text:
MathSciNet bibliographic data MR1777031 (2001e:46025) 46B25 (43A46 46A22 46E30 47B38) Godefroy, G.; Kalton, N. J.; Li, D. Operators between subspaces and quotients of \$L\sp 1\$$L\sp 1$. Indiana Univ. Math. J. 49 (2000), no. 1, 245–286. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


text:
RSA Key and RSA Security

Is there any relationship between the RSA keypair standard and RSA security?

If not, can anyone explain what RSA stands for (so I can keep the two seperate)?

-
Did you read the two articles? –  DrLecter Jul 9 at 18:08

In both cases (the company name and the algorithm name), the letters “RSA” stand for the initials of the surnames of Ron Rivest, Adi Shamir, and Leonard Adleman… $R$ivest, $S$hamir, and $A$dleman.

Wrapping it up in short:

• In 1977, they (Ron Rivest, Adi Shamir, and Leonard Adleman) first publicly described an algorithm which was named after them and is commonly known as the RSA public key cryptography algorithm.

• In 1982, they (Ron Rivest, Adi Shamir, and Leonard Adleman) founded a company to offer cryptographic products and services (including products and services related to their “RSA public key cryptography algorithm”. They named their company RSA Security.

Actually, there’s not much to be confused about. RSA is short for the names $R$ivest, $S$hamir, and $A$dleman. The individual context will point either to the company, or the cryptosystem.

Thinking about it, there’s only one moment where I can imagine confusion to kick in: when you forgot to concentrate while reading… as there is a bit of a difference between “RSA Security” (which points to the company) and papers talking about ”RSA’s security” (which means “the security of the RSA algorithm”).

Anyway, the articles you’ve linked to yourself already explain all that and more. Therefore, I can’t help asking myself the same question DrLecter already asked in a comment: “Did you read the two articles?”.

-

format: plain


text:
Ticket #5785 (closed enhancement: fixed)

Opened 4 years ago

Wiki pages (plug-ins) that should be deleted

Reported by: Assigned to: rjollos rjollos low TracHacks normal otaku42 0.11

Description

I'm opening this ticket to start listing wiki pages for plug-ins that should be deleted from trac-hacks. It would be good to clean up the trac-hacks a little bit since it is getting a bit bloated and has many dead pages.

-----Original Message-----
From: Michael Renzmann [mailto:mrenzmann@otaku42.de]
Sent: Monday, August 31, 2009 3:31 AM
To: Ryan J Ollos
Cc: th-users@lists.trac-hacks.org
Subject: Re: [th-users] Removing invalid hacks/pages

> The list of plug-ins on trac-hacks is fairly huge, and it make it
> difficult to find something useful when there lots of useless stuff.

True. I have something like that on my to-do list, but was short of time
to actually take care of it. So your help is more than welcome, thanks!

> If an admin was willing to delete some pages, I would make the effort to
> keep track of worthless pages I come across and send them along.

Yes, please do, and post the list here. I won't be able to weed these
pages/Subversion subdirectories out at once, but I'll try to take care of
it as quickly as possible.

Bye, Mike

Change History

09/05/09 23:58:00 changed by rjollos

• priority changed from normal to low.
• owner set to otaku42.
• component changed from SELECT A HACK to TracHacks.

(follow-up: ↓ 11 ) 09/06/09 00:39:30 changed by rjollos

On a related issue, I cleaned up all the tickets that did not have a Component specified, as can be seen in this report.

09/09/09 21:28:00 changed by rjollos

I'm taking a more obvious approach and tagging all of the adandoned pages with abandoned and removing the version tags so that the listing don't appear on the main page.

[[ListTagged(abandoned)]]

08/19/10 00:01:37 changed by rjollos

• cc set to otaku42.
• owner changed from otaku42 to rjollos.

I'll go ahead and remove the pages listed here since I now have elevated permissions.

08/19/10 00:13:46 changed by rjollos

Looks like I chose a bad name for tagging the pages, since some of those listed as 'abandoned' are possibly relevant projects that are just not developed in a long time. I left those in place and only pages for "Plugins" that have no source code and are obviously spam or projects that were never started and are quite old.

(follow-up: ↓ 10 ) 08/19/10 00:21:46 changed by rjollos

Looks like I don't have permission to remove the svn directories for the following wiki pages that have been removed, so I'll leave this ticket open:

• AutoTrac
• ChangeLogPlugin
• ChartingModulePlugin
• DmsSorucePlugin
• DragAndDropPlugin
• GooTestPlugin
• LuntbuildTracIntegrationPluginIntegration
• MarketPlugin
• ModuleAnkurIntegration
• ScompiProjectPlugin
• SuportTicketPlugin
• SvnRevisionPlugin
• TracGitPlugin
• TracNav
• WantedPagesMacro
• WikiPageNamePlugin

08/19/10 00:32:25 changed by rjollos

On the main page, under WikiStart#Miscellany, the following replacement should be made: ChangeLogPlugin -> ChangeLogMacro.

(in reply to: ↑ 8 ; follow-up: ↓ 17 ) 08/19/10 07:34:02 changed by otaku42

Looks like I don't have permission to remove the svn directories

You now have read/write permission to the complete repository. Please go ahead and remove the directories of the mentioned plugins, and please also remove the according components from the ticket system.

Feature request to the author of the TracHacksPlugin (/me looks in the mirror): add something that allows to delete hacks when necessary. This should reverse all changes that had been done when the hack has been registered.

(in reply to: ↑ 3 ; follow-up: ↓ 12 ) 08/19/10 21:52:47 changed by hasienda

On a related issue, I cleaned up all the tickets that did not have a Component specified, as can be seen in this report.

I did it again tonight for another 16 tickets.

(in reply to: ↑ 11 ) 08/19/10 22:05:54 changed by rjollos

I did it again tonight for another 16 tickets.

Thanks, I deleted a bunch of those spam tickets you flagged. Nice idea to add [SPAM] to the head of the summary. I'll then see them in my RSS feed of the timeline and delete them.

08/20/10 09:56:19 changed by rjollos

(In [8375]) Removing obsolete project directories. Refs #5785.

08/20/10 10:00:31 changed by rjollos

(In [8376]) Removing obsolete project directories. Refs #5785.

08/20/10 10:35:13 changed by rjollos

(In [8377]) Restoring autotracplugin which I accidentally deleted in [8375] while cleaning up some empty projects. Refs #5785.

08/20/10 10:37:01 changed by rjollos

• status changed from new to closed.
• resolution set to fixed.

You now have read/write permission to the complete repository. Please go ahead and remove the directories of the mentioned plugins, and please also remove the according components from the ticket system.

Thanks! SVN directories and Ticket Components have been removed.

(in reply to: ↑ 10 ) 08/20/10 10:44:44 changed by rjollos

Feature request to the author of the TracHacksPlugin (/me looks in the mirror): add something that allows to delete hacks when necessary. This should reverse all changes that had been done when the hack has been registered.

Btw, do we have an Admin wiki page that lists steps to do common tasks that may not yet have been implemented in a script such as:

• Reassign an adopted project to the new maintainer
• Remove a project (as discussed in this ticket)

If not, perhaps I'll make such a page, since its nice to have the steps listed for reference each time the task comes up ... I tend to make fewer mistakes that way.

I'm also not completely clear on how I might need to update BadContent when spam is detected, so a short description of that could be added as well.

08/22/10 08:28:29 changed by rjollos

(In [8408]) Removing project that has been integrated into the AnnouncerPlugin project. Refs #5785.

(follow-ups: ↓ 20 ↓ 21 ) 08/23/10 03:26:25 changed by AdrianFritz

• status changed from closed to reopened.
• resolution deleted.

Author shows ChecklistPlugin is deprecated (2 years ago, and proposed TracFormsPlugin instead).

Also as you previously pointed at ticket:5785#comment:9, still pending to update section WikiStart#Miscellany (replace ChangeLogPlugin -> ChangeLogMacro).

(in reply to: ↑ 19 ) 08/23/10 03:57:12 changed by rjollos

Author shows ChecklistPlugin is deprecated (2 years ago, and proposed TracFormsPlugin instead).

I'd like to cleanup and get rid of old plugins that aren't useful, but I have reservations about deleting even deprecated projects. The only projects I have deleted so far were entirely empty, or the result of project renames.

I'll wait for Michael to advise on how to proceed, but at present I think the best we can do with a project like that one is to remove the 0.11 tag so that it doesn't appear on WikiStart, which is what I did just now.

Also as you previously pointed at ticket:5785#comment:9, still pending to update section WikiStart#Miscellany (replace ChangeLogPlugin -> ChangeLogMacro).

Good point. I'm not able to edit WikiStart as far as I can tell, but let's leave the ticket open until Michael is able to.

(in reply to: ↑ 19 ) 01/23/12 06:42:12 changed by rjollos

• status changed from reopened to closed.
• resolution set to fixed.

Also as you previously pointed at ticket:5785#comment:9, still pending to update section WikiStart#Miscellany (replace ChangeLogPlugin -> ChangeLogMacro).

Fixed now.

Add/Change #5785 (Wiki pages (plug-ins) that should be deleted)

Change Properties


text:
# CFD-Wiki talk:Format and style guide

## Vector/Matrix formating

I (JasonD) propose that we edit a paragraph/blurb here on this until it comes to some steady state and then either go the forum with it or just move it to the style guide.

Here's what we seem to agree on:

• n-tuples (nx1 arrays) need to differentiated from matrices
• Elements of arrays/vectors are lower case with subscripts

Here are some open questions:

• Exact way to differentiate n-tuples:
• \vec
• \mathbf - this will avoid confusion with vectors in $R^3$
• "Best" (for sake of consistency) letter to use as solution n-tuple
• x is used in most of the linear algebra literature, so I (JasonD) think this would be best for the linear systems section.

### Samples

• $Ax = b$
• $A\,x = b$
• $A\vec{x} = \vec{b}$
• $A\,\vec{x} = \vec{b}$
• $A\mathbf{x} = \mathbf{b}$
• $A\,\mathbf{x} = \mathbf{b}$


text:
### About language, learning and tomatoes

[2] The maths behind the plants: Running the algorithm
Published: 03 Jun 2019

Last week we explained the Rescorla-Wagner rule and set out to learn which fluid (or maybe a combination of both fluids) is most likely to yield tomatoes. We will calculate the first few learning weights by hand, to give you a good understanding of how the machinery works.

