tag:blogger.com,1999:blog-8912760531275005722024-03-13T11:08:25.408-07:00Cognitive DissonancesLearning and teaching physics through cognitive conflicts:
science puzzles and paradoxes becoming challenging teasersCognitive Dissonanceshttp://www.blogger.com/profile/15069035350995446947noreply@blogger.comBlogger25125tag:blogger.com,1999:blog-891276053127500572.post-44758192836828575722023-04-09T07:23:00.002-07:002023-04-09T07:32:07.737-07:00Why is there a conserved quantity for every continuous global symmetry?<p></p><div style="text-align: center;"><a href="https://commons.wikimedia.org/wiki/File:EmmyNoether_MFO3096.jpg#/media/File:EmmyNoether_MFO3096.jpg"><img alt="EmmyNoether MFO3096.jpg" height="307" src="https://upload.wikimedia.org/wikipedia/commons/1/13/EmmyNoether_MFO3096.jpg" width="320" /></a></div><br /><p></p>
<p style="text-align: justify;">One of the most important theorems in physics is the theorem that states:</p><p style="text-align: center;">"To every differentiable global symmetry of a physical system there corresponds a conservation law"</p><p style="text-align: justify;">This theorem is known as the <i>first Noether's theorem</i>, in honor of the great mathematician Emmy Noether, who proved it in 1915 in the context of classical mechanics (both relativistic and non-relativistic, but not quantum). By the way, Noether is part of the group of leading scientists and professors in her field who <a href="https://www.elsaltodiario.com/paradoja-jevons-ciencia-poder/seguiremos-haciendo-matematicas-en-gotinga">lost their jobs due to the intolerance of the Nazis when they came to power</a>, as they immediately passed a law preventing Jews and Communists from working in universities and public institutions. This happened before the Holocaust and the Second World War. It is important to remember this so that it does not happen again.</p><p style="text-align: justify;">This relationship between symmetries and conservation laws established by Noether is one of the most powerful ideas that human beings have ever had. Conservation laws are a very useful tool for finding out how the quantities of a physical system change over time. Knowing that there are physical quantities that do not change allows us to write equations where the unknowns are the quantities that do change. We can then use the quantities that do not change to find out how the other quantities change.</p><p style="text-align: justify;">On the other hand, the symmetries of a physical system are related to its aesthetic aspect. For example, a sphere is beautiful because, no matter how you rotate it, it remains the same. Noether's theorem thus relates beauty to usefulness in physics in a certain way. Pragmatism and aesthetics go hand in hand.</p><p style="text-align: justify;">However, for the physics student, it is not immediately evident that a continuous symmetry implies a conserved quantity. <span style="text-align: left;">Apparently, they are two things that have nothing to do with each other. What is the reason for this relationship?</span></p><p style="text-align: justify;">However, today we know that the world is not classical, but quantum, and that classical mechanics is nothing more than an approximation of the behavior of physical systems in a certain limit. Therefore, Noether's original proof does not serve us for the fundamental laws of nature. Does Noether's theorem still hold in quantum mechanics?</p><p style="text-align: justify;">These are the two questions we are going to answer in this post.</p><p style="text-align: justify;"><br /></p><p style="text-align: justify;"><span></span></p><a name="more"></a><h3 style="text-align: justify;"><span><a name='more'></a></span>Continuous transformations in quantum mechanics</h3><p>Quantum mechanics has forever changed <a href="http://divulgamadrid.blogspot.com/p/la-mecanica-cuantica-es-la-teoria-que.html" target="_blank">our conception of the physical world</a>, the foundations of <a href="https://github.com/dmvaldman/library/blob/master/computer%20science/Aaronson%20-%20Quantum%20Computing%20Since%20Democritus.pdf" target="_blank">information theory</a> and even our conception of the <a href="http://divulgamadrid.blogspot.com/2020/02/algebra-geometria-matematicas.html" target="_blank">mathematical objects</a>. An example of the latter is what happens with transformation groups. The way a group $G$ of transformations acts on a physical system can be very complicated. The physical system can be highly nonlinear and have a highly nontrivial geometric structure. However, the <i>principle of superposition of quantum states</i> tells us that in quantum mechanics, the states in which any physical system can be found are elements of a vector space $\mathcal{H}$, since any linear combination of several quantum states represents another possible quantum state for the system. </p>
<p>This means that for each element $g$ of the group $G$, when it acts on the physical system, it corresponds to a linear transformation $\pi(g)$, an operator that acts on this vector space of quantum states. Linear transformations are much easier to study. Moreover, these transformations must be unitary, since they are the only ones that preserve the total probability, as the sum of the probabilities of all possible events in quantum mechanics must always be equal to one, no matter how many transformations we apply to the physical system.</p>
<p>The set of all these linear transformations on the space of quantum states is said to form a representation of the group $G$ on the space $\mathcal{H}$. More specifically, a <i>representation</i> is a function $\pi$ that associates, to each element $a$ of a group, its corresponding linear transformation $\pi(a)$ in a vector space, such that the structure of the group is respected, that is, it is the same to compose two transformations and find the representation of the result, as to compose the representations of each transformation. Mathematically, it is written as:</p>
<div>$$ \pi(b) \cdot \pi(a)=\pi(ba)$$</div>
<p>It is evident that mapping all the elements of a group to the identity transformation in a vector space is trivially a representation, which is called the <i>trivial or scalar representation</i>. What is interesting for mathematicians is to study the non-trivial representations of the different groups that faithfully reproduce their structure. The dimension of the vector space $\mathcal{H}$ on which the operators $\pi(g)$ act is called the <i>dimension of the representation</i> $\pi$. It is also interesting from the mathematical point of view to study which are the <i>irreducible representations</i> of each group $G$. These are the ones that have no sub-representations, that is, there is no proper vector subspace $\mathcal{H}^\prime \subset \mathcal{H}$ such that $\pi$ acting on $\mathcal{H}^\prime$ is a representation. This is because, given two representations $\pi_1$ and $\pi_2$ acting on two vector spaces $\mathcal{H}_1$ and $\mathcal{H}_2$, it is always possible to construct the direct sum representation $\pi_1 \oplus \pi_2$ that acts on the direct sum vector space $\mathcal{H}_1 \oplus \mathcal{H}_2$ generated by the set of vectors that arise from combining the set of vectors of a basis of $\mathcal{H}_1$ with that of a basis of $\mathcal{H}_2$. In matrix notation:</p>
<div>$$ \pi(a) = \begin{pmatrix}\pi_1(a) & 0 \\ 0 & \pi_2(a)\end{pmatrix} $$</div>
<p>In physics, however, what is interesting is to know in each specific case under which representations the quantum states of each physical system transform. It is said that the set of all these linear transformations on the quantum state space forms a representation of the group $G$ on the space $\mathcal{H}$. More specifically, a <i>representation</i> is a function $\pi$ that associates with each element $a$ of a group, its corresponding linear transformation $\pi(a)$ in a vector space, so that the structure of the group is respected, that is, it is the same to compose two transformations and find the representation of the result, than to compose the representations of each transformation. Mathematically, it is written</p>
<div>$$ \pi(b) \cdot \pi(a)=\pi(ba)$$<br /></div>
<div>
It is evident that mapping all elements of a group to the identity transformation in a vector space is trivially a representation, which is called <i>trivial or scalar representation</i>. What is interesting for mathematicians is to study the non-trivial representations of the different groups that reproduce their structure faithfully. The dimension of the vector space $\mathcal{H}$ on which the operators $\pi(g)$ act is called the <i>dimension of the representation</i> $\pi$. It is also interesting from a mathematical point of view to study the <i>irreducible representations</i> of each group $G$. These are those that have no sub-representations, that is, where there is no proper vector subspace $\mathcal{H}^\prime \subset \mathcal{H}$ such that $\pi$ acting on $\mathcal{H}^\prime$ is a representation. This is because, given two representations $\pi_1$ and $\pi_2$ acting on two vector spaces $\mathcal{H}_1$ and $\mathcal{H}_2$, it is always possible to construct the direct sum representation $\pi_1 \oplus \pi_2$ that acts on the direct sum vector space $ \mathcal{H}_1 \oplus \mathcal{H}_2$ generated by the set of vectors that arises from joining the set of vectors of a basis of $ \mathcal{H}_1 $ with that of a basis of $\mathcal{H}_2$. In matrix notation:
</div>
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<p>
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<div style="text-align: justify;">In physics, however, <b>what is interesting is knowing in each concrete case under which representations the quantum states of each physical system transform</b>. Since all these representations have to be unitary, the operators $\pi(g)$ must all be unitary, that is, their adjoint must be equal to their inverse. Such representations of the group $G$ are called <i>unitary representations</i>. As unitary transformations in a vector space conserve the scalar product, vectors that were orthogonal remain so after that transformation. This makes any reducible representation writable as a direct sum of irreducible representations, although it may be difficult to find the basis in which it is expressed in block diagonal form, as in the previous matrix.</div>
<p></p>
<p style="text-align: justify;"><br /></p>
<h3 style="text-align: justify;">Lie algebra of the U(1) group</h3>
<div style="text-align: justify;"><br /></div>
<div style="text-align: justify;">Since every element of U(1) can be written in the form $e^{i\theta}$, with $\theta$ a real parameter, the imaginary unit $i$ is said to be the <i>generator</i> of the U(1) group. This terminology is due to the fact that every element of the group infinitesimally close to the neutral element (identity transformation) can be written as</div><div style="text-align: center;">$ e^{i\theta} \sim 1+i\theta $</div><div style="text-align: justify;">since $\theta$ is small, which can be reached by adding quantities proportional to the imaginary unit. Analogously, since this element corresponds to the operator $e^{-iL_z\theta}$ in the presentation $\pi$, in which the group acts on the vector space of quantum states, the antihermitian operator $X=-iL_z$ is said to be the generator of the action of the U(1) rotation group on the space of states. Since physicists prefer to work better with quantum mechanical observables, which are hermitian operators, we call the generator of the rotations, abusing language, $L_z$, that is, the <b>angular momentum</b> $L_z$ <b>is the generator of the action of the U(1) rotation group on the space of states</b>.</div>
<div style="text-align: justify;"><br /></div>
<div style="text-align: justify;">Mathematicians normally use a different language. U(1) is a <i>Lie group</i>, a set that, in addition to having a group structure, has a differential manifold structure, in this case of real dimension 1. The tangent space to this manifold at the point where the neutral element is located has real dimension 1, and is generated by the imaginary unit $i$. That is, every vector of that tangent space is of the form $i\theta$. This tangent space is called the Lie algebra of the U(1) group. <b>The imaginary unit</b> $i$ <b>is therefore the generator of the Lie algebra of the U(1) group</b>, which is denoted as u(1) and is simply the real line (or rather, the imaginary line). The unitary representation $\pi$ is nothing more than a function between the Lie group U(1) and the Lie group of unitary transformations that act on the state space of the quantum system. As such, it must have a derivative $\pi^\prime$, which is nothing more than a linear function that goes from the Lie algebra of the group U(1) to the tangent space at the identity of the Lie group of unitary transformations of the state space:</div><div style="text-align: justify;">
<p style="text-align: center;">$ \pi^\prime (i\theta)=\theta \frac{d}{d\theta^\prime} \pi(e^{i\theta^\prime}) \arrowvert_{\theta^\prime=0}=\theta \frac{d}{d\theta^\prime} e^{-iL_z\theta^\prime}\arrowvert_{\theta^\prime=0}=-iL_z\theta $</p>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiz6xndOZbmawB2veL8K1N0GEw7wOIDfCxETLzNEZ0pLveFwUWffkMzMVjFyXeCAOhWkDa4UAyd7LrK2FYXavlTqGTNAC9tLXd5Kr2_ZRftMfZ6j2LIido9ReWSRAYYbNBZTLiy-A1hSo0C/s1600/002U1liealgebra.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="362" data-original-width="610" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiz6xndOZbmawB2veL8K1N0GEw7wOIDfCxETLzNEZ0pLveFwUWffkMzMVjFyXeCAOhWkDa4UAyd7LrK2FYXavlTqGTNAC9tLXd5Kr2_ZRftMfZ6j2LIido9ReWSRAYYbNBZTLiy-A1hSo0C/s1600/002U1liealgebra.png" /></a></div>
<p><br /></p>
<h3>Lie algebras and their representations</h3>
<p><br /></p>
<p>This is something that not only happens with the group U(1), but with any other Lie group, only that in general the Lie algebra of the group will be a real vector space of dimension greater than 1, generated by operators $X$ that do not commute with each other in the general case of a non-abelian group. In particular, the <i>Lie algebra of the group</i> $G$ is defined as the space of matrices $X$ such that</p>
<p style="text-align: center;">$ e^{tX}\in G $</p>
<p>for any real number $t$. It is important to note that not all elements of $G$ need to be expressible in this way, nor does it necessarily happen that each $t$ gives a different element of $G$. What does happen is that <b>each vector of the Lie algebra</b> $X$ <b>defines a path in</b> $G$ <b>that passes through the identity element</b> (when $t=0$) <b>with velocity vector</b></p>
<p style="text-align: center;">$ \frac{d}{dt} \left( e^{tX} \right) \arrowvert_{t=0}=X $</p>
<p>What we have seen happen with representations of the U(1) group also holds for representations of other groups G, even if they are not unitary.
