)) uncountable?

How does the countable union of countable sets help in establishing that cf(card(R)) is uncountable? Without CH we don't know that sets smaller than R are countable - so maybe we can get R with a countable union of uncountable sets which still are smaller than R. --SirJective (84.151.224.202) 23:03, 2 January 2006 (UTC)

It follows from König's Theorem that k < cf(2k) for any infinite cardinal k, so in particular the cardinality of the continuum has uncountable cofinality. I don't see how the countable union of countable sets is relevant here, and I suspect this is a mistake based on assuming CH (as I've seen mistakes of this nature on Wikipedia before). --Zundark 14:26, 3 January 2006 (UTC)
Thanks for pointing out the proof to me, I agree with your suspicion - assuming CH clearly rules out card(R) = aleph_omega, which was the question I had when I found this page ;). I edited the article accordingly. --SirJective (129.187.111.179) 13:47, 11 January 2006 (UTC)

## For any infinite well-orderable cardinal number κ?

Isn't every cardinal well-ordered? In particular, isn't 0 a least element of every cardinal? -lethe talk + 12:49, 31 January 2006 (UTC)

Hmm... maybe what is meant is not that cardinals need not be well-founded, but rather that they need not be totally ordered. If we don't assume AC, that is. -lethe talk + 12:51, 31 January 2006 (UTC)
By a "well-orderable cardinal", it means a cardinal k such that any set of cardinality k can be well-ordered. This is true of all cardinals, of course, but the proof requires AC. (As with many articles, this one is a mess as regards AC - sometimes mentioning when it is needed, and at other times silently assuming it.) --Zundark
Yeah, I was thinking of cardinals as some least ordinal, and well ordering of ordinals is built into their definitions. But of course, that definition of cardinals relies on AC. And the proof of well ordering in the general case also requires AC. I knew that stuff, but the reason I was momentarily confused was because well-ordering really means two things: least element, and totalness of ordering. I somehow thought that sentence meant that you need AC to prove every cardinal has a least element, which is certainly not true. But I was just being dumb. Anyway, thanks for the clarification. -lethe talk + 14:30, 31 January 2006 (UTC)

## every limit cardinal is singular

The article says: "one can prove that for a limit ordinal δ ${\displaystyle \mathrm {cf} (\aleph _{\delta })=\mathrm {cf} (\delta )}$". But I have a problem: ℵδ is a limit cardinal iff δ is a limit ordinal. And I think ℵδ is always greater than δ, while cf(δ) ≤ δ. Therefore, this statement implies that no inaccessible cardinal exists. Somethings not right. Or am I mucking it up? -lethe talk + 08:51, 1 February 2006 (UTC)

Hmm.. let's go through that a little more carefully. Let δ be a limit ordinal. Then ℵδ is a limit cardinal. Then, according to the article, cf(ℵδ) = cf(δ). cf(δ) ≤ δ, so cf(ℵδ) ≤ δ. If ℵδ is regular, then we have ℵδ ≤ δ. But I claim that ℵλ is strictly greater than λ, for all λ.
For a successor ordinal λ+1, ℵλ+1 = union {all ords of cardinality ℵλ} ≥ ℵλ > λ (this last by inductive assumption). Since ℵλ+1 ≥ λ, it must also be greater than λ+1, since ℵλ+1 is a limit ordinal, ℵλ+1 > λ ==> ℵλ+1 > λ+1.
Now if λ is a limit ordinal, ℵλ = union {ℵκ: κ<λ} > union {κ: κ<λ} = λ.
OK, I think that might be a problem. I use the fact that if ℵκ > κ for all κ<λ (inductive hypothesis), then I can take the union of both sides and have union {ℵκ: κ<λ} > union {κ: κ<λ}. But that operation probably isn't valid. For instance, n+1 > n for all finite n, but if you take the union, you get ω > ω.
So maybe it's possible to have ℵλ = λ if λ is a limit ordinal? Maybe this equality will hold for the inaccessible cardinals? -lethe talk + 09:38, 1 February 2006 (UTC)
There are many ordinals ${\displaystyle \delta }$ for which ${\displaystyle \delta =\aleph _{\delta }}$. For example, let ${\displaystyle \kappa _{0}}$ be any cardinal number, and for each positive integer ${\displaystyle n}$ let ${\displaystyle \kappa _{n}=\aleph _{\kappa _{n-1}}}$, then let ${\displaystyle \delta }$ be the supremum of all the ${\displaystyle \kappa _{n}}$. --Zundark 09:42, 1 February 2006 (UTC)
Wow. Easy enough to see, but still surprising. I added a section about aleph fixed points to aleph number. Since I'm a newbie to the topic, maybe you can just glance it over? Thanks again for your help. -lethe talk + 10:54, 1 February 2006 (UTC)

