strategy - Can you Turn MI to MU?

Using the letters M, I, and U, which can be used to create words. Can you turn MI into MU using in each step one of the following rules:

• 
  
  

  Insert a U to the end of a word ending in I. e.g: MI becomes MIU.

  
  


• 
  

  Re-insert the letters after the M e.g: MIU to MIUIU.

  
  


• 
  

  Replace the letters III with a U. e.g: MUIIIU to MUUU.

  
  


• 
  

  Get rid of letters UU. For example: MUUU to MU.

  
  
  

Can you use the 4 rules to change the word MI into MU? If so how many does it take?

Note: Please use spoilers.

Answer

This is the MU-Puzzle from one of my favourite books Gödel, Escher, Bach.

The solution is here:

No. It is impossible. The number of Is in the string will never be evenly divisible by 3. You need it to be so that rule 3 can reduce it to 0 and thus acheive the goal. We start with 1 I. We may only double it in the hopes of making it a multiple of 3 - no power of 2 is a multiple of 3 - nor is any number not already a multiple of 3 by any power of 2 divisible by 3. Since we start at 1, it cannot be done.

homework and exercises - Calculating quantum partition functions

...By quantizing we the get the following Hamiltonian operator

$$\hat{H}=\sum_{\mathbf{k}}\hbar \omega(\mathbf{k})\left(\hat{n}(\mathbf{k})+\frac{1}{2} \right)$$ where $\hat{n}(\mathbf{k})=\hat{a}^{\dagger}(\mathbf{k})\hat{a}(\mathbf{k})$ is the number operator of oscillator mode $\mathbf{k}$ with eigenvalues $n_{\mathbf{k}}=0,1,2,\dots$.

Using the quantum canonical ensemble show that the internal energy $E(T)$ is given by>

$$E(T)=\langle H \rangle = E_0 + \sum_{\mathbf{k}}\frac{\hbar \omega(\mathbf{k})}{e^{\beta\hbar \omega(\mathbf{k})}-1}$$

where $E_0$ is the sum of ground state energies of all the oscillators.

I started this by calculating the partition function

$$\begin{align} Z &= \sum_{\Gamma}e^{-\beta \mathcal{H}(\Gamma)} \\ &= \sum_{\Gamma}e^{-\beta (\sum_{\mathbf{k}}\hbar \omega(\hat{n}(\mathbf{k})+\frac{1}{2}))} \end{align}$$ ($\Gamma$ is a microstate of the system)

but I cannot see the thought process behind evaluating these, particularly with respect to the summations. This is a common problem I have found.

I would then go on to use $E=-\frac{\partial \ln Z}{\partial \beta}$

Answer

Quantum mechanically the general expression you want for the partition function is $$ Z = \mathrm{Tr} \left( \mathrm{e}^{-\beta H} \right),$$ where $\mathrm{Tr}$ means the trace (i.e. sum over micro-states). Now you can use the fact that the modes are independent, so that quantum Boltzmann operator $\mathrm{e}^{-\beta H}$ factorises into a product. This means that you can evaluate the trace over each oscillator mode separately: $$ Z = \mathrm{Tr} \left( \mathrm{e}^{-\beta H} \right) = \mathrm{Tr} \left( \prod_\mathbf{k}\mathrm{e}^{-\beta H_\mathbf{k}}\right) = \prod_\mathbf{k} \mathrm{Tr}_\mathbf{k} \left( \mathrm{e}^{-\beta H_\mathbf{k}}\right) = \prod_\mathbf{k} Z_\mathbf{k} $$ where $\mathrm{Tr}_\mathbf{k}$ means the trace over only the Hilbert space of mode $\mathbf{k}$, and $$H_\mathbf{k} = \hbar\omega(\mathbf{k})\left(\hat{n}(\mathbf{k}) + \frac{1}{2}\right).$$ Now $\mathrm{Tr}_\mathbf{k}$ means simply averaging over all the possible states in the Hilbert space, which you might as well choose to be the eigenstates of the number operator $\hat{n}(\mathbf{k})\lvert m_\mathbf{k}\rangle= m_\mathbf{k} \lvert m_\mathbf{k}\rangle$, with $m_\mathbf{k} = 0,1,2,\ldots$. So you have to evaluate $$ Z_\mathbf{k} = \mathrm{Tr}_\mathbf{k} \left( \mathrm{e}^{-\beta H_\mathbf{k}}\right) = \sum_{m_\mathbf{k}=0}^\infty \langle m_\mathbf{k} \rvert \mathrm{e}^{-\beta H_\mathbf{k}} \lvert m_\mathbf{k} \rangle. $$

inert gases - Potential vs Kinetic Energy of Particles in Gas

"In the gas phase, the molecules are freely moving particles traveling through space, where the kinetic energy associated with each particle is greater than the potential energy of intermolecular forces."

Qualitatively, this makes perfect sense. The particles have are moving very quickly which trumps any attractive forces.

However, I don't quite understand the energetics of such a situation. This motion is where the kinetic energy term comes from. On the other hand, the tendency for the particles to attract is represented by a potential energy term. Two particles that are attracted to one another in close proximity would have a positive potential energy, right? How do we know that the kinetic energy term has to be greater than the potential energy term?

EDIT IN RESPONSE TO ANSWER

Thought I would post this in case anyone else was confused. This is my rationalization of the answer.

This has cleared up a lot for me. This also has to be why solids vibrate in place. For the following, assume two particles in one dimension. Assume the particles are at some finite distance from each other, each with no KE. Call this distance d. We can define the PE to be 0 at this distance, d. Since they attract each other, they fall towards one another. The attractive force is applied across the distance between them, hence work. The amount of work (or force * dist) to bring the particle to a certain velocity is its kinetic energy. In other words, all of the PE is converted to KE. Assuming the particles then “collide,” elastically of course, the particles reverse directions. They are now moving away, each with some KE. They still attract one another, however. The attractive force will apply itself across some distance. Well the amount of work to stop them will be equal to their KEs. This will happen at a distance at the distance d. At this point, there will not be enough KE to keep moving away from one another. They fall back towards one another and the process repeats. This is the vibration. In a gas, there is not enough attraction to stop the moving away from one another. In other words, the KE outweighs the PE.

Answer

A correction, the potential describing the inter-molecular force is negative. Have a look at http://en.wikipedia.org/wiki/Lennard-Jones_potential. If both molecules have a separation distance that puts them between 1 and 2 on the horizontal axis of the attached figure ,that means their potential energy is negative. Hence their total energy is

:

Total energy = kinetic energy (positive by definition) + potential energy (sign depends on the sign of the potential)

So if they are between 1 and 2 on the horizontal axis in the attached figure, the previous addition will become subtraction. In order to make the molecules free their total energy has to be positive which means their kinetic energy has to exceed their potential energy.

electromagnetism - Derivation of the Biot-Savart Law

Can someone provide a derivation of the Biot-Savart law for electromagnetic induction? To be clear, $$ d\vec{B}~=~\frac{\mu_0}{4\pi}\frac{I d\vec{\ell}\times \vec{r}}{r^3}. $$

Is there a simple way to compute the magnetic field at a point between two Helmholtz coils, if the radii of the coils are the same and the current through each coil is the same?

quantum mechanics - Berry curvature in one dimensional parameter space

Berry curvature is defined by

$$\Omega_{n,\mu\nu}(R) = \frac{\partial A_{n,\nu}}{\partial R^{\mu}} - \frac{\partial A_{n,\mu}}{\partial R^{\nu}} $$ where $R$ is an parameter in hamiltonian. I think it is well defined if the dimension of parameter $R$ is more than 2. But what if the dimension of parameter $R$ is 1? Is it possible to say something about Berry curvature if the dimension of parameter is 1? If the dimension is 1, I think even the closed path integral for Berry phase is not well derfined.

electromagnetic radiation - Do two beams of light attract each other in general theory of relativity?

In general relativity, light is subject to gravitational pull. Does light generate gravitational pull, and do two beams of light attract each other?

Answer

The general answer is "it depends." Light has energy, momentum, and puts a pressure in the direction of motion, and these are all equal in magnitude (in units of c = 1). All of these things contribute to the stress-energy tensor, so by the Einstein field equation, it is unambiguous to say that light produces gravitational effects.

However, the relationship between energy, momentum, and pressure in the direction of propagation leads to some effects which might not otherwise be expected. The most famous is that the deflection of light by matter happens at exactly twice the amount predicted by a massive particle, at least in the sense that in linearized GTR, ignoring the pressure term halves the effect (one can also compare it a naive model of a massive particle at the speed of light in Newtonian gravity, and again the GTR result is exactly twice that).

Similarly, antiparallel (opposite direction) light beams attract each other by four times the naive (pressureless or Newtonian) expectation, while parallel (same direction) light beams do not attract each other at all. A good paper to start with is: Tolman R.C., Ehrenfest P., and Podolsky B., Phys. Rev. 37 (1931) 602. Something one might worry about is whether the result is true to higher orders as well, but the light beams would have to be extremely intense for them to matter. The first order (linearized) effect between light beams is already extremely small.

quantum mechanics - Meaning of $int phi^dagger hat A psi :mathrm dx$

While analysing a problem in quantum Mechanics, I realized that I don't fully understand the physical meanings of certain integrals. I have been interpreting:

• 
  

  $\int \phi^\dagger \hat A \psi \:\mathrm dx$ as "(square root of) probability that a particle with state $|\psi\rangle$ will collapse to a state $|\phi\rangle$ when one tries to observe the observable corresponding to $\hat A$"

  
  


• 
  

  $\int \phi^\dagger \psi \:\mathrm dx$ as "(square root of) probability that a particle with state $|\psi\rangle$ will collapse to a state $|\phi\rangle$".

