acoustics - Do we hear sounds differently on the highest mountains?

Some searching gives that above 6,000 meter altitude the air density is less than half of that at sea level. Speed of sound is about 15-20% slower and "acoustic impedance" seems to change too.

Do humans notice these differences in sound? Does the pitch of tones change noticeable?

Answer

There more sides to this scenario that you're considering. Firstly, if we are assuming that the temperature is the same at sea level and on the high mountains, then the speed of sound doesn't actually change, as a constant temperature will take care of the air pressure-density ratio. $$c = \sqrt{\kappa \frac{p}{\rho}} $$ Where $p$: static air pressure, $\rho$: air density and $\kappa$ the adiabatic index $c_p/c_v$.

Again the statement being: The static air pressure and the density of air are proportional at the same temperature, meaning the ratio $p/\rho$ is always constant, on a high mountain or even on sea level altitude.

So in the scenario you're describing, if $T$ is taken as constant, then the speed of sound doesn't change, but its intensity does, as the density of the air is much lower on top of mountains, rough approximation for the intensity would be $$I \propto p v \propto \omega² c \rho$$ Where $v$ is the speed of air molecules, $\omega$ sound frequency. So one thing is for sure, you will need to shout much louder on a mountain, for people further ahead to be able to hear you.

Furthermore, if $T$ is changing, then ($p/\rho=constant$) doesn't hold anymore, so $c_{air}$ changes. There are rough approximates relating the speed of sound to $T$. In a crude manner: $$c_{air} = 331.3 \frac{m}{s} \sqrt{1+\frac{\theta}{273.15}}$$ Where $\theta$ is the air temperature in °C. Such estimate gives e.g. $60 \frac{cm}{s}$ change of speed of sound for $1 °C$ change of temperature. Further scenarios:

At $-20°C$: $c_{air} \approx 319 \frac{m}{s}$

At $0°C$:$c_{air} \approx 331 \frac{m}{s}$

At $20°C$: $c_{air} \approx 343 \frac{m}{s}$

At $100°C$: $c_{air} \approx 387 \frac{m}{s}$

Next logical step would be to consider the change in wavelength of the sound, when $c_{air}$ changes. For this you have the general formula $c = \lambda f $

So $\lambda$ changes with $T$ as the speed of sound changes and in case of a flute e.g. the length of the vibrating air column doesn't change, so when $c_{air}$ changes due to $T$ fluctuations, then the sound frequency $f$ changes (or pitch of tone as you call it). But since we don't use flutes to speak, this doesn't concern us, so in order to conclude, in a mountain, the intensity I, speed of sound $c$ and the sound wavelength $\lambda$ change (the last two only hold for $T$ varying) but not the pitch.

homework and exercises - Finding net magnetic and electric force on charged particle

This is from my textbook, it is not an assigned problem, but I want to understand.

It says:

Consider the situation in the figure, in which there is a uniform electric field in the x direction and a uniform magnetic field in the y direction. For each example of a proton at rest or moving in the x, y, or z direction, what is the direction of the net electric and magnetic force on the proton at this instant?

I believe I need to use the equation

$$ F_{net}=(q\overrightarrow{E})+q(\overrightarrow{v} \times \overrightarrow{B}) $$

But I'm not sure exactly how. We've just started to learn about this, and I want to get a head start. Could anyone put me on the right track?

Answer

First of all, I think you might have written the equation for the net force incorrectly: $$\vec{F}_{net} = q\vec{E} + q(\vec{v} \times \vec{B})$$ The second term is $q(\vec{v} \times \vec{B})$ and not $q(\vec{E} \times \vec{B})$.

From the first term of the force equation ($q\vec{E}$), we can see that the electric field will try push the proton parallel to it (so the proton will be pushed a bit in the $\hat{x}$-direction).

Note that the second term of the equation ($q(\vec{v} \times \vec{B})$) is perpendicular to the velocity (direction of motion) of the proton. You might remember this from earlier: when a force acts perpendicular to the motion of a body, the force acts centripetally - that is, the body starts revolving in a circle. Therefore, the magnetic force is a centripetal force.

So we have two ways the proton can be pushed: the electric field pushes it in the $\hat{x}$-direction and the and magnetic field (when the proton is moving) tries to get the proton to revolve around the axis perpendicular to the direction of motion ($\hat{v}$) and the magnetic field ($\hat{B}$) (remember, the second term is a cross product).

The proton gets pushed in both ways at the same time (in the first example, it doesn't move at first, but then starts to move as the electric field pushes it, so the magnetic force appears too). It might be a bit hard to visualize, but I'll talk about the first example: the proton is both first pushed in the $\hat{x}$-direction (so its velocity is non-zero) and then undergoes the centripetal force (from the magnetic field) which causes it to start revolving around the x-axis. However, just before it dips under the z-axis, the proton stops moving (the electric field caused it to decelerate), causing the electric field to accelerate it again, restarting the process while causing the proton to move a net displacement along the z-axis (since it never rotated back to the origin).

_{(source: physics-animations.com)}

(Note that the directions in the animation are not the same as in the second example).

Can you start to visualize how it will move in the other examples?

quantum mechanics - Approximate expression for the ground state of hopping Hamiltonian

In second quantization, the Hamiltonian describing the hopping process between two neighboring sites is given ($N$ - number of particles and $M$ - number of sites) by:

$$\hat{\mathcal H} = J\sum\limits_{\langle i,j \rangle}\hat{a}^{\dagger}_{i}\hat{a}_{j}$$

It can be diagonalized using Fourier series

$$\hat{a}_{i} = \frac{1}{\sqrt{M}}\sum\limits_{\mathbf{k}} \hat{b}_{\mathbf{k}}e^{-i\mathbf{k}\cdot\mathbf{R}_i}$$

The ground state is given by

$$|\mathrm{GS}\rangle = \frac{1}{\sqrt{N!}}\left( \hat{b}^{\dagger}_{\mathbf{k} = \mathbf{0}}\right)^N | 0 \rangle \approx C\ e^{\sqrt{N}\hat{b}^{\dagger}_{\mathbf{k} = \mathbf{0}}}| 0 \rangle$$

How is the approximation justified ($C$ ensures normalization)?

I calculated a few quantities using both states. First of all both states can be represented using site creation operators $\hat{a}_i$ in the following way: $$|\rm GS_{1}\rangle = \frac{1}{\sqrt{N!}}\left[\frac{1}{\sqrt{M}}\sum\limits_{i=1}^{M}\hat{a}_{i}^{\dagger} \right]^{N} |0\rangle$$ $$|\rm GS_2 \rangle = \prod\limits_{i=1}^{M}e^{\sqrt{\frac{N}{M}}(\hat{a}^{\dagger}_i - \hat{a}_i)}|0\rangle$$ so the first state is $\rm SU(M)$ coherent state while the second one is the product of Glauber coherent states. A few quantities of interest:

1. 
   

   Energy - expectation value of the Hamiltonian: $$\langle \hat{\mathcal H} \rangle_{1} = 2JN$$ $$\langle \hat{\mathcal H} \rangle_{2} = 2JN$$

   
   
   


2. 
   

   Particle density and fluctuations: $$\langle \hat{n}_{i} \rangle_{1} = \frac{N}{M},\ \ \ \langle \Delta \hat{n}_{i}^{2} \rangle_{1} = \frac{N}{M}\left(1 - \frac{1}{M} \right)$$ $$\langle \hat{n}_{i} \rangle_{2} = \frac{N}{M},\ \ \ \langle \Delta \hat{n}_{i}^{2} \rangle_{2} = \frac{N}{M}$$

   
   


3. 
   

   Total number of particles and fluctuations: $$\langle \hat{N} \rangle_{1} = N,\ \ \ \langle \Delta \hat{N}^{2} \rangle_{1} = 0$$ $$\langle \hat{N} \rangle_{2} = N,\ \ \ \langle \Delta \hat{N}^{2} \rangle_{2} = N$$

   
   

As @NorberSchuch mentioned in his answer below particle density fluctuations are practically the same and coincide in the thermodynamic limit.

Answer

The basic idea is that the term with the highest weight in the exponential series is exactly the desired term. Further, all of the weight is in terms with closeby particle number, and small fluctuations in this number do not matter in many cases.

(Note: I will omit the $k=0$ subscript and just write $b^\dagger$ in the following for convenience.)

First, we have that
$$ e^{\sqrt{N}b^\dagger} |0\rangle = \sum \frac{\sqrt{N}^k}{k!} (b^\dagger)^k |0\rangle\ . $$ Let us now study which term in the series has the highest weight. To this end, we have to make sure that the corresponding states have the same normalization. This is, if we define a normalized state $$ |k\rangle = \frac{(b^\dagger)^k}{\sqrt{k!}}|0\rangle\ , $$ each of which appears in the series with a probability $$ |c_k|^2 = \frac{N^k}{k!}\ . $$ This is an (unnormalized) Poisson distribution, whose mean is exactly at $k=N$, as desired.

Note that this does in fact not mean that the two states are close in any distance measure (in fact, they aren't). However, it tells us that most of the weight in the sum is in states with $|k-N|\le c\sqrt{N}$, as the variance of the Poisson distribution is $N$. Since in many applications, we in fact care about the number of particles per site (which is an intensive quantity and well-defined as the system size goes to infinity), and any such fluctuation in $k$ will vanish as $N\rightarrow\infty$, both states will have the same particle density in the thermodynamic limit, and thus describe the same physics.

wordplay - Bridge words 2 - word pairs linked by their front and back

I made a puzzle earlier (Bridge words), and thought I'd make another (encouraged by a request from @VictorHenry).

