No physicist/astrophysicist I; All I know about the Chandrasekhar limit is that it apparently limits the mass a star may survive, beyond which it degenerates to a neutron star, or a black-hole.
Does something similar apply to a black-hole? Is there an upper limit to black-hole mass? Can it acquire infinite mass? Is there a limit beyond which it may no longer acquire mass? Does a black-hole die? Is there a corresponding time-frame applicable?
The canonical momentum is defined as
$p_{i} = \frac {\partial L}{\partial \dot{q_{i}}} $, where $L$ is the Lagrangian.
So actually how does $p_{i}$ transform in one coordinate system $\textbf{q}$ to another coordinate system $\textbf{Q}$ ?
http://en.wikipedia.org/wiki/Hamiltonian_mechanics#Charged_particle_in_an_electromagnetic_field
When dealing with the Hamiltonian of the electromagnetic field, the derivation of $p_{j} = m \dot{x_j} + eA_j$ on the above link is usually written as $\textbf{p} = m \textbf{v} + e\textbf{A}$
but the derivation is based on using Cartesian coordinates, does it mean that $\textbf{p}$ is really a vector? If we are using another general coordinates, say, spherical coordinates, can we still have $\textbf{p} = m \textbf{v} + e\textbf{A}$ ? If no, I think the form of Hamiltonian in electromagnetic field
$H = \frac{(\textbf{p} - e\textbf{A})^2}{2m} + e\phi$
will only be valid in Cartesian coordinates. In any other coordinates, $H$ carries a different form!
Any comments are appreciated.
Hamilton's principle states that a dynamic system always follows a path such that its action integral is stationary (that is, maximum or minimum).
Why should the action integral be stationary? On what basis did Hamilton state this principle?
The Drude Model helped me quite a lot to visualize how current could flow in a circuit. However, there is still a point that I cannot grasp in the explanation given by some people when they talk about energy. When considering a round trip of one electron along a circuit, it is usually said that :
The electron gains energy when going through the battery and loses that energy along the wire as it collides with the lattice of the conductor.
My problem with this reasoning is that it is implying that the electron has more energy when it leaves the battery than anywhere else in the circuit. However, according to the Drude model, when the electrons collide with the lattice, they indeed lose kinetic energy but they regain kinetic energy as soon as they start accelerating thanks to electric fields produced along the wire. This causes electrons to have an average drift speed. In other words, their kinetic energy is constant on average, which is at first glance, in contradiction with the reasoning given above. What am I missing here?
Answer
The battery sets up an electric field in the external circuit which all the mobile electrons feel in all the external circuit. This means that there is a force on all the mobile electrons in the external circuit.
So these mobile electrons are accelerated by the electric field and gain kinetic energy from the electric field which is maintained by the battery.
The accelerated electron then interact with the vibrating lattice ions and on average there is a transfer of energy from the mobile electrons to the lattice ions is the ions vibrate more.
After the interaction with the lattice ions the mobile electrons are accelerated again thus gaining kinetic energy, interacting with the lattice ions etc.
So on top of the random thermal motion of the mobile electrons there is a drift velocity away from the negative terminal of the battery and towards the positive terminal of the battery as they gain kinetic energy from the electric field and then lose some to the lattice ions.
Within the battery the chemical process moves electrons from the positive terminal to the negative terminal thus giving the electron electric potential energy from the chemical energy of the reacting chemicals in the battery.
The key point is that outside the battery all the mobile electrons drift round the circuit from the negative terminal to the positive terminal of the battery. They continuously lose electric potential energy, gain kinetic energy and then lose the kinetic energy on colliding with the lattice ions which vibrate more - the conductor's temperature is increased which is called "ohmic heating".
In Bose-Einstein condensation, the chemical potential is less than the ground state energy of the system($\mu<\epsilon_g$). But why does the massless boson such as photon have zero chemichal potential($\mu=0$)?
image courtesy of http://www.learncbse.in/
Why is the saturation current the same, as you vary the frequency of incident light but keep its intensity constant?
If intensity is a measure of the energy of the incident beam, the product of the number of packets of light and the energy of each packet must be constant. So, if you increase the frequency, the energy of each packet must increase but the number of packets must decrease to keep the intensity constant. So, the decrease in the number of packets should correspond to the decrease in the number of electrons being ejected from the emitter plate. If the above discussion talks about taking place in unit time, shouldn't the saturation current decrease with increase in frequency, keeping intensity constant?
Note: Another thread I found on this website hinted at the intensity talked about here is the number of energy packets. But I'm not sure how this is a good way of defining intensity, especially when "comparing and contrasting" this quantized view with that of a classical wave.
My professor, introducing Heisenberg uncertainty principle, started from the Fourier transform and the classical uncertainty for waves.
He told about the localized impulsive wave $\delta(x)$ which has defined position but total uncertainty of impulse (its Fourier transform is composed of every possible momentum). On the other hand, a wave of defined impulse is a monochromatic wave, which spreads over the entire position axis and doesn't have a proper localization.
I'm perfectly comfortable those considerations, but then, out of noting, he writes
$$\Delta x \: \Delta k \geq 1/2$$
From this it's easy to derive the Heisenberg principle, but I can't understand where the previous formula comes from.
Does it come from Fourier transform properties, from the properties of optical waves, or from something else?
Answer
The Heisenberg Uncertainty Principle has two distinct aspects:
One is the identification of matter as a wave and, in particular, the relationship between a particle's momentum $p$ and its wavelength $\lambda$ through de Broglie's relationship $p=h/\lambda$. This is the crucial bit of physical input.
The second one is purely mathematical, and it's the relationship $\Delta x\, \Delta k\geq 1/2$. This is a general fact about waves and their Fourier transforms, and in a signal-processing context it's known as the bandwidth theorem.
In general, the bandwidth theorem is a bit hard to state precisely - or rather, there are multiple valid slightly different ways to state it, depending on exactly how you define the terms that appear in it and the classes of functions you're considering. However, in all its incarnations it is simply a fundamental fact of the theory of Fourier transforms.
As an example, if you have a complex-valued function $f(x)$ normalized to $\int_{-\infty}^\infty |f(x)|^2\:\mathrm dx=1$ and you define the position uncertainty as $$ \Delta x=\sqrt{\int_{-\infty}^\infty x^2 \: |f(x)|^2\:\mathrm dx - \left(\int_{-\infty}^\infty x \: |f(x)|^2\:\mathrm dx\right)^2} $$ the Fourier transform as $$ \tilde f(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-ikx}f(x)\:\mathrm dx, $$ and the wavevector uncertainty as $$ \Delta k=\sqrt{\int_{-\infty}^\infty k \: |\tilde f(k)|^2\:\mathrm dk - \left(\int_{-\infty}^\infty k \: |\tilde f(k)|^2\:\mathrm dx\right)^2} ,$$ then the uncertainty relation $$\Delta x\:\Delta k\geq \frac12$$ holds at least for all continuously differentiable $f$ such that $f'$, $\hat xf$ and $\hat k\tilde f$ are in $L_2$ (example proof). The uncertainty principle does hold for broader classes of functions, at least in a moral sense, but as I said there are multiple valid variants and it's a pain to list them all. However, for any suitable class of (generalized) functions, and definitions of the uncertainties, as long as the left-hand side's uncertainty product makes sense then it will have some sort of lower bound of order unity.
I was wondering why Faraday's law of induction and Maxwell-Ampere's law (without sources) are not totally symmetric in the sense that Maxwell-Ampere's law has a $\epsilon_0 \mu_0$ term on the right (in SI units) while Faraday's law doesn't, as symmetry is an important feature in most physical laws.
\begin{align} \nabla\times\mathbf E&=-\frac{\partial\mathbf B}{\partial t} \\ \nabla\times\mathbf B&=\color{blue}{\mu_0\varepsilon_0}\frac{\partial\mathbf E}{\partial t} \end{align}
A popular reference book states the reason being "that we use SI units". Can anyone tell me how using a particular unit can affect the symmetry of physical laws written in their mathematical form?