In the first trial (row 1 in Table 4) the cue pot is present, and the outcome was NO_TOMATO. This means that the second condition (2) from the three learning situations above is satisfied: there is positive evidence for the outcome, and we need to calculate the sum of weights $$\sum{w_{.j}}$$ from the cue to the outcome. In the first step, this sum can only be zero because learning is only commencing. Next, still as per the formula in (2), we subtract this zero from one $$(1 - 0)$$, yielding 1, and multiply that resulting 1 with the learning rate (speed) - $$\gamma$$, which we fix to a conveniently small value of 0.01. Hence, we calculate$$0.01 \times 1 = 0.01$$. This $$0.01$$ is the value of the connection weight between the pot and NO_TOMATO after the first update, as reflected in Table 5 below. For the remaining cues and outcomes, the situation is as follows. If red and blue liquids are not present, then condition (1) from the above listed learning situations applies and nothing happens. Because the cue pot is present on this trial, but the outcome TOMATO is not, we need to apply rule (3) for negative evidence or: $$\gamma(0 - \sum{w_{kj}}) = 0.01(0 - 0) = 0$$. In sum, all the remaining cells in the first row of Table 4 contain zeros.

In the second trial, the cues pot, red, blue, and the outcome TOMATO are present. For this, we need to apply rule (2): $$\Delta w^{t} = \gamma(1 -\sum{w_{.j}})$$, and then we also need to update: $$w_{ij}^{t+1} = w_{ij}^{t}+ \Delta w^{t}$$. As this is the very first “appearance” of the outcome TOMATO, the sum of all weights ($$\sum{w_{.j}}$$) is zero, and thus, in the case of the cue pot, for example, we have: $$\gamma(1 - \sum{w_{kj}}) = 0.01(1 - 0) = 0.01 \times 1 = 0.01$$. The same applies to the two remaining cues, red and blue. We can see that after the second update, the three respective cells in the second row of the Table 5 (pot and TOMATO; red and TOMATO; blue and TOMATO), contain exactly those values. The outcome NO_TOMATO was not present on the second learning trial and that means we need to apply the negative evidence (or error-correcting) equation to capture the change: $$\Delta w^{t} =\gamma(0 -\sum w_{.j})$$, and then update the connection weights with: $$w_{ij}^{t+1} = w_{ij}^{t} + \Delta w^{t}$$. Now, recall that NO_TOMATO has already been "experienced" on the first learning trial, which means that the sum of weights for the currently present cues (pot, red, blue) will not necessarily be equal to zero. Instead it is $$\sum{w_{.j} = 0.01}$$, which is a consequence of having the pot present in the first trial and gaining a weight of 0.01 there. Our change thus becomes: $$\Delta w^{t} = \gamma(0 -\sum{w_{.j}}) = 0.01(0 - 0.01) = 0.01 \times (-0.01) = -0.0001$$. And from that we can update the values for all cues as follows: for both the red and the blue, which have not previously occurred, the update is the same as their weights in the current step ($$w_{ij}^{t}$$), i.e. they have zero values, and we have: $$w_{ij}^{t+1} =w_{ij}^{t} + \Delta w^{t} = 0 + (-0.0001) = -0.0001$$; the pot gets: $$w_{ij}^{t+1} = w_{ij}^{t} + \Delta w^{t} = 0.01 + (-0.0001) = 0.0099$$. This is shown in the second row of Table 5.

We can now proceed and calculate the value of the weights after the third learning trial. The third learning trial combines pot and red paired with the outcome TOMATO. Let's get the sums of the weights ($$\sum{w_{.j}}$$) for the two possible outcomes, TOMATO and NO_TOMATO. We can get some help from Table 5 which does all the book-keeping for us, having all weights aligned. The connection weight for both pot and red to TOMATO is now 0.01 after trial 2, which gives $$\sum{w_{.j}} = 0.01 + 0.01 = 0.02$$. Analogously, the sum of weights for the outcome NO_TOMATO is $$\sum{w_{.j}} = 0.0099- 0.0001 = 0.0098$$. Now, given that in the third learning trial pot, red, and TOMATO are present, we need to apply the positive evidence update rule (2) in those cases, the no change update rule (1) for all combinations with the absence of blue liquid, and the negative evidence update rule (3) in all remaining cases (i.e., pot and NO_TOMATO, and red and NO_TOMATO). Let's list all results, separately:

 Cue and Outcome Change: $$\Delta w^{t}=\gamma(1-\sum(w_{.j}))$$ Update: $$w_{ij}^{t+1}=w_{ij}^{t}+\Delta w^{t}$$ pot $$\rightarrow$$ TOMATO $$0.01(1-0.02)=0.0098$$ $$0.01+0.0098=0.0198$$ red $$\rightarrow$$ TOMATO $$0.01(1-0.02)=0.0098$$ $$0.01+0.0098=0.0198$$ pot $$\rightarrow$$ NO_TOMATO $$0.01(0-0.0098)=-0.000098$$ $$0.0099-0.000098=0.009802$$ red $$\rightarrow$$ NO_TOMATO $$0.01(0-0.0098)=- 0.000098$$ $$-0.0001-0.000098=-0.000198$$

These values are listed in Table 5, together will all the values for all other Trials.

Table 5
 Trial pot red blue Outcome Outcome Outcome TOMATO NO_TOMATO TOMATO NO_TOMATO TOMATO NO_TOMATO 1 0.0000 0.0100 0.0000 0.0000 0.0000 0.0000 2 0.0100 0.0099 0.0100 -0.0001 0.0100 -0.0001 3 0.0198 0.0098 0.0198 -0.0002 0.0100 -0.0001 4 0.0194 0.0197 0.0194 0.0097 0.0100 -0.0001 5 0.0291 0.0195 0.0194 0.0097 0.0197 -0.0003 6 0.0286 0.0293 0.0194 0.0097 0.0192 0.0095 7 0.0381 0.0289 0.0289 0.0093 0.0192 0.0095 8 0.0376 0.0385 0.0289 0.0093 0.0186 0.0191 9 0.0469 0.0381 0.0383 0.0088 0.0186 0.0191 10 0.0561 0.0376 0.0474 0.0084 0.0186 0.0191 11 0.0648 0.0369 0.0562 0.0077 0.0274 0.0185 12 0.0739 0.0364 0.0562 0.0077 0.0365 0.0179 13 0.0822 0.0358 0.0645 0.0071 0.0448 0.0173 14 0.0810 0.0452 0.0645 0.0071 0.0436 0.0268 15 0.0895 0.0447 0.0731 0.0066 0.0436 0.0268 16 0.0882 0.0540 0.0731 0.0066 0.0422 0.0361 17 0.0866 0.0634 0.0715 0.0160 0.0422 0.0361 18 0.0953 0.0624 0.0715 0.0160 0.0509 0.0351 19 0.0943 0.0718 0.0715 0.0160 0.0509 0.0351 20 0.0929 0.0807 0.0715 0.0160 0.0495 0.0440

Petar/Dagmar


text:
### Areas and Ratios

Do you have enough information to work out the area of the shaded quadrilateral?

### Napoleon's Hat

Three equilateral triangles ABC, AYX and XZB are drawn with the point X a moveable point on AB. The points P, Q and R are the centres of the three triangles. What can you say about triangle PQR?

### Plane to See

P is the midpoint of an edge of a cube and Q divides another edge in the ratio 1 to 4. Find the ratio of the volumes of the two pieces of the cube cut by a plane through PQ and a vertex.

# Rarity

##### Age 16 to 18 Challenge Level:

Ping sent this solution from Thailand.

(1) If $5^m=4^m 2^n$, then $5 = 2^{2m + n}$ which is impossible as 2 and 5 are prime so there are no positive integer solutions $m$ and $n$ of this equation.

(2) We have $a^m d^n=c^n b^m$. As $a$ and $b$ are coprime, we get $a^m|c^n$. Because $c$ and $d$ are coprime, so $c^n|a^m$. This means $a^m=c^n$. Similarly, $b^m=d^n$.

If $a^m=c^n$ and $b^m=d^n$, then obviously $(a/b)^m=(c/d)^n$.

This implies that $a^m$ and $c^n$ have the same prime factors. Write $a = p_1^{u_1}...p_k^{u_k}$ and $c = p_1^{v_1}...p_k^{v_k}$ and for all $j$ we have $mu_j = nv_j$ so that $${u_1\over v_1} = {u_2\over v_2} = ... = {u_k\over v_k} = {m\over n}.$$ Similarly for $b$ and $d$. This is a very special necessary relationship between $a$ and $c$ and also between $b$ and $d$ so solutions rarely occur to the equation: $$\left({5\over 4}\right)^m = \left({2\over 1}\right)^n.$$ We now show this is a sufficient condition. Conversely suppose $a = p_1^{u_1}...p_k^{u_k}$ and $c = p_1^{v_1}...p_k^{v_k}$ and $${u_1\over v_1} = {u_2\over v_2} = ... = {u_k\over v_k}.$$ We call this common ratio ${n\over m}$ then $u_jm=v_jn$ for all $j$ and $a^m = b^n$. Similarly if corresponding ratios of the powers of the prime factors of $b$ and $d$ are constant and also equal to ${n\over m}$ then $b^m=d^n$ giving $$\left({a\over b}\right)^m = \left({c\over d}\right)^n.$$


text:
Kickstart 2020 with new opportunities! - hours only!Up to 80% off on all courses and bundles.-Close
Introduction
Quartiles
Percentiles
Quantiles
Standard Deviation and Variance
10. Sample standard deviation

## Instruction

The formula for the standard deviation (SD) we gave you:

SD=\sqrt{\frac{1}{n}((x_1-\overline{x})^2+(x_2-\overline{x})^2+\ldots+(x_n-\overline{x})^2)}

is in fact the formula for population standard deviation. It should only be used when we have the whole dataset and not just a sample of it (for example, if we're computing the standard deviation of quiz results for all students in our class).

SD=\sqrt{\frac{1}{n-1}((x_1-\overline{x})^2+(x_2-\overline{x})^2+\ldots+(x_n-\overline{x})^2)}

As we said before, all these formulas are heuristics. People have found that the sample standard deviation works slightly better for sample data—that is, it gives more accurate results.

External tools for calculating standard deviation (such as the ones in Excel) usually have two different functions: one for population standard deviation and one for sample standard deviation. In Excel, the function for calculating population standard deviation is called STDEVP. The function for sample standard deviation is called STDEV.