Find a pair of "bridge words" to fill in the gaps of the sentences below. Bridge words are words where the last $n\geq3$ letters of the first word are the same as the first $n$ letters of the second word. The words must then both be more than $n$ letters long ("hotshots-hotshots" would be a valid bridge pair ($n=4$), but "here-here" would not be).

For example:
Q: Sometimes we torture people. Sometimes we don't. We ______ our ______.
A: We intersperse our persecution.

Okay, here goes:

1. Good climbers leave the _______ ________.


2. I caught _____ of the person _______ in the pool.


3. She lives far out on the ______, ______ from society.


4. I could see that the glamorous lady was ______ _______ that caught the light.



5. We've ______ all the evidence and concluded that alcoholism is _______.


6. The dentist used an odd ______ to save information about her patient's ______.


7. I was sick in the West Indies, but in the ______ I was the ______.


8. I take a coat and dress out of the wardrobe, ______ the _______, and return them.


9. Grimm's _________ was left ______ after the queen guessed his name.


10. The king said to the regent: "I'll be back tomorrow, so you must _____ _______."

Hint:

$n$ is equal to 5 or 6.

Answer

Hooray! These seem harder but I'm making progress. Going to go ahead and post this to keep track of which ones I've solved so far.

1. Good climbers leave the _______ ________.

mountain untainted

2. I caught _____ of the person _______ in the pool.


3. She lives far out on the ______, ______ from society.


4. I could see that the glamorous lady was ______ _______.


5. We've ______ all the evidence and concluded that alcoholism is _______.

gathered/hereditary

6. The dentist used an odd ______ to save information about her patient's ______.

filename/enamel

7. I was sick in the West Indies, but in the ______ I was the ______.

Antilles/illest

8. I took two coats out of the wardrobe to ______ the two _______.

launder / undercoats

9. Grimm's _________ was left ______ when the queen guessed his name.

Rumpelstiltskin/tsking

10. The king said to the regent: "I'll be back tomorrow, so you must _____ _______."

govern overnight

quantum mechanics - Does electromagnetic field collapse the wave function of charged particles?

In an electron double slit experiment, let's put two charged plates behind the slits in an attempt to move the pattern up and down on the the screen.

What will happen? Will it just shift the interference pattern on the screen or washes it out completely?

If it washes it out, what's the minimal field that doesn't affect the pattern? Since I don't believe the electron double slit experiment was performed in an environment where all fields were exactly zero, but they still managed to get the pattern.

thermodynamics - Squared d'Alembert Operator

I was investigating an equation suggested by Arun and Sivaram (but with some investigation) does not originate with them,

$$\frac{\partial R}{\partial t} = \alpha \nabla^2R$$

From this, I wondered if a Minkowski space analogue would exist such that the wave equation is non-linear

$$\frac{\partial R}{\partial t} = \alpha \Box R = \alpha \partial^{\mu}\partial_{\mu}R = \alpha g^{\mu \nu}\partial_{\nu}\partial^{\mu}R$$

For those interested, the Ricci flow [is] the heat equation for a Riemannian manifold. I was interested enough that I searched for similar approaches and did in fact find one:

https://en.wikipedia.org/wiki/Relativistic_heat_conduction

Except they have used a notation to ''highlight the scalar property'' of the d'Alembert operator in this context by squaring the operator $$\Box^2$$. Doing so allows you to write it in the hyperbolic form

$$\frac{1}{c^2} \frac{\partial^2 R}{\partial t^2} + \frac{1}{\alpha}\frac{\partial R}{\partial t} = \nabla^2 R$$

This allows us to write

$$\frac{\partial R}{\partial t} = \alpha \Box^2 R = -\frac{\alpha}{c^2}\frac{\partial^2 R}{\partial t^2} + \alpha \nabla^2 R$$

I am not very fond though of this notation where we square the operator. The question really revolves around the use of this notation. Is it really pertinent? By understanding the generalization of a covariant derivative in terms of the Laplacian is the d'Alembert operator allows us to look at the RHS operator ''as though it was like'' the presence of that covariant derivative. The Ricci scalar curvature is just

$$R \equiv R_{ijkl}g^{ik}g^{jl}$$

It's not impossible to consider it as I had first expected as a form satisfying the covariant derivative, but in a simpler more precise form is

$$^{g}\nabla R = \nabla R$$

Which is a notation for the covariant derivative acting on $$R$$ with respect to some Riemannian metric $$g$$. https://arxiv.org/pdf/1504.02910.pdf

electromagnetism - What is the physical process (if any) behind magnetic attraction?

I understand that the electromagnetic force can be described as the exchange of virtual photons. I also understand that it's possible for virtual photons, unlike their real counterparts, to have mass (for various complicated reasons.)

I'm quite new to all this, so I'm finding it hard to believe that photon exchange is powerful enough to move macroscopic objects (or hold up a train), so I'm convinced I'm looking at it the wrong way.

So specifically:

1. Are the virtual photons just a useful way to imagine what's actually happening, or are actual photons being exchanged?


2. Can the force applied to magnetically attracted/repelled macroscopic objects be described purely in the QED sense of virtual particle exchange, or is it the result of something else we can point to, or is it voodoo from the metaphorical aether?


3. Can force simply be spontaneously created by a field interacting with objects within it (as far as I can tell that's not what fields are/do)

And very generally:

• Where does the force to move macroscopic objects originate (at the most fundamental level)?

Answer

I don't know where you get your intuition about how strong a virtual particle can push or pull. Why would virtual particles push any less than real ones?

1. Virtual particle exchange is mathematically equivalent to other formulations of the theory. I don't know how to answer your question, because it is meaningless in logical positivism. If two ways of calculating something give the same answer, it is a meaningless question to ask which of the two is really going on underneath it all.


2. The force between two magnets is due to virtual particle exchange, it is also due to the field, because the virtual particle exchange just gives the classical field. But you can also just say "two magnets feel a force" in the 17th century style, without saying anything more. There is no reason to reduce the force between magnets to some pushes and pulls of material objects, because material objects are made out of fields too (although of a different kind, with different statistics). Your intuition is backwards--- nonlocal pushing and pulling is the norm, not the exception.This is not voodoo because you can measure it, and anything you can measure is real.


3. Classical fields are defined by the forces they create on objects, so a field can push and pull. The reason magnets attract is that they create a field which pushes them. If you think of a quantum field, then it is virtual particle exchange, but this is equivalent for magnets to just a field pushing and pulling.

thermodynamics - Is the Boltzmann brain problem really clearly established as a problem?

Starting from the assumption of a cosmology in which Boltzmann brains dominate over evolved ones, it is not immediately obvious to me that there is a real problem, since only Boltzmann brains indistinguishable from evolved brains will be posing the question in the first place.

In other words, the vast majority of Boltzmann brains will not be troubled by the paradox because their memories will not be consistent with a stable/large physical universe, and so they would never ask such a question in the first place.

So we must only consider the subset of Boltzmann brains whose experiences would be so consistent with a universe like ours that they would pose the question "why are we not Boltzmann brains?" And in such cases, even though we may be Boltzmann brains, our experiences would be indistinguishable from evolved brains.

Is this a criticism of the Boltzmann brain paradox that has been dispatched with, or is it a legitimate cause to be suspicious of whether it is really a paradox at all?

quantum mechanics - Number of Nodes in energy eigenstates

I have a question from the very basics of Quantum Mechanics. Given this theorem:

For the discrete bound-state spectrum of a one-dimensional potential let the allowed energies be $E_1

What is the physical interpretation for the number of nodes in the concrete energy eigenstate? I understand that the probability of finding the particle in the node point is $0$ for the given energy. However, why does the ground state never have a node? or why does every higher energy level increments number of nodes precisely by 1?

Answer

I guess there is not that much to grasp, unless you can really understand dark spots on an electron diffraction pattern. Very roughly explanation would be to interpret wave functions of a particle in a potential well as "standing waves", or as two interfering waves reflected from the walls of the well. Increasing the energy leads to higher harmonics, which leads to additional nodes. Nodes' numbering is the same as in the case of a classical string.

nuclear physics - Gamma Ray LASER Theory and Technology

I am aware that a similar question has been asked by someone else in the past, but in a very general form. Due to the physics interest and technology, in this question I put emphasis on the detail of the physics part and the question becomes very specific.

We know the power of visible-light-laser and the effects it can have on matter: Industry, Medicine, Military, Entertainment and many other applications.

Given the large amount of energy of $\gamma$-photons, one can extrapolate and see the new applications of LASER designed to amplify $\gamma$-light, emitted by nuclear isomers, which might be called “GLASER” (gamma light amplification by stimulated emission of radiation).

The trick is to excite nuclei to a metastable state, to achieve “population inversion” and then stimulate them to decay simultaneously. The excitation can be achieved either by soft neutron bombardment or by synchrotron irradiation, in order to excite the nuclei into an angular momentum state that does not match the one of the ground state. These nuclei can remain at that state for sufficiently long time, so population inversion can be achieved.

QUESTION:

How can the nuclei be stimulated to decay and emit their $\gamma$-photon so that to achieve GLASER? This could depend on how the nuclei are excited in their isomeric states to begin with.

lateral thinking - How did the man hang himself?

_{This may be simple and fairly well known, but I couldn't find it anywhere here, and felt it deserved a place in the ranks of puzzles}

A man is found dead, hanging in an empty barn having committed suicide. The base of the barn was 20 feet by 30 feet, and the ceiling was 10 feet above the ground. The rope was 4 feet from the roof to the man's neck, and the man's feet were hanging 1 foot above the ground, and the man was 5 feet from where the rope was around his neck to his toes. The man was hanging from the exact center of the barn. Aside from the man and the rope, there were no other objects in the barn, and the door to the barn was closed and locked from the inside, so no one had come in or out since the deed was done. The barn was structurally sound, with no holes or windows to be found anywhere. The roof was also water tight, but the ground beneath the man was damp. The ladder used to tie the rope up was found outside around the corner (guess he didn't want those that found him to have to move it later). There were no supports or beams along the ceiling of the barn.