  
  

All integrals are over all space here, and all $\phi$s and $\psi$s are normalized.

Now, I realize that interpretation of the first integral doesn't really make sense when put side by side with the second one.

For example, when $|\phi\rangle$ is an eigenstate of $\hat A$, I get two different expressions for "the square root of probability that one will get the corresponding eigenvalue of $\psi$ when one tries to observe the observable corresponding to $\hat A$". As far as I can tell, $\int \phi^\dagger \hat A \psi \:\mathrm dx \neq\int \phi^\dagger \psi \:\mathrm dx $, even if $|\phi\rangle$ is an eigenstate of $\hat A$. I am inclined to believe that the second integral is the correct answer here (it comes naturally when you split $|\psi\rangle$ into a linear combination of basis vectors). But I am at a loss as to the interpretation of the first integral.

So, my question is, what are the physical interpretations of $\int \phi^\dagger \hat A \psi \:\mathrm dx$ and $\int \phi^\dagger \psi \:\mathrm dx$?

(While I am familiar with the bra-ket notation, I have not used it in this question as I don't want to confuse myself further. Feel free to use it in your answer, though)

Answer

I'll start with the second one. $\int\phi^\ast\psi\,\mathrm{d}x$ is, as Chris says in the comments, the scalar (or dot) product of $\phi$ and $\psi$. In the Dirac notation, it is written as $\langle\phi|\psi\rangle$ and it gives the overlap of the two wavefunctions. In other words, it gives the probability amplitude (i.e., what you call square root of probability) that starting from $|\psi\rangle$ we will measure the state to be $|\phi\rangle$.

The first integral works in pretty much the same way; it's just the scalar product of $|\phi\rangle$ with $\hat{A}|\psi\rangle$. This means that you first apply $\hat{A}$ to the state $|\psi\rangle$ and then measure its overlap with $|\phi\rangle$. But this is not the same as calculating the amplitude probability of measuring $|\phi\rangle$ after the measurement of $\hat{A}$ is performed on $|\psi\rangle$. If you measured $\hat{A}$, that would be described by a set of projectors $\hat{\Pi}_i = |\chi_i\rangle\langle\chi_i|$, where $|\chi_i\rangle$ is an eigenvector of $\hat{A}$. The state $|\psi\rangle$ collapses to $|\chi_i\rangle$ with probability $p_i = |\langle\chi_i|\psi\rangle|^2$ which then collapses to $|\phi\rangle$ with probability $q_i = |\langle\phi|\chi_i\rangle|^2$. The overall probability is then $P = \sum_i p_i q_i$, which is not the same as $|\langle\phi|\hat{A}|\psi\rangle|^2$, as you can see if you write $\hat{A}|\psi\rangle = \sum_i \chi_i|\chi_i\rangle\langle\chi_i|\psi\rangle$, with $\chi_i$ being the eigenvalues of $\hat{A}$.

Edit:

The interpretation of $\hat{A}|\psi\rangle$ can be divided into several cases:

1. 
   

   If $\hat{A}$ is unitary, $\hat{A}|\psi\rangle$ can be understood as a time evolution of $|\psi\rangle$ governed by a Hamiltonian $\hat{H}$ fulfilling $\hat{A} = \exp(i\hat{H}t)$.

   
   


2. 
   

   If $\hat{A}$ is not unitary, but is trace-decreasing (i.e., $\langle\psi|\hat{A}^\dagger\hat{A}|\psi\rangle \le \langle\psi|\psi\rangle$ for each $|\psi\rangle$) it can describe some probabilistic evolution $|\psi\rangle\to|\varphi_i\rangle$ with probability $p_i\le 1$.

   
   


3. 
   

   The general case cannot be, I believe, interpreted generally. It requires renormalization (as does case 2) but this time the norm cannot be interpreted as the probability. Thus, it depends on the specific operator used.

   
   

As daaxix pointed out in the comments, the expression $\langle\phi|\hat{A}|\psi\rangle$ can be understood as a form of generalized mean value. Assuming $\hat{A}$ is Hermitian and writing $|\psi\rangle = \sum_k c_k|\chi_k\rangle$, $|\phi\rangle = \sum_k d_k|\chi_k\rangle$, we have $$ \langle\phi|\hat{A}|\psi\rangle = \sum_{k,l} c_k d_l^\ast \langle\chi_l|\hat{A}|\chi_k\rangle.$$ Due to orthogonality of $|\chi_k\rangle$, we can write $\langle\chi_l|\hat{A}|\chi_k\rangle = \delta_{kl} \langle\chi_k|\hat{A}|\chi_k\rangle$, so we have $$ \langle\phi|\hat{A}|\psi\rangle = \sum_k c_k d_k^\ast \langle\chi_k|\hat{A}|\chi_k\rangle. $$ This gives a sum of the expectation values $\langle\chi_k|\hat{A}|\chi_k\rangle$ (i.e., the eigenvalues $\chi_k$), with weight coefficients given by the expansion of the vectors $|\psi\rangle$, $|\phi\rangle$, namely $c_k d_k^\ast$.

riddle - My first is in the kill zone, but not in no man's land

My first is in the kill zone, but not in no man's land.
My second is in a fifth reef knot, but not in rope or string.

My third is in loneliness, but not in unison or simultaneity.
My fourth follows a formula, but isn't in a code.
My fifth

What am I?

Answer

You are ...

... the question "Equine musicians" by Rand al'Thor.

My first is in the kill zone, ...

The thing we are looking for is made up of numbers or digits, because each of the first three lines has a numeral hidden in the first phrase:

    My first is in the kill z[one], but not in no man's land.
    My second is in a fif[th ree]f knot, but not in rope or string.
    My third is in l[one]liness, but not in unison or simultaneity.

(The second line was edited. It had "fourth reef knot" before, which would lead to two numbers.)

My fourth follows a formula, but isn't in a code.

What follows a formula might be a number, too: Formula One.

My fifth

This should probably also indicate a number. There's nothing after "The fifth", so the number is probably zero. (Rand al'Thor has revealed that this is the right answer in a comment. I had considered it, but didn't know what to do with the whole number.)

Okay, now what?

We've got a five-digit number. This number (or maybe sequence of digits) can mean a lot of things.

Here at Puzzling Stack Exchange, it can identify a question. For example, if you look at the address bar right now, you'll see the 5-digit number 61639. Replacing that with the number we found brings us to "Equine musicians".

That question is supposed to be a hint for [another question] of Rand al'Thor's. I have seen "Equine musicians" not long ago: It was in the list of related links in the sidebar to the "Strange sorceresses" question.

Entropy and Crystal Growth

I was reading about growing single crystals and I'm a little confused about this -

In most crystal growing processes, a "seed crystal" is used, and the rest of the material crystallizes on the seed crystal (for example the czochrlaski process).

But thinking about entropy, wouldn't it be more favourable for the vapour/liquid to not crystallize and hence maintain a larger disorder?

A similar process occurs in supersaturated vapour as well, where a small liquid droplet acts as a "seed" for more condensation to occur. So I suspect I'm missing something, but I can't figure out what.

Answer

Indeed this is a handy counter example to people who, not understanding the second law, claim that evolution is impossible on the basis that entropy decreases (nevermind that by this misunderstanding life itself would be impossible as well). The growing crystal is not a closed system: it exchanges energy and matter with the surrounding environment, and this can lead to a local entropy decrease if it is compensated by an entropy increase of the environment. The thermodynamic potential that is truly minimized, taking these exchanges into account, is the free energy. (Note that there are slightly different definitions of the free energy which apply to slightly different types of process, but they are all morally the same.)

electromagnetism - Magnetic monopoles and special relativity

I was thinking about magnetism as a product of special relativity and the result of this approach to the magnetic monopoles. So if magnetism is a product of electricity(like electricity from another point of view),then why do we need monopoles to exist?I know that many theories predict the existence magnetic monopoles but i am referring specifically to the concept of classical relativistic magnetism and magnetic monopoles,so do not give me answers that are mainly based on what other theories predict.

EDIT: why do we need electric AND magnetic monopoles to describe electromagnetism if the two are the same thing from another moving frame of reference?And if we do not NEED magnetic monopoles,why is there even a place for them to exist in relativistic electromagnetism?

EDIT: I know that the mathematics of the theory allow for magnetic monopoles, but the essence of the question is the following:
If I work from one frame of reference and change to any other frame of reference, there are no sources of magnetism that can be related to magnetic monopoles?

Answer

The idea that magnetism is a side-effect of electricity is deeply mistaken. The sooner you forget about that idea, the better.

Electricity and magnetism are the two aspects of a single phenomenon, electromagnetism. Read that sentence again: They are not two aspects of electricity, they are two aspects of electromagnetism. Electricity does not cause magnetism and magnetism does not cause electricity, but rather special relativity joins them together, just as it joins space and time into spacetime. (See also my related answer here).

For most electromagnetic phenomena, there is no frame of reference in which it is purely electric, or purely magnetic. It is always a mix of the two, although it's a different mix in different frames of reference. For example, there is no frame of reference in which a refrigerator magnet has no magnetic field. In some frames it will also have an electric field, but a magnetic field is there too.