</p><div style="text-align: justify;">In this case, we have that $\pi \left( e^{tX} \right) $ <b>is a path in the space of linear transformations of the vector space on which the representation acts, which also passes through the identity transformation</b> when $t=0$, and that it has as velocity vector at that point</div>$$ \pi^\prime(X)= \frac{d}{dt} \pi \left( e^{tX} \right) \arrowvert_{t=0} $$<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjW7CvbWxREvjlxfVQDM2xcNgbKycouGPp3yENCV2m8xbJy-6vCW0QK8KXbbV7yonWLRPr8VtE4tkzapMv6EUoeZhFbc7I0qVMkzn_tAX6YfaR1tRU3kVs9Xe44ydyYpT9Y3XyGaGNPaNEV/s1600/003liealgebrarep.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="285" data-original-width="568" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjW7CvbWxREvjlxfVQDM2xcNgbKycouGPp3yENCV2m8xbJy-6vCW0QK8KXbbV7yonWLRPr8VtE4tkzapMv6EUoeZhFbc7I0qVMkzn_tAX6YfaR1tRU3kVs9Xe44ydyYpT9Y3XyGaGNPaNEV/s1600/003liealgebrarep.png" /></a></div><div style="text-align: justify;">Then we say that $\pi^\prime$ is the <i>representation of the Lie algebra</i> of the group G. Note that $\pi^\prime$ is uniquely determined by the representation of the Lie group $\pi$. Furthermore, it can be easily shown that</div><div style="text-align: center;">$ e^{t\pi^\prime(X)}=\pi \left( e^{tX} \right) $</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">The fact that the Lie algebra is a vector space over the field of real numbers allows us to define a special representation of the Lie group, called the <i>adjoint representation</i>, in which the element $g$ acts on the Lie algebra operator $X$ by the following operation</div>$$ (Ad(g))(X)=gXg^{-1} $$<div style="text-align: justify;">In the case of the group U(1), being abelian, the adjoint representation is trivial. This means that the angular momentum operator $L_z$ does not change when we perform a rotation around the $Z$ axis. However, if the group $G$ is non-abelian, then in general the generators $X$ of the group will change when a transformation belonging to $G$ is performed. If that transformation has been generated by the operator $Y$, then the infinitesimal version of that change, which is also an element of the Lie algebra, is</div><div style="text-align: center;">$ \frac{d}{dt} (e^{tY}Xe^{-tY}) \arrowvert_{t=0}=[Y,X] $</div><div style="text-align: justify;">which is equal to the commutator of $Y$ with $X$. Therefore, this commutator measures how $X$ changes infinitesimally under a transformation generated by $Y$.
<div><br /></div><div>Therefore, the Lie algebra representation of the group $G$ that corresponds to the adjoint representation $Ad$ of the group is:</div>
$$
ad(Y)(X)=[Y,X]
$$
<div>And this is why mathematicians call this commutator the <i>Lie bracket</i>. Only when both operators commute, the transformation generated by one of them leaves the other invariant and, moreover, we know that the observables associated with both operators can take well-defined values simultaneously. But the non-Abelian character of the group $G$ means that these commutators are not generally zero. The structure of the group $G$ gives rise to a Lie bracket structure on its tangent space, which is the Lie algebra of the group $G$. And this is why this tangent space is called the Lie algebra: in addition to having a vector space structure, it has an associated Lie bracket.</div>
<div><br /></div>
<div>The corresponding formulas for the representation $\pi$ of $G$, which acts on the quantum state space, are:</div><div style="text-align: center;">$
\pi^\prime(gXg^{-1})=\pi(g)\pi^\prime (X) (\pi (g))^{-1}
$</div><div style="text-align: center;">$
\pi^\prime ([Y,X])=[ \pi^\prime (Y) , \pi^\prime (X) ]
$</div><div>Therefore, the representation $\pi^\prime$ inherits the same commutator structure as the Lie algebra it represents.</div>
<div><br /></div>
<h3>The Noether theorem in quantum mechanics</h3>
<div><br /></div>
<div>The quantum state of the system changes over time, so that the probability amplitudes of the different mechanical-quantum events will change, but the probabilities must still add up to one. In addition, the temporal evolution of two mutually exclusive mechanical-quantum states (perpendicular to each other) must give rise to two other states that are also mutually exclusive, and this evolution must be linear, by the second principle of superposition of quantum states. The mechanical-quantum states will therefore change according to a certain unitary transformation whose representation in the state space we call $\hat{U}(t)$. We can understand this unitary transformation as a representation of the group of real numbers (the time line) acting on the vector space of the quantum states of the system.</div>
<div><br /></div>
<div>If we call $\psi(t_o)$ the wave function of the quantum system at a given instant $t_o$, the result of applying the evolution operator $\hat{U}(t)$ to the wave function is:</div><div style="text-align: center;">$
\psi(t+t_o)=\hat{U}(t) \psi(t_o)
$</div><div>On the other hand, the Taylor expansion of the function $\psi(t+t_o)$ around $t=0$ is:</div><div style="text-align: center;">$
\psi(t+t_o)=\sum_{n=0}^\infty \frac{1}{n!}\frac{d^{(n)}\psi}{dt}(t_o)t^n=e^{t\frac{d}{dt_o}}\psi(t_o)
$</div><div>which coincides with the exponential function if we accept that there can be a derivative operator in the exponent. The conclusion is that the evolution operator in quantum mechanics can be written as an exponential of a Hermitian operator:</div><div style="text-align: center;">$
\hat{U}(t)=e^{-\frac{i}{\hbar}\hat{H}t}
$</div><div>where $\hat{H}=i\hbar\frac{d}{dt_o}$ is called the <i>Hamiltonian operator </i> of the system. Therefore, the Hamiltonian operator (multiplied by $-i$) is a vector belonging to the Lie algebra corresponding to a representation of the group of temporal translations on the vector space of quantum states of the system, that is, the generator of these temporal translations in that representation. Note that in the case of temporal evolution, there is a different convention for signs than the one we have used for rotations and is used for the rest of transformations. With this last convention, the generator of temporal translations is not $H$, but $-H$. Therefore, the generator vector of the Lie algebra is $X=+iH$, while for rotations $X=-iL_z$. The reason for doing this for the Hamiltonian is due to <a href="https://divulgamadrid.blogspot.com.es/2018/02/relatividad-cuadrivectores.html" target="_blank">the signature of space-time in relativity</a>, where the sign for the component of the four-vector corresponding to energy is opposite to the sign for the components corresponding to momentum. This makes the unitary operator that converts the wave function $\psi(t_0)$ into $\psi(t_0-t)$ to be<br /><div style="text-align: center;">$ e^{+itH}$</div><div>However, the temporal evolution operator that transforms the wave function $\psi(t_0)$ into $\psi(t_0+t)$ must then be</div><div style="text-align: center;">$\hat{U}(t)=e^{-\frac{i}{\hbar}\hat{H}t}$</div><div>In quantum mechanics, the generators of unitary transformations are Hermitian operators that physically represent observables. In this case, the observable that represents the Hamiltonian is energy. The eigenvalues of this operator are therefore the possible values of the system's energy. If the Hamiltonian of a system is time-independent, then the eigenvectors of the Hamiltonian, being also of the evolution operator, only change under a temporal evolution with a phase factor</div><div style="text-align: center;">$ e^{-\frac{i}{\hbar}Et}=\cos(Et/\hbar)-i\sin(Et/\hbar),$</div><div>where $E$ is the corresponding eigenvalue. The temporal dependence of these states, which are called <i>stationary</i>, is that of a classical harmonic oscillator with angular frequency $\omega=E/\hbar$. As they remain eigenvectors of the Hamiltonian with the same eigenvalue, the system's energy remains the same. It is a conserved quantity.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">Now consider the Lie group $G$ generated by both rotations around the $Z$ axis and temporal evolution of the quantum system. The action of this group on the system's state space is given by a representation in which each rotation angle $\theta$ and each time $t$ elapsed corresponds to a unitary operator</div>$$ U(\theta, t)=e^{-i\theta L_z-itH}$$<div style="text-align: justify;">Therefore, in this case the Lie algebra of the group is a real vector space of dimension 2, and is generated, at least in its representation $\pi^{prime}$ on the state space, by the anti-Hermitian operators $-iL_z$ and $-iH$. For a general quantum system, rotations around the $Z$ axis do not commute with temporal translations, so the group $G$ is non-Abelian. This is reflected in its Lie algebra in that the generator of rotations and the generator of temporal translation do not commute, so their corresponding representations on the system's state space, $L_z$ y $H$, do not commute.
<div>This physically means that for most quantum systems one can imagine, these two observables cannot be well defined simultaneously. If this is the case, when performing a rotation, instead of considering that the quantum state changes, we can consider that the Hamiltonian of the system changes according to the adjoint representation:</div>
<div>$$<br />H^\prime= e^{-i \theta L_z} H e^{+i \theta L_z}<br />$$</div>
<div>Thus, if this rotation is infinitesimal:</div>
<div>$$<br />H^\prime = H -id\theta [L_z, H]<br />$$</div>
<div>On the other hand, when evolving with time, instead of considering that the system's state changes, we can consider that the angular momentum operator changes in the form:</div>
<div>$$<br />L_z(t)= e^{itH} L_z e^{-itH}<br />$$</div>
<div>$L_z(t)$ is called the angular momentum operator in the Heisenberg picture. After an infinitesimal time, this operator changes in the form:</div>
<div>$$<br />\frac{dL_z}{dt} (t)=-i [L_z,H]<br />$$</div>
<div>Therefore, <b>the commutator</b> $[L_z,H]$ <b>simultaneously measures two things: how energy changes when performing a rotation on the system, and how angular momentum changes when the system evolves with time</b>. What happens in those physical systems in which the Hamiltonian is invariant under rotations around the $Z$ axis? In that case, as $H$ and $L_z$ commute, it also happens that the angular momentum operator in the Heisenberg picture does not change with time. That is, <b>quantum systems with symmetric Hamiltonians under rotations are the same as those in which angular momentum is conserved</b>. We thus obtain the quantum-mechanical version of Noether's theorem.</div>
<div></div>
<div>It is evident that what we have done with rotations also holds for the rest of differentiable transformations. If the quantum system has the symmetry of being invariant under all elements of a Lie group $G$, then the unitary operators that represent the action of $G$ on the quantum states commute with the temporal evolution operator. </div>
<div>This makes the generators of these unitary representations, which are Hermitian operators, also commute with the evolution operator and the Hamiltonian. The physical quantities represented by these Hermitian operators will therefore also be conserved quantities. In addition, by Schur's lemma, the Hamiltonian, when commuting with the operators that implement the unitary representations of that symmetry group, has to be proportional to the identity operator within each representation. This means that all states in the same representation of that symmetry group are degenerate and have the same energy. Therefore, when a quantum system has a symmetry, the vector space of all states that have the same energy will always be a representation of that symmetry group. Moreover, that representation, being unitary, will always be a direct sum of irreducible representations of the group G. This idea has allowed for a surprising connection between the theory of group representations and modular functions to be found, a connection for which, by the way, string theory has been necessary, which is not only a fundamental tool for every theoretical physicist who wants to stay up-to-date in their discipline today but also for any mathematician.</div><div><br /></div><div><h3>The Noether theorem in classical mechanics</h3><div><br /></div><div>In classical mechanics, each state of the system is characterized by a point of coordinates $(q,p)$ in the phase space. The phase space can be very complicated, even singular, and studying how the transformations made to the physical system act on this space is much more complicated than in quantum mechanics, where the state space is linear.</div><div>Therefore, the easiest way to prove the Noether theorem in classical mechanics is to try to emulate what is done in quantum mechanics, linearizing the problem in some way. The technique that needs to be applied here is a standard technique in modern mathematics: if we work not with a space but with the functions in that space, then transformations on that space become much more tractable.</div><div>In classical physics, any function $f(q,p)$ represents a specific observable. If the Hamiltonian of the system were equal to the observable $f(q,p)$, then the state of the system would evolve over time according to the Hamilton equations</div><div>$$\dot{q}=\frac{\partial f}{\partial p}$$</div><div>$$\dot{p}=-\frac{\partial f}{\partial q}$$</div><div>which can be written in terms of the Poisson bracket as</div><div>$$\dot{g}=\{ g,f \}$$</div><div>where</div><div>$$\{ g,f \}= \frac{\partial g}{\partial q} \frac{\partial f}{\partial p} - \frac{\partial g}{\partial p} \frac{\partial f}{\partial q} $$</div></div><div><br /></div><div>When the observable $f(q,p)$ is not the Hamiltonian of the system, we can interpret these equations as the equations that tell us how the classical state of the physical system changes when we perform a transformation generated by the observable $f(q,p)$ on it. </div>
<div>
<p>For example, the observable $p$ generates the transformation</p><p><span style="text-align: center;">$\frac{dq}{da}=\{ q,p \}=1 \Rightarrow q(a) = q(0) + a$</span></p>
<p>
which is nothing but a translation in the $q$ coordinate. Physically, therefore, the Poisson bracket $\{ g,f \}$ tells us how the observable $g$ changes in classical mechanics when we perform on the physical system the transformation generated by the observable $f$. Mathematically, it is said that the observable $f$ acts as the moment map of a vector field in the phase space. This is nothing but a pedantic way of saying that to each function $f(q,p)$ corresponds a vector field $X_f$ which associates to each function $g(q,p)$ another function
</p>
<p>
$$X_f(g)=\{ g,f \}$$
</p>
<p>
This vector field gives us the velocity vector (careful, "velocity" in the phase space) associated with the trajectories in the phase space that give us how the physical system evolves when applying the transformation generated by $f$. This transformation is what physicists know as "canonical transformation", since it preserves the Poisson brackets.