As this course only deals with descriptive statistics, we'll use the phrase "standard deviation" to mean "population standard deviation". Note that in other sources, the term may be used differently.


text:
Rule of Transposition/Formulation 2/Proof 1

Theorem

$\vdash \paren {p \implies q} \iff \paren {\neg q \implies \neg p}$

Proof

Proof of Forward Implication

By the tableau method of natural deduction:

$\vdash \paren {p \implies q} \implies \paren {\neg q \implies \neg p}$
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies q$ Assumption (None)
2 2 $\neg q$ Assumption (None)
3 1, 2 $\neg p$ Modus Tollendo Tollens (MTT) 1, 2
4 1 $\neg q \implies \neg p$ Rule of Implication: $\implies \II$ 2 – 3 Assumption 2 has been discharged
5 $\paren {p \implies q} \implies \paren {\neg q \implies \neg p}$ Rule of Implication: $\implies \II$ 1 – 4 Assumption 1 has been discharged

$\blacksquare$

Proof of Reverse Implication

By the tableau method of natural deduction:

$\vdash \left({\neg q \implies \neg p}\right) \implies \left({p \implies q}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $\neg q \implies \neg p$ Assumption (None)
2 2 $p$ Assumption (None)
3 2 $\neg \neg p$ Double Negation Introduction: $\neg \neg \II$ 2
4 1, 2 $\neg \neg q$ Modus Tollendo Tollens (MTT) 1, 3
5 1, 2 $q$ Double Negation Elimination: $\neg \neg \EE$ 4
6 1 $p \implies q$ Rule of Implication: $\implies \II$ 2 – 5 Assumption 2 has been discharged
7 $\left({\neg q \implies \neg p}\right) \implies \left({p \implies q}\right)$ Rule of Implication: $\implies \II$ 1 – 6 Assumption 1 has been discharged

$\blacksquare$

Law of the Excluded Middle

This proof depends on the Law of the Excluded Middle, by way of Double Negation Elimination.

This is one of the axioms of logic that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.

However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom.

This in turn invalidates this proof from an intuitionistic perspective.


text:
## DMPG '19 G1 - Camera Calibration Challenge

View as PDF

Points: 12 (partial)
Time limit: 2.5s
Memory limit: 128M

Authors:
Problem type

One of the recommendations made this year was for Kirito to make a pre-recorded opening speech for the DMPG that would be played at the satellite sites.

To achieve this, Kirito borrowed AvaLovelace's camera and digital art expertise. While fooling around with learning about the camera's features, he realized that he accidentally messed up the camera's exposure correction! Panicking, he recalls what she taught him about exposure:

A photo can be represented as a grid of by pixels, and the pixel in row and column has a brightness , which can be any real number from to inclusive. If you average the brightnesses of all the pixels in a typical image, the result is called the proper exposure.

Most digital cameras have an exposure correction feature. By choosing a correction constant and multiplying all the pixel brightnesses in an image by , a darker or brighter image can be obtained. When applying a correction constant, if any pixel brightnesses become greater than 1, those values are "clipped" and reduced to 1.

Armed with this knowledge, Kirito knows that to re-calibrate the camera, he has to answer queries:

What is the correction constant necessary for the proper exposure of this image to be ?

Since he would prefer not to work with floating-point numbers, for each query , he would like to know the smallest integer such that applying the correction constant to the image results in a proper exposure greater than or equal to .

#### Input Specification

The first line of input will contain space separated integers, and .
The next lines will each contain space-separated integers, the pixel brightnesses multiplied by .
This will be followed by a single integer, .
The next integers will each contain a single integer, multiplied by .

#### Output Specification

lines, where the th line contains the smallest possible that will result in a proper exposure greater than or equal to .

#### Sample Input 1

2 3
360000 304000 120000
408000 312000 960000
1
480000

#### Sample Output 1

1250000

#### Sample Input 2

2 3
480000 580000 560000
380000 400000 480000
3
120000
480000
360000

#### Sample Output 2

250000
1000000
750000


text:
# Matlab; How to specify Coupon frequency for Interest Rate Swap

I'm trying to price an interest rate swap and would like to change the default coupon payment frequency from 1 a year to 2 or 4 a year. I'm using

Price = swapbyzero(RateSpec, LegRate, i, Maturity, 'Principal', Principal);

and tried

Price = swapbyzero(RateSpec, LegRate, i, Maturity, 'Principal', Principal,'Period',2);

where 'Period' is a valid option when pricing bond using

prbyzero

I'm using matlab 2013a

-

I don't know what is supported by MatLab (I use Java to do such stuff :-).

But in case you do not find a solution from the swapbyzero function you mentioned I can suggest a workaround:

• Value a swap with the annual fix frequence.

Given that it is a payer swap (pays the fixed leg), correct the value by:

• Substract the value of an annual fix coupon bond and
• Add the value of the desired frequence fix coupon bond.

PS: Note that the MatLab function is likely not appropriate to value a swap in a "realistic" way, since it looks like multi-curve (OIS discounting) is not considered here (I wonder if term structure is considered?), see also Rationale for OIS discounting for collateralized derivatives?

-

Name-Value pair option 'LegReset' [n n] where n is the frequency

-
Is this the answer? –  chrisaycock May 27 '13 at 15:07
Question was also asked on Stack Overflow, and this was the answer accepted. stackoverflow.com/questions/16764495/… –  Phil H May 31 '13 at 13:28


text:
## Recursos de colección

#### Project Euclid (Hosted at Cornell University Library) (202.106 recursos)

The Annals of Probability

1. #### Stochastic control for a class of nonlinear kernels and applications

Possamaï, Dylan; Tan, Xiaolu; Zhou, Chao
We consider a stochastic control problem for a class of nonlinear kernels. More precisely, our problem of interest consists in the optimization, over a set of possibly nondominated probability measures, of solutions of backward stochastic differential equations (BSDEs). Since BSDEs are nonlinear generalizations of the traditional (linear) expectations, this problem can be understood as stochastic control of a family of nonlinear expectations, or equivalently of nonlinear kernels. Our first main contribution is to prove a dynamic programming principle for this control problem in an abstract setting, which we then use to provide a semimartingale characterization of the value function. We...

2. #### A class of globally solvable Markovian quadratic BSDE systems and applications

Xing, Hao; Žitković, Gordan
We establish existence and uniqueness for a wide class of Markovian systems of backward stochastic differential equations (BSDE) with quadratic nonlinearities. This class is characterized by an abstract structural assumption on the generator, an a priori local-boundedness property, and a locally-Hölder-continuous terminal condition. We present easily verifiable sufficient conditions for these assumptions and treat several applications, including stochastic equilibria in incomplete financial markets, stochastic differential games and martingales on Riemannian manifolds.

3. #### Random walks on the random graph

Berestycki, Nathanaël; Lubetzky, Eyal; Peres, Yuval; Sly, Allan
We study random walks on the giant component of the Erdős–Rényi random graph $\mathcal{G}(n,p)$ where $p=\lambda/n$ for $\lambda>1$ fixed. The mixing time from a worst starting point was shown by Fountoulakis and Reed, and independently by Benjamini, Kozma and Wormald, to have order $\log^{2}n$. We prove that starting from a uniform vertex (equivalently, from a fixed vertex conditioned to belong to the giant) both accelerates mixing to $O(\log n)$ and concentrates it (the cutoff phenomenon occurs): the typical mixing is at $(\nu\mathbf{d})^{-1}\log n\pm(\log n)^{1/2+o(1)}$, where $\nu$ and $\mathbf{d}$ are the speed of random walk and dimension of harmonic measure on...

4. #### Lattice approximation to the dynamical $\Phi_{3}^{4}$ model

Zhu, Rongchan; Zhu, Xiangchan
We study the lattice approximations to the dynamical $\Phi^{4}_{3}$ model by paracontrolled distributions proposed in [Forum Math. Pi 3 (2015) e6]. We prove that the solutions to the lattice systems converge to the solution to the dynamical $\Phi_{3}^{4}$ model in probability, locally uniformly in time. Since the dynamical $\Phi_{3}^{4}$ model is not well defined in the classical sense and renormalisation has to be performed in order to define the nonlinear term, a corresponding suitable drift term is added in the stochastic equations for the lattice systems.

5. #### An $L^{p}$ theory of sparse graph convergence II: LD convergence, quotients and right convergence

Borgs, Christian; Chayes, Jennifer T.; Cohn, Henry; Zhao, Yufei
We extend the $L^{p}$ theory of sparse graph limits, which was introduced in a companion paper, by analyzing different notions of convergence. Under suitable restrictions on node weights, we prove the equivalence of metric convergence, quotient convergence, microcanonical ground state energy convergence, microcanonical free energy convergence and large deviation convergence. Our theorems extend the broad applicability of dense graph convergence to all sparse graphs with unbounded average degree, while the proofs require new techniques based on uniform upper regularity. Examples to which our theory applies include stochastic block models, power law graphs and sparse versions of $W$-random graphs.

6. #### Quenched invariance principle for random walks with time-dependent ergodic degenerate weights

Andres, Sebastian; Chiarini, Alberto; Deuschel, Jean-Dominique; Slowik, Martin
We study a continuous-time random walk, $X$, on $\mathbb{Z}^{d}$ in an environment of dynamic random conductances taking values in $(0,\infty)$. We assume that the law of the conductances is ergodic with respect to space–time shifts. We prove a quenched invariance principle for the Markov process $X$ under some moment conditions on the environment. The key result on the sublinearity of the corrector is obtained by Moser’s iteration scheme.

7. #### Dimensional improvements of the logarithmic Sobolev, Talagrand and Brascamp–Lieb inequalities

Bolley, François; Gentil, Ivan; Guillin, Arnaud
In this work, we consider dimensional improvements of the logarithmic Sobolev, Talagrand and Brascamp–Lieb inequalities. For this, we use optimal transport methods and the Borell–Brascamp–Lieb inequality. These refinements can be written as a deficit in the classical inequalities. They have the right scale with respect to the dimension. They lead to sharpened concentration properties as well as refined contraction bounds, convergence to equilibrium and short time behavior for the laws of solutions to stochastic differential equations.

8. #### Random planar maps and growth-fragmentations

Bertoin, Jean; Curien, Nicolas; Kortchemski, Igor
We are interested in the cycles obtained by slicing at all heights random Boltzmann triangulations with a simple boundary. We establish a functional invariance principle for the lengths of these cycles, appropriately rescaled, as the size of the boundary grows. The limiting process is described using a self-similar growth-fragmentation process with explicit parameters. To this end, we introduce a branching peeling exploration of Boltzmann triangulations, which allows us to identify a crucial martingale involving the perimeters of cycles at given heights. We also use a recent result concerning self-similar scaling limits of Markov chains on the nonnegative integers. A motivation...

9. #### An upper bound on the number of self-avoiding polygons via joining

Hammond, Alan
For $d\geq2$ and $n\in\mathbb{N}$ even, let $p_{n}=p_{n}(d)$ denote the number of length $n$ self-avoiding polygons in $\mathbb{Z}^{d}$ up to translation. The polygon cardinality grows exponentially, and the growth rate $\lim_{n\in2\mathbb{N}}p_{n}^{1/n}\in(0,\infty)$ is called the connective constant and denoted by $\mu$. Madras [J. Stat. Phys. 78 (1995) 681–699] has shown that $p_{n}\mu^{-n}\leq Cn^{-1/2}$ in dimension $d=2$. Here, we establish that $p_{n}\mu^{-n}\leq n^{-3/2+o(1)}$ for a set of even $n$ of full density when $d=2$. We also consider a certain variant of self-avoiding walk and argue that, when $d\geq3$, an upper bound of $n^{-2+d^{-1}+o(1)}$ holds on a full density set for the counterpart in...

