electromagnetic radiation - Has someone measured that in the far field of radio waves, the electric and magnetic field oscillate synchronously?

That in near field both fields oscillating 90° displaced is obvious since in antenna rod a lot of electrons will be accelerated and and this oscillation of electric field is escorted by alignment of electrons magnetic dipole moments. Thus oscillation of magnetic field - if strong enough - produces the next oscillation of electric field and so on.

But somehow this two components of EM field were going over to a state - called far field - there both fields oscillation at the same time. In know that Hertz produces sparks and registered this modulation of EM radiation some meters away.

Does he or someone else measure the synchronicity of the two components with amplitudes not shifted 0° to each other and of course both perpendicular to the direction of propagation and not shifted by 90° like in the near field?

Near field https://upload.wikimedia.org/wikipedia/commons/e/e6/Photon_Spin_%2B1.PNG

Far field https://upload.wikimedia.org/wikipedia/commons/3/35/Onde_electromagnetique.svg

Edit because of the comment from KyleKanos: What I'm asking is, has someone measured, that at one moment of time the peak of magnetic component is in the same distance from source as the peak at the electric field. That should be not easy because this peaks are moving with c.

Answer

Google search for "antenna far field magnetic field measurement" does not match any result. Always the electric field is measured and from this is concluded about magnetic field. I could not see that there is any measurement of electric and magnetic fields at the same point to show that both fields in phase.

hamiltonian formalism - Book about classical mechanics

I am looking for a book about "advanced" classical mechanics. By advanced I mean a book considering directly Lagrangian and Hamiltonian formulation, and also providing a firm basis in the geometrical consideration related to these to formalism (like tangent bundle, cotangent bundle, 1-form, 2-form, etc.).

I have this book from Saletan and Jose, but I would like to go into more details about the [symplectic] geometrical and mathematical foundations of classical mechanics.

Additional note: A chapter about relativistic Hamiltonian dynamics would be a good thing.

Answer

Structure and Interpretation of Classical Mechanics (table of contents) certainly deserves mention. It might not have as much differential geometry as you'd like, though they have a followup article titled Functional Differential Geometry.

newtonian mechanics - Newtons Cradle, Collision Theory

It has come to the point in my computing program now where I have 5 swinging pendulums that are all modified at once by slider values. These values are drawn onto the from and passed through the class pendulum. To change the values the update button is run, and their is a default option to use a setup that works. The values are also shown on screen through some simple labels. The zero button sets all values (except length) to zero.
The values for each pendulum drawn is:

• String length


• angle


• Acceleration (angular)


• Angluar velocity


• Damping or friction


• Gravity

My calcuations for these values: (Bob = the weight at the end of the string, and there are X and Y Co-ordiantes for it.

bobX = originX + (Math.Sin(angle) * length);

        bobY = originY +(Math.Cos(angle) * length);

Caclulate Acceleration:

Angular Acceleration = (-1 * gravity / length) * Math.Sin(angle); 

Increment Velocity:

        Angular Velocity += Angluar Acceleration;

Increment Angle

        angle += AngularVelocity;

Friction action, pendulum eventually 0's

        AngularVelocity*= damping;

Now my next task is to make the pendulums collide "realistically" and recreate a newtons cradle type effect. I've looked into elastic collision solutions on particle e.g. here is an excellent example: https://stackoverflow.com/questions/345838/ball-to-ball-collision-detection-and-handling

I've also looked at the raw physics behind it here but my brain couldn't get to grips with the raw physics side of things.

So I'm wondering if anybody could help me figuring out the theory for the collisions.
My current idea is when the co-ordinates meet the velocity will simply change polarity e.g. +10 to -10 in the opposing pendulum. Am I correct here? If you are into code I've posted the code on stack overflow here!

Answer

There is something called impulse $J$ which represents a very high force over a short period of time resulting in a step change in velocity.

For a particle object $\Delta v = \frac{J}{m}$. When two objects collide and equal and opposite impulse acts on both of them resulting in

$$\begin{aligned} \Delta v_1 &= \frac{J}{m_1} & \Delta v_2 & = \frac{-J}{m_2} \end{aligned} $$

So how do you find what the impulse $J$ is and hence the changes in speed?