So in conclusion, magnetism is a fundamental physical phenomenon in its own right, not merely a funny way of talking about certain aspects of electricity. For that reason, it's possible for magnetism to contain phenomena that cannot be directly extrapolated from everyday electricity plus SR.

PS: If there are no magnetic monopoles in a certain reference frame, then there are no magnetic monopoles in ANY reference frame. Conversely, if there ARE magnetic monopoles in one reference frame, then there are magnetic monopoles in every reference frame.

lateral thinking - Man buys item, throws it away on his way out

I thought of this lateral thinking puzzle. Hopefully, it's not too broad and hasn't been asked before.

A man walks into a store, buys an item for $1, and throws it away along with the receipt on his way out. What item did he buy?

Notes:

• His decision to throw the item away on his way out was completely justified.


• He did not put anything in his mouth between buying the item and throwing it away.

EDIT: To make this question less broad, here are some notes that I should've had in the beginning:

• He bought the item because he wanted the item.


• "Throw away" means he put the item and the receipt in the trash can.


• The item was reasonably priced at 1 USD in 2018.


• He threw the item away because his perception of the item's intrinsic value changed.

Answer

It was

A scratch card. And presumably not a winning one.

newtonian mechanics - Force when acceleration is zero

i tried to find it in books but i was unable to do so and i am not able to understand this and I couldn't think of a better example. i repeat. its just an example. if i punch someone's face with my arm having constant velocity, it will mean that the acceleration of my arm is zero. F=ma, therefore force exerted by my arm is zero. Now if the force applied is zero the why does the person whom i punched experiences pain. please help. thanks.

Answer

For simplicity's sake, let's start with punching a concrete wall. Let's assume you are not strong enough to actually damage the wall. So (again for simplicities sake) if your fist is traveling at 1 m/s, when it hits the wall it will be moving zero, so it DECELERATES as part of the process of colliding with the wall, and the force of that deceleration will be applied to your hands (possibly breaking your knuckles). The mass of the wall is not relevant; this is why your hand would take the same damage from a smooth rock as it would from punching a large lead weight, it's just about the deceleration of your hand.

If you start slowing down your hand slightly before you punch the wall, your hand will have to do less deceleration when it collides, that's why your knuckle bones would be less broken. The same would be true if you were moving slower to begin with.

Now let's say you're trying to punch Neo from the Matrix. He sees you coming and dodges backwards, and coincidentally he happens to quickly accelerates enough (which takes energy from his muscles) so that he is moving his face backwards at the same speed that your fist is moving. Your fist will never actually hit him, so there will be no force applied to your hand.

Now let's say you're trying to punch the guy that stole your girlfriend. The same deceleration will occur. Some of the force from the deceleration of your fist at collision will go into breaking his jaw. Some of the force from the deceleration will go into moving his head back (if his head moves, your fist and his face receive less damage, because the force of your punch went into moving his head backwards). And some of the deceleration force will still go into hurting your hand.

electric circuits - Work done by magnetic field on current carrying conductor

Suppose a wire of length L carrying a current I is kept in a uniform magnetic field B perpendicular to the current. The force on the wire will be IBL and work done by magnetic force when wire moves a distance d along the force will be IBLd. But magnetic force cannot do any work on a moving charged particle and hence total work done on all particles by magnetic force should be zero. Where does the work IBLd come from?

Answer

The work comes from the battery that is driving the current through the wire.

Even if the wire were stationary, the battery would be supplying work at a rate $I^{2}R$. But with the wire moving, the battery would need to be supplying extra work at a rate $\mathscr{E}I$ in order to overcome the emf generated by the moving wire.

Now, $\mathscr{E}$ is equal to the rate at which the wire cuts magnetic flux so $\mathscr{E}=BLv$ (in which $v=\frac{d}{t}$), so the extra rate of doing work has to be $\mathscr{E} I=BLvI=BLdI / t $. And this is equal to the rate of mechanical work done on the wire!

refraction - How phase velocity is related to deflection angle?

Refractive index is a function of velocity in the medium.How is this related to deviation angle inside the medium?

Answer

Provided that the electron & the atomic beams also exhibit refraction,it seems that this is a particle's property.Deflection angle is proportional to particle's mass/size for specific medium.Photon behaves as particle in this effect.Mass is given by de Broglie equation:m=hv/c^2 , v=frequency

general relativity - Retarded potential in gravitational field?

Is there a retarded potential concept in gravitational field similar to electromagnetic radiation?

Landau's ambiguous statement about the existence of inertial frames

Landau writes "It is found, however, that a frame of reference can always be chosen in which space is homogeneous and isotropic and time is homogeneous."

Does he mean that we can prove the existence of an inertial frame or does he want to say that it is assumed by doing enough number of experiments?

Can we start with some axioms and definitions of properties of space and time and then deduce the existence of such a frame in which space is homogeneous and isotropic and time is homogeneous?

Answer

I believe this is just a restatement of the first Newton's law.

electricity - traditional transformer design improvement

The traditional tranformers suffer a lot of losses in various forms, some of them are as follows : 1. Flux leakage 2. Eddy currents 3. Hysteresis

Figure :

But as we are taught induction we come across the attached figure of two inductors placed coaxially various times. My point is that if we would actually place inductors in this way, we would have the following benefits :

Note : taking inner coil as primary

1. NO NEED OF CORE : As the flux of inner coil passes from outer on its own, no core is needed for transferring flux from primary to secondary


2. NO FLUX LEAKAGE : As entire flux of inner inductor passes from outer, no flux is leaked.


3. NO EDDY CURRENTS: As no core is used no eddy currents are generated.


4. NO HYSTERESIS : Similarly due to absense of core, no loss due to hysteresis is encountered.

So what really is the problem in creating transformers in such a way to reduce losses much more than the current status ?

All relevant opinions and comments appreciated

My original post : http://thinklo.blogspot.in/2013/10/improving-transformers.html

Answer

Well a problem with your concept is that an "air core" transformer can produce only a limited magnetic flux density; limited by the amount of current you can run through it.

If your transformer is for power conversion/transmission, your scheme would be extremely inefficient at power line frequencies. It can be shown that the efficiency is maximized when the "copper losses" are equal to the "iron losses" or "core losses if you prefer.

Since, your transformer has no iron core, there are very low to near zero core losses, but the copper losses (wire resistance) is very high.

With an "iron" core, you can get magnetic fields of 10-15,000 Gauss (is that 1-1.5 Tesla), because of the high permeability, so you can use fewer turns of thicker wire, so less resistance, and copper losses, but more core losses, eddy currents, hysteresis losses etc.

The design of efficient and cost effective transformers, is a very complex discipline.

Sometimes it can get slightly insane. Problem with iron cores, is that they saturate somewhere in the 15-20,000 Gauss region, so if you want to go higher in flux density than that, you have to get rid of the iron. Someone did that once on an electromagnet for an accelerator. They used just two turns of wire, each a foot in diameter (the wire, not the coil). They put 800 Volts across those two turns and it drew 6 million Amps; but gave them twice the field they could get with iron. Guys name was Marcus Oliphant, in Australia. They called his machine (which worked) the white Oliphant.

scattering - What is the physical interpretation of imaginary terms in the neutron optical potential model?

I'm doing a computational project on neutron scattering and I've found that if you simulate (n,n) collisions on Uranium 238, the program predicts resonances not found in experimental data.

The optical model includes imaginary surface or volume potential terms which supposedly damp out these resonances.

How should an imaginary potential be interpreted in this context?

waves - Why is Huygens' principle only valid in an odd number of spatial dimensions?

Apparently Huygens' principle is only valid in an odd number of spatial dimensions:

Why is this?

[EDIT] This is somewhat perplexing, since AFAIK it's pretty common to teach freshmen about double- and single-slit diffraction using a two-dimensional analysis and invoking Huygens' principle. Does this work only because there's an ignored third axis of translational symmetry?

I wonder if it's possible to gain insight by making a grid and doing sort of a finite-element analysis.

newtonian mechanics - Concept of mass

In classical mechanics, there seems to be a need to distinguish between inertial and gravitational mass. Some texts show how the concept of mass may be defined with some mathematical rigor. There is also an equivalence principle.

I would like to know if the concept of mass in Newton's "Principia" is not expressed in a much simpler manner.

edit:Newton on Mass and Force. By Ed Dellian, Bogenstr. 5, D-14169 Berlin.

What does Newton say about “mass”? Let us read the “Definitio 1” which opens the Prin- cipia: “Quantitas materiae est mensura eiusdem orta ex illius densitate et magnitudine conjunctim.” That is: The quantity of matter is that measure of it which arises from its density and volume conjointly.

So Newton's mass is simply "quantity of matter" obtained through density x volume (density => weight concept). Had we defined our unit mass with a fixed volume of water, then any object's mass could easily be measured using a scale against a certain volume of water - the volume of water would give the mass. This manner of definition could be taken as a fundamental definition of mass; it may seem to be just gravitational mass.

With a definition of mass done, then force is defined as simply: mass x acceleration. We may not need any discussion of "inertial mass".

word - Like totally amazing interchangeable sister outfit accessory swapping or whatever

OMG, me and my sister are like total BFFs and we share clothes like all the time! We soooo have the same sense of style and we borrow stuff from each other like every single second!