</p>
<p>As the Poisson bracket is antisymmetric, bilinear, and also satisfies the Jacobi identity, mathematically, it is said that the Poisson bracket is the Lie bracket that assigns the Lie algebra structure to the space of functions in the phase space. In fact, this Lie algebra, which has infinite dimension, was historically the first Lie algebra that was studied, although it is clearly more complicated than the Lie algebras we have worked with before in quantum mechanics, since in quantum mechanics we have the linearity of the Hilbert space, but not here.</p>
<p>The application that associates to each $f$ the vector field $-X_f$ is a homomorphism between the Lie algebra of functions in the phase space and the Lie algebra of vector fields in the phase space. But note that this homomorphism is not injective, since adding a constant to $f$ does not change the corresponding $X_f$. Nor is it surjective, because not all vector fields in the phase space can be written in the form</p>
<p>
$$X_f(g)=\{ g,f \}$$
</p>
<p>The physical requirement that the Poisson bracket be equal to a derivative forces this Lie bracket to have an extra property: it must obey the Leibniz product rule for derivatives. This makes, at least for polynomial functions in the phase space, the Poisson bracket determined solely by its values on linear functions, since the calculation of all others is reduced to those of linear functions by applying the Leibniz rule. These values of the Poisson brackets of the linear functions define a symplectic form in the dual phase space (the space generated by the coordinates $q$ and $p$)</p>
<p>
$$\Omega (g,f) = \{ g,f \}$$
</p>
<p>
Note that $\Omega (g,f)$ is an antisymmetric and nondegenerate bilinear map that endows the dual phase space with the structure of a symplectic space. And this is why mathematicians call canonical transformations <i>symplectomorphisms</i>. By preserving the Poisson brackets, they also preserve the symplectic form defined from them.</p></div><div style="text-align: justify;">Once we have all the machinery of Lie algebras in place, we see that the same thing happens in classical mechanics as in quantum mechanics, since the vector field</div><div><div style="text-align: justify;">$$ X_H(g)=\{ g,H \} $$</div><div style="text-align: justify;">gives us the velocity vector associated with the trajectories in the phase space that tell us how the physical system evolves when we apply the transformation generated by the Hamiltonian $H$, that is, it tells us how the physical quantity $g$ evolves with time. But at the same time, this same quantity also coincides, due to the antisymmetry of the Poisson bracket, with how the Hamiltonian $H$ changes as we change the system according to the transformation generated by $g$.</div></div><div style="text-align: justify;">$$ -X_g(H)=\{ g,H \} $$<br /></div><div style="text-align: justify;">If the Poisson bracket between $H$ and $g$ is zero, it happens at the same time that the transformation generated by $g$ leaves the Hamiltonian invariant and that $g$ does not change with time. This is the content of the original theorem proved by Noether, although she did not do it this way, and which we now understand as a particular case of the corresponding theorem in quantum mechanics applied in the limit where the system can be approximated as classical. Recall that in this limit, the relationship between Poisson brackets and commutators is:<br /></div><div style="text-align: justify;">$$ \widehat{\{f,g\}}=-\frac{i}{\hbar}[\hat{f},\hat{g}] $$<br /></div><div style="text-align: justify;"><br /></div><h3 style="text-align: justify;">Conclusion</h3><div style="text-align: justify;"><br /></div><div style="text-align: justify;">While in quantum mechanics the possible states of the system, which give us maximum knowledge about it, although not complete in the classical sense due to the uncertainty principle, form a Hilbert space, which is a linear vector space. This linearity makes transformations on the system much easier to handle than in quantum mechanics, where states of maximum knowledge, which are also states of complete knowledge, are points in a phase space that can be very complicated and even singular. It is a space with much fewer restrictions than the Hilbert space of quantum states. Transformation groups are, for this reason, much easier to handle in quantum mechanics than in classical mechanics.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">The original article by Noether, where she demonstrates the relationship between differentiable symmetries of a system and conservation laws in classical mechanics, is difficult to read. But in quantum mechanics, it is trivial to see this relationship between conservation laws and symmetries. This relationship comes from the formula $[H,g]=0$, which, being antisymmetric, can be understood in two ways. On the one hand, it can be treated as the equation that gives us the temporal derivative in the Heisenberg picture of the operator $g$. Being zero means that $g$ is a conserved quantity.
<div><p>On the other hand, that formula can be understood as the infinitesimal change of the Hamiltonian under the transformations generated by $g$. When it is zero, we can say that $g$ generates a symmetry. That is why, whenever there is a symmetry, its generator $g$ is conserved over time, and whenever an observable is conserved over time, that observable generates a symmetry.</p></div><div><p>In the classical limit, this proof still holds, but the commutators are replaced by Poisson brackets, which are more complicated structures. Quantum mechanics makes the proof of Noether's theorem more direct and simple. The linearity of the Hilbert space of quantum states makes everything much more elegant. The greater beauty of quantum mechanics compared to classical mechanics is also evident in Noether's theorem.</p></div><div><i>About the author: Sergio Montañez Naz has a PhD in physics and is a public high school teacher in the Community of Madrid.</i></div><div><i><br /></i></div><div>This post is a translation of the post <i><a href="http://divulgamadrid.blogspot.com/2022/07/teorema-noether.html" target="_blank">Por qué a toda simetría continua le corresponde una cantidad conservada</a></i> using ChatGPT. Some proofreading is still needed, but most sentences are accurate.</div><div><br /></div><div><h3>Bibliographic references</h3><ul><li>Stillwell J. (2008), <i>Naive Lie theory</i>, Undergraduate Texts in Mathematics, Springer-Verlag.</li><li>Tapp K. (2016), <i>Matrix groups for undergraduates</i>, second ed., American Mathematical Society.</li><li>Woit P. (2017): <i>Quantum Theory, Groups and Representations. An Introduction</i>. Springer International Publishing.</li></ul></div>
</div></div></div></div><p></p></div><p></p>Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com0tag:blogger.com,1999:blog-891276053127500572.post-64623187068026415432018-07-04T02:36:00.000-07:002018-07-04T02:36:00.152-07:00Do all chlorine atoms have the same mass?Mathematical induction is a mathematical proof technique. It is essentially used to prove that a property P(n) holds for every natural number n, i.e. for n = 1, 2, 3, and so on. The method of induction requires two cases to be proved:<br />
<br />
<ol>
<li>The first case, called the base case, proves that the property holds for the number 1.</li>
<li>The second case, called the induction step, proves that, if the property holds for one natural number n, then it holds for the next natural number n + 1.</li>
</ol>
These two steps establish the property P(n) for every natural number n = 1, 2, 3, ...<br />
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<iframe allowfullscreen="" class="giphy-embed" frameborder="0" height="270" src="https://giphy.com/embed/HPA8CiJuvcVW0" width="480"></iframe></div>
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<span style="font-size: x-small;">Mathematical induction can be informally illustrated by reference to the sequential effect of falling dominoes. (<a href="https://giphy.com/gifs/dominos-HPA8CiJuvcVW0">via GIPHY</a>)</span></div>
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Do all chlorine atoms have the same mass?<br />
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<ol>
<li>First we establish a base case for one chlorine atom (n=1). The case with just one chlorine atom is trivial. If there was only one chlorine atom in the universe, then clearly all chlorine atoms in that universe would have the same mass.</li>
<li>We then prove that if in any random set of n chlorine atoms every atom had the same mass, then in any random set of n+1 chlorine atoms every atom must also have the same mass. First, exclude the last chlorine atom and look only at the first n chlorine atoms; all these have the same mass since n chlorine atoms always have the same mass. Likewise, exclude the first chlorine atom and look only at the last n chlorine atoms. These too, must also have the same mass. Therefore, the first chlorine atom in the group has the same mass as the chlorine atoms in the middle, who in turn have the same mass as the last chlorine atom. Hence the first chlorine atom, middle chlorine atoms, and last chlorine atom have all the same mass, and we have proven that: If n chlorine atoms have the same mass, then n+1 chlorine atoms will also have the same mass.</li>
</ol>
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We already saw in the base case that the rule ("all chlorine atoms have the same mass") was valid for n=1. The inductive step showed that since the rule is valid for n=1, it must also be valid for n=2, which in turn implies that the rule is valid for n=3 and so on. Thus in any group of chlorine atoms, all chlorine atoms must have the same mass. We have proof that in any universe, no matter how many chlorine atoms exist, all chlorine atoms must have the same mass.<br />
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Nevertheless, it is an experimental fact that it is false that all the atoms of the same element have the same mass. For instance, chlorine's atomic mass of 35.5 a.m.u. is an average of the masses of the different isotopes of chlorine. This is calculated by working out the relative abundance of each isotope. For example, in any sample of Chlorine 25% will be Cl-37 and 75% Cl-35, so there are chlorine atoms with different (35 a.m.u. and 37 a.m.u.) masses.<br />
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<span style="text-align: justify;">Please, explain your reasoning. You can post your attempted answers in the comment box below. Please, do not use Facebook, Twitter or Instagram to give your answers.</span>Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com0tag:blogger.com,1999:blog-891276053127500572.post-37971908498107251782018-06-30T02:30:00.003-07:002018-06-30T02:30:59.545-07:00Can an atom have a half-integer number of neutrons?<br />
<div style="text-align: center;">
<a href="https://commons.wikimedia.org/wiki/File:Prout_William_painting.jpg#/media/File:Prout_William_painting.jpg"><img alt="Prout William painting.jpg" height="200" src="https://upload.wikimedia.org/wikipedia/commons/5/52/Prout_William_painting.jpg" width="187" /></a></div>
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<span style="font-size: x-small;">William Prout, by Henry Wyndham Phillips, 1820 - 1868 - From a miniature by Henry Wyndham Phillips, Public Domain, </span><a href="https://commons.wikimedia.org/w/index.php?curid=3671497"><span style="font-size: x-small;">Lin</span>k</a></div>
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Prout's hypothesis was an early 19th-century attempt to explain the existence of the various chemical elements through a hypothesis regarding the internal structure of the atom. In 1815 and 1816, the English chemist William Prout published <a href="http://web.lemoyne.edu/~giunta/PROUT.HTML" target="_blank">two papers</a> in which he observed that the atomic weights that had been measured for the elements known at that time appeared to be whole multiples of the atomic weight of hydrogen. He then hypothesized that the hydrogen atom was the only truly fundamental object, which he called "protyle", and that the atoms of other elements were actually groupings of various numbers of hydrogen atoms.<br />
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Prout's hypothesis was an influence on Ernest Rutherford, and that is the reason why he suggested in 1920 the name "protons" por the positive particles that live in the atomic nuclei. The name "proton" comes from the suffix "-on" for particles, added to the stem of Prout's word "protyle". Later, the English physicist Sir James Chadwick <a href="https://www.nobelprize.org/nobel_prizes/physics/laureates/1935/" target="_blank">discovered the neutron</a>. Both particles, proton and neutron, have almost the same mass (1 a.m.u.), which is much bigger that the electron mass, and that is why the elements known at Prout's time were measured to be whole multiples of the atomic mass of hydrogen, which is 1 a.m.u.<br />
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Nevertheless, nowadays we know that chlorine's atomic mass is 35.5 a.m.u. Since we know that each chlorine atom has 17 protons inside its nucleus, does it mean that chlorine atoms have a half-integer number of neutrons?<br />
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<span style="text-align: justify;">Please, explain your reasoning. You can post your attempted answers in the comment box below. Please, do not use Facebook, Twitter or Instagram to give your answers.</span>Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com0tag:blogger.com,1999:blog-891276053127500572.post-70259008905507779852017-12-03T12:30:00.000-08:002018-06-30T02:04:58.143-07:00Why do nonmetals have both positive and negative oxidation numbers?The chemical elements can be broadly divided into metals and nonmetals according to their tendency to loose or gain electrons:<br />
<ul>