It was determined that the man was too weak to jump and pull himself into the noose, or to climb onto the roof with just his hands. How did he manage to hang himself?

_{Change log: No windows, no holes in the roof. Roof is water tight. Door to the barn was locked from the inside. Damp ground was found under the man}

Answer

Canonical Answer:

The man stood on a block of ice which has since melted.

---

Non-canonical Answers:

1. The man stood on his horse who then took off and is out wandering the fields. (This is in the same category as the ice: "Man stood on $X$ which has since removed itself through means common to things of its type.")

2. If it wasn't for the explicitly stating the ladder was used to hang the rope and implying the general weakness of the man, the man could have thrown the rope over the rafters, tied the ends together, and then climbed into the rafters. He then unties the rope, attaches it to the rafters, and ties a noose. He slips the noose over his head and jumps down.

3. Again, this might be ruined by the weakness, but he could tie it in the rafters with the ladder and then climb the rope, tie the noose, slip it on, and let go of the rope.

4. There is a powerful electromagnet under the floor set on a timer. The man was lifted up his shoes laced with natural magnets oriented to the opposite pole, put the noose around his neck, and waited for the magnet to turn off.

5. If it didn't state that it was suicide, I'd say that someone else hung him up there for cattle thievin'.

6. He was a depressed wizard with the power of levitation.

7. The rope is sentient and strangled its abusive master.

8. The rope was tied around the rafters before the barn was raised. He stood there as his friends and family raised the barn, lifting his body into the air.

9. He used the ladder. It is stated that the ladder was put away but not that the man was the one who put it away. That conclusion is only guessed at. The first person on the scene put the ladder away to make it seem like a murder as they were embarrassed that the man had committed suicide.

---

Answers from the comments:

A. He climbed the ladder around the corner of the barn (where it was later found) to the roof, then jumped through the hole in the roof. – Paul Draper

B. The man was too weak to jump into the noose himself, but he put Flubber on the floor (or his shoes) and jumped on that. – Paul Draper

C. Ropes with natural fibers can shrink 5-10% when wet. He tied the rope around his neck, and leaky roof cause the once 10ft rope to become 9ft. – Paul Draper

D. The man dug a pit, tied the noose outside it, and jumped into the pit. – Paul Draper

Another Riley Riddle ®

Asking a Riley Riddle to pass the time :P

My prefix is a car,

My infix is a band,

My suffix is a degree.

I live on the ground.

What am I?

Note: "I live on the ground" means that I actually live on the ground. And all of my type lives on the ground.

Hint #1:

I live.

Answer

Inspired by LorentzB answer

Cabbage

My prefix is a car

Cab

My infix is a band

Abba

My suffix is a degree

Bage "bachelor of agricultural engineering"

I live on the ground

Obvious

quantum mechanics - What does it actually mean "to define a field" in QFT?

First of all: how does one define one operator in a Hilbert space? This is just a mathematics question and the answer is simple: we have a Hilbert space at hand $\mathcal{H}$, then we define a function $A : D(A)\subset \mathcal{H}\to \mathcal{H}$ that is linear, $D(A)$ being its domain.

Then we must define the function. In other words, we must say how $A$ acts on $D(A)$. This means that we must say what $A|\psi\rangle$ is for each $|\psi\rangle \in D(A)$. Usually we do so establishing a rule in terms of a general $|\psi\rangle$, perhaps using a basis, or sometimes we can do so in an indirect manner.

As for definining a function, this happens in all mathematics: to define a function, we need a set, and then we define the function on some subset of this set. So it is not possible to define a function, unless we know beforehand: (i) the set $\mathcal{H}$, (ii) the domain $D(f)\subset \mathcal{H}$ and (iii) the range $\mathcal{H}'$.

In the case of Hilbert spaces, $\mathcal{H}$ is a known Hilbert space to the problem, $D(f)$ is the domain of the operator and $\mathcal{H}'=\mathcal{H}$. Let us call $\mathcal{L}(\mathcal{H})$ the set of all operators on the Hilbert space $\mathcal{H}$.

This is just mathematics. Now, if we want to define a field of operators on spacetime, what do we need? Well, following this logic (which is the standard math, not anything fancy) we need a function $\phi : M\to \mathcal{L}(\mathcal{H})$. But, wait a minute, to build this function we need to say how it acts. In other words, for each event $x\in M$ we must say what $\phi(x)$ is.

Fine, so how do we say what is $\phi(x)$? It is an operator in $\mathcal{H}$. Hence to define $\phi(x)$ we need to say how it acts on $\mathcal{H}$, in its domain $D(\phi(x))$. In other words, we need to specify for each $x\in M$ what is the action $\phi(x)|\psi\rangle$ for each $|\psi\rangle \in \mathcal{H}$, otherwise we haven't defined $\phi$ at all.

One can argue that quantum fields must be viewed as operator-valued distributions, although I'm still unsure if this is the standard approach, but anyway, the problem remains and it is the same thing. To define the quantum field $\phi(f)$ we must say how it acts $\phi(f)|\psi\rangle$ for each test function $f$ and each $|\psi\rangle\in \mathcal{H}$.

And of course, we need $\mathcal{H}$. Although this is less important since all Hilbert spaces of same dimension are isomorphic. On the other hand, truly defining $\phi$ is vital.

Now comes the question: what do physicists do in QFT? They pick one scalar field $\phi(x)$ and say: "ok, now we just make $\phi(x)$ become operators obeying $[\phi(x),\pi(y)]=i\delta(x-y)$ and we are done". Even more is done! One writes $\phi(x)$ in terms of other operators $a(p)$, which are also not known and relate to the commutation relation. Then you imagine: "fine, the next step is naturally to define these operators" and nothing is done, all operators are left undefined with the commutation relations proposed. How can one have commutation relations and not have the operators?

In quantum mechanics I can even accept that. The state space $\mathcal{E}$ is there by postulate, the observables are there by postulate, and we assume all of them (we can even invoke Stone-von Neumann theorem if we want to be more rigorous). In QFT we have a field, so we need the functional dependency $\phi(x)$, and neither $x\mapsto \phi(x)$ nor $|\psi\rangle \mapsto \phi(x)|\psi\rangle$ are ever defined.

In that sense I'm really confused. What does actually mean for physicists in the context of QFT to define a field? How can one work with operator-valued fields (distributions) when one just says that commutation relations are obeyed without ever defining the fields and operators? What is actually going on here?

Answer

You don't need to define objects if you assume they exist. I know that sounds silly but that's what axioms do for us - they hand us things without the needs to construct or justify those things, in mathematics and physics alike. Basically, in quantum field theory we assume that we are handed the operator-valued fields acting on some Hilbert space. That's the first two Wightman axioms: States are rays in Hilbert space $H$ and a "field" is a distribution with values in the space of operators on $H$. Nothing more is known in a generic QFT - we cannot describe $H$ explicitly, and the action of the fields is likewise mysterious to us in the general case.

One might say that that is precisely the thing that makes QFT fundamentally more difficult from quantum mechanics with finitely-many degrees of freedom. Thanks to the Stone-von Neumann theorem, just assuming that there exist operators $x,p$ fulfilling the canonical commutation relations on some space allows us to know that this space is unitarily equivalent to $L^2(\mathbb{R}^n)$ and $x$ and $p$ act my multiplication and differentiation, meaning we can view states as wavefunctions, etc. QM in finitely many degrees of freedom is concrete in the sense that we can explicitly write down the space of states and the operators acting on them. But we don't define $x$ and $p$ to act in this way a priori, it is the SvN theorem that grants us the power to do so.

The proper axiomatization of this attitude, that is assuming the existence of disembodied "operators" with commutation relations without any definite space they act on, is to axiomatize quantum theories as the theory of certain linear functionals (states, expectation values) on $C^\ast$-algebras. an abstract $C^\ast$-algebra does not act on anything, and states are merely linear functionals on it that correspond to taking expectation values. This is an "intrinsic" view of quantum mechanics, assuming nothing more than the structure of the operators as an algebra - there's no Hilbert space, no operators acting on anything, nothing, so from this point of view, the question of how to "define" a quantum field just looks silly - you write down the generators of the algebra and their relations, and that's it (modulo defining the Banach structure on it). The contact to the more familiar world of Hilbert spaces is made through the notion of $C^\ast$-representations and in particular the GNS construction.

In QFT, that is to say quantum mechanics with infinitely many degrees of freedom, we lack the power of Stone and von Neumann. There are uncountably many unitarily inequivalent representations of the CCR (this is one aspect of Haag's theorem), so we cannot write down any specific action of the field on some specific space just by assuming the space and the fields exist. So we construct explicitly the free field (which you should view as first defining the $a_p,a_p^\dagger$ as the ladder operators on a Fock space, then putting together the field), and do clever tricks to somehow gain knowledge about the interacting theories out of this, e.g through the LSZ formula, which really is the cornerstone of the canonical formalism of QFT.

Now, if you're worried how physicists "define" the fields associated to certain Lagrangians, then the LSZ formula is the closest you get to an answer - through LSZ, you get all the vacuum expectation values or "Wightman functions", and the point of the Wightman axioms is precisely that they allow the Wightman reconstruction theorem to hold that states that the n-point functions are sufficient to reconstruct the fields. Now, physical theories are unfortunately rarely known to be Wightman theories, but this is the roadmap for how you'd rigorously hope to define quantum fields in the Wightman approach.