We don't even care, like, who was the last to borrow what from who or whatever. We can work anything into our outfits no matter what and make like a totally new outfit every time! And we always look so cute!

Seriously, the one thing is, we like never ever ever wear the same outfit twice. That would be so embarrassing you could just die! You know what I mean?

Anyway, check it out:

Take sisters CASTLE & OATHS through 6 more outfits
by lending/borrowing one "accessory" (letter) at a time


------------------------------------


Solution:


CASTLE          OATHS
  |           .   |
  |         .     |
  |        S      |
  |      .        |
  v    .          v

CASTLES         OATH
  |    .          |
  |      .        |
  |        L      |
  |          .    |
  v            .  v
CASTES          LOATH
  |    .          |
  |      .        |
  |        E      |

  |          .    |
  v            .  v
CASTS           LOATHE
  |           .   |
  |         .     |
  |        O      |
  |      .        |
  v    .          v
COASTS          LATHE
  |    .          |

  |      .        |
  |        S      |
  |          .    |
  v            .  v
COATS           LATHES
  |    .          |
  |      .        |
  |        O      |
  |          .    |
  v            .  v

CATS            LOATHES

(At this point, I typically remind solvers that my puzzles only use well-known words, but these tandem chains are difficult to construct without a few near-slang words. So be prepared for an occasional ding, ting, zing, bling, cred, or dang.)

See if you can help these 5 sister pairs change their outfits:

1.  Take sisters SAND & PILE through 7 more outfits by lending/borrowing one "accessory" (letter) at a time


2.  Take sisters TRIED & GINS through 7 more outfits by lending/borrowing one "accessory" (letter) at a time



3.  Take sisters BEERS & TAPS through 7 more outfits by lending/borrowing one "accessory" (letter) at a time


4.  Take sisters SCORED & CUP through 8 more outfits by lending/borrowing one "accessory" (letter) at a time


5.  Take sisters COOTS & BATS through 8 more outfits by lending/borrowing one "accessory" (letter) at a time

---

Epilogue:

Extra super double credit to @osdavison @Brandon_J and @Dolgubon for coming up with longer chains than I had prescribed!

See the sequel here

Answer

Completed answer:

SAND & PILE (I can get 8, one more than expected)

AND & PILES --> LAND & PIES --> LANDS & PIE --> LADS & PINE --> LEADS & PIN --> LEAD & PINS --> LED & PAINS --> SLED & PAIN

TRIED & GINS (7, as expected) - partial credit to OmegaKrypton - please go upvote his answer!

TIED & GRINS --> TIE & GRINDS --> TIES & GRIND --> TIDES & GRIN --> IDES & GRINT --> IDE & GRINTS --> NIDE & GRITS --> NIDES & GRIT

BEERS & TAPS (7) - osdavison has an independent solution with 11 words; please go upvote it.

BEES & TRAPS --> BEETS & RAPS --> BESETS & RAP --> BESTS & REAP --> BEASTS & REP --> BEATS & REPS --> BETS & REAPS

SCORED & CUP (8, as expected) - credit to Gareth McCaughan - please go upvote his answer!

CORED & CUPS --> CRED & COUPS --> CURED & COPS --> CURD & COPSE --> CURDS & COPE --> CUDS & COPER --> CUD & COPERS --> CRUD & COPES

COOTS & BATS (8, as expected) - PartyHatPanda has an independent solution - go upvote it!

COTS & BOATS --> COSTS & BOAT --> COASTS & BOT --> COAST & BOTS --> CAST & BOOTS --> CASTS AND BOOT --> CATS & BOOST --> CAT & BOOSTS

cosmology - Estimating age of the universe by Hubble's law?

Hubble's law states that $v=Hx$

Age of the universe is calculated by $T= \frac{x}{v} = \frac{1}{H}$

but the velocity is not constant; it changes with distance, so I think that this equation cannot be applied because simply the velocity is NOT uniform

$v$ should have been replaced with $\frac{\mathrm{d}x}{\mathrm{d}t}$, so it gives this differential equation $\frac{\mathrm{d}x}{\mathrm{d}t}=Hx$.

Nevertheless, when this differential equation is solved, and we substitute for initial condition ($x=0$ when $t=0$), it appears that there are no solutions, because it will be an exponential function, which can't take the value $0$ ever.

What is wrong with my understanding?

Answer

You're actually pretty close to the correct method for estimating the age of the Universe, but $H$ is not constant with time, it is $H=H(t)$.

One of the many ways of writing the equation to solve is:

$$t(z) = \frac{1}{H_0}\int_z^\infty \frac{dz}{(1+z)E(z)}$$

Here $z$ is redshift; $z\rightarrow\infty$ at the Big Bang, and $z=0$ now, so if you integrate from $0$ to $\infty$ you get $t(0)$ , the age of the Universe.

$E(z)$ is a function describing the relative content of the Universe at different redshifts, one example of $E(z)$ is the one for the $\rm\Lambda CDM$ model:

$$E(z) = \sqrt{\Omega_{\Lambda,0}+(1-\Omega_0)(1+z)^2+\Omega_{m,0}(1+z)^3+\Omega_{r,0}(1+z)^4}$$

The $\Omega$ are density parameters, which you can read more about here or here. The $0$ subscripts indicate that these should be the values measured at $z=0$, so these are constants and you can write the age of the Universe in terms of them (or you could, if that integral had an analytic solution - I don't suggest trying to solve it by hand, unless you make some simplifying assumptions or approximations first!).

The reason that $t(0)\approx1/H_0$ works reasonably well is that the Universe has been expanding at very nearly the rate it is expanding at now for the majority of the time since the Big Bang. So approximating $v$ as constant at the present value actually gets you pretty close to the right answer.

logical deduction - You have one question to tell whether the number I'm thinking of is 1, 2, or 3

This is an interesting puzzle which was passed to me by a friend some time ago. I do know the answer, but will refrain from self-answering on this to see where it goes.

I'm thinking of a number: 1, 2, or 3. You may ask me one question, which I will answer to the best of my ability. I may not, however, tell you my number or any codified version of my number. I can only answer yes, no, maybe, I don't know, etc.

How do you tell the number I'm thinking of?

Answer

Would it work if I ask:

"I'm thinking of a number: either 0 or 1. Is the sum of our numbers greater than 2?" I could figure out the answer if that's a question I'm allowed to ask.

If your number is 3, then you would say "yes" because no matter what you add, it would be true.

If your number is 2, you would say "I don't know". If my number was 0, then the answer would be no, but if my number is 1 the answer is yes. So you don't know.

If your number is 1, then you would respond "no". No matter what number you add, the number will never be greater than 2.

quantum field theory - Is ghost-number a physical reality/observable?

One perspective is to say that one introduced the ghost fields into the Lagrangian to be able to write the gauge transformation determinant as a path-integral. Hence I was tempted to think of them as just some auxiliary variables introduced into the theory to make things manageable.

But then one observes that having introduced them there is now an extra global $U(1)$ symmetry - the "ghost number"

• 
  

  Hence hasn't one now basically added a new factor of $U(1)$ to the symmetry group of the theory? How can the symmetry of the theory depend on introduction of some auxiliary fields?

  
  


• 
  

  Now if one takes the point of view that the global symmetry has been enhanced then the particles should also lie in the irreducible representations of this new factor. Hence ghost number should be like a new quantum number for the particles and which has to be conserved!

  
  
  


• 
  

  But one sees that ghost field excitations are BRST exact and hence unphysical since they are $0$ in the BRST cohomology.

  
  

I am unable to conceptually reconcile the above three ideas - the first two seem to tell me that the ghost-number is a very physical thing but the last one seems to tell me that it is unphysical.

• 
  

  At the risk of sounding more naive - if the particles are now charged under the ghost number symmetry then shouldn't one be able to measure that in the laboratory?

  
  



• 
  

  Lastly this ghost number symmetry is a global/rigid $U(1)$ symmetry - can't there be a case where it is local and needs to be gauged?

  
  

Answer

The mystery here should disappear once one realizes that the BRST complex -- being a dg-algebra -- is the formal dual to a space , namely to the "homotopically reduced" phase space.

For ordinary algebras this is more familiar: the algebra of functions $\mathcal{O}(X)$ on some space $X$ is the "formal dual" to $X$, in that maps $f : X \to Y$ correspond to morphisms of algebras the other way around $f^* : \mathcal{O}(Y) \to \mathcal{O}(X)$.

Now, if $X$ is some phase space, then an observable is simply a map $A : X \to \mathbb{A}$. Dually this is a morphism of algebras $A^* : \mathcal{O}(\mathbb{A}) \to \mathcal{O}(X)$. Since $\mathcal{O}(\mathbb{A})$ is the algebra free on one generator, one finds again that an observable is just an element of $\mathcal{O}(X)$.

(All this is true in smooth geometry with the symbols interpreted suitably.)

The only difference is now that the BRST complex is not just an algebra, but a dg-algebra. It is therefore the formal dual to a space in "higher geometry" (specifically: in dg-geometry). Concretely, the BRST complex is the algebra of functions on the Lie algebroid which is the infinitesimal approximation to the Lie groupoid whose objects are field configurations, and whose morphisms are gauge transformations. This Lie groupoid is a "weak" quotient of fields by symmetries, hence is model for the reduced phase space.