<li>Atoms that belong to metallic elements tend to loose electrons. When they loose electrons, they become cations, positive ions with a charge that equals the number of electrons they have lost. That number is given by the oxidation number. For instance, sodium's oxidation number is +1, while calcium's oxidation number is +2.</li>
<li>On the other hand, atoms that belong to nonmetallic elements tend to gain electrons, so they become anions, ions with a negative charge that equals the number of electrons they have gain. For instance, fluorine tends to gain one electron and becomes F<sup>-</sup>. That is why it has oxidation number -1. </li>
</ul>
But, as we can see in the following periodic table, most nonmetals have both positive and negative oxidation number:<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhPOT2ZAkdG3vx04J1qE-7vLJ8xRmhUdhcIVrvnQRG_LxsYIm-VGxi2nRaDtQ9dPsp9z8qGey4Gxz7tw4SBtX9rw5CjNEBMCSh_-ZcGjfK6UQv8u9DN9pNqvZDLcxMqbxrkLp0thyS0a6YY/s1600/OxidationNumbers.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="843" data-original-width="1600" height="210" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhPOT2ZAkdG3vx04J1qE-7vLJ8xRmhUdhcIVrvnQRG_LxsYIm-VGxi2nRaDtQ9dPsp9z8qGey4Gxz7tw4SBtX9rw5CjNEBMCSh_-ZcGjfK6UQv8u9DN9pNqvZDLcxMqbxrkLp0thyS0a6YY/s400/OxidationNumbers.png" width="400" /></a></div>
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Why do nonmetals have both positive and negative oxidation number if they always tend to gain electrons?<br />
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<span style="text-align: justify;">Please, explain your reasoning. You can post your attempted answers in the comment box below. Please, do not use Facebook or Twitter to give your answers.</span></div>
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Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com0tag:blogger.com,1999:blog-891276053127500572.post-65903801954821072612017-12-02T10:44:00.000-08:002017-12-03T12:27:56.154-08:00Why do metals seem colder although they have the same temperature?<div style="text-align: center;">
<a href="https://commons.wikimedia.org/wiki/File:Mercury_Thermometer.jpg#/media/File:Mercury_Thermometer.jpg"><img alt="Mercury Thermometer.jpg" height="200" src="https://upload.wikimedia.org/wikipedia/commons/5/57/Mercury_Thermometer.jpg" width="91" /></a></div>
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<span style="font-size: xx-small;">By <a href="https://commons.wikimedia.org/wiki/User:Anonimski" title="User:Anonimski">Anonimski</a> - <span class="int-own-work" lang="en">Own work</span>, <a href="https://creativecommons.org/licenses/by-sa/3.0" title="Creative Commons Attribution-Share Alike 3.0">CC BY-SA 3.0</a>, <a href="https://commons.wikimedia.org/w/index.php?curid=29726755">Link</a></span></div>
<span style="text-align: justify;">All the objects that have been inside your room for more than one hour are at room temperature. That is because heat flows from hotter to colder objects, so if you put a cold object inside the room, heat will flow to it until it reaches room temperature.</span><br />
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If we touch a piece of metal that is inside the room it feels cold. But when we touch the other objects of the room they don't feel as cold. Why is that? Why do metals seem colder than the other objects in the room although they have the same temperature?</div>
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Please, explain your reasoning. You can post your attempted answers in the comment box below. Please, do not use Facebook or Twitter to give your answers.</div>
<br />Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com2tag:blogger.com,1999:blog-891276053127500572.post-45679800748431947992017-09-03T11:28:00.000-07:002017-09-04T00:25:32.791-07:00What is the difference between a mixture and a compound?According to most textbooks:<br />
<ul>
<li>a compound is an entity consisting of two or more atoms, commonly from different chemical elements, which associate via chemical bonds.</li>
<li>On the other hand, a mixture is a material made up of two or more different substances which are mixed but are not combined chemically.</li>
</ul>
So the difference is that a mixture refers a physical combination of substances, whereas a compound refers to a chemical combination.<br />
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<a href="https://commons.wikimedia.org/wiki/File:SaltInWaterSolutionLiquid.jpg#/media/File:SaltInWaterSolutionLiquid.jpg"><img alt="SaltInWaterSolutionLiquid.jpg" height="200" src="https://upload.wikimedia.org/wikipedia/commons/8/89/SaltInWaterSolutionLiquid.jpg" width="105" /></a></div>
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<span style="font-size: xx-small;">By <a href="https://commons.wikimedia.org/wiki/User:Chris_73" title="User:Chris 73">Chris 73</a> / <a class="external text" href="http://commons.wikimedia.org/">Wikimedia Commons</a>, <a href="http://creativecommons.org/licenses/by-sa/3.0" title="Creative Commons Attribution-Share Alike 3.0">CC BY-SA 3.0</a>, <a href="https://commons.wikimedia.org/w/index.php?curid=11084">Link</a></span></div>
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<br />
But these definitions do not say anything unless we establish the difference between a chemical and a physical combination and, according to the same textbooks:<br />
<ul>
<li>a chemical process is a method or means of somehow changing one or more compounds,</li>
<li>whereas physical changes are changes affecting a substance, but not its chemical composition, because they do not change chemical bonding.</li>
</ul>
As you probably have already realized, the definitions are circular! We put two substances together. If the process is not chemical, what we obtain is a mixture, not a compound. But we defined a non-chemical process as the one where the compounds are still the same compounds, but mixed. Who is Alice? She is Bob's cousin. And who is Bob? Alice's cousin. We still do not know who is Alice!<br />
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Are you able to give a definition of compounds and mixtures that is not circular? How can we define chemical process without saying that a chemical process is different from a physical one in that compounds change?<br />
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Notice the distinction cannot come from the physical properties if the substance, because the physical properties of a mixture may differ from those of the components. In addition, evolved or absorbed heat cannot be the solution because, both in chemical reactions and in mixtures, heat is either evolved (an exothermic process) or absorbed (an endothermic process).<br />
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<span style="text-align: justify;">Please, explain your reasoning. You can post your attempted answers in the comment box below. Please, do not use Facebook or Twitter to give your answers. </span>Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com2tag:blogger.com,1999:blog-891276053127500572.post-5637224324839755732017-08-30T12:27:00.000-07:002017-08-30T12:59:27.868-07:00Why don't protons in a nucleus repel each other?<div style="text-align: justify;">
The atomic nucleus is the small, dense region consisting of protons and neutrons at the center of an atom.</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjo9ALk6maITfx2MFJ4B20JU_5ahByekQEZxMiXR5gVVYl3WeFvyJ1l3F7r_hbXfuOrwZComjCCpb_PL_hVdVqNW2akI3r425RbbbYBP3O4s5hSCJc5E8ufGRJxBj33zGsfBMFV61I3GEKS/s1600/nucleus.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="312" data-original-width="373" height="166" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjo9ALk6maITfx2MFJ4B20JU_5ahByekQEZxMiXR5gVVYl3WeFvyJ1l3F7r_hbXfuOrwZComjCCpb_PL_hVdVqNW2akI3r425RbbbYBP3O4s5hSCJc5E8ufGRJxBj33zGsfBMFV61I3GEKS/s200/nucleus.jpg" width="200" /></a></div>
<span style="font-size: x-small;">A model of the atomic nucleus showing it as a compact bundle of the two types of nucleons:</span></div>
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<span style="font-size: x-small;">protons (red) and neutrons (blue).</span></div>
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<span style="font-size: xx-small;">By <a class="new" href="https://commons.wikimedia.org/w/index.php?title=User:Marekich&action=edit&redlink=1" title="User:Marekich (page does not exist)">Marekich</a> - <span class="int-own-work" lang="en">Own work</span> (vector version of PNG image), <a href="http://creativecommons.org/licenses/by-sa/3.0" title="Creative Commons Attribution-Share Alike 3.0">CC BY-SA 3.0</a>, <a href="https://commons.wikimedia.org/w/index.php?curid=21701588">Link</a></span></div>
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The atomic number Z of a chemical element is the number of protons found in the nucleus of an atom. Since neutrons are neutral while protons are positively charged with a charge that is, in absolute value, equal to the electron charge, the atomic number is identical to the charge number of the nucleus. In an uncharged atom, the atomic number is also equal to the number of electrons, and that is why the atomic number uniquely identifies a chemical element. For instance, Z=1 is called hydrogen and Z=6 carbon.</div>
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The world nuclide is referred to a 'species of nucleus', characterized by its number of protons Z, its number of neutrons N, and its nuclear energy state. Identical nuclei belong to one nuclide, for example each nucleus of the carbon-13 nuclide is composed of 6 protons and 7 neutrons. On the other hand, the members of the group of all the nuclides of the same elements are called isotopes. That is, the nuclides with equal atomic number, i.e., of the same chemical element but different neutron numbers, are called isotopes of that element.</div>
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Stable nuclides are nuclides that are not radioactive and so (unlike radionuclides) do not spontaneously undergo radioactive decay. It has been found that there are 80 elements with one or more stable isotope. For instance, carbon has 15 known isotopes, from carbon-8 to carbon-22, of which carbon-12 and carbon-13 are stable.</div>
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But, because of the fact all the protons have the same charge and are very closed to one another in the nucleus, one expects that they repel each other by a strong electric force, a force that must be much stronger than the force acting between the nucleus and the surrounding electrons. This force should make the nucleus explode. Therefore, no nuclide with more than one proton should be stable, that is, the only stable element should be hydrogen!<br />
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Please, explain your reasoning. You can post your attempted answers in the comment box below. Please, do not use Facebook or Twitter to give your answers. </div>
Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com3tag:blogger.com,1999:blog-891276053127500572.post-44518125928438337082017-07-02T07:51:00.001-07:002017-07-30T13:31:15.877-07:00Why do astronauts float weightless in the International Space Station?<div class="separator" style="clear: both; text-align: center;">
<a href="https://commons.wikimedia.org/wiki/File:International_Space_Station_after_undocking_of_STS-132.jpg#/media/File:International_Space_Station_after_undocking_of_STS-132.jpg" style="margin-left: 1em; margin-right: 1em;"><img alt="A rearward view of the International Space Station backdropped by the limb of the Earth. In view are the station's four large, gold-coloured solar array wings, two on either side of the station, mounted to a central truss structure. Further along the truss are six large, white radiators, three next to each pair of arrays. In between the solar arrays and radiators is a cluster of pressurised modules arranged in an elongated T shape, also attached to the truss. A set of blue solar arrays are mounted to the module at the aft end of the cluster." height="127" src="https://upload.wikimedia.org/wikipedia/commons/0/04/International_Space_Station_after_undocking_of_STS-132.jpg" width="200" /></a></div>
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<span style="font-size: xx-small;">By NASA/Crew of STS-132 - </span><a class="external free" href="http://spaceflight.nasa.gov/gallery/images/shuttle/sts-132/hires/s132e012208.jpg" rel="nofollow" style="font-size: x-small;">http://spaceflight.nasa.gov/gallery/images/shuttle/sts-132/hires/s132e012208.jpg</a><span style="font-size: xx-small;">
(</span><a class="external free" href="http://spaceflight.nasa.gov/gallery/images/shuttle/sts-132/html/s132e012208.html" rel="nofollow" style="font-size: x-small;">http://spaceflight.nasa.gov/gallery/images/shuttle/sts-132/html/s132e012208.html</a><span style="font-size: xx-small;">), Public Domain, </span><a href="https://commons.wikimedia.org/w/index.php?curid=10561008" style="font-size: x-small;">Link</a></div>
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<span style="background-color: white; color: #161616; font-family: "helvetica neue" , "helvetica" , "arial" , sans-serif; font-size: 15px;">We have seen on TV that astronauts float weightless in the International Space Station (ISS), and during spacewalks. In fact, t</span><span style="background-color: white; color: #161616; font-family: "helvetica neue" , "helvetica" , "arial" , sans-serif; font-size: 15px;">he ISS serves as a microgravity research laboratory in which crew members conduct experiments in biology, physics and other fields. Microgravity </span><span style="color: #161616; font-family: "helvetica neue" , "helvetica" , "arial" , sans-serif;"><span style="font-size: 15px;">is more or less a synonym of weightlessness and zero-g (zero gravitational field strength).</span></span><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgLrwLIZUXX8HG7kuBB6C3o8th0mfCF7a8hmuaJ-Qj7gAWPqhLGVLmbZaE9XUPEPvpXN4Hg01xTUW2EnLEnwZ7UISY8mv37rAfq3V3grQneMp1PJUHXpL1WfOYcRCdMmeaiPwESURrFZKQB/s1600/floatingattheiss.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="180" data-original-width="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgLrwLIZUXX8HG7kuBB6C3o8th0mfCF7a8hmuaJ-Qj7gAWPqhLGVLmbZaE9XUPEPvpXN4Hg01xTUW2EnLEnwZ7UISY8mv37rAfq3V3grQneMp1PJUHXpL1WfOYcRCdMmeaiPwESURrFZKQB/s1600/floatingattheiss.gif" /></a></div>