In the abstract setting, though, what you need to define is not the fields as functions themselves, but their $C^\ast$-algebra. And given a collection of classical fields and a Lagrangian, that gives you a notion of commutation relations by taking the CCR/CAR between the fields and their canonical momenta, and therefore an algebra. So in the abstract setting there's not that much to define, after all, even in the case of quantum fields.

logical deduction - Duel between the knight and the dragon

A dragon and a knight live on an island where the only sources of fresh water are a pond (containing ordinary water) and six wells, numbered 1 to 6. Each well's water is completely indistinguishable from the pond water, but contains a magical poison that has no immediate symptoms, yet suddenly kills the drinker about an hour after drinking.

To make things complicated still, each well contains a different variety of poison. If a drinker who has been poisoned by a well drinks the water from another well, the result depends on the relative numbers. If the second well has a higher number then both poisons will eliminate each other and the drinker will be cured. If the second well has the same number or a lower one, it is as though the drinker had only drunk from the first well. For example, if you drink from well 1 and then from well 4 you will not be poisoned, but if you drink from well 4 and then well 1 you will be poisoned as though you had drunk only from well 4.

As a result of these rules, water from well 6 can cure poison from any of the other wells, but, when drunk by someone who is not poisoned, is incurably lethal. Furthermore, while wells 1-5 are a short walking distance from each other, well 6 is located at the top of an unclimbable mountain on the island that the knight cannot reach but the dragon can fly onto very quickly.

This sets the stage for the following puzzle: both knight and dragon understand these rules completely, know the numbering of each of the wells, and each want the other dead. Being evenly matched in combat, they arrange a special sort of duel. Each secretly fills a glass of water from one of the island's sources, then meets the other in a field, where they exchange glasses, and drink. They may seek water from any of the island's sources that they can personally reach before and after.

Is it possible for either the knight or the dragon to ensure it will survive this duel?

Note: drinking the same poison twice in a row (or 20 times in a row) has exactly the same effect as drinking the poison once.

Answer

The knight can ensure survival

by drinking from well 4 first, and 3 + 5 after. If the dragon uses pond water, or 1, 2, 3 or 4, the last drink from 5 will neutralize the others (4, ? and 3). If the dragon gives 5 or 6, it will neutralize the first '4', then 3 will be neutralized by 5.

As AeJey said, the dragon can ensure survival by

Drinking from well 5 before, and 6 after the duel.

general relativity - How do spatial curvature and temporal curvature differ?

While looking at the metrics of different spacetimes, i came across the "Ellis wormhole", with the following metric:

$$c^2d\tau^2=c^2dt^2-d\sigma^2$$

where

$$d\sigma^2=d\rho^2+(\rho^2+n^2)d\Omega^2$$

I note that the temporal term has a constant coefficient. The Wikipedia article mentions:

There being no gravity in force, an inertial observer (test particle) can sit forever at rest at any point in space, but if set in motion by some disturbance will follow a geodesic of an equatorial cross section at constant speed, as would also a photon. This phenomenon shows that in space-time the curvature of space has nothing to do with gravity (the 'curvature of time’, one could say).

So this metric would not result in any "gravitational effects".

Looking at the Schwarzschild metric:

$$c^2d\tau^2=(1-\frac{r_s}{r})c^2dt^2-(1-\frac{r_s}{r})^{-1}dr^2-r^2(d\theta^2+\sin^2\theta d\phi^2)$$

Here we have a non-constant coeffcient for the first component. And this metric clearly has an attractive effect on particles, e.g. it's geodesics have the tendency towards $r\rightarrow0$.

1. 
   

   Does that mean the gravitational effect comes primarily from a "curvature of time" and not from spatial curvature? I assume part of the answer has to do with the motion through time being dominant for all but the fastest particles?

   
   


2. 
   

   Is the spatial curvature the primary cause of the visual distortion, e.g. the bending of light paths, in these metrics?

   
   



3. 
   

   I'm getting the picture that temporal curvature primarily affects objects moving fast through time (static and slow objects), and spatial curvature primarily affects objects moving fast through space (photons). Is this a good picture or completely wrong?

   
   


4. 
   

   If the spacetime around an "Ellis wormhole" is purely spatial, does that mean the faster i move (through space), the more i would feel the attraction and also second order effects like tidal forces?

   
   


5. 
   

   Are there physical metrics, e.g. valid solutions for the EFE which only have temporal curvature but no spatial curvature? Would such an object behave like a source of gravity, without the gravitational lensing?

   
   


6. 
   
   

   If such objects would be valid, would that mean you could pass them unharmed or even unnoticed at high speeds (moving fast through space), but would be ripped to pieces if you are moving slowly (moving fast through time)?

   
   

Answer

You need to be cautious about treating a time curvature and spatial curvature separately because this split is not observer independent. In some cases a metric can be written in coordinates where the $dt^2$ term is $c^2$ (or unity in geometric units) but this is just a choice of coordinates.

If you take, for example, the FLRW metric then we usually write it as:

$$ ds^2 = -dt^2 + a(t)\left(dx^2 + dy^2 + dz^2\right) $$

where $t$, $x$, $y$ and $z$ are the comoving coordinates. However it can also be written using conformal coordinates as:

$$ ds^2=a(\eta)^2(-d\eta^2+dx^2+dy^2+dz^2) $$

It's the same metric, describing the same spacetime geometry, but in one case the time coordinate looks as though it is curved while in the other case it looks as if it is flat. Both metrics are perfectly good descriptions of the geometry and we choose whichever version happens to be most convenient for our purposes.

But back to your question: the trajectory of a freely falling particle, i.e. its geodesic, is given by the geodesic equation:

$$ \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} = -\Gamma^\alpha_{\,\,\mu\nu}U^\mu U^\nu \tag{1} $$

In this equation $\mathbf x$ is the position $(t,x,y,z)$ of the particle in spacetime, $\mathbf U$ is the four velocity and the symbols $\Gamma^\alpha_{\,\,\mu\nu}$ are the Christoffel symbols that describe the curvature of the spacetime. You can think of this as a kind of equivalent to Newton's second law in that it relates the second derivative of position to the curvature.

Suppose we consider a stationary particle (stationary in our coordinates that is). Since the particle is stationary in space the components of the four velocity $U^x = U^y = U^z = 0$ and only $U^t$ is non-zero. In that case the geodesic equation (1) simplifies to:

$$ \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} = -\Gamma^\alpha_{\,\,tt}U^t U^t \tag{2} $$

Calculating the Christoffel symbols is a huge pain unless you have a copy of Mathematica to hand, but you can usually find them by Googling as indeed is the case for the Ellis wormhole (NB that link is a PDF) and the only non-zero Christoffel symbols are (I'll list them all in case the link above breaks):

$$\begin{align} \Gamma^\rho_{\theta\theta} &= -\rho \\ \Gamma^\rho_{\phi\phi} &= -\rho\sin^2\theta \\ \Gamma^\theta_{\theta \rho} = \Gamma^\theta_{\rho\theta} &= \frac{\rho}{n^2+\rho^2} \\ \Gamma^\theta_{\phi\phi} &= -\sin\theta\cos\theta \\ \Gamma^\phi_{\phi \rho} = \Gamma^\phi_{\rho\phi} &= \frac{\rho}{n^2+\rho^2} \\ \Gamma^\phi_{\phi\theta} &= \Gamma^\phi_{\theta\phi} = cot \theta \end{align}$$

Note that all the symbols $\Gamma^\alpha_{tt}$ are zero, so our geodesic equation (2) becomes:

$$ \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} = 0 $$

Or in other words in the Ellis wormhole a stationary particle remains stationary.

But even this result needs to be treated with some care because you have to understand your coordinates to interpret it. To show this consider the FLRW metric that I referred to above. I won't go through the details but you can do exactly the same calculation for the FLRW metric and reach the same conclusion:

$$ \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} = 0 $$

But remember that in the FLRW metric the coordinates are comoving coordinates, not the coordinates that you or I use when e.g. measuring the distances to distant galaxies, and the comoving coordinates are moving relative to everyday coordinates (which is why distant galaxies are moving and indeed accelerating relative to us). Even when we find that in a particular coordinate system a stationary particle remains stationary, this doesn't mean we would actually observe a stationary object to remain stationary.

(Though as it happens in the Ellis wormhole spacetime you and I would observe that a stationary object remains stationary.)

I think this addresses your questions 1 to 4. As for your questions 5 and 6, as it happens I asked exactly the same question in What makes a coordinate curved? and the answer is that at least two principal curvatures must be non-zero. So you cannot find a geometry/coordinate system where the curvature is only in the time coordinate.

thermodynamics - Are all diffusion-like processes described as wave-like in relativity-compatible formulations?

Citing from Wikipedia's article on relativistic heat conduction:

For most of the last century, it was recognized that Fourier equation (and its more general Fick's law of diffusion) is in contradiction with the theory of relativity, for at least one reason: it admits infinite speed of propagation of heat signals within the continuum field. [...] To overcome this contradiction, workers such as Cattaneo, Vernotte, Chester, and others proposed that Fourier equation should be upgraded from the parabolic to a hyperbolic form,

$$\frac{1}{C^2}\frac{\partial^2 \theta}{\partial t^2} +\frac{1}{\alpha}\frac{\partial \theta}{\partial t}=\nabla^2\theta$$ also known as the Telegrapher's equation. Interestingly, the form of this equation traces its origins to Maxwell’s equations of electrodynamics; hence, the wave nature of heat is implied.

It appears to me that the PDEs describing any other diffusion process –for instance, the Fokker–Planck equation for Brownian motion– will also assume an infinite speed of propagation. Then, if my intuition is correct, they'll be incompatible with SR, and will have to be "upgraded" to hyperbolic, wave-like equations.

