riddle - Small and resilient

At the same time I can be low or high,
I can be found where the ground meets the sky.

I may have a bald spot on top of my head,
but pick a fight with me, and you may be dead.

Slow moving, slow growing, I eventually change,
but I'll always have friends out on the range.

What am I?

Part 2 of the series.

Answer

You are

A mountain

At the same time I can be low or high, I can be found where the ground meets the sky.

Mountains can be seen on the horizon; some of them are low (usually older mountains, which have eroded), some high (usually younger ones)

I may have a bald spot on top of my head, But pick a fight with me and you may be dead.

Snow-capped peaks are something you see on mountains high enough - these resemble a "bald spot" on the top of someone's head. But people who climb mountains—seen as a challenge or fight—could "lose the fight" often die.

Slow moving, slow growing, I eventually change, But I'll always have friends out on the range.

Mountains change very slowly, from parts of the earth moving as well as erosion; mountains are found in "ranges", that is, a bunch of them all together.

word - A Resurrected Riley Riddle

I made a new Riley Riddle to make up for the laziness in my previous Riley Riddle, also inspired by @KevinL's Riley Riddle.

My prefix won't stand, but it cracks at the bottom.

My suffix has heads in the clouds, not forgotten.

My infix is land, but no flowers can blossom.

My body has legs in a shroud, then wings open.

What am I?

The answer is $9$ letters, and the title is a clue. For the last part, I can include the knowledge tag... but as a hint, I won't — once you solve for the affixes, you will most likely know of the last line.

Answer

Is this a

BUTTERFLY?

My prefix won't stand, but it cracks at the bottom.

Haha, you said BUTT!! (Although it can't stand by itself, haha you said buttcrack!)

My suffix has heads in the clouds, not forgotten.

Although we'd like to forget about a FLY, we can't because waiter, they're in my soup again.

My infix is land, but no flowers can blossom.

This would be SURF AND TURF, where if flowers blossom you're doing it wrong.

My body has legs in a shroud, then wings open.

This is the BUTTERFLY coming out of its chrysalis and opening its wings, a sort of resurrected caterpillar. (What, you were expecting a fourth joke?) ;)

space - How would you move without gravity?

I'm wondering how would you move without gravity? You would still have air pressure at 1Atm.

Would you "swim" in the air or would you have do something else?

Answer

The answer is: with great difficulty.

In a space ship there should be hand holds and footholds and a grid of ropes crisscrossing and you would move by pulling and pushing yourself: muscle power.

Air is too thin to get a reactive force to propel yourself effectively by swimming.

Once a motion is started, stopping will have to be taken into account too.

special relativity - Can we accelerate photons to travel at speed of $1.5c$? Or any $xc$ when $x > 1$?

1. 
   
   

   Can we move photons to travel at speed greater than speed of light in vacuum? If no then why? For example: I want to send some photons at 1.5c. Is that possible?

   
   


2. 
   

   How a photon of light can go from 0 to 67 million miles per hour instantaneously. What is the motive force that propels the photon? Where does the energy come from?

   
   


3. 
   

   Why is $c$ "limited" to 186,000mps?

   
   


4. 
   

   Why is the speed of light finite?

   
   
   


5. 
   

   What prevents the light from traveling even faster than it does?

   
   

Classical Field Theory - Continuum limit in forming the Lagrangian density and the elasticity modulus

I have been looking at taking the continuum limit for a linear elastic rod of length $l$ modeled by a series of masses each of mass $m$ connected via massless springs of spring constant $k$. The distance between each mass is $\Delta x$ which we use to express the total length as $l=(n+1)\Delta x$. The displacement from the equilibrium position is given by $\phi(x,t)$.

The discrete Lagrangian in terms of the $i$th particle $\mathscr L$ is composed as follows,

\begin{equation} \mathscr L=\frac{1}{2}\sum _{i=1}^{n}m\dot \phi _i^2-\frac{1}{2}\sum ^n _{i=0}k(\phi_{i+1}-\phi _i)^2 \end{equation}

At this point we take the continuum limit such that the number of masses in the fixed length of rod tends to infinity and correspondingly the inter-particle distance tends to zero. It is fruitful to multiply top and bottom by $\Delta x$ such that we can define two quantities that remain constant during this limit namely the linear density ($\mu=m/\Delta x$) and the elastic modulus ($\kappa=k\Delta x$).

\begin{equation} \mathscr L=\frac {1}{2} \sum _{i=1}^{n}\Delta x\bigg(\frac{m}{\Delta x}\bigg)\dot {\phi} _i^2-\frac {1}{2} \sum _{i=0}^{n}\Delta x(k\Delta x)\bigg(\frac{\phi _{i+1}-\phi _i}{\Delta x}\bigg)^2 \end{equation}

It is easy to see why the linear density remains constant since both the number of masses per unit length increases while simultaneously the unit length decreases.

However my question is regarding the elastic modulus, I fail to see how it remains constant in this limit.

The argument goes as follows; Since the extension of the rod per unit length is directly proportional to the force exerted on the rod the elastic modulus being the constant of proportionality. The force between two discreet particles is $F_i=k(\phi _{i+1}-\phi _i)$, the extension of the inter particle spacing per unit length is $(\phi _{i+1}-\phi _i)/\Delta x$. Therefore (HOW) $\kappa=k\Delta x$ is constant. Its easy to relate them but why is it constant!?!

Answer

Honestly, I think this is one of those cases where you should just accept it and push on. This 'derivation' is really nothing more than a pedagogical device to make field theory seem somewhat natural to students with a background in classical mechanics.

What we are trying to do is to take the continuum i.e. $N\to \infty$ limit of the following Lagrangian:

$$L_N=\frac{1}{2} \Biggl(\sum_{i=1}^N\Delta x \frac{m}{\Delta x} \dot{\phi_i}^2-\sum_{i=1}^{N-1}\Delta x\ k\Delta x \biggl[\frac{\phi_{i+1}-\phi_i}{\Delta x}\biggr]^2\Biggr) $$

define $\mu=\frac{m}{\Delta x}$ and $Y=k\Delta x$

Clearly, for a continuum limit, we get infinitely many particles, so the total kinetic energy of the system should diverge... unless we impose (or put in by hand, as they call it), that $\mu$ remains constant, not $m$. Similarly, it is obvious that the equilibrium force of each spring $F=k \Delta x$ should vanish... unless we impose that $k\Delta x$ is constant when we take our limit. With these ad-hoc assumptions, and replacing the discrete index $i$ with a continuous spatial coordinate, we get

$$L\equiv \lim_{N\to \infty}L_N=\frac{1}{2}\int_0^l \mathrm{d}x \biggl(\mu\dot \phi^2 -Y(\nabla\phi)^2\biggr)$$ This gives us the right action for a free, massless, scalar field \begin{align*}S[\phi]&=-\frac{Y}{2}\int_0^l \mathrm{d}x\ \mathrm{d}t \biggl(-\frac{\mu}{Y}\dot \phi^2+(\nabla \phi)^2\biggr)\\ &=-\frac{\mu c^2}{2} \int_0^l \mathrm{d}x\ \mathrm{d}t \biggl(-\frac{1}{c^2}(\partial_t\phi)^2+(\nabla\phi)^2\biggr) \hspace{2cm}c=\sqrt{\frac{Y}{\mu}}\\ &=-\mu c^2\int_0^l\mathrm{d}^2x\ \frac{1}{2}\eta^{\mu\nu}\partial_\mu\phi\partial_\nu\phi\end{align*}

The definition of $c$ is the standard one for the speed of longitudinal waves, and as one can see this Lagrangian is also reminiscent of the action for a relativistic point particle (especially the prefactor). This is, of course, a very nice result, so we can be happy about the way we took our limit, even if we had to make some ad-hoc assumptions.

statistical mechanics - Gross "temperature" of a globular cluster

Globular clusters can be very large, which means we can do statistics about the stars in them. And that means we can try matching their star-as-particle potential/kinetic energy distribution against a Boltzmann distribution, which might mean that a suitable globular cluster has a temperature; however a quick search on the internet seems to evoke only stellar surface temperatures, even though I'm sure the preceding is not at all original.

Can anyone quote off-hand typical cluster temperatures or cite such a calculation in a freely-available paper?

This related question was a simple yes/no, or at least its answers are; here star systems come up, but temperature is never mentioned.

Answer

A typical velocity dispersion in a globular cluster is 10 km/s. For a typical 1 solar mass subgiant in an old globular, then equating the kinetic energy to $3kT/2$, we get $T = 5\times 10^{60}$ K.

Doesn't seem that helpful really...

The concept of temperature is only ever applied in a relative sense - i.e. some component is hotter than another. Can't say I've ever seen absolute temperatures used. An example would be the concept of a negative heat capacity, whereby if energy is lost from a cluster by stellar evaporation (or by the hardening of a binary system), the stars gain kinetic energy. i.e Energy is lost but the "temperature" increases. Exactly the same thing happens in a contracting gas cloud or star.