10. #### Path-dependent equations and viscosity solutions in infinite dimension

Cosso, Andrea; Federico, Salvatore; Gozzi, Fausto; Rosestolato, Mauro; Touzi, Nizar
Path-dependent partial differential equations (PPDEs) are natural objects to study when one deals with non-Markovian models. Recently, after the introduction of the so-called pathwise (or functional or Dupire) calculus [see Dupire (2009)], in the case of finite-dimensional underlying space various papers have been devoted to studying the well-posedness of such kind of equations, both from the point of view of regular solutions [see, e.g., Dupire (2009) and Cont (2016) Stochastic Integration by Parts and Functional Itô Calculus 115–207, Birkhäuser] and viscosity solutions [see, e.g., Ekren et al. (2014) Ann. Probab. 42 204–236]. In this paper, motivated by the study of...

11. #### Size biased couplings and the spectral gap for random regular graphs

Cook, Nicholas; Goldstein, Larry; Johnson, Tobias
Let $\lambda$ be the second largest eigenvalue in absolute value of a uniform random $d$-regular graph on $n$ vertices. It was famously conjectured by Alon and proved by Friedman that if $d$ is fixed independent of $n$, then $\lambda=2\sqrt{d-1}+o(1)$ with high probability. In the present work, we show that $\lambda=O(\sqrt{d})$ continues to hold with high probability as long as $d=O(n^{2/3})$, making progress toward a conjecture of Vu that the bound holds for all $1\le d\le n/2$. Prior to this work the best result was obtained by Broder, Frieze, Suen and Upfal (1999) using the configuration model, which hits a barrier...

12. #### Nonbacktracking spectrum of random graphs: Community detection and nonregular Ramanujan graphs

Bordenave, Charles; Lelarge, Marc; Massoulié, Laurent
A nonbacktracking walk on a graph is a directed path such that no edge is the inverse of its preceding edge. The nonbacktracking matrix of a graph is indexed by its directed edges and can be used to count nonbacktracking walks of a given length. It has been used recently in the context of community detection and has appeared previously in connection with the Ihara zeta function and in some generalizations of Ramanujan graphs. In this work, we study the largest eigenvalues of the nonbacktracking matrix of the Erdős–Rényi random graph and of the stochastic block model in the regime...

13. #### Existence conditions of permanental and multivariate negative binomial distributions

Eisenbaum, Nathalie; Maunoury, Franck
Existence conditions of permanental distributions are deeply connected to existence conditions of multivariate negative binomial distributions. The aim of this paper is twofold. It answers several questions generated by recent works on this subject, but it also goes back to the roots of this field and fixes existing gaps in older papers concerning conditions of infinite divisibility for these distributions.

14. #### Existence conditions of permanental and multivariate negative binomial distributions

Eisenbaum, Nathalie; Maunoury, Franck
Existence conditions of permanental distributions are deeply connected to existence conditions of multivariate negative binomial distributions. The aim of this paper is twofold. It answers several questions generated by recent works on this subject, but it also goes back to the roots of this field and fixes existing gaps in older papers concerning conditions of infinite divisibility for these distributions.

15. #### Existence conditions of permanental and multivariate negative binomial distributions

Eisenbaum, Nathalie; Maunoury, Franck
Existence conditions of permanental distributions are deeply connected to existence conditions of multivariate negative binomial distributions. The aim of this paper is twofold. It answers several questions generated by recent works on this subject, but it also goes back to the roots of this field and fixes existing gaps in older papers concerning conditions of infinite divisibility for these distributions.

16. #### Existence conditions of permanental and multivariate negative binomial distributions

Eisenbaum, Nathalie; Maunoury, Franck
Existence conditions of permanental distributions are deeply connected to existence conditions of multivariate negative binomial distributions. The aim of this paper is twofold. It answers several questions generated by recent works on this subject, but it also goes back to the roots of this field and fixes existing gaps in older papers concerning conditions of infinite divisibility for these distributions.

17. #### The vacant set of two-dimensional critical random interlacement is infinite

Comets, Francis; Popov, Serguei
For the model of two-dimensional random interlacements in the critical regime (i.e., $\alpha=1$), we prove that the vacant set is a.s. infinite, thus solving an open problem from [Commun. Math. Phys. 343 (2016) 129–164]. Also, we prove that the entrance measure of simple random walk on annular domains has certain regularity properties; this result is useful when dealing with soft local times for excursion processes.

18. #### The vacant set of two-dimensional critical random interlacement is infinite

Comets, Francis; Popov, Serguei
For the model of two-dimensional random interlacements in the critical regime (i.e., $\alpha=1$), we prove that the vacant set is a.s. infinite, thus solving an open problem from [Commun. Math. Phys. 343 (2016) 129–164]. Also, we prove that the entrance measure of simple random walk on annular domains has certain regularity properties; this result is useful when dealing with soft local times for excursion processes.

19. #### The vacant set of two-dimensional critical random interlacement is infinite

Comets, Francis; Popov, Serguei
For the model of two-dimensional random interlacements in the critical regime (i.e., $\alpha=1$), we prove that the vacant set is a.s. infinite, thus solving an open problem from [Commun. Math. Phys. 343 (2016) 129–164]. Also, we prove that the entrance measure of simple random walk on annular domains has certain regularity properties; this result is useful when dealing with soft local times for excursion processes.

20. #### The vacant set of two-dimensional critical random interlacement is infinite

Comets, Francis; Popov, Serguei
For the model of two-dimensional random interlacements in the critical regime (i.e., $\alpha=1$), we prove that the vacant set is a.s. infinite, thus solving an open problem from [Commun. Math. Phys. 343 (2016) 129–164]. Also, we prove that the entrance measure of simple random walk on annular domains has certain regularity properties; this result is useful when dealing with soft local times for excursion processes.

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text:
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2021-09-14, 20:21   #3102
EdH

"Ed Hall"
Dec 2009
Adirondack Mtns

7·577 Posts

Quote:
 Originally Posted by bur How can the percentage of uniques be found during sieving? By doing a manual filtering step on the relations? Impressive progress anyway, about the same timeframe I need for a C167...
remdups4 is what I use. It can be found in the ggnfs package. (ggnfs/contrib/remdups/remdups4.c)

Edit: Depending on which sieving package you are using, you will need to gather all the relations into a single file and:
Code:
[z]cat <file> | ./remdups4 1000 ><uniquerels>

Last fiddled with by EdH on 2021-09-14 at 20:24


text:
$$\require{cancel}$$

# 3.7: Energy Diagrams

An energy diagram provides us a means to assess features of physical systems at a glance. We will examine a couple of simple examples, and then show how it can be used for more advanced cases in physics and chemistry. It's important to understand that there is no new physics in here – what we have learned so far now is simply represented diagrammatically, making it easier in some cases to see the "big picture" of a physical system.

## Elements of Energy Diagrams

First of all, it should be noted that we will be confining ourselves to energy diagrams for 1-dimensional motion. This dimension will be represented by the horizontal axis, and the vertical axis will have units of energy. Secondly, the physical systems represented by energy diagrams will involve only one (conservative) force acting on an object.

Construction of an energy diagram entails first graphing the potential energy function for the conservative force on the axes. Note that potential energy function includes an arbitrary additive constant, which means that the entire graph can be moved up or down on the vertical axis as much as one likes without changing the physical system at all. There is one common convention that is followed regarding the height of the graph on the vertical axis, which we will see below, but it should be remembered that this is only convention, and doesn't change any of the physical properties of the system.

The graph of the potential energy function could apply to any object under the influence of this conservative force. To represent a specific system, the diagram also needs to indicate the total mechanical energy of the system, and this is done with a horizontal line with the correct height on the vertical axis.

That's all there is to drawing these diagrams. The real value comes from interpreting them, which we will discuss in the context of a couple of simple examples.

## Two Simple Examples

Let's look at the energy diagrams for the two conservative forces we have dealt with so far... gravity and the elastic force.

### Gravity

If we choose the arbitrary constant $$U_o$$ for the gravitational potential energy to be zero, we have as a graph of the potential energy function a straight line that passes through the origin. We then include a horizontal line to represent the total energy of the particular system (which I will label as $$E_{tot}$$). Now for interpretation...

Figure 3.7.1 – Energy Diagram for Object Influenced by Gravity Near Earth's Surface

The position of the object (which in this case is the height above some defined zero point), is the value along the horizontal axis. For every position of the object, there is a corresponding value of its potential energy, given by the height of the $$U\left(y\right)$$ graph above (if positive) or below (if negative) the horizontal axis. The total (mechanical) energy of this system is conserved (i.e. it is the same for every position of the object), which explains why the total energy graph is a horizontal line. For a given position, the gap between the total energy line and the potential energy line equals the kinetic energy of the object, since the sum of this gap and the height of the potential energy graph is the total energy.

We can also interpret the intersection point of the total energy and the potential energy graphs. At this point, the total energy equals the potential energy, which means the object has no kinetic energy – i.e. the object is at rest at this position. How can an object under the influence of only gravity be at rest? It can be for just an instant, when it reaches the peak of its flight. Therefore the value on the horizontal axis corresponding to this intersection point is the highest elevation the object can reach. Note that for heights (horizontal axis values) greater than this, the potential energy is greater than the mechanical energy, which would require a negative kinetic energy. This is of course impossible, and we call this the forbidden region of the diagram, as we will never find the system in one of these states. As time passes, when the object reaches the intersection point, it must have done so from the allowed region, which means that when the object comes to rest here, is reverses its direction of motion. Consequently, this position is often referred to as the turnaround point.

There is one other nugget of information we can extract from this diagram, though in this particular case it is fairly trivial. If we evaluate the negative of the slope of the potential energy graph at the point where the object is at some moment, we know the force acting on the object at that moment. In the case of gravity, the force is the same everywhere:

$F_y = -\dfrac{dU}{dy} = -\dfrac{d}{dy} \left(mgy\right) = -mg$

[Note: There is no need for partial derivatives here, as we are only dealing with one-dimensional potential energy functions.]

If we change the arbitrary constant, the only quantities that change in the entire picture are the potential energy and total energy. Every physically-observable quantity (kinetic energy, turnaround point, and force) remains unchanged. This may not be immediately apparent, but looking at the graph it is easy to see:

Figure 3.7.2 – Redefined Zero Point for Gravitational Potential Energy

Interestingly, the fact that this potential is a straight line means that a shift of the graph up or down by an additive constant is equivalent to redefining the origin. This is easily seen by noting that this graph can also be viewed as the previous graph shifted to the left by $$y_o$$, where $$mgy_o=U_o$$. So for this simple case, changing the zero point of potential energy is equivalent to changing the position which we call the origin.