Typically a rule is placed for collision saying that the final relative velocity is proportional in magnitude to the initial relative velocity, but in an opposite direction. Namely:

$$ \left( \left( v_1 + \Delta v_1 \right) - \left( v_2 + \Delta v_2 \right) \right) = -\epsilon \left( v_1 - v_2 \right) $$

This only applies for the components of velocity along the contact normal direction (along the impulse direction). The constant $\epsilon$ is called the coefficient of restitution and it is $\epsilon=0$ for a plastic contact (sticking) and $\epsilon=1$ for an elastic contact (bouncing) and everything in-between.

Combine the above to get

$$ \boxed{ J = (\epsilon+1) \left(v_2-v_1\right) \left( \frac{1}{m_1}+\frac{1}{m_2}\right)^{-1}}$$

cipher - The mystery of NOC +10, Part 2

Background

Since March 6th, 2015, a YouTube channel called Noc +10 has been uploading puzzle videos about once a week. Each video begins with the same introduction, followed by some form of clues which can be decoded to point to a hidden video's YouTube ID.

The hidden videos have had a nautical theme, possibly related to Naval Operation Concept 2010, but haven't presented any clear puzzles, aside from whatever meta puzzle or story is being told.

This question provides a writeup of the first five videos, and it's primary answer solves the fifth to reveal the sixth.

After that, three very strange videos that broke the pattern were posted, but they contained no apparent puzzles. I choose to ignore them for the time being.

7. Fourth contact - March 30, 2015

0101 0101 0000 0111

The seventh video, which is numbered sequentially in binary with all the others, begins with the NOC +10 logo and the same music playing in the background. At the halfway point, it switches to displaying the text ISL 4 9 xEFTvZ while playing a new song, identified as "Air On the G string", August Wilhelmj's arrangement of a piece by J. S. Bach.

As before, we can expect the text displayed to decode to a eleven-alphanumeric-character YouTube video ID.

According to Wikipedia, the song is most notable for Wilhelmj's arrangement, which allowed him to play the entire piece on just the G string of his violin. I suspect that that means that the letter G is a major factor in the decoding, although I'm not sure how.

I've tried shifting the letters by 'G' (like with the Caesar cipher in the first video), as well as several operations on the ASCII values involving G, but nothing has led to an active video ID.

---

Can you crack the last video's code and find the ID of the next one?

---

Other links:

Answer

The mystery has finally been solved!
The answer is:

hOH49oVWKmQ

Explanation:
The encrypted text is actually "lSL 4 9 xEFTvZ" and not "ISL 4 9 xEFTvZ" (the difference is in the first letter, which is lower-case L instead of upper-case i).
The code is then easily decrypted shifting back the first block of letters by 4, the second block by 9 and keeping intact the two numbers.

The result is:

hOH49oVWKmQ

As you can see, the main difficulty was in the interpretation of the first character (absolutely impossible to tell from the video). The cipher itself wasn't too hard (just two Gronsfeld ciphers).

Optics and Refraction

A monochromatic red ray of light, enters a beaker of water , $\gamma = \gamma ' / \mu $ , $\mu $ is the refractive constant.

The wavelength should reduce, (taking $\mu$ as $1.33$ ), so the light should become green, but we see it as red light. Why?

Answer

This is an important question because it gets at the concept of color, and what that really means. Mitchell is correct that when you observe scattering from a laser beam sent through a beaker of water, you are really only seeing the photons which have left the water ($\mu\approx1.33$) and the beaker (glass $\mu\approx1.5$), traveled through the air ($\mu\approx1.00029$), entered your eye ($\mu\approx1.33$), and were absorbed by your retina. So which color do you see? And does it change if you repeat the experiment underwater or use oil ($\mu\approx1.5$) in the beaker instead?