So this means that an observable on the space formally dual to a BRST complex $V^\bullet$ is a dg-algebra homomorphism $A^* : \mathcal{O}(\mathbb{A}) \to V^\bullet$. Here on the left we have now the dg-algebra which as an algebra is free on a single generator which is a) in degree 0 and b) whose differential is 0. Therefore such dg-morphisms $A^*$ precisely pick an element of the BRST complex which is a) in degree 0 and b) which is BRST closed.

This way one recovers the definition of observables as BRST-closed elements in degree 0. In other words, the elements of higher ghost degree are not observables.

electromagnetic radiation - How can the Huygens-Fresnel principle be derived from the Maxwell equations?

The Huygens-Fresnel principle states that every point to which a luminous disturbance reaches becomes a source of a spherical wave. I have been trying to understand this considering a infinite screen with a microscopic hole $dS$ on which a plane wave is incident, but I see no obvious way to describe the resulting propagating wave.

How can it be derived from the Maxwell equations?

forces - Confusion with $F=-nabla V$, $F$ conservative

I am rather confused by the relationship $F=-\nabla V$.

If a pen drops from a height it loses potential energy so $\nabla V$ is negative. From the above equation this means that the gravitational force acting on the pen is positive, but isn't it acting in the negative direction (downwards)?

quantum mechanics - What is the physical meaning of the sum of two non-commuting observables?

Scenario: ${\mathcal A}$ and ${\mathcal B}$ are two observables. Mathematically we model them by two Hermitian operators $A\colon H \to H$ and $B\colon H \to H$ on a separable Hilbert space. Physically they correspond to experiments $E_A$ and $E_B$, whose results are values in $Spec(A)$ and $Spec(B)$; repetitions produce value distributions on these spectra, expectation values, variances and higher momenta. The mathematical operator $A+B$ also is Hermitian. So let us look for an experiment which corresponds to this operator and let us study its expectation value in state $\varphi$.

Naive approach: Let us try pair-experiments. Assume we have a black box producing samples of state $\varphi$. Take a sample of the state, do experiment $E_A$ and get result $a$. Sample the state again, do experiment $E_B$ and get result $b$. Call the sum $a+b$ the result of the pair-experiment.

If $Spec(A) = \{ a_1, a_2 \}$ and $Spec(B) = \{b_1, b_2\}$ then the pair experiment has spectrum $\{ a_1 + b_1, a_1 + b_2, a_2 + b_1, a_2 + b_2 \}$. Obviously the pair-experiment has to be described in $H \otimes H$ and with a completely different observable. Details are straight forward, but we have no experiment for $A + B$. :-(

Second attempt: Let us assume that ${\mathcal A}$ and ${\mathcal B}$ are compatible and $A$ and $B$ commute. Then we can do the following: Sample the state once, on that sample do experiments $E_A$ and $E_B$ in whatever sequence, receive sequence independent values $a$ and $b$ and add them. Mathematically all is good. $A$ and $B$ share an eigenbasis, the spectrum of $A + B$ is the sum of the eigenvalues (belonging to the same shared eigenspace). Expectation values work out as expected. :-)

Now my question: $A + B$ still is a Hermitian operator, even if $A$ and $B$ do not commute. So I still am curious to which experiment this operator belongs to.

Note: In case of the product $A \cdot B$, the operator $A\cdot B$ is no longer Hermitian if the operators do not commute, and this makes it impossible for me to ask that question for the product. My question would break the preconditions of the formalism. But in $A + B$ the formalism allows to pose this question...

Update: In consequence of some comments I will try to specify my question more clearly: What is the physical meaning of the sum of two observables?

Obviously the "sum of two observables" is not the "sum of the values of the two observables". Assume that observable $A$ may have the values $2$ or $3$ and assume that observable $B$ may have the values $100$ or $200$ then the observable $A+B$ does not have the values $102$, $103$, $202$ or $203$ as a simple, naive approach might suggest or as an understanding of "sum of the values of the two observables" might suggest.

With this intuition failing, I would like to get an understanding of the physical meaning of $A + B$ starting from an understanding of $A$ and $B$.

Update 2: Adjusted the description of the pair experiment to a less misleading form.

thermodynamics - How to find this probability with the microcanonical ensemble?

When dealing with isolated systems we are dealing with the microcanonical ensemble. In that case, we suppose that each individual microstate has the same probability.

So if $\Omega(E)$ is the number of states of energy $E$, since each has the same probability, we must have

$$P(E) = p \Omega(E),$$

where $p$ is the individual probability. Then we have

$$\frac{1}{p} = \sum_{E}\Omega(E)$$

the total number of microstates.

Now, suppose the system is composed of two parts $A$ and $B$ separated by a totally isolating wall.

In that case, consider the process on which the wall becomes diathermal. Thus energy exchange is allowed, until equilibrium is reached.

I'd like to understand how could one find the probability $P(E)dE$ that the energy of $A$ lies between $E$ and $E+dE$ after the process.

I mean, $A$ and $B$ are not isolated. In that case, the microcanonical ensemble techniques don't work for them, just for the composite system. But the energy of the composite system doesn't change.

So what should be done in a situation like this? How this probability can be found?

particle physics - Is "quantum superposition" just a fancy way of saying that a system is in one state or another with some probabilities?

When we say that an electron is in a quantum state that is a linear combination of two eigenstates, one with the probability of 75% and the other 25%, what is actually happening?

1. Is the electron "really" somehow magically in these two eigenstates at once?

OR

2. The electron is in only one of these eigenstates at a time point, we just can't tell which one it is without observing it. And when we observe the electron many times, we find it in one eigenstate in 75% of the time, and in the other eigenstate in 25% of the time, because the electron switches between these two states and just spends three times more time in one eigenstate than the other?

Obviously, I think the second case is correct.

But which case is correct? The 1st or the 2nd?

My reactions to your possible answers:

---

If the 1st case is correct: That's insane, thank you!

---

If the 2nd case is correct: Then, isn't the principle of superposition just an "assumption" that should not be taken literally? That is, the electron is not at two eigenstates at once, but one of them at a certain time point, and we just have no way of finding out which state, before observing it? So we just say that it is in an eigenstate X with a probability P(X), such that the sum of probabilities for all eigenstates equal to 1?

I could also, for example, say: I don't know where my friend Max currently is. He could be at school with a probability of 75% and at a bar with a probability of 25%. We all know that Max is not at school and the bar simultaneously, and we can just call him to find out where he is. But before calling him, for the sake of being able to conduct our calculations, we can "assume" that he is at school and the bar simultaneously, with the assigned probabilities, but we actually know that we just made this assumption to be able to carry on with our calculations about Max.

Isn't it just a fancy way of saying that something is in state X with P(X) and state Y with P(Y), and it is in only but only one of these states at all times, however, we just can't find out which state before observing that something, but can merely say with which probability we're likely to find it at a certain state?

Then why is everyone so surprised about the superposition principle?

lateral thinking - Older twin's birthday is one day after younger's

I have been asked this puzzle in an interview. There was not a lot of time left so I couldn't solve it quickly and the interviewer kind of let it go as he had to wrap up the interview

Foo and Bar are twin brothers. Foo is older than Bar by 5 minutes. Foo celebrates his birthday every year on March 16th and Bar celebrates his birthday on March 15th. Explain the scenario.

rotational dynamics - Rotation matrix of Euler's equations of rotation relative to inertial reference frame

I was playing with simulation of Euler's equations of rotation in MATLAB,

$$ I_1\dot{\omega}_1 + (I_3 - I_2)\omega_2\omega_3 = M_1, $$

$$ I_2\dot{\omega}_2 + (I_1 - I_3)\omega_3\omega_1 = M_2, $$

$$ I_3\dot{\omega}_3 + (I_2 - I_1)\omega_1\omega_2 = M_3, $$

where $I_1$, $I_2$ and $I_3$ are the principal moments of inertia of the rigid body, $\omega_1$, $\omega_2$ and $\omega_3$ are the angular velocities around the axes of these moments of inertia, $\dot{\omega}_i$ denotes the time derivative of the angular velocity $\omega_i$ and $M_i$ denotes the external torque applied along the axis of $\omega_i$.

I would also like to find the rotation of the body, but the equations above have a rotating reference frame, so finding the rotation does not seem to have a simple answer. In order to express the rotation of the body I would think that a rotation matrix as a function of time, $R(t)$, would be a good option, such that a point $\vec{x}_0$ on the rigid body can be relocated in a inertial frame of reference with,

$$ \vec{x}(t) = R(t) \vec{x}. $$

This rotation matrix should change over time as the body rotates, but any two rotations can be combined into one effective rotation by multiplying the two rotation matrices. Thus the rotation matrix after a discrete time step should be,

$$ R_{n+1} = D R_n, $$

where $D$ is the rotation during that time step.

Any rotation matrix (I believe with the exception of reflections) can be written as,

$$ R = \begin{bmatrix} \cos\theta\!+\!n_x^2(1\!-\!\cos\theta) & n_xn_y(1\!-\!\cos\theta)\!-\!n_z\sin\theta & n_xn_z(1\!-\!\cos\theta)\!+\!n_y\sin\theta \\ n_xn_y(1\!-\!\cos\theta)\!+\!n_z\sin\theta & \cos\theta\!+\!n_y^2(1\!-\!\cos\theta) & n_yn_z(1\!-\!\cos\theta)\!-\!n_x\sin\theta \\ n_xn_z(1\!-\!\cos\theta)\!-\!n_y\sin\theta & n_yn_z(1\!-\!\cos\theta)\!+\!n_x\sin\theta & \cos\theta\!+\!n_z^2(1\!-\!\cos\theta) \end{bmatrix}, $$

where $\vec{n}=\begin{bmatrix}n_x & n_y & n_z\end{bmatrix}^T$ is the axis of rotation and $\theta$ the angle of rotation.