If this were a general rule, would we have, for instance, a relativistic wave equation for Brownian motion? It appears unlikely... Is there, then, any example of diffusion-like/dispersive equation whose form "survives" into a relativity-compatible description?

Edit:

I'll add a broader reformulation of the question, as suggested by a @CuriousOne comment:

Can we find a first order equation that models the finite velocity limits or are we automatically being thrown back to second order equations? Is there a general mathematical theorem at play here about the solutions of first vs. second order equations?

thermodynamics - Is entropy violated inside black holes and worm holes?

Do the laws of thermodynamics hold true everywhere in universe ?

What about black holes and worm holes ?

Answer

No, black holes do not violate the 2nd Law of Thermodynamics.

Imagine that we want to violate the 2nd Law of Thermodynamics by throwing some volume of ideal gas into a black hole. This would seem to violate the 2nd law because when it is outside the black hole the ideal gas contributes some calculable amount of entropy to the total entropy of the universe. Once the gas crosses the event horizon of the black hole, however, it is hidden from all observers who are outside the black hole. We conclude then that in this process \begin{equation} dS_{universe} < 0. \end{equation} That is, we have decreased the entropy of the universe. This is a big problem and we need a way to resolve it. The answer is to assume that black holes carry entropy.

An important mathematical theorem in General Relativity is the Area Theorem, which states that the surface area of a black hole's event horizon can never decrease (given reasonable physical assumptions). So \begin{equation} dA \geq 0. \end{equation} In the early 1970s, Jacob Bekenstein had the insight that he could relate the area of the event horizon of a black hole to another quantity that we know can never decrease: entropy. By dimension analysis, Bekenstein wrote down the equation for black hole entropy: \begin{equation} S_{BH}=\alpha\frac{k_BAc^3}{G\hbar}, \end{equation} where $\alpha$ is some dimensionless constant (which Hawking proved is equal to exactly $1/4$).

Now, let's return to the problem of throwing an ideal gas into a black hole. We know that the mass of the black hole will increase, hence the area of the event horizon will increase. If, for example, our ideal gas also carries charge and angular momentum, then the charge and angular momentum of the black hole will change as well. We can relate the change in the area of the event horizon $dA$, the change in the mass of the black hole $dM$, the change in the charge of the black hole $dQ$, and the change in the angular momentum of the black hole $dJ$ via \begin{equation} \frac{\kappa}{8\pi}dA=dM + \Phi dQ - \Omega_H dJ, \end{equation} where $\kappa$ is the surface gravity of the black hole, $\Phi$ is the electrostatic potential of the black hole horizon, and $\Omega_H$ is the angular velocity of the event horizon. This equation should look familiar. It is the 1st Law of Black Hole Thermodynamics, and it is directly analogous to the ordinary 1st law.

We can write down a generalized 2nd Law of Thermodynamics:

\begin{equation} S_{tot} = S_{universe} + S_{BH}, \end{equation} so that $dS_{tot} \geq 0$, even when we consider processes involving black holes.

This is far from the end of the story, and many interesting questions abound. For example, the entropy of a thermodynamic system is usually explained as characterizing the number of accessible microstates of the system. If black holes have a well defined entropy, then it should be possible to express this entropy as the logarithm of the number of accessible black hole microstates. In 1995, Andrew Strominger and Cumrun Vafa showed that by counting certain states of black holes in string theory, they could correctly reproduce the Bekenstein-Hawking entropy of those special types of black holes.

astrophysics - How can a black hole produce sound?

I was reading this article from NASA -- it's NASA -- and literally found myself perplexed. The article describes the discovery that black holes emit a "note" that has physical ramifications on the detritus around it.

Sept. 9, 2003: Astronomers using NASA’s Chandra X-ray Observatory have found, for the first time, sound waves from a supermassive black hole. The “note” is the deepest ever detected from any object in our Universe. The tremendous amounts of energy carried by these sound waves may solve a longstanding problem in astrophysics.

The black hole resides in the Perseus cluster of galaxies located 250 million light years from Earth. In 2002, astronomers obtained a deep Chandra observation that shows ripples in the gas filling the cluster. These ripples are evidence for sound waves that have traveled hundreds of thousands of light years away from the cluster’s central black hole.

“The Perseus sound waves are much more than just an interesting form of black hole acoustics,” says Steve Allen, of the Institute of Astronomy and a co-investigator in the research. “These sound waves may be the key in figuring out how galaxy clusters, the largest structures in the Universe, grow.”

Except:

• Black holes are so massive that light, which is faster than sound, can't escape.


• Sound can't travel in space (space has too much, well, space)


• It's a b-flat?

So: How can a black hole produce sound if light can't escape it?

quantum mechanics - Is there a minimum possible rotation?

Quantum mechanics brought to us concepts as the Planck length and the Planck time — i.e. the shortest measurable length and the shortest measurable interval of time it makes sense talking about.

By analogy, is there such a thing as a Planck angle, i.e. an angle whose amplitude is so small that no theoretically known improvement in measurement instruments could measure an angle narrower than that?

Answer

I'm not sure your first sentence is right: Planck length and time arise from natural units wherein all the fundamental physical constants are taken to be unity. Thus the notions of Planck Length and Time simply arise from the definition of a particularly convenient system of units.

As for what these units have to do with quantum mechanics and physics in general is answered, for example, by the Plank Length Wikipedia page:

"There is currently no proven physical significance of the Planck length."

Likewise for the Planck time. Some as yet experimentally unvalidated theories ascribe a physical significance to these lengths. In any case, it is widely believed in Physics that future theories - particularly of quantum gravity - will show how hitherto unknown behaviors peculiar to very small length / time intervals arise.

As for whether there is such an analogy for angle, that is just as much in question as for length / time scale significance.

Note, on an unrelated topic, that there is a relationship between angle and quantization: the compactness of the angle space (the compactness of the circle) is what gives rise to the quantization of angular momentum. Angular momentum in physics comes in discrete, countable units, where linear momentum, arising from the noncompact real line domain for position, does not and can take on any value. See This Physics SE question for more details.

quantum field theory - Where do negative powers of $f_pi$ in the hadronic amplitudes come from?

According to Peskin and Scrhroeder the pion decay constant $f_\pi$ is defined via the following matrix element $$\left\langle0|j^{\mu5a}(x)|\pi^b(q)\right\rangle=-if_\pi \delta^{ab} q^\mu e^{-iqx}$$ while $j^{\mu5a}=\bar{Q}\gamma^\mu\gamma^5\tau^a Q$ with $Q$ the up-down quark doublet $Q=(u, d)^T$. The point is, that intuitively the definition seems to imply that $f_\pi$ is a kind of amplitude for a pion to behave like a pair of quarks and vice-versa.

This intuition works, for example, in the leptonic decay of a charged pion, where the amplitude $\mathcal{M}(\pi^+\to l^+\nu_l)\propto f_\pi G_F$. The first multiplier I interpret here as the amplitude for the pion, a bound state of quarks, to decouple into two free quarks, and the Fermi constant $G_F$ then describes how those two quarks decay into a lepton pair.

My question is the following. How can I understand the appearance of the negative powers of $f_\pi$ in some other hadronic amplitudes? For example, take the $\pi^0\to2\gamma$ decay (which bothers me most, practically) with the amplitude $$\mathcal{M}(\pi^0\to2\gamma)\propto \frac{e^2}{f_\pi}$$

One can visualize this process via the Feynman diagram where the pion first splits into a quark pair, then one of the quarks emit photon and, after, annihilate with the other quark emitting the second photon. The factor $e^2$ is natural since there were two photons emitted. But I would expect the $f_\pi$ factor to appear in the numerator, just as in the case with a leptonic decay. What is the reason here that it is in fact the negative power of $f_\pi$ in the correct formula?

Answer

How to compute pion interactions in general? Since pions are (pseudo)goldstone bosons, the procedure is following: starting from lagrangian which contains quark fields $q(x)$ (and suppose that you've integrated out gluon sector), you need to extract pion degrees of freedom from them, namely $$ q(x) \equiv U\tilde{q}(x), $$ where $$ U = e^{i\gamma_{5}\frac{\pi_{a}t_{a}}{f_{\pi}}} $$ There you just need to subtitute nonzero VEV $\langle |\bar{\tilde{q}}\tilde{q} |\rangle$.

From the described picture follows that the basic object of pion effective field theory is $U$, which depends on argument $\frac{\pi}{f_{\pi}}$.

In general the relevant effective lagrangian for free pions which is needed to answer your question has the form $$ \tag 1 S = \int d^{4}x\left(\frac{f_{\pi}^{2}}{4}\text{Tr}\left[ \partial_{\mu}U\partial^{\mu}U^{+}\right] - f_{\pi}^{2}m_{\pi}^{2}\text{Tr}\left[ U^{\dagger} + U - 2\right] + ...\right) + N_{c}\Gamma_{WZ}, $$ Here $$ \Gamma_{WZ} \equiv \frac{i}{240 \pi^{2}}\int d^{5}x\epsilon^{ijklm}\text{Tr}\left[U\partial_{i}U^{-1}U\partial_{j}U^{-1}U\partial_{k}U^{-1}U\partial_{l}U^{-1}U\partial_{m}U^{-1}\right] $$ is the Wess-Zumino term, which captures all the information about anomalies of underlying theory, and dots represent higher derivative terms and mixed terms.

Note the presence of $f_{\pi}^{2}$ at the first term in $(1)$ (it is needed for canonical form of pions kinetic term) and an absense of such quantity in the Wess-Zumino term (since in fact it is just the number).

