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A Series of Lectures on Fine-Tuning in Biology

A recent lecture and a series of interviews has been posted online where cosmologist George F.R. Ellis discusses the issue of fine-tuning in biology at considerable length and in considerable detail. Of course, the larger theme here is that to discuss and understand things like Darwinian evolution properly, one needs to have an understanding of the underlying physics, as it is laws of physics that allow life to emerge and for Darwinian evolution to occur in the first place. Here are the lectures:

Misinformation on Sikhism

Even though this posting is a bit different than my usual ones and is outside the scope of this blog, I thought that as a Sikh myself, I have stayed too silent on several issues with regards to the Sikh community, and certain principles of the Sikh religion, that are seemingly unknown to those both inside and outside of the Sikh community. I will address here some common misconceptions and misinformation about Sikhs that are both being spread inside and outside of the Sikh community.

1. Sikhism is NOT a hybrid of Islam and Hinduism: Sikhism is a unique religion with a unique origin beginning with the teachings of the first Sikh Guru, Guru Nanak Dev Ji. Guru Nanak formed the religion to be uniquely different from Hinduism and Islam, as he opposed many of the practises common in those religions.
2. Sikhs are not to cut their hair or trim their beards. This is perhaps the most common fact that is blatantly missed by those members of the Sikh community that wish to cut their hair and propagate this misinformation to defend their actions as being accepted in Sikhism. This is a very wrong ideology for several reasons:      Guru Gobind Singh Jee, the 10th Guru of the Sikhs explicitly described the form of a Sikh in Persian:

Ik Onkaar Sri Waheguru Jee Kee Fateh || Sri Mukhvaak PaatShaahee Dasvee||

Nishaanay Sikhi Ee Haroof Panj Kaaf|| Hargiz Na Baashad Ee Panj Muaf ||

Karra Kaardo Kachh Kanghaa Bida || Bila Kesh Haych Asat Jumleh Nishaa ||

Haraf Haae Kaat Asat Ee Panjkaaf || Bi Daanand Baavar Na Goyam Khilaaf ||

HukaaJaamat Halaalo Haraam || Baachishay Hinaa Kardaroo Sayaam Faam||

Note that the usual argument that this is for the Khalsa, and not Sikhs, is patently false, as Guru Sahib explicitly says “Nishaanay Sikhi”, and not “Nishaanay Khalsa”. This point is therefore a moot point.

Guru Gobind Singh Sahib also in his Hukam to the Afghanistan Sikh sangat said:

Tusi Khande da Amrit Panja to lena

Kachh, Kirpan da visah nahee karna

Sarb Loh da kara hath rakhna

Dono vakat kesa dee palna karna

Sarbat sangat abhakhia da kutha

Khave naheen, Tamakoo na vartana

Bhadni tatha kanya-maran-vale so mel na rakhe

Meene, Massandei, Ramraiye ki sangat na baiso

Gurbani parhni…Waheguru, Waheguru japna

Guru kee rahat rakhnee

Sarbat sangat oopar meri khushi hai.
Patshahi Dasvi

Jeth 26, Samat 1756

These are not my opinions. This is explicitly the command of Guru Gobind Singh Jee. Further, the aforementioned passages also explicitly state that Sikhs are not to drink alcohol or smoke. It is truly puzzling and embarrassing why drinking alcohol has become somewhat synonymous with Sikhs, particularly, in the Punjab region.

Further, Bhai Desa Singh Jee explicitly writes the following on trimming beards:

dhaarrhaa mushh sir kaes banaaee || hai eih dhrirrh jih prabhoo razaaee || maett razaaee j sees mu(n)ddaavai || kahu thae jag kaisae har paavai

It could not be more clear, that Sikhs are to emphatically not cut their hair or trim their beards.

3. Sikhs do not celebrate Diwali. There are many Sikhs around the world that insist on celebrating Diwali, the Hindu festival of lamps. Some have even justified this action by conflating the Bandhi Chorr divas related to Sri Guru Hargobind Sahib with the Diwali day. This is also wrong for the following reasons:

For years now, I have heard this constant story of how it is acceptable for Sikhs to celebrate Diwali as per the Hindu traditions of lighting lamps, etc… Further, Raagis and Bhai Sahibs in Gurdwaras have conflated Bandhi Chorr Diwas with lighting lamps as per Hindu Diwali traditions. They further support these ideas with the supposed Vaar from Bhai Gurdas Jee, in which they ironically only mention and repeat the first line! : “deewaalee dee raath dheevae baaleean”. Of course, just by reading this line, it would suggest that the aforementioned actions are justified. But taking one line completely out of context leads one to these conclusions. A full reading of Bhai Gurdas Jee’s Vaar on the Diwali matter which given the timeframe is also a historical first-hand account suggests that Sikhs are to practice completely the opposite and in fact, lighting lamps is contrary to Gurmat. The full Vaar’s transliteration is below:
Vaars Bhai Gurdaas 19-6

diwali dee raath dheevae baaleeani

thaarae jaath sanaath a(n)bar bhaaleean

fulaa(n) dhee baagaath chun chun chaaleean

theerathh jaathee jaath nain nihaaleean

har cha(n)dhuree jhaath vasaae ouchaaleean

The essence of this Vaar is in every line after the first. Namely, in the third, fourth, and fifth lines, Bhai Gurdas Jee compares those that celebrate Diwali by lighting lamps akin to those who go on long pilgrimages to find God, and to those who search for God by worshipping the stars, or things in nature, etc… All contrary to Gurmat by a simple reading of Japjee Sahib! Indeed, Bhai Sahib Jee in the last line clearly states that a person of Gurmat does not practice any of these things, which he declares to be temporary and pointless.

So, there you have it. A simple reading of the full Vaar changes the entire context of the “importance” of Diwali in Sikhism. I doubt many Sikhs will read this posting with sincerity, but someone has to speak the truth.

4. Sikhs Do Not Eat Meat: Sikhs most certainly do not eat meat. Despite this, many Sikhs continue to insist that eating meat is permissible as long as it is not Halal, etc. This is also wrong for the following reasons:

The Sikh Gurus including the Guru Granth Sahib Jee (the present living Guru of the Sikhs) very explicitly discuss how eating meat is not for Sikhs, some examples below:

1. Guru Granth Sahib – Page 1374 – “Kabeer Khoob Khaanaa Keecharee Jaa Mai Amrit Lon, Heraa Roti Kaaranay Galaa Kataavai Kaun”.

2. Guru Granth Sahib – Page 140 – “Jay Rat Lagay Kaparay Jaamaa Hoay Paleet, Jo Ray Peeveh Maansaa Tin Kio Nirmal Cheet”.

3. The essence of why a Sikh cannot be satisfied with “Jhatka” and simply opposed to Halal is due to Guru Naanak Dev Ji, Page 468 on SGGSJ: “Daaiaa Jaanay Jee Kee Kichh Pun Daan Karay”

Several other key points are as follows:

Anpas Ko Munwar Kar Thaapoh, Kaa Ko Kaho Kasaaee. (SGGS 1103)

5. “Bed Kateb Kaho Mat Jhoothhay, Jhoothhaa Jo Na Bichaarey.

Jo Sabh Meh Ek Khudai Kahat Ho,To Kio Murghi Maarey” (SGGS 1350)

6. “Rojaa Dharey, Manaavey Mlah, Svaadat Jee Sanghaarey.

Aapaa Deldi Avar Nahin Dekhey,Kaahey Kow Jhakh Maarey” (SGGS 1375)

7. “Kabir Jee Jo Maareh Jor Kar,Kaahtey Heh Ju Halaal.

Daftar Daee Jab Kaadh Hai, Hoegaa Kaun Havaal” (SGGS 1375)

8. “Kabir Bhaang, Machli, Surapaan Jo Jo Praanee Khahey.

Tirath, Barat, Nem Kiaye Te Sabhay Rasaatal Jahey” (SGGS 1376)

9. “Kabir Khoob Khaana Khichri, Ja Meh Amrit Lon

Heraa Rotee Kaarney Galaa Kataavey Kon” (SGGS 1374)

These are examples/Hukams explicitly from Sri Guru Granth Sahib Jee going back to the days of Bhagat Kabeer Jee, to Guru Naanak Sahib, all the way to Guru Gobind Singh Sahib. This is factual evidence that your claims are incorrect, and any such claims made by Sikhs that only “Halal” is forbidden, is simply wrong, as these Sikhs do not have an answer/simply ignored such aforementioned verses from Guru Granth Sahib Jee, because of a desire to not leave meat.

5. Women in Sikhism

In Sikhism, women are completely equal to men, there is zero tolerance for those men that would treat women poorly or lower than them. Indeed, according to Sri Guru Granth Sahib Jee, women are to be treated higher than men:

From woman, (man) is born; within woman, (man) is conceived; to woman he is engaged and married. With woman

(man) establishes friendship; through woman, the future generations come. When woman dies, (man) seeks another

woman; because of woman, (man) becomes related (to other people – ਲੈਣ-ਦੇਣ ਦੇ ਸਾਰੇ ਸੰਸਾਰਕ ਬੰਧਾਨੁ, etc.). From her,

(even) kings are born; so why call her bad? From woman, woman is born; without woman, there would be no one at all.  (sggs 473).

6. Sikhs and Rakhdi/Raksha Bandhan Ceremony

There are many Sikhs around the world that insist on Rakhdi/Raksha Bandhan ceremony. Sikhs absolutely are not supposed to participate in this festival. On Raksha Bandhan, sisters tie a rakhi (sacred thread) on her brother’s wrist. This symbolizes the sister’s love and prayers for her brother’s well-being, and the brother’s lifelong vow to protect her.

This very concept implies that women are weaker than men and therefore need their protection. This is in direct conflict with Guru Nanak Dev Ji’s philosophy regarding women as described above, and very much against the Khalsa Rehat, where women and men take amrit equally! Therefore, the idea that women should tie a thread to their brother’s wrist for “protection” is very much against the concept of equality, and ignores the roles played by Sikh women in history that didn’t need any man for protection. Some examples are:

• Mata Gujri Jee
• Mata Ganga Dev Jee
• Mai Bhago
• Mata Jito Jee
• Bibi Rajni
• Mata Kishan Kaur

And many, many others!

Conclusions:

The purpose of this piece was simply to stop the misconceptions related to the aforementioned points from spreading.

Further, the majority of us live in free societies, where freedom of religion is the expectation and the norm. I am a big believer in this concept, and fully appreciate having such a law in existence. This of course means that people are free to practice their religion as they see fit. However, while this principle must absolutely be firm, I question what it means for someone to call themself a Sikh on one hand and disagree with the tenets that the Sikh gurus established. This is more ironic because Sikh means “disciple”. So, you can call yourself a Sikh, but if you don’t follow the principles of the Sikh religion, what “religion” are you following?

Mathematical Origins of Life

The purpose of this post is to demonstrate some very beautiful (I think!) mathematics that arises form Darwinian evolutionary theory. It is a real shame that most courses and discussions dealing with evolution never introduce any type of mathematical formalism which is very strange, since at the most fundamental levels, evolution must also be governed by quantum mechanics and electromagnetism, from which chemistry and biochemistry arise via top-down and bottom-up causation. See this article by George Ellis for more on the role of top-down causation in the universe and the hierarchy of physical matter. Indeed, my personal belief is that if some biologists and evolutionary biologists like Dawkins, Coyne, and others took the time to explain evolution with some modicum of mathematical formalism to properly describe the underlying mechanics instead of using it as an opportunity to attack religious people, the world would be a much better place, and the dialogue between science and religion would be much more smooth and intelligible.

In this post today, I will describe some formalism behind the phenomena of prebiotic evolution. It turns out that there has been a very good book by Claudius Gros and understanding evolution as a complex dynamical system (dynamical systems theory is my main area of research), and the interested reader should check out his book for more details on what follows below.