For an infinitesimal time step the rotation, $D$, has the axis of rotation equal to the normalized angular velocity vector and the angle of rotation equal to the magnitude of the angular velocity vector times the times step. Do I need to use the angular velocity vector in the rotating or inertial reference frame for this?

---

I was not completely sure if I understood the notation of the answer of David Hammen, so I performed a simple test. The test involves applying two rotations. Initially the two reference frames are lined up, where x, y and z axes in the figures for the inertial reference frame always point to the left, right and up respectively, while for the rotating reference frame the x, y and z axes are represented by $\vec{e}_1$, $\vec{e}_2$ and $\vec{e}_3$ respectively. The first rotation is 90° around the x axis:

                       

Because in both reference frames the rotation is around the x axis, thus the rotation matrices of this rotation are,

$$ R_{I:0\to 1} = R_{R:0\to 1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & -1\\ 0 & 1 & 0 \end{bmatrix}, $$

where the subscripts $I$ and $R$ stand for the inertial and rotational reference frames respectively.

The next rotation is 90° around the z axis or $\vec{e}_2$:

                       

The two reference frames now differ, thus the rotation matrices of this rotation are,

$$ \begin{array}{rrrrrr} R_{I:1\to 2} = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} & & & & & R_{R:1\to 2} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \end{array}. $$

The final orientation should look like this:

                       

The corresponding total rotation matrix is,

$$ R_{0\to 2} = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}. $$

When combining the two rotations in both reference frames, then the total rotation matrix can be obtained with the following order of multiplications,

$$ R_{0\to 2} = R_{R:0\to 1} R_{R:1\to 2} = R_{I:1\to 2} R_{I:0\to 1}, $$

thus it appears that the following is true,

$$ R_{n+1} = R_n D_R = D_I R_n. $$

---

Since Euler's equations use the rotational reference frame, thus $D_R$ will be used. The rotation, $D_R$, using the general equation for a rotation matrix, can be simplified, by using linear approximation since $dt$ is assumed to be small, to,

$$ R(t+dt) = R(t) \begin{bmatrix} 1 & -\omega_3dt & \omega_2dt \\ \omega_3dt & 1 & -\omega_1dt \\ -\omega_2dt & \omega_1dt & 1 \end{bmatrix} = R(t) \left( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0 \end{bmatrix}dt \right) $$

Or in terms of the time derivative,

$$ \dot{R}(t) = \frac{R(t+dt)-R(t)}{dt} = R(t) \begin{bmatrix} 0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0 \end{bmatrix}. $$

Question: When numerically integrating this, together with Euler's equation of rotation, is there a way to ensure that the determinant of $R$ remains equal to one (otherwise $\vec{x}(t)$ will also be scaled)?

newtonian gravity - Why is heavier object more reluctant to get falling down?

Is it because of the upward force that stops the object? for example-

newtonian mechanics - Intuitively why time period of shm do not depend on displacement from mean position, ie. amplitude

Why simple harmonic motion time period equation is independent from displacement from mean position?

waves - Why can we distinguish different pitches in a chord but not different hues of light?

In music, when two or more pitches are played together at the same time, they form a chord. If each pitch has a corresponding wave frequency (a pure, or fundamental, tone), the pitches played together make a superposition waveform, which is obtained by simple addition. This wave is no longer a pure sinusoidal wave.

For example, when you play a low note and a high note on a piano, the resulting sound has a wave that is the mathematical sum of the waves of each note. The same is true for light: when you shine a 500nm wavelength (green light) and a 700nm wavelength (red light) at the same spot on a white surface, the reflection will be a superposition waveform that is the sum of green and red.

My question is about our perception of these combinations. When we hear a chord on a piano, we’re able to discern the pitches that comprise that chord. We’re able to “pick out” that there are two (or three, etc) notes in the chord, and some of us who are musically inclined are even able to sing back each note, and even name it. It could be said that we’re able to decompose a Fourier Series of sound.

But it seems we cannot do this with light. When you shine green and red light together, the reflection appears to be yellow, a “pure hue” of 600nm, rather than an overlay of red and green. We can’t “pick out” the individual colors that were combined. Why is this?

Why can’t we see two hues of light in the same way we’re able to hear two pitches of sound? Is this a characteristic of human psychology? Animal physiology? Or is this due to a fundamental characteristic of electromagnetism?

Answer

This is because of the physiological differences in the functioning of the cochlea (for hearing) and the retina (for color perception).

The cochlea separates out a single channel of complex audio signals into their component frequencies and produces an output signal that represents that decomposition.

The retina instead exhibits what is called metamerism, in which only three sensor types (for R/G/B) are used to encode an output signal that represents the entire spectrum of possible colors as variable combinations of those RGB levels.

homework and exercises - Find the speed of two balls joined by a string and the impulse in the string when the string goes taut

The question I'm trying to answer:

"Two balls, 4kg and 6kg, are joined by a light inextensible string which is initially slack. The lighter ball is projected away from the heavier mass at $12ms^{-1}$. Find the final speed of both balls and the impulse in the string when it becomes taut."

From the description, I assumed that the ball which is not being projected remains in rest until the string becomes taut. I'm not sure if this assumption is right or not (I don't have solutions) but I think it is by simply imagining the situation. Now, to find the velocity of the balls when the string becomes taut we can use the conversation of momentum:

$$4(0) + 6(12)=(4+6)v$$ $$v=\frac{72}{10}=7.2ms^{-1}$$

So the speed of both balls is $7.2ms^{-1}$.

Now, finding the impulse:

$$I=(m_1 + m_2)v - mu=(4+6)7.2 - 6(12)=0kgms^{-1}$$.

Now, I know the impulse can't be zero so I'm wondering where my logic has failed me.

Answer

The impulse equals the change in momentum of one of the balls. You have calculated the change in total momentum, ie final momentum of system minus initial momentum of system. This will always be zero if momentum is conserved.

However, even with the correct calculation of impulse your solution is wrong. You assume that the balls have the same speed after the string becomes taut. This is equivalent to a completely inelastic collision. Real strings do not behave like this. They behave like springs which are extremely stiff. Springs conserve energy; within their elastic limit strings are observed to do the same. The "collision" between the balls, mediated by the taut string, is elastic.

So in addition to conservation of momentum you also need to apply conservation of kinetic energy. The simplest way of doing the latter is to make the relative speed of separation equal the relative speed of approach.

See my answer to Force Transfer Between two bodies linked by a rope.

geometry - Tiling rectangles with T pentomino plus rectangles

Inspired by Polyomino Z pentomino and rectangle packing into rectangle

Also in this series: Tiling rectangles with F pentomino plus rectangles

Tiling rectangles with N pentomino plus rectangles

Tiling rectangles with U pentomino plus rectangles

Tiling rectangles with V pentomino plus rectangles

Tiling rectangles with W pentomino plus rectangles

Tiling rectangles with X pentomino plus rectangles

The goal is to tile rectangles as small as possible with the T pentomino. Of course this is impossible, so we allow the addition of copies of a rectangle. For each rectangle $a\times b$, find the smallest area larger rectangle that copies of $a\times b$ plus at least one T-pentomino will tile. Examples shown, with the $1\times 1$ or the $1\times 2$, you can tile a $3\times 3$ as follows:

Now we don't need to consider $1\times 1$ or $1\times 2$ any longer as we have found the smallest rectangle tilable with copies of T plus copies of $1\times 1$ and $1\times 2$.

There are at least 10 more solutions. I tagged it 'computer-puzzle' but you can certainly work some of these out by hand. The larger ones might be a bit challenging.

Answer

I assume these are the remaining ones you found as well; for 2x3

a simple 7x5 = 35 one:

for 2x2

a 10x8 = 80 one

for 1x5:

11x10 = 110

for 2x4:

14x22 = 308

for 2x5:

12x15 = 180

for 3x4:

18x19 = 342

and finally a large one for 3x5

which uses the same central figure formed by the Ts as the 1x5.
28x40 = 1120

quantum field theory - Renormalization condition: why must be the residue of the propagator be 1

In on-shell (OS) scheme, one of the renormalization conditions is that the propagator, say, a scalar theory

$$\frac{1}{p^2+m^2-\Sigma(p^2)-i\epsilon}$$

must have a unit residue at the pole of physical mass $p^2=-m^2$. Some textbooks say this is to make sure the propagator behaves like a free field propagator near the pole. But why?

Answer

The OS condition that $$ \frac{\partial\Sigma}{\partial p^2}|_{p^2=-m^2} = 0 $$ implies that the residue in the propagator remains equal to one.

Suppose that we used a different renormalization scheme in which our counter-terms contain no finite parts (e.g. MS scheme). In the OS scheme, we removed finite parts which were logarithmic in our artificial regularization scale $\mu$. In our new choice, the propagator might have a residue, say $R$.

This residue manifests itself in an irritating way; the field will be re-normalized such that $\phi = \frac{1}{\sqrt{R}} \phi_B$. In the LSZ formula, however, external lines contribute factors $R$ (from the KG equation cancelling the propagators). So external scalar lines contribute a factor $\sqrt{R}$ in the MS scheme.