Answer
Maxwell's equations in vacuum are symmetric bar the problem with units that you have identified. In SI units $$ \nabla \cdot {\bf E} = 0\ \ \ \ \ \ \nabla \cdot {\bf B} =0$$ $$ \nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}\ \ \ \ \ \ \nabla \times {\bf B} = \mu_0 \epsilon_0 \frac{\partial {\bf E}}{\partial t}$$
If we let $\mu_0=1$, $\epsilon_0 =1$ (effectively saying we are adopting a system of units where $c=1$, then these equations become completely symmetric to the exchange of ${\bf E}$ and ${\bf B}$ except for the minus sign in Faraday's law. They are symmetric to a rotation (see below).
If the source terms are introduced then this breaks the symmetry, but only because we apparently inhabit a universe where magnetic monopoles do not exist. If they did, then Maxwell's equations can be written symmetrically. We suppose a magnetic charge density $\rho_m$ and a magnetic current density ${\bf J_{m}}$, then we write $$ \nabla \cdot {\bf E} = \rho\ \ \ \ \ \ \nabla \cdot {\bf B} = \rho_m$$ $$ \nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t} - {\bf J_m}\ \ \ \ \ \ \nabla \times {\bf B} = \frac{\partial {\bf E}}{\partial t} + {\bf J}$$
With these definitions, Maxwell's equations acquire symmetry to duality transformations. If you put $\rho$ and $\rho_m$; ${\bf J}$ and ${\bf J_m}$; ${\bf E}$ and ${\bf H}$; ${\bf D}$ and ${\bf B}$ into column matrices and operate on them all with a rotation matrix of the form $$ \left( \begin{array}{cc} \cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{array} \right),$$ where $\phi$ is some rotation angle, then the resulting transformed sources and fields also obey the same Maxwell's equations. For instance if $\phi=\pi/2$ then the E- and B-fields swap identities; electrons would have a magnetic charge, not an electric charge and so on.
Whilst one can argue then about what we define as electric and magnetic charges, it is an empirical fact at present that whatever the ratio of electric to magnetic charge (because any ratio can be made to satisfy the symmetric Maxwell's equations) all particles appear to have the same ratio, so we choose to fix it that one of the charge types is always zero - i.e. no magnetic monopoles.
I mention all this really as a curiosity. It seems to me that the real symmetries of Maxwell's equations only emerge when one considers the electromagnetic potentials.
e.g. if we insert $B = \nabla \times {\bf A}$ and $E= -{\bf \nabla V} - \partial {\bf A}/\partial t$ into our Ampere's law $$\nabla \times (\nabla \times {\bf A}) = \frac{\partial}{\partial t} \left({\bf -\nabla V} - \frac{\partial {\bf A}}{\partial t}\right) +{\bf J}, $$ $$-\nabla^2 {\bf A} +\nabla(\nabla \cdot {\bf A}) = -\nabla \frac{\partial V}{\partial t} - \frac{\partial^2 {\bf A}}{\partial t^2} + {\bf J}.$$ Then using the Lorenz gauge $$\nabla \cdot {\bf A} + \frac{\partial V}{\partial t} = 0$$ we can get $$ \nabla^2 {\bf A} - \frac{\partial^2 {\bf A}}{\partial t^2} + {\bf J} = 0$$ A so-called inhomogeneous wave equation. A similar set of operations on Gauss's law yields $$ \nabla^2 V - \frac{\partial^2 V}{\partial t^2} + \rho= 0$$
These remarkably symmetric equations betray the close connection between relativity and electromagnetism and that electric and magnetic fields are actually part of the electromagnetic field. Whether one observes $\rho$ or ${\bf J}$; ${\bf E}$ or ${\bf B}$, is entirely dependent on frame of reference.
If the fine structure constant is different in different parts of the universe, then what would happen if we travelled to those regions?
(I realise this is completely impossible as they are impossibly distant, so this is just a thought experiment).
If we hop in our spaceship and travel to a region where the fine structure constant has a greater magnitude, then will we be subject to those physical effects? Or will we, as beings composed of matter from our current region, still be subject to our original value of the fine structure constant?
Answer
If we move to a region where the fine structure constant is different, as we see by measuring the electrons and protons in this region, we would feel the new value of the fine structure constant also, gradually, as we moved to the new region. The reason is because the electrons in our bodies are quantum-mechanically indistinguishable from the electrons in the distant matter, and if they interacted differently, you would be able to tell them apart. This is the reason that a variation of the fine structure constant must be due to a new field which gradually varies from point to point, like a quintessence field. Such an interaction is non-renormalizable so it is difficult to make a model for it, so one should be skeptical.
This is part of a series of puzzles where you must identify the name of a video game
Previous puzzle is here: Name the Game #1
Turn the tap and open the valve
You won't have to face her by yourself
Something new comes out, oh it's blue!
Go for a bounce, you know you want to
Name the video game
Answer
I think the answer is:
Portal 2
Turn the tap and open the valve
Not 100% sure what this means, but Valve is the company who published Portal 2. Turning the tap could refer to the new gels which kinda look like they're coming out of faucets.
You won't have to face her by yourself
Unlike Portal, when you face GladOS in Portal 2, you have Wheatley alongside you.
Something new comes out, oh it's blue! Go for a bounce, you know you want to
One of the additions made in Portal 2 are various gels throughout the facility. One of them, the repulsion gel, is blue and allows you to bounce around.
I've read a little bit about zero-energy states, but I just don't get it. I'm just starting to study quantum mechanics and, at least for all the potentials I've seen until now (the most popular ones, like the infinite well or the harmonic oscillator), there is a certain minimum value $E_0\neq0$.
So the question is: given the time-independent Schrodinger's equation $$-\frac{\hbar}{2m}\frac{\partial^2\Psi(x)}{\partial x^2}+V(x)\Psi(x)=E\Psi(x)$$
is it possible, for any potential $V(x)$, that $E=0$? If it is, What would an example of that potential be?
Answer
Yes. For example, consider the harmonic oscillator potential $V(x)$, where the ground state has energy $\hbar \omega / 2$. Then the ground state of the potential $V(x) - \hbar \omega / 2$ has zero energy.
This works because in quantum mechanics, like in classical mechanics, absolute energies don't matter. You can always add or subtract constants.
However, there's a more interesting question I think you wanted to ask instead, which is, is it possible to have a state with energy lower than the minimum of $V(x)$? The answer is no, because in such a case, $\partial^2 \Psi / \partial x^2$ will always be positive, and you can check that there is no normalizable $\Psi(x)$ that satisfies this. So particles always sit a little higher than the bottom of a well. This can also be thought of as a consequence of the uncertainty principle.
For example, why is the semileptonic $B$ decay $B \to X\ell\nu$ inclusive?
I can't find any definition of these frequently used terms, strange.
To be efficient, a phase-matching condition has to be fulfilled in many nonlinear optical processes. For instance, the phase-matching requirement for second-harmonic generation is
$k_{2\omega}=2k_{\omega}$ or $\Delta k = k_{2\omega}-2k_{\omega}=0$
It is often said that this is equivalent to momentum conservation. However, even if $\Delta k \neq 0$, the process still takes place - although with lower efficiency and a finite coherence length $L = \frac{\pi}{\Delta k}$.
How can the conversion process still occur while momentum is not conserved? Is there momentum transfer to the medium? I guess not, because in many nonlinear processes only virtual photons participate. Do the photons 'borrow' momentum to make the jump? In other words, how does this work?
Answer
There is typically considered to be an uncertainty which softens the matching condition. In the case of momentum, the momentum state is only as well defined as the spatial extent of the interaction allows it to be. If the interaction length is given by $L$, which we can take to be an approximate measure of the position uncertainty $\Delta x$, then the corresponding momentum uncertainty is \begin{equation} \Delta p \ge \frac{\hbar}{\Delta x} \end{equation} so that there is a corresponding uncertainty in $\Delta k = \Delta p / \hbar$, which gives (up to some factors) $\Delta k \sim (\Delta x)^{-1}$. The finitude of the system size, either in time or space, means that the determination of the Fourier coefficients has a certain wiggle room (in $\omega$ or $\mathbf{k}$, respectively).
If the number of fermions is $n$, we expect the quantity $(-1)^n$ to be conserved, i.e., $n$ never changes between even and odd. This is known as conservation of statistics. In the normal context of particles with the properties traditionally expected for fermions and bosons, it can be derived from conservation of angular momentum. If $n$ were to change from, say, even to odd, then the total angular momentum of the system would have to change from an integer to a half-integer.