We can for simplicity consider a quasispecies as a system of macromolecules that have the ability to carry information, and consider the dynamics of the concentrations of the constituent molecules as the following dynamical system:

$\boxed{\dot{x}_{i} = W_{ii}x_{i} + \sum_{j \neq i}W_{ij}x_{j} - x_{i} \phi(t)}$,

where $x_{i}$ are the concentrations of $N$ molecules, $W_{ii}$ is the autocatalytic self-replication rate, and $W_{ij}$ are mutation rates.

From this, we can consider the following catalytic reaction equations:

$\boxed{\dot{x}_i = x_{i} \left(\lambda_{i} + k_i^j x_j - \phi \right)}$,

$\boxed{\phi = x^{k}\left(\lambda_{k} + \kappa_k^j x_j\right) }$,

$x_i$ are the concentrations, $\lambda_i$ are the autocatalytic growth rates, and $\kappa_{ij}$ are the transmolecular catalytic rates. We choose $\phi$ such that

$\boxed{\dot{C} = \sum_i \dot{x}_i = \sum_i x_i \left(\lambda_i + \sum_j \kappa_{ij}x_{j} \right) - C \phi = (1-C)\phi}$.

Clearly:

$\lim_{C \to 1} (1-C)\phi = 0$,

that is, this quick calculation shows that the total concentration $C$ remains constant.

Let us consider now the case of homogeneous interactions such that

$\kappa_{i \neq j} = \kappa$, $\kappa_{ii} = 0$, $\lambda_i = \alpha i$,

$\boxed{\dot{x}_{i} = x_{i} \left(\lambda_i + \kappa \sum_{j \neq i} x_{j} - \phi \right)}$,

which becomes

$\boxed{\dot{x}_i = x_i \left(\lambda_i + \kappa - \kappa x_i - \phi\right)}$.

This is a one-dimensional ODE with the following invariant submanifolds:

$\boxed{x_{i}^* = \frac{\lambda_i + \kappa - \phi}{\kappa}}$,

$\boxed{x_i^* = 0, \quad \lambda_i = N \alpha}$.

With homogeneous interactions, the concentrations with the largest growth rates will dominate, so there exists a $N^*$ such that $1 \leq N^* \leq N$ where

$\boxed{x_i^* = \frac{\lambda_i + \kappa - \phi}{\kappa}, \quad N^* \leq i \leq N}$,

$\boxed{0, \quad 1 \leq i < N^*}$.

The quantities $N^*$ and $\phi$ are determined via normalization conditions that give us a system of equations:

$\boxed{1 = \frac{\alpha}{2\kappa} \left[N(N+1) - N^*(N^* - 1)\right] + \left[\frac{\kappa - \phi}{\kappa}\right] \left(N + 1 - N^*\right)}$,

$\boxed{0 = \frac{\lambda_{N^*-1} + \kappa - \phi}{\kappa} = \frac{\alpha(N^* - 1)}{\kappa} + \frac{\kappa - \phi}{\kappa} }$.

For large $N, N^*$, we obtain the approximation

$\boxed{N - N^* \approx \sqrt{\frac{2 \kappa}{\alpha}}}$,

which is the number of surviving species.

Clearly, this is non-zero for a finite catalytic rate $\kappa$. This shows the formation of a hypercycle of molecules/quasispecies.

These computations clearly should be taken with a grain of salt. As pointed out in several sources, hypercycles describe closed systems, but, life exists in an open system driven by an energy flux. But, the interesting thing is, despite this, the very last calculation shows that there is clear division between molecules $i = N^*, \ldots N$ which can be considered as a type of primordial life-form separated by these molecules belonging to the environment.

Sikhs in the Army

There has been much noise in recent days of Sikhs wanting to enlist in the U.S. Army and being unable to do so because they insist on following their religious mandates of uncut hair, an untrimmed beard, and covering their hair with a turban. The recent news stories are summarized here:

http://www.newsday.com/opinion/sikh-student-faces-the-army-s-catch-22-editorial-1.9627206

As usual, a myriad of folks have come out chiding the Sikh person for not conforming by disobeying his religion and removing his hair and beard. Such comments have been supported with questions that have typically come up whenever similar stories have come in the media of how Sikhs are expected to wear gas masks, how can they perform in combat operations without a helmet, etc… The point is that all of these issues are in fact, non-issues and are completely unfounded. There is much literature on Sikhs’ performance in the various wars throughout history and their success in said wars with turbans and beards, so I will not pursue that discussion here. However, I would like to take this opportunity to point out a level of hypocrisy  which people who oppose Sikhs in the military on the aforementioned grounds seem to have a penchant for. Here are just a few examples of countless numbers of White, American soldiers fighting in US conflicts with beards and in some cases, long hair. Why weren’t the same questions and objections raised in their cases?

Here are just some examples of Sikhs fighting and giving their lives in World War II for the Allied forces. As you can see, there are no technical issues of turbans and beard arising in these cases:

“…..It is a matter of regret that due to the obsession of the present times people are distorting the superior religious and social values, but those who wish to preserve them with respect, we should appreciate them as well as help them. Sikhs do need our help for such a cause and we should give it happily. Those who know the Sikh history, know England’s relationship with the Sikhs and are aware of the achievements of the Sikhs, they should persistently support the idea of relaxation to Sikhs to ride a motorbike with their turbans on, because it is their religious privilege.”

“…British people are highly indebted and obliged to Sikhs for a long time. I know that within this century we needed their help twice and they did help us very well. As a result of their timely help, we are today able to live with honour, dignity, and independence. In the war, they fought and died for us, wearing the turbans. At that time we were not adamant that they should wear safety helmets because we knew that they are not going to wear them anyways and we would be deprived of their help. At that time due to our miserable and poor situation, we did not force it on them to wear safety helmets, why should we force it now? Rather, we should now respect their traditions and by granting this legitimate concession, win their applaud.”

So, we should all strive to be a bit more educated in this regard. These exemptions have been granted to White, American soldiers in the past, and even if not in a formal sense, there is an eerie silence coming from the same critics of Sikhs wanting to join the military when non-Sikhs have also not conformed to these rules as the above pictures clearly demonstrate.

A Universe from Nothing

Many people have obviously been reading Lawrence Krauss’ infamous book: “A Universe from Nothing: Why There is something Rather Than Nothing”

Krauss and many other physicists continuously engage in this type of low-level philosophy with the ironic goal of diminishing the value of philosophy using “science”.

This paper dissects all the arguments in Krauss’ book and shows from a mathematical standpoint that like others who make similar arguments, they are not grounded in actual physics and are extremely flawed. One therefore concludes, that such arguments are not based in science, but in bad philosophy.

The paper: A Universe from Nothing

A teaser: This is what “nothing” actually looks like, well one depiction of it anyways:
But this is not nothing, it is something, where did this structure come from? Krauss actually ignores the question entirely in his book, which is very strange.

The famed cosmologist George Ellis also has discussed Krauss’ book in one of his talks, here is the link for that:

A video making some of the arguments in the above paper easier to understand can be found here:

Sikhs and Diwali

For years now, I have heard this constant story of how it is acceptable for Sikhs to celebrate Diwali as per the Hindu traditions of lighting lamps, etc… Further, Raagis and Bhai Sahibs in Gurdwaras have conflated Bandhi Chorr Diwas with lighting lamps as per Hindu Diwali traditions. They further support these ideas with the supposed Vaar from Bhai Gurdas Jee, in which they ironically only mention and repeat the first line! : “deewaalee dee raath dheevae baaleean”. Of course, just by reading this line, it would suggest that the aforementioned actions are justified. But taking one line completely out of context leads one to these conclusions. A full reading of Bhai Gurdas Jee’s Vaar on the Diwali matter which given the timeframe is also a historical first-hand account suggests that Sikhs are to practice completely the opposite and in fact, lighting lamps is contrary to Gurmat. The full Vaar’s transliteration is below:

Vaars Bhai Gurdaas

diwali dee raath dheevae baaleeani
thaarae jaath sanaath a(n)bar bhaaleean
fulaa(n) dhee baagaath chun chun chaaleean
theerathh jaathee jaath nain nihaaleean
har cha(n)dhuree jhaath vasaae ouchaaleean

The essence of this Vaar is in every line after the first. Namely, in the third, fourth, and fifth lines, Bhai Gurdas Jee compares those that celebrate Diwali by lighting lamps akin to those who go on long pilgrimages to find God, and to those who search for God by worshipping the stars, or things in nature, etc… All contrary to Gurmat by a simple reading of Japjee Sahib! Indeed, Bhai Sahib Jee in the last line clearly states that a person of Gurmat does not practice any of these things, which he declares to be temporary and pointless.

So, there you have it. A simple reading of the full Vaar changes the entire context of the “importance” of Diwali in Sikhism. I doubt many Sikhs will read this posting with sincerity, but someone has to speak the truth!

Reply to Recent NYT Article: “God, Darwin, and My College Biology Class”

I recently came across the article/op-ed in the NYTimes, titled, “God, Darwin, and My College Biology Class”, http://www.nytimes.com/2014/09/28/opinion/sunday/god-darwin-and-my-college-biology-class.html?_r=0

In this article, it is stated: “Since Darwin, however, we have come to understand that an entirely natural and undirected process, namely random variation plus natural selection, contains all that is needed to generate extraordinary levels of non-randomness. Living things are indeed wonderfully complex, but altogether within the range of a statistically powerful, entirely mechanical phenomenon.”

When it is stated that “contains all that is needed to generate extraordinary levels of non-randomness” is factually not accurate, for one is making the mistake that many reductionists make by assuming that all complexity arises from bottom-up causation alone, and are completely ignoring the effects of top-down causation. The reason is as follows: Lower levels of complexity are necessarily governed by uncertainties due to quantum mechanics, it is not clear how these quantum uncertainties transition to a classical state. Mathematically, these uncertainties that are at the heart of the random variation that is cited are governed in a Hilbert, L^2 Lebesgue integrable space. Classical systems, determined by phase space manifolds have these probabilistic domains in the cotangent bundle of the manifold. The article is essentially saying that the cotangent bundle determines the phase space and not the other way around which is not correct. Further, there remains the unsolved issue of how the quantum fluctuations become classical (unless you follow the untestable many worlds route, which has major problems – see S D Hsu Modern Physics Letters A27: 1230114 (2012) for one interesting comment, and the writings of Sudarsky.

On the other side, top-down causation via cosmology and Einstein’s equations seed the correct conditions for dynamical Darwinian evolution to take place to begin with, for some reason, the author completely leaves this out.

Without a question, the author is an expert an evolutionary biology, but I am afraid he has looked through these issues through a very narrow lens, which does not do the issue a full and complete justification, and indeed, is responsible for much of the discomfort with evolutionary theory that is described so accurately and well in the article.

In the mean time, I would humbly suggest that the interested reader look at the following articles which describe Darwinian evolution in a more complete context as a function of emergence and complexity through physics which underlies biology. Also, one should see the work of Denis Noble, http://musicoflife.co.uk who advocates for a dynamical systems-based view of biological systems, which I personally believe to be correct, as it is much more mathematically and physically sound compared to standard evolutionary theory. This YouTube video of a lecture from the noted cosmologist GFR Ellis also sums up the problem with the reductionist view of evolutionary biology: http://youtu.be/nEhTkF3eG8Q

Title:
Laws, Causation and Dynamics at Different Levels
Authors:
Butterfield, Jeremy
Publication:
eprint arXiv:1406.4732
Publication Date:
06/2014
Origin:
ARXIV
Keywords:
Physics – History and Philosophy of Physics, Physics – Popular Physics
Comment:
29 pages, 3 figures; Interface Focus (Royal Society London), volume 2, 2012, pp. 101-114; doi:10.1098/rsfs.2011.0052
Bibliographic Code:
2014arXiv1406.4732B
Abstract

I have two main aims. The first is general, and more philosophical (Section 2). The second is specific, and more closely related to physics (Sections 3 and 4). The first aim is to state my general views about laws and causation at different ‘levels’. The main task is to understand how the higher levels sustain notions of law and causation that ‘ride free’ of reductions to the lower level or levels. I endeavour to relate my views to those of other symposiasts. The second aim is to give a framework for describing dynamics at different levels, emphasising how the various levels’ dynamics can mesh or fail to mesh. This framework is essentially that of elementary dynamical systems theory. The main idea will be, for simplicity, to work with just two levels, dubbed ‘micro’ and ‘macro’ which are related by coarse-graining. I use this framework to describe, in part, the first four of Ellis’ five types of top-down causation.