The key to understanding why we always see red light in this experiment rather another color is due to the nature of your photoreceptors, which (like most light absorbers) are not concerned with the wavelength of the light. They instead respond to the frequency of the light, $\nu$ (or photon energy $E=h\nu$), which remains unchanged as light propagates from one (linear) medium to the next. The reason for this is that there are certain energies, determined by quantum mechanics, which are assigned to the electronic states in the absorbing material (your retina, in this case). In order to conserve energy, the photon needs to have the right energy to be absorbed, promoting the material from a low energy state to a high energy state. Wavelength has nothing to do with it. Indeed, when we say "color" in reference to light, we are often actually speaking of the frequency rather than wavelength. When we do refer to a wavelength as a color, it is always with an implicit assumed refractive index, usually $\mu=1$.

quantum mechanics - Heisenberg Uncertainty Principle scientific proof

Heisenberg's uncertainty principle states that:

$$\sigma(x)\sigma( p_x )\ge \frac {\hbar}{2}.$$

What is the scientific proof of this principle? Operators Uncertainty

Answer

The uncertainty principle, in the variance formulation, states that in any quantum state $|\rangle$, the quantity

$$\langle (p-

)^2 \rangle \langle (x-\langle x\rangle)^2\rangle \ge {\hbar^2 \over 4} $$

To understand why shifting p and x by their expected value and squaring gives the squared uncertainty, see this answer.

The proof is by noting the following

$$ |\langle \psi | \eta \rangle| \le \sqrt{ ||\psi||^2 ||\eta||^2}$$

This is the statement that the dot-product of two vectors is less than the product of their lengths. It is called the "Cauchy Schwartz inequality". For the special case above, defining the operators $P= p-\langle p\rangle$ and $Q=x-\langle x\rangle$ (and squaring both sides),

$$ ( \langle P Q \rangle )^2 \le \langle PP\rangle\langle QQ\rangle $$

Where to see that the above is an instance of Cauchy Schwarz, take:

$$ |\psi\rangle = P|\rangle$$ $$ |\eta\rangle = Q|\rangle$$ While the product PQ can be decomposed into a real and imaginary part

$$ PQ = {1\over 2} (PQ+QP) + {1\over 2} (PQ-QP) $$

The first part is imaginary, because if you take the Hermitian conjugate, it changes sign. The second part is real (this is ultimately because P and Q are real, i.e. Hermitian). The expected value of PQ squared is the square of the imaginary and real parts separately

$$ (\langle P Q \rangle)^2 = {1\over 4} (\langle [P,Q]\rangle)^2 + {1\over 4}(\langle PQ+QP)\rangle)^2 $$

Since both square things are positive, this means that the left hand side is bigger than one quarter the square of the commutator. The commutator is unchanged by the shifting,

$$ [P,Q] = [p,x] = \hbar $$

So that

$$ \langle P^2 \rangle \langle Q^2\rangle \ge (\langle PQ \rangle)^2 \ge {1\over 4} (\langle [P,Q] \rangle)^2 = {\hbar^2 \over 4} $$

The proof is usually given in one line, as directly above, where the Cauchy Schwarz step (first inequality), the imaginary/real part decomposition (second inequality) and the shifted canonical commutation relations (last equality) are assumed internalized by the reader.

This proof appears on Wikipedia, it is used in all QM books, but perhaps this explanation is clearer.

quantum field theory - What is the meaning of the size of a particle in QFT?

I have often seen people refer to the size of a particle being at most a given value, or a particle being a point particle, in the context of quantum field theory. Examples are the Wikipedia entry on the electron, where it says

Observation of a single electron in a Penning trap shows the upper limit of the particle's radius is $10^{-22}$ meters. Also an upper bound of electron radius of $10^{-18}$ meters can be derived using the uncertainty relation in energy.

or the accepted answer to the related question Do electrons have shape?, which starts

As far as we know the electron is a point particle

this answer to a physics stackexchange question which mentions

an upper bound on the [electron] radius of $10^{-22} m$

or in the particle adventure by the particle data group

We don't know exactly how small quarks and electrons are; they are definitely smaller than $10^{-18}$ meters, and they might literally be points, but we do not know.

However, in QFT a particle is a quantum of excitation of a field (see e.g. the concept of a particle in QFT, which doesn't explicitly talk about size or point particles), and it is not so clear what would be the meaning of the concept of size.

I have also heard "probing a system at a length scale" in these contexts, like in the aforementioned answer this answer in which it says

What they actually list in that reference is not exactly a bound on the electron's size in any sense, but rather the bounds on the energy scales at which it might be possible to detect any substructure that may exist within the electron. Currently, the minimum is on the order of 10 TeV, which means that for any process occurring up to roughly that energy scale [...], an electron is effectively a point. This corresponds to a length scale on the order of $10^{-20}$ m, so it's not as strong a bound as the Dehmelt result.