In the context of tachyons, there's an odd twist, however. The spin-statistics theorem is based on the assumption that the field has to commute at points that are spacelike in relation to one another. This commutation relation wouldn't hold for tachyons, so they get a free pass on spin-statistics. In fact Feinberg, in the classic paper that introduced the term "tachyon," found that the most plausible possibility was that tachyons would be spin-0 fermions (Feinberg 1967).
Feinberg says on p. 1099, discussing selection rules,
[...]there are restrictions following from the conservation of statistics. I shall simply assume here that if we assign a number $+1$ for bosons and $-1$ to fermions and multiply these numbers for a multiparticle system that these products are conserved in any transition.[16]
Footnote 16 says,
See, for example, Greenberg and Messiah (Ref. 10) where this is proven, however, under assumptions that may be invalid for tachyon theories.
Given Feinberg's assumption, we have two separate conservation laws, conservation of statistics and conservation of angular momentum, and the result is that if tachyons are unique in combining integer spin with Fermi statistics, they have a special conservation law that only applies to them: if $t$ is the number of tachyons, then $(-1)^t$ is conserved. This means, for example, that you can't produce just one tachyon, and you also can't have a process that produces a tachyon plus an electron, even though in normal field theory it's perfectly OK to produce two unlike fermions.
Does conservation of statistics really have some independent logical status, or is Feinberg just making an assumption that could be false, but that might make the field theory he's constructing easier to work with or more familiar? The understanding of tachyons in QFT has progressed a lot since 1967, so I wonder if this issue is better understood today than it was then.
G. Feinberg, "Possibility of Faster-Than-light Particles". Phys. Rev. 159 no. 5 (1967) 1089. Copy available here (may be illegal, or may fall under fair use, depending on your interpretation of your country's laws).
This is not a duplicate. I am not asking whether anything can escape a BH, I understand nothing can. My question is whether the potential barrier in the definition of Quantum tunneling can be a BH EH? Does the EH qualify as a potential barrier?
I have read this question:
Can a particle tunnel from inside a black hole?
But there is no answer.
In principle, can energy "tunnel" directly out of a black hole? If not, why not?
Now the answers mention Hawking radiation. Not the type where particle-antiparticle gets created, one inside the EH, one outside. I am talking about the Hawking radiation where it is about Quantum tunneling.
Yet, I read everywhere on this site that nothing can escape the BH.
Please see here:
https://arxiv.org/abs/hep-th/9907001
From wiki:
Quantum tunnelling or tunneling (see spelling differences) is the quantum mechanical phenomenon where a subatomic particle passes through a potential barrier. Quantum tunneling is not predicted by the laws of classical mechanics where surmounting a potential barrier requires enough potential energy.
Now the trick here is, that even though the EH would be a potential barrier, still, nothing can tunnel through it from inside, because nothing can move outwards from inside. Am I correct? Is the EH a potential barrier? Is it just that the direction of the particles can never be outwards?
Question:
Can this potential barrier be the EH of a BH?
If not, why is the EH different then a potential barrier in the definition of Quantum tunneling?
Answer
A hand waving answer refers once more to the basic tunneling phenomenon as described here
Please note the dark energy level line. Tunneling happens at the same energy level, it is only probabilities that allow the existence of the wavefunction at a measurable level beyond the barrier.
Gravity has not been quantized , only effective quantizations exist, but the arguments should work. The black hole at the quantum level by definition is a potential well and the deeper in the well the higher the binding energy , in quantum mechanics.
By construction of the horizon, there are no energy levels outside the horizon at the same level as the energy levels inside the horizon, so the probability of tunneling is zero by mathematical arguments. There is no energy level outside the "horizon barrier" to which a particle inside the barrier could tunnel, as per the figure above.
Hypothetically: tunneling should be a part of the merging of the horizons in the merging of black holes, as in the LIGO event. There should be similar energy levels in the approaching black holes , and if quantum mechanics holds there would be tunneling during the approach to merge.
My first child was a rebel. He is estranged from us.
My second used to be much closer to me, but we are losing touch.
My third is similar in temperament to my second, but smarter, although sometimes he lets my second persuade him into doing things for him.
My fourth is quite a snappy dresser, and is very popular.
My fifth is the strongest, and the most help around the home.
My sixth and seventh are quite similar, but my seventh is friendlier. My sixth and fifth don't get on well.
My eighth likes to keep an air of mystery about herself.
I keep my baby closest to me. He can be very temperamental.
What is my name?
PART ONE: Answer the question in the riddle above.
PART TWO: Explain how all the pieces of information (I count roughly 16 of them) fit in.
(I have a suspicion that the "Who am I?" might be guessed fairly quickly; the real challenge to this riddle is in explaining what all the different parts mean. There isn't any fluff and it does all mean something. Be sure to explain your answer fully!)
For your answer to be complete, you must complete both part one and part two.
Hint for part two:
This is a description of our family. All the descriptions of my children are talking about my children themselves, not strangers who share their names.
The critics against Copernicus and Galileo argued that, if the earth moved, then, a heavy body that was dropped from a very high tower should fall to the west of the foot of the tower. Galileo argued, however, that this body would fall slightly to the east. My first guess is that the body would fall directly under the point in which it was dropped. Who is right?
Answer
Who is right?
Galileo.
This sounds a little like a homework problem. Is it? If so it should be labelled as such.
The sun rises in the east and sets in the west. Thus the earth is rotating "to the east". The earth ostensibly has a constant rotational velocity $\omega$ of about one revolution per 24 hours. That means a point at the base of the tower has a linear velocity "to the east" of $$ v_{tower}=R_e\omega\;, $$ where $R_e$ is the radius of the earth.
And a point at the top of the tower (of height $H$) has a velocity "to the east" of $$ v_{body}=(R_e+H)\omega\;, $$ which is also the initial velocity of the dropped body.
You should be able to work out the rest on your own.
Note: I will expand this question with more specific points when I have my own internet connection and more time (we're moving in, so I'm at a friend's house).
This question is broad, involved, and to some degree subjective.
(I started out as a physics-only student, but eventually decided to add a mathematics major. I am greatly interested in mathematics; the typical curriculum required for physics students is not deep or thorough enough; mathematics is more general (that means work!); and it only requires a few more classes. Naturally, I enjoy mathematics immensely.)
This question asks mainly of undergraduate-level study, but feel free to discuss graduate-level study if you like.
Please do not rush your answer or try to be comprehensive. I realize the StackOverflow model rewards quick answers, but I would rather wait for a thoughtful, thorough (on a point) answer than get a fast, cluttered one. (As you probably know, revision produces clear, useful writing; and a properly-done comprehensive answer would take more than a reasonable amount of time and effort.) If you think an overview is necessary, that is fine.
For a question this large, I think the best thing to do is focus on a specific area in each answer.
Update: To Sklivvz, Cedric, Noldorin and everyone else: I had to run off before I could finish, but I wanted to say I knew I would regret this; I was cranky and not thinking clearly, mainly from not eating enough during the day. I am sorry for my sharp responses and for not waiting for my reaction to pass. I apologize.
Re: Curricula:
Please note that I am not asking about choosing your own curriculum in college or university. I did not explicitly say that, but several people believed that was my meaning. I will ask more specific questions later, but the main idea is how a physics student should study mathematics (on his or her own, but also by choosing courses if available) to be a competent mathematician with a view to studying physics.
I merely mentioned adding a mathematics major to illustrate my conclusion that physics student need a deeper mathematical grounding than they typically receive.
And now I have to run off again.
If I'm not mistaken, Schrödinger was influenced to look at wave equations because of de Broglie's assertion about particles having a wavelength. He started with the Hamiltonian equation which is related to principle of least action. What I don't get is how he ended up with an equation that describes a probability distribution.
Maybe I have a poor understanding of the equation, but starting with a deterministic function relating to principle of least action and ending up with a probability function seems to be like trying to model the orbit of the moon and ending up with an equation that accurately models the mating pattern of monarch butterflies. What I mean by this is that the evolution of the Hamiltonian and Lagrangian equations from the principle of least action, then everything from Newtonian mechanics to Relativity from those equations seems like a very neat and traceable path. Going from a wave equation to a probability density seems to have a logical gap (at least in my mind).
So my question is: what is the intuitive connection (if any) that clarifies/justifies how this leap is made?