In response to discussion in the comments - globular clusters are very old, usually many times their two-body relaxation timescales. They come into a (virial) equilibrium, where the speed distribution of the stars should be approximately Maxwellian. Yes, the fast moving tail will escape, just like fast moving molecules escape the Earth's atmosphere. But the rate of escape is roughly 1% per relaxation timescale. There is plenty of time for the pseudo-temperature to gradually adjust to such a slow change, meaning that the concept of a temperature can still be used.

Below I show the 1D radial velocity distributions seen in the core of globular cluster M4 (taken from Sommariva et al. 2009). The data are split into bins according to the position in a colour-magnitude diagram (a proxy for stellar mass). The blue lines are Gaussian fits (what you would expect for a Maxwellian velocity distribution) and they are pretty good.The instrumental resolution is of order 0.2 km/s for these observations, so that is a negligible contributor, but there will be some outliers due to binary motion.

The main effect of stellar escape is that it invalidates the concept of a global temperature (for a similar reason). Clusters cannot be isothermal. The velocity dispersion (and pseudo temperature) must (and is measured to) decrease with radius, otherwise the stars at large radii would escape (see below; the 1D velocity dispersion vs radius for a globular cluster $\omega$ Cen, from Scarpa et al. 2003).

fluid dynamics - Time taken to reach efflux velocity

By combining Bernoulli's Equation and equation of continuity, we can derive efflux velocity as:

$$u = \sqrt{2gh\left(\frac{{A_1}^{2}}{{A_1}^{2}-{A_2}^{2}}\right)}$$

where $A_1$ is area of the open surface and $A_2$ is the area of the hole.

But my question is how long does it take to achieve this velocity?

Let's say the hole was initially closed, and suddenly it opened, fluid particles will get pushed and accelerate from initially being at rest till they reach this steady velocity. How long does this take? I don't think it can be assumed to be instantaneous?

wordplay - Diary of a hairdresser

Today I went to cut my hair, and the hairdresser told me a weird story.

Yesterday, I had a group of people come to cut their hair. Together, they embodied the one type of hair I hate.
A celebrity came in first, and brightly asked me to trim a bit past the middle.
A player came in next, and pretendedly asked me for a number one.
A glutton grunted at me to ask me to cut off a bit at the front.
Finally, a fellow worker stylishly asked me to cut off everything but the ends.

---

This story is based on a fictional event and taken from there...

Part of an upcoming metapuzzle

Answer

She hates

STRAIGHT hair.

---

A celebrity came in first, and brightly asked me to trim a bit past the middle.

STAR (clued by celebrity/brightly)

A player came in next, and pretendedly asked me for a number one.

ACTOR (clued by player/pretendedly)

A glutton grunted at me to ask me to cut off a bit at the front.

PIG (clued by glutton/grunted)

Finally, a fellow worker stylishly asked me to cut off everything but the ends.

HAIRSTYLIST (clued by fellow worker/stylishly)

biophysics - What happens to body chemistry at the speed of light?

Assume that I'm traveling at the speed of light in one direction. My brain is also traveling at the speed of light in that direction. Presumably there is at least one receptor site in my brain that is oriented in such a way that particles would need to travel faster than the speed of light to reach their destination receptor.

What would happen to those particles? Would my brain cease to function because these particles couldn't reach their destination?

Answer

You’re thinking of velocity addition the “usual” way: if you’re in a bus going 30 mph and you throw a ball forward at 10 mph then the ball will be going 40 mph according to someone outside the bus. But it turns out that this isn’t quite correct when you’re going at speeds ~10% or more the speed of light: the velocities have to add so that nothing goes faster than light. Instead of using the usual formula,

$$ v_\mathrm{total} = v_1 + v_2 , $$

we use the relativistic formula

$$ v_\mathrm{total} = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}} . $$

(To be clear, the relativistic formula is always correct; it’s just that when $v_1$ and $v_2$ are small—as they always are in everyday life—the denominator of this fraction will be basically 1, and so the formula reduces to the simpler one that we’re more used to.)

So the answer to your question is that it’s not a problem for a human to be going close to the speed of light. As sb1 pointed out, the electrons in your brain don’t know or care how fast your body is moving; as long as your entire body is moving at the same speed, all of your biological processes will continue exactly as before, because your body is at rest with respect to itself.

The other objection you raised was that the electrons in your brain—which are moving very quickly relative to your body—would be moving faster than $c$. As you can see from the second formula above, velocities always add so that their sum is less than $c$: even if your body is travelling at $0.99c$ (relative to someone at rest) and it contains an electron going at $0.99c$ (relative to your body, and in the same direction), that person at rest will measure the electron as moving at $0.99994c$.

antimatter - How are pions created?

I have listened to a lecture explaining the strong force which is actually nuclear force and colour force.

And the pions are the force carrier of the nuclear force. And basically pions are created from two quarks (a quark and an anti-quark). But my question is, where do these two quarks come from?

Answer

We don't know where quarks come from, anymore than we know where electrons come from. They are supplied by nature, as far as we know, as the basic elementary particles that the world around us is built upon. In the same way, we don't yet know why the speed of light is 300,000 m/s or what Dark Matter or Dark Energy are.

We have a theory, called the Standard Model, which is based on experimental evidence. In turn, the Standard Model is based around Quantum Field Theory, which provides theoretical predictions that we can test using instruments such as the LHC.

However, to make the Standard Model work, we have to simply accept the properties of around 20 parameters, such as the strength of interaction of photons with electrons. We have no real idea why these parameters have the values they do, we can only measure them and use them in our predictions.

From Wikipedia Strong Interaction

An animation of the nuclear force (or residual strong force) interaction between a proton and a neutron. The small colored double circles are gluons, which can be seen binding the proton and neutron together. These gluons also hold the quark-antiquark combination called the pion together, and thus help transmit a residual part of the strong force even between colorless hadrons.

Quarks use gluons, which are massless, to exchange the color force, which keeps 3 quarks together to form neutron and protons (baryons), and pions(mesons) use gluons to keep a quark and antiquark together. BUT, the force carried by gluons has a very short range, always staying inside baryons and mesons. The force between protons and neutrons, which keeps the nucleus together, is carried by mesons. If you read the above link (slowly :) ) it will give you the full picture.

homework and exercises - Is magnetic force non-conservative?

If magnetic field is conservative, then why not the magnetic force?

My professor thinks it is non conservative but he couldn't explain to me why?

quantum mechanics - Part of a Wigner theorem

I was trying to understand why there should exist operator in Hilbert space to correspond to any symmetry transformation and found about Wigner's theorem. In it, I can see that any transformed vector in Hilbert space can be represented as a linear combination of transformed base vectors leading to something like, action of any operator on the specific state can be expressed as the action of this operator on the base states. How this proves existence of the operator I do not see. It seems to me that this operators is linear or antilinear but this is given it actually exists. Am I missing something? Can anyone point out part of the theorem which proves pure existence and not the properties of the operator? So to add something, symmetry only tells us ho a ray is transformed into another ray, it does not tell us which of the vectors of the old ray went into which of the new one. Where in the Wigner theorem can we see proven that such a connection can be established?

general relativity - Exact meaning of radial coordinate of the Schwarzschild metric

In this answer as well as on Wikipedia the radial coordinate of the Schwarzschild metric is described as follows:

...the r co-ordinate is the value you get by dividing the circumference of the circle by 2π.

This circumvents (no pun intended) the problem of talking about the path outwards from the singularity, but what about the circle itself: does it have the length we would measure when flying around it (however this is done) or is it a virtual circle assuming flat space, i.e. no mass at the center.

Answer

The circumference is the distance you would measure if you laid out a (very long!) tape measure along a circle centred on the black hole. Or if piloted your spaceship round the black hole it's the distance your navigation computer would record.

Extracting these distances is less work then you might think. Suppose we use polar coordinates so $r$ is the distance from the black hole, $\theta$ is the latitude and $\phi$ is the longitude. The metric is then:

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

Suppose we trace out a circle round the equator so $dt = dr = d\theta = 0$ and $\theta = \pi/2$, then the metric simplifies to:

$$ ds^2 = r^2d\phi^2 $$

or:

$$ ds = rd\phi $$

As we go round the equator $\phi$ goes from zero to $2\pi$, so integrating to get the distance travelled gives:

$$ \Delta s = r\int_0^{2\pi} d\phi = 2\pi r $$

To show that the $r$ coordinate is not simply the radial distance let's use the same technique to measure the distance from $r = R$ to the singularity $r = 0$. We'll use a radial line at constant $t$, $\theta$ and $\phi$, so $dt = d\theta = d\phi = 0$, and substituting these into the metric gives:

$$ ds = \frac{dr}{\sqrt{1-\frac{r_s}{r}}} $$

so the proper distance from $r=R$ to the centre is:

$$ \Delta s = \int_0^R \frac{dr}{\sqrt{1-\frac{r_s}{r}}} $$

which is most definitely not just $R$. I won't evaluate the integral here because I've already done it in my answer to How much extra distance to an event horizon?. The result isn't particularly illuminating.

Can cellular automata be reconcilied with quantum mechanics?

CAs are deterministic representations of the universe, which, according to the Bell's inequality are not entirely accurate. Cells interact "locally" (only with the closest neighbours), while quantum entanglement proposes the opposite.