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<span style="background-color: white; color: #161616; font-family: "helvetica neue" , "helvetica" , "arial" , sans-serif; font-size: 15px;">Nevertheless, the ISS maintains an orbit with an altitude h of between 330 and 435 km so, </span><span style="background-color: white; color: #161616; font-family: "helvetica neue" , "helvetica" , "arial" , sans-serif; font-size: 15px;">according to </span><span style="color: #161616; font-family: "helvetica neue" , "helvetica" , "arial" , sans-serif;"><span style="font-size: 15px;">Newton's law of universal gravitation, the gravitational field strength g in the ISS is:</span></span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjoqjuG80JPOCA95HTN52VlMcxdhMqiCfQytMesHAseSZDto23eD8fv3le0VSkv7NoHk-UaqgLYCHOtvzpgwDzED1lmLr03dtyLsalSmd8UjwXDZwScaGdHsvdCTMJTIAUS-n4yf5e_dqDd/s1600/CodeCogsEqn.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="41" data-original-width="199" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjoqjuG80JPOCA95HTN52VlMcxdhMqiCfQytMesHAseSZDto23eD8fv3le0VSkv7NoHk-UaqgLYCHOtvzpgwDzED1lmLr03dtyLsalSmd8UjwXDZwScaGdHsvdCTMJTIAUS-n4yf5e_dqDd/s1600/CodeCogsEqn.gif" /></a></div>
<span style="color: #161616; font-family: "helvetica neue" , "helvetica" , "arial" , sans-serif;"><span style="font-size: 15px;">where G is Newton's constant, </span></span><span style="color: #161616; font-family: "helvetica neue" , "helvetica" , "arial" , sans-serif; font-size: 15px;">M mass of Earth and R is the radius of Earth.</span></div>
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<span style="color: #161616; font-family: "helvetica neue" , "helvetica" , "arial" , sans-serif;"><span style="font-size: 15px;"><br /></span></span>
<span style="color: #161616; font-family: "helvetica neue" , "helvetica" , "arial" , sans-serif;"><span style="font-size: 15px;">Taking into account that the gravitational field strength at the surface of the Earth is g=9.8 m/s^2, we conclude that things and a</span></span><span style="color: #161616; font-family: "helvetica neue" , "helvetica" , "arial" , sans-serif; font-size: 15px;">stronauts inside the ISS weigh only a 10% less than they do on Earth! The weight is almost the same! </span><span style="color: #161616; font-family: "helvetica neue" , "helvetica" , "arial" , sans-serif;"><span style="font-size: 15px;">How is this possible?</span></span></div>
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<span style="color: #161616; font-family: "helvetica neue" , "helvetica" , "arial" , sans-serif;"><span style="font-size: 15px;"><br /></span></span></div>
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<span style="color: #161616; font-family: "helvetica neue" , "helvetica" , "arial" , sans-serif;"><span style="font-size: 15px;">Please, explain your reasoning. You can post your attempted answers in the comment box below. Please, do not use Facebook or Twitter to give your answers.</span></span></div>
Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com2tag:blogger.com,1999:blog-891276053127500572.post-61069482382156558302017-06-11T06:58:00.000-07:002017-06-14T01:20:32.826-07:00Preparation for the International Physics Olympiad (IPhO): Electromagnetism<div>
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The IPhO syllabus includes:</div>
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<b>2.3 Electromagnetic fields</b></div>
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<ul>
<li style="text-align: justify;"><b>2.3.1 Basic concepts: </b>Concepts of charge and current; charge conservation and Kirchhoff's current law. Coulomb force; electrostatic field as a potential field; Kirchhoff's voltage law. Magnetic B-field; Lorentz force; Ampere's force; Biot-Savart law and B-field on the axis of a circular current loop and for simple symmetric systems like straight wire, circular loop and long solenoid.</li>
<li style="text-align: justify;"><b>2.3.2 Integral forms of Maxwell's equations:</b> Gauss' law (for E- and B-fields); Ampere's law; Faraday's law; using these laws for the calculation of fields when the integrand is almost piecewise constant. Boundary conditions for the electric field (or electrostatic potential) at the surface of conductors and at infinity; concept of grounded conductors. Superposition principle for electric and magnetic fields; uniqueness of solution to well- posed problems; method of image charges.</li>
<li style="text-align: justify;"><b>2.3.3 Interaction of matter with electric and magnetic fields;</b> Resistivity and conductivity; differential form of Ohm's law. Dielectric and magnetic permeability; relative permittivity and permeability of electric and magnetic materials; energy density of electric and magnetic fields; ferromagnetic materials; hysteresis and dissipation; eddy currents; Lenz's law. Charges in magnetic field: helicoidal motion, cyclotron frequency, drift in crossed E- and B-fields. Energy of a magnetic dipole in a magnetic field; dipole moment of a current loop.</li>
<li style="text-align: justify;"><b>2.3.4 Circuits</b>: Linear resistors and Ohm's law; Joule's law; work done by an electromotive force; ideal and non-ideal batteries, constant current sources, ammeters, voltmeters and ohmmeters. Nonlinear elements of given V-I characteristic. Capacitors and capacitance (also for a single electrode with respect to infinity); self-induction and inductance; energy of capacitors and inductors; mutual inductance; time constants for RL and RC circuits. AC circuits: complex amplitude; impedance of resistors, inductors, capacitors, and combination circuits; phasor diagrams; current and voltage resonance; active power.</li>
</ul>
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Members of the Spanish National Team can download the course notes here:</div>
</div>
<ul>
<li><a href="https://drive.google.com/file/d/0Bwj1kvSuqjehR3N6c3NtT0NZOHc/view?usp=sharing" target="_blank">Campo electrostático</a>: </li>
<li><a href="https://drive.google.com/file/d/0Bwj1kvSuqjehX1R3dEFOQkRuZkk/view?usp=sharing" target="_blank">Corriente eléctrica</a></li>
<li><a href="https://drive.google.com/file/d/0Bwj1kvSuqjehV3FlWmcyVGFIOUk/view?usp=sharing" target="_blank">Campo magnetostático</a></li>
<li><a href="https://drive.google.com/file/d/0Bwj1kvSuqjehZm8xdG1qUUV5cGs/view?usp=sharing" target="_blank">Campo electromagnético</a></li>
</ul>
Some cognitive conflicts involving Electromagnetism:<br />
<ul>
<li><a href="http://cdissonances.blogspot.com.es/search/label/Electromagnetism" target="_blank">Cognitive Dissonances: Electromagnetism</a></li>
</ul>
It is also useful to follow the IPhO's Study Guide by Jaan Kalda<br />
<div style="orphans: 2; text-align: start; text-decoration-color: initial; text-decoration-style: initial; text-indent: 0px; widows: 2;">
<ul>
<li><a href="http://www.ioc.ee/~kalda/ipho/electricity-circuits.pdf" target="_blank">Study guide: Circuits</a></li>
</ul>
Here you can find the solutions to some of the problems:<br />
<ul>
<li>Pr 3: [IPhO1996Theory1] <a href="https://drive.google.com/file/d/0Bwj1kvSuqjehbUI1eG9XQ2JYZFE/view?usp=sharing" style="font-family: "Times New Roman";" target="_blank">Problem</a><span style="font-family: "times new roman";"> </span><a href="https://drive.google.com/file/d/0Bwj1kvSuqjehcy03MVhRUVRkMXM/view?usp=sharing" style="font-family: "Times New Roman";" target="_blank">Solution</a></li>
<li>Pr 69: [IPhO1994Theory2] <a href="https://drive.google.com/file/d/0Bwj1kvSuqjehWkNPWnZRbndaS0U/view?usp=sharing" style="font-family: "times new roman";" target="_blank">Problem Solution</a></li>
<li>[IPhO1983Theory2] <a href="https://drive.google.com/file/d/0Bwj1kvSuqjehQXlodlpzR1VxRW8/view?usp=sharing" style="font-family: "times new roman";" target="_blank">Problem</a><span style="font-family: "times new roman";"> </span><a href="https://drive.google.com/file/d/0Bwj1kvSuqjehZ3A2blpiR3p5ZUE/view?usp=sharing" style="font-family: "times new roman";" target="_blank">Solution</a></li>
<li>[IPhO2010Theory1] <a href="https://drive.google.com/file/d/0Bwj1kvSuqjehVkY4N3FJMHNuME0/view?usp=sharing" style="font-family: "times new roman";" target="_blank">Problem</a><span style="font-family: "times new roman";"> </span><a href="https://drive.google.com/file/d/0Bwj1kvSuqjehT3p6MTR0Wl9RVnM/view?usp=sharing" style="font-family: "times new roman";" target="_blank">Solution</a></li>
<li>[IPhO2001Theory1c] <a href="https://drive.google.com/file/d/0Bwj1kvSuqjehS0xkRWFERVZYNWM/view" target="_blank">Problem</a> <a href="https://drive.google.com/file/d/0Bwj1kvSuqjehS0xkRWFERVZYNWM/view" target="_blank">Solution</a></li>
</ul>
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<div style="text-align: center;">
<a href="https://drive.google.com/file/d/0Bwj1kvSuqjehS0xkRWFERVZYNWM/view" target="_blank">Back to</a><a href="http://cdissonances.blogspot.com.es/p/preparation-for-international-physics.html"> the main page</a></div>
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<a href="http://cdissonances.blogspot.com.es/p/preparation-for-international-physics.html"><img border="0" data-original-height="239" data-original-width="834" height="55" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhUTCjZcJ0S5_cbu5YOSrNjMSKLvGM55MgJ_fu-f63c27GHNs8DBM7pBHDxwg5Xy0BmSaWUVe5FlTX4ubOSWP7I0JynCqSRjGfUDR13uGF50gNdaMpk09Jd_SLQpANqbbZsagLh__lPVLlY/s200/IPhO.jpg" width="200" /></a>
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Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com0tag:blogger.com,1999:blog-891276053127500572.post-91612239178395034312017-06-11T06:33:00.001-07:002017-06-11T23:57:58.416-07:00Preparation for the International Physics Olympiad (IPhO): Relativity<div style="text-align: justify;">
The IPhO Syllabus includes:</div>
<ul>
<li style="text-align: justify;"><b>2.5 Relativity:</b> Principle of relativity and Lorentz transformations for the time and spatial coordinate, and for the energy and momentum; mass-energy equivalence; invariance of the spacetime interval and of the rest mass. Addition of parallel velocities; time dilation; length contraction; relativity of simultaneity; energy and momentum of photons and relativistic Doppler effect; relativistic equation of motion; conservation of energy and momentum for elastic and non-elastic interaction of particles.</li>
</ul>
Members of the Spanish National Team can download the notes here:<br />
<ul>
<li><a href="https://drive.google.com/file/d/0Bwj1kvSuqjehUTdZRG5PMi1ablE/view?usp=sharing" target="_blank">Introducción a la relatividad especial</a></li>
</ul>
To know more:<br />
<ul>
<li><a href="http://divulgamadrid.blogspot.com.es/p/cursos.html">Taller de Relatividad para estudiantes de bachillerato</a>.</li>
</ul>
Some cognitive conflicts involving Relativity:<br />
<ul>
<li><a href="http://cdissonances.blogspot.com.es/search/label/Relativity" target="_blank">Cognitive Dissonances: Relativity</a></li>
</ul>
It is also useful to follow the IPhO's Study Guide by Siim Ainsaar:<br />
<ul>
<li><a href="http://www.ioc.ee/~kalda/ipho/Ainsaar_Relativity.pdf" style="font-family: "times new roman";" target="_blank">A Glimpse into the Special Theory of Relativity</a></li>
</ul>
Here you can find the solutions to some of the problems:<br />
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<li>[IPho2006Theory2] <a href="https://drive.google.com/file/d/0Bwj1kvSuqjehNElHYmt6TWVlNnc/view?usp=sharing" style="font-family: "times new roman";" target="_blank">Problem</a><span style="font-family: "times new roman";"> </span><a href="https://drive.google.com/file/d/0Bwj1kvSuqjehdlZBZEJLdk9MdmM/view?usp=sharing" style="font-family: "times new roman";" target="_blank">Solution</a></li>
<li>[IPhO2003Theory3] <a href="https://drive.google.com/file/d/0Bwj1kvSuqjehVDB5SHFvdXFfZkk/view?usp=sharing" target="_blank">Problem</a> <a href="https://drive.google.com/file/d/0Bwj1kvSuqjehR0VGdEN2RGFxX3c/view?usp=sharing" target="_blank">Solution</a></li>
<li>[IPhO1994Theory1] <a href="https://drive.google.com/file/d/0Bwj1kvSuqjehTTU1NzFTcnhyOWs/view?usp=sharing" target="_blank">Problem Solution</a></li>
</ul>
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<a href="http://cdissonances.blogspot.com.es/p/preparation-for-international-physics.html" style="font-family: "times new roman"; text-align: center;"></a><a href="http://cdissonances.blogspot.com.es/p/preparation-for-international-physics.html">Back to the main page</a></div>
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Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com0tag:blogger.com,1999:blog-891276053127500572.post-85390551494167859362017-06-11T05:25:00.001-07:002017-06-16T14:26:03.234-07:00Preparation for the International Physics Olympiad (IPhO): Thermodynamics and Statistical Physics<div style="text-align: justify;">
The IPhO Syllabus includes:</div>
<div style="text-align: justify;">
<br /></div>
<div>
<div style="text-align: justify;">
<b>2.7 Thermodynamics and statistical physics</b></div>
</div>
<div>
<div>
<ul>
<li style="text-align: justify;"><b>2.7.1 Classical thermodynamics:</b> Concepts of thermal equilibrium and reversible processes; internal energy, work and heat; Kelvin's temperature scale; entropy; open, closed, isolated systems; first and second laws of thermodynamics. Kinetic theory of ideal gases: Avogadro number, Boltzmann factor and gas constant; translational motion of molecules and pressure; ideal gas law; translational, rotational and oscillatory degrees of freedom; equipartition theorem; internal energy of ideal gases; root-mean-square speed of molecules. Isothermal, isobaric, isochoric, and adiabatic processes; specific heat for isobaric and isochoric processes; forward and reverse Carnot cycle on ideal gas and its efficiency; efficiency of non-ideal heat engines.</li>