Answer
Intro:
Ján Lalinsky's answer outlines the standard story, which basically follows Max Born's recount in his 1954 Nobel lecture, but the details are a bit more complicated. For instance, to say that Schroedinger simply followed de Broglie's wave idea in deriving his equation is to strip the basis of his original approach to a cryptic phrase that removes its entire logic. And then we are left with the mystery of the Schroedinger equation that, unintuitively, "pops out of nowhere" .
In a way, Schroedinger himself is partly responsible for this view, due to the way he structured and submitted for publication his four seminal papers of 1926, "Quantisation as a Problem of Proper Values", parts I-IV (for English reprints see for instance Stephen Hawking's collection "The Dreams That Stuff Is Made Of"). His celebrated equation first appears, as a time-independent version for an electron in a central electric field, in Eq.(5) of Part I. Contrary to common standards, the justification given there is indeed stripped down to bare essentials: a variational formulation, eq.(2), based on a Hamilton-Jacobi equation, eq.(1). This is probably why the "derivation" is commonly equated to more or less glorified handwaving. But the detailed basis for Schroedinger's approach - both formal and intuitive - is in fact the subject of Part II, where the stated purpose is
"to throw more light on the general correspondence which exists between the Hamilton-Jacobi differential equation of a mechanical problem and the "allied" wave equation, i.e. equation (5) of Part I […]"
As Schroedinger points out himself,
"So far we have only briefly described this correspondence on its external analytical side by the transformation (2), which is in itself unintelligible, and by the equally incomprehensible transition from the equating to zero of a certain expression to the postulation that the space integral of the said expression shall be _stationary_$^1$." ( Footnote $^1$ proceeds to explain that "The procedure […] was only intended to give a provisional, quick survey of the external connection between the wave equation and the Hamilton-Jacobi equation," where "$\psi$ is not actually the action function of a definite motion in the relation stated in (2) of Part I," but "On the other hand the connection between the wave function and the variation problem is of course very real: the integrand of the stationary integral is the Lagrange function for the wave process.")
To answer the question:
The 2nd paragraph of Part II then proceeds to outline Schroedinger's starting point, which he says was in fact Hamilton's own "starting-point for his theory of mechanics, which grew out of his [Hamilton's] Optics of Non-homogeneous Media":
"Hamilton's variational principle can be shown to correspond to Fermat's Principle for a wave propagation in configuration space (q-space), and the Hamilton-Jacobi equation expresses Huygens' Principle for this wave propagation. Unfortunately this powerful and momentous conception of Hamilton is deprived, in most modern reproductions, of its beautiful raiment as a superfluous accessory, in favour of a more colorless representation of the analytical correspondence."
This analogy between dynamics in configuration space and wave optics in inhomogeneous media is, by all means, the guiding intuition behind Schroedinger's approach.
What he does technically is to reformulate Hamilton's approach with help from a method developed by Hertz. The latter basically equips the configuration space with a non-Euclidian metric defined by the kinetic energy, $ds^2 = 2 {\bar T}(q_k, {\dot q}_k) dt^2$ (another footnote reveals that the problem was thoroughly studied by Felix Klein as early as 1891, and was well-known to Sommerfeld). The dynamics is now described not by paths in the (multidimensional) configuration space, but by the "wave-like propagation" of equi-action surfaces normal to these paths. It is this "propagation" that is analogous to that of optical wavefronts in an inhomogeneous medium, where "paths" correspond to ray optics, while wavefronts obviously correspond to wave ("undulatory") optics. Schroedinger then pushes the conjecture that this analogy must be complete. That is, he assumes that just as the necessity of wave optics follows from the breakdown of ray optics for short paths of large curvature (interference and diffraction around obstacles smaller than or comparable to the wavelength), in exactly the same way the breakdown of classical mechanics, also for short paths of large curvature, calls for a wave mechanics based on equi-action wavefronts. He then infers the simplest "wave mechanics" on configuration space that has as "ray limit" the usual classical mechanics.
But, he does not draw on de Broglie to do so. In fact, he doesn't have to. He first derives de Broglie's relation between particle velocity and (configuration space) wave group by using Planck's relation $E = h\nu$ in the equation for the action wavefronts. Nevertheless, he does emphasize the consistency of his approach with de Broglie's hypothesis of "phase waves".
As for the statistical interpretation, it is noteworthy that Schroedinger concluded Part IV with Section §7, On the Physical Significance of the Field Scalar (wavefunction). He observed that $\psi\psi^*$ must be "a kind of weight-function in the system's configuration space", which indicates how likely the system is to be found in that particular configuration. He actually comes very close to stating the statistical interpretation himself:
"[…] we may say that the system exists, as it were, simultaneously in all positions kinematically imaginable, but not "equally strongly" in all."
Furthermore, Schroedinger notes that, consistent with the weight interpretation, the configuration space integral of the weight function is conserved, and the weight function itself satisfies a continuity equation with the current density we all know. From this he concludes that charge density must satisfy a continuity equation as well, and electric charge is therefore conserved.
So all that was eventually left for Max Born, was to draw the well-known conclusion.
What does a Galilean transformation actually mean? I'm having trouble defining the equation for displacement shifts $x'=x-vt$. Does it mean that to any event $C$ the displacement in the primed coordinate system is the displacement to $C$ minus the velocity times time of the primed coordinate system? If so then don't both coordinate systems have to be at the same event at $t = 0$? I don't see this specified anywhere I look. Also, how does it directly show invariance of the euclidean distance between two points?
Answer
You have three question marks in your question. Addressing each of them separately, in chronological order;
1. How is the electric field set up in a wire by a battery so that we can assume that potential at a node is constant? 2. What happens to the electric field across a resistance which makes voltage 'drop'?
Assume the following circuit:
The battery sets up some electric field throughout the wire, with a component along (parallel to) the wire (otherwise, there is no current). Zooming into the wire, 
Here I display only the component of the electric field and electron velocity that is along the wire. As the electron moves a distance d, the work done on it by the electric field is Eqd. This work is non-zero. In other words, electric potential energy of the electron varies with the distance it travels along the wire - and thus so does potential. How, then, is potential on a node the same?
Now, add a resistive load:
![[1]: https://i.stack.imgur.com/dEb](https://i.stack.imgur.com/cQyC0.png)
The potential at all points on the wire above the battery is 10V, below it - 0V. Considering the above - how is this so?
NOTE: I am aware that a non-ideal wire has resistance. My question lies with what resistance has to do with a difference in potentials in the first place - and why voltage drop is not CAUSED BY a charge's distance from the terminals of the battery.
Exact duplicate it is.
Answer
First, to avoid confusion and off-topic threads, replace your "wire" with a resistive conductor such as a heating element, a carbon rod, etc. (In other words, don't let your wire be melted or vaporized before a discussion even begins!)
Q: Inside the above non-ideal conductor, why is the e-field the same everywhere, regardless of electron position?
A: it's because electric circuits are based on electrostatics, where all the patterns of currents are determined by surface charge. A wire is very different than the insulating gap between the plates of a capacitor.
With currents in any closed circuit, a conductor will automatically develop a dynamically-adjusting pattern of surface-charge, where this charge distributes itself in such a way to guarantee that the conductor's internal e-field (inside each straight section of the conductor) is constant, and is directed axially along the conductor. The uniform current in a wire is caused by the pattern of e-fields produced by the gradient of surface-charge on that wire. The particular pattern of surface-charge arises from a complicated give-and-take feedback process, where the charge is determined by patterns of resistance and current-density in the conductor (and, the current-density in turn is determined by the patterns of surface-charge!) Simple calculations don't exist for such things, so in order to view/understand the charge-distribution and circuit-fields, we'd normally use FEMM Finite-Element modelling software designed for E&M. (That, or run away screaming! Go hide yourself under Ohm's Law and simple waveguide-mechanics, where the full complexity of the actual real-world physics is traditionally ignored.)
Note: the e-field in any curved sections of your conductors won't be constant. Instead, the surface charge will distribute itself so that any charges drifting inside the conductor are forced to take a curved path which follows the conductor.
An excellent and detailed treatment of this topic is available on the undergrad textbook site for MATTER AND INTERACTIONS:
.pdf: A Unified Treatment of Electrostatics and Circuits, 1999 Chabay/Sherwood
In introductory electricity courses, electrostatic and circuit phenomena are usually treated as separate and unrelated. By emphasizing the crucial role played by charges on the surfaces of circuit elements, it is possible to describe circuit behavior directly in terms of charge and electric field. This more fundamental description of circuits makes it possible to unify the treatment of electrostatics and circuits.