Title:
The arrow of time and the nature of spacetime
Authors:
Ellis, George F R
Publication:
eprint arXiv:1302.7291
Publication Date:
02/2013
Origin:
ARXIV
Keywords:
General Relativity and Quantum Cosmology, Physics – History and Philosophy of Physics
Comment:
56 pages, 7 figures
Bibliographic Code:
2013arXiv1302.7291E
Abstract

This paper extends the work of a previous paper [arXiv:1208.2611] on the flow of time, to consider the origin of the arrow of time. It proposes that a past condition’ cascades down from cosmological to micro scales, being realized in many microstructures and setting the arrow of time at the quantum level by top-down causation. This physics arrow of time then propagates up, through underlying emergence of higher level structures, to geology, astronomy, engineering, and biology. The appropriate space-time picture to view all this is an emergent block universe (EBU’), that recognizes the way the present is different from both the past and the future. This essential difference is the ultimate reason the arrow of time has to be the way it is.

Title:
Recognising Top-Down Causation
Authors:
Ellis, George F R
Publication:
eprint arXiv:1212.2275
Publication Date:
12/2012
Origin:
ARXIV
Keywords:
Physics – Classical Physics, Nonlinear Sciences – Adaptation and Self-Organizing Systems, Physics – History and Philosophy of Physics
Comment:
11 pages, 2 figures, 2 tables. 2nd prize in FQXI essay competition
Bibliographic Code:
2012arXiv1212.2275E
Abstract

One of the basic assumptions implicit in the way physics is usually done is that all causation flows in a bottom up fashion, from micro to macro scales. However this is wrong in many cases in biology, and in particular in the way the brain functions. Here I make the case that it is also wrong in the case of digital computers – the paradigm of mechanistic algorithmic causation – and in many cases in physics, ranging from the origin of the arrow of time to the process of state vector preparation. I consider some examples from classical physics, as well as the case of digital computers, and then explain why this is possible without contradicting the causal powers of the underlying microphysics. Understanding the emergence of genuine complexity out of the underlying physics depends on recognising this kind of causation.

Title:
Evolutionary Transitions and Top-Down Causation
Authors:
Imari Walker, Sara; Cisneros, Luis; Davies, Paul C. W.
Publication:
eprint arXiv:1207.4808
Publication Date:
07/2012
Origin:
ARXIV
Keywords:
Nonlinear Sciences – Adaptation and Self-Organizing Systems, Quantitative Biology – Other Quantitative Biology
Comment:
8 pages, 4 figures; Proceedings of Artificial Life XIII (2012) p. 283-290
Bibliographic Code:
2012arXiv1207.4808I
Abstract

Top-down causation has been suggested to occur at all scales of biological organization as a mechanism for explaining the hierarchy of structure and causation in living systems. Here we propose that a transition from bottom-up to top-down causation — mediated by a reversal in the flow of information from lower to higher levels of organization, to that from higher to lower levels of organization — is a driving force for most major evolutionary transitions. We suggest that many major evolutionary transitions might therefore be marked by a transition in causal structure. We use logistic growth as a toy model for demonstrating how such a transition can drive the emergence of collective behavior in replicative systems. We then outline how this scenario may have played out in those major evolutionary transitions in which new, higher levels of organization emerged, and propose possible methods via which our hypothesis might be tested.

Title:
On the limits of quantum theory: Contextuality and the quantum-classical cut
Authors:
Ellis, George F. R.
Affiliation:
AA(Mathematics Department, University of Cape Town, South Africa)
Publication:
Annals of Physics, Volume 327, Issue 7, p. 1890-1932.
Publication Date:
07/2012
Origin:
ELSEVIER
(c) 2012 Elsevier Inc.
DOI:
10.1016/j.aop.2012.05.002
Bibliographic Code:
2012AnPhy.327.1890E
Abstract

This paper is based on four assumptions: 1. Physical reality is made of linearly behaving components combined in non-linear ways. 2. Higher level behaviour emerges from this lower level structure. 3. The way the lower level elements behaves depends on the context in which they are embedded. 4. Quantum theory applies to the lower level entities. An implication is that higher level effective laws, based on the outcomes of non-linear combinations of lower level linear interactions, will generically not be unitary; hence the applicability of quantum theory at higher levels is strictly limited. This leads to the view that both state vector preparation and the quantum measurement process are crucially based on top-down causal effects, and helps provide criteria for the Heisenberg cut that challenge some views on Schrödinger’s cat.

Title:
Top-Down Causation and Autonomy in Complex Systems
Authors:
Juarrero, Alicia
Affiliation:
AA(Emeritus, Prince George’s Community College)
Publication:
Downward Causation and the Neurobiology of Free Will, Understanding Complex Systems. ISBN 978-3-642-03204-2. Springer Berlin Heidelberg, 2009, p. 83
Publication Date:
00/2009
Origin:
SPRINGER
Keywords:
Physics
(c) 2009: Springer Berlin Heidelberg
DOI:
10.1007/978-3-642-03205-9_5
Bibliographic Code:
2009dcnf.book…83J
Abstract

Evolutionary evidence shows that complex dynamical systems become increasingly self-directed and decoupled from merely energetic forces over time. In this paper I analyze these transformations, concentrating on changes in the type of top-down causation that characterizes such self-organized and autopoietic pro cesses. Specifically, I show that the top-down selection criteria of these systems makes some of them autonomous, and that because once evolution reaches humans the criteria according to which voluntary actions are selected are semantic and symbolic – and can be self-consciously chosen – human self-direction constitutes a form of strong autonomy that can arguably be considered “free will.”

Title:
Top-Down Causation and the Human Brain
Authors:
Ellis, George F. R.
Affiliation:
AA(Mathematics Department, University of Cape Town)
Publication:
Downward Causation and the Neurobiology of Free Will, Understanding Complex Systems. ISBN 978-3-642-03204-2. Springer Berlin Heidelberg, 2009, p. 63
Publication Date:
00/2009
Origin:
SPRINGER
Keywords:
Physics
(c) 2009: Springer Berlin Heidelberg
DOI:
10.1007/978-3-642-03205-9_4
Bibliographic Code:
2009dcnf.book…63E
Abstract

A reliable understanding of the nature of causation is the core feature of science. In this paper the concept of top-down causation in the hierarchy of structure and causation is examined in depth. Five different classes of top-down causation are identified and illustrated with real-world examples. They are (1) al gorithmic top-down causation; (2) top-down causation via nonadaptive information control; (3) top-down causation via adaptive selection; (4) top-down causation via adaptive information control; and (5) intelligent top-down causation (i.e., the effect of the human mind on the physical world). Recognizing these forms of causation implies that other kinds of causes than physical and chemical interactions are effective in the real world. Because of the existence of random processes at the bottom, there is sufficient causal slack at the physical level to allow all these kinds of causation to occur without violation of physical causation. That they do indeed occur is indicated by many kinds of evidence. Each such kind of causation takes place in particular in the human brain, as is indicated by specific examples.

Title:
Top-Down Causation by Information Control: From a Philosophical Problem to a Scientific Research Program
Authors:
Auletta, G.; Ellis, G. F. R.; Jaeger, L.
Publication:
eprint arXiv:0710.4235
Publication Date:
10/2007
Origin:
ARXIV
Keywords:
Quantitative Biology – Other Quantitative Biology
Comment:
Revised version to meet referee’s comments, and responding to a paper by Wegscheid et al that was not mentioned in the previous version. 23 pages, 9 figures
Bibliographic Code:
2007arXiv0710.4235A
Abstract

It has been claimed that different types of causes must be considered in biological systems, including top-down as well as same-level and bottom-up causation, thus enabling the top levels to be causally efficacious in their own right. To clarify this issue, important distinctions between information and signs are introduced here and the concepts of information control and functional equivalence classes in those systems are rigorously defined and used to characterise when top down causation by feedback control happens, in a way that is testable. The causally significant elements we consider are equivalence classes of lower level processes, realised in biological systems through different operations having the same outcome within the context of information control and networks.

Title:
Physics and the Real World
Authors:
Ellis, George F. R.
Affiliation:
AA(Mathematics Department, University of Cape Town)
Publication:
Foundations of Physics, Volume 36, Issue 2, pp.227-262
Publication Date:
02/2006
Origin:
CROSSREF; SPRINGER
Keywords:
Physics, emergence, causality
DOI:
10.1007/s10701-005-9016-x
Bibliographic Code:
2006FoPh…36..227E
Abstract

Physics and chemistry underlie the nature of all the world around us, including human brains. Consequently some suggest that in causal terms, physics is all there is. However, we live in an environment dominated by objects embodying the outcomes of intentional design (buildings, computers, teaspoons). The present day subject of physics has nothing to say about the intentionality resulting in existence of such objects, even though this intentionality is clearly causally effective. This paper examines the claim that the underlying physics uniquely causally determines what happens, even though we cannot predict the outcome. It suggests that what occurs is the contextual emergence of complexity: the higher levels in the hierarchy of complexity have autonomous causal powers, functionally independent of lower level processes. This is possible because top-down causation takes place as well as bottom-up action, with higher level contexts determining the outcome of lower level functioning and even modifying the nature of lower level constituents. Stored information plays a key role, resulting in non-linear dynamics that is non-local in space and time. Brain functioning is causally affected by abstractions such as the value of money and the theory of the laser. These are realised as brain states in individuals, but are not equivalent to them. Consequently physics per se cannot causally determine the outcome of human creativity, rather it creates the possibility space allowing human intelligence to function autonomously. The challenge to physics is to develop a realistic description of causality in truly complex hierarchical structures, with top-down causation and memory effects allowing autonomous higher levels of order to emerge with genuine causal powers.

Title:
Emergence and Dissolvence in the Self-organisation of Complex Systems
Authors:
Testa, Bernard; Kier, Lemont B.
Publication:
Entropy, vol. 2, Issue 1, p.1-25
Publication Date:
03/2000
Origin:


text:
# torch.split¶

torch.split(tensor, split_size_or_sections, dim=0)[source]

Splits the tensor into chunks. Each chunk is a view of the original tensor.

If split_size_or_sections is an integer type, then tensor will be split into equally sized chunks (if possible). Last chunk will be smaller if the tensor size along the given dimension dim is not divisible by split_size.

If split_size_or_sections is a list, then tensor will be split into len(split_size_or_sections) chunks with sizes in dim according to split_size_or_sections.

Parameters
• tensor (Tensor) – tensor to split.

• split_size_or_sections (int) or (list(int)) – size of a single chunk or list of sizes for each chunk

• dim (int) – dimension along which to split the tensor.

Example:

>>> a = torch.arange(10).reshape(5,2)
>>> a
tensor([[0, 1],
[2, 3],
[4, 5],
[6, 7],
[8, 9]])
>>> torch.split(a, 2)
(tensor([[0, 1],
[2, 3]]),
tensor([[4, 5],
[6, 7]]),
tensor([[8, 9]]))
>>> torch.split(a, [1,4])
(tensor([[0, 1]]),
tensor([[2, 3],
[4, 5],
[6, 7],
[8, 9]]))

format: plain


text:
27

# Questions

3
1
vote
0
295
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### birational equivalence and mirror CYs

jan 21 11 at 18:11 J Verma 1,7401426

2
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1
393
views

### is complex moduli space of a Calabi - Yau Kahler

jan 4 11 at 20:47 Igor Rivin 31.1k33771

2
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193
views

### Axiomatic treatment of A-model and B-model in sense of Atiyah

mar 1 11 at 20:47 J Verma 1,7401426

1
1
vote
1
453
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### Hirzebruch surfaces

jun 16 11 at 20:04 Dmitri 16.5k13677

0
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100
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### Equivariant $\hat{A}$ - genus of a spin manifold

feb 20 at 4:47 J Verma 1,7401426

1
0
0
202
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### topological B model

nov 29 10 at 6:43 J Verma 1,7401426

0


text:
#2044
Paras Mehta
Keymaster

Yes, it is possible. The matrix to use would be a beta matrix. If a latent factor V is defined by other latent variables U_1 to U_p, a (p by 1) beta matrix with one of the paths fixed to 1.0 will do the trick.