<li style="text-align: justify;"><b>2.7.2 Heat transfer and phase transitions:</b> Phase transitions (boiling, evaporation, melting, sublimation) and latent heat; saturated vapour pressure, relative humidity; boiling; Dalton's law; concept of heat conductivity; continuity of heat flux.</li>
<li style="text-align: justify;"><b>2.7.3 Statistical physics:</b> Planck's law (explained qualitatively, does not need to be remembered), Wien's displacement law; the Stefan- Boltzmann law.</li>
</ul>
</div>
</div>
Members of the Spanish National Team can download the notes here:<br />
<ul>
<li><a href="http://drive.google.com/file/d/0Bwj1kvSuqjehYmN6LXd6YlU4UU0/view?usp=sharing_eil&ts=594427e1" target="_blank">Introducción a la termodinámica y a la física estadística</a></li>
</ul>
Some cognitive conflicts involving Thermodynamis and Statistical Physics:<br />
<ul>
<li><a href="http://cdissonances.blogspot.com.es/search/label/Thermodynamics%20and%20Statistical%20Physics" target="_blank">Cognitive Dissonances: Thermodynamics and Statistical Physics</a></li>
</ul>
It is also useful to follow the IPhO's Study Guide by Jaan Kalda:<br />
<ul>
</ul>
<div>
<ul>
<li><a href="http://www.ioc.ee/~kalda/ipho/Thermodyn.pdf" target="_blank">Study guide for IPhO-s: Thermodynamics</a></li>
</ul>
Here you can find the solutions to some of the problems:<br />
<ul>
<li>Pr 1 & 7: [IPhO1996Theory1] <a href="https://drive.google.com/file/d/0Bwj1kvSuqjehbUI1eG9XQ2JYZFE/view?usp=sharing" target="_blank">Problem</a> <a href="https://drive.google.com/file/d/0Bwj1kvSuqjehcy03MVhRUVRkMXM/view?usp=sharing" target="_blank">Solution</a></li>
<li>Pr 6 & 34: [IPhO1992Theory3] <a href="https://drive.google.com/file/d/0Bwj1kvSuqjehYk9GOGhGRG41Nmc/view?usp=sharing" target="_blank">Problem Solution</a></li>
<li>Pr 24: [IPhO1989Theory1] <a href="https://drive.google.com/file/d/0Bwj1kvSuqjehb2RaUUNkRXYwZjA/view?usp=sharing" target="_blank">Problem Solution</a></li>
<li>[IPhO2007Theory] <a href="https://drive.google.com/file/d/0Bwj1kvSuqjehejVmXzgyMlE5M1k/view" target="_blank">Problem</a> <a href="https://drive.google.com/file/d/0Bwj1kvSuqjehYUFwdFp5UlJaQzA/view" target="_blank">Solution</a></li>
</ul>
</div>
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Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com0tag:blogger.com,1999:blog-891276053127500572.post-86198488993558857312016-09-26T13:30:00.000-07:002016-09-26T13:30:17.208-07:00Why do not the sizes of Venus and Mars as viewed from Earth change during the course of the year?<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiQjfTIO7NdBh0v45PM73an9xDSQWntkYL7UFUstJ6O_-w4A0YULDwT0bu0BWeD4HvVejvDCrviyCtD9A9v0omy-2w9LTDu9hxkhHSMbgcF9kcZMeAPWfgUsWCo2NocD4Olen1Sby0nJani/s1600/145px-Venus-real_color.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiQjfTIO7NdBh0v45PM73an9xDSQWntkYL7UFUstJ6O_-w4A0YULDwT0bu0BWeD4HvVejvDCrviyCtD9A9v0omy-2w9LTDu9hxkhHSMbgcF9kcZMeAPWfgUsWCo2NocD4Olen1Sby0nJani/s1600/145px-Venus-real_color.jpg" /></a></div>
<br />
<div style="text-align: center;">
<span style="font-size: x-small;">Venus, by NASA or Ricardo Nunes - </span><a class="external free" href="http://www.astrosurf.com/nunes/explor/explor_m10.htm" rel="nofollow" style="font-size: small;">http://www.astrosurf.com/nunes/explor/explor_m10.htm</a><span style="font-size: x-small;">, Public Domain, </span><a href="https://commons.wikimedia.org/w/index.php?curid=338424" style="font-size: small;">https://commons.wikimedia.org/w/index.php?curid=338424</a></div>
<br />
<div style="text-align: justify;">
Just before his death, in 1543, Nicolaus Copernicus published in his book <i>On the Revolutions of the Celestial Spheres</i> a Heliocentric model of the universe, that is, a model of the universe that placed the Sun rather than the Earth at the center of the universe. This is considered a major event in the history of science, triggering the Copernican Revolution and making an important contribution to the Scientific Revolution.</div>
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<br />
<div style="text-align: center;">
<a href="https://commons.wikimedia.org/wiki/File:Geoz_wb_en.svg#/media/File:Geoz_wb_en.svg"><img alt="Geoz wb en.svg" height="290" src="https://upload.wikimedia.org/wikipedia/commons/thumb/3/33/Geoz_wb_en.svg/1200px-Geoz_wb_en.svg.png" width="320" /></a></div>
<div style="text-align: center;">
<span style="font-size: x-small;">By Original image by <a class="new" href="https://commons.wikimedia.org/w/index.php?title=User:Nikolang&action=edit&redlink=1" title="User:Nikolang (page does not exist)">Niko Lang</a>
SVG version by <a class="mw-redirect" href="https://commons.wikimedia.org/wiki/User:Booyabazooka" title="User:Booyabazooka">User:Booyabazooka</a> - <span class="int-own-work" lang="en">Own work</span>, <a href="http://creativecommons.org/licenses/by-sa/2.5" title="Creative Commons Attribution-Share Alike 2.5">CC BY-SA 2.5</a>, <a href="https://commons.wikimedia.org/w/index.php?curid=5502912">https://commons.wikimedia.org/w/index.php?curid=5502912</a></span></div>
<br />
<br />
<div style="text-align: justify;">
According to Copernicus' model, since the E<span style="text-align: justify;">arth circulates the Sun in an orbit outside that of Venus and inside that of Mars, the apparent size of both Venus and Mars should change appreciably during the course of the year. This is because when the Earth is around the same side of the sun as one of those planets it is relatively close to it, whereas when it is on the opposite side of the sun to one of them it is relatively distant from it. When the matter is considered quantitatively, as it can be within Copernicus's own version of his theory, the effect is a sizeable one, with a predicted change in apparent diameter by a factor of about eight in the case of Mars and about six in the case of Venus.</span></div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
On the other hand, according to the Ptolemaic system (the Geocentric model) Venus and Mars should not change appreciably during the course of the year because its epicyclical motion implies only a small change in distance from the Earth.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
However, when the planets are
observed carefully with the naked eye, no change in size can
be detected for Venus, and Mars changes in size by no more
than a factor of two. This gives us strong evidence for the Geocentric model and refutes the Heliocentric model! How is this possible?</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
Please, explain your reasoning. You can post your attempted answers in the comment box below. Please, do not use Facebook or Twitter to give your answers.</div>
Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com6tag:blogger.com,1999:blog-891276053127500572.post-18108603687324494862016-09-19T13:00:00.000-07:002016-09-19T13:00:35.206-07:00Something happens with the lightIf we split a collimated light beam by using a half-silvered mirror, then the two resulting beams (A and B) have exactly the same intensity. Since the light is made of photons, that means that half of the photons go through path A, and the other half through path B.<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjiY_963_kaxSR_bIqT0tx_ChgxBO-Tg7FAZB2ia9jYWubU3ByI67br4r1F1HNpfepow7ovFF9ubb2Ov-hOU0Mc2Z64xO5D8KCBIHFMV8ynD97x0I1Da0ZgZkuiR6X9NbX-k_ao_2qsz2ba/s1600/beamsplit.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjiY_963_kaxSR_bIqT0tx_ChgxBO-Tg7FAZB2ia9jYWubU3ByI67br4r1F1HNpfepow7ovFF9ubb2Ov-hOU0Mc2Z64xO5D8KCBIHFMV8ynD97x0I1Da0ZgZkuiR6X9NbX-k_ao_2qsz2ba/s1600/beamsplit.jpg" /></a></div>
If we now reflect both beams by a mirror and the two beams then pass a second half-silvered mirror and enter two detectors as explained in the picture:<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjXSXRe7D2cH__MZc8Gy5qG47J4Sf5iNUt0SNlD_Ck_-XONQLcuOjMUnsT35OthbYVHvA2x6J-7Rxc2pxjZ0A7v1W7gO8BR89CTooBVYC6UrDokXAwDEtlF1AJsvOsAVJq7nRGDjeIzIA77/s1600/mzi.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="226" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjXSXRe7D2cH__MZc8Gy5qG47J4Sf5iNUt0SNlD_Ck_-XONQLcuOjMUnsT35OthbYVHvA2x6J-7Rxc2pxjZ0A7v1W7gO8BR89CTooBVYC6UrDokXAwDEtlF1AJsvOsAVJq7nRGDjeIzIA77/s320/mzi.jpg" width="320" /></a></div>
then we expect the A beam to be split into two beams. We will call them A1 and A2. A1 goes to dectector 1, while A2 goes to detector 2. Each one contains 50% of A-photons, that is, 25% of the photons of the original light beam:<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg9EEfKrfmxn4G0MXyVpUlIptBfkH7-DOtDZxMwR9p3uDTm6ySa6_ObnPzOdOQIsBDUXIkKNAVTwiR9ekLRS07cRDN9cM_m-wI44xVpJUp8AZgspEf9M8923R8pxH30IrJxDF-BJFtjqant/s1600/mzi2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="226" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg9EEfKrfmxn4G0MXyVpUlIptBfkH7-DOtDZxMwR9p3uDTm6ySa6_ObnPzOdOQIsBDUXIkKNAVTwiR9ekLRS07cRDN9cM_m-wI44xVpJUp8AZgspEf9M8923R8pxH30IrJxDF-BJFtjqant/s320/mzi2.jpg" width="320" /></a></div>
On the other hand, we also expect the B beam to be split into two beams. We will call them B1 and B2. B1 goes to dectector 1, while B2 goes to detector 2. Each one contains 50% of B-photons, that is, 25% of the photons of the original light beam:<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjQ64oNSJG7kDMRpuWbLgVcRCDEPyPSOMMgTdaLO15nIbKfXODr2cMT_RISTEcurYWMvmrPWMnaXEIdra4bXdXeeEzMK9jWqBez74_JXfQuE0lHXlbhLtWewim8k7bGuKP1EQ5a5uYN4Q3W/s1600/mzi3.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="226" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjQ64oNSJG7kDMRpuWbLgVcRCDEPyPSOMMgTdaLO15nIbKfXODr2cMT_RISTEcurYWMvmrPWMnaXEIdra4bXdXeeEzMK9jWqBez74_JXfQuE0lHXlbhLtWewim8k7bGuKP1EQ5a5uYN4Q3W/s320/mzi3.jpg" width="320" /></a></div>
So the amount of photons that should arrive to detector 1 is 25% + 25% = 50%, and the same for detector 2:<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjhxPX6HCb7TA3U4fY46Ct3JzSMWHXTCg6DJweHOd-QQ513LiqkoY7WPVVk_KFA4LcKbVuziJcQ6am4i6I99MUnDMcOZXqSP3jItw3dM_tlcD8_J4dFO4jMZiTrZV-w375O4_MBYzplGRPM/s1600/mzi4.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="226" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjhxPX6HCb7TA3U4fY46Ct3JzSMWHXTCg6DJweHOd-QQ513LiqkoY7WPVVk_KFA4LcKbVuziJcQ6am4i6I99MUnDMcOZXqSP3jItw3dM_tlcD8_J4dFO4jMZiTrZV-w375O4_MBYzplGRPM/s320/mzi4.jpg" width="320" /></a></div>
Nevertheless, once we have carried out the experiment, what we found is that 100% of photons arrive to detector 2 and no photon arrives to detector 2!<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgefNBQrD98S1jiNvmfA2HhKJ5GDU59iDAOaDFQVaCFFnTedaT05VHH93k10aKd7NZAdWzfAAhD-w1KXCavMXDfmZHFm1pvgQTDCDi9KLEdxuci-jFA3-ktxXikSRfs5n0FmS8ebSWB6le0/s1600/mzi5.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="226" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgefNBQrD98S1jiNvmfA2HhKJ5GDU59iDAOaDFQVaCFFnTedaT05VHH93k10aKd7NZAdWzfAAhD-w1KXCavMXDfmZHFm1pvgQTDCDi9KLEdxuci-jFA3-ktxXikSRfs5n0FmS8ebSWB6le0/s320/mzi5.jpg" width="320" /></a></div>
Moreover, what is even more puzzling, if we obstruct channel A (or B, it does not matter), then we detect the same number of photons in detector 2 as the number detected in detector 1. Are you able to figure it out? Try it!<br />
<br />
<span style="text-align: justify;">Please, explain your reasoning. You can post your attempted answers in the comment box below. Please, do not use Facebook or Twitter to give your answers.</span><br />
<br />Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com7tag:blogger.com,1999:blog-891276053127500572.post-74078498481547036882016-09-12T13:00:00.000-07:002016-09-12T13:00:12.170-07:00Why do ice cubes melt faster in fresh water than in salt water?<div style="text-align: justify;">
The <i>melting point</i> of a solid is the temperature at which it changes state from solid to liquid at atmospheric pressure. When considered as the temperature of the reverse change from liquid to solid, it is referred to as the <i>freezing point</i>.</div>
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<br /></div>
<div style="text-align: justify;">
The freezing point of a solvent is depressed when another compound is added, meaning that a solution has a lower freezing point than a pure solvent. This phenomenon is used in technical applications to avoid freezing, for instance by adding salt or ethylene glycol to water. If you live in a place that has lots of snow and ice in the winter, then you have probably seen the highway department spreading salt on the road to melt the ice.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
Now, let us consider the following experiment:</div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj2j2O1cPx_WPzkNdQwwLHjQhakfNCO4oBbnJB-Wmi63x3V69OnSXzguadTsIzcgxOvZR3hbny2t5AC8mejrY2D-OzEGGXYBDa4F4KzttLXiYTK_a_D_QWDPUipfI0fhKFDIaCxBsp1w2ma/s1600/2icecubes.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="142" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj2j2O1cPx_WPzkNdQwwLHjQhakfNCO4oBbnJB-Wmi63x3V69OnSXzguadTsIzcgxOvZR3hbny2t5AC8mejrY2D-OzEGGXYBDa4F4KzttLXiYTK_a_D_QWDPUipfI0fhKFDIaCxBsp1w2ma/s320/2icecubes.gif" width="320" /></a></div>
<div style="text-align: justify;">
</div>
<ol>
<li>Make two almost identical ice cubes.</li>
<li>Mix 1 teaspoon of salt in an 8 oz. cup of water. This will be our salt water cup.</li>
<li>Fill a 8 oz. cup with water, but with no salt added. This will be our fresh water cup</li>
<li>Place one ice cube into each cup simultaneously. Which ice cube do you predict would melt the fastest?</li>
</ol>
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<div style="text-align: justify;">
Naively, one would think that, according to the previous information, since salt lowers the freezing/melting point of water, the ice cube in the salt water cup should melt the fastest.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
Nevertheless, if you carry out the experiment, it leaves no doubt. The ice cube in the fresh water cup melts faster!</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