The short formal answer on your question is following. In order to describe the mentioned processes - $\pi^{+} \to l^{+} \nu_{l}, \pi^{0} \to ll^{+}, \pi^{0} \to 2\gamma$ - we need to elongate the derivatives in $(1)$. The in turns out that the first two processes are mediated by the kinetic term in chiral effective field theory action $(1)$, which has $f_{\pi}^{2}$ factor in the front, while the general contribution into process $\pi \to \gamma\gamma$ is made by the Wess-Zumino term, which hasn't any dimensional factors in front of it. The amplitudes for the first two processes are proportional to $f_{\pi}$, while for the latter it is proportional to $f_{\pi}^{-1}$.

My statement that after gauging and adding the lepton part such action contains information about processes $\pi \to \mu \bar{\nu}_{\mu}$ and $\pi \to \gamma \gamma$.

When gauging the first terms of $(1)$, we simply modify partial derivatives $\partial_{\mu}$ to $$ \partial_{\mu}U \to \partial_{\mu} - iR_{\mu}U + iUL_{\mu}, $$ where $$ L \equiv g_{2}\frac{W_{a}\tau_{a}}{2} + g_{1}W^{0}\begin{pmatrix} \frac{1}{6} & 0 \\ 0 & \frac{1}{6}\end{pmatrix}, \quad R \equiv g_{1}W^{0}\begin{pmatrix}\frac{2}{3} & 0 \\ 0 & -\frac{1}{3} \end{pmatrix} $$ (in general, $g_{i}$ are matrices of constants (CKM matrix elements)).

We then may extract $\pi^{\pm}W^{\mp}$ vertex from the kinetic term of $(1)$ (here $W^{\pm} \equiv \frac{1}{\sqrt{2}}(W_{1} \mp iW_{2})$). Since, as I've written above, pion fields as the goldstone phase are always in combination $\frac{\pi}{f_{\pi}}$ and by taking into account that there are $f_{\pi}^{2}$ in the front of it, we have that $\pi W$ term is proportional to $f_{\pi}$.

Next, gauging of the Wess-Zumino term is harder (see an answer here). But now is only one important thing - the degree of $f_{\pi}$, so we immediately may give the result: the part of gauged WZ term which contains one pion field is inversely proportional to $f_{\pi}$.

homework and exercises - Properties of Fluids-theoritical confusion

I am having a very basic confusion on how we calculate the height of atmosphere when we assume that the density does not change with altitude(density remains 1.29 kg/m$^3$).

I want to know why we say that in this case that the atmospheric pressure is $$P=\rho gh$$

I think my confusion here is if there is any pressure exerted on the atmosphere itself!

Because we can say in this question that $$ P_{\text{on the atmosphere}}-P_{\text{by the atmosphere}}=\rho gh $$

Then if pressure on the atmosphere is 0 then my confusion is gone :) Otherwise please explain the same.

By the way the answer to the above question is about 8Kms.

Answer

Perhaps you're looking for the formula $$\nabla p = \rho \mathbf g$$ that relates the pressure gradient at any point with an acceleration field, which is usually taken to be gravity alone in many practical cases. If both $\rho$ and $\mathbf g$ are constant, then $\mathbf g$ comes from a potential $V=-\mathbf g\cdot\mathbf r$, so that $p= p_0+\rho\mathbf g\cdot\mathbf r$, where $p_0$ is any integration constant. Since $\mathbf g$ is pointing downward, the inner product $\mathbf g\cdot\mathbf r$ reduces to just $-g\Delta h$, where $\Delta h$ is assumed to be positive when measured w.r.t to a "ground" level and going upward. Hence the pressure profile at constant density and gravity is $$p = p_0 - \rho g h,$$ where $p_0$ is the pressure at $h=0$ (e.g. the pressure at sea level). As you go up, $h$ increases and therefore the pressure decreases, which physically corresponds to the fact that there is a shorter column of fluid at higher heights. Clearly, when there is no more fluid on top the $p$ must be zero, which leads to the equation $$p_0 = \rho g h$$ in $h$, and the solution is the total height of fluid.

Photoelectric effect, low frequency light

Let's say we have a emitter, emitting light that has frequency f, less than the threshold frequency of a metal.

If you leave light shining onto that metal, for long enough, does the energy of the individual photons accumulate, on the electrons, so eventually they will ionize, or does this not happen? What am I missing?

Answer

For simplicity let's consider the photoelectric effect in a thin metal foil:

The first step in the photoelectric effect is when a photon strikes an electron in the metal and transfers all its energy to it. The electron energy is now equal to the photon energy $h\nu$. If this energy is greater then the work function $\phi$ the electron can escape the metal and will emerge with a kinetic energy:

$$ \tfrac{1}{2}mv^2 = h\nu- \phi $$

However the $h\nu \lt \phi$ the electron will in effect bounce off the metal-air interface back into the metal:

and the electron will start rattling around inside the metal. The trouble is that the metal has some resistance to the motion of electrons and the electron will very quickly lose its energy and come to a halt. By very quickly I mean less than a nanosecond.

So if a second photon strikes the electron before the electron has slowed to a halt, and while the electron is travelling in the right direction then yes the second photon could add enough energy to eject the electron. So in that case we would have photoelectrons ejected by absorbing two photons.

However this process is very unlikely as the two photons would have to be absorbed within a very short time. In practice the rate at which photoelectrons are ejected by two (or more) photon absorption is very slow though it can be observed in special cases. For example the paper Double-Quantum Photoelectric Emission from Sodium Metal by M. C. Teich, J. M. Schroeer, and G. J. Wolga, Phys. Rev. Lett. 13, 611, 1964 reports observation of exactly this effect in sodium.

electromagnetism - Induced motional EMF where the wire is stationary and the field source is moving?

If a wire is moving in a magnetic field $B$, there is a Lorentz force acting on both the postive and negative charges, separating them to create an electric field, which is a great explanation to aid my understanding of how ($- \epsilon$) is induced.

What if I changed the dynamics of this system, making the wire stationary $$\therefore v = 0$$

And moved the magnetic field source(solenoid,magnet,etc...) in the same direction of the previous case(i.e same $v$). Will that change the direction of the Lorentz force acting on the charges? Or is it the same?

My initial assumption, is that they are relative, leading to the same results. Yet not so sure when considering Lenz's law it confuses me further.

Answer

The principle of relativity: The laws of physics are the same in all inertial frames of reference. Since the Lorentz force is a valid law of physics, it will not change when we pass from one reference frame to another.

First frame, wire is moving. There is no $\mathbf E$ field. Lorentz force $\mathbf F = q\mathbf v\times\mathbf B$. Apparently, you were OK with this frame.

Second frame, you are moving with wire. If you transform the electromagnetic field to the new frame, you will get: $ \mathbf E' = \mathbf v\times\mathbf B $, where $\mathbf v$ is the velocity of your frame of reference with respect to the first frame of reference (provided $\mathbf v$ is perpendicular to $\mathbf B$). And: $\mathbf B' = \mathbf B$.

Its wise to notice, this transformations of the electromagnetic field from one frame to another are Galilean. They are not relativistic. So, they are only valid when the velocity of your frame $v$ is far less than $c$. If you want, the complete relativistic transformations can be seen here.

Now, apply lorentz force to the fields in your frame: $$ \mathbf F = q(\mathbf E' + \mathbf v'\times\mathbf B') = q\mathbf E' $$

where $\mathbf v'$ is the velocity of the wire with respect your frame of reference (ie, zero. After all, you are moving with the wire in this frame, so your relative velocity is zero).

quantum field theory - Irreducible Representations Of Lorentz Group

In Weinberg's The Theory of Quantum Fields Volume 1, he considers classification one-particle states under inhomogeneous Lorentz group. My question only considers pages 62-64.

He define states as $P^{\mu} |p,\sigma\rangle = p^{\mu} |p,\sigma\rangle $, where $\sigma$ is any other label. Then he shows that, for a Lorentz Transformation : $$P^{\mu}U(\Lambda)|p,\sigma\rangle = \Lambda^{\mu}_{\rho} p^{\rho}U(\Lambda)|p,\sigma\rangle $$ Therefore: $$U(\Lambda)|p,\sigma\rangle = \sum_{\sigma'} C_{\sigma' \sigma}(\Lambda,p)|\Lambda p,\sigma'\rangle.$$ Then he wants to find $C$ in irreducible representations of the inhomogeneous Lorentz group. For any $m$ he chooses a $k$ such that $k^{\mu}k_{\mu} = - m^2$. Then defines express $p$'s with mass m, according to $p^{\mu} = L^{\mu}_{v}(p)k^v$.

Then he defines $$|p,\sigma\rangle = N(p)U(L(p))|k,\sigma\rangle$$ (where $N(p)$ are normalization constants). I didn't understand this last statement. Is $\sigma$ an eigenvalue of the corresponding operator, or just a label? I mean, if $J |k,\sigma \rangle = \sigma |k,\sigma\rangle $ then is it true, $J |p,\sigma\rangle = \sigma |p,\sigma\rangle$. If so how can we say that if $$U(\Lambda)|k,\sigma\rangle = \sum_{\sigma'} C_{\sigma' \sigma}(\Lambda,k)|\Lambda k=p,\sigma'\rangle$$

Thanks for any help. First pages of these notes on General Relativity from Lorentz Invariance are very similar to Weinberg's book.

Answer

For Poincaré algebra there are (as far as I know) two different approaches to find its representations. In the first approach one begins from a finite dimensional representation of (complexified) Lorentz algebra, and using it one constructs a representation on the space of some fields on Minkowski space. Representation so obtained is usually not irreducible and an irreducible representation is obtained from it through some differential equation. E.g. space of massive Dirac fields satisfying Dirac equation form an irreducible representation of Poincaré group (added later : last statement is not quite correct).