So, in layman terms, what changes should be made in cellular automata in order to -if this is even possible- make them represent the universe realistically?

Answer

The main question is how do you map the CA to reality? You need to say how you describe an experimental situation in terms of the CA variables. If the map is such that an atom is described by a local clump of automata variables, and a far-away atom is described by another local clump of automata variables far away, it is flat out impossible to reproduce quantum mechanics, even in a crude way. This type of model is thoroughly ruled out by Bell's inequality violation.

But there is no requirement that the map between atomic observables and CA variables is local. If you imagine that the CA is on the surface of a holographic screen (as t'Hooft often liked to draw), then any one atom can be described by gross properties of essentially all the CA variables, nonlocally, while another atom far away is also described by a different property of all the CA variables together, so that they are always interacting. But it is concievable that statistically, those properties of the CA that describe each atom individually look like they obeying a wavefunction time evolution.

This type of thing is very hard to rule out, at least, I don't know how you would show that this sort of thing can't reproduce quantum mechanics to the extent that it has been measured.

This is something I wonder about off and on. Is it possible, even in principle, to find a CA with a physical number of variables, on the order of the cosmological horizon area divided by the Planck area, which reproduces the observed predictions of quantum mechanics by a horrendously nonlocal identification between the properties of objects and the CA variables?

It is certainly impossible to reproduce all of quantum mechanics with a model of this sort. Shor's algorithm for factoring 10,000 digit numbers will certainly fail, because there aren't enough bits and operations in the CA to do the factoring. But we haven't built a quantum computer of this size yet, so that this may be seen as a safe prediction of all such models--- that quantum computers will fail at a certain not-so-enormous number of qubits.

So it is impossible to reproduce full QM, but it might be possible to reproduce a cheap QM, which matches the cheap QM we have observed to date. You must remember that every time we verify the prediction of QM, we are not in a regime where it is doing an exponential computation of large size, precisely because if it were, we wouldn't be able to compute the consequences to compare with experiment in the first place.

The nonlocality can be in space and time together. For previous answers regarding related stuff, see here: Consequences of the new theorem in QM?

quantum field theory - MVH amplitudes and the unitarity method

In the last 5 years there has been a silent revolution in QFT called the unitarity method and the Maximum Violating Helicity (MVH) Amplitudes that basically consist an alternative way to obtain the same amplitudes that you would obtain by the Lagrangian formalisms and the Feynmann diagrams without breaking the bank in supercluster computing time.

Now, there is little known about why/how this method works in the theoretical aspect. Are there interesting things that might result from this in the theoretical domain, or it just happens to be an extremely convenient calculation tool? What are the main insights that the existence of these identities give us?

If you feel that a summary of the state of the art is all that can be offered for now, feel free to give that as an answer.

fluid dynamics - Why, when one opens 1 car window, does that noise occur?

When you're driving and you open 1 car window, say the front one, there comes a horrible noise, but when you open another window just the slightest bit, this noise goes away (I'm sure most people know what I'm talking about then I mention this 'noise').

1. Why does this noise occur?


2. Why does it go away when another window is slightly opened?

(Not sure about the tag).

Answer

The car is behaving like a closed pipe, so you get a resonance set up. There's a Wikipedia article here, but for once the Wikipedia article isn't that great, so there's another better article here. I imagine you (like most of us) will at some point have discovered you can make a sound by blowing across the top of an opened bottle, and it's the same thing happening in your car with the open window acting like the opening in the bottle. Since your car is much bigger than a bottle the resonance frequency is uncomfortably low.

When you open a second window you get an air current flowing through the car and this destroys the resonance.

newtonian mechanics - Marvin the Martian vs. the Death Star: how much energy will they actually need to disintegrate the Earth?

According to a detailed analysis by Dave Typinski, Marvin the Martian’s Illudium Q-36 Explosive Space Modulator will require $1.711 \cdot 10^{32}~\text{J}$ to shatter the Earth into a gravitationally unbound primordial dust cloud. However, this energy is over 30% lower than the gravitational binding energy derived from the average density model estimated at $2.24 \cdot 10^{32}~\text{J}$. Which seems irrational: if the Earth were in a lower energy state at a uniform density distribution, how did its density gradient evolve in the first place – wouldn’t it be evolving to a homogenous density distribution in that case?

Curtis Sexton and Wikipedia seem to agree with my reasoning on this point: according to Curtis Saxton’s calculations the Death Star’s beam weapon output would have to exceed $2.4 \cdot 10^{32}~\text{J}$ to impart escape speed to all of the matter comprising an Earth-like planet.

They can’t both be right, so who wins? My money’s on the Death Star, but I haven’t found the mistake in Dave’s calculations for Marvin yet. Does anyone here have the analytical chops to see where he went wrong, or alternatively, where I’ve spaced out?

Here's Dave's mathematical estimate for Marvin:

First he offers an equation for compiling the layers of various densities from the PREM data into a sum of

where $r$ is the radial distance from the Earth’s center and

the various constants are noted below:

$i$. Layer; Height $h_i$ (m); $a_i$ ($kg \cdot m^{-5}$); $b_i$ ($kg \cdot m^{-4}$); $c_i$ ($kg \cdot m^{-3}$)

1 Inner core; 1.2215×10$^{6}$; -2.1773×10$^{-10}$ 1.9110×10$^{-8}$ 1.3088×10$^{4}$

2 Outer core; 3.4800×10$^{6}$; -2.4123×10$^{-10}$; 1.3976×10$^{-4}$; 1.2346×10$^{4}$

3 D'' layer; 3.6300×10$^{6}$; 0.00; -5.0007×10$^{-4}$; 7.3067×10$^{3}$

4 Lower Mantle; 5.7010×10$^{6}$; -3.0922×10$^{-11}$; -2.4441×10$^{-4}$; 6.7823×10$^{3}$

5 Inner transition zone 1; 5.7710×10$^{6}$; 0.00; -2.3286×10$^{-4}$; 5.3197×10$^{3}$

6 Inner transition zone 2; 5.9710×10$^{6}$; 0.00; -1.2603×10$^{-3}$; 1.1249×10$^{4}$

7 Outer transition zone; 6.1510×10$^{6}$; 0.00; -5.9706×10$^{-4}$; 7.1083×10$^{3}$

8 Low velocity zone & lid; 6.3466×10$^{6}$; 0.00; 1.0869×10$^{-4}$; 2.6910×10$^{3}$

9 Inner crust; 6.3560×10$^{6}$; 0.00; 0.00; 2.9000×10$^{3}$

10 Outer crust; 6.3680×10$^{6}$; 0.00; 0.00; 2.6000×10$^{3}$

11 Ocean; 6.3710×10$^{6}$; 0.00; 0.00; 1.0200×10$^{3}$

Using the real density distributions for each shell layer:

And calculating for each point $h$ at any given radial distance from the outside in, with the inner and upper bounds $h_1$ and $h_2$ respectively, gives:

So for each $i^{th}$ layer:

And for each point at a radial distance of $r$ we have:

where $ΔU_0$ = 0 and the indexed constants for each piece of the function are described in the table above.

This yields:

-1.711×10$^{32}~\text{J}$

Many thanks to anyone who can explain how this came out wrong!

mathematics - Number Theory class

A magician said to an audience member:

• Pick a number,


• Double it,


• Add $7$,


• Multiply it by $5$,


• Subtract the number you started with,


• Remove any non-zero digit from the answer,

and then tell me the remaining digits in any order.

The audience member said

$6$ and $8$.

And the magician announced,

The digit you removed was $3$, right?

The magician is right! Why? Justify!

homework and exercises - How to derive an equation for the mass of a pendulum bob?

How to derive an equation for the mass of a pendulum bob?

If anyone knows of a resource that shows this (I've looked) or could explain how to derive an equation for the mass of a pendulum would be much appreciated.

EDIT: I'm talking about a simple pendulum. Some more context may be given with the following example:

Some mass $M$ is release from rest at a horizontal position. $M$ reaches the bottom of its path (so directly under the pivot) at a velocity of 2.2m/s. I understand that I can derive for instance, the length of the rope attached to the bob as $L = v^2/2g$. So I can find the length of the rope. Now with no further information is it possible to derive an equation that can be used to describe the mass of this bob?

quantum mechanics - Free particle propagation amplitude calculation

I have a quick calculational question.

In Peskin and Schroeder, Chapter 2, they want to look at the amplitude for a particle to propagate between two arbitrary points, $x$ and $x_0$, in an arbitrary amount of time t given the Hamiltonian $\sqrt{p^2c^2+m^2c^4}$. To do this, we look at the inner product of the time-evolved particle that was at $x_0$ with the $x$ eigenket: $\langle \vec{x}|\exp(-i\hat{H}t/\hbar)|\vec{x_0}\rangle$, correct?

If we insert two complete sets of momenta, the integral becomes (give or take factors of $\hbar$):

$\frac 1 {(2\pi)^3}\int d^3p\; \exp(-it\sqrt{p^2c^2+m^2c^4}/\hbar)\exp(i\vec{p}\cdot(\vec{x}-\vec{x_0}))$

I'm not quite sure where to proceed from here... Peskin and Schroeder somehow reach:

$\frac 1 {2\pi^2 |x-x_0|}\int_0^\infty dp\;p\sin(p|x-x_0|)\exp((-it\sqrt{p^2c^2+m^2c^4}/\hbar)$.