Why do ice cubes melt slower in salt water? Please, explain your reasoning. You can post your attempted answers in the comment box below. Please, do not use Facebook or Twitter to give your answers.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
I will give you a clue: repeat the experiment, but this time, after you place the ice cubes in the cups, wait 30 seconds and add a couple of drops of food coloring to each cup without disturbing the water in the cups.</div>
Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com5tag:blogger.com,1999:blog-891276053127500572.post-54340406716693842062016-09-05T13:00:00.000-07:002016-09-07T00:43:06.311-07:00Why does the death of a living being affect the decay of carbon-14? <div style="text-align: justify;">
Carbon-14 is a radioactive isotope of carbon with an atomic nucleus containing 6 protons and 8 neutrons. Carbon-14 decays into nitrogen-14 through beta decay:</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEispvx6w7KrfLCBx6WLjfv7Wp0cKNyVCgOk8xhdty1tF4WsCu-zopRCTVLZLyp1AfDtmKk-tQkix-ufoId1z2QDqgc9vzw5yA6u74cjqgUIGD6M-IJEUmoLieN_lLbS5G0yn1gIg0E29U_d/s1600/carbon14.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="91" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEispvx6w7KrfLCBx6WLjfv7Wp0cKNyVCgOk8xhdty1tF4WsCu-zopRCTVLZLyp1AfDtmKk-tQkix-ufoId1z2QDqgc9vzw5yA6u74cjqgUIGD6M-IJEUmoLieN_lLbS5G0yn1gIg0E29U_d/s320/carbon14.gif" width="320" /></a></div>
<div style="text-align: justify;">
By emitting a beta particle (an electron, e<sup>-</sup>) and an electron antineutrino (ν<sub>e</sub>), one of the neutrons in the carbon-14 nucleus changes to a proton and the carbon-14 nucleus becomes the stable (non-radioactive) isotope nitrogen-14.</div>
<div style="text-align: justify;">
<div>
<div>
The equation governing the decay of a radioactive isotope is</div>
</div>
<div style="text-align: center;">
$$ N=N_0 e^{-\frac{t}{\tau}}$$</div>
<div>
<div>
where N<sub>o</sub> is the number of atoms of the isotope in the original sample (at time t = 0, when the organism from which the sample was taken died), and N is the number of atoms left after time t. On the other hand, the mean-life τ is the average or expected time a given atom will survive before undergoing radioactive decay.</div>
<div>
Since the amount of carbon-14 inside a piece of wood or a fragment of bone decrease as the carbon-14 undergoes radioactive decay, measuring the amount of carbon-14 in a sample provides information that can be used to calculate when the animal or plant died. The mean-life of carbon-14 is 8267 years, so the equation above can be rewritten as:</div>
</div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi4tH6teIykc7e1ttfT2837Z115W8rvhZBd0oozaVKkW8SjUTy6WhaLKZP4lVxE6THa0K4r62zqHn-5VqfeFKWSb2a30P0mzhWE0PGUVmlsoc_vx5dOcdCsH2MbYq7JBr5cvTGleHcZ-gbo/s1600/radioactivedecay.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="46" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi4tH6teIykc7e1ttfT2837Z115W8rvhZBd0oozaVKkW8SjUTy6WhaLKZP4lVxE6THa0K4r62zqHn-5VqfeFKWSb2a30P0mzhWE0PGUVmlsoc_vx5dOcdCsH2MbYq7JBr5cvTGleHcZ-gbo/s320/radioactivedecay.gif" width="320" /></a></div>
<div>
Nevertheless, radioactive decay is a process that takes place inside the nucleus, so nor a change of temperature neither chemical reactions affect radioactive decay. Carbon-14 atoms inside a living being are decaying after and before the living being dies. So why is this method used efficiently to measure when the living being died? How do we know N<sub>o</sub>, the amount of carbon-14 the living being had at the moment it died, if carbon-14 was also decaying when the plant or the animal was alive?<br />
<br />
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<div style="margin: 0px;">
Please, explain your reasoning. You can post your attempted answers in the comment box below. Please, do not use Facebook or Twitter to give your answers.</div>
</div>
</div>
</div>
Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com0tag:blogger.com,1999:blog-891276053127500572.post-22428556394798445982016-08-29T13:00:00.000-07:002016-09-07T00:43:29.752-07:00Does the expansion of space apply in the solar system?<div style="text-align: justify;">
The prevailing cosmological model for the universe accounts for the fact that the universe expanded from a very high density and high temperature state, and that nowadays the expansion is even accelerating. This is an expansion of space, that is, the increase of the distance between two distant parts of the universe with time. It is an intrinsic expansion whereby the scale of space itself changes. This is different from other examples of expansions and explosions in that, as far as observations can ascertain, it is a property of the entirety of the universe rather than a phenomenon that can be contained and observed from the outside.</div>
<div>
<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg-I7SlMmExBg75fvIAVp0ZSJDaepup2fruFv54yOwmfYTIgSq560S-Dkf1xqbyfbm18unomOdU2cMeXAUtXUKM2un1jJZ_dk_Zwhzmsa9AVZp5yflSWMLnWOLp-mwmrqncJ3Xi8nMavxzt/s1600/640px-CMB_Timeline300_no_WMAP.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="206" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg-I7SlMmExBg75fvIAVp0ZSJDaepup2fruFv54yOwmfYTIgSq560S-Dkf1xqbyfbm18unomOdU2cMeXAUtXUKM2un1jJZ_dk_Zwhzmsa9AVZp5yflSWMLnWOLp-mwmrqncJ3Xi8nMavxzt/s320/640px-CMB_Timeline300_no_WMAP.jpg" width="320" /></a></div>
<div style="text-align: center;">
<br /></div>
<div style="text-align: center;">
By NASA/WMAP Science Team - Original version: <a class="external text" href="http://map.gsfc.nasa.gov/media/060915/index.html" rel="nofollow">NASA</a>; modified by <a href="https://commons.wikimedia.org/wiki/User:Kaldari" title="User:Kaldari">Ryan Kaldari</a>, Public Domain, https://commons.wikimedia.org/w/index.php?curid=11885244</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
Since it is an intrinsic expansion, it is natural to think that the planets in our solar system are expanding with time, as universe is. Moreover, our measurement devices should be expanding too. But, taking into account that a measurement is the assignment of a number to a characteristic of an object by comparing with other objects, why were we able to measure the expansion of the universe if our devices are expanding too?</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
Does the expansion of space apply to the objects inside our solar system?</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
Please, explain your reasoning. You can post your attempted answers in the comment box below. Please, do not use Facebook or Twitter to give your answers.<br />
<div>
<br /></div>
<div style="text-align: justify;">
</div>
</div>
Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com3tag:blogger.com,1999:blog-891276053127500572.post-51569394103916398852016-08-22T13:00:00.000-07:002016-09-07T00:43:43.367-07:00Why are cosmic ray muons decaying more slowly than predicted?<div style="text-align: justify;">
Muons are unstable elementary particles. They are heavier than electrons and neutrinos but lighter than all other matter particles. They decay via the weak interaction. A muon decays most commonly to an electron, an electron antineutrino, and a muon neutrino:</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgONQ1TclrKs1gkMkMxQZvkbcSXKobJUbJA1slNqg6PkeHdTT7qBjiJlDsUQpRWkkrCO1VoE-6I37mNcUrdCp4orMV6REp2QQQymkk-hd1c_NhwHJspiKcL4k6LDFhmR5vAvSRfyvqttjAK/s1600/muondecay.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="43" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgONQ1TclrKs1gkMkMxQZvkbcSXKobJUbJA1slNqg6PkeHdTT7qBjiJlDsUQpRWkkrCO1VoE-6I37mNcUrdCp4orMV6REp2QQQymkk-hd1c_NhwHJspiKcL4k6LDFhmR5vAvSRfyvqttjAK/s200/muondecay.gif" width="200" /></a></div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
The mean lifetime, τ = 1/Γ, of the muon is (2.1969811±0.0000022 ) µs. That means that every 2.19698 µs the population of muons is reduced by a factor e=2.71828.</div>
<div style="text-align: justify;">
An experiment compared the population of cosmic-ray-produced muons at the top of a mountain, whose height is 2 km, to that observed at sea level. Those muons were traveling at 0.95c, where c is the speed of light, so they arrive to the sea level t=7 µs later. At the top of the mountain the measured population was N<span style="font-size: xx-small;">o</span>=563 muons per hour. Therefore, according to the decay law, the expected population of muons at the sea level should be:</div>
<div style="text-align: justify;">
$$ N=N_0 e^{-\frac{t}{\tau}}=23$$</div>
<div style="text-align: justify;">
muons. Nevertheless, 413 muons where measured, so the muon sample at the sea level was only moderately reduced! The muons were decaying about 10 times slower!</div>
<div style="text-align: justify;">
Are you able to explain this anomaly? Try it!<br />
<br />
<br />
<div style="-webkit-text-stroke-width: 0px; color: black; font-family: "Times New Roman"; font-size: medium; font-style: normal; font-variant-caps: normal; font-variant-ligatures: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: justify; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px;">
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<div style="margin: 0px;">
Please, explain your reasoning. You can post your attempted answers in the comment box below. Please, do not use Facebook or Twitter to give your answers.</div>
</div>
</div>
Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com0tag:blogger.com,1999:blog-891276053127500572.post-87410827173611555472016-08-16T13:00:00.000-07:002016-09-07T00:44:06.902-07:00Why did the hunter wound the bear?<div style="text-align: justify;">
A hunter is located 100 meters due South from a bear. Then he goes East 100 meters. After that, he looks to the North and he shoots to the North, wounding the bear. Nevertheless, we know that the bear did not move!</div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjehyphenhyphennyp4R-IZWg3y5te24OLxN6HYuNYaN0yDfoonR5wruCKb2Wpot_cAoBAx59C8p6bFZ5rnWUI7NtvfpbEGL3KsxY1__VsHSTsvRrpgBZkJFFtrNA3tl7QhA5KsYo-Tv_4tAE-7KWKRsN/s1600/hunter.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="302" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjehyphenhyphennyp4R-IZWg3y5te24OLxN6HYuNYaN0yDfoonR5wruCKb2Wpot_cAoBAx59C8p6bFZ5rnWUI7NtvfpbEGL3KsxY1__VsHSTsvRrpgBZkJFFtrNA3tl7QhA5KsYo-Tv_4tAE-7KWKRsN/s320/hunter.gif" width="320" /></a></div>
<div style="text-align: center;">
<br /></div>
<div style="text-align: justify;">
You will be even more puzzled after realizing that the question is:</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: center;">
<i>What color is the bear?</i></div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
<br />
<div style="-webkit-text-stroke-width: 0px; color: black; font-family: "Times New Roman"; font-size: medium; font-style: normal; font-variant-caps: normal; font-variant-ligatures: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: justify; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px;">
</div>
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<div style="margin: 0px;">
Please, explain your reasoning. You can post your attempted answers in the comment box below. Please, do not use Facebook or Twitter to give your answers.</div>
</div>
</div>
Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com5tag:blogger.com,1999:blog-891276053127500572.post-55204680633112688892016-07-20T13:00:00.000-07:002016-07-20T14:27:42.598-07:00The Moon is getting further away from Earth. Where does this extra energy come from?All bounded orbits where the gravity of a central body dominates are elliptical in nature. In the case of the Moon orbiting the Earth, the eccentricity of the ellipse is so small (0.055) that it is almost a circle:<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiLLIKTWnqjFwBfFEWXVEXTpr04VT4DcDs2oH8bcHPf29tQ2g6VYxT755PoIgE1Ueh0Tus6R-GE1w9MRD7KKhBru89oRBw0OgHpUSgmakuqZIH5qI2-DTQODqu3S_QzsGsDlf4KzzMtqHXz/s1600/Moon.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="268" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiLLIKTWnqjFwBfFEWXVEXTpr04VT4DcDs2oH8bcHPf29tQ2g6VYxT755PoIgE1Ueh0Tus6R-GE1w9MRD7KKhBru89oRBw0OgHpUSgmakuqZIH5qI2-DTQODqu3S_QzsGsDlf4KzzMtqHXz/s320/Moon.gif" width="320" /></a></div>
Therefore, the gravitational force <b><i>F<sub>g</sub></i></b> that the Earth exerts on the Moon is perpendicular to Moon's velocity <b><i>v</i></b>, so it is a centripetal force <b><i>F<sub>c</sub></i></b>, making the trajectory of the Moon bend:
<br />
$$
F_{g}=F_{c} \\<br />
\frac{GMm}{r^2}=\frac{mv^2}{r}
$$ where <i>G</i> is Newton's constant, <i>M</i> is Earth's mass, <i>m</i> is Moon's mass and <i>r</i> is the radius of the orbit.<br />
<br />
This implies that the kinetic energy of the Moon is<br />
$$<br />
K=\frac{1}{2}mv^2=\frac{GMm}{2r}<br />
$$ which is smaller than the absolute value of the potential energy<br />
$$<br />
U=-\frac{GMm}{r}<br />
$$ So the mechanical energy of the Moon is<br />
$$<br />
E=-\frac{GMm}{2r}<br />
$$<br />
<br />
We know that at the time of its formation, the Moon sat much closer to the Earth, a mere 22,500 km away, compared with the 402,336 km between the Earth and the Moon today. So the Moon is getting further away from Earth, now at the rate of 3.78 cm per year. Nevertheless, according to the last equation, a larger <i>r</i> means that the Moon has more energy every year. Is its energy non conserved? Who is giving energy to the Moon?Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com4tag:blogger.com,1999:blog-891276053127500572.post-38318157793257451052016-07-18T13:00:00.000-07:002016-09-23T08:42:38.347-07:00Can an object exceed the speed of light if we push it for enough time?<div style="text-align: justify;">
The second Newton's law of motion establishes that, in an inertial reference frame, the vector sum of the forces <b><i>F</i></b> on an object is equal to the mass <i>m</i> of that object multiplied by the acceleration vector <b><i>a</i></b> of the object. That means that if we apply a constant force on a body without friction, then the object will move with constant acceleration, increasing its speed by the same amount every second.