Another approach is to find (irreducible, unitary) Hilbert space representation of identity component of Poincaré algebra by so called "Little group method". This is what Weinberg is doing in pages 62-64 in volume 1 of his QFT book. Idea of this approach is following --

In momentum space fix a hyperboloid $S_m=\{p|p^2=m^2,p_0 \geq 0\}$ corresponding to a given (nonnegative) mass $m$. (note : here I am using signature $(1,-1,-1,-1)$)

Choose a 4-momentum $k$ on $S_m$. Let $G_k$ be the maximal subgroup of (the identity component) of the Lorentz group such that $G_k$ fixes $k$. i.e. for each Lorentz transformation $\Lambda\in G_k$ we have $\Lambda k=k$. $G_k$ is called little group corresponding to 4-momentum $k$.

Let $V_k$ be a fixed finite dimensional irreducible representation of $G_k$ (or double cover of $G_k$)$^{**}$. Fix a basis of this vector space $|k,1\rangle,|k,2\rangle,\ldots,|k,n\rangle$ where $n$ is (complex) dimension of $V_k$ {note that $k$ is a fixed vector, and not a variable.}

Now for every other $p\in S_m$ introduce a vector space $V_p$ which is spanned by the basis $|p,1\rangle,|p,2\rangle,\ldots,|p,n\rangle\;.$

Hilbert space representation of (the identity component of) the Poincaré group is now constructed by gluing these vector spaces $V_p$'s together. This is done as follows :-

i) Define $H$ to be direct sum of $V_p$'s.

ii) For every $p\in S_m$ fix a Lorentz transformation $L_p$ that takes you from $k$ to $p$, i.e. $L_p(k)=p$. Also fix a number $N(p)$ (this is used for fixing suitable normalization for the basis states). In particular, take $L_k=I$.

iii) Define operator $U(L_p)$ corresponding to $L_p$ on $V_k$ as :-

$U(L_p)|k,\sigma\rangle =N(p)^{-1}|p,\sigma\rangle,\:\sigma=1,\ldots,n\tag1$

This only defines action of $L_p$'s on subspace $V_k$ of $H$. But in fact this definition uniquely extends to the action of whole of (identity component of) Poincaré group on the whole of $H$ as follows --

Suppose $\Lambda$ be ANY Lorentz transformation in the identity component of the Lorentz group, and $|p,\sigma\rangle$ be any basis state. Then (all the following steps are from Weinberg's book):

\begin{align}U(\Lambda)|p,\sigma\rangle &= N(p) U(\Lambda) U(L_p)|k,\sigma\rangle\,\,\,\,\,\, \textrm{using def. (1)}\\ &= N(p) U(\Lambda.L_p)|k,\sigma\rangle \,\,\,\, \textrm{(from requiring}\,\, U(\Lambda) U(L_p)=U(\Lambda.L_p))\\ &= N(p) U(L_{\Lambda p}.L_{\Lambda p}^{-1}.\Lambda.L_p)|k,\sigma\rangle\\ &= N(p) U(L_{\Lambda p})U(L_{\Lambda p}^{-1}.\Lambda.L_p)|k,\sigma\rangle\;.\end{align}

Now note that $L_{\Lambda p}^{-1}.\Lambda.L_p $ is an element of $G_k$ {check it} and $V_k$ is irreducible representation of $G_k$. So $U(L_{\Lambda p}^{-1}.\Lambda.L_p)|k,\sigma\rangle$ is again in $V_k$; and from (1) we know how $U(L_{\Lambda p})$ acts on $V_k$; thus we know what is $U(\Lambda)|p,\sigma\rangle\;.$

Summarizing, the idea of little group method is to construct irreducible Hilbert space representations of the identity component of Poincare group starting from finite dimensional irreducible representations of the Little group corresponding to a fixed four momenta.

---

$^{**}$ If $V_k$ is not a proper representation of $G_k$ but is a representation of the double cover $\mathcal{G}_k$ of $G_k$ then we'll also need to specify a section $G_k\to \mathcal{G}_k$ of the covering map so that we know how $G_k$ acts on $V_k$.

special relativity - Does action really have to be Lorentz-invariant in SR?

From Landau & Lifshitz The Classical Theory Of Fields it is said:

To determine the action integral for a free material particle (a particle not under the influence of any external force), we note that this integral must not depend on our choice of reference system, that is, it must be invariant under Lorentz transformations.

This seems understandable. But in comments on this answer Ján Lalinský says that "there is no good physical reason why action should be invariant". Further he suggests another Lagrangian than that given in L&L, namely, if we denote L&L Lagrangian as $L_0$, then the example could be $L_0+Cv_x$, which is clearly anisotropic. Clearly, the equations of motion must not change with this Lagrangian, because $Cv_x$ is a total time derivative (of $Cx$).

On the other hand, in this answer Luboš Motl says that "the invariance of the action follows from special relativity – and special relativity is right (not only) because it is experimentally verified. [snip] If $S$ depended on the inertial system, so would the terms in the equations $\delta S=0$, and these laws of motion couldn't be Lorentz-covariant".

How could I connect L&L and Luboš's arguments with Ján's example? Can both sides be simultaneously right? They seem to contradict each other.

riddle - Hints Among Confusion

The Roman gods hid treasure
In robes of rose and green
A stone lay at the center
And gold was in between
Fetched by a dog named rover
Lee in "Show Me" was seen
Sir, move if you are clever

And tell me what I mean.

The solution is a single English word.

Answer

I think this juicy rhyme concerns a

mango

The Roman gods hid treasure

Roman gods hides mango.

In robes of rose and green : A stone lay at the center : And gold was in between,

It's peel is red and green, stone seed in the center and the golden pericarp in between.

Fetched by a dog named rover

Hidden in 'dog named' backwards (fetched) is mango.

Lee in "Show Me" was seen

"Show Me" is the motto of Missouri state abbreviation MO → Ang (Lee) in MO gives mango!

Sir, move if you are clever : And tell me what I mean.

Sir - man, move - go → mango

Title: Hints Among Confusion

An anagram (confusion) of Among gives Mango

newtonian gravity - What is my real weight?

My weighting machines notes my weight to be 65. Should I read it 65N or 65kg.

PS: I only need a correct comment.

This question is different, since, I know very clearly what mass and weight are. But very part is that generally students (like me) and people are much more confused that what are they actually measuring in their weight, after they had passed their class 9 and learnt about gravity.

This question can directly solve such query.

Answer

The term "mass" is an intrinsic property of any body, and doesn't depend on external factors. The term "weight" is a force, i.e. it measures how much a mass is accelerated. Your mass is $m = 65\,\mathrm{kg}$. Your weight on Earth, which accelerates you at $g = 9.8\,\mathrm{m}\,\mathrm{s}^{-2}$, is $$ w \equiv mg = 65\,\mathrm{kg}\times 9.8\,\mathrm{m}\,\mathrm{s}^{-2} = 637\,\mathrm{N}. $$ Similarly, your weight on the Moon is $65\,\mathrm{kg}\times 1.6\,\mathrm{m}\,\mathrm{s}^{-2} = 104\,\mathrm{N}$, and in deep space it's $65\,\mathrm{kg}\times 0\,\mathrm{m}\,\mathrm{s}^{-2} = 0\,\mathrm{N}$, i.e. you're weightless.

Because any acceleration gives you weight, you don't need a massive planet; if you want to visit the ISS, the acceleration of your spacecraft (reaching $29\,\mathrm{m}\,\mathrm{s}^{-2}$, according to NASA) makes you weigh $1950\,\mathrm{N}$ at liftoff, in addition to the $637\,\mathrm{N}$ that Earth makes you weigh. When you get to the ISS, Earth's gravity isn't much weaker than at the ground. There, $g=8.7\,\mathrm{m}\,\mathrm{s}^{-2}$, so if it were hovering above Earth, you'd weigh $565\,\mathrm{N}$. However, since the ISS is in free fall around the Earth, your acceleration with respect to the ISS is $0\,\mathrm{m}\,\mathrm{s}^{-2}$, and you're weightless.

So technically, saying "I weigh 65 kg" is wrong. Instead you should say "My mass is 65 kg". But don't do that. It'll only get you in trouble.

rotational dynamics - Why does torque point perpendicular to direction of the motion?

I have an intuition problem calculating torque using the cross product formula. As for example let the magnitude of the force be 50 lbs and length of the wrench be one foot and you are exerting force in a clockwise motion and the angle you apply the force 60 degrees. This is an example so I can ask my question. Using the right hand rule the torque points perpendicular to the force you are applying to the bolt. In this case since the sine of 60 degrees is about .86 it would be (.86)(50) foot lbs. How can the bolt turn clockwise if the force is concentrated perpendicular to where it needs to turn? The cross product formula demands the torque be perpendicular. Obviously my mistake but I don't see where.

Answer

How can the bolt turn clockwise if the force is concentrated perpendicular to where it needs to turn?

Because that force is perpendicular to the direction towards the rotation-centre. Not to the turning direction. The bolt does indeed turn in the same way as the force pulls it.

When you define a torque vector direction, you have a problem. You can't define a vector direction as something that turns around. The direction must be along a straight line. So instead of choosing the torque "turn", we could choose the torque axis as the vector direction.

Have a look at this picture:

The axis is vertical through the bolt along the two upwards/downwards arrows. If you choose to define the torque vector direction along this axis, all fits. We just have to remember that choice.

Torque is: $$\vec \tau = \vec F \times \vec r$$

The force vector $\vec F$ times the vector towards the rotation-centre $\vec r$ gives the torque vector. The result of a cross-product is mathematically a vector pointing vertically upwards, so this fits perfectly to that choice. The torque vector $\vec \tau$ that you get from this calculation has the torque magnitude but the torque-axis direction.