To get here, however, it seems that you would have to assume that $p$ points in the same direction as $x-x_0$ to get spherical symmetry or something. Why can we do this? Aren't we summing over all possible momenta?

Thanks!

Answer

Work with spherical coordinates in momentum space and choose $z$-axis in $p$-space along $\vec{x}-\vec{x_0}$. Then $$\vec{p}\cdot(\vec{x}-\vec{x_0})=p|x-x_0| \cos\theta$$

Then your integral $$ \frac 1 {(2\pi)^3}\int d^3p \exp(-it\sqrt{p^2c^2+m^2c^4}/\hbar)\exp(i\vec{p}\cdot(\vec{x}-\vec{x_0})) \\ = \frac 1 {(2\pi)^3}\int_0^{\infty} p^2 dp \int_0^{2\pi} d\phi \int_{-1}^{1} d(\cos\theta) \exp(ip|x-x_0| \cos\theta) \exp(-it\sqrt{p^2c^2+m^2c^4}/\hbar) $$

Now do the $\phi$ and $\cos\theta$ integrals to get $$ \frac 1 {4\pi^2 \, i \, |x-x_0|}\int_0^{\infty} p dp \Big[ \exp(ip|x-x_0|) - \exp(-ip|x-x_0|) \Big] \exp(-it\sqrt{p^2c^2+m^2c^4}/\hbar) $$

Recognize that the quantity in the square brackets is $2i\sin(p|x-x_0|)$ to get your answer.

quantum mechanics - Is double-slit experiment dependent on rate at which electrons are fired at slit?

I am a mathematician and I am studying string theory. For this purpose I studied quantum theory. After reading Feynman's book in which he described the double-slit experiment (Young's experiment) I was wondering if I send one electron per day or per month (even more), could I see the interference pattern?

Answer

Yes, the interference pattern will occur, although you'll have to wait a while to be able to see it. As long as the average arrival time between photons is markedly greater than the travel time from slit to detector, the actual rates don't matter - each photon interacts with the slits by itself.

This URL shows such an experiment, in which a laser beam was so attenuated that the separation between photons was in the kilometer range, while the target-detector distance was in the meter range, and an image intensifier was used to detect photon positions. After about 500,000 photons had been detected, the result was

visible light - Only sea water appears blue in color, why this is not happening in river water?

Is the salt in the water the reason for scattering sunlight into blue?

Answer

The reason is not salt water. Large masses of water seem to be blue, both oceans and swimming pools.

Water absorbs most of the red wavelength. Most materials have their color from their surface level valence electron's absorption and re-emission processes. In water, it is not like that, because most of the color is caused by vibration of the molecules. Because of the photons get inelastically scattered (instead of just being absorbed), part of their energies get transformed into the vibrational energies of the molecules.

Now when a photon interacts with an atom, three things can happen:

1. 
   

   elastic scattering, Rayleigh scattering, when the photon keeps its energy and changes the angle

   
   


2. 
   

   inelastic scattering, the photon gives part of its energy to the atom and changes the angle

   
   


3. 
   

   absorption, the photon gives all its energy to the atom, and the absorbing electron moves to a higher energy level.

   
   

In the case of large masses of water, three things cause the color:

1. 
   

   elastic scattering, Rayleigh scattering, on the surface of the water, sunlight is reflected and appears to be blue from certain angles

   
   


2. 
   

   inelastic scattering, part of the photons' energies transfers into the energies of the vibrational energies of the molecules. This means, that because of Hydrogen bonding, the red color wavelength photons will be shifted to blue color wavelengths.

   
   


3. 
   
   

   water can absorb and re-emit light too, but because of the high ratio of refraction and reflection in water, the ratio of photons absorbed is low.

   
   

This gives a body of water when looking through it a blue color. But then comes the question, why is a body of water when looking at it (not through) blue? Light scattering by suspended matter is required for the color in this case, and the blue light needs to return to the surface to be visible. Such suspended matter can also shift the scattered light to green color, often seen in rivers. The answer to the question of why rivers appear green instead of blue is because of those three factors have different effects on different types of masses of water.

Pools appear usually more colorless, but still blue, and not green. This is because the effects come in this order of importance:

1. 
   

   reflection of the bottom of pool is dominant

   
   


2. 
   
   

   reflected light from the surface (blue) is less dominant, and vibrational shifts are less dominant than the color of the bottom

   
   


3. 
   

   reflected light from suspended matter (green) is less dominant

   
   


4. 
   

   angle of observation less dominant

   
   

Oceans appear bluer because:

1. 
   

   reflected light from the surface is dominant and vibrational shifts

   
   


2. 
   

   angle of observation (and depth) is important because oceans look blue from far away, when deep enough (so that the bottom is not reflecting)

   
   


3. 
   

   reflected light from suspended matter is not so dominant because oceans tend to be clear

   
   

Rivers, shallow lakes:

1. 
   

   Reflection from suspended matter (green) is important because shallow lakes and rivers tend to have more sand and dust then oceans and pools

   
   


2. 
   

   Reflection of the bottom is important because they are shallow, and the bottom is not blue (more gray like sand)

   
   


3. 
   
   

   angle of reflection is important because lakes from far away and certain angles appear bluer when the bottom is not visible and surface reflection becomes dominant

   
   


4. 
   

   Vibrational shift become less dominant because the suspended matter is more dominant and the water is not deep enough

   
   

rotational kinematics - Proof of constant angular velocity in rigid body motion

I'm studying rigid body motion on Landau but I'm having troubles to understand this proof of the fact that the angular velocity $\vec{\Omega}$ is constant for a rigid body.

My doubt is about the two last equations in the last two lines of the text. If I use the two I get

$\vec{V'}=\vec{V}+(\vec{\Omega}-\vec{\Omega'})\times \vec{r'} +\vec{\Omega} \times\vec{a}$

How are (31.3) derived from this?

Answer

You have

$$ \boldsymbol{v} = \boldsymbol{V'} + \boldsymbol{\Omega'} \times \boldsymbol{r'} $$

and

$$ \boldsymbol{v} = \boldsymbol{V} + \boldsymbol{\Omega} \times (\boldsymbol{r'}+\boldsymbol{a}) $$

You collect the $\boldsymbol{r'}$ terms

$$ \boldsymbol{v} = \boldsymbol{V'} + \boldsymbol{\Omega'} \times \boldsymbol{r'} = \left( \boldsymbol{V}+ \boldsymbol{\Omega} \times \boldsymbol{a} \right) + \left( \boldsymbol{\Omega} \times \boldsymbol{r'} \right) $$

which is solved uniquely when

$$ \begin{align} \boldsymbol{\Omega'} & = \boldsymbol{\Omega} \\ \boldsymbol{V'} &= \boldsymbol{V}+ \boldsymbol{\Omega} \times \boldsymbol{a} \end{align} $$

newtonian mechanics - Is it possible for the Normal Force to do work?

Can a normal force do work?

Note: work = $Fd\cos(\theta)$, where $F$ is the magnitude of the force, $d$ is the magnitude of the displacement, and $\theta$ is the angle between the two.

Attempt at an answer:

My textbook's answer:
No the normal force does not cause an object to be displaced, it's perpendicular to the direction of motion.

My logic:

But what if the displacement is in a vertical direction, so now the normal force with is at either 0 degrees or 180 degrees? For example, in an elevator if it is accelerating upwards doesn't the normal force do work? Or for example, when someone jumps up he pushes down on the ground, then by Newton's third law the ground pushes back up, so that's a normal force that causes him to be displaced.

Which answer is correct and if my examples are wrong what is wrong in them?

Answer

In the elevator example, you are correct. The elevator floor does work on you as you stand in a rising elevator.

In the jumping example, the floor is not doing work. After all, it is just a floor, it has nowhere to get the energy from to do work on you. (The floor of the elevator gets it from the motor running the elevator.) When you jump, the force from the floor on your shoes increases, but there is no displacement between the floor and your shoes until your shoes leave the floor, and at that point there's no force any more. Your torso is moving upwards while you jump, but there's no force from the floor on your torso, so the floor isn't doing work. All the energy of a jump comes from internal energy, not external work.

In fact, the floor does a bit of negative work when you jump, taking energy away from you, because it deflects a bit, moving down as you jump up.

When you lift an object off a table, the table does a tiny bit of work on the object for a similar reason. When the object sits on the table, the table deforms a bit. As you lift, the table goes back to its original shape, so there's a small bit of force*distance work being done. The deflection is usually so small as to be unnoticeable. On a trampoline it is much larger; enough to make jumping a very different experience.

quantum mechanics - What is the usefulness of the Wigner-Eckart theorem?

I am doing some self-study in between undergrad and grad school and I came across the beastly Wigner-Eckart theorem in Sakurai's Modern Quantum Mechanics. I was wondering if someone could tell me why it is useful and perhaps just help me understand a bit more about it. I have had two years of undergrad mechanics and I think I have a reasonably firm grasp of the earlier material out of Sakurai, so don't be afraid to get a little technical.

