About Gandhi Viswanathan

Professor of Physics, UFRN

Brazilian scientists turn down medals in repudiation of President Bolsonaro

Source:

https://en.mercopress.com/2021/11/08/brazilian-scientists-turn-down-medals-in-repudiation-of-president-bolsonaro


A group of Brazilian scientists who were awarded the National Order of Scientific Merit have decided to turn down the honours after President Jair Bolsonaro chose to remove one of them from the list.

The decision came in solidarity after Bolsonaro refused to give the award to a researcher who had conducted a study on the ineffectiveness of chloroquine against the coronavirus. Chloroquine is a drug against malaria which Bolsonaro has defended throughout the COVID-19 pandemic.

In a public letter released this weekend, 21 researchers from different universities and scientific centres have renounced the National Order of Scientific Merit, after the head of state excluded Marcus Vinícius Guimaraes Lacerda of the Oswaldo Cruz Foundation medical research centre (Fiocruz) and Adele Schwartz Benzaken, director of Fiocruz for the Amazon.

A study by Lacerda shows how the use of chloroquine, a drug against malaria and lupus, is not only completely useless in patients with coronavirus, but if administered in higher doses it can also cause arrhythmias to people with heart conditions.

Benzaken, for her part, served as director of the department in charge of analyzing and investigating the AIDS disease and viral hepatitis of the Ministry of Health before being fired with the arrival of Bolsonaro to power.

Both had been included in the list of those who would receive the award this year by the Ministry of Science and Technology.

“As scientists, we do not tolerate how denialism in general, peer harassment and recent cuts in federal budgets for science and technology have been used as tools to roll back the important advances made by the Brazilian scientific community in the last decades,” the researchers said in a letter published by Folha de Sao Paulo.

Bolsonaro, who since the beginning of COVID-19 has minimized its effects and even recommended chloroquine as a medication, removed the two scientists from the list included in the decree published Saturday in an extraordinary issue of the Official Gazette.

The scientists resigning their medals addressed their letter to the Ministry of Science and Technology: “We consider our presence on the list gratifying and we are very honoured with the possibility of receiving one of the highest recognitions that a scientist can receive in the country, but the tribute offered by a government that not only ignores science and actively boycotts the recommendations of specialists are not compatible with our careers.”

According to an investigation by Congress, the Government’s omissions in the face of the pandemic contributed to making Brazil the second country with the most deaths in the world, with more than 600,000 victims, and the third with more cases with 21.8 million infections.

The parliamentary committee that investigated the situation accused the president of nine crimes, including crimes against humanity, violation of sanitary measures and irregular use of public money.

O voto impresso é uma boa ideia?

Fonte: Folha de São Paulo:

https://www1.folha.uol.com.br/colunas/joel-pinheiro-da-fonseca/2021/08/o-voto-impresso-e-uma-boa-ideia.shtml


Por Joel Pinheiro da Fonseca, Economista, mestre em filosofia pela USP.

2.ago.2021 às 23h15

Os céticos sobre as urnas eletrônicas são crédulos demais na segurança da puração humana de cédulas de papel


 

Vou ignorar o lado político e me ater aos méritos do voto  impresso. A pergunta é: o voto impresso tornará nossas eleições mais seguras e transparentes? A resposta é não. E trará problemas adicionais.

Dois argumentos são utilizados a favor do voto impresso. Um deles vê com desconfiança a totalização dos votos que o TSE faz depois de receber os resultados das cerca de 500 mil urnas espalhadas pelo país. Esse receio é infundado.

Ao término da votação, cada urna já imprime um boletim com os votos de cada candidato. Se a totalização do TSE divergisse da soma desses boletins impressos, seria facílimo de mostrar. Até hoje, nunca aconteceu.

A dúvida legítima é quanto à segurança de cada urna individual. Será que ela está contabilizando os votos corretamente? Ao contrário da totalização feita pelo TSE, essa contagem de cada urna não pode, atualmente, ser conferida de
forma independente.

O voto impresso daria uma possibilidade de recontagem que não depende do que a própria urna diz. O problema é que os defensores do voto impresso, tão céticos quanto à segurança da urna eletrônica, são crédulos demais na segurança da apuração humana de milhões de cédulas de papel. O Brasil tem longa experiência
com a apuração manual: processo demorado e sempre sujeito a fraudes, que de fato ocorriam.