Simplest case is the ‘hierarchical random-intercept’:
http://xxm.times.uh.edu/learn-xxm/three-level-hierarchical-model-with-observed-and-latent-variables-at-multiple-levels/

Here is an example of random-slopes for a latent DV.
http://xxm.times.uh.edu/learn-xxm/two-level-confirmatory-factor-analysis-with-a-random-slope/

In each case, a factor is defined by other latent factors.

Best,

Paras


text:
Home

>

English

>

Class 11

>

Chemistry

>

Chapter

>

Equilibrium

>

Explain liquid-vapour equilibr...

# Explain liquid-vapour equilibrium.

Text Solution

Solution : Process : We consider the example of a transparent box carrying a U-tube with mercury (manometer). Drying agent like anhydrous calcium chloride (or phosphorus penta-oxide) is placed for a few hours in the box. By doing this air in the box is free from vapour (moisture). (see fig.)
 Now, after removing the drying agent by tilting the box on one side, a watch glass (or petri dish) containing water is quickly placed inside the box.

 Observation : It will be observed that the mercury level in the right limb of the manometer slowly increases and finally attains a constant value.
 Assumption : The observation shows that (i) In the beginning the pressure inside the box increases. (ii) After some time pressure inside the box becomes constant. (iii) The volume of water in the watch glass decreases.

 Description : Initially there was no water vapour (or very less) inside the box. As water evaporated the pressure in the box increased due to addition of water molecules into the gaseous phase inside the box, [H_2O to H_2O_"(vapour)"] So, the rate of evaporation is constant.
 The rate of increase in pressure decreases with time due to condensation of vapour into water.[H_2O_((g)) to H_2O_((l))] While water is converting from gases to liquid form, atoms are increasing. After some time pressure become constant. There is equilibrium in box as rate of evaporation and rate of condensation become equal.
 Rate of evaporation (H_2O_((l)) to H_2O_((g))) = Rate of condensation (H_2O_((g)) to H_2O_((l))) .....(i)
 therefore at the equilibrium stage H_2O_((l)) hArr H_2O_"(vapour)"
 At equilibrium the pressure exerted by the water molecules at a given temperature remains constant and is called the equilibrium vapour pressure of water (or just vapour pressure of water), vapour pressure of water increases with temperature.
 Water and water vapour are in equilibrium position at atmospheric pressure (1.013 bar) and at 100^@C in a close vessel. The boiling point of water is 100^@C at 1.013 bar pressure. For any pure liquid at one atmospheric pressure the temperature at which the liquid and vapours are at equilibrium is called normal boiling point.

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text:
# Increasing the space between two rows?

This is possible duplicate of increasing space between two rows. But because of two many math expression there, I could not possibly understand it. Regrets.

I would like to increase the space between two rows so that table can be filled manually using pen or pencil. Here is MWE.

\documentclass[a4paper]{article}

%% Language and font encodings
\usepackage[english]{babel}
\usepackage[utf8x]{inputenc}
\usepackage{booktabs}
%% Sets page size and margins
\usepackage[a4paper,top=3cm,bottom=2cm,left=3cm,right=3cm,marginparwidth=1.75cm]{geometry}
\begin{document}
\begin{table}
\centering
\begin{tabular}{p{5cm} p{5cm}}
\toprule
Item & Quantity \\
\midrule
Widgets & \dotfill \\
\midrule
\end{tabular}
\end{table}
\end{document}

• Widgets & \dotfill \\[10pt] Gadgets & \dotfill \\  which uses the optional argument to \\  or, alternately, \midrule &\\ Widgets & \dotfill \\ &\\ Gadgets & \dotfill \\ \midrule  which manually adds blank rows – Steven B. Segletes Oct 4 '16 at 11:36
• @ Steven B. Segletes it increased the space between 2nd and 3rd row, but not between first and second row. – Vaibhav Oct 4 '16 at 11:47
• Here is what I did: \begin{table} \centering \begin{tabular}{p{5cm} p{5cm}} \toprule Item & Quantity \\ [10pt] \midrule Widgets & \dotfill \\ [10pt] Gadgets & \dotfill \\ \midrule \end{tabular} \end{table} \end{document} – Vaibhav Oct 4 '16 at 11:47

Modify arraystretch:

\documentclass{article}
\usepackage{booktabs}
\begin{document}
{ % begin box to localize effect of arraystretch change
\renewcommand{\arraystretch}{3.0}
\begin{table}
\centering
\begin{tabular}{p{5cm} p{5cm}}
\toprule
Item & Quantity \\
\midrule
Widgets & \dotfill \\
\midrule
\end{tabular}
\end{table}
} % end box
\end{document}

• Though the problem is it involves all the tables in a document. – Vaibhav Oct 4 '16 at 12:02
• Right @Vaibhav, and I edited to localize the change. But it still adds space in unwanted areas, e.g., the header row. The other answers are much better! – Homer White Oct 4 '16 at 12:22

You can use \addlinespace from booktabs:

\documentclass[a4paper]{article}

%% Language and font encodings
\usepackage[english]{babel}
\usepackage[utf8x]{inputenc}
\usepackage{booktabs}
%% Sets page size and margins

\usepackage[a4paper,top=3cm,bottom=2cm,left=3cm,right=3cm,marginparwidth=1.75cm]{geometry}

\begin{document}

\begin{table}
\centering
\begin{tabular}{p{5cm} p{5cm}}
\toprule
Item & Quantity \\
\midrule
Widgets & \dotfill \\
\midrule
\end{tabular}
\end{table}

\end{document}

Here I show 3 ways:

1. the optional argument to \\[]

3. Add a custom strut to lines where needed.

Approach 1 will not work following a \modrule.

\documentclass[a4paper]{article}

%% Language and font encodings
\usepackage[english]{babel}
\usepackage[utf8x]{inputenc}
\usepackage{booktabs}
%% Sets page size and margins
\usepackage[a4paper,top=3cm,bottom=2cm,left=3cm,right=3cm,marginparwidth=1.75cm]{geometry}
\begin{document}
%\begin{table}
\centering
Use optional argument to \verb|\\|, but can't be used for line after a
\verb|\midrule|

\begin{tabular}{p{5cm} p{5cm}}
\toprule
Item & Quantity \\
\midrule
Widgets & \dotfill \\[10pt]
\midrule
\end{tabular}
%\end{table}

\bigskip

\begin{tabular}{p{5cm} p{5cm}}
\toprule
Item & Quantity \\
\midrule
&\\
Widgets & \dotfill \\
&\\
\midrule
\end{tabular}

\bigskip

\def\mystrut{\rule{0pt}{2\normalbaselineskip}}
\begin{tabular}{p{5cm} p{5cm}}
\toprule
Item & Quantity \\
\midrule
Widgets\mystrut & \dotfill \\


text:
Technologically Embodied
Geometric Functions
Complex Multiplication Overview

In this multi-part activity you will investigate the multiplication of two complex numbers v and w using a combination of algebra and geometry. First, remember that you can express w as w=(x_w + i*y_w).

(Although one more commonly sees expressions like v = (a + b*i) or w = (c + d*i), in the current explorations we choose to connect the real and imaginary components of a complex number to its horizontal and vertical coordinates on the complex plane. We will also make significant use of the modulus (distance from the origin) and the argument (angle from the positive real axis) of a complex number, which we will represent as (r_w, theta_w) to emphasize the connection to polar coordinates.)

Because w=(x_w + i*y_w), you can use the distributive property to write v*w = v*(x_w + i*y_w) = v*x_w + v*i*y_w.

In this expanded form, x_v and y_v are both real numbers. This enables you to use geometry on the complex plane to multiply complex numbers, based on three geometric techniques:

• geometrically multiplying a complex number by i (to construct v*i),
• geometrically multiplying a complex number by a real number (to construct v*x_w and v*i*y_w), and
• geometrically adding two complex numbers (to construct the sum v*x_w + v*i*y_w).

In the following activities, you will develop these three techniques and use them to find an elegant way to multiply two complex numbers.

• Part 1: What happens when you dilate a complex number by a scale factor like 3, 0.5, or -2? Describe the result numerically.
• Part 2: What happens when you rotate a complex number by an angle such as pi/2, pi, or 3 pi/2? Describe the result numerically. How can you use your results to multiply a complex number by i?
• Part 3: What happens when you translate a complex number by the vector corresponding to a second complex number? Describe the result numerically. [Not yet available]
• Part 4: Start with two complex numbers and apply what you’ve learned to construct their product: v·w = v*x_w + v*i*y_w
• Part 5: Starting with the construction of v*w in Part 4 based on algebra, find a much simpler way to multiply two complex numbers by using just two steps: the dilate-rotate method. Then investigate whether this method is commutative with respect to v and w? (That is, does w*v = v*w for complex numbers?) Second, is the composition itself commutative? (That is, does rotate-dilate give the same result as dilate-rotate?)

• Release Information and Rights

Update History:


text:
$$\require{cancel}$$

# 4.5: Uniform Circular Motion

Learning Objectives

• Solve for the centripetal acceleration of an object moving on a circular path.
• Use the equations of circular motion to find the position, velocity, and acceleration of a particle executing circular motion.
• Explain the differences between centripetal acceleration and tangential acceleration resulting from nonuniform circular motion.
• Evaluate centripetal and tangential acceleration in nonuniform circular motion, and find the total acceleration vector.

Uniform circular motion is a specific type of motion in which an object travels in a circle with a constant speed. For example, any point on a propeller spinning at a constant rate is executing uniform circular motion. Other examples are the second, minute, and hour hands of a watch. It is remarkable that points on these rotating objects are actually accelerating, although the rotation rate is a constant. To see this, we must analyze the motion in terms of vectors.

## Centripetal Acceleration

In one-dimensional kinematics, objects with a constant speed have zero acceleration. However, in two- and three-dimensional kinematics, even if the speed is a constant, a particle can have acceleration if it moves along a curved trajectory such as a circle. In this case the velocity vector is changing, or $$\frac{d\vec{v}}{dt}$$ ≠ 0. This is shown in Figure $$\PageIndex{1}$$. As the particle moves counterclockwise in time $$\Delta$$t on the circular path, its position vector moves from $$\vec{r}(t)$$ to $$\vec{r}(t + \Delta t)$$. The velocity vector has constant magnitude and is tangent to the path as it changes from $$\vec{v}$$(t) to $$\vec{v}\left(t + \Delta t\right)$$, changing its direction only. Since the velocity vector $$\vec{v}(t)$$ is perpendicular to the position vector $$\vec{r}$$(t), the triangles formed by the position vectors and $$\Delta \vec{r}$$, and the velocity vectors and $$\Delta \vec{v}$$ are similar. Furthermore, since

$|\vec{r}(t) | = | \vec{r} (t + \Delta t)| \nonumber$

and

$| \vec{v} (t)| = | \vec{v} (t + \Delta t)|, \nonumber$

the two triangles are isosceles. From these facts we can make the assertion

$\dfrac{\Delta v}{v} = \dfrac{\Delta r}{r}$

or

$\Delta v = \dfrac{v}{r} \Delta r.$

We can find the magnitude of the acceleration from

$a = \lim_{\Delta t \rightarrow 0} \left(\dfrac{\Delta v}{\Delta t}\right) = \frac{v}{r} \left(\lim_{\Delta t \rightarrow 0} \dfrac{\Delta r}{\Delta t}\right) = \frac{v^{2}}{r} \ldotp$

The direction of the acceleration can also be found by noting that as $$\Delta$$t and therefore $$\Delta \theta$$ approach zero, the vector $$\Delta \vec{v}$$ approaches a direction perpendicular to $$\vec{v}$$. In the limit $$\Delta t → 0,$$ $$\Delta \vec{v}$$ is perpendicular to $$\vec{v}$$. Since $$\vec{v}$$ is tangent to the circle, the acceleration $$\frac{d \vec{v}}{dt}$$ points toward the center of the circle. Summarizing, a particle moving in a circle at a constant speed has an acceleration with magnitude

$a_{c} = \frac{v^{2}}{r} \ldotp \label{4.27}$

The direction of the acceleration vector is toward the center of the circle (Figure $$\PageIndex{2}$$). This is a radial acceleration and is called the centripetal acceleration, which is why we give it the subscript $$c$$. The word centripetal comes from the Latin words centrum (meaning “center”) and petere (meaning to seek”), and thus takes the meaning “center seeking.”