From that I infer that he means that a point particle is a particle with no substructure (whatever that exactly means). A possibly non-point particle may look pointlike up to a given energy scale, which seems plausible enough. To translate this into a length scale, you convert to the correct dimensions using factors of $c$ and $\hbar$.

Is it really an energy scale we are talking about when referring to the size and is it just a manner of speech to call this an upper bound on the size of the particle? Does it have any meaning as an actual size?

Finally this answer to a somewhat related question states that

Pointlike is a technical term that refers to the fact that in the standard model, the Lagrangian is a function of fields at the same point (rather than of integrals over fields in some small neighborhood of this point ...)

which seems to suggest that the size of a particle might be defined as the distance up to which the values of the field influence the value of the Lagrangian density at a point. Is that a sensible interpretation of the size? Is it equivalent to the other possible meaning (as the length corresponding to the energy below which no substructure can be detected)?

logical deduction - Three statements, who is saying the truth?

Bob: Steve is lying

Steve: Bob is lying

Fred: Steve is lying or Bob is lying

"or" is inclusive

How would I approach this question? I honestly have no idea. I was thinking of making a truth table and assuming that each of them is saying the truth once and then going through the statements, but i'm not sure how I would handle the "or" bit within a truth table.

free fall - Gravity and bottomless pits

Assuming that someone is theoretically able to make a hole through the center of a large planet, and then jumps down the hole, what will happen? Given my understanding of gravity and energy, my estimate is that in the absence of any resistive forces, the person would fall right through to the other side, with at least part of their body coming through, and would continue oscillating like this. If there were resistive forces, such as air resistance, the person would oscillate similar to a spring winding down, eventually stopping in the middle of the planet. Is this what would happen, or are my calculations off somewhere?

Answer

If you use Gauss's theorem for the gravitational field of the Earth at radius $R$ and assume it has uniform density $\rho$. $$ g(R) = -GM(R)/R^2 = -4\pi R G \rho /3,$$ where $M(R)$ is the mass inside radius $R$, and is given by $4\pi R^3 \rho/3$.

Thus the force the body feels towards the centre of the Earth (neglecting air resistance - see below) is proportional to $R$ and thus $$\frac{d^2 R}{dt^2} = -\frac{4\pi G \rho}{3} R,$$ which is a simple harmonic motion equation with an oscillation frequency of $(4\pi G\rho/3)^{1/2}$.

If I assume an average $\rho =5000\ kg/m^3$, this means an oscillation period of 88 minutes.

If you really did hollow out such a tunnel, it would quickly fill with air. You would reach a terminal velocity of order 200 km/h at best. Travelling at this sort of speed as you went through the centre of the Earth would mean you would only rise by a height on the other side given approximately by $$ \frac{1}{2} m v^2 = m \int_0^{H} g(R)\ dR = m \frac{2\pi G \rho H^2}{3} $$ $$ H = \left(\frac{3}{4\pi G \rho}\right)^{1/2} v$$ Putting in some numbers then gives $H \simeq 50\ km$. So nowhere near emerging at the antipodes.

thermodynamics - Ideal gas reversible cycle

So I got this cycle on a quiz. As of my understanding, in the transformation 1-2 there is also received heat and given heat ( sorry if that's not the english exprimation, not my first language). Now I have a question:

As I understand it, there is a given point M on 1-2 until which heat is received, and from M to 2 it is given. You can solve for it with the general equation p=aV + b. Anyways, why is it imperative for the received heat to be first and not the other way around? Is it because of the 2nd law of thermodynamics? And if so, could you explain it for this case?

Thank you in advance!

Answer

No. It's the result of applying the first law of thermodynamics.

The equation for the straight line between the initial and final volumes is $$P=P_0\left(3-\frac{V}{V_0}\right)\tag{1}$$

From the 1st law of thermodynamics for this reversible process, we have: $$dU=nC_vdT=dQ-PdV$$or $$dQ=nC_vdT+PdV\tag{2}$$Combining this with the ideal gas law gives:$$dQ=\frac{C_v}{R}d(PV)+PdV\tag{3}$$ Substituting Eqn. 1 into Eqn. 3 gives: $$dQ=\frac{C_v}{R}P_0\left(3-2\frac{V}{V_0}\right)dV+P_0\left(3-\frac{V}{V_0}\right)dV\tag{4}$$ At $V=V_0$, this becomes $$dQ=P_0\left(\frac{C_v}{R}+2\right)dV$$which is greater than zero. At $V=2V_0$, Eqn. 4 yields: $$dQ=P_0\left(1-\frac{C_v}{R}\right)dV$$Since $C_v/R>1$, dQ is negative at $V=2V_0$. So dQ changes sign between the initial volume and the final volume.