Quite a bit of circuit-physics is revealed in this paper about E&M misconceptions and textbook failure, Sefton 2002, Understanding Electricity and Circuits: What the Text Books Don't Tell You
Another source is R. Morrison's textbook "The Fields of Electronics: understanding electronics using basic physics." Find a cheap used copy, or get free pdf downloads from library or university locations using this Wiley online link.
Morrison complains about the near-complete lack of physics resources for learning far more circuit-theory than found in basic intro classes, see http://web.archive.org/web/20120315092836/http://www.ralphmorrison.com/Ralph_Morrison/An_Open_Letter.html
Also see numerous essays at http://web.archive.org/web/20120519181208/http://www.ralphmorrison.com/Ralph_Morrison/Short_papers_on_interference_topics.html
Solve the nonogram.
(Click for a larger version.)
Answer
Final answer is
or perhaps I should say
PSE.
I confess that I only bothered going this far by strict logical inference
before allowing myself to make the obvious assumptions about the structure of the solution, leading to this (where white and yellow should now be treated as equivalent):
whereupon it's clear how to get to the solution at the top.
Hi friends I am starting to characterize a system in which we will plan to measure the response of a sample of liquid to radio waves. In order to do this I am exciting two antennas with the ports of an Agilent two-port network analyzer as shown in the image below.
So far I see no significant difference in the data when the container of liquid is present or absent between the two antennas. The data shows vey high values for S11 and S22, and low values for S12 and S21 mostly independent of the sample presence.
May I know if I should expect some difference due to the transmission media, or what I might test or change in the setup in order to become more sensitive to the sample?
Answer
I've added your images back to your question, slightly modified, I hope you don't mind. Now you might consider those arrows I've added, then go back and check to see if you are using ~950 MHz antennas (which are $\lambda/4$ and will resonate at integer multiples) and then ask if that's really the best thing to use. Away from resonance, they will naturally reflect most of the power.
You are mostly measuring the bandwidth of the antennas, except near the resonance where they can effectively radiate.
The network analyzer will measure the whole system, and so unless you use really good antennas that are nearly perfectly matched, mostly you will see reflections between the coax and the antenna connected to it.
An excellent, isolated antenna would dip down much deeper than -20 dB, but for a cheap WiFi antenna that's not bad, it means that roughly 99% of the power in the coax is leaving the driven antenna. However, at that point, still only 1% of the power is being received by the adjacent antenna and coupled into the other cable, so most of the power is probably going into free space.
What you can try is to focus on only the frequency where the antenna resonates naturally, which look like about 950 MHz. Scan only around there, with many small steps in frequency in order to make a smooth curve.
I think that if you use patience and study this narrow range, (say 800 to 1100 MHz or so) you might really see a difference between sample-in and sample-out.
However, the difference might be related to the effect of the dielectric sample on the antenna resonance, so you can also repeat the experiment with your sample near one antenna but away from the other; to the left of the left antenna, and to the right of the right antenna.
This may give you a better understanding of the limitations of your current set-up, and some ideas how to improve it. I'd recommend reading further as well.
Good luck, and have fun!
Note: This puzzle was inspired by the one here, by Mike Q.
Every square in the grid above, when the puzzle is complete, has a number between 1 and 9 in it and either is shaded or is not. Each 3x3 square in the grid contains exactly 4 shaded squares, forming an L, T, or S tetromino, which can be rotated or mirrored and can be placed anywhere in the square. Note that while the yellow squares in the given grid are confirmed shaded, the other squares all may or may not be shaded- no square in the given grid is confirmed unshaded yet.
The tetrominoes additionally follow the rule that each row and column of 3x3 squares contains exactly one L, T, and S tetromino, and no two squares contain the same tetromino unless one is rotated or mirrored relative to the other (for example, these two:
_#_ __#
_#_ __#
##_ _##
would not be allowed, but these three:
_#_ _#_ ___
_#_ _#_ ###
##_ _## __#
would).
The shaded squares follow the additional rules that they form a contiguous path with no 2x2 squares shaded, and that any unshaded region (that is, any collection of adjacent unshaded squares) must be directly adjacent to the edge of the grid (the following, for example, would not be allowed, as the center region is not adjacent to the edge.).
###__
#_##_
#__##
###__
Finally, the numbers follow the normal rules of sudoku, with the addition that the four numbers in each tetromino must add up to 20, and that no two tetrominoes share the same set of four numbers.
Some hints:
The possible combinations of four numbers that add up to 20 are:
(1 2 8 9) (1 3 7 9) (1 4 6 9) (1 4 7 8)
(1 5 6 8) (2 3 6 9) (2 3 7 8) (2 4 5 9)
(2 4 6 8) (2 5 6 7) (3 4 5 8) (3 4 6 7)
The rule on L/T/S tetrominoes means:
One diagonal must have three of the same tetromino, while the other must have one of each.
Regarding which squares in a 3x3 square are shaded, it's worth noting:
If a corner square is shaded, the opposite corner cannot be.
and:
If two squares on opposite sides are shaded, the square between them must be shaded as well.
Finally, regarding which set of four numbers goes in each tetromino:
Note that only four out of the twelve sets of numbers contain a 5, only two lack a 1 and a 2, and only three lack both a 3 and a 6.
An extra hint, since there aren't any answers yet:
You should be able to tell straight away that the center and lower-right sections contain either T or L tetrominoes, since they have a line of three shaded squares in a row. This divides possible solutions into four categories (numbering the sections with the equivalent on the number pad): 5T-3T, 5T-3L, 5L-3T, and 5L-3L. The second of these can have the shading completely determined using only the shading rules, and then be quickly eliminated using the sudoku/add-to-20 rules, while the third reduces to two almost-complete shadings depending on the orientation of the 3T. Some brute force is required, but by paying attention to where 9s have to go and how many sections have neither a 1 nor a 2 shaded, both can be eliminated. The techniques you use here should help figure out the rest.
Since there are still no answers, here's one last hint. If there are still no answers a week or so from now, I'll go ahead and post the solution.
The last possibility of the previous hint reduces to three cases depending on the lower-center and lower-right regions. If the upper-left corner of the lower-center region is shaded, and the upper-right corner of the lower-right region is not, the center-left region cannot be shaded properly, and if both are shaded the center-right cannot. If the center-left has unshaded upper-right corner, the region's (and most of the puzzle's) shading is determined by whether it contains a T or an S tetromino. In the former case, the upper-center region quickly becomes impossible to number once a few of the 9's are figured out while in the latter the upper-left proves impossible to number. This confirms that both the center and the lower-right regions (and thus the upper-left) must contain T tetrominoes. It is subsequently easy to determine that the upper-right region must contain an S tetromino, which determines most of the shading. After this, which of two sets of numbers go in the center-left region, which of two possible shadings the upper-left region has, and which (if any) of the regions have neither 1 nor 2 shaded determine the remainder of the puzzle.
Answer
It took me quite some time, and I've used all hints to start, but I believe this should do:
It seems I managed to make all incorrect guesses first, so (even without OP) I'm convinced that the solution is unique.