### decipherability of unicode aleph symbol

You-all are using a character (in "think ℵδ") which just looks like a square-box to me. And I do not think that my browser is defective. Why? If it is supposed to be an ${\displaystyle \aleph }$, then you should use $\aleph$ to create it. JRSpriggs 12:25, 19 March 2006 (UTC)

It's not that your browser is defective, but rather that your system doesn't have a unicode enabled font with that symbol. As far as using math tags, I never use math tags for inline math text. I'm fighting hard not to revert your changes, I've seen others do the same revert. But I try to remind myself that not everyone has a capable computer. Please do consider getting some unicode fonts though. -lethe talk + 14:06, 19 March 2006 (UTC)
Even if I can find a better unicode font than I have or get a Hebrew language package, that would only solve the problem for ME. Surely there are many other readers out there who only see square boxes when you-all use those characters. JRSpriggs 07:18, 21 March 2006 (UTC)
It has a codepoint of x2135, part of the block of "letterlike symbols". This is a different glyph from the Hebrew letter. But anyway, if you don't have it, you don't have it. I didn't revert you, but I don't agree with your argument. If we insist on everything being readable in every single browser, we'll never be able to move forward. -lethe talk + 08:18, 21 March 2006 (UTC)
For what it's worth, I use IE 6 SP 2, and Firefox 1.5.0.1, and I see the "you don't have this character" box in IE, but I see the aleph-lookalike glyph in Firefox. I hypothesize that it's so much a font thing, since both programs surely access the same pool of fonts, as a browser thing.151.200.185.98 02:32, 22 March 2006 (UTC)

Like the user at 151.200.whatever (who is probably my nephew Edward), I could not read the small alephs in Internet Explorer, but now that I am using Firefox (without any change in fonts), they are clear. JRSpriggs 09:02, 25 March 2006 (UTC)

I guess I was wrong about this being a unicode font issue. I guess the real issue is that Internet Explorer is broken. I don't like making Wikipedia (or the web in general) cater to the bugs of Microsoft, but it's unrealistic to expect to be able to not support IE when it constitutes 90% of the market or whatever. It's unreasonable to tell everyone who wants to view wikipedia to get a different web browser, right? -lethe talk + 09:22, 25 March 2006 (UTC)
Yes, I feel that we must support IE by avoiding the use of those small alephs. Now that I have down-loaded Firefox, I checked that the characters which I replaced were in fact alephs, which they were as I had inferred from the context. JRSpriggs 10:33, 26 March 2006 (UTC)

## I've reorganized the article a bit

I've reorganized the article a bit. I moved the definition of regular and singular into the lead. I added some content to the "Examples" section (please double check my additions), and moved that section to follow immediately after the lead (I think it helps to have examples as soon as possible). And I titled the remaining content "Properties". Comments? Paul August 16:51, 21 February 2006 (UTC)

## New section on cofinality of well-ordered sets

This section was copied with minor changes from the Wikipedia article on ordinal numbers. JRSpriggs 07:48, 18 March 2006 (UTC)