### Elastic Force

We take precisely the same steps to draw the energy diagram for a mass on a spring, but there are some differences, such as two forbidden regions and a different slope for every position, and there is one additional feature for this potential that doesn't exist for the case of gravity: an equilibrium point.

Figure 3.7.3 – Energy Diagram for Object Influenced by Elastic Force

The two forbidden regions arise here because the spring has a maximum stretch and a maximum compression that result in potential energy equaling the total energy. Regions of potential energy confined by two turnaround points like this are often referred to as potential wells. Clearly the slope of the potential energy curve is different everywhere, which reflects the fact that the force by the spring is different for every position the mass can have.

An equilibrium point occurs whenever the slope vanishes (at maxima, minima, and inflection points in the potential energy curve) – there are simply places where the force vanishes. For the spring, it is the position where the spring is neither stretched nor compressed from its natural length. This particular equilibrium is referred to as a stable equilibrium for the following reason: If the object is at rest at this point, and it is given a small nudge in either direction, the resulting force acts to bring the object back to its original position. We can see this here, because the slope on the (+) side of the equilibrium point is positive, which means the force is in the negative direction. The force on the (–) side of the equilibrium point similarly acts back toward the equilibrium. Forces that create stable equilibrium like this are called restoring forces.

It should be clear that any minimum in the potential energy curve will lead to a stable equilibrium, but what about maxima? In this case, the forces that result from small displacements away from the equilibrium point act to push or pull the object farther from its starting point. This type of equilibrium is therefore referred to as an unstable equilibrium. Inflection points lie between parts of the curve that are concave (stable) on one side and convex (unstable) on the other, and the resulting equilibrium is referred to as a meta-stable equilibrium.

As with the case of gravity, shifting the entire curve up or down by an amount $$U_o$$ doesn't change any of the physics, including the position of the equilibrium point. Unlike the gravity case, shifting the entire curve left or right (changing the definition of the origin) is not equivalent to the addition of an additive constant to the potential energy curve. But in both cases, the physics is unchanged by the positioning of the curve relative to either of the axes.

## Bound States of Two Particles

While our models of terrestrial gravity and the elastic force are useful, we have to keep in mind that force interactions are between two objects, and the energy diagrams we have drawn appear to involve only a single object. The leap to discussing two objects is not a difficult one. Instead of being a position along the $$x$$ or $$y$$ axis, the one dimension of freedom becomes the separation of the two objects, which we represent with the variable $$r$$. If we can express the force as a function of the separation of the two objects interacting, then we can express the potential energy as a function of that variable as well, and voilà – we can draw an energy diagram. For the sake of interpreting such diagrams correctly, we have to keep in mind that the horizontal axis represents a separation, rather than a position, which leads to a big difference from the energy diagrams we created above – the horizontal axis has no negative values. It also should be remembered that these diagrams only relate motion between the particles along the line joining them. If we want to include motion around each other (as in the case of an orbit), we require more information than we can get from the energy diagram.

When we look at the universe in both the microscopic and macroscopic realm, we see countless examples of forces holding systems together. Gravity interactions between the sun and the planets keeps them in orbit. Electromagnetic interactions between protons and electrons holds them together in atoms, and electromagnetic interactions between atoms bind them into molecules. Systems of two bodies that possess too little total energy to escape each other's attractive force are in what is called a bound state. For the purposes of this section, we will confine ourselves to discussing bound states between microscopic particles, and save the discussion of orbits of celestial bodies for the chapter on gravitation.

Bound states between atoms and molecules in our universe are quite extraordinary. Here is how the famous Nobel laureate Richard Feynman put it:

If, in some cataclysm, all of scientific knowledge were to be destroyed, and only one sentence passed on to the next generations of creatures, what statement would contain the most information in the fewest words? All things are made of atoms—little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling upon being squeezed into one another. In that one sentence ... there is an enormous amount of information about the world.

From our earlier discussion, we know that this means that forces between atoms (and between molecules, which are clusters of atoms) must be restoring in nature, with an equilibrium separation. Indeed this does tell us a lot about the shape of the potential energy curve – it must contain a local minimum. But we know even more than this. Clearly a separation distance of zero is not possible (the particles can't occupy the same space), and this is assured by an ever-increasing force as they continue to get closer. This means that the potential energy curve gets steeper and approaches infinity as the separation distance gets closer to zero (i.e. the graph gets closer to the vertical axis). We know one other thing: The force between the two particles gets weaker as they get farther apart, and drops to zero in the limit as their separation goes to infinity. This means that the potential energy curve "flattens out" to a horizontal line as the distance from the vertical axis goes to infinity. Amazingly, this seemingly insignificant amount of information gives us a general shape for the interaction potential of two particles.

Figure 3.7.4 – Shape of Inter-Particle Interaction Potentials

The horizontal axis has been intentionally left out here, because this is not something we can determine. Recall that this potential energy curve can be raised or lowered an arbitrary amount without changing the physics, so we are actually free to place the horizontal axis (the point of zero energy) wherever we like. It is here that we come to the convention alluded-to at the start of this section: It is generally agreed to place the horizontal axis at the position that the potential energy curve reaches as $$r \rightarrow \infty$$. That is, it is usually agreed that the potential energy of the interaction vanishes when the particles are separated by an infinite distance (see Figure 3.7.5).

One nice consequence of the $$U\left(r \rightarrow \infty \right) \rightarrow 0$$ is that it gives us a simple rule-of-thumb to determine whether or not the two particles in this potential are bound to each other. If the total energy of the system is positive (i.e. the horizontal line representing the total energy is above the $$r$$-axis), then that means that when the two particles are moving away from each other, the graphs never intersect to give a "forbidden region," and they just keep moving apart – they are not bound. If the total energy is negative, then the total energy horizontal line intersects the potential energy graph in two places, giving two turnaround points, keeping the particles within a range of separations.

Figure 3.7.5 – Bound and Unbound States

All of the physical interpretations we came up with above apply to this function as well, though it might be a bit confusing at first, since for much of the graph the potential energy is negative. But note:

• The kinetic energy for a specific position is still always positive, and equals the gap between the point on the curve and the total energy line.
• The force between the particles is still the negative of the slope, which means it is attractive (seeks to make $$r$$ smaller) when the slope is positive, and is repulsive (seeks to make $$r$$ bigger) when the slope is negative.
• The equilibrium and turnaround points are defined as the bottom of the dip and the intersection points of the total energy line and potential energy curve, respectively.

When two particles are bound to each other, to break the bond, energy must be added to the system. That is, the total energy line must be moved up until it is above the $$r$$-axis. Systems of particles must receive additional energy to break chemical bonds, and energy only comes out of chemical reactions in which new bonds are formed. For some reason, belief of the exact reverse is a very common misconception.

A model of the potential energy that works very well for two neutrally-charged atoms or molecules was constructed in 1924 by John Lennard-Jones. This Lennard-Jones potential has a simple-but-surprising form (yes, those powers are not typos!):

$U\left(r \right) = \epsilon \left[ {\left( \dfrac{r_o}{r} \right)}^{12} - 2 {\left( \dfrac{r_o}{r} \right)}^6 \right]$

Example $$\PageIndex{1}$$

Show that in the Lennard-Jones potential, $$r_o$$ is the separation of the particles when they are at equilibrium, and $$\epsilon$$ is the depth of the potential well (expressed as a positive value).

Solution

The minimum of this curve occurs at the point where the derivative vanishes, so:

$0 = \dfrac{dU}{dr} = \epsilon \dfrac{d}{dr} \left[ {\left( \dfrac{r_o}{r} \right)}^{12} - 2 {\left( \dfrac{r_o}{r} \right)}^6 \right] = 12{\left( \dfrac{r_o}{r} \right)}^{11} - 12 {\left( \dfrac{r_o}{r} \right)}^5 \;\;\; \Rightarrow \;\;\; r = r_o \nonumber$

The value of the potential at the minimum is $$U \left(r=r_o \right)$$, which one can see from a simple plug-in is in fact equal to $$\left(- \epsilon \right)$$.

Example $$\PageIndex{2}$$

A system of two particles bound in a Lennard-Jones potential requires that an addition of energy equal to $$\frac{3}{4}\epsilon$$ in order to become unbound. Find the turnaround points of this bound system in terms of $$r_o$$.

Solution

The potential energy equals to the total energy at the turnaround points, and the total energy is given to be $$-\frac{3}{4}\epsilon$$ (i.e. the energy to unbind the particles is the amount needed to bring the total energy to zero), so:

$-\frac{3}{4}\epsilon = \epsilon \left[ {\left( \dfrac{r_o}{r} \right)}^{12} - 2 {\left( \dfrac{r_o}{r} \right)}^6 \right] \;\;\; \Rightarrow \;\;\; -\frac{3}{4}= \left[ x^2 - 2x \right], \;\;\;\;\; where\;\;x \equiv {\left( \dfrac{r_o}{r} \right)}^6 \nonumber$

Now solve the quadratic for $$x$$:

$4x^2 -8x +3 = 0 \;\;\; \Rightarrow \;\;\; x = \dfrac{8 \pm \sqrt{8^2 - 4\left(4\right)\left(3\right)}}{2\left(4\right)} = \frac{3}{2} \;\; or \;\; \frac{1}{2} \nonumber$

Now solve for the two r values:

${\left( \dfrac{r_o}{r} \right)}^6 = x \;\;\; \Rightarrow \;\;\; r = x^{-\frac{1}{6}} r_o \;\;\; \Rightarrow \;\;\; r_{min}=\boxed{0.93 r_o}, \;\;\; r_{max}=\boxed{1.12 r_o} \nonumber$

## Modeling Bonds as Springs

It is quite common to model chemical bonds as springs, but this seems like a strange practice, given that the potential energy function looks like Equation 3.7.2, which is nothing like the potential energy function of a spring. Surprisingly, a spring can nevertheless act as a reasonable replacement for the Lennard-Jones potential when the total energy is deep in the well. Certainly the curve at the bottom of the well resembles the parabolic curve of the elastic potential, but one might argue that the bottom of any concave curve will resemble the elastic potential energy curve. It turns out that this argument is completely correct – every smooth concave curve can be approximated by the parabolic curve of the elastic potential energy!

How well this approximation works depends upon the range of values we confine ourselves to. If we look at the whole curve, then obviously the approximation of the Lennard-Jones potential with a parabola breaks down badly. But as we narrow-down our view to a smaller and smaller range near the bottom of the well, this approximation gets better.