Dark energy converted to electricity

Possible Duplicate:
Dark Energy saving the world

In theory, dark energy be used to generate electricity. Given its apparent abundance, this could provide rather a lot (perhaps too much) clean energy. Has any research been conducted on this or any other practical applications for dark energy? (other than repelling galaxies ;)

Answer

Dark energy is not really an energy in the common sense and there absolutely is no project to produce electricity via dark energy. Dark energy is most probably a mistake in our calculations. Even if it is not so, the nature of dark energy is currently so far from our grasp that there is no way, even theoretically, to generate energy from dark energy yet.

electromagnetism - When is the force null between parallel conducting wires?

Consider two long wire with negligible resistance closed at one end of the resistance R (say a light bulb), and the other end connected to a battery (DC voltage). Cross-sectional radius of each wire is in $x = 100$ times smaller than the distance between the axes of the wires.

Question:

At which value of resistance R (in ohms) the resultant force of the interaction of parallel conductors disappears?

Answer

The magnetic force should be straightforward, you probably have a formula for it, or you can use the field strength from one wire, cross the current in the other to compute the force. The charge is more difficult, there you have two oppositely charged cyllinders separated by a distance. You can need to figure out the voltage difference between them as a function of the charge denisty per unit length of the system. As a first approximation assume the charge is evenly distributed around the circumference of your wire, then you should be able to compute the voltage difference between the two wires by integrating the electric field between them. I'm not going to do this (if it really is homework, you need to do it yourself). You should be able to use symmetry to reduce the amount of detailed math you need. If you are really ambitious, you could calculate the effect of uneven charge distribution on each wire, i.e. there will be more charge on the surface closer to the other wire than on the outside -or at least show the approximate size of the correction factor. Should be a fun exercise.

Can anyone analyze this perpetual machine that a Norwegian built?

Take a look at this video:

http://www.youtube.com/watch?v=tlx2PgESXhs

It shows a machine that a Norwegian artist built. It seemingly keeps going forever. They filmed it over 3 days and it kept going at the same time.

Can someone analyze the energy losses in this machine?

I'm seeing several losses at the moment:

1. The ball is losing energy due to friction while going around the track.


2. The pendulums are losing energy due to air friction.

What other sources of energy losses can you notice?

How often does nuclear fusion occur within the human body?

I'm just curious. I figure atoms fuse occasionally just by chance, like quantum tunneling or rogue waves. Is this true? If so, any idea how often?

Answer

This'll be a very rough order of magnitude estimate, but as you'll see it's good enough.

Suppose that two hydrogen atoms bump into each other. In order to fuse, the nuclei have to tunnel to within about a nuclear distance of $10^{-15}$ m of each other. The tunneling probability is something like $e^{-(2mE)^{1/2}L/\hbar}$, where $E$ is the energy gap, $m$ is the particle mass, and $L$ is the distance. The distance is of order $10^{-10}$ m (a Bohr radius) and the energy is about an MeV (the electrical potential energy of two protons right next to each other. I work out the numbers to get a probability of about $e^{-20000}$.

You'd next have to multiply that by the number of "chances" (number of times two atoms collide with each other). That's a large number by ordinary standards, but it's not exponentially large in the same way that the probability is exponentially small. Say you've got $10^{29}$ atoms in you, and each one collides with something else $10^{10}$ times per second. Then the number of chances per second is a mere $10^{39}$. I made up that number $10^{10}$ out of nowhere, but whatever it is, it's not $10^{10^4}$, which is what it'd have to be for there to be any significant probability.

So it never happens.

word problem - Guess The Object - What Am I Riddle (Part 5)

Since part 4 is dragging it's feet, What am I?

Puny mortals, I control your very lives!
When I'm angry, you cease motion,

but yield to my envy,
Without me you descend into chaos,
I am never wrong, only long
Turn from me at your peril.