Let’s investigate some examples that illustrate the relative magnitudes of the velocity, radius, and centripetal acceleration.

Example $$\PageIndex{1}$$: Creating an Acceleration of 1 g

A jet is flying at 134.1 m/s along a straight line and makes a turn along a circular path level with the ground. What does the radius of the circle have to be to produce a centripetal acceleration of 1 g on the pilot and jet toward the center of the circular trajectory?

Strategy

Given the speed of the jet, we can solve for the radius of the circle in the expression for the centripetal acceleration.

Solution

Set the centripetal acceleration equal to the acceleration of gravity: 9.8 m/s2 = $$\frac{v^{2}}{r}$$.

Solving for the radius, we find

$r = \frac{(134.1\; m/s)^{2}}{9.8\; m/s^{2}} = 1835\; m = 1.835\; km \ldotp$

Significance

To create a greater acceleration than g on the pilot, the jet would either have to decrease the radius of its circular trajectory or increase its speed on its existing trajectory or both.

Exercise 4.5

A flywheel has a radius of 20.0 cm. What is the speed of a point on the edge of the flywheel if it experiences a centripetal acceleration of 900.0 cm/s2?

Centripetal acceleration can have a wide range of values, depending on the speed and radius of curvature of the circular path. Typical centripetal accelerations are given in Table $$\PageIndex{1}$$.

Table $$\PageIndex{1}$$: Typical Centripetal Accelerations
Object Centripetal Acceleration (m/s2 or factors of g)
Earth around the Sun 5.93 x 10-3
Moon around the Earth 2.73 x 10-3
Satellite in geosynchronous orbit 0.233
Outer edge of a CD when playing 5.75
Jet in a barrel roll (2-3 g)
Roller coaster (5 g)
Electron orbiting a proton in a simple Bohr model of the atom 9.0 x 1022

## Equations of Motion for Uniform Circular Motion

A particle executing circular motion can be described by its position vector $$\vec{r}(t)$$. Figure $$\PageIndex{3}$$ shows a particle executing circular motion in a counterclockwise direction. As the particle moves on the circle, its position vector sweeps out the angle $$\theta$$ with the x-axis. Vector $$\vec{r}(t)$$ making an angle $$\theta$$ with the x-axis is shown with its components along the x- and y-axes. The magnitude of the position vector is $$A = |\vec{r}(t)|$$ and is also the radius of the circle, so that in terms of its components,

$\vec{r} (t) = A \cos \omega \hat{i} + A \sin \omega t \hat{j} \ldotp \label{4.28}$

Here, $$\omega$$ is a constant called the angular frequency of the particle. The angular frequency has units of radians (rad) per second and is simply the number of radians of angular measure through which the particle passes per second. The angle $$θ$$ that the position vector has at any particular time is $$\omega$$t.

If $$T$$ is the period of motion, or the time to complete one revolution ($$2 \pi\, rad$$), then

Velocity and acceleration can be obtained from the position function by differentiation:

$\vec{v} (t) = \frac{d \vec{r} (t)}{dt} = -A \omega \sin \omega t \hat{i} + A \omega \cos \omega t \hat{j} \ldotp \label{4.29}$

It can be shown from Figure $$\PageIndex{3}$$ that the velocity vector is tangential to the circle at the location of the particle, with magnitude A$$\omega$$. Similarly, the acceleration vector is found by differentiating the velocity:

$\vec{a} (t) = \frac{d \vec{v} (t)}{dt} = -A \omega^{2} \cos \omega t \hat{i} - A \omega^{2} \sin \omega t \hat{j} \ldotp \label{4.30}$

From this equation we see that the acceleration vector has magnitude A$$\omega^{2}$$ and is directed opposite the position vector, toward the origin, because $$\vec{a}$$(t) = −$$\omega^{2} \vec{r}$$(t).

Example $$\PageIndex{2}$$: Circular Motion of a Proton

A proton has speed 5 x 106 m/s and is moving in a circle in the xy plane of radius r = 0.175 m. What is its position in the xy plane at time t = 2.0 x 10−7 s = 200 ns? At t = 0, the position of the proton is 0.175 m $$\hat{i}$$ and it circles counterclockwise. Sketch the trajectory.

Solution

From the given data, the proton has period and angular frequency:

$T = \frac{2 \pi r}{v} = \frac{2 \pi (0.175\; m)}{5.0 \times 10^{6}\; m/s} = 2.20 \times 10^{-7}\; s \nonumber$

$\omega = \frac{2 \pi}{T} = \frac{2 \pi}{2.20 \times 10^{-7}\; s} = 2.856 \times 10^{7}\; rad/s \ldotp \nonumber$

The position of the particle at t = 2.0 x 10−7 s with A = 0.175 m is

\begin{align*} \vec{r} (2.0 \times 10^{-7}\; s) & = A \cos \omega (2.0 \times 10^{-7}\; s) \hat{i} + A \sin \omega (2.0 \times 10^{-7}\; s) \hat{j}\; m \\[4pt] & = 0.175 \cos (2.856 \times 10^{7}\; rad/s) (2.0 \times 10^{-7}\; s) \hat{i} + 0.175 \sin (2.856 \times 10^{7}\; rad/s) (2.0 \times 10^{-7}\; s) \hat{j}\; m \\[4pt] & = 0.175 \cos (5.712\; rad) \hat{i} + 0.175 \sin (5.172\; rad) \hat{j}\; m \\ & = 0.147 \hat{i} - 0.095 \hat{j}\; m \ldotp \end{align*}

From this result we see that the proton is located slightly below the x-axis. This is shown in Figure $$\PageIndex{4}$$.

Significance

We picked the initial position of the particle to be on the x-axis. This was completely arbitrary. If a different starting position were given, we would have a different final position at t = 200 ns.

## Nonuniform Circular Motion

Circular motion does not have to be at a constant speed. A particle can travel in a circle and speed up or slow down, showing an acceleration in the direction of the motion.

In uniform circular motion, the particle executing circular motion has a constant speed and the circle is at a fixed radius. If the speed of the particle is changing as well, then we introduce an additional acceleration in the direction tangential to the circle. Such accelerations occur at a point on a top that is changing its spin rate, or any accelerating rotor. In Displacement and Velocity Vectors we showed that centripetal acceleration is the time rate of change of the direction of the velocity vector. If the speed of the particle is changing, then it has a tangential acceleration that is the time rate of change of the magnitude of the velocity:

$a_{T} = \frac{d |\vec{v}|}{dt} \ldotp \label{4.31}$

The direction of tangential acceleration is tangent to the circle whereas the direction of centripetal acceleration is radially inward toward the center of the circle. Thus, a particle in circular motion with a tangential acceleration has a total acceleration that is the vector sum of the centripetal and tangential accelerations:

$\vec{a} = \vec{a}_{c} + \vec{a}_{T} \ldotp \label{4.32}$

The acceleration vectors are shown in Figure $$\PageIndex{5}$$. Note that the two acceleration vectors $$\vec{a}_{c}$$ and $$\vec{a}_{T}$$ are perpendicular to each other, with $$\vec{a}_{c}$$ in the radial direction and $$\vec{a}_{T}$$ in the tangential direction. The total acceleration $$\vec{a}$$ points at an angle between $$\vec{a}_{c}$$ and $$\vec{a}_{T}$$.

Example $$\PageIndex{3}$$: Total Acceleration during Circular Motion

A particle moves in a circle of radius r = 2.0 m. During the time interval from t = 1.5 s to t = 4.0 s its speed varies with time according to

$v(t) = c_{1} - \frac{c_{2}}{t^{2}}, c_{1} = 4.0\; m/s, c_{2} = 6.0\; m \cdotp s \ldotp$

What is the total acceleration of the particle at t = 2.0 s?

Strategy

We are given the speed of the particle and the radius of the circle, so we can calculate centripetal acceleration easily. The direction of the centripetal acceleration is toward the center of the circle. We find the magnitude of the tangential acceleration by taking the derivative with respect to time of |v(t)| using Equation \ref{4.31} and evaluating it at t = 2.0 s. We use this and the magnitude of the centripetal acceleration to find the total acceleration.

Solution

Centripetal acceleration is

$v(2.0\; s) = \left(4.0 - \dfrac{6.0}{(2.0)^{2}}\right) m/s = 2.5\; m/s \nonumber$

$a_{c} = \frac{v^{2}}{r} = \frac{(2.5\; m/s)^{2}}{2.0\; m} = 3.1\; m/s^{2} \nonumber$

directed toward the center of the circle. Tangential acceleration is

$a_{T} = \Big| \frac{d \vec{v}}{dt} \Big| = \frac{2 c_{2}}{t^{3}} = \frac{12.0}{(2.0)^{3}} m/s^{2} = 1.5\; m/s^{2} \ldotp \nonumber$

Total acceleration is

$|\vec{a}| = \sqrt{3.1^{2} + 1.5^{2}} m/s^{2} = 3.44\; m/s^{2}$

and $$\theta$$ = tan−1 $$\left(\dfrac{3.1}{1.5}\right)$$ = 64° from the tangent to the circle. See Figure $$\PageIndex{6}$$.

Significance

The directions of centripetal and tangential accelerations can be described more conveniently in terms of a polar coordinate system, with unit vectors in the radial and tangential directions. This coordinate system, which is used for motion along curved paths, is discussed in detail later in the book.


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Time and Work Questions

Q:

50 men can build a tank in 40days, but though they begin the work together, 5 men quit every ten days. The time needed to build the tank is ?

 A) 50 days B) 48 days C) 47.5 days D) 49 days

Explanation:

50 men can build a tank in 40 days
Assume 1 man does 1 unit of work in 1 day
Then the total work is 50×40 = 2000 units

50 men work in the first 10 days and completes 50×10 = 500 units of work
45 men work in the next 10 days and completes 45×10 = 450 units of work
40 men work in the next 10 days and completes 40×10 = 400 units of work
35 men work in the next 10 days and completes 35×10 = 350 units of work
So far 500 + 450 + 400 + 350 = 1700 units of work is completed and

remaining work is 2000 - 1700 = 300 units

30 men work in the next 10 days. In each day, they does 30 units of work.

Therefore, additional days required = $\inline \fn_jvn \small \frac{300}{30}=10$

Thus, total 10+10+10+10+10 = 50 days required.


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# What are the products of the reaction between fluorine and ammonia?

What are the products of the reaction between fluorine and ammonia? Is it the same as with chlorine and ammonia, which depend upon which reactant is taken in excess?

Fluorine reacts with ammonia to give dinitrogen and hydrogen fluoride as the major product.

The Formation of Dinitrogen Tetrafluoride in the Reaction of Fluorine and Ammonia - J. Am. Chem. Soc. 1959, 81 (23), 6338-6339 - states that dinitrogentetrafluoride accounts for 6% of the yield while the major reaction is:

$\ce{3F2 + 2NH3 → N2 + 6HF}$

The same article quotes in its first paragraph that the result remained the same irrespective of the concentration of the reactants. (In the case of chlorine it matters which reactant is taken in excess)

• I wonder if the N2F2 produced is just the product of NF3 degredation... Feb 15, 2015 at 14:12

Actually, there are three types of reaction occuring between ammonia and fluorine which are given here. They all depend on reaction conditions and concentration of reactants. All the reactions are dangerous and produce volatile compounds. Reaction should be done under expert supervision. Please do not try this if you new in chemistry field.