É possível hackear as urnas eletrônicas? Teoricamente sim. Até hoje, contudo, ninguém conseguiu. Há mais de 30 camadas de segurança, e a Justiça Eleitoral realiza testes periódicos. O código-fonte do programa das urnas é plenamente auditável. Seriam necessários hackers agindo em milhares de urnas uma a uma, posto que elas não estão conectadas à internet e nem entre si.

É mais fácil fraudar a apuração manual do que hackear as urnas eletrônicas. Ou seja, o método para conferir o voto eletrônico é menos confiável que o próprio voto eletrônico. Adicionou-se um elo à corrente, mas esse elo é mais fraco.

Quem tem dois números não tem nenhum. Se houver discrepância entre o eletrônico e o papel, o que acontece? Não saberemos se o problema estava na urna eletrônica ou na contagem dos papeizinhos. Pela lógica, teríamos que ficar com o eletrônico, mais seguro, e descartar o papel (ou seja, a recontagem no papel seria inútil). Na prática, é claro, teremos o palco armado para a judicialização e a incerteza, com toda a instabilidade que isso traz.

Os defensores do voto impresso respondem que essas cédulas impressas serão mais seguras que as velhas cédulas manuais, pois terão certificação digital. Ora, mas então essa mudança nas eleições só vai transferir nossa “fé” do sistema da urna eletrônica para o sistema de certificação digital das cédulas, ambos igualmente inacessíveis aos olhos do cidadão.

Quem duvida sem provas das urnas eletrônicas também poderá duvidar da certificação eletrônica que garante a idoneidade das cédulas impressas.

As eleições brasileiras funcionam muito bem. A votação é rápida, a apuração ocorre no mesmo dia. Não há qualquer indício de que fraudes na urna eletrônica jamais tenham ocorrido. Todas as alegações em contrário são falsificações, como as mostradas por Bolsonaro em sua live.

Se os deputados aprovarem o voto impresso, teremos uma votação mais demorada (mais passos para o eleitor, mais problemas técnicos das impressoras), mais cara (custo de impressoras, técnicos, apuradores e sistema de certificação) e mais passível de contestações, sem ganho relevante de segurança.

 

Brazil’s Bolsonaro has undermined science during an epic public-health crisis

Scientists in Brazil are facing particularly hard times.  I thus consider it timely that Nature has featured a News article about the situation.  Click on the link below to read the article directly from Nature.com :

‘We are being ignored’: Brazil’s researchers blame anti-science government for devastating COVID surge

Researchers say that President Jair Bolsonaro’s administration has undermined science during an epic public-health crisis.

Derivation of the Fermi-Dirac and Bose-Einstein distributions

Below we derive in standard textbook fashion the Fermi-Dirac and Bose-Einstein distributions from the perspective of the grand canonical ensemble, which in a way is the natural ensemble to consider.

Consider a system of {N} identical quantum particles. Let {n_j} be the number of particles occupying the single-particle states {|j\rangle}. Let {\epsilon_j} be the energy for the state {|j\rangle}. The occupation number {n_j} is limited to {n_j=0,~1} for fermions due to the Pauli exclusion principle. For bosons the occupation numbers can be arbitrarily large.

The total energy is thus given by

\displaystyle E= \sum_{j} \epsilon _j n_j ~, \ \ \ \ \ (1)

and

\displaystyle N= \sum_j n_j ~. \ \ \ \ \ (2)

Our goal is to calculate the mean occupation number {\overline{ n_j}} for fermions and bosons when they are in thermodynamic equilibrium, at temperature {T} and with chemical potential {\mu}.

1. The grand canonical partition function

For fixed chemical potential {\mu}, temperature parameter {\beta} and volume {V}, the grand canonical partition function is given by

\displaystyle \Xi(\beta, \mu, V) = \sum _{\rm all~states} \exp [-\beta (E-\mu N)] ~. \ \ \ \ \ (3)

In the situation previously described, we can write it as

\displaystyle \begin{array}{rcl} \Xi &=& \sum _{n_1} \sum _{n_2} \sum _{n_3} \ldots \exp \bigg[ \sum_j -\beta (n_j \epsilon_j - \mu n_j)\bigg] \\ &=& \sum _{n_1} \sum _{n_2} \sum _{n_3} \ldots \prod_j \exp [ -\beta (n_j \epsilon_j - \mu n_j) ] \\ &=& \bigg[\sum _{n_1} \exp [ -\beta (n_1 \epsilon_1 - \mu n_1) ] \bigg] \bigg[\sum _{n_2} \exp [ -\beta (n_2 \epsilon_1 - \mu n_2) ] \bigg] \ldots \end{array}

Hence,

\displaystyle \Xi = \prod_j ~\sum _{n_j} \exp [-\beta n_j (\epsilon_j - \mu)] ~. \ \ \ \ \ (4)

In other words, {\Xi} can be factored as product over a sum over occupation numbers for the {j}-th single-particle state. The latter sum is easy to calculate for bosons and even easier for fermions.