Answer

I will not get into theoretical details -- Luboš ad Marek did that better than I'm able to.
Let me give an example instead: suppose that we need to calculate this integral:

$\int d\Omega (Y_{3m_1})^*Y_{2m_2}Y_{1m_3}$

Here $Y_{lm}$ -- are spherical harmonics and we integrate over the sphere $d\Omega=\sin\theta d\theta d\phi$.

This kind of integrals appear over and over in, say, spectroscopy problems. Let us calculate it for $m_1=m_2=m_3=0$:

$\int d\Omega (Y_{30})^*Y_{20}Y_{10} = \frac{\sqrt{105}}{32\sqrt{\pi^3}}\int d\Omega \cos\theta\,(1-3\cos^2\theta)(3\cos\theta-5\cos^3\theta)=$

$ = \frac{\sqrt{105}}{32\sqrt{\pi^3}}\cdot 2\pi \int d\theta\,\left(3\cos^2\theta\sin\theta-14\cos^4\theta\sin\theta+15\cos^6\theta\sin\theta\right)=\frac{3}{2}\sqrt{\frac{3}{35\pi}}$

Hard work, huh? The problem is that we usually need to evaluate this for all values of $m_i$. That is 7*5*3 = 105 integrals. So instead of doing all of them we got to exploit their symmetry. And that's exactly where the Wigner-Eckart theorem is useful:

$\int d\Omega (Y_{3m_1})^*Y_{2m_2}Y_{1m_3} = \langle l=3,m_1| Y_{2m_2} | l=1,m_3\rangle = C_{m_1m_2m_3}^{3\,2\,1}(3||Y_2||1)$

$C_{m_1m_2m_3}^{j_1j_2j_3}$ -- are the Clebsch-Gordan coefficients

$(3||Y_2||1)$ -- is the reduced matrix element which we can derive from our expression for $m_1=m_2=m_3=0$:

$\frac{3}{2}\sqrt{\frac{3}{35\pi}} = C_{0\,0\,0}^{3\,2\,1}(3||Y_2||1)\quad \Rightarrow \quad (3||Y_2||1)=\frac{1}{2}\sqrt{\frac{3}{\pi}}$

So the final answer for our integral is:

$\int d\Omega(Y_{3m_1})^*Y_{2m_2}Y_{1m_3}=\sqrt{\frac{3}{4\pi}}C_{m_1m_2m_3}^{3\,2\,1}$

It is reduced to calculation of the Clebsch-Gordan coefficient and there are a lot of, tables, programs, reduction and summation formulae to work with them.

quantum mechanics - Slit screen and wave-particle duality

In a double-slit experiment, interference patterns are shown when light passes through the slits and illuminate the screen. So the question is, if one shoots a single photon, does the screen show interference pattern? Or does the screen show only one location that the single photon particle is at?

Answer

The answer is yes to both questions: yes, the screen does show one location for one particle and yes, the accumulated picture after repeating the experiment many, many times does show the interference pattern.

There is a set of beautiful pictures and a video of the double slit experiment in one-particle-per-time mode that can be found here (the experiment is with electron but conceptually there is no difference).

visible light - Is black a color or absence of color?

Is black a color or absence of color? When there is no light, everything is black, so how can we say that black is a color?

Answer

Black is the absense of color, which our brain visualizes to us in some way. I can't really say that I know how you percieve black. But it IS a fact that when no photons reach the photoreceptor cells in your eye, your brain will get a distinct value from it, and that value will be different than when it receives red or blue or any other wavelength of light.

The value that your brain reads from your photoreceptor cell will be different for black than for blue - but it DOES get a value. I guess that value would be 0, but it might just as well be 1 or 42 - who knows? :-) So in that sense, I would say that black is indeed a color.

In physics, it is a bit different. It makes no sense talking about colors in physics. In physics, you talk about wavelengths as a continous spectrum of wavelengths from very low frequency to very high frequency. Either there is a wavelength, or there is no wavelength.

Humans have named certain ranges of these wavelengths as colors red, green, blue, violet etcetera - because it makes it easier to communicate.

There is no range of wavelengths in the spectrum that has the name "black" - and in physics "black" means that there is no wave at all. At the same time, there is no wavelength that has the name "white", because white is a combination of all color frequencies at the same time. Our eyes give white another distinct value, in the same way as black.

No object (except perhaps black holes) are completely black. Warm objects radiate low frequency light, and can never be "black". Your photoreceptor cells might give your brain the same value as for black, for many differenct physical colors - but no objects are actually black.

So the short version of the answer is:

Your brain probably observes black the same way it observes any other color - as a certain "value" read off a photoreceptor cell - and in that sense, it is a color. In physics, nothing is black.

gravity - How do gravitons impact on general relativity?

As I'm reading about GR a lot lately, I was wondering: how do gravitons (if they exist ofc) impact the general relativity?

Since in GR, when we look at particles moving in space-time, we are only looking from geometric point of view so to say. Since gravity is represented with curvature of space-time via Einstein equations, we don't say gravity is a force that influences on bodies, gravity is just curvature affecting the bodies.

So if there is a graviton (gravitons) which would be mediators of gravity as a force within or not within the Standard Model, how would this be reconciled with the view of gravity as a curvature of space and time?

I would guess that this kinda question was asked by some scientists and answered, but I never really read anything on it. I don't even remember seeing gravitons mentioned in standard books about GR.

Are there any explanations about it?

Answer

How do gravitons impact on general relativity? [...] I would guess that this kinda question was asked by some scientists and answered...

It's been asked but never answered satisfactorily. The full impact on general relativity would be that it would become a theory of quantum gravity. Nobody has ever been able to construct a satisfactory theory of quantum gravity. We have various guesses, and we can reason by analogy with other fields such as the electromagnetic field, but basically not much is known.

logical deduction - A couple of couples

Two couples have babies simultaneously in different hospitals. They don't know the expected gender, and each have one child which is F or M. How can they arrange in advance, given a guess each on birth, that at least one couple will correctly guess the gender of the other's child.

_{Source: This is based on Ed Felten's Coin toss}

Answer

I am interpreting

that each couple can wait to see what gender their own child is before making a guess about the other couple's child. And that "each have one child which is F or M" to exclude twins and intersex.

Then it seems this works:

The first couple guesses that the second couple's baby has the same gender as their own, and the second couple guesses that the first couple's baby has the opposite gender as their own? Exactly one guess will be correct.

How electric field can be less than force per unit test charge?

I have been going through some physics objective questions and struck with this one Q:A negatively charged metallic ball is supported on a rigid insulating stand. We wish to measure the electric field E at a point P in the same horizontal level as that of the metallic ball. To do so, we put a positive charge q and measure F/q. The electric field at the point P is
a) =F/q
b) <(F/q)
c) >(F/q)
d) none of these.
Guess what?. I chose (a) and its wrong. (b) is correct. I am squeezing my brain. How can that be correct.? Please give detailed explanation.

Answer

The field strength is only $F/Q$ if the test charge is infinitesimally small so it cannot affect the metal ball.

For example, suppose the metal ball is uncharged in which case the fortce should be zero. A positive charge will polarise the ball and create an attractive force even though the charge on the ball is zero. In the case of a charged ball the positive charge will polarise the ball and increase the atttractive force.

electrostatics - Is charge point-like or a smear?

Coulomb gave the law for the force between two static charges while considering them to be points in space. But the differential form of Gauss' Law talks about charge densities, a thing possible only if charges are smeared out in space.

Even Feynman addresses to the problem in his lectures when he says that on solving for the electrostatic energy in the field of a point charge we get infinity as the limit.

So do we know now that whether charges are point-like or smeared out?

quantum mechanics - Where do symmetries in atomic orbitals come from?

It is well established that:

'In quantum mechanics, the behavior of an electron in an atom is described by an orbital, which is a probability distribution rather than an orbit.

There are also many graphs describing this fact: http://en.wikipedia.org/wiki/Electron: (hydrogen atomic orbital - one electron) In the figure, the shading indicates the relative probability to "find" the electron, having the energy corresponding to the given quantum numbers, at that point.

My question is: How do these symmetries shown in the above article occur? What about the 'preferable' axis of symmetries? Why these?

Answer

The hydrogen atom is spherically symmetric, so for any solution of the Schrödinger equation for the hydrogen atom, any rotation of that solution must also be a solution. If you do the math on how to rotate a solution, it turns out that the solutions with a particular energy $E_n$ fall into groups labeled by an integer $l < n$. The integer $l$ is physical: $\hbar^2 l(l+1)$ is the magnitude squared of the angular momentum. Within each group, rotating the solution gives you a new solution in the same group. These two facts are of course connected: a rotation can't change the length of a vector.

One can show that each group contains $2l+1$ independent solutions, in that any solution $|n,l\rangle$ where the energy is $E_n$ and the angular momentum $\hbar^2 l(l+1)$ can be written as a sum $$|n,l\rangle = \sum_{m=-l}^l c_m |n,l,m\rangle$$ (I apologize for the somewhat poor notation.)