1. $\ce{2NH3 + 3F2 ->[\Delta] 6HF + N2}$ (mentioned)
2. $$\ce{2NH3 + 6F2 → 6HF + 2NF3}$$

Ammonia react with fluorine to produce hydrogen fluoride and nitrogen(III) fluoride. The reaction takes place at the low temperature.

1. $$\ce{4NH3 + 3F2 → 3NH4F + NF3}$$

Ammonia react with fluorine to produce ammonium fluoride and nitrogen(III) fluoride. Ammonia should be aqueous. In this reaction, the catalyst can be metallic copper. Impurities: tetrafluorohydrazine $\ce{N2F4}$.

The reaction is this $$\ce{5 F2 {(g)} + 2 NH3 {(g)} -> N2F4 {(g)} + 6 HF {(g)}}$$ And it doesn’t depend upon which reactant is taken in excess; I believe it’s because fluorine is just too reactive.

• "And it doesn’t depend upon which reactant is taken in excess." I find that hard to believe. Can you provide a source? As worded your statement implies that reactive 1 mol of F2 with 100 mol of ammonia will still form N2F4. May 15, 2015 at 12:19
• N2F4 is a minor product , major product is always N2 Jan 11, 2020 at 14:02

When fluorine reacts with an excess of ammonia it forms $\ce{NH4F}$. When fluorine reacts with a limited supply of ammonia it forms $\ce{HF}$. Both these reactions yield nitrogen ($\ce{N2}$) gas.


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# Galaxy And Mass Assembly (GAMA): the wavelength dependence of galaxy structure versus redshift and luminosity

Kennedy, R, Bamford, SP, Baldry, IK, Häußler, B, Holwerda, BW, Hopkins, AM, Kelvin, LS, Lange, R, Moffett, AJ, Popescu, CC, Taylor, EN, Tuffs, RJ, Vika, M and Vulcani, B (2015) Galaxy And Mass Assembly (GAMA): the wavelength dependence of galaxy structure versus redshift and luminosity. Monthly Notices of the Royal Astronomical Society, 454 (1). pp. 806-817. ISSN 0035-8711

## Abstract

We study how the sizes and radial profiles of galaxies vary with wavelength, by fitting Sersic functions simultaneously to imaging in nine optical and near-infrared bands. To quantify the wavelength dependence of effective radius we use the ratio, $\mathcal{R}$, of measurements in two restframe bands. The dependence of Sersic index on wavelength, $\mathcal{N}$, is computed correspondingly. Vulcani et al. (2014) have demonstrated that different galaxy populations present sharply contrasting behaviour in terms of $\mathcal{R}$ and $\mathcal{N}$. Here we study the luminosity dependence of this result. We find that at higher luminosities, early-type galaxies display a more substantial decrease in effective radius with wavelength, whereas late-types present a more pronounced increase in Sersic index. The structural contrast between types thus increases with luminosity. By considering samples at different redshifts, we demonstrate that lower data quality reduces the apparent difference between the main galaxy populations. However, our conclusions remain robust to this effect. We show that accounting for different redshift and luminosity selections partly reconciles the size variation measured by Vulcani et al. with the weaker trends found by other recent studies. Dividing galaxies by visual morphology confirms the behaviour inferred using morphological proxies, although the sample size is greatly reduced. Finally, we demonstrate that varying dust opacity and disc inclination can account for features of the joint distribution of $\mathcal{R}$ and $\mathcal{N}$ for late-type galaxies. However, dust does not appear to explain the highest values of $\mathcal{R}$ and $\mathcal{N}$. The bulge-disc nature of galaxies must also contribute to the wavelength-dependence of their structure.

Item Type: Article This is a pre-copyedited, author-produced PDF of an article accepted for publication in Monthly Notices of the Royal Astronomical Society following peer review. The version of record Galaxy And Mass Assembly (GAMA): the wavelength dependence of galaxy structure versus redshift and luminosity is available online at: http://dx.doi.org/doi:10.1093/mnras/stv2032 astro-ph.GA; astro-ph.GA Q Science > QB Astronomy Astrophysics Research Institute Oxford University Press Author 05 Oct 2015 09:29 10 Sep 2017 01:05 http://researchonline.ljmu.ac.uk/id/eprint/2118


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# Accelerator Physics (physics.acc-ph)

• This paper will detail changes in the operational paradigm of the Fermi National Accelerator Laboratory (FNAL) magnetron $H^{-}$ ion source due to upgrades in the accelerator system. Prior to November of 2012 the $H^{-}$ ions for High Energy Physics (HEP) experiments were extracted at ~18 keV vertically downward into a 90 degree bending magnet and accelerated through a Cockcroft-Walton accelerating column to 750 keV. Following the upgrade in the fall of 2012 the $H^{-}$ ions are now directly extracted from a magnetron at 35 keV and accelerated to 750 keV by a Radio Frequency Quadrupole (RFQ). This change in extraction energy as well as the orientation of the ion source required not only a redesign of the ion source, but an updated understanding of its operation at these new values. Discussed in detail are the changes to the ion source timing, arc discharge current, hydrogen gas pressure, and cesium delivery system that were needed to maintain consistent operation at >99% uptime for HEP, with an increased ion source lifetime of over 9 months.
• The magnetron-style $H^{-}$ ion sources currently in operation at Fermilab use piezoelectric gas valves to function. This kind of gas valve is sensitive to small changes in ambient temperature, which affect the stability and performance of the ion source. This motivates the need to find an alternative way of feeding H2 gas into the source. A solenoid-type gas valve has been characterized in a dedicated off-line test stand to assess the feasibility of its use in the operational ion sources. $H^{-}$ ion beams have been extracted at 35 keV using this valve. In this study, the performance of the solenoid gas valve has been characterized measuring the beam current output of the magnetron source with respect to the voltage and pulse width of the signal applied to the gas valve.
• Fermilab is committed to upgrade its accelerator complex to support HEP experiments at the intensity frontier. The ongoing Proton Improvement Plan (PIP) enables us to reach 700 kW beam power on the NuMI neutrino targets. By the end of the next decade, the current 400 MeV normal conducting LINAC will be replaced by an 800 MeV superconducting LINAC (PIP-II) with an increased beam power >50% of the PIP design goal. Both in PIP and PIP-II era, the existing Booster is going to play a very significant role, at least for next two decades. In the meanwhile, we have recently developed an innovative beam injection and bunching scheme for the Booster called "early injection scheme" that continues to use the existing 400 MeV LINAC and implemented into operation. This scheme has the potential to increase the Booster beam intensity by >40% from the PIP design goal. Some benefits from the scheme have already been seen. In this paper, I will describe the basic principle of the scheme, results from recent beam experiments, our experience with the new scheme in operation, current status, issues and future plans. This scheme fits well with the current and future intensity upgrade programs at Fermilab.
• Background The nuclear structure of the cluster bands in $^{20}$Ne presents a challenge for different theoretical approaches. It is especially difficult to explain the broad 0$^+$, 2$^+$ states at 9 MeV excitation energy. Simultaneously, it is important to obtain more reliable experimental data for these levels in order to quantitatively assess the theoretical framework. Purpose To obtain new data on $^{20}$Ne $\alpha$ cluster structure. Method Thick target inverse kinematics technique was used to study the $^{16}$O+$\alpha$ resonance elastic scattering and the data were analyzed using an \textitR matrix approach. The $^{20}$Ne spectrum, the cluster and nucleon spectroscopic factors were calculated using cluster-nucleon configuration interaction model (CNCIM). Results We determined the parameters of the broad resonances in \textsuperscript20Ne: 0$^+$ level at 8.77 $\pm$ 0.150 MeV with a width of 750 (+500/-220) keV; 2$^+$ level at 8.75 $\pm$ 0.100 MeV with the width of 695 $\pm$ 120 keV; the width of 9.48 MeV level of 65 $\pm$ 20 keV and showed that 9.19 MeV, 2$^+$ level (if exists) should have width $\leq$ 10 keV. The detailed comparison of the theoretical CNCIM predictions with the experimental data on cluster states was made. Conclusions Our experimental results by the TTIK method generally confirm the adopted data on $\alpha$ cluster levels in $^{20}$Ne. The CNCIM gives a good description of the $^{20}$Ne positive parity states up to an excitation energy of $\sim$ 7 MeV, predicting reasonably well the excitation energy of the states and their cluster and single particle properties. At higher excitations, the qualitative disagreement with the experimentally observed structure is evident, especially for broad resonances.
• The paper discusses an advanced level information system to support educational, research and scientific activities of the Department "Electrophysical Facilities" (DEF) of the National Research Nuclear University "MEPhI" (NRNU MEPhI), which is used for training of specialists of the course "Physics of Charged Particle Beams and Accelerator Technology".
• Multiphoton processes in undulators with plane polarized magnetic field are considered. It is shown that the use of strong magnetic fields in the undulator, for beams with relatively low energy makes it possible to increase substantially the frequencies of the amplified electromagnetic waves without noticeably decreasing the gain. The absorption, emission probabilities and the gain are calculated.
• A high-energy muon collider scenario requires a final cooling system that reduces transverse emittance to ~25 microns (normalized) while allowing longitudinal emittance increase. Ionization cooling using high-field solenoids (or Li Lens) can reduce transverse emittances to ~100 microns in readily achievable configurations, confirmed by simulation. Passing these muon beams at ~100 MeV/c through cm-sized diamond wedges can reduce transverse emittances to ~25 microns, while increasing longitudinal emittances by a factor of ~25. Implementation will require optical matching of the exiting beam into downstream acceleration systems.
• In a partnership with SLAC National Accelerator Laboratory (SLAC) and Jefferson Lab, Fermilab will assemble and test 17 of the 35 total 1.3 GHz cryomodules for the Linac Coherent Light Source II (LCLS-II) Project. These devices will be tested at Fermilab's Cryomodule Test Facility (CMTF) within the Cryomodule Test Stand (CMTS-1) cave. The problem of magnetic pollution became one of major issues during design stage of the LCLS-II cryomodule as the average quality factor of the accelerating cavities is specified to be $2.7 \times 10^{10}$. One of the possible ways to mitigate the effect of stray magnetic fields and to keep it below the goal of 5 mGauss involves the application of low permeable materials. Initial permeability and magnetic measurement studies regarding the use of 316L stainless steel material indicated that cold work (machining) and heat affected zones from welding would be acceptable.
• An "ultimate" high intensity proton source for neutrino factories and/or muon colliders was projected to be a ~4 MW multi-GeV proton source providing short, intense proton pulses at ~15 Hz. The JPARC ~1 MW accelerators provide beam at parameters that in many respects overlap these goals. Proton pulses from the JPARC Main Ring can readily meet the pulsed intensity goals. We explore these parameters, describing the overlap and consider extensions that may take a JPARC-like facility toward this "ultimate" source. JPARC itself could serve as a stage 1 source for such a facility.