So, whilst this choice in the OS scheme is somewhat arbitrary, it's convenient, because external scalar lines contribute a factor 1.

I'm trying to learn these points myself, so hopefully someone can expand/correct my answer where necessary...

How to make a monopole magnet?

I want to create a monopole magnet. Is this practically / theoretically possible?

fluid dynamics - What really allows airplanes to fly?

What aerodynamic effects actually contribute to producing the lift on an airplane?

I know there's a common belief that lift comes from the Bernoulli effect, where air moving over the wings is at reduced pressure because it's forced to travel further than air flowing under the wings. But I also know that this is wrong, or at best a minor contribution to the actual lift. The thing is, none of the many sources I've seen that discredit the Bernoulli effect explain what's actually going on, so I'm left wondering. Why do airplanes actually fly? Is this something that can be explained or summarized at a level appropriate for someone who isn't trained in fluid dynamics?

(Links to further reading for more detail would also be much appreciated)

Answer

A short summary of the paper mentioned in another answer and another good site.

Basically planes fly because they push enough air downwards and receive an upwards lift thanks to Newton's third law.

They do so in a variety of manners, but the most significant contributions are:

• The angle of attack of the wings, which uses drag to push the air down. This is typical during take off (think of airplanes going upwards with the nose up) and landing (flaps). This is also how planes fly upside down.


• The asymmetrical shape of the wings that directs the air passing over them downwards instead of straight behind. This allows planes to fly level to the ground without having a permanent angle on the wings.

Explanations showing a wing profile without an angle of attack are incorrect. Airplane wings are attached at an angle so they push the air down, and the airfoil shape lets them do so efficiently and in a stable configuration.

This incidence means that even when the airplane is at zero degrees, the wing is still at the 5 or 10 degree angle.

-- What is the most common degree for the angle of attack in 747's, 757's, and 767's

Any object with an angle of attack in a moving fluid, such as a flat plate, a building, or the deck of a bridge, will generate an aerodynamic force (called lift) perpendicular to the flow. Airfoils are more efficient lifting shapes, able to generate more lift (up to a point), and to generate lift with less drag.

--Airfoil

optics - Fourier transform of two pulses of light

I have laser beam path that fires two pulses of light in a gaussian distribution, so the intensity graph over time is two identical gaussians separated by a distance $t_0$. In other words, a gaussian convolved with two $\delta$ functions equidistant from the origin. When I take the Fourier transform of this, I get a graph with beats, but I'm having trouble figuring out what this graph means physically. When we feed this beam through a spectrometer, we are trying to find out if we should see fringes, and if so, what their frequency is. Any help would be appreciated.

Answer

This is a great question because you have to think carefully about what you experiment is actually doing to answer it.

As in Andreas H's answer, you may in principle (and in some practice) see the intensity comb (the "beats" as you call them) at the output of grating spectrometer. But in practice it won't always be like this and you have to analyse your kit a bit more deeply to find out whether you will see the beats or a smooth Gaussian on your image plane.

First and foremost, the laser's coherence time must be greater than the time interval spanned by the two pulses. Otherwise the two pulses' spectrums may tend to add incoherently and you may simply see the spectrum for one pulse alone but simply double in intensity. The laser's and modulation system's physics will have something to say here. If the pulses are made by pulsing the laser's supply current, or by Q-switching, then you will just get to pulses comprising mutually incoherent photons, so you do see simply a continuous Gaussian as the spectrum. If the pulses arise from mode locking, or by passive modulation of a continuous wave, long coherence time laser, then the comb spectrum will in principle be seen at the output.

Now let's think about the grating. Firstly, we need to assume our imaging system for the grating has to be of low enough aberration that the grating spectrometer's output truly is the spatial Fourier transform (Fraunhofer diffraction) of the input aperture (i.e. the grating lit by the time-pulsed field). If there is aberration, or if the output is not truly a far field image, then the output will be some more-compicated-than-simply-the-Fourier-transform diffraction integral (a Fresnel diffraction integral, for instance) of the input aperture, which may mask the comb, so that you may tend to see more like a smooth Gaussian spectrum.

Moreover, the grating's spectrometer abilities arise from the spatial frequency upconversion imposed on a spatial field distribution by multiplying this distribution by an infinitely wide comb function. This spatial frequency upconversion depends on the light's wavelength, so the grating diffracts different wavelength components at different Bragg angles and thus splits spectrum. In practice, the grating is of finite width, i.e. multiplied by some compactly supported gating function. In the Fraunhofer diffraction, this modification by the gating function corresponds to a convolution with the gating function's Fourier transform. So you theoretical Fourier transform of the form $\exp\left(-\frac{x^2}{2\,\sigma^2}\right) \cos(\alpha\,x\,\Delta t)$ get convolved with something like a $\mathrm{sinc}$ function and thus the combs can get smeared into one another, particularly if the time separation between the input pulses is very big. Here $x$ is the sideways distance from the output spectrum's centre, $\alpha$ is set by the system's dimensions and delays and $\Delta t$ is the delay between the two input pulses. The output intensity can thus look more like $\exp\left(-\frac{x^2}{2\,\sigma^2}\right)$.

Lastly, the combs can be very very finely spaced, so, depending on the spatial resolution of your CCD array or whatever sensor you use in the spectrometer, the imaging sensor might impose a spread-out kernel that smears over the combs. In this case, you will also only tend to see a continuous Gaussian spectrum.

electromagnetism - Curl of an electromagnetic wave

I can't understand the concept of the curl of an electromagnetic wave.

$$ \nabla \times E = -\frac{\partial \textbf{B}}{\partial t} $$

All of the examples I find show a current through a conductor, or that paddle wheel in water which I fail to see the distinction of that with an E-M wave. What I am trying to understand is the curl of say a laser beam light.

So lets say I have this sin wave which I plot in MATLAB which represents a section of a laser beam propagating through free space (We shall say it is the net product of $E_x$ and $E_y$):

How do you take the curl of it? Where do you take the curl of it. eg the whole beam, a certain section? Can I do this in MATLAB to see visualization of the concept of curl?

I mainly seek descriptive or pictorial answers.

Answer

Let us deal with this on two parts: (1) How do you take the curl?

The E-field you describe has the form ${\bf E} = E_0 \sin (kz-\omega t)\, {\bf i}$, where the latter unit vector assumes the wave is polarised along the x-axis, while the wave travels along the z-axis. (You can equally polarise it along the y-axis if you choose). If I define it in this way, then the E-field only has an x-component. i.e. $E_{y}=E_{z}=0$

The curl of the wave can be evaluated as described in the answer by JamalS, so in this case, as $E_{y}=E_{z}=0$, then the partial derivatives of these components are also zero and there are only two possible non-zero terms in the curl.

$$\nabla \times {\bf E} = {\bf j} \frac{\partial E_{x}}{\partial z} - {\bf k} \frac{\partial E_{x}}{\partial y}\, .$$ Because we chose to have the wave travelling along the z-axis then $E_{x}$ is not a function of $y$, so the second term is zero and $$\nabla \times {\bf E} = {\bf j} \frac{\partial E_{x}}{\partial z} = kE_{0} \cos (kz -\omega t)\, {\bf j}\, ,$$ perpendicular to the E-field and the direction of motion of the wave, but changing direction with time.

OK, that's the Maths, but (2) how to "visualise" or deduce without doing the Maths? I have to admit to struggling with this one. The paddle wheel analogy is always the one I use. A field with a non-zero curl will make the paddle wheel turn and the axis of rotation points in the direction of the curl.

If I assume that the MATLAB plot you show has z along the horizontal axis, then you can imagine the E-field as vertical arrows of size proportional to the E-field strength at that instant in time. Imagine placing a paddle wheel at some point along the axis. In general, the E-field on one side of the wheel will be of a different strength to the E-field on the other side of the wheel and hence it will rotate around an axis perpendicular to z. If we then roll the clock forward to some later instant of time, the situation could reverse with the E-field now stronger on the other side and the wheel rotates in the opposite direction - i.e. a cosinusoidally varying curl in a direction perpendicular to the wave motion and perpendicular to the E-field.

Here's an attempt to show this. The two plots show the instantaneous E-field strengths and directions at two instants of time. I add a paddle wheel which sits in the field at the horizontal positions shown. In the top plot the wheele rotates clockwise. Some time later the field has changed as shown in the bottom plot and the wheel would rotate anti-clockwise.

electromagnetism - What is the purpose of the Maxwell Stress Tensor?

In the calculation of the forces acting on a charge/current distribution, one arrives at the Maxwell stress tensor:

$$\sigma_{ij}=\epsilon_0 E_iE_j + \frac{1}{\mu_0} B_iB_j -\frac{1}{2}\left(\epsilon_0E^2+\frac{1}{\mu_0}B^2\right)$$

In the case of electrostatics, this element of the stress tensor denotes the electromagnetic pressure acting in the $i$ direction with respect to a differential area element with its normal pointing in the $j$ direction. Equivalently, we can replace "electromagnetic pressure" with "electromagnetic momentum flux density" in order to "make sense". With this mathematical construction, assuming a static configuration, the total force acting on a bounded charge distribution $E$ is given by

$$(\mathbf{F})_i=\oint_{\partial E} \sum_{j}\sigma_{ij} da_j $$

Where $da_j$ is the area element pointing in the $j$ direction (e.g. $da_{3}=da_z=dxdy$).