The mean occupation numbers can then be calculated as a weighted average for the grand canonical ensemble:

\displaystyle \begin{array}{rcl} \overline{n_j} &=& \frac{ \sum _{n_1,n_2\ldots} n_j \exp \bigg[ \sum_j -\beta (n_j \epsilon_j - \mu n_j)\bigg] } { \sum _{n_1,n_2\ldots} \exp \bigg[ \sum_j -\beta (n_j \epsilon_j - \mu n_j)\bigg] } \\ &=& \frac {1}{\Xi} \frac{\partial \Xi}{\partial(- \beta \epsilon_j)} \end{array}

so that

\displaystyle \overline{n_j} = -\frac{1 }{ \beta } \frac{\partial \log \Xi}{\partial\epsilon_j} ~. \ \ \ \ \ (5)

Substituting for {\Xi} from (4) we get

\displaystyle \begin{array}{rcl} \overline{n_j} &=& -\frac{1 }{ \beta } \frac{\partial}{\partial\epsilon_j} \log \left[ \prod_k ~\sum _{n_k} \exp [-\beta n_k (\epsilon_k - \mu)] \right] \\ &=& -\frac{1 }{ \beta } \frac{\partial}{\partial\epsilon_j} \sum_k \log \left[ \sum _{n_k} \exp [-\beta n_k (\epsilon_k - \mu)] \right] \\ &=& \sum_k -\frac{1 }{ \beta } \frac{\partial}{\partial\epsilon_j} \log \left[ \sum _{n_k} \exp [-\beta n_k (\epsilon_k - \mu)] \right]~. \end{array}

Unless {k=j} the derivative vanishes, so that

\displaystyle \overline{n_j} = -\frac{1 }{ \beta } \frac{\partial }{\partial\epsilon_j} \log \left[ \sum _{n_j} \exp [-\beta n_j (\epsilon_j - \mu)] \right] ~. \ \ \ \ \ (6)

2. Fermi-Dirac statistics

For fermions {n_j} cannot be greater than 1. So

\displaystyle \sum _{n_j} \exp [-\beta n_j (\epsilon_j - \mu)] = 1+ \exp [-\beta (\epsilon_j - \mu)]~. \ \ \ \ \ (7)

Hence,

\displaystyle \begin{array}{rcl} \overline{n_j} &=& -\frac{1 }{ \beta } \frac{\partial \log }{\partial\epsilon_j} (1+ \exp [-\beta (\epsilon_j - \mu)])\\ &=& -\frac{1 }{ \beta } \frac{ -\beta \exp [-\beta (\epsilon_j - \mu)]} {(1+ \exp [-\beta (\epsilon_j - \mu)])}\\ &=& \frac{ \exp [-\beta (\epsilon_j - \mu)]} {(1+ \exp [-\beta (\epsilon_j - \mu)])} ~. \end{array}

Finally we get

\displaystyle \overline{n_j} = \frac{1 } {\exp [\beta (\epsilon_j - \mu)]+1} ~. \ \ \ \ \ (8)

3. Bose-Einstein Statistics

For bosons we have

\displaystyle \sum _{n_j} \exp [-\beta n_j (\epsilon_j - \mu)] = \sum _{n_j} (\exp [-\beta (\epsilon_j - \mu)])^{n_j} \ \ \ \ \ (9)

which is a geometric series.