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## Convergence groups are Fuchsian groups.(English)Zbl 0785.57004

Between 1927 and 1942, Nielsen wrote a series of long papers of diffeomorphisms of surfaces. In recent years, Thurston completed Nielsen’s theory of surface diffeomorphisms looking at it from the more general point of view of Teichmüller theory and modular groups. The last paper of the above series appeared in 1942 and concerned a problem which became later known as the Nielsen realization problem: given a finite group $$G$$ of isotopy classes of diffeomorphisms of a surface $${\mathcal F}$$ (a finite subgroup of the mapping class group $$\pi_ 0\text{ Diff}({\mathcal F})\cong \text{Out }\pi_ 1{\mathcal F}$$), can one choose representatives realizing $$G$$ as a group of diffeomorphisms of the surface? Nielsen’s methods give a solution to this problem if there exists a “simple axis” for $$G$$: an isotopy class of a simple closed curve whose images under $$G$$ give disjoint isotopy classes. The method is as follows: suppose $${\mathcal F}$$ is covered by the hyperbolic plane $${\mathbf H}^ 2=\text{Int } D^ 2$$ (the interior of the 2-disk) and let $$F$$ be the covering group which is a Fuchsian group (a discrete group of Möbius transformations). Purely algebraically, the group $$G$$ defines an extension $$1\to F\hookrightarrow E\to G\to 1$$, and by lifting representatives of elements of $$G$$ to $$D^ 2$$ and restricting to $$S^ 1=\partial D^ 2$$ one gets a well defined action of $$E$$ on $$S^ 1$$ (but not on $$D^ 2$$ where only $$F$$ acts). The problem then is to extend the action of $$G$$ on $$S^ 1$$ to an action on $$D^ 2$$ extending the given action of $$F$$. If there exists a simple axis for $$G$$ one may first extend to it preimages in $${\mathbf H}^ 2$$ (a collection of disjoint geodesics) and then use some inductive procedure.
The present work can be considered as the anlogue of Nielsen’s paper in the more difficult case where no simple axis exists. By a careful geometric analysis of the situation, again an extension of $$E$$ from $$S^ 1$$ to $$D^ 2$$ is constructed thus giving the first completely constructive solution of the Nielsen realization problem (the first complete solution has been given by Kerckhoff in 1983 using methods of Thurston from Teichmüller theory). However, the present paper is formulated in a more general setting which has other important applications. In fact, one starts with a “convergence group” $$E$$ acting on the circle $$S^ 1$$; this generalizes the action of a Fuchsian group on $$S^ 1=\partial D^ 2$$ maintaining some of its essential features: for any sequence of distinct elements of $$E$$ there exist $$x,y\in S^ 1$$ and a subsequence $$f_ i$$ such that on $$S^ 1-\{x,y\}$$, $$f_ i\to y$$ and $$f_ i^{-1}\to x$$ uniformly on compact sets. The main result of the paper states that a convergence group is conjugate in $$\text{Homeo}(S^ 1)$$ to the restriction of a Fuchsian group (this result has been obtained also by Casson and Jungreis using different 3- dimensional methods). For example, a group $$E$$ acting on $$S^ 1$$ and containing the restriction of a Fuchsian group as a subgroup of finite index is a convergence group, thus the solution of the Nielsen realization problem. The other major application is the solution of the Seifert fiber space conjecture: if $$M$$ is a compact orientable irreducible 3-manifold with infinite fundamental group containing an infinite cyclic normal subgroup $${\mathbf Z}$$ then $$M$$ is a Seifert fiber space (this is part of Thurston’s geometrization conjecture for 3-manifolds). The solution depends on a result of Mess who showed that the quotient $$\pi_ 1 M/{\mathbf Z}$$ is (basically) an $$S^ 1$$-convergence group. By the main result of the present paper, the quotient is isomorphic to a Fuchsian group, therefore $$\pi_ 1 M$$ is isomorphic to the fundamental group of a Seifert fiber space and by a result of Scott $$M$$ is homeomorphic to a Seifert fiber space.
This is certainly an impressive work bringing finally to a conclusion some old work of Nielsen.

### MSC:

 57M50 General geometric structures on low-dimensional manifolds 20F65 Geometric group theory 57N10 Topology of general $$3$$-manifolds (MSC2010) 30F99 Riemann surfaces 57N05 Topology of the Euclidean $$2$$-space, $$2$$-manifolds (MSC2010) 57R50 Differential topological aspects of diffeomorphisms
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Computer Engineering and Applications ›› 2021, Vol. 57 ›› Issue (21): 263-269.

### Remote Sensing Military Target Detection Algorithm Based on Lightweight YOLOv3

QIN Weiwei, SONG Tainian, LIU Jieyu, WANG Hongwei, LIANG Zhuo

1. 1.School of Nuclear Engineering, Rocket Force University of Engineering, Xi’an 710025, China
2.School of Artificial Intelligence, OPtics and ElectroNics（iOPEN）, Northwestern Polytechnical University, Xi’an 710072, China
3.China Academy of Launch Vehicle Technology, Beijing 100076, China
• Online:2021-11-01 Published:2021-11-04

### 基于轻量化YOLOv3的遥感军事目标检测算法

1. 1.火箭军工程大学 核工程学院，西安 710025
2.西北工业大学 光电与智能研究院，西安 710072
3.中国运载火箭研究院，北京 100076

Abstract:

In the process of intelligent missile penetration, detecting enemy anti-missile positions from massive remote sensing image data has great application value. Due to the limited computing power of the missile-borne deployment environment, this paper designs a remote sensing target detection algorithm that takes into account lightweight, detection accuracy and detection speed. A typical remote sensing military target data set is produced, and the data set is clustered and analyzed by the K-means algorithm. The MobileNetV2 network is used to replace the backbone network of the YOLOv3 algorithm to ensure the lightweight and detection speed of the network. A lightweight and efficient channel coordinated attention module and a target rotation invariance detection module suitable for remote sensing target characteristics are proposed, and they are embedded in the detection algorithm to improve the detection accuracy on the basis of network lightweight. Experimental results show that the accuracy rate of the algorithm in this paper reaches 97.8%, an increase of 6.7 percentage points, the recall rate reaches 95.7%, an increase of 3.9 percentage points, the average detection accuracy reaches 95.2%, an increase of 4.4 percentage points, and the detection speed reached 34.19 images per, and the network size is only 17.5?MB. The results show that the algorithm in this paper can meet the comprehensive requirements of intelligent missile penetration.


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# Area of Hollow Cylinder: Defining Terms and Cross Sectional Area

Glide to success with Doorsteptutor material for NEST Mathematics: fully solved questions with step-by-step explanation- practice your way to success.

A hollow cylinder is one which is empty from inside and has some difference between the internal and external radius.

Tubes, circular buildings, straws these are all examples of a hollow cylinder.

## Area of a Hollow Cylinder

### Defining Terms

Let be the outer radius of the given cylinder and be its inner radius and ‘h’ be its height.

be the outer circumference and be the inner circumference.

and be the outer and inner surface areas respectively.

t be the thickness of the cylinder

### Area of Hollow Cylinder Formulas

The Circumference of a circle (C) is given by:

therefore,

The Lateral Surface Area for a cylinder is:

therefore,

the external curved surface area

the internal curved surface area

Thus, Lateral Surface Area of a hollow cylinder

### Cross Sectional Area

Let A be the area of a cross-section of a hollow cylinder,

for a circle, therefore,

for the area enclosed by

for the area enclosed by

for the cross-sectional area of hollow cylinder

Total Surface Area of a Hollow Cylinder:

Example: Find (in ) the curved surface area of a hollow cylinder with thickness 2 cm external radius 8 cm and height is 20 cm.

Solution:

Let the external radius, internal radius , and height of the hollow cylinder

Thickness

Height

Here using formula,

Curved surface area of hollow cylinder

Sum of the 8 and 6,

Here 14 divided by 7

Example 2: The external surface area of a hollow cylinder is . Its internal diameter is 16 cm and height 15 cm. Find the curved surface area.

Solution:

Here given, external surface area of a hollow cylinder

Formula of surface area of cylinder

Height

Values are put in the equation (1),

Both the side cancel.

2 and 15 take the other side,

First, we divide 540 by 15.

Curved Surface area

Put the value,


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Showing two splitting fields are different

I am trying to do problem 3.28 from the Algebra questions on this site. It says the following:

How would you find the Galois group of $x^3+2x+1$? Adjoin a root to $\mathbb Q$. Can you say something about the roots of $x^3+3x+1$ in this extension?

I can easily find that the Galois group of $x^3+2x+1$ is $S_3$. Let $\theta$ be a root of $x^3+2x+1$. The second question is a bit vague, but I assume the intended answer is that $x^3+3x+1$ has no roots in $\mathbb Q(\theta)$. Indeed, if $x^3+3x+1$ had a root in $\mathbb Q(\theta)$, then it would have a root in the splitting field of $x^3+2x+1$. However, since this splitting field is Galois, this would imply that all of the roots of $x^3+3x+1$ lie in the splitting field of $x^3+2x+1$. We can also see easily that the Galois group of $x^3+3x+1$ is $S_3$, so this would mean that the splitting fields of $x^3+2x+1$ and $x^3+3x+1$ are actually the same.

So my question is:

How can we show that the splitting fields of $x^3+2x+1$ and $x^3+3x+1$ are different fields?

• Look at their discriminants. The discriminant of the first is $-59$ and of the second is $-135$. Therefore, $\sqrt{59}i$ and $\sqrt{15}i$ belong to their splitting fields, respectively. If the fields were the same then its degree over $\mathbb{Q}$ would be multiple of $4$, since it would contain the extension by the roots of $x^2+15$ and then the further extension by the roots of $x^2+59$, which doesn't split in the first. – YAlexandrov Mar 9 '18 at 14:33
• Ahh yes that works. Thank you! – Colin Defant Mar 9 '18 at 14:44
• @YAlexandrov It may be something you would rather not do, but IMHO you fleshing out that comment to an answer would be welcomed. Your call, of course. – Jyrki Lahtonen Mar 9 '18 at 21:49


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# 1 Minus 1 Equals What? (Fun with sums and limits)

By Alex Beal
August 5, 2011

Here’s a cool problem I came across when reviewing for one of my calculus exams:

You begin with the numbers 0 through 1. Every iteration removes 1/3 from the remaining segments. As stated above, the expression for the total amount removed after m iterations is:

The first iteration removes 1/3. The second removes 1/3 from the remaining two segments. The remaining segments are 1/3 long, and 1/3 of that is 1/9. We do that once for each remaining segment, so 2/9 is removed. And so on.

What if we iterate an infinite number of times? Then we have a convergent geometric series. Finding the sum is easy:

So, what’s the total amount or length of numbers removed? 1. That’s the entire length. It seems like we’ve removed the entire segment. But wait a minute. If we look at the image of the original question above, we see that after every iteration, there are 2n+1 segments remaining. When n=0, there are 2 segments. When n=1, there are 4 segments, etc. As we remove more and more, 2n+1 approaches infinity. In other words, the limit does not exist, and the number of segments goes to infinity.

So, even though we’ve seemingly removed the entire length, there are an infinite number of segments remaining.


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Limited access

If:

$$W=\left\{\begin{pmatrix} x_1 \\\ x_2 \\\ x_3\end{pmatrix}\in\mathbb{R}^3\ \biggr|\ \ x_1+x_2+x_3=1\right\}$$

...then which of the following statements are TRUE?

Select ALL that apply.

A

$W$ is not a subspace of $\mathbb{R}^3$ because it doesn't contain the zero vector

B

$W$ is not a subspace of $\mathbb{R}^3$ because it is not closed under vector addition

C

$W$ is not a subspace of $\mathbb{R}^3$ because it is not closed with respect to scalar multiplication

D

$W$ is a subspace of $\mathbb{R}^3$ since it is the solution set of a linear system of equations

Select an assignment template


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# Cos 30 Degrees

In trigonometry, the cosine function is defined as the ratio of the adjacent side to the hypotenuse. If the angle of a right triangle is equal to 30  degrees, and then the value of cosine at this angle i.e., the value of Cos 30 degree is in a fraction form as √3/2. It is noted that the value of Sec 30 will be reciprocal of Cos 30 value.

## Cos 30 Value

Cos 30 degrees is written as cos 30° and has a value in fraction form as √3/2.

Cos 30° = √3/2

Cos 30° = √3/2 is an irrational number and equals to 0.8660254037 (decimal form).

Therefore, the exact value of cos 30 degrees is written as 0.8660 approx.

√3/2 is the value of Cos 30° which is a trigonometric ratio or trigonometric function of a particular angle.

## Cos 30

Another alternative form of Cos 30° is pi/6 or π/6 or Cos 33 (⅓)g

 Form Formula Value Trigonometric ratio √3 / 2 0.8660254037 Circular system pi/6 or π/6 0.8660254037 Centesimal system Cos 33 (⅓)g 0.8660254037

### Cos 30 Degrees Proof

Now that you know the value of Cos 30 degrees, let’s explore how to derive this value.