</div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjyGwz94V1BIKO_OVN4jwpw7u-d_6UeTKnjfz1AESg97JGXByUT3Q2jP0X-X55i5AEUgfgJ8UYhWKBI2cvrY4RlaHRKO_8LEpnuvaawyRqalEaxin6PI4E4m7__dTTwo5tz8w8xFjQ0_JVz/s1600/SecondLaw.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="163" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjyGwz94V1BIKO_OVN4jwpw7u-d_6UeTKnjfz1AESg97JGXByUT3Q2jP0X-X55i5AEUgfgJ8UYhWKBI2cvrY4RlaHRKO_8LEpnuvaawyRqalEaxin6PI4E4m7__dTTwo5tz8w8xFjQ0_JVz/s200/SecondLaw.jpg" width="200" /></a></div>
<div style="text-align: justify;">
The acceleration is proportional to the force, so if the net force is 100 Newtons and the mass is 2 kilograms, the acceleration will be 50 meters per second every second. But if the force is 2 N, the body will increase its speed by 1 m/s every second. Notice that this is not a huge acceleration. Nevertheless, if we keep pushing and wait for 300000000 seconds (9.5 years) the object will move faster that light.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
But we know that nothing can exceed the speed of light. This is a well-established law of nature whose confirmation has become routine in current particle accelerators.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
Try to find the solution to this contradiction!<br />
<div style="-webkit-text-stroke-width: 0px; color: black; font-family: "Times New Roman"; font-size: medium; font-style: normal; font-variant-caps: normal; font-variant-ligatures: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: justify; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px;">
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<div style="margin: 0px;">
Please, explain your reasoning. You can post your attempted answers in the comment box below. Please, do not use Facebook or Twitter to give your answers.</div>
</div>
</div>
Sergio Montañezhttp://www.blogger.com/profile/08247791768896074379noreply@blogger.com0tag:blogger.com,1999:blog-891276053127500572.post-28278697283690282142016-07-13T13:00:00.000-07:002016-10-03T12:33:36.660-07:00Is hydrogen atom in bifluoride connected by two covalent bonds?<div style="text-align: justify;">
In chemistry, a valence electron is an electron that is associated with an atom, and that can participate in the formation of a chemical bond. In a single covalent bond, both atoms in the bond contribute one valence electron in order to form a shared pair. For instance, hydrogen atoms have one valence electron, while oxygen atoms have two. That is why a water molecule contains one oxygen and two hydrogen atoms that are connected by singles covalent bonds:</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://commons.wikimedia.org/wiki/File:Water_molecule_3D.svg#/media/File:Water_molecule_3D.svg" style="margin-left: auto; margin-right: auto;"><img alt="Space filling model of a water molecule" height="171" src="https://upload.wikimedia.org/wikipedia/commons/1/1c/Water_molecule_3D.svg" width="200" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td class="tr-caption" style="font-size: 12.8px;"><span style="font-size: xx-small;">By Dbc334 (first version); Jynto (second version) - File:Water-3D-vdW.png, Public Domain, https://commons.wikimedia.org/w/index.php?curid=1332739</span></td></tr>
</tbody></table>
</td><td class="tr-caption"><br /></td></tr>
</tbody></table>
So the Lewis structure of water is H-O-H.<br />
<br />
But let us consider a more complex example: bifluoride. Bifluoride is an inorganic anion with the chemical formula [HF<sub>2</sub>]<sup>−</sup>.<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://commons.wikimedia.org/wiki/File:Hydrogendifluoride-3D-vdW.png#/media/File:Hydrogendifluoride-3D-vdW.png" style="margin-left: auto; margin-right: auto;"><img alt="Hydrogendifluoride-3D-vdW.png" height="128" src="https://upload.wikimedia.org/wikipedia/commons/7/73/Hydrogendifluoride-3D-vdW.png" width="200" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="font-size: xx-small;">Public Domain, https://commons.wikimedia.org/w/index.php?curid=1496287</span><br />
<div>
<span style="font-size: 12.8px;"><br /></span></div>
</td><td class="tr-caption"><br /></td></tr>
</tbody></table>
<div style="text-align: justify;">
It is not a strange anion. Some [HF<sub>2</sub>]<sup>-</sup> salts are common, examples include potassium bifluoride (KHF2) and ammonium bifluoride ([NH4][HF2]).</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
As shown in the figure above, the structure of the anion is symmetric, with the hydrogen situated in the mid-point of the F-F distance. In addition, the H-F contacts in this ion are very short, like in covalent bonds, and the corresponding H.F interactions are also strong enough to be classified as covalent bonds. So the corresponding Lewis structure of the anion should be [F-H-F]<sup>-</sup>. But, how is this possible? Hydrogen atoms have one valence electron. They cannot be connected by two covalent bond!</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
Are you able to give an explanation? Accept the challenge!<br />
<br />
<br />
<div style="-webkit-text-stroke-width: 0px; color: black; font-family: "Times New Roman"; font-size: medium; font-style: normal; font-variant-caps: normal; font-variant-ligatures: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: justify; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px;">
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<div style="margin: 0px;">
Please, explain your reasoning. You can post your attempted answers in the comment box below. Please, do not use Facebook or Twitter to give your answers.</div>
</div>
</div>
Cognitive Dissonanceshttp://www.blogger.com/profile/15069035350995446947noreply@blogger.com1tag:blogger.com,1999:blog-891276053127500572.post-50672787231824336192016-07-11T13:00:00.000-07:002016-07-11T13:07:56.956-07:00Is expansion of gases a violation of inertia law?<div style="text-align: justify;">
The first Newton's law (also called <i>inertia law</i>) says that, when viewed in an inertial reference frame, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a net force.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
Let us suppose that we have a box with a wall splitting it in two halves. The left half contains a rest gas (no wind inside), while the right half is empty. Suddenly, we remove the wall. Because of the fact that gases expand to fill their containers, we know that the gas of the left half will move to the right side in order to fill the whole box:</div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiQgreIFN9bDA12KSGKNTCgLMRgZQum6eE_xIIvcqEJAxago22H8xDA8JV5-32-Oi5hWuw8p6LqrXzk8HQoq5Loolw_haOkSmEt-Al9H9xs3GbZu33Rr4m0WCZLi4VQ6_6udGXRLqCiqL6O/s1600/gasinaboximage036.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="107" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiQgreIFN9bDA12KSGKNTCgLMRgZQum6eE_xIIvcqEJAxago22H8xDA8JV5-32-Oi5hWuw8p6LqrXzk8HQoq5Loolw_haOkSmEt-Al9H9xs3GbZu33Rr4m0WCZLi4VQ6_6udGXRLqCiqL6O/s400/gasinaboximage036.gif" width="400" /></a></div>
<div style="text-align: justify;">
But there is no net force acting on the gas, so why has it moved to the right half of the box? Do gases violate the inertia law?</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
Are you able to give an explanation? Accept the challenge!</div>
Cognitive Dissonanceshttp://www.blogger.com/profile/15069035350995446947noreply@blogger.com3tag:blogger.com,1999:blog-891276053127500572.post-48790472491961088662016-07-04T14:37:00.001-07:002016-07-04T15:06:58.050-07:00Can we obtain extra surface by assembling the pieces in a different way?Picture A and B show different arrangements made of similar shapes in slightly different configurations:<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://commons.wikimedia.org/wiki/File%3AMissing_square_puzzle-AB.png" style="margin-left: auto; margin-right: auto;" title="By Krauss (Own work) [CC BY-SA 4.0 (http://creativecommons.org/licenses/by-sa/4.0)], via Wikimedia Commons"><img alt="Missing square puzzle-AB" height="277" src="https://upload.wikimedia.org/wikipedia/commons/thumb/6/6d/Missing_square_puzzle-AB.png/512px-Missing_square_puzzle-AB.png" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">By Krauss (Own work) [CC BY-SA 4.0 (http://creativecommons.org/licenses/by-sa/4.0)], via Wikimedia Commons</td></tr>
</tbody></table>
So the question is: where does the extra one unit square come from?<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://commons.wikimedia.org/wiki/File%3AMissing_Square_Animation.gif" style="margin-left: auto; margin-right: auto;" title="By Trekky0623 at English Wikipedia (Transferred from en.wikipedia to Commons.) [Public domain], via Wikimedia Commons"><img alt="Missing Square Animation" height="126" src="https://upload.wikimedia.org/wikipedia/commons/8/8c/Missing_Square_Animation.gif" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">By Trekky0623 at English Wikipedia (Transferred from en.wikipedia to Commons.) [Public domain], via Wikimedia Commons</td></tr>
</tbody></table>
Are you able to solve this puzzle?Cognitive Dissonanceshttp://www.blogger.com/profile/15069035350995446947noreply@blogger.com4tag:blogger.com,1999:blog-891276053127500572.post-33553347623311342902016-07-04T09:24:00.002-07:002016-09-07T00:45:04.764-07:00If nuclear fusion is the reverse of fission, why is energy released in both processes?Nuclear power is the use of nuclear reactions that release nuclear energy to generate heat. There are basically two ways to release energy from nuclei:<br />
<ul>
<li>nuclear fission, which is either a nuclear reaction or a radioactive decay process in which the nucleus of an atom splits into smaller parts (lighter nuclei). This released energy is the one that is frequently used in steam turbines to produce electricity in nuclear power plants.</li>
</ul>
<div style="text-align: center;">
</div>
<div style="text-align: center;">
<div class="separator" style="clear: both; text-align: center;">
</div>
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</div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgxWgh26_JBz9HM3n7NPxZlqhwYbqTsGulAmwW5ek58h5QJPsXZnY2QnZhLgXV07yE6hGK-iwNO4UcPyyyFFY5zfHG9ZXBkVxj9VqIVKp1pKVkeSEVQAa6-w8AVVPFIA2GtMQw-H6V1P_9O/s1600/1200px-Nuclear_fission.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgxWgh26_JBz9HM3n7NPxZlqhwYbqTsGulAmwW5ek58h5QJPsXZnY2QnZhLgXV07yE6hGK-iwNO4UcPyyyFFY5zfHG9ZXBkVxj9VqIVKp1pKVkeSEVQAa6-w8AVVPFIA2GtMQw-H6V1P_9O/s320/1200px-Nuclear_fission.gif" width="205" /></a></div>
<span style="font-size: x-small;">Public Domain, https://commons.wikimedia.org/w/index.php?curid=486924</span></div>
<ul>
<li>nuclear fusion, which is a reaction in which two atomic nuclei fuse to form a heavier nucleus. Nuclear fusion reactors are not yet economically viable, but this technology is currently under research and it could become viable in a few decades.</li>
</ul>
<br />
<div style="text-align: center;">
<a href="https://commons.wikimedia.org/wiki/File:Deuterium-tritium_fusion.svg#/media/File:Deuterium-tritium_fusion.svg"><img alt="Deuterium-tritium fusion.svg" height="200" src="https://upload.wikimedia.org/wikipedia/commons/thumb/3/3b/Deuterium-tritium_fusion.svg/1200px-Deuterium-tritium_fusion.svg.png" width="178" /></a></div>
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<span style="font-size: x-small;">By <a class="new" href="https://commons.wikimedia.org/w/index.php?title=User:Wykis&action=edit&redlink=1" title="User:Wykis (page does not exist)">Wykis</a> - <span class="int-own-work" lang="en">Own work</span>, based on <a class="extiw" href="https://en.wikipedia.org/wiki/File:D-t-fusion.png" title="w:File:D-t-fusion.png">w:File:D-t-fusion.png</a>, Public Domain, https://commons.wikimedia.org/w/index.php?curid=2069575</span></div>
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If splitting a nucleus into two smaller nuclei releases energy, it seems that combining two smaller nuclei into one larger nucleus would require energy, not release it, because it is the inverse process. So, why can we obtain energy from both processes?<br />
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Are you able to resolve this cognitive conflict?<br />
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Please, explain your reasoning. You can post your attempted answers in the comment box below. Please, do not use Facebook or Twitter to give your answers.</div>
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Cognitive Dissonanceshttp://www.blogger.com/profile/15069035350995446947noreply@blogger.com2tag:blogger.com,1999:blog-891276053127500572.post-9742591157987574152016-07-04T08:15:00.003-07:002016-07-19T05:31:51.276-07:00Why are cognitive dissonances useful for teaching?<div style="text-align: justify;">
Psychologists define cognitive dissonance as the mental stress or discomfort experienced by an individual who holds two or more contradictory beliefs, ideas, or values at the same time, performs an action that is contradictory to one or more beliefs, ideas, or values, or is confronted by new information that conflicts with existing beliefs, ideas, or values. It has been found that an individual who experiences a cognitive dissonance tends to become psychologically uncomfortable, and is motivated to try to reduce this inconsistency [Festinger, L. (1957).<span class="c3"> <i>A Theory of Cognitive Dissonance</i></span>. California: Stanford University Press].
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That is why psychologists have incorporated cognitive dissonance into models of basic processes of learning, notably constructivist models [Ausubel, David P.,Novak, J.D.,Hanesian, H. (1978) <span class="c3"><i>Educational Psychology: A Cognitive View</i></span> (2ª ed.). New York: Holt, Rinehart and Winston.] [Ausubel, David.P. (2000). <span class="c3"><i>The Acquisition and Retention of Knowledge</i></span>. Dortrecht, Netherlands: Kluwer.].</div>
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In these models teachers gather information about students’ existing ideas and use this information to design activities that foster dissonances in their minds by increasing their awareness of conflicts between students’ prior beliefs and new information (e.g., by requiring students to defend prior beliefs and confronting them with unexpected experimental results or different ideas). Then, teacher guides students to find by themselves correct explanations that resolve the conflict.<br />
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<a href="http://cdissonances.blogspot.com.es/2016/07/if-nuclear-fusion-is-reverse-of-fission.html" target="_blank"><img border="0" height="264" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg84NIZpha6SVP5vOmzMSlFmeJoazUtbUlTBKJAQ9qGVm_u5gOj9qL5BRwrecqDgNRh8eiiTLtfC7nzTdo7XqKIiK7xpI92-qkqiyWUdaDIApMGSMhlgf_SVc7ZcP1jhjENVdeOB_C2Xkdh/s320/tweet.gif" width="320" /></a></div>
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Notice that with this methodology no external reward is needed. Cognitive dissonances are enough to increase students' enthusiasm for educational activities. Some researchers have concluded that students who are forced to attribute their work to this intrinsic motivation came to find the task genuinely enjoyable [Aronson, E. (1995). <span class="c3"><i>The Social Animal</i></span>. New York: W.H. Freeman and Co.].</div>
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Once students have found by themselves the correct explanation that resolve the cognitive dissonance, a conceptual change has happened. Now, students do not see the world with the same eyes as before, and this change will remain forever, so they keep easily what they have learned even several years before. In fact, it has been shown that teaching methodologies based on cognitive dissonances significantly increase learning in science and reading [Guzzetti, B.J.; Snyder, T.E.; Glass, G.V.; Gamas, W.S. (1993). "Promoting conceptual change in science: A comparative meta-analysis of instructional interventions from reading education and science education". <span class="c3"><i>Reading Research Quarterly</i></span> 28: 116–159.].</div>
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Therefore, we can say that we have found a magic wand that will make the students learn science! Nevertheless, things are not so easy. The really difficult part of this process is to find the suitable cognitive dissonances that make the students learn every specific chunk of scientific knowledge. And this is a too technical task to be left to psychologists! Scientists should do it. That is why I have created this web site, to make a database of questions/interventions that could help teachers to foster dissonances in student minds. Of course, since we are looking for cognitive conflicts with students beliefs, the way we can foster these conflicts depends strongly on the students. So,</div>
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<li class="c0 c4" style="text-align: justify;">If you are a teacher, you have to select the most suitable question/intervention in each case. <a href="http://cdissonances.blogspot.com.es/p/how-to-collaborate.html" target="_blank">You can also collaborate with this project</a> by sharing the questions/interventions that you use to foster dissonances in your students.</li>
<li class="c0 c4" style="text-align: justify;">If you are a student, enjoy trying to <a href="http://cdissonances.blogspot.com/" target="_blank">resolve the conflicts</a>. You can post your trials in the comment box below each puzzling question. Some weeks after each publication we will include the explanations that eliminate the inconsistency. </li>
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I hope you found this web site useful!</div>
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Sergio Montañez<br />
Founder of <i>Cognitive Dissonances</i></div>
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Cognitive Dissonanceshttp://www.blogger.com/profile/15069035350995446947noreply@blogger.com0