The reason this works is related to some amazing mathematics: Any smooth function of a single variable can be written as a series (often infinite) of powers of that variable, with each term multiplied by a different constant:

$f\left( x \right) = a_o + a_1 x + a_2 x^2 + a_3 x^3 + \dots$

From this perspective, it's clear that every function has a bit of the $$x^2$$ function in it. For the Lennard-Jones potential, we can define $$x$$ in the expansion above as $$\left(1-\dfrac{r}{r_o}\right)$$, which is a measure of how far the particle separation is from the equilibrium. For example, if the separation distance $$\left(r\right)$$ is 90% of the equilibrium distance $$\left(r_o\right)$$, then $$\left(1-\dfrac{r}{r_o}\right)=0.1$$, and for $$r$$ equal to 99% of $$r_o$$, $$\left(1-\dfrac{r}{r_o}\right)=0.01$$.

In the spring model, it is precisely the distance that the spring is stretched or compressed from the equilibrium that determines the potential energy. So if we write the Lennard-Jones potential as an expansion in powers of $$\left(1-\dfrac{r}{r_o}\right)$$, we can more easily compare it to a spring potential energy:

$\epsilon \left[ {\left( \dfrac{r_o}{r} \right)}^{12} - 2 {\left( \dfrac{r_o}{r} \right)}^6 \right] = U\left( r \right) = a_o + a_1 \left(1-\dfrac{r}{r_o}\right) + a_2 \left(1-\dfrac{r}{r_o}\right)^2 + a_3 \left(1-\dfrac{r}{r_o}\right)^3 + \dots$

To use a spring to model this function, we want the series to look like a quadratic, so we need the contributions of the terms in the series with powers greater than 2 to be small compared to the terms before them. Well, if we restrict ourselves to values of $$r$$ that are close to $$r_o$$ (i.e. consider particles that are separated by a distance close to the equilibrium separation, which means the total energy is quite low), then the difference $$\left(1-\dfrac{r}{r_o}\right)$$ is a small number less than 1. The more times we multiply this small number by itself, the smaller the result, so higher powers provide ever-smaller contributions to the sum of the series.

All that remains if we are going to use the elastic potential energy to approximate the Lennard-Jones (or any other) potential is to figure out how to determine the equivalent "spring constant." Fortunately we have a nice trick for doing this. We can compare the way in which the spring constant enters the potential energy function to the series expansion, and determine what number we need from the series expansion to find the effective spring constant:

$\frac{1}{2}k_{eff}x^2 = a_2x^2 \;\;\; \Rightarrow \;\;\; k_{eff}=2a_2$

Okay, so we know where to find the effective spring constant – we express the actual potential as an infinite series, then look at the constant that multiplies the quadratic term in the expansion, and multiply it by two. But this still means we need to come up with the series... or does it? Consider the following useful mathematical trick: First, take two derivatives of Equation 3.7.3:

$f\left( x \right) = a_o + a_1 x + a_2 x^2 + a_3 x^3 + \dots \;\;\; \Rightarrow \;\;\; \dfrac{df}{dx} = 0 + a_1 + 2a_2 x + 3a_3 x^2 + \dots \;\;\; \Rightarrow \;\;\; \dfrac{d^2f}{dx^2} = 0 + 0 + 2a_2 + 6a_3 x + \dots$

Now evaluate this second derivative at the point $$x=0$$. All of the terms after the constant term then vanish, leaving what we are looking for:

$\left.\dfrac{d^2f}{dx^2}\right|_{x=0} = 2a_2 = k_{eff}$

It is left as an exercise for the reader to show that the effective spring constant for the Lennard-Jones potential deep within the well comes out to be: $$72\dfrac{\epsilon}{r_o^2}$$. [Note that instead of taking two derivatives with respect to $$x$$ and evaluating at $$x=0$$, where $$x \equiv 1- \dfrac{r}{r_o}$$, we can perhaps more easily do the equivalent calculation of taking the derivatives with respect to $$r$$ and evaluate at $$r=r_o$$.]


text:
Scintillation Properties of a Cs2NaCeBr6 Crystal under X-ray, $gamma$-ray and Proton Irradiations

Abstract
In this paper, the crystal growth and scintillation properties of aCs$_{2}$NaCeBr$_{6}$ crystal are presented. The crystal was grownwith the Bridgman method by using a two-zone transparent furnace.The investigation of the scintillation properties of the crystalwere performed under excitation by X-rays, $gamma$-rays and protonsat room temperature. When the Cs$_{2}$NaCeBr$_{6}$ crystal wasexcited by X-rays and protons, the emission spectrum of the crystalwas observed in the range of 380 $sim$ 460 nm, which is due to the5d $o$ 4f transitions of Ce$^{3+}$. The decay times of theCe$^{3+}$ emission excited by $^{137}$Cs 662 keV $gamma$-rays. wereabout 140 ns (94 \%) and 880 ns (6 \%). The energy resolution of thecrystal was obtained to be 6.7 \% for the $^{137}$Cs 662 keV$gamma$-rays and the absolute light yield of theCs$_{2}$NaCeBr$_{6}$ crystal was about 25,000 phs/MeV.
Keywords: Bridgman, Cs$_{2}$NaCeBr$_{6}$, Proton, Scintillation

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Indexed/Covered by


text:
# The Special Relativity

December 2013
Jaroslav Kopernicky
Jan Onderco

## Abstract

The purpose of this research is to analyze uniform circular motion of a steel ball pushed by two springs against force sensors with a limited freedom of motion in a box that is mounted on a wheel in a moving reference frame and in the ground reference frame by using Galilean transformation. Before a reader dismisses the subject as a straightforward problem that is well understood by the current physics the authors ask a simple question.
What happened to the centrifugal acceleration of a free body in a cycloid motion? Why it is not considered together with the tangential and centripetal accelerations? This is an underlying motto of this research, please, keep reading to find out.

## Introduction

The first postulate of the Special Relativity says:

The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems of coordinates in uniform translatory motion. OR: The laws of physics are the same in all inertial frames of reference.
From Wikipedia[1]

The following lines are my explanation of what Jaroslav Kopernicky pointed out to me in one of our discussions.

The question: Does the first postulate apply to the tangential and radial accelerations and their vector sum of an object moving in an uniform circular motion observed from two inertial frames of reference when the outer reference frame combines the uniform circular motion with a uniform translatory motion.

For example if we take a horizontally oriented wheel, its axle is perpendicular to the ground and we keep it moving at:

• constant angular velocity (uniform circular motion) fixed to the ground
• constant angular velocity (uniform circular motion) on a moving train/car – straight line, constant velocity, the vertically oriented wheel axle is attached to the moving frame

What measurements do we get for the tangential and radial acceleration components from an accelerometer mounted on the wheel? The accelerometer would be a steel ball in a box mounted on the wheel. The ball has a limited freedom of motion in two horizontal directions, it’s not attached to the wheel and it does not form one rigid body together with the wheel. There are force sensors mounted inside of the box walls. The steel ball is being pushed by two adjustable springs onto the force sensors. The force sensors change their resistance based on how much force is being applied. The change of the resistance leads to a voltage change, then there is an A/D converter, the data is being captured by a computer and transmitted to the world and the force readings can then be translated into the accelerations and/or kinetic energy changes. I guess a picture is worth a thousand words.

Figure 1: The steel ball box with FlexiForce® sensors

Figure 2: A top view of our hovercraft with the steel ball box on the wheel

It’s obvious that the ground being our fixed reference frame with no velocity and a train/car/hovercraft moving in a straight line at a constant velocity are inertial frames of reference. The rotating wheel is an experiment within our inertial reference frames.
Based on the first postulate of the Special Relativity we are not supposed to detect uniform motion within the inertial reference frame without any signal from outside. If we succeed then we can state that the first postulate is not valid.

Here comes Jaroslav Kopernicky’s observation and discovery. The steel ball on the wheel on the train/car/hovercraft is moving in a cycloid motion when observed from the ground reference frame which brings acceleration and deceleration of a body with its own inertia into the picture! This is a top view of a cycloid that describes our experiment:

Figure 3: The Cycloid

## Nomenclature summary

$\cap$  –  cycloid subscript referencing cycloid equations and values in the ground inertial reference frame
$\circ$  –  circle subscript referencing circle equations and values in the moving inertial reference frame
$\theta$  –  the angle of the rotation of the circle in radians
$\omega$  ;  the angular velocity of the circle
$d\theta=dt$  ;  $t$ corresponds to the angle $\theta$ therefore $\omega=1$ in our plots
$v$  –  the velocity of the moving frame; $v=1$ in our plots
$r$  –  the radius of the wheel; $r=1$ in our plots
$c$  –  the speed of light

## The analysis

A circle of radius r, consisting of the points (x, y);

In the ground reference frame – Eq.1?
$x_\cap=vt-r\sin (\omega t)$
$y_\cap=r-r\cos (\omega t)$
The textbook cycloid equations.
In the moving reference frame – Eq.1O
$x_\circ=-r\sin (\omega t)$
$y_\circ=r-r\cos (\omega t)$
The textbook circle equations.

The first derivatives are the velocities;

In the ground reference frame – Eq.2?
$v_{x\cap}$$\space v-r\omega\cos(\omega t)$
$v_{y\cap}$$\space r\omega\sin(\omega t)$
In the moving reference frame – Eq.2O
$v_{x\circ}$$\space -r\omega\cos(\omega t)$
$v_{y\circ}$$\space r\omega\sin(\omega t)$

The and are components of the tangential velocity vector and the calculation of the magnitude is a straightforward vector sum.

In the ground reference frame – Eq.3?
In the moving reference frame – Eq.3O

Now we are at one of the most intriguing and puzzling crossroads in physics. There are three ways to continue our analysis of the cycloid accelerations:
1 – calculate the kinetic energy of a particle as observed in the ground and moving reference frames.
2 – calculate the derivative of to obtain the magnitude of the acceleration of such a particle.
3 – calculate the derivatives of and and combine them to obtain the magnitude of the acceleration.

## 1. The kinetic energy analysis

“The kinetic energy EK of the particle is defined as
Eq.4
and is the total work which must be done on the particle to bring it from a state of rest to a velocity v”.[2]

The kinetic energy of the steel ball in both inertial reference frames is;

In the ground reference frame – Eq.5?

or

This is a combination of the kinetic energy of circular motion plus time-varying velocity due to cycloidal motion in the ground reference frame.
In the moving reference frame – Eq.5O

Textbook result for the kinetic energy of particle moving in a constant circular motion.

The kinetic energy change predicts some work to be done on the ball in the ground reference frame as a function of time changing velocity, in other words we expect force magnitude changes based on the work-energy principle but we do not expect any force magnitude changes in the moving reference frame since the steel ball moves in a uniform circular motion therefore there is no work done on the ball from the moving reference frame point of view.
This is a discrepancy between the “two inertial reference frames” and it’s a clear contradiction to the first postulate of the Special Relativity.
Many physicists expressed their opinion how the Special Relativity is straightforward and simple. The three steps: position, velocity, kinetic energy that undermine the first postulate of SR are arguably even more simple.