We will study the two approaches to derive it.

1. Theoretical approach
2. Practical approach

Theoretical approach

Knowing the property that length of an opposite side is half of the length of the hypotenuse – Right triangle property for an angle equal to 30 degrees in the given right triangle.

As we know two sides those are the length of the opposite side and hypotenuse but the third side is unknown i.e adjacent side. We need to find this side to find the value of Cos 30 degrees.

We will use the Pythagorean theorem to find the value of this third side

In triangle AOB,

AO 2 = AB2 + AB2

Hypotenuse = d

Opposite side = d/2

d2 = (d/2)2 + AB2

d2 = d2/4 + AB2

d2 – d2/4 = AB2

(4d2 – d2)/4 = AB2

3d2/4 = AB2

Sqrt (3) x d/2 = AB

AB/d = √(3) / 2

As d is the length of the opposite side, so

Length of adjacent / length of hypotenuse = √3 / 2

The value of angle BOA in the given right triangle is pi/6 and the given ratio as per the definition of cosine ratio represents the value of Cos 30 degrees

Cos 30° =Length of adjacent/length of hypotenuse = √3/2 = 0.8660254037

Practical Approach

Another way to calculate the value of Cos 30 degree is through a practical approach. So if we make a right triangle using constructions, Cos pi/6 can be calculated.

Step 1: Mark a point P on the plane and horizontally draw a line on it.

Step 2: With the help of protractor makes an angle of 300 with the centre as P and taking baseline as the drawn line in step 1.

Step 3: Use the ruler to draw a line to make an angle of 30 degrees.

Step 4: Using compass draw an arc on the line of angle 30 degrees with any length. Mark that point as Q.

Step 5: From Q draw a line perpendicular to the base (horizontal line). Mark it as R (Intersecting point of the perpendicular line to the base.)

Now after drawing a right-angled triangle POQ, we can calculate the value of Cos 30 degrees.

Cos 30° =Length of the adjacent/length of the hypotenuse = OP/OQ.

In the above construction, the length of the side OP is taken as 7.5 cm and the length of the adjacent side is unknown. But if we measure it with a ruler, It comes out to be 6.5 cm.

Cos 30° = OP/OQ = 6.5/7.5 = 0.866666…

Stay tuned to BYJU’S to learn math in the most fun and engaging way.

 Related Links Cos 60 Degree Cos 0 Degree Cosine Function Cosine Rule


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# GFTSHP - Editorial

Setter: Pranav Dev
Tester: Abhinav Sharma, Aryan
Editorialist: Lavish Gupta

Simple

Sorting

# PROBLEM:

Chef wants to impress Chefina by giving her the maximum number of gifts possible.

Chef is in a gift shop having N items where the cost of the i^{th} item is equal to A_{i}.
Chef has K amount of money and a 50 \% off discount coupon that he can use for at most one of the items he buys.

If the cost of an item is equal to X, then, after applying the coupon on that item, Chef only has to pay {\bf \left\lceil \frac{X}{2} \right\rceil} (rounded up to the nearest integer) amount for that item.

Help Chef find the maximum number of items he can buy with K amount of money and a 50 \% discount coupon given that he can use the coupon on at most one item.

# EXPLANATION:

Let M be the maximum number of items that Chef will buy, and let \{i_1, i_2, \ldots i_M\} be the list of items that costs minimum among all possible list of items of size M. Also, let S = \sum_{\alpha = 1}^{M} A_{i_{\alpha}}

Observation 1: Chef will always use the discount coupon on the most expensive item. This is because the amount of money that Chef will pay will be S - discount, and as the price increases, the discount increases.

Observation 2: The list of items will have the M lowest priced items. To prove this, let us introduce a new item j in the list of items and remove i_\alpha such that A_{i_\alpha} < A_j. We can make 4 cases on {discount on A_{i_\alpha}, discount on A_{j}} and in each case, we can see that the original list results in less overall cost.

So we can first sort the items in the increasing order of price, and for each i such that 1 \leq i \leq N, check if we can take first i elements by applying discount on the i^{th} item. The largest feasible i will be our answer.

# TIME COMPLEXITY:

O(N \cdot \log{N}) for each test case.

# SOLUTION:

Editorialist's Solution
#define ll long long
#define fo(i , n) for(ll i = 0 ; i < n ; i++)
#include<bits/stdc++.h>
using namespace std ;

void solve()
{
ll n , k ;
cin >> n >> k ;

ll arr[n] ;
for(int i = 0 ; i < n ; i++)
cin >> arr[i] ;
sort(arr , arr+n) ;

ll sum = 0, ans = 0;

for(ll i = 0 ; i < n ; i++)
{
ll new_cost = (arr[i] + 1)/2 ; // ceil division
if(sum + new_cost <= k)
{
ans++ ;
sum += arr[i] ; // updating sum for next iteration
}
else
break ;
}

cout << ans << '\n' ;
return ;
}

int main()
{

ll t = 1 ;
cin >> t ;
while(t--)
{
solve() ;
}

cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";

return 0;
}

Can you please explain why are you using new_cost as 50% discount for again and again? Because in the question, its given that we can use that coupon for at most one item.

2 Likes

If you review the items in ascending order, if you can afford the next item at full price from your remaining budget, it is always acceptable to pay that to get the maximum gifts solution.

Why? Because either you can afford the next more expensive item at half-price, or you can’t, but there’s no benefit to taking the current affordable item at half price because it won’t make it easier to afford subsequent items.

What is that line which is written just before return 0 statement.
What is its role, what is that syntax??

tc  = int(input())
for i in range(tc):
next_line = input()
n, k = list(map(int, next_line.split()))
next_line = input()
prices = list(map(int, next_line.split()))
sorted_prices = sorted(prices)
cost = 0
j = 0
while j < n:
if cost + sorted_prices[j] > k:
break
cost += sorted_prices[j]
j += 1

if cost + round(sorted_prices[j] / 2) <= k:
j += 1
print(j)

I am getting the Runtime Error (NZEC) for 2 of the 3 tasks for this solution. Any idea why this error could occur in the above code.

#include <iostream>
#include<algorithm>
using namespace std;
int Discount(int a)
{
float x=a/2;
if(x-(a/2)<0.5) return (a/2);
else return ((a/2)+1);
}
int ITEMS(int a[],int n,int k)
{
int i,ans=0;
for(i=0;i<n;++i)
{
if(k>a[i])
{
k=k-a[i]; ans++;
}
else
{
if(k>=Discount(a[i]))
{
++ans;
break;
}break;
}
}
return ans;
}
int main()
{
int t; cin>>t;
while(t–)
{
int n,k; cin>>n>>k;
int a[n];

for(int i=0;i<n;++i) cin>>a[i] ;

sort(a,a+n);
cout<<ITEMS(a,n,k)<<endl;
}
return 0;
}

can anyone say the mistake in the code?

look at the solution carefully… we are just checking the discount for the current item…we are actually adding a[i] in sum and not new_cost which means we are providing discount only once

1 Like

O yes… Now I got it. I was doing the same approach using C, but got TLE for the last task. Maybe my sorting algo was taking too much time

It is the cerr command which is used in this case to print the time taken (of course this won’t be treated as output) but just to be safe people want to check so that they won’t get TLE.

/*#include <bits/stdc++.h>

using ll = long long;
using namespace std;

#define endl “\n”

void solveTestCase()
{
ll N, K;
cin >> N >> K;
ll A[N];
for (int i = 0; i < N; i++)
cin >> A[i];

sort(A, A + N);

int i = 0;
while (K > 0 && i < N)
{
if (A[i] > K)
{
if ((A[i] / 2) + 1 <= K)
{
i++;
}
break;
}
K -= A[i];
i++;
}
cout << i << endl;

}

int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
int testCases;
cin >> testCases;
while (testCases–)
{
solveTestCase();
}

return 0;

}
*/
Why it is giving WA in first 2 testcases and passed in third Testcase ??

1 Like

In the case when all items can be purchased, your final if statement will attempt to check beyond the end of available items. You could instead check this condition within the if statement that is inside the while loop

(A[i] / 2) + 1 will give the wrong answer for checking discount on even-priced items.

@joffan but the max number of gifts available in the shop is equal to n which in the while loop I am checking. So it shouldn’t check beyond it.

After your while loop, you are not checking whether j < n or not.
Maybe accessing sorted_prices[n] resulted in array index out of bounds error.

#include <bits/stdc++.h>
using namespace std;

int main() {
long long int t;
cin>>t;
while(t--)
{
long long int n,k,count=0,sum=0;
cin>>n>>k;
vector <long long int> v;
for(long long int i=0;i<n;i++)
{
long long int l;
cin>>l;
v.push_back(l);
}
sort(v.begin(),v.end());
long long j=accumulate(v.begin(),v.end(),0);
if(j<=k)
{
cout<<n<<endl;
}
else
{
for(long long int i=0;i<v.size();i++)
{
sum=sum+v.at(i);
if(sum<=k)
{
count++;
}
else
{
sum=sum-v.at(i);
sum=sum+round((double)v.at(i)/2);
if(sum<=k)
{
count++;
}
i=v.size();
}
}
cout<<count<<endl;
}
}
return 0;
}

I am getting the desired output but on submitting the above code WA is being returned can anyone please tell me what’s wrong with the above code.

Thanks

As I said, the test I’m talking about is after the while loop. Your index has (potentially) gone out of range.

I’m wondering if you tried the sample input for the problem. The second case given there allows all items to be purchased full price, which breaks your code.

def main():

testcases=int(input())

for test in range(testcases):

n,k=map(int,input().split())

l=list(map(int,input().split()))

l.sort()

index=0

s=0

count=0

for i in range(n):

s+=l[i]

if s>k:
break

count+=1

s=s-l[i]

if count==n:

print(n)

else:

a=ceil(l[i]/2)

if s+a<=k:

print(count+1)

else:

print(count)

My solution in Python

@joffan @adithya5768 Thanks a lot. Able to figure the issue now.
But I am still getting WA for few test cases. Below is the updated code.

tc  = int(input())
for i in range(tc):
next_line = input()
n, k = list(map(int, next_line.split()))
next_line = input()
prices = list(map(int, next_line.split()))
sorted_prices = sorted(prices)
cost = 0
j = 0
while j < n:
if cost + sorted_prices[j] > k:
break
cost += sorted_prices[j]
j += 1

if (j < n):
if (cost + round(sorted_prices[j] / 2) <= k):
j += 1
print(j)


text:
# Chapter 6 - Section 6.3 - Exponential Functions - 6.3 Assess Your Understanding: 25

$\approx 149.952$

#### Work Step by Step

Use a calculator to evaluate the expression, and then round-off the answer to three decimal places: $125e^{0.026(7)} \approx 149.952$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
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# Trident Revision Mod 1.17.1 (A New Way to Get Tridents)

Minecraft Trident Revision Mod 1.18.1, 1.17.1 introduces a new way for players to earn the fabled Trident weapon without having to rely entirely on chance. The Trident has always been the most sought-after weapon in the game since it is not only strong but it can also be enchanted with an attribute that allows the trident to return to the player, making them feel like actual Aquaman. The likelihood of obtaining a valuable Trident, on the other hand, is totally random, as they can only be found underwater, in random chests scattered across the overworld. Players may make them with the Revision by following some of the most sensible recipes.

 Trident Revision Mod 1.17.1 (A New Way to Get Tridents)

Features:

• A new way to get the Trident has been introduced.

Pictures:

 Trident Revision Mod 1.17.1 (A New Way to Get Tridents)

 Trident Revision Mod 1.17.1 (A New Way to Get Tridents)

 Trident Revision Mod 1.17.1 (A New Way to Get Tridents)

 Trident Revision Mod 1.17.1 (A New Way to Get Tridents)

 Trident Revision Mod 1.17.1 (A New Way to Get Tridents)