Recall that for {|r|<1} we can write

\displaystyle (1-r) \sum_{n=0}^\infty r^n = 1~, \ \ \ \ \ (10)

so that

\displaystyle \sum_{n=0}^\infty r^n = {1 \over 1-r}~, \ \ \ \ \ (11)

from which we get

\displaystyle \sum _{n_j} \exp [-\beta n_j (\epsilon_j - \mu)] = {1 \over 1 - \exp [-\beta (\epsilon_j - \mu)]}~. \ \ \ \ \ (12)

Substituting, we get

\displaystyle \begin{array}{rcl} \overline{n_j} &=& -\frac{1 }{ \beta } \frac{\partial \log }{\partial\epsilon_j} \left[ {1 \over 1 - \exp [-\beta (\epsilon_j - \mu)]} \right] \\ &=& -\frac{1 }{ \beta } \frac{\partial \log }{\partial\epsilon_j} \big[ 1 - \exp [-\beta (\epsilon_j - \mu)] \big]^{-1} \\ &=& \frac{1 }{ \beta } \frac{\partial \log }{\partial\epsilon_j} \big[ 1 - \exp [-\beta (\epsilon_j - \mu)] \big] \\ &=& \frac{1 }{ \beta } \bigg[ { \beta \exp [-\beta (\epsilon_j - \mu)] \over 1 - \exp [-\beta (\epsilon_j - \mu)]} \bigg] \\ &=& { \exp [-\beta (\epsilon_j - \mu)] \over 1 - \exp [-\beta (\epsilon_j - \mu)]} . \end{array}

Finally we get

\displaystyle \overline{n_j} = \frac{1 } {\exp [\beta (\epsilon_j - \mu)]-1} ~. \ \ \ \ \ (13)

4. Summary

The Fermi-Dirac and Bose-Einstein distributions can be written as

\displaystyle \overline{n(\epsilon)}= \frac{1 } {\exp [\beta (\epsilon - \mu)] \pm 1} ~, \ \ \ \ \ (14)

with the plus sign for Fermi-Dirac statistics and the minus sign for Bose-Einstein statistics.

A neat identity for the determinant of a power of (1+A)

Here we prove a slight variation on the well known identity for a complex matrix A:

\displaystyle {\displaystyle \det (I+A) =\sum _{k=0}^{\infty } {\frac {1}{k!}}\left(-\sum _{j=1}^{\infty }{\frac {(-1)^{j}}{j}}{\rm tr} \left(A^{j}\right)\right)^{k} ~.} \ \ \ \ \ (1)

Here I is the identity matrix. We prove below the following for \alpha \in \Bbb R \setminus \{ 0\}:

\displaystyle {\displaystyle \det [\hspace{1mm}(I+A)^\alpha\hspace{1mm}] =\sum _{k=0}^{\infty } {\frac {1}{k!}}\left(-\alpha\sum _{j=1}^{\infty }{\frac {(-1)^{j}}{j}}{\rm tr} \left(A^{j}\right)\right)^{k} ~.} \ \ \ \ \ (2)

The above formulas are valid in an analytic sense whenever the sum on the right converges. But independently of convergence they are always true if interpreted in terms of formal power series.

First, recall that

\displaystyle \det (\exp(A)) = \exp({\rm tr} (A))~. \ \ \ \ \ (3)

It is actually quite easy to prove (3). We can proceed as follows in 3 steps: (i) The equation clearly holds for diagonal matrices; (ii) since the determinant and trace are invariant with respect to similarity transformations, the formula is also clearly true for all diagonalizable matrices; (iii) finally, the diagonalizable matrices are dense in the set of all square matrices and the determinant and trace are continuous functions, which means that the formula must hold true at all limit points, i.e. for all matrices.

If we replace {A} with {\log (I+A)^\alpha } we get

\displaystyle \det [\hspace{1mm}(I+A)^\alpha\hspace{1mm}] = \exp\big[{\rm tr} [\alpha \log (1+A)]\big]~. \ \ \ \ \ (4)

Recall the Mercator series

\displaystyle \log (1+ x) = \sum_{ j =1}^\infty (-1)^{ j -1} \frac {x^ j } j ~.\ \ \ \ \ (5)

Substituting we get

\displaystyle \det [\hspace{1mm}(I+A)^\alpha\hspace{1mm}] = \exp\bigg[\alpha~{\rm tr} \bigg( \sum_{ j =1}^\infty (-1)^{ j -1} \frac {A^ j } j \bigg) \bigg] = \exp\bigg( {-\alpha} \sum_{ j =1}^\infty \frac{(-1)^{ j }}{ j } {\rm tr}({A^ j })\bigg) ~.\ \ \ \ \ (6)

The claim (2) follows from expanding the exponential in its well-known Taylor series.

{\square}