This decomposition is based on choosing a particular axis, and taking each state to depend on the angle $\varphi$ around this axis as $e^{im\varphi}$. The appearance of axes of symmetry in these plots is due to this choice of axis and particular decomposition. With another choice of axis, which is the same as a rotation, the states will be mixed.

The bottom line is that it's not each solution -- wavefunction -- that needs to be spherically symmetric, but the total set of solutions.

quantum computer - No cloning theorem and exclusive-or (XOR) operator

According to IBM's website,

[...]where we would [classically] have done an assignment (x=y), we instead initialize the target (x=0) and use exclusive or (x^=y).

This sounds like x is a copy (clone) of y, however cloning is impossible in quantum mechanics. What's going on?

Answer

This is a standard trick in quantum computing: to allow reversibility when computing some function $f$, you keep all the inputs and you encode the result in an initially "blank" qubit set initially to $|0\rangle$. What you're describing is the simplest case: one single input qubit, and $f$ equal to the identity function.

Thus, you want your gate to take the state $|0\rangle|0\rangle$ (where the first qubit is the input and the second is blank) to $|0\rangle|0\rangle$ (i.e. keep the first one and compute on the second one), and to take $|1\rangle|0\rangle$ into $|1\rangle|1\rangle$. As the IBM website points out, one can do this with a CNOT gate controlled on the first qubit.

The no-cloning theorem, however, is not broken. Consider what happens when you input some general state $|\psi\rangle=a|0\rangle+b|1\rangle$: by linearity, the output will be $$a|0\rangle|0\rangle+b|1\rangle|1\rangle.$$ This is an entangled state, and definitely not equal to the cloned state $|\psi\rangle|\psi\rangle$.

The way to interpret this is that the standard classical logic gates - assignment, OR, AND, etc. - are indeed implementable exactly as in (reversible) classical computing for the computational basis, but their behaviour on a general state is then fixed by linearity and must then be examined to see what it comes out as. Thus "assignment" works on the computational basis but not for a superposition state. Other gates like AND or OR don't even make much sense as such if their inputs are in (separable) superpositions or even entangled.

electromagnetic radiation - Are there weak force waves?

In the same way as there are electromagnetic and gravitational waves that update the information on their respective field, is there an analogue for the weak and strong forces?

Answer

No, there are no weak or strong waves in the sense as there are for electromagnetic or gravitational waves.

The electromagnetic and gravitational waves are classical objects, they are possible vacuum solutions to the classical equation of motion for the field strength of the respective force, and can be radiated by objects charged under the respective force. But the weak and the strong force have no analogous classical limits - the weak force is suppressed by a factor $\mathrm{e}^{-\mu r}$ due to the mass of the W- and Z-bosons and thus very different from the EM or gravitational force, and it doesn't make sense to speak of a classical limit of the strong force because gluons and quarks are confined - there are no net charges under the strong force on a classical level, hence the strong force just vanished from our description.

In other words, the weak and the strong force are, in some sense, "fully quantum" in that their importance to our world comes completely from their quantized description, and a classical description does not make sense for them, thus we cannot speak of a classical concept such as a wave for the weak and strong forces.

electromagnetism - Is there a travelling speed of for electric field? If yes, what is it?

Consider two circuits containing a battery, bulb, switch and conducting wires but of length 1 meter in one case and 1000KM in other. When switched on at the same time both the bulbs glow instantaneously. I think it is because, electric filed travels from positive terminal of battery to negative terminal through wire almost instantaneously.

My question is there any speed for this electric field? If yes, what is it? By any modification in this circuit is it possible to induce any delay in glowing the bulb?

Answer

There are two different questions at work here, that you've kind of mashed together. The first question is "What is the speed at which a change in the electric field propagates?" The answer to that is the speed of light. In QED terms, the electromagnetic interaction that we see as the electric field is mediated by photons, so any change in an established field (say, due to shifting the position of the charge creating the field) won't be felt by a distant object until enough time has passed for a photon from the source to make it to the observation point.

The second question is "What is the speed of propagation of electric current?" This speed is slower than the speed of light, but still on about that order of magnitude-- the exact value depends a little on the arrangement of wires and so on, but you won't be far off if you assume that electrical signals propagate down a cable at the speed of light.

This relates to electric field in that the charge moving through a circuit to light a light bulb has to be driven by some electric field, so you can reasonably ask how that field is established, and how much time it takes. Qualitatively, the necessary field is established by excess charge on the surface of the wires, with the surface charge being generally positive near the positive terminal of a battery and generally negative near the negative terminal, and dropping off smoothly from one to the other so that the electric field is more or less piecewise constant (that is, the field is the same everywhere inside a wire, and the field is the same everywhere inside a resistor, but the two field values are not the same).

When the circuit is first connected, there is a rapid redistribution of the charge on the surface of the wires which establishes the surface charge gradients that drive the steady-state current that will eventually do whatever it is you want it to do. The time required to establish the gradients and settle in to the steady-state condition is very fast, most likely on the order of nanoseconds for a normal circuit.

There's a good discussion of the business of how, exactly, charges get moved around to drive a current in the textbook that we use for our introductory classes, Matter and Interactions, by Chabay and Sherwood. It doesn't go into enough detail to let you calculate the relevant times directly, but it lays out the basic science pretty well.

(It's a textbook for a first-year introductory physics class, so it sweeps a lot of condensed matter physics under the metaphorical rug-- there's no discussion of band structure or surface modes, or any of that. It's fairly solid conceptually, though, at least according to colleagues who know more about those fields than I do.)

Two electrons exchanging a photon

An electron can emit a real photon, which can be absorbed by another electron. Thus two electrons can exchange a real photon. However, when two electrons repulse, the math of their interactions is described by a virtual photon. Thus it seems that two electrons can exchange either a real or virtual photon (at least conceptually). What conditions define whether the photon that two electrons exchanged was real or virtual?

Answer

An electron can emit a real photon,

Real photon means it can be measured/detected. For example in this single photon at a time experiment, the dots are measured photons, and the photons are real.

which can be absorbed by another electron.

Electrons do not absorb photons. They interact with photons. For example in Compton scattering:

A photon interacts with the electron, the electron becomes off shell because it absorbs part of the four momentum of the incoming photon and a lower energy photon leaves.

Thus two electrons can exchange a real photon.

Wrong . If the second photon hits another electron, the process will be repeated. It is not an exchange but a scattering.

quantum mechanics - Sine function as a momentum eigenfunction

I'm confused as to whether the sine function can be technically considered an eigenfunction for momentum operator.

Once the sine function is decomposed, it can be decomposed as a linear sum of two eigenfuntions for the momentum operator since $$\sin(kx)=\frac{1}{2i}\left[e^{ikx}+e^{-ikx}\right] \, .$$ Applying the momentum operator on each of these functions gives $hi$ and $-hi$ for the momentum values. However, applying the momentum operator on $\sin(kx)$ itself clearly shows that it is not an eigenfunction one cannot recover the original sine function after applying the operator. Why is sine not an eigenfunction despite the fact that it is a linear sum of two eigenfunctions?

Also, for a particle in a box, one can use its wavefunction (a sine function $\sqrt{2/L} \sin(n\pi x/L)$) to calculate the average momentum as 0. If sine is not an eigenfunction for momentum, how can we calculate the average for the momentum observable?

pattern - What is an Inflated Word™?

_{This is in the spirit of the What is a Word/Phrase™ series started by JLee with a special brand of Phrase™ and Word™ puzzles.}

If a word conforms to a special rule, I call it an Inflated Word™.

Use the following examples below to find the rule.

And, if you want to analyze, here is a CSV version:

Inflated Words™,Not Inflated Words™
BAYOU,SWAMP
DIARY,JOURNAL
EASILY,SIMPLY
EATERY,DINER
FLAY,WHIP

FORAY,PILLAGE
MAY,APRIL
NEATH,UNDER
PRAY,IMPLORE
SOAPY,SUDSY
SOOTY,DIRTY
TUSK,HORN
WHEY,CURDS
SHOYU,SOY

Answer

The answer is:

If you remove all the vowels from the words (not including Y) they still make a word.

And the extra inflated:

Even if you remove the Y, it makes a word

Work and Kinetic energy rely on each other which came first?

In all the places I've looked kinetic energy is derived from the definition of work, but I don't understand how the definition of work was developed with out the concept of kinetic energy. My question really is:

What is the logic behind the definition of work without referring to kinetic energy?

OR

What is the logic behind the definition of kinetic energy without referring to work?

Answer

Imagine for a moment that you don't know anything about work or energy, you only know Newtons second law that introduces the notion of a force and that it is related to acceleration. You also know that there are different kind of forces like gravity or the electric force. Then, by some ingenious strike of yours you say lets multiply force by distance and lets call the resulting quantity energy (or work). You then set forth to explore how this quantity behaves in various situations, for example you could compute the energy of an accelerated particle. Because this will happen really often you call the result kinetic energy. Then imagine a body, subject to gravitation and you compute the energy again and call the result potential energy. You get the idea. In our derivation we set up force as the first principle and build the notion of energy on top of it.