Consider the following theory comprising of $n$ bosons and $n$ fermions (along with their conjugates) on a Riemannian Manifold, with arc length parameter $t$ (section 10.4.1, Mirror Symmetry by Vafa et al.):$$L=\frac{1}{2}g_{ij}\dot{\phi}^i\dot{\phi}^j+\frac{i}{2}g_{ij}\bigl(\bar{\psi}^iD_t\psi^j-D_t\bar{\psi}^i\psi^j\bigr)-\frac{1}{4}R_{ijkl}\psi^i\psi^j\bar{\psi}^k\bar{\psi}^l\tag{1}$$with the fermion covariant derivative:$$D_t\psi^i=\partial_t\psi^i+\dot{\phi}^j\Gamma^i_{jk}\psi^k\tag{2}$$I am having a problem deriving the following conjugate momenta for $\phi^i$ and $\psi^m$:$$p_m=\frac{\partial L}{\partial \dot{\phi}^m}=g_{mj}\dot{\phi}^j\tag{3}$$$$\pi_{\psi^m}=ig_{mj}\bar{\psi}^j\tag{4}$$ Here are the details of the issues that I'm having:
Problem: From the fact that the momentum conjugate to $\bar{\psi}^i$ has not been given, I deduce that it must be zero, so I try to use an integration by parts to absorb the two terms enclosed within the brackets in $(1)$ in to one single term. The result is the following Lagrangian:$$L=\frac{1}{2}g_{ij}\dot{\phi}^i\dot{\phi}^j+ig_{ij}\bar{\psi}^iD_t\psi^j-\frac{1}{4}R_{ijkl}\psi^i\psi^j\bar{\psi}^k\bar{\psi}^l\tag{5}$$ Inflicting the partial derivatives I get the following result:$$\pi_{\psi^m}=\frac{\partial L}{\partial \dot{\psi}^m}=ig_{mj}\bar{\psi}^j\tag{6}$$$$\pi_{\bar{\psi}^m}=\frac{\partial L}{\partial \dot{\bar{\psi}}^m}=0\tag{7}$$So far I have the fermionic momenta correct. The issue arises when I try to compute the bosonic momentum:$$p_m=\frac{\partial L}{\partial \dot{\phi}^m}=g_{mj}\dot{\phi}^j+i\Gamma_{jkm}\bar{\psi}^j\psi^k\tag{8}$$Clearly there is an additional non vanishing term which makes $(8)$ differ from $(3)$. Alternatively, should we (leaving aside the fermionic momenta for a while) work with $(1)$ thinking that two such terms may cancel, we get:$$p_m=\frac{\partial L}{\partial \dot{\phi}^m}=g_{mj}\dot{\phi}^j+\frac{i}{2}\bigl(\partial_jg_{km}-\partial_kg_{jm}\bigr)\bar{\psi}^k\psi^j\tag{9}$$Again this is non vanishing term. Is the fermionic covariant derivative to be treated independent of $\dot{\phi}^m$ so that it is killed by the derivative operator $\frac{\partial}{\partial \dot{\phi}^m}$? Or is $(3)$ a typing error? Or is it something else? Kindly help out.
Answer
Yes, OP's eq. (8) for the canonical momentum is correct while eq. (3) is indeed an typo on the top of p. 208 in Ref. 1. See also point 6 in my Phys.SE answer here.
Calculating the fermionic canonical momentum is subtle for reasons mention in point 2 of the same answer.
References:
Two people play a game. The first one (A) picks one number from 1 to 10 and then the second one (B) also picks a number from 1 to 10 and adds it to the first number; the next round continues in the same way. Whoever gets to 45 is winner.
For example, A->4 and B->5 (4+5=9) A->3 (9+3=12) ... B->3(42+3=45) B is winner.
If A picks 3 in their first move, which number B can choose that definitely win the game for B?
1 2 3 4 5 6 7 8 9 10
Answer
9
Reasoning:
B's goal is to be able to answer any A's move with a move which guarantees B the victory. If B's first move is 9, the current sum is 12, and the sum of numbers left to 45 is 33, which is divisible by 11. After that whatever number x A picks, B answers with a number 11-x (which is always possible), increasing the current sum by 11 and reaching 45 on the fourth round.
What is the most comprehensive and up-to-date review of Higher Spin Theory (in particular, in AdS space)?
Answer
It really depends if you are interested in the 2+1 dimensional or the 3+1 dimensional theory.
Under the condition that you mean 3+1 dimensions:
Nonlinear higher spin theories in various dimensions - X. Bekaert, S. Cnockaert, Carlo Iazeolla, M.A. Vasiliev is a standard reference
Elements of Vasiliev theory - V.E. Didenko, E.D. Skvortsov for an introduction
Additionally there is a whole special issue of "Journal of Physics A" that includes reviews (also for 2+1 dimensional Holography (Gaberdiel and Gopakumar) and 2+1 dimensioanl black holes (Ammon et al)): https://iopscience.iop.org/1751-8121/46/21
You are standing on a roof in Israel at the top of the Ladder. You notice that there is a small package taped to the top rung of the ladder. You pick it up and take it to your hotel.
Once inside, you open the package. A blank white piece of paper flutters out of the package: 
You also see a small scrap of paper. It says
The image above has two prime numbers associated with it. Multiply them together and you have the key for the cryptic-clue that comes with this paper.
You then see another scrap of paper. It reads:
Dhjsn spcfe sdjtnjnn bypvnk ju uie ptsbod mvsm pf wjat(8)
You then see another scrap of paper. It says:
Your clue may be found halfway down the Trail of Tears
Where is the next clue, and which country do you have to go to. Also, where and what is the "Trail of Tears"
Answer
I figured out the country and have a guess to the location.
Key solution
The first thing I noticed is the picture is 397 x 431 px, which are both prime numbers. This gives us the key 171107.
Cipher answer
Decoding as a Vigenère cipher with the key we get
Cairn loved swimming around in the island full of pits(8)
Decoded Cipher meaning
The cipher is referring to the country Pitcairn Islands.
Trail of Tears
Google maps seems to recognize a Trail of Tears on Pitcairn Island (including a street view), but I can't seem to find any other information about it. It seems just to be the name of a trail on the island so maybe this is the exact location.
Answer
As dmckee says in his comment - Population III stars have no metals (a tiny bit of lithium and beryllium), but they are not "pure hydrogen stars", they still have the big bang fraction of Helium.
Taking the second part of your question first. These "stars" will last for ever. Their final fate is to become a completely degenerate ball of helium, supported by electron degeneracy pressure. They will never become hot enough to fuse anything beyond deuterium, hydrogen and lithium. Instead they will slowly fuse all their hydrogen, but because they are fully convective they will be able to resupply the core with fresh fuel until it is all gone. This scenario is discussed by Laughlin, Bodenheimer & Adams (1997) in the context of a solar composition - they suggest this phase takes around $10^{13}$ years for a star just above the minimum mass for hydrogen burning. Once the fuel is exhausted, the star will continue to cool and will contract a little more until its electrons become completely degenerate. At which point it can continue to cool, but now at constant radius since the pressure becomes independent of temperature. Such a star (a helium white dwarf) will just sit and cool forever, until their protons decay, or something else externally happens to them or the universe.
The answer to the first part must come from theoretical models. The canonical work by Chabrier & Baraffe (1997) find a "minimum mass" for hydrogen burning (which they define as that mass where hydrogen burning can supply the stars luminosity after 1 billion years) is about $0.072M_{\odot}$ for a star with solar metallicity, but rises to about $0.083M_{\odot}$ for stars which have less than a tenth of the solar metallicity. Earlier work by Baraffe et al. (1995) determined a limit of $0.09M_{\odot}$ for a metallicity 30 times less than the Sun. The review of Burrows et al. (2001) says that the minimum mass for a "zero metallicity" population III star is $0.092M_{\odot}$. This limit originates from modelling work performed by Saumon et al. (1994).
I am not aware of anything more recent, nor of anything more recent that tackles the entirely hypothetical question of a pure hydrogen star. However, one could wave ones hands and say that the virial theorem tells us that the central temperature is proportional to $\mu M/R$, where $\mu$ is the mean atomic mass in the gas. If hydrogen burning starts at a fixed central temperature, then the mass at which that occurs will be proportional to $\mu^{-1}$. If we remove the helium from the mixture, then $\mu$ goes down from around 0.6 to 0.5. So this very crude argument (it implicitly assumes a perfect gas and that the envelope opacity and hence radius is not changed significantly) might suggest the minimum mass for hydrogen burning rises to about $0.6\times 0.092/0.5 = 0.11M_{\odot}$.
Further edit: Your revised question now asks about fusion, rather than hydrogen fusion. Well, deuterium burning can occur at much lower masses - probably about 15 Jupiter masses.
Is it true that a glass window, that has been placed in a wall for about 10 years or more, is thicker on the bottom than on the top? I can vaguely remember my physics teacher saying that this was true. So if this is true, how is this possible?
Answer
The observation that old windows are sometimes found to be thicker at the bottom than at the top is often offered as supporting evidence for the view that glass flows over a timescale of centuries. The assumption being that the glass was once uniform, but has flowed to its new shape, which is a property of liquid. However, this assumption is incorrect; once solidified, glass does not flow anymore. The reason for the observation is that in the past, when panes of glass were commonly made by glassblowers, the technique used was to spin molten glass so as to create a round, mostly flat and even plate (the crown glass process, described above). This plate was then cut to fit a window. The pieces were not, however, absolutely flat; the edges of the disk became a different thickness as the glass spun. When installed in a window frame, the glass would be placed with the thicker side down both for the sake of stability and to prevent water accumulating in the lead cames at the bottom of the window. Occasionally such glass has been found thinner side down or thicker on either side of the window's edge, the result of carelessness during installation.