As long as you remember this choice - this definition - all is good. Everytime you hear "the direction of the torque is horizontal", you know that this is only the axis of the torque; the torque (the turn) is then upright.

electromagnetism - Electric field energy of two parallel moving charges at relativity speeds

Assume two particles of the same charge $q$ moving in parallel trajectories on the $x,y$ plane. The distance between them is $d$ and it's defined in the $y$ axis. The problem is to calculate the energy of the electric field when the speed of the charges is relative to the speed of light.

I would say that the energy of the electric field should be given by $$U_E = {\epsilon_0 \over 2} \int |E|^2 dV ~,$$ where $dV$ is for integration in volume $V$.

Also I would say that the field of the particles at time $t$ is defined by the position of the particles at a time $t'$ before $t$. The field is a long equation one can find if he uses the Liénard-Wiechert potentials.

And here I arrive to my problem: The expression of the electric field is too complicated and I have $E=E_1 +E_2$ by the two charges. So it seems that a direct calculation of the integral would be quite the situation.

So, the question is: Is there something I am missing that could simplify the calculations or is there a simple answer that I don't see? If no, and the answer must come from the above integral, is there some easy way to calculate (don't solve the integral, please. Just give a direction- mostly I ask if there is a physics admission that would simplify the problem).

Thank you.

Note: I will give here the expression for the fields:

$$\bar E(\bar r ,t ) = {q \over 4 \pi \epsilon_0} {i \over (\bar i \cdot \bar u)^3 } [(c^2 - v^2) \bar u + i \times (u \times a)] $$ and

$$\bar B( \bar r ,t )={1 \over c} i \times \bar E( \bar r, t) $$ where, $i$: the distance between the charge at $t'$ and a point calculating the field at $t$.

$v$: the speed of the particle

$\bar u =c \hat i - \bar v $

$a$: the acceleration, here zero (0)

---

EDIT

Note2 (6/25/2015): The solution given to this particular problem by my professor at the university goes as follows:

The electric energy is given by $U_E =\frac{1}{2} qV_1 + \frac{1}{2} q V_2 $. Because the trajectories are parallel the potentials are equal to a specific point in space so: $U_E =qV $. Also, because the speeds are very high (relativistic) we take the Liénard-Wiechert electric potential. Thus the electric energy is: $U_E ={ q^2 \over e \pi \epsilon_0 d } {1 \over \sqrt{1-{u^2 \over c^2}} }$

I disagree with this approach. Because we have two charges moving, they emit both electric and magnetic fields. So the electric field as a function of the above-mentioned potential is given : $E= - \nabla V - {\partial \bar A \over \partial t} $, and the vector potential or its time derivative isn't zero. So we cannot just write that the energy of the electric field is equal to charge times the electric potential.

Is there something I haven't understood or is this approach truly mistaken?

Again, thank you for your consideration and your help.

Answer

Constantine, take a look at what Minkowski said in Space and Time:

"In the description of the field caused by the electron itself, then it will appear that the division of the field into electric and magnetic forces is a relative one with respect to the time-axis assumed; the two forces considered together can most vividly be described by a certain analogy to the force-screw in mechanics; the analogy is, however, imperfect."

Note how he talked about THE field, singular. There's only one field, and it's the electromagnetic field $F_{\mu\nu}$. This field is associated with two forces, and to understand why, you should combine typical depictions of the electric field and the magnetic field to depict the electromagnetic field like this:

Note how the resulting "spinor" is reminiscent of vector fields? Now imagine you have two of them, with no initial relative motion. How are they going to move? Remember this: co-rotating vortices repel, counter-rotating vortices attract. They're going to move like this, linearly apart:

The linear "electric" force E between them is given by Coulomb's law, see Wikipedia where you can read this: "an electric field is a vector field that associates to each point in space the Coulomb force experienced by a test charge". But note that there is no actual electric field present, merely a linear force resulting from the interaction of two electromagnetic fields. These fields can also cause rotational force, because if you sling one spinor past the other, they don't just repel linearly, they go round one another too, something like this:

Now you can also see rotational "magnetic" force B, and you can perhaps appreciate why we talk of rot which is short for rotor. But note that the spinors have to have some initial motion relative to each other for this, and not just linearly towards each other or away from each other. Hence your two electrons only move linearly apart. Their motion relative to you doesn't change their motion relative to each other. They still move linearly apart even though you're passing them at some relativistic speed. Or they're passing you at some relativistic speed. And there is no electric field, just electric force resulting from $F_{\mu\nu}$ field interactions, so the energy associated with this is the integral of the force x distance. Then you chuck in the Lorentz factor for the relativistic motion with respect to you. What your professor said sounds right to me.

word - A Smuggled List of S'S

Smuggled this strange list from [redacted] 's place. Maybe he is making a puzzle. Who is the victim?

• Scarlet covers


• Swindler's imprecation


• Sour disordered equipment



• Smash journal


• Sinful America


• Surgeon oath


• Storms carport

_{Working for Generalist Countdown - anagram}

Answer

What are the items on the list?

Synonyms of the words on the list can be anagrammed to a tag on Puzzling SE plus one letter. For example: "Scarlet covers" are "red lids", which is an anagram of "riddle" plus "s". The complete list is:

Scarlet covers: red lids — riddle + s
Swindler's imprecation: cheat's dammit — mathematics + d
Sour disordered equipment: acidic unglued tool — logical deduction + u
Smash journal: plow diary — wordplay + i
Sinful America: evil USA — visual + e
Surgeon oath: Dr vow — word + v
Storms carport: typhoons garage — steganography + o

Who is the victim?

If we assume that the third item is correct and that the second item yields an extra d, we can anagram the extra letters to Deusovi. (Thanks, jafe, for spotting that. I had a rather less convincing guess first.)

general relativity - Geometric interpretation of Electromagnetism

For gravity, we have General Relativity, which is a geometric theory for gravitation.

Is there a similar analog for Electromagnetism?

Answer

1. Kaluza-Klein theory. This is similar to General Relativity, but instead of three space dimensions plus time, there are four space dimensions plus time. The fourth dimension is cyclic, and satisfies some symmetry conditions. The electromagnetic potential appears as the components of the metric in the fourth space dimension. It is usually rejected on the grounds that we can't see the fourth space dimension, or that it is made too small to be seen. In fact, the symmetry conditions along this dimension make it indistinguishable, and moving along it is equivalent to a gauge transformation. So, this is the only evidence predicted by the theory, no matter how large we make the cyclic dimension. Which leads us to

2. Gauge theory. As mentioned by DImension10 Abhimanyu PS, electromagnetism can be described by a gauge theory whose gauge group is $U(1)$; the electromagnetic potential becomes a connection, and the electromagnetic field the curvature associated to the connection. It is in fact the symmetry group of the fourth dimension in Kaluza-Klein theory. For mathematicians, a gauge theory is described in terms of principal bundles, which, if the gauge group is $U(1)$, are in fact 4+1 dimensional spaces, satisfying symmetry conditions like in the Kaluza-Klein theory. So, mathematically, they are equivalent, although there are variations of the Kaluza-Klein theory which cannot be described by a standard gauge theory.

3. Rainich-Misner-Wheeler theory. There is a way to obtain electromagnetism from geometry, in the 4d spacetime of General Relativity. Rainich was able to give in 1925 necessary and suficient conditions that spacetime is curved in a way which corresponds to the electromagnetic field. By Einstein's equation, the spacetime curvature is related to the field. So, Rainich decided to see if one can obtain the electromagnetic field from the curvature, using Einstein's equation. He found some necessary and sufficient conditions for the Ricci tensor, which are of algebraic and differential nature. This works for source free electromagnetism. There is an ambiguity, given by the Hodge duality between the electric and the magnetic fields, for the source free Maxwell equations. So, basically, the field is recovered up to a phase factor called complexion. The idea was rediscovered by Misner and Wheeler three decades later, who combined it with the wormholes of Einstein and Rosen. They interpreted the ends of the wormholes as pairs of electrically charged particles-antiparticles. The electromagnetic field, in this view, doesn't need a source, since the field lines go through the wormhole. While this idea may seem bizarre, it allowed to obtain "charge without charge", and to fix the undetermined phase factor. This model of particles had some issues, for instance it couldn't explain the spin, and Misner and Wheeler abandoned it.

material science - Why is paper more frangible when it is wet?

My four-year-old daughter asked me why paper tends to fall apart when wet, and I wasn't sure. I speculated that the water lubricates the paper fibers so that they can untangle and separate more easily, but I really wasn't sure.

Answer

Forever_a_Newcomer is on the right lines, but it's not like water dissolving salt.

Paper is mostly made from cellulose fibres (depending on the type there may also be filers and glazes like clay). Cellulose molecules bristle with hydroxyl (OH) groups, and these form hydrogen bonds with each other. It's these hydrogen bonds that make the individual fibres stiff, and also hold the fibres together.

Water is also full of OH bonds, obviously since it's H$_2$O, and the water molecules form hydrogen bonds with the hydroxyl groups on the cellulose, which breaks the hydrogen bonds that cellulose molecules form with each other. There are two results from this: firstly the cellulose fibes in the paper become floppy, because their internal hydrogen bonds are broken, and secondly the fibres separate from each other more easily. The combination of these two effects makes paper easier to tear apart when wet.

Most organic materials show similar behaviour. For example cotton is also easier to tear when wet (cotton is also made mostly from cellulose). Also hair becomes floppier and more easily damaged when wet, though the effect is less pronounced because hair contains fewer hydrogen bonds than cellulose fibres.