It now so happens that you could go the other way round, i.e. you can set up the concept of energy as your first principle and define kinetic energy by 0.5mv^2 and the potential of a gravitating body and from this derive the concept of a force by energy/distance. Most modern theories go this way but both approaches are valid.

I hope that helps.

---

Edit: The important thing with both approaches is that we observe the thing that we call energy to be conserved in nature (It is a very strong observation since so far we've never found any experiment to violate it. We call it the 1. law of thermodynamics). We could e.g. define energy = force times distance ^2. We would get another result for kinetic energy and potential energy of course. But then we won't find that energy is conserved

particle physics - Is the Standard model an effective field theory (EFT)?

I've seen both positive and negative answers to this question, though most part of the community seem to agree it can be said it is an EFT up to the electroweak scale. My question is: What are the main arguments from each side? References are welcome too.

Answer

Well, the Standard Model is definitely an effective field theory description of physical reality since it neglects both gravity and neutrino masses.

It's also an effective QFT in the sense that there's a sensible definition of it on the lattice. (Not trivial, because there are chiral fermions.)

But I think what you're really asking is: Does the specific QFT we currently call the Standard Model have a continuum limit? That's an open question in theoretical physics. I think most theorists lean towards "no", because hypercharge and quartic Higgs couplings have beta functions with the wrong sign in the perturbative regime. Absent a miracle, this means that those sectors have Landau poles, making it impossible to find a continuum limit with interactions in the IR. (A Landau pole means that if you adjust the lattice coupling as you refine the lattice in order to keep the long-distance physics fixed, you'll see the lattice coupling run off towards infinity at a finite lattice scale.)

But this isn't proof: the perturbative approximation becomes unreliable as one approaches the region where the Landau pole would happen. Numerical computation with lattice $\phi^4$ theory also suggests that there's a Landau pole. However, those calculations aren't under analytic control; we don't know that we're not missing something.

On the other hand, I don't know of any compelling evidence that the Standard Model might not have a Landau pole.

fluid dynamics - Terminal velocity of a steel ball in water

I am investigating on the terminal velocity of steel balls moving in water. I used the balls with different diameters, such as 3.17mm, 6.02mm, etc. And I used a full-filled 1L graduated cylinder. I dropped the ball below the water surface. And I calculate the terminal velocity by a program called Logger Pro.

Theoretically, the terminal velocity of steel ball with diameter 3.17mm in water is about 37m/s. But in my experiment, it is only 0.7m/s. The difference is really huge.

And also, I got a graph of terminal velocity as a function of square of radius of steel ball. It is a straight line, but it does not go through origin.

Can anyone help me with that?

Answer

Stokes Law is not going to apply in this situation because the water flow around the ball will be turbulent not laminar. The way to see this is to calculate the Reynold's number. For a sphere this is approximately given by:

$$ Re \approx \frac{\rho_wdv}{\mu} $$

If we feed in $\rho_w = 1000$ kg/m$^3$, $d = 0.00317$ m, $v = 37$ m/s and $\mu = 0.001$ Pa.s then we get $Re \approx 117290$. A Reynold's number of greater than one means the flow is turbulent, and since Re for a $3.17$ mm ball travelling at $37$ m/s is far greater than one we conclude that the flow is turbulent and Stoke's law doesn't apply.

In a turbulent regime the viscous drag is given by:

$$ F_d = \tfrac{1}{2}\rho C_d A v^2 $$

where $\rho$ is the density of the fluid (i.e. water), $A$ is the frontal area of the object, $v$ is the velocity and $C_d$ is a dimensionless parameter called the drag coefficient. The value of $C_d$ depends on the Reynold's number (so it isn't really a constant) but it's generally of order one.

word - What is the world?

Thought I'd get back into puzzle-making by giving Fortnightly Topic Challenge #22 a go.

---

As you work through one of your countless puzzle books, you find a sheet of paper inside with a poem on it. It's been a while, you think – almost a year in fact. You suppose your brother was just too busy chasing down his dreams to be trying to vex the puzzle addict that you are. Is this some sort of twisted Christmas present? You take a closer look...

The World

On the surface, the world seems to give you a choice:
Take the head of an ostrich? A grasshopper's tail?
Maybe eggs of a fish left beheaded on sale!
Is that wonder I hear coming from your posh voice?

But beneath the appearance, it looks to you old:
There you see a hyena's true heart on the ground,

With a donkey's hind feet lying back and around,
And the murmured assent is unstarted, untold.

At its core, though, there lies a most wonderful street
Where the secrets of Burgess's Toad are revealed,
And the wings of a swift are high over the field;
Just a second of time makes it all but complete.

What is your brother trying to tell you?

---

Note: don't read the poem too closely; a good number of words (especially rhyming words) are there only for the rhyme/meter.

Answer

In the poem, ...

... each stanza refers to some letters, defined several times in each of the four lines.

The first stanza ...

... encodes OR: a choice (this or that?); the head of an ostrich and a grasshopper's tail; roe reversed and with the first letter removed; a homophone of awe.

The second stanza ...

encodes YE: an old form of you; a hyena's true heart – I guess hyenas have their hearts towards the left side like humans too; The donkey's hind feet reversed; and aye without its first letter.

The third stanza ...

... encodes ST: an abbreviation for Street; the secrets of Burgess's Toad; the wings of the swift; s (second) and t (time)

The final answer ...

... is obtained from these letters. They could be treated as a simple anagram, but Gareth McCaughan has found how to construct the solution from the first words in each stanza:

OR is At the surface.

YE is Beneath the appearance.
ST is At its core.

The pairs are nested so that ST is the inner core, YE the intermediate layer and OR the outer layer:

(O (Y (ST) E) R).

A very appropriate construction for an OYSTER!

Finally, the title ...

... is a reference to a proverb, as Ankoganit points out in the comments below: My brother wanted to tell me that the world is my oyster.

black holes - Would the universe get consumed by blackholes because of entropy?

Since the total entropy of the universe is increasing because of spontaneous processes, black holes form because of entropy (correct me if I'm wrong), and the universe is always expanding, would the universe be consumed by black holes or would its area increase too quickly for that to happen?

Answer

Entropy isn't a force that causes things to happen.

But anyway, the answer is no. Not all matter in the universe is expected to eventually collapse into black holes. See Adams and Laughlin, http://arxiv.org/abs/astro-ph/9701131 , section VD.

Note also that black holes eventually evaporate, so when matter collapses into black holes, the result is that it eventually gets recycled into Hawking radiation, most of which is photons.

quantum mechanics - Periodic boundary condition on a Wave Function of a Particle in a Box

Until now, solving the Schrodinger Equation for a particle in a box was relatively easy because the boundaries conditions imposed zero value on the wave function at the boundaries. But now I must find the normalized wave function of the same problem imposing just these periodic boundaries conditions:

$$Y(x,y,z)=Y(x+L,y,z),\\ Y(x,y,z)=Y(x,y+L,z),\\ Y(x,y,z)=Y(x,y,z+L).$$

I got stuck in the normalization process. Before, using the boundary condition (one dimension) $Y(0)=Y(L)=0$, I could get just one constant to solve for, in

$$Y(x)= A\sin{kx} + B\cos{kx}.$$

Applying the conditions above, I get

$$Y(x)= A\sin{kx}\quad \text{where}\quad k=\pi n/L$$

which is easy normalize. But now, with this periodic boundary condition, $Y(0)=Y(L)$. How could I find it?

Answer

When imposing a periodic boundary condition, the amplitude of the wavefunction at coordinate $x$ must match that at coordinate $x+L$, so we have:

$$\Psi(x)=\Psi(x+L)$$

In your previous 'particle in a box' scenario, you mention that the general form of the wavefunction is given by a linear combination of sine and cosine with complex coeficients. It might be helpful to remember that this can also be expressed as an exponential with the form:

$$\Psi(x)\propto e^{ikx}$$

Hopefully that should get you off the starting block.

quantum field theory - Haag's theorem and practical QFT computations

There exists this famous Haag's theorem which basically states that the interaction picture in QFT cannot exist. Yet, everyone uses it to calculate almost everything in QFT and it works beautifully.

1. 
   

   Why? More specifically to particle physics: In which limit does the LSZ formula work?

   
   


2. 
   

   Can someone give me an example of a QFT calculation (of something measurable in current experiments, something really practical!) in which the interaction picture fails miserably due to Haag's theorem?

   
   

quantum field theory - Is it possible that the Big Bang was caused by virtual particle creation?

As far as I understand, it is understood that throughout the universe there exists, what is known as, a quantum field from which, due to its fluctuations, temporary (pairs of) virtual particles continuously appear in a random, unpredictable (or should I say probabilistic?) fashion.

For my idea to have any viability, I am assuming that this quantum field is an intrinsic part of reality, that existed "before" or rather at the time of the Big Bang already. Is this a correct assumption? Or is it believed that this quantum field was "formed" at the time, or perhaps even later than, the moment of the Big Bang? I hope not.