The charge conjugation operator $C$ reverses the charge of a state. But it may or may not convert a particle to its antiparticle. For example, consider a neutrino which is charge-neutral and left-handed while its antiparticle is also charge-neutral but right-handed. Therefore, charge conjugation is not sufficient to produce the antineutrino from a neutrino but CP is. $P$ is responsible for changing the chirality.
So is it not $CP$ instead of $C$ that is responsible for changing a particle to its antiparticle?
Is this fact related to the questions here and here?
I have no idea what is, no logic ... but probably I am missing something 
Billy the gardener uses a grey lego bolt to plant beans inside the timber ring, Sallah!
Who is not in the garden?
Hint:
A fellowship of nine
Answer
Based mostly on the answer by Otaia (I changed some of the clues and arrived at a different answer.. Then based on more hints I changed more clues and arrived at his original answer... )
Aragorn is not in the garden.
The sentence refers to The Lord of the Rings.Billy - Pippin, played by Billy Boyd
the gardener - Samwise
Grey - Gandalf the Grey
Lego - Legolas
Bolt - Merry, portrayed by Dominic Monaghan, who also played the Bolt in X-Men Origins (Credit user3453281)
Beans - Boromir, portrayed by Sean Bean
Timber - Frodo, played by Elijah Wood*
Sallah - played by John Rhys-Davies, who also played Gimli* Though on an interesting note I found that many believe that the 'orn' at the end of Aragorn's name means 'tree'.) Which could make Timber refer to Aragorn and make Frodo the answer. I however think that is more of a stretch than timber referring to Frodo.
Why in this video does the 2nd black hole appears to change size and appear larger the farther away it gets? How can you see behind it? http://www.youtube.com/watch?v=ENd8Sz0AFOk
Answer
Black holes bend light passing them and this means they act as a lens. The phenomenon is called gravitational lensing. The way black holes bend light is different to the way a conventional lens, for example in a magnifying glass, bends light and as a result there can be some very odd visual effects. In this case the lensing by the black hole at the front is magnifying the black hole at the back and distorting it into an Einstein ring.
Working out exactly what the gravitational lensing does is exceedingly complicated. If anyone is interested some details of the calculation shown in this image are described in this paper.
Inside the core of a star thermonuclear fusion reaction fuses hydrogen atom into helium releasing massive heat/light and energy.When a black hole eats up enough stars and gases it devours itself by releasing leftover gases and radiation {nothing but energy} called Quasar.
In both the cases there's massive amount of energy involved.I keep hearing the word "Pure Energy" instead of just "Energy". What is pure energy? How is it different from just energy?
Answer
Please note: I am not an astronomer - astronomers, please call me out if I am mistaken.
'Pure energy' is one of those terrible science journalism phrases that usually refers to hard electromagnetic radiation. Energy is a property of light and matter and not a substance in itself. In nuclear fusion, nucleons combine to form bound states with lower total energies than their separate components and the difference is mainly released as high energy photons and neutrinos.
Black holes do not devour themselves when they get too big. A quasar is an active galactic core, where a supermassive black hole at the center of a galaxy is accreting huge amounts of matter. This releases a lot of radiation in two ways - the accretion disk gets very hot and glows in the x-ray region, whilst the characteristic polar flows of a black hole are streams of gas/plasma moving at relativistic speeds which will light up whatever they collide with.
I have questions about rotation.
There is a sphere in space. I can apply a force to cause the sphere to rotate around a central axis. An infinite number of possible central axes can be drawn.
Can I apply a force and then another force such that the sphere will rotate around 2 different central axes at the same time? I think yes.
Is there an upper limit to how many different axes of rotation a sphere can have at the same time? Or do various axes (all axes?) somehow cancel out or add up, like linear vector addition - even though 3 different forces contributed to my linear motion the net effect on me can be expressed by a single vector.
If 1 is true, and there are no external influences (whatever force got sphere rotating has stopped) will motion of the sphere change such that rotation is just around just one axis over time?
Answer
Can I apply a force and then another force such that the sphere will rotate around 2 different central axes at the same time?
No, this is not the case. Any rigid body, at any time, can only be rotating about one instantaneous axis of rotation. If you apply additional torques this axis can shift, but there's no such thing as having more than one axis of rotation.
Now, that said, if the body is asymmetric, like, say, a slab of wood, then you can think about spinning it quickly about its long axis and then more slowly about an axis orthogonal to that, but even then that's an illusion: at any given time, the block is undergoing an instantaneous rotation about a single axis, with the funky property that this axis will shift position with respect to both the body and the inertial laboratory frame.
In general, the rotational motion of the body is described by the direction $\hat{\mathbf n}$ of this axis and the angular velocity $\omega$ of the rotation, which get combined into a single vector $\boldsymbol{\omega}=\omega \hat{\mathbf n}$ for convenience. In the absence of torques, this angular velocity vector is not conserved; instead, the body rotates with constant angular momentum $$ \mathbf L=I\boldsymbol\omega, $$ where $I$ is the moment of inertia matrix for the body; the rotational motion also conserves the rotational kinetic energy $E=\frac12 \boldsymbol{\omega}\cdot \mathbf L=\frac12 \boldsymbol{\omega}\cdot I \boldsymbol{\omega}$. That's about all that you can say in the general case, though if you move to a body-fixed frame you can analyze the motion a bit more understandably: there the angular momentum moves about (because the frame is not inertial) but it conserves both the energy and the total angular momentum $L^2$, which confines it to well-defined curves as described previously here and here on this site.
For the specific case of a sphere, then yes - when free of torques, both $\mathbf L$ and $\boldsymbol\omega$ will stay constant and the sphere will rotate with constant angular velocity about a fixed axis.
Whether you can get the first couple of Maxwell equations from a variational principle? In the second volume of the Landau theoretical physics said that it is impossible.
Answer
The Maxwell Lagrangian is given by,
$$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$
where $F_{\mu\nu}$ is the field-strength of the gauge field, or alternatively may be interpreted as the curvature of a $U(1)$ Lie algebra valued connection, $A_{\mu}$. By applying the variational principle we obtain,
$$\partial_\mu F^{\mu\nu}=0$$
in vacuum. In terms of the electric and magnetic fields,
$$\nabla \cdot \vec{E}=0 \quad \quad \partial_t \vec{E}=\nabla \times \vec{B}$$
we recover two of Maxwell's equations. Notice, in differential form language, $F=dA$, i.e. the curvature is an exact form, and all exact forms are also closed under the operation of exterior differentiation, i.e.
$$dF=d^2 A=0$$
Converting the above expression to a tensor equation, using the standard definition,
$$d\omega^{(n)}_{a_1 \dots a_n}=\frac{1}{n!} \left( \partial_{[a_1} \omega_{\dots a_n]}\right)$$
recovers the tensor form of the Bianchi identity,
$$\partial_\lambda F_{\mu\nu}+\partial_\mu F_{\nu\lambda}+\partial_\nu F_{\lambda\mu}=0$$
from which the two remaining Maxwell equations follow:
$$\nabla \cdot \vec{B}=0\quad \quad \partial_t \vec{B}=-\nabla\times \vec{E}$$
Recall, given the spin connection $\omega$, by Cartan's second structure equation, the curvature form is,
$$\mathcal{R}=d\omega + \omega \wedge \omega$$
However, the Lie group $U(1)$ is Abelian, and the structure constants vanish, hence the above simplifies,
$$\mathcal{R}=d\omega$$
which is completely analogous to the definition of the electromagnetic field strength. Other gauge groups may not possess the same field-strength. For example, in quantum chromodynamics, $SU(3)$ is non-Abelian, and the extra term does not vanish; in tensor form:
$$G_{\mu\nu}^a=\partial_\mu A^a_\nu - \partial_\nu A^a_\mu + gf^{a}_{bc}A^{b}_\mu A^{c}_\nu$$
I feel more tired walking 1 km than cycling 1 km at the same speed. However when cycling I am moving the extra weight of the cycle along!
Answer
This is one of those cases where the physics definition of "work done" does not match your experience.