## Error in Properties?

The second sentence, "Moreover, any cofinal subset of B whose cardinality is equal to the cofinality of B is well-ordered, and these sets are all order isomorphic.", in the section "Properties" seems (to me) to be incorrect. Let us take B to be the Integers with the usual order. Their cardinality is aleph-null. And so is the cardinality of the obvious co-final subset which is the natural numbers (which are well-ordered). But one could also choose the cofinal subset to be the integers themselves (after all, B is cofinal in itself, right?), since they also have cardinality aleph-null which is the cofinality. And the integers are not well-ordered, contrary to the sentence. Also the integers are not order-isomorhpic to the natural numbers, again contrary to the sentence. JRSpriggs 12:16, 19 March 2006 (UTC)

OK. I see that I was failing to read the second sentence in the context of the first sentence, "If A admits a totally ordered cofinal subset, then we can find a subset B which is well-ordered and cofinal in A.", which says that that B is well-ordered. So the integers would not be a counter-example, since they are not well-ordered. Still, the second sentence is hard to understand. Could it be expanded to make it clearer? JRSpriggs 07:13, 21 March 2006 (UTC)

## Merge "cofinal" with "cofinality"?

about merging cofinality with cofinal (mathematics), I say absolutely. support. -lethe talk + 20:30, 6 July 2006 (UTC)

I have removed the overlap between cofinality and cofinal (mathematics). Now I am not sure about the merge anymore. I think that enough could be said about the properties of cofinal subsets to justify separate articles. --Tobias Bergemann 20:48, 6 July 2006 (UTC)

I'm a bit of a mergist by heart. So whether or not we could fit them logically into separate articles is not as relevant to me as whether a single merged article would be too long. I think the answer to the latter question is "no", and hence I support the merger. I also note that having separate articles for the noun and the adjective is a bad policy. But your point is taken, when we use the word "cofinality" we are usually talking about only a least cardinality (and only apropos a cardinal number), where as with the word "cofinal" we may talk about any set inside any poset. The latter usage is different. I still say merge though. But let's hear what others think as well. -lethe talk + 21:24, 6 July 2006 (UTC)
If we do not merge, then I think that "Cofinal (mathematics)" should be renamed "Cofinal subset". JRSpriggs 06:47, 7 July 2006 (UTC)
I agree. — Tobias Bergemann 07:45, 7 July 2006 (UTC)
I'm not necessarily against a merge, but if they're merged I feel strongly that the article should be at cofinality, which is the more used concept. --Trovatore 14:57, 7 July 2006 (UTC)

## Merge "regular cardinal" into "cofinality"?

Right now there is a certain asymmetry between the content on regular cardinals and regular ordinals: regular ordinal redirects to cofinality while regular cardinal has its own (very short) article. Maybe regular cardinal should be merged into cofinality. — Tobias Bergemann 11:49, 7 July 2006 (UTC)

No, I think regular cardinal should be expanded, and regular ordinal should redirect there. There's a lot to be said about regular cardinals that goes far beyond their cofinality. --Trovatore 14:58, 7 July 2006 (UTC)

## Is this right?

"This definition of cofinality relies on the axiom of choice, as it uses the fact that every non-empty set of cardinal numbers has a least member." - This is a way to avoid the axiom of choice.

From Axiom of choice:

For certain infinite sets X, it is also possible to avoid the axiom of choice. For example, suppose that the elements of X are sets of natural numbers. Every nonempty set of natural numbers has a least element, so to specify our choice function we can simply say that it takes each set to the least element of that set. This gives us a definite choice of an element from each set and we can write down an explicit expression that tells us what value our choice function takes. Any time it is possible to specify such an explicit choice, the axiom of choice is unnecessary.