What I would like to know is, what is the advantage of introducing such an object? I have yet to see a problem where this has any real utility. Sure, we can now relate the net force on a charge distribution to the E&M fields on the surface, but are there any problems where that is really better than just straight up calculating it? In an experiment, does one ever really measure the E&M fields on the boundary of an apparatus to calculate the net force?

Answer

but are there any problems where that is really better than just straight up calculating it?

In the first place, it allows us to formulate general theorem of conservation of momentum in macroscopic electromagnetic theory. Suppose some amount matter is enclosed inside an imaginary surface $\Sigma$ in vacuum, no matter is present on the boundary itself but field may be non-zero anywhere. In a simplified terms, the theorem is

rate of change of total momentum in a region of space bounded by imaginary surface $\Sigma$ in vacuum = surface integral of Maxwell tensor $\sigma$ over $\Sigma$

or formally

$$ \frac{d}{dt}\bigg(\mathbf P_{matter} + \mathbf P_{field} \bigg) = \oint_{\Sigma}d\Sigma_{i} \sigma_{ij}. $$

but are there any problems where that is really better than just straight up calculating it?

The Maxwell tensor is also utilizable in calculation of total EM force on a solid object, for example the force on electrically polarized body in external electric field or force on a magnet near metal body or another magnet. Such calculations can be done with pencil and paper for highly symmetrical systems like dielectric cylinder in charged parallel-plate capacitor or two magnet cylinders facing each other with their bases.

newtonian mechanics - Why does the phase velocity of a string attached to springs depend on the wave length?

A string can be described by the wave equation $$ (\partial_t^2 - \partial_x^2) \, \varphi = 0 $$ while a string attached to a spring (i.e. with a harmonic restoring force) at each location is described by the wave equation with dispersion term: $$ (\partial_t^2 - \partial_x^2 + m^2) \, \varphi = 0 $$

An important difference between the two systems is that without a dispersion term ($m=0$) all plane wave solutions have exactly the same phase velocity. In contrast, if $m\neq 0$ the phase velocity depends on the wave length.

While it is straightforward to derive this mathematically, I'm struggling to understand why this is the case. Is there some intuitive way to understand why for the string with restoring force the phase velocity ($v = \omega /k)$ depends on the wave length?

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Background:

If we consider an ordinary string and focus on one specific point on it, say one maximum, it will travel during the time interval $\Delta t$ the distance $\Delta x$. This fact is independent of the wave length of the wave in question. Thus the phase velocity is always $v = \frac{\Delta x}{\Delta t}$.

In contrast, for a string attached to springs, this is no longer the case. During a fixed time interval $\Delta t$, the maximum of a short wave travels a different distance $\Delta x'$ than then maximum of a long wave. This means that the phase velocity of short and long waves is different: $$v' = \frac{\Delta x'}{\Delta t} \neq = \frac{\Delta x''}{\Delta t}= v''$$

So the answer must have something to do with the fact that for a long wave there are more springs involved per wave form. However, the whole issue is still far from obvious for me. In particular, I'm not even sure how to determine (without calculating anything) whether the phase velocity of a long wave is faster or slower than for a short wave.

Answer

The extraneous springs cause faster vertical oscillations on top of what is already caused by the non-dispersive medium (the rope), thereby propagating the waves faster and making the dispersion relation non-linear.

For large $\lambda$, the spring energy dominates the oscillation because the string is stretched in a rather smooth way, so less contribution from potential energy of the stretched string, and we have $\omega \sim \sqrt{\chi/\mu} \sim m$. In the case of high $k$, the string is quite distorted and the stretched string contributes much more to the oscillation than the extraneous springs, and one therefore retrieves the case of linear dispersion $\omega \sim k$ as in the case of string without springs.

The string in the case $m=0$ can be modeled as a 1D chain of masses (mass $\mu$ each), with horizontal spacing $a$, neighbouring masses connected to each other by massless strings under tension $T$. In order to simulate waves on a rope in the continuum limit, one then needs to take the limit $\mu \to 0$ and $a \to 0$ such that $\mu/a$ tends to the mass density of the rope, and speed of wave propagation, $c = \sqrt{\frac{T}{\mu/a}}$, is finite. In your case $c=1$, so we can choose $T = \mu/a$. For small, smooth vertical oscillations of the chain, this system is then described by the wave equation without the mass term. (eg. cf. Greiner, Classical Mechanics; or David Morin, Classical Mechanics)

Propagating wave-like solutions are then guaranteed by Bloch's theorem $-$ in the limit $a \to 0$, the size of the unit cell shrinks to zero and we've pure exponential (sine/cosine) waves as solutions.

$m=0$

In the case $m=0$, the equilibrium state of the system about which small oscillations are considered could be translated in the vertical direction as a whole with no energy cost and that would result in a new equilibrium state. The choice of a definite ground state then implies that we start with a symmetry broken ground state, thereby inevitably running into masless Goldstone bosons for $k\to 0$. With this one can imagine a dispersion of the form, $\omega_{k} = A k+B k^2+\dots \ $ about $k=0$, which will satisfy the wave equation only for linear dispersion, leading to dispersion-free waves.

$m\neq0$

In the other case with finite mass $m$, one needs to attach springs of stiffness $\chi = \mu m^2$ to each mass $\mu$, from say below, in order to produce the mass term in the wave equation for vertical oscillations. The ground state is now unique and we do not have Goldstone modes anymore. In fact, even in the limit $T\to 0$ when the wave effects almost cease to exist, the system still needs a finite energy cost $\omega^*\gtrsim m$ to be excited (as pointed out already by @Fabian). This means $\omega_{k}/k$ diverges as $k \to 0$ and thus the phase velocity clearly must be wavelength dependent.

The continuum Lagrangian can be diagonalized in momentum space yielding normal modes, i.e. $$ L = \sum_{k} \frac{1}{2} \bigg[\dot{\phi}_{-k}\dot{\phi}_{k}-(k^2+m^2)\phi_{-k}\phi_{k} \bigg],$$ the normal mode frequencies simply being $\omega_{k}^2 = k^2 + m^2$.

In QFT, the $m=0$ and $m\neq0$ case would correspond to the massless and massive Klein-Gordon fields respectively; the latter would require at least the rest energy $m$ to create a particle in the momentum state $k=0$, consistent with the dispersion relation.

Now, suppose conversely the dispersion relation was linear $-$ then the wave would be photon like and it would take vanishingly small energy energy to create photons at arbitrarily small $k$, which the extraneous springs will simply not allow in the model with springs and masses.

Finally, note that we can apply classical virial theorem on the discrete system of springs and masses in mode with wave vector $k$ as the potential energy function is a homogeneous function of degree $n=2$, and consider the limit $a\to0$, yielding \begin{align} \langle T \rangle_k = \frac{n}{2} \langle U \rangle_k = \langle U \rangle_k \\ \implies \omega_{k}^2 = k^2 + m^2. \end{align}

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Further Edit

Following your edit: The fact that the phase velocity is faster for a long wavelength wave has nothing to do with the fact that there are more springs per wavelength.

In fact, if you just compute the energy of the string+spring system per wavelength, you will get that the kinetic energy contribution per wavelength is $\frac{1}{4} \mu A^2 \omega_k^2$, the potential energy of the stretched string per wavelength is $\frac{1}{4} \mu A^2 k^2$ and that of the springs is $\frac{1}{4} \mu A^2 m^2$, where $A$ is the amplitude of the wave in mode $k$. So, per wavelength everything is really independent of the wavelength.

However, from the virial theorem we know that the average potential energy must be equal to the average kinetic energy, which produces the non-linear dispersion quoted above.

I still think the most interesting conceptual part is the emergence of Goldstone modes in a string (w/o springs) for which $\omega_k \to 0$ as $k\to 0$ $-$ which is itself a consequence of broken symmetry. In a crystal which has broken translational symmetry, these Goldstone modes correspond to acoustic modes that cost vanishingly little energy for $k\to 0$ (sound waves). In a superfluid, these correspond to Bogolons. In ferromagnets, these Goldstone modes are the magnons.

The specific form of the dispersion law requires a knowledge of the potential energy functional of the system. In the case with strings the potential energy contribution is $\Big(\frac{\partial y}{\partial x}\Big)^2$, which in momentum space looks like $k^2 \vert y_k\vert^{2}$ and yields $\omega_k^2 \propto k^2$. If the potential energy had another form, eg $\propto C \Big(\frac{\partial^2 y}{\partial x^2}\Big)^2$, it would look like $C k^4 \vert y_k\vert^2$ in momentum space and yield a dispersion $\omega_k^2 \sim k^4$ (this is the non-relativistic dispersion formula and will eventually spread wavepackets). However, in all these cases the common theme is the emergence of a Goldstone mode.

The case of the string with extraneous springs breaks precisely this theme by forcing a unique groundstate (corresponding to the unique state when all the springs are unstretched) and creating an excitation gap $m$. For long-wavelength modes ($k \gg m$) this means $\omega_k \sim m$ (i.e. one still needs to excite the spring), and for short wavelength ($k \gg m$), $\omega_k \sim k$ to excite mode $k$ (like in case of a free string). The interpolation formula $\omega_k = \sqrt{k^2+m^2}$ does exactly this.

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p.s. Thanks to Prof. Stone for the Pantograph Drag reference.