These are unprecedented claims and it’s a perplexing situation that demands experimental proof to resolve this puzzle. How come nobody noticed it before? The answer is hidden in the Eq.5?!

When condition
Cnd.1
is true then the kinetic energy in the ground reference frame is close to being the same as the kinetic energy in the moving reference frame because being orders of magnitude less than results in contributing very little to the velocity change therefore in this case we can write
Eq.6
The equations Eq.6 and Eq.5O are equal. No force magnitude changes are expected in both inertial reference frames.

When condition
Cnd.2
is true then being orders of magnitude less than results in contributing very little to the velocity change therefore in this case we can write
Eq.7
The kinetic energy is constant in the ground reference frame and no force magnitude changes are expected.

An observer in the moving reference frame has two options.

If the velocity can be determined by a signal from outside Cnd.2O-1

then the observer will come to the same conclusion as the observer in the ground reference frame – Eq.7.

If the velocity can not be determined Cnd.2O-2

then the observer in the moving reference frame sees only rotation and the Eq.5O applies.
Nevertheless no force magnitude changes are expected for the observer in the moving reference frame in both instances.

When condition
Cnd.3
then the Eq.5? and Eq.5O are not equal. The accelerometer based on the steel ball does not ‘care’ about any reference frames. It reports only one set of data to both observers, in the ground reference frame and in the moving reference frame.
What data do we get? Is there a change in the force magnitude as expected per Eq.5? or there is no change as per Eq.5O?

The best real life examples that point to the correct answer are related to one type of the critical speed that rises from rotating systems with an imbalance in them that are in a linear motion when . An observed vibration as also seen in our experiments implies that the data is expected to correlate with Eq.5?. The ground reference frame is the preferred one in this case.
Another excellent experiment/proof might be satellite flyby anomalies (Wikipedia)[5]. The Earth’s and the satellite speeds in the Solar system are in the same range therefore the satellite accelerations are also a function of the Earth’s and the satellite speeds and an angle between the Earth’s and satellite orbital planes.

Our experiment

Video: The experiment setup[4]

The steel ball – 226g is placed 0.14m from the wheel center.
The circumference is;

Our force sensors voltage range 0V – 5V represents force range from 0N – 20N.
Assuming the first postulate of the Special Relativity is valid we expect the centrifugal force prior to the motion being equal to the centrifugal force during the uniform translatory motion with the exception of a noise.

—– RUN #1 —–
[V] [0.01s]
Chart 1: No wheel rotation, only linear motion

The hovercraft speed 0.5m/s (10V on the pulley power supply that pulls our hovercraft).
Blue line – the wheel tangential force sensor readings
The linear motion starts at around 8.7s; there is no motion prior to that point in time; the box is at the front of the hovercraft therefore the radial force sensors registered the force spike; the linear motion ends at around 18.5s; the bump in the middle of the track (limited options for the experiment track) is clearly visible but otherwise it’s just a small noise.

Video Run #1: The linear motion, no wheel rotation[4]

—– RUN #2 —–
[V][0.01s]
Chart 2: Low angular velocity of the wheel and linear motion

The hovercraft speed 0.5m/s (10V on the pulley power supply that pulls our hovercraft).
Blue line – the wheel tangential force sensor readings
The frequency of rotation is 0.36 rps ~ 0.32m/s
The linear motion starts at around 8.4s and ends at around 18.4s; the wobble comes from imbalanced wheel; the chart shows that there is very little difference between the stationary rotation and the rotation during the linear motion. We can see only a noise.
The RUN #2 is in the range of the Cnd. 3 of the kinetic energy analysis when . The inertia of the steel ball though does not generate the centrifugal force that is big enough and detectable over the noise.

Video Run #2: The linear motion, the slow wheel rotation[4]

—– RUN #3 —–
[V][0.01s]
Chart 3: Medium angular velocity of the wheel and linear motion

The hovercraft speed 0.5m/s (10V on the pulley power supply that pulls our hovercraft).
Blue line – the wheel tangential force sensor readings
The frequency of rotation is 0.59 rps ~ 0.52m/s
The linear motion starts at around 5.7s and ends at around 15.7s; the chart still does not show much difference between the stationary rotation and the rotation during the hovercraft linear motion.
The RUN #3 is in the range of the Cnd. 3 of the kinetic energy analysis when . The inertia of the steel ball still does not generate the centrifugal force that is big enough and detectable over the noise.

Video Run #3: The linear motion, the medium wheel rotation[4]

—– RUN #4 —–
[V][0.01s]
Chart 4: High angular velocity of the wheel and linear motion

The hovercraft speed 0.5m/s (10V on the pulley power supply that pulls our hovercraft).
Blue line – the wheel tangential force sensor readings
The frequency of rotation is 1.11 rps ~ 0.98m/s
The linear motion starts at around 6.4s and ends at around 16.4s; the selected chart areas point to a voltage dips that are not supposed to be there if the first postulate of the Special Relativity was true and readings would show noise filled line similar to the one prior to the linear motion starting at 6.4s as it is the case in the previous runs.
The explanation of our measurement follows in the next chapter and the Figure 14 is the best fit to show what is happening.
Our force sensors voltage range 0V – 5V represents force range from 0N – 20N. The chart shows we operate in 0.1N to 1N range and this is well above any Coriolis force effect observable in Toronto, Ontario, Canada.
The RUN #4 is in the range of the Cnd. 3 of the kinetic energy analysis when . The inertia of the steel ball does generate the centrifugal force that is big enough and detectable over the noise.

Video Run #4: The linear motion, the fast wheel rotation[4]

## 2. The derivative of v_{mag} analysis

Now we are back at the crossroads and let us investigate the derivative of .
The equations Eq.3? and Eq.3O give us a starting point. They show that the cycloid tangential velocity magnitude in the ground frame changes, it is a function of . The circle velocity magnitude in the moving frame is constant.

Figure 4: The cycloid and the tangential velocity magnitudes plotted on the same graph for convenience

This is a very important finding because velocity derivatives are not going to be equal as well. If the tangential velocity magnitude changes in the ground frame then what is the tangential acceleration magnitude in the ground frame? What is the tangential acceleration magnitude in the moving frame?

In the ground reference frame Eq.8?

In the moving reference frame Eq.8O

Here is a plot with the tangential acceleration magnitudes.

Figure 5: The cycloid, the tangential velocity magnitude and the tangential acceleration magnitude

The points 0, 2?, 4?, … represent cusps. The acceleration is undefined here. The reason is a change from the acceleration -1 (the maximum deceleration) to the maximum acceleration 1. This can happen only through a singularity. The acceleration changes to deceleration at the ?, 3?, 5?, … points.

The cycloid tangential acceleration is a function of t as it is described in a previous paragraph and the circular motion has 0 tangential acceleration. Let us rename to and to .

In the ground reference frame Eq.8?
In the moving reference frame Eq.8O

The trajectories in both reference frames are curved so in order to obtain the full picture we need to analyze the radial acceleration as well.

The radial acceleration is responsible for the velocity vector rotation. The radial acceleration keeps the velocity vector always tangential to the curved trajectory, it is a function of the velocity and the radius.
Eq.9
Radius is defined by a curvature of the trajectory.
Eq.10
The curvature[2] is
Eq.11

The cycloid curvature is Eq.12?

The circle curvature is Eq.12O

In the ground reference frame Eq.13?

In the moving reference frame Eq.13O

Let us recap the equations 3, 8, 13.

In the ground reference frame
Eq.3?, Eq.8?, Eq.13? and new Eq.14?

Eq.14?

In the moving reference frame
Eq.3O, Eq.8O, Eq.13O and new Eq.14O

Eq.14O

The in Eq.14? is the total net cycloid acceleration in the ground reference frame. The in Eq.14O is the total net circle acceleration in the moving reference frame.

The cycloid acceleration in the ground reference frame is

Eq.15?

We come to a conclusion that both accelerations, cycloid and circular are the same. It appears that cycloid tangential and radial accelerations complement each other. When the tangential acceleration decreases then the radial increases and the other way around as it can be seen in the Figures 8 – 14. Their vector sum ends up in the same vector as we have in the circular motion. It appears that based on Eq.14? and Eq.14O being equal there is a conflict with the kinetic energy analysis equations Eq.5? and Eq.5O. How to solve this puzzle? The answer is the inertia!

Anybody advocating Eq.14? and Eq.14O being equal has to answer the following question. Where is the centrifugal acceleration? Why is it omitted? As Jaro Kopernicky points out the inertia is inherent in the mass. It can not be ignored!

The centrifugal and centripetal accelerations are in 90o angle to the tangential acceleration. They do not contribute to the work done on the ball but both forces are present, they are inseparable.
If we imagine our steel ball rolling on a table then the gravity and the table with its legs pushing back against the gravity do not perform any work but they are inseparable as well. Any change in the kinetic energy is a result of the horizontal component of a force applied on the steel ball.
Applying the same logic we come to a conclusion that the cycloid kinetic energy change in the ground reference frame comes from the tangential acceleration only, therefore there is no discrepancy between the kinetic energy analysis and the derivative of analysis. The analysis shows the difference between the acceleration/force in both inertial reference frames in accordance with the kinetic energy analysis once the inertia of the free body is properly accounted for and undermining the first postulate of the Special Relativity as well.

Rigid body, string and gravity

The Figure 6 shows direction of the tangential, the centripetal and the centrifugal accelerations.

Figure 6: The cycloid accelerations for a rigid body

The rigid body holds point B in place and it does not allow it to accelerate along the direction of the tangential acceleration.

If the ball/body B was attached to the center with a string or it was flying through a space around a planet then there are only two forces acting on the ball/body. It is:

• the string or gravity from the center
• the centrifugal force caused by the inertia

The end result is the tangential acceleration that will increase the velocity for the ball on the string and it’s trajectory will become more prolate/spiral. The body flying through a space will increase its velocity and its trajectory will be elliptic.

The Figure 7 shows direction of the tangential and the centrifugal accelerations. There is nothing that can hold the ball/body back therefore they will speed up along the tangential acceleration.

Figure 7: The cycloid accelerations for a string and gravity

Prolate and Curtate Cycloids

Let us see what is happening when the trajectory is not a perfect cycloid.
The plots when the velocity of the moving frame is greater than the angular velocity of the rotating circle which is equal to 1m/s.