—Preceding unsigned comment added by Drysh (talkcontribs)

What is your method of making an "explicit choice" and so avoiding the axiom of choice? JRSpriggs 02:45, 17 August 2006 (UTC)
IMHO it should be changed to: "This definition of cofinality avoids using the axiom of choice, as it relies on the fact that every non-empty set of cardinal numbers has a least number." - The first emplies it uses the axiom of choice, but actualy it avoids using it because you are making an explicit choice. - Drysh 19:00, 17 August 2006 (UTC)
No, it relies on the Axiom of Choice. The explicit choice is the choice of the least member of a non-empty set of cardinals. You are right that we don't need the Axiom of Choice to pick this least member if it exists. The trouble is that without the Axiom of Choice it may not exist. --Zundark 20:09, 17 August 2006 (UTC)

As a matter of fact, quoting axiom of choice again, "Trichotomy: If two sets are given, then they either have the same cardinality, or one has a smaller cardinality than the other." is one of the theorems which are equivalent to the axiom. In other words, without the axiom there are pairs of cardinal numbers which are incomparable -- neither is smaller than the other nor are they the same. JRSpriggs 03:48, 18 August 2006 (UTC)

This depends on your definition of a cardinal. You can define cardinals such that cardinals are well-ordered independent of choice, but the drawback is that it is then possible to have sets with undefined cardinality. —Preceding unsigned comment added by 24.196.91.135 (talk) 00:11, 13 July 2010 (UTC)

## Arithmetics of cofinality

Are there any rules to compute the cofinality of the sum, product or power of ordinals? For example, what is the cofinality of ${\displaystyle \omega ^{\omega +42}+\omega ^{2}+\omega \cdot 3\,}$? Albmont (talk) 20:25, 3 April 2009 (UTC)

Yes. One uses the fact that the binary operations of ordinal arithmetic are continuous and strictly increasing in the right argument. Thus the cofinality of ${\displaystyle \omega ^{\omega +42}+\omega ^{2}+\omega \cdot 3\,}$ is the same as the cofinality of ${\displaystyle \omega \cdot 3=\omega +\omega +\omega \,}$ which is the same as the cofinality of ${\displaystyle \omega \,}$ which is just ${\displaystyle \omega \,}$ itself.
However, if the right argument is zero or a successor (such as the "3" in your example), then this does not work because continuous functions are λ-continuous for any infinite regular ordinal λ, but neither 0-continuous nor 1-continuous in general. (But multiplication is 0-continuous; and addition is 1-continuous.) JRSpriggs (talk) 05:40, 4 April 2009 (UTC)

## Relying on axiom of choice

It is written : "This definition of cofinality relies on the axiom of choice, as it uses the fact that every non-empty set of cardinal numbers has a least member" Isn't it always true by the well ordering of the ordinals that every set of ordinals has a minimum ? Why do one need the axiom of choice for that ? — Preceding unsigned comment added by 79.182.240.117 (talk) 16:30, 30 June 2012 (UTC)

Without the axiom of choice there are non-wellorderable sets. Their cardinal numbers do not have initial ordinals. So you cannot sort them that way, JRSpriggs (talk) 20:00, 30 June 2012 (UTC)


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# 0 to 2A variable current source circuit

I'm trying to build a 0 to 2A variable current source following the Linear app note.

Here's the schematic:

The circuit is a difference amplifier (LT1995) providing gain to the voltage developed across a sense resistor. The amplifier output voltage is fed into a inverting integrator (LTI1880) with a RC network (R = 1k, C = 10 nF). The output from the integrator is fed into a low pass RC filter (R = 100, C = 10 nF) and that voltage is placed on the gate of a P Power MOSFET (IRF9530).

The source of the IRF9530 is tied to the "low side" of the sense resistor (terminal going into the input P1 on the LTI1995 in figure). The drain of the FET I have tied to ground for testing but I plan to have an inductive coil load in the future.

The output current in my measurements does not scale as $\dfrac{V_{in}}{5 \times R_s}$. This is my goal for this circuit.

With no load attached (drain left floating), I provide voltage on the input (Vin) pin the amplifier output will develop a positive output as it should but the integrator will quickly integrate down to the negative rail (-15V). The IRF9530 is operating with a very large (-15V) gate to source voltage in this case.