Assuming the former, and also understanding that those pairs of virtual particles typically annihilate each other almost instantaneously, but sometimes actually create real particles — for instance in Hawking radiation — is it a far stretch to think that the Big Bang was possibly started by the highly improbable (but in an "infinite" timespan of underlying reality, likely to occur), cataclysmic event of a huge amount of virtual particle all appearing, either "at the same time", or in such a sequential manner, that they couldn't annihilate each other anymore and were destined to actually form large quantities of real particles?

special relativity - What would an observer see if he/she flew toward a clock at relativistic speeds?

If an observer approaches a clock at a significant fraction of the speed of light, would they see the clock's hands moving at a faster or slower than usual rate?

I figure there are two competing effects at play - time dilation and diminishing distance.

Answer

You can see this as an example of the relativistic Doppler shift (for equations, see eg: http://en.wikipedia.org/wiki/Relativistic_Doppler_effect ).

The hands of the clock are moving with some angular frequency and you are moving with a velocity v towards the clock. It follows that the frequency you are seeing will be higher, thus the clocks hands will move faster.

Conceptually this makes sense. Suppose a picture of the clock is emitted each second. Since you're moving towards the clock, you will pick up one of those pictures more often than once per second, thus making the clock seem to run faster.

electromagnetism - Why does the divergence of the Poynting vector have energy flux density?

The poynting vector is defined as

$\vec{S}=\mu_{0}^{-1}\vec{E}\times \vec{B}$

Taking the divergence of the poynting vector, one arrives at

$\vec{\nabla} \cdot \vec{S}=-\frac{\partial u}{\partial t}=0$

after some algebraic manipulation.

Note $u$ is the electromagnetic energy density.

The claim is that the $\vec{\nabla} \cdot \vec{S}$ is an energy flux density.

How do I see this is true?

Thanks in advance.

statistical mechanics - Evaluating low-temperature dependence of the BCS gap function

How does one go about evaluating the behavior of the BCS gap $ \Delta = \Delta(T) $ for $ T \to 0^+ $ under the weak coupling approximation $ \Delta/\hbar\omega_D \ll 1 $?

In Fetter & Walecka, Quantum theory of Many-Particle Systems, Prob. 13.9 it is said that the starting point is $$\tag{1} \ln\frac{\Delta_0}{\Delta} = 2\int_0^{\hbar\omega_D}{\frac{\mathrm d\xi}{\sqrt{\xi^2+\Delta^2}}\frac{1}{e^{\beta\sqrt{\xi^2+\Delta^2}}+1}}, $$ which I have no problem deriving from the theory, but I can't find a way to actually evaluate this integral even under the approximations $ \hbar\omega_D \to \infty $, $ \Delta \approx \Delta_0 $ (in the RHS) and $ \beta\Delta \to \infty $. [Of course $ \beta = (k_BT)^{-1} $ and $ \Delta_0 = \Delta(T = 0)$.] I have tried several approaches, used different Taylor-expansions and changes of variables, but I am simply stuck.

For the record, the expected behavior is supposed to be $$\tag{2} \Delta(T) \sim \Delta_0\left(1 - \sqrt{\frac{2\pi}{\beta\Delta_0}}e^{-\beta\Delta_0}\right) .$$

EDIT: Just leaving it here for the posterity. I found a more complete way to tackle this integral; specifically, under the WC approximation one has $$ \int_0^{+\infty}{\frac{\mathrm d x}{\sqrt{x^2 + 1}}\frac{e^{-\beta\Delta\sqrt{x^2 + 1}}}{1 + e^{-\beta\Delta\sqrt{x^2 + 1}}}} = \int_1^{+\infty}{\frac{\mathrm d y}{\sqrt{y^2 - 1}}\frac{e^{-\beta\Delta y}}{1 + e^{-\beta\Delta y}}} = $$ $$ = \sum_{k=1}^{+\infty}\int_1^{+\infty}{\frac{\mathrm d y}{\sqrt{y^2 - 1}} (-1)^{k+1}e^{-k\beta\Delta y}} = \sum_{k=1}^{+\infty}(-1)^{k+1}\int_0^{+\infty}{\mathrm d t\; e^{-k\beta\Delta \cosh{t}}} = \sum_{k=1}^{+\infty}(-1)^{k+1} K_0(k\beta\Delta), $$

$ K_0 $ being the 0-order modified Bessel function of the second kind, whose asymptotic behavior is known and may be used to solve the problem in a relatively clean way (and even find the corrections at higher orders, which are $ \in O(e^{-\beta\Delta}(\beta\Delta)^{-k - 1/2}) $. Cf. Abrikosov, Gorkov, Dzyaloshinski, Methods of Quantum Field Theory in Statistical Phyisics, 1963. Pagg. 303-304.

Answer

Hints:

1. 
   

   Define difference $\delta:=\Delta-\Delta_0$. Deduce from $|\delta|\ll |\Delta_0|$ that the lhs. of eq. (1) is $$\tag{A}\text{lhs}~\approx~ -\frac{\delta}{\Delta_0}.$$

   
   



2. 
   

   Substitute $\xi=x\Delta $ in the integral on the rhs. of eq. (1). Deduce using $\hbar \omega_D \gg \Delta$ that the rhs. is $$\tag{B} \text{rhs}~\approx~ \int_{\mathbb{R}} \! \frac{dx}{\sqrt{1+x^2}} \frac{1}{e^{\beta\Delta \sqrt{1+x^2}}+1}. $$

   
   


3. 
   

   Deduce from $\beta\Delta\gg 1$ that we can simplify the rhs. further to a Gaussian integral $$\tag{C} \text{rhs}~\approx~ \int_{\mathbb{R}} \! dx~ e^{-\beta\Delta (1+\frac{1}{2}x^2)}~=~\sqrt{\frac{2\pi}{\beta\Delta}}e^{-\beta\Delta} . $$ Such arguments are closely related to the method of steepest descent.

   
   


4. 
   

   Deduce eq. (2).

   
   

measurements - How is a second measured? And why is it measured that way?

The earth's rotation around the sun isn't exactly 24 hours. It off by some seconds which becomes somewhere around 6 hours per year and 1 day in 4 years(leap year), which brings the question why didn't we modify the measurement of 1 second ever so slightly so as to avoid leap years altogether.

Well how is 1 second measured exactly? Wikipedia says

the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom. [1]

Well, why is it measured that way? Is there any technical reason like it is easy to measure and can be standardized easily? Or is it really possible to modify the measurement of time?

Answer

why didn't we modify the measurement of 1 second ever so slightly so as to avoid leap years altogether.

The rotation of earth and the revolution of earth around the sun is not at all synchronized. The Earth really rotates 365.24219647 times during each revolution (at 1992, this ratio changes slightly every year, the tropical year gets roughly a half second shorter each century); so even if we fixed the definition of time to the revolution of earth around sun, we will still need a leap year every 4 years (what we wouldn't need would be leap seconds).

Another reason is because precise time measurement would become incomparable. Since the period of rotation of earth (i.e. tropical year) isn't constant, if we used the definition of second to exactly match the period of revolution, then whenever you want to specify a precise duration of time, you'll also have to specify which year that definition of second is taken from, and you'll need a table that records the length of second of each year.

Is there any technical reason ... ?

Yes, because with the proper equipments anyone, anytime can take a caessium-133 atom, put it under the specified condition and measure the same second, and it won't have yearly change like second from earth rotation/revolution would. As far as we know, the frequency of caessium-133 in the 1978 should be the same as the frequency of another caessium-133 in 2049.

quantum mechanics - Why electrons can't radiate in their atoms' orbits?

It's an old-new question (I found only one similar question with unsatisfactory (for me) answer: Where did Schrödinger solve the radiating problem of Bohr's model?)

It's strange for me how all books simply pass by such an important question, and mentioning strange and mathematically unsupported reasons such as:

• 
  

  orbits are stationary (while as I know this is just idealization, there is no stationary orbits in reality even for Hydrogen)

  
  



• 
  

  electrons are actually not localized due to uncertainty principle, thus they have no acceleration (while obviously in a non-spherically symmetric orbits a kind of "charge acceleration distribution" always exist)

  
  


• 
  

  vacuum fluctuations play a major role (according to QED).

  
  

I'm not interested in how Bohr or Schroedinger explained it, I want to see a rigorous proof with QM, QED or maybe even the standard model as whole. I would like to see how this question was closed.

Answer

This question can be answered in the simple framework of non-relativistic quantum mechanics. The electron's electromagnetic charge's density and current — which are the source of the classical electromagnetic field — are given by the electron's probability density and current distributions $$\rho (t,x)=\psi^*(t,x)\,\psi(t,x)\,$$ $$j(t,x)\propto \psi^*(t,x)\,\nabla\psi(t,x)-\psi(t,x)\,\nabla\psi^*(t,x)\,.$$ As in a stationary state $\psi(t,x)=e^{-i\omega\, t}\,\phi(x)$, neither the density nor the current depend on time and therefore they don't emit electromagnetic energy, according to Maxwell equations with $\rho$ and $j$ as sources.

However, when one takes into account the quantum nature of the electromagnetic field, the probability of radiating a photon (quantum of the electromagnetic field) by an atom in a stationary state is different from zero due to the phenomenon of spontaneous emission.