Figure 8: The velocity 1.1m/s Red-cycloid; Black-tangential acceleration; Green-radial acceleration

Figure 9: The velocity 1.2m/s Red-cycloid; Black-tangential acceleration; Green-radial acceleration

Figure 10: The velocity 1.5m/s Red-cycloid; Black-tangential acceleration; Green-radial acceleration

Figure 11: The velocity 2.0m/s Red-cycloid; Black-tangential acceleration; Green-radial acceleration

The plots when the velocity of the moving frame is smaller than the angular velocity of the rotating circle which is equal to 1m/s.

Figure 12: The velocity 0.9m/s Red-cycloid; Black-tangential acceleration; Green-radial acceleration

Figure 13: The velocity 0.8m/s Red-cycloid; Black-tangential acceleration; Green-radial acceleration

Figure 14: The velocity 0.5m/s Red-cycloid; Black-tangential acceleration; Green-radial acceleration

When we observe the Figure 14 and the area from 2.5 up to 3.7 we can see a maximum deceleration around 2.6 and maximum acceleration around 3.6 when a point on the wheel travels through ‘the loop’ at ?. This explains the Chart 4 readings.

## 3. The derivatives of v_x and v_y analysis

Now we are back at the crossroads and let us investigate the derivative of and for both reference frames.
The equations Eq.2? and Eq.2O give us a starting point.

In the ground reference frame – Eq.16?
$a_{x\cap}$$\space r\omega^2\sin(\omega t)$
$a_{y\cap}$$\space r\omega^2\cos(\omega t)$
In the moving reference frame – Eq.16O
$a_{x\circ}$$\space r\omega^2\sin(\omega t)$
$a_{y\circ}$$\space r\omega^2\cos(\omega t)$

Yes, the equations are equal but having seen the kinetic energy analysis now we know that these equations can be used only in cases when $v$ and $r\omega$ are not in the same range.

## The conclusion

The steel ball – accelerometer with the limited freedom of motion does not have a clue about the circular motion or the translatory motion of inertial reference frame. It follows the cycloid tangential and radial accelerations. These are as real as they can get. The centrifugal force as the consequence of the inertia and the centripetal force of the wheel cancel each other and the tangential force change is the reason for the kinetic energy fluctuation from 0 at the 0, 2?, … points to its maximum at the ?, 3?, … points in the case of an ideal cycloid motion.
Jaroslav Kopernicky’s claim seems to be proven. He calls the tangential acceleration changes a mobiloscopic effect. We have a device that can detect an uniform translatory motion!
The first postulate of the Special Relativity is proven to be false.

The logical consequence is that there is always a higher/preferred frame of reference up to the first absolute frame of reference. The Earth orbits the Sun; The Sun orbits the center of the Milky Way; The Milky Way flies in a cluster with other galaxies;…
The Universe spins around an axes.
The Big Bang started with a huge spin.

Inertial Tendencies of the Massive Body
Following the Path of the Cycloid.

Inertia, accordingly to Newton’s views, appears inherent and independent for every single particle.
This assumption raises an importance of the behavior – inertial changes – of the body having mass, moving on the cycloid path.
One of early examples was the toy carousel equipped with two pendulums positioned diagonally on the carousel. When the rotating carousel moves in the plane of rotation, pendulums of equal length and weight fly in different levels; One advancing with the movement of the carousel flies higher, while one moving momentarily in the opposite direction flies lower. Each pendulum moves on its own cycloid path.
When carousel rotates stationary, pendulums fly at the same level.
The mercury switch replacing the pendulum makes an intermittent (On – Off) connection when the rotating carousel moves. But when carousel rotates stationary – not moving laterally – the switch keeps only one position.
This phenomenon provides possibilities of indication and measurements of the uniform movement, its speed and direction. It also has potential to indicate and explain observations in astronomy (zone convenient for life) and geology (tides, prediction of earthquakes, etc).
This Jano Onderco’s analysis and connected experiments are welcomed contributions to the further understanding of the phenomenon.

Jaroslav J. Kopernicky
January 16th, 2014

If there is any traction to Jaroslav Kopernicky’s Mobiloscopic Effect, perhaps it can be explained as follows. It has been surmised that a rotating body produces its own reference frame. Since the rotation is at all times around a single point even though that point may be moving with constant velocity with respect to the external world, there is a radially-directed acceleration. Even if this rotating body is moving with constant or time-varying motion with respect to another (inertial) frame, at every instant it is producing its own (new) “accelerated” reference frame. Therefore, the turntable on the track is producing at every instant, a local “accelerated” frame with its own radially-directed acceleration. The Mobiloscopic system is composed of an “accelerated” frame moving with respect to an inertial frame. This may allow the detection of constant motion of the “accelerated” frame with reference to the inertial frame. The term “accelerated” is in quotes here because it is not clear whether a rotating body actually produces a true non-inertial/accelerated reference frame. In space, a rotation in one direction has to be countered by a similar rotation in the opposite direction, accounting for moment of inertia. Whether one rotation cancels the creation of an “accelerated” frame by the opposite rotation is not clear. Notwithstanding, due to the rotationally-derived nature of the “accelerated” frame, the detected signal of constant motion will be cyclic in nature and possibly cycloidal as predicted by Kopernicky.
Jan Onderco has put considerable thought and effort into understanding the dynamics and energetics of this deceptively simple mechanical system. The detection of constant linear motion in this way would have profound implications technologically as well as scientifically and philosophically if true.

George Hathaway
April 22nd, 2014

## Thank you note to Hathaway Consulting Services

George and Caryl
Thank you for all your exceptional support and help.
This project would not see the light of day without your generosity.
It is hard to find words to express our gratitude.

Thank you very much,
Jaro & Jano

## References

[1]Wikipedia – Postulates of special relativity; from http://en.wikipedia.org/wiki/Postulates_of_special_relativity
[2] J.L.Meriam, L.G.Kraige Engineering Mechanics Volume 2 Dynamics 7th edition, Page 159, 2012 John Wiley & Sons, Inc.
[3] 18.02A Topic 22: Vector derivatives: velocity, curvature; from http://web.mit.edu/dvp/18.01A/topic22.pdf
[4] The reference for the PDF readers to our web page http://theelectromagneticnatureofthings.com/the-special-relativity/
[5]Wikipedia – Flyby anomaly; from http://en.wikipedia.org/wiki/Flyby_anomaly

Thank you, to all the people behind the GNU Octave software, FlexiForce® Sensors, Sparkfun accelerometers and Raspberry Pi.


text:
Last month the U.S. renewable energy sector generated more energy than coal-fired plants, the first time this has happened. Renewable energy includes hydro, biomass, wind, solar and geothermal. The U.S. Energy Information Administration statistics project renewable energy will sporadically top coal again this year and next. Natural gas replaced coal in 2015 as the most popular source of power in the country. Some of the renewables’ impact is due to seasonal issues because some coal plants are taken off-line for maintenance in spring and fall when demand is low. It’s forecasted that renewable energy will produce 18% of U.S. electricity in 2019 and nearly 20% in 2020. Imagine what it would look like if there was more emphasis on renewable energy sources.

https://www.mnn.com/earth-matters/energy/blogs/renewable-energy-track-top-coal-first-time-us

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Asymptotically, you should be able to interchange the role of $Y_t$ and $X_t$ and estimate lets say $$X_t =\alpha^* + \beta^* Y+ u^*_t$$ where $\alpha^* = -\alpha/\beta$ and $\beta^*= 1/\beta$ and still get cointegrated relationship. However, crucial caveat here is that this holds only asymptotically and only for $R^2$ close to unity. In finite samples you ...

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I know this is a old post, but since I found it and I know the answer I might as well share it. I would go further than the first component, but it also depends on how much of the variance is actually explained within the component. There is two approaches that I use when determining which components I am using. The rule of thumb is to use and component ...

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The reason for that is that 2SLS does not solve omitted variable bias without its cost. First, for every omitted variable you would have to find suitable instrument. Finding instruments is incredibly hard so why would you purposely create more omitted variable bias? Unless you are trying to challenge yourself for fun or doing it as a practice exercise for ...

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1 - The normal form does not match the word problem. The profit of the DC firm when the other firm plays C is 1, not 2. 2 - Say $p$ is the probability given to C by player 1, then in equilibrium, player 2 is indifferent between his two strategies. Hence, $p - (1-p) = p$ which only has one solution. 3 - You need to specify what you mean by cooperation. ...

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Hint on the non-existence of mixed strategy NE: Suppose the row player uses a mixed strategy that plays C with probability $p$ and DC with probability $1-p$. What's the column player's best response to such a strategy? Does the response depend on the value of $p$? Based on this, what can you say about the optimality of any mixed strategy that assigns ...

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Introductory Econometrics: A modern approach by Wooldridge Econometric Theory and Methods - Davidson and MacKinnon Econometric Analysis of Cross Section and Panel Data. by Jeffrey Wooldridge Mostly Harmless Econometrics Microeconometrics - Trivedi

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This is simply saying that the falling returns are usually correlated with high variance (i.e. sharp movement in prices). This is not very surprising... you can see from the graph of S&P vs VIX below. Whenever the stock market plunges, volatility shoots up. What @heh said was right. You need to compare assets that are more or less substitutable. For ...

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I am tempted to deviate from others' answer and suggest that the best Econometrics textbook for someone with a strong enough mathematical background might not a "mathematically-minded" econometrics textbook, but rather a textbook that focuses on "empirical methods and economics research", the best of which at the moment seems to remain: Mostly Harmless ...

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Depending on the context this is sometimes called Jackknife resampling In statistics, the jackknife is a resampling technique especially useful for variance and bias estimation. The jackknife pre-dates other common resampling methods such as the bootstrap. The jackknife estimator of a parameter is found by systematically leaving out each ...

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\begin{align*} \mathbb{E}\left[ \epsilon_t \vert X \right] &= \mathbb{E}\left[\epsilon_t \vert X_1, \dots, X_t, \dots, X_n \right]\\ &= \mathbb{E} [Y_t - X_t'\beta^0 \, \vert X_1, \dots, X_t, \dots, X_n] \\ &= \mathbb{E} [Y_t - X_t'\beta^0 \, \vert X_t] \quad \text{since $\{Y_t,X_t\}$ is iid.} \\ &=\mathbb{E}\left[ \epsilon_t \vert X_t \right]...

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The ar.ols() function in R defaults to order $10log_{10} (N)$. This implies that, to catch the 30th time lag, you need at least 1000 observations. That seems like a lot, but bear in mind that you are then running OLS on 30 features (lags t-1 to t-30). All else equal, it is not clear why 30 is the magic number for you here, and you can't just drop the more ...

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They will simply be sucked up into one variable. To "fix" this, you could simply drop all of Brazil's characteristics and leave a constant in: $\ln(EXP_D) = c + \beta_1 \ln(GDP_D) + \beta_2 \ln(POP_D) + \ldots$ Edit Just understood the question from the comments below. If you have only a cross-sectional data, you cannot distinguish the effects of GDP, ...

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