When the drain was tied to ground, and the voltage at the input was tied to ground the the FET gate voltage was -2.2V.

Can someone provide a better picture of the operation of the circuit and places where perhaps I'm assembling this wrong. I have prototyped the circuit on the breadboard.

Datasheets LTI1995: http://cds.linear.com/docs/en/datasheet/1995fb.pdf

• Can you provide a schematic of your circuit? – Ricardo May 17 '14 at 20:07
• Hi Ricardo, Tanks for response. I have attached a schematic of the entire circuit on imgur: i.imgur.com/nOcO8Vi.png – shraken May 17 '14 at 23:44
• If you're trying to build the circuit exactly as in the app note, then there was no need to get the schematic out. Sorry, my bad. In any case, I've inlined the schematic into the question for you. – Ricardo May 18 '14 at 0:55

Your circuit is apparently working fine, you just don't realize it.

With the drain left floating, the MOSFET cannot pull any current, and therefor the integrator keeps providing a lower and lower gate drive, trying to draw current.

With the drain grounded and the reference at 0 volts, the integrator goes just low enough to start pulling a tiny amount of current and satisfies the loop.

What you now need to do is get a set of power resistors, such as 10 ohm / 25W, 20 ohm / 25 watt, etc. and use them as loads. From the voltage across your load resistors you can determine the current through them.

However, before you do that, you need desperately to provide a heatsink for your MOSFET. Its' peak power dissipation at a current of 2A will be nearly 30 watts, worst case (for very low load resistances).

You also need to make sure that your shunt resistor is of high enough power. If you're following the app note religiously and using a .2 ohm resistor, it needs to be a 1-watt or better unit, since at 2 amps it will dissipate (2 x 2 x .2) = .8 watts.

You would also do well to reconsider your power wiring. The contacts on a solderless breadboard may not handle 2 amps well.

Finally, you need to make damn sure you can't provide a reference voltage greater than 2 volts.

And finally finally, I suggest you do a good deal of testing with resistors, in order to convince yourself that the circuit really is working. I suspect that you may find it doesn't work nearly as well when you give it a highly inductive load like a coil. It may work, and it may not, since the effects of the inductance will tend to counteract the effects of the loop integrating capacitor, and the result may be instability.

Otherwise, congratulations.

• Hi WhatRoughBeast, Thanks for response and considerations. I have a programmable resistor instrument that can sink 150W so i'll experiment with sweeping the load resistance up to the 500 ohm limit. I have a good heatsink for the MOSFET that I am using on a prototype PCB. Yea that's one of my primary concerns in regards to driving a large coil. First step is to verify that it can drive a programmable 0 to 2 A current into a fixed 50-ohm load. – shraken May 18 '14 at 1:46
• Keep in mind that, for a 50 ohm load, you will not be able to drive more than 0.3 amps into it with your 15 volt power supply. (15 / 50 = .3). If you want to run the whole 2A range, set the load to ~7.5 ohms. – WhatRoughBeast May 18 '14 at 4:00
• Could someone tell me what the integrator is for? I'd think that direct connection of in amp and mosfet gate would do. Or is it preventing oscillation? – Nicolas D May 19 '14 at 17:09

The work-around I ended up using for this circuit was getting rid off the LPF formed by the 100-ohm and 10 nF capacitor. There must be instability in the loop for when the LPF is included the gate is pulled down to the negative (-15 Volt rail).

With the LPF removed the expected functionality is achievable. The current delivered to the load doesn't follow the equation exactly but it's linear and good enough.

Thanks for the feedback and support. I ran into another problem where output of a DAC is unable to drive the voltage reference (REF). I ended up putting a buffer between and this fixed it.

An alternative topology I found in an Analog Device app note is: http://www.analog.com/static/imported-files/application_notes/AN-968.pdf

Anyways I appreciate the responses. Cheers!