If you move an object that is subject to gravity along a horizontal surface, physics tells us the only work done is the work done to accelerate the object, and overcome the force of friction.
Biologically, "walking" is a complex action that involves many muscles contracting and stretching. But muscles are not a reversible system (unlike a spring) - work done contracting is not returned when they stretch again (incidentally this is why some animals like kangaroos have highly elastic tendons... this greatly improves the efficiency of their jumping). So if you do a deep squat, returning to the same position, you will have expended (chemical) energy, even though you "did no net work" in the physics sense. Walking involves continuous (small) changes in the height of your center of mass, and so a lot of "shallow squats". Even if you could walk smoothly without bouncing your center of mass up and down, your legs will bend and stretch as you absorb the shocks of the road (your leg has to be straight when it is placed in front of you, and bend when it is directly underneath you - or you have to move your center of gravity).
When you look at cycling, you don't have to carry your entire weight on your legs - the only work you need to do is work required to overcome the small rolling friction (at walking speed) and air drag (if you go a little faster). Also, your center of mass stays at a constant height, so there is no energy lost in "bouncing" so much.
If you look at calories burned, this is confirmed. Riding a fast mile on a bicycle you will burn about 50 kcal, and much less if you go more slowly; running a mile will burn about 130 kcal (depending on how heavy you are - this is for 170 pound runner).
UPDATE
In Umberger et al, "A model of human muscle energy expenditure", Computer Methods in Biomechanics and Biomedical Engineering, 2003 Vol. 6 (2), pp. 99–111 the authors give a detailed model of energy expenditure of different muscles, showing clearly that load, not just extension, play a big role. And obviously when you carry your entire weight, you are carrying more of the load. They include the following diagram showing what muscles are loaded during what part of the walking cycle:
It occurred to me that the earliest "bicycle" that I am aware of was the velocipede, a contraption that allowed one to "walk" while part of the body weight was carried by the "bike". This immediately reduced the effort required for locomotion and provides further evidence for the above.
Image from that article:
Given the double slit experiment, send one photon through the open slits, then stop. With just that one photon traveling through both slits it will appear as a point on the screen. If the screen was ultra sensitive would we see the typical diffraction pattern in addition to the single point?
Say I fall into the event horizon of a black hole. As I cross the black hole, I would appear to outside onlookers to freeze in time, and would never move from that point again. In my perspective, time would seem to pass normally, so I would immediately fall into the black hole. But how? If an onlooker was to stay there and look at me frozen in time, I would stay frozen to them forever, even when the universe and time itself had ended. So my question is, how can I ever fall into the black hole if by any onlookers perspective I never do?
It is generally accepted that the radius of convergence of perturbation series in quantum field theory is zero.
E.g. 't Hooft in "Quantum Field Theory for Elementary Particles":
"The only difficulty is that these expansions will at best be asymptotic expansions only; there is no reason to expect a finite radius of convergence.”
Or Jackiw in "The Unreasonable Effectiveness of Quantum Field Theory":
"Quantum field theoretic divergences arise in several ways. First of all, there is the lack of convergence of the perturbation series, which at best is an asymptotic series."
The main argument for this comes from Dyson. (See: Dyson, Divergence of perturbation theory in quantum electrodynamics, Phys.Rev. 85 (1952) 631–632).
Now, in "Can We Make Sense Out of Quantum Chromodynamics?" t' Hooft introduced a transformation (now called t' Hooft transformation, or t' Hooft renormalization scheme) that replaces the coupling constant $g$ with $g_R$ such that $\beta(g_R) =a_1 g_R+ a_2 g_R^2$. In other words, the beta function for the new parameter $g_R$ has only two terms and not infinitely many as the series for $g$.
Isn't this in contradiction with the claim that the radius of convergence is zero? Or is the series for $g_R$ still divergent although it only contains two terms? If yes, where can we observe the divergence after the t' Hooft transformation?
I thought this changes the way series of numbers are looked at
1
1 1
2 1
1 2 1 1
Write down the next three lines
Answer
The answer is this:
111221
312211
13112221
The first number specifies the quantity of digits of the set above it and the second number specifies what the digit is. The second line is 11 it is saying that the line above it is one one. The third line states that the line above it is two ones. The fourth line is saying there is one two and one one.
How can I prove Thevenin's and Norton's theorem? Thévenin's theorem can be used to convert any circuit's sources and impedances to a Thévenin equivalent.
Answer
How can I prove Thevenin's and Norton's theorem?
Here's an outline - you can fill in the dots.
Measure the voltage (with an ideal voltmeter) between any two nodes of an arbitrary linear circuit. Call this voltage the open circuit voltage $V_{OC}$ since there is zero current through an ideal voltmeter.
Then, place an ideal ammeter across the same two nodes and measure the current. Call this current the short circuit current $I_{SC}$ since there is zero voltage across an ideal ammeter.
Now, you have the voltage when there is zero current and the current when there is zero voltage.
So, you have two points on the current-voltage (IV) plot for these two nodes. As you know, it takes two points to uniquely identify a line in this plane and the equation for this line is
$$V(I) = V_{OC} - \frac{V_{OC}}{I_{SC}}I$$
Can you take it from here?
Update 1 to respond to a comment.
@AlfredCentauri, How do we prove that the characteristic is a straight line?
If the circuit is linear, the superposition theorem holds and, thus, any circuit voltage or current can be expressed as a sum of terms, each term involving one source and equal to the circuit voltage or current due to that source alone, i.e., the result obtained by zeroing all other sources.
In the previous section, we measured the voltage across two terminals of a circuit and labelled that voltage $V_{OC}$.
By superposition, and assuming a DC circuit, if we connect a current test source across these terminals such that $I = I_S$, the voltage across the terminals is given by
$$V = V_{OC} - R_{EQ}I_S$$
where $R_{EQ}$ is the equivalent resistance between the terminals when all the circuit sources are zeroed (with all circuit sources zeroed, there is just a resistor network between the terminals).
Thus, by superposition, we know that for a linear DC circuit, there is, between any two terminals, an equivalent circuit with identical terminal characteristics: a voltage source with voltage $V_{OC}$ in series with a resistor with resistance $R_{EQ}$
And, since there is just one line between the two measured points in the I-V plane found in the previous section, it follows that
$$R_{EQ} = \frac{V_{OC}}{I_{SC}} $$
Update 2:
To verify the validity of my arguments above, I found a formal proof of Thevenin's theorem in one of my undergrad textbooks: "Fundamentals of Circuits, Electronics, and Signal Analysis" by Kendall L. Su.
I'll excerpt and paraphrase this proof found in the appendix A.1 on page 568
To simplify the proof, we shall assume that the network in question is excited by an independent current source [$I_{sN}$] at its terminal pair (terminals a and b in Figure A.1). Furthermore we shall assume that the network contains $N-1$ independent current sources and $M$ independent voltage sources, and a number of LTI elements, including LTI controlled sources.
Since the network is LTI, the superposition property prevails. That is to say, $V_{ab}$ is a linear combination of all the strengths of the independent sources. This fact can be expressed analytically as
$$V_{ab} = \sum_{j=1}^{N-1}Z_{Nj}I_{sj} + \sum_{k=1}^{M}H_{Nk}E_{sk} + Z_{NN}I_{sN} = V_{OC} + Z_{NN}I_{sN}$$
$$V_{OC} = \sum_{j=1}^{N-1}Z_{Nj}I_{sj} + \sum_{k=1}^{M}H_{Nk}E{sk}$$
$$Z_{NN} = \frac{V_{ab}}{I_{sN}}, E_{sk}=0, I_{sj}=0$$
...
The circuit of Figure A.2 [a voltage source with voltage $V_{OC}$ in series with $Z_{NN}$ connected between terminals a and b where the source $I_{sN}$ is connected] has exactly the relationship described by
$$V_{ab} = V_{OC} + I_{sN}Z_{NN}$$
The current source $I_{sN}$ cannot tell the difference between [the actual circuit and the equivalent circuit]. Hence the two circuits are equivalent electrically.
I was reading a textbook. I found that it was mentioned the speed of sound increases with increase in temperature. But sound is a mechanical wave, and it travels faster when molecules are closer.
But an increase in temperature will draw molecules away from each other, and then accordingly the speed of sound should be slower. How is it possible that the speed of sound increases if temperature increases? What is the relation of speed of sound and temperature?