Hint:

In the city you see me every day.

Answer

A traffic light: angry (red) means stop, envy (green) means go

Why are muons considered to be "elementary particles" in the Standard Model?

According to this article, a muon decays into one electron and two neutrinos.

According to this article, elementary particles or fundamental particles are particles "whose substructure is unknown, thus it is unknown whether it is composed of other particles." I have also seen somewhere that it is a particle that cannot be reduced into other constituent particles.

While perhaps not a sure thing, seems like the decay indicates that the muon may be just a composite particle, perhaps consisting of one electron and two neutrinos?

Based on this, why does the muon fit with the above definition of an elementary or fundamental particle?

I realize there are much more complicated, historical reasons as to why it was included in the Standard Model, but this question is just related so how it fits (or doesn't fit) the stated definition above.

It seems to me that we really can only get solid evidence of elementary vs. composite when we smash the particles together and see what comes out and compare that to all the masses, energies and momentum before and after? Until we do that with muons, how can we know with much certainty?

And perhaps we'll have a better answer with a Muon collider: http://en.wikipedia.org/wiki/Muon_collider/ http://map.fnal.gov/

To that point, seems that electrons may not be fundamental after all: https://www.sciencedaily.com/releases/2016/04/160404111559.htm

Answer

Indeed, we can't know for sure if muons are elementary are not. In this sense, the situation is like the 1950's, where we had a zoo of mesons and hadrons, but didn't yet know they were made of quarks. We wouldn't have direct confirmation of quarks for decades to come, like how we have no indications of muon compositeness now.

Despite this, the quark model was accepted, and all attempts to make muons composite have not been. There are many scientific reasons to reject muon substructure:

• Predictive power. Your model basically says that all the charged leptons are 'really' electrons or positrons with bound neutrinos. This makes no direct predictions of new particles or processes. In contrast, the quark model predicted whole families of new particles.


• Theoretical simplicity. You need to postulate a new force to keep the neutrinos bound to the electrons, since the weak force is certainly not enough to do it. This was okay for the quark model, because we knew there was a force we didn't understand at the time (the strong nuclear force), but we have no indications for such a force. This 5th force should significantly affect the motion of neutrinos, but we haven't seen any such thing.


• Direct observation. Since muons are fairly light, whatever force is binding the electrons and neutrinos together must be fairly weak (since the binding energy $E = mc^2$ is low). This indicates that we should have been able to tear apart a muon into its constituent parts by now, or at least excite its energy levels. This hasn't been observed.


• Occam's razor. There's no need to postulate composite particles to explain decays. For example, excited atoms can decay to ground-state atoms and photons. This process can be described simply and exactly by coupling the atom to the electromagnetic field; it doesn't require a photon to be 'inside' the excited atom. Worse, there are alternative decay channels which emit two photons instead! The 'photon inside' picture isn't even self-consistent.

In short, there's no guarantee that muons aren't composite. But there are many compelling experimental and theoretical reasons suggesting so.

A riddle about unusual terms and history

Forty five before it was.

Now a tribe but once did buzz.
I'm mostly desert with a little e.
Was proposed but never would be.
Though quite religious I'm historic too.
And followed by industry, book, and news.

What am I?

Hint

forty five as in 1896.

p.s. A similar riddle is used in another of my puzzles (where in the world am I supposed to go). But I thought that it deserved a puzzle all to itself.

Answer

You're

Deseret

Forty five before it was.

It was proposed instead of Utah, the 45th state

Now a tribe but once did buzz.

LDS (Jaredites?)

I'm mostly desert with a little e.

Deseret

Was proposed but never would be.

Indeed, Utah came to be instead

Though quite religious I'm historic too.

LDS again

And followed by industry, book, and news.

Deseret Industries, Deseret Book, Deseret News

no computers - Some-what-not-with-standing word connections

What words can fill the boxes?   Each pair of linked words forms an uncapitalized unhyphenated word when joined end-to-start according to their left-to-right or top-to-bottom order in the link.   Some boxes indicate the lengths of their mystery words.

               _____                   For example
     friend---|_____|---shape         -------------        friend---SHIP---shape
                 |  \_____                                            | \
                 |  |__4__|           Linked words:                   |  YARD
               __|__   \                                              |    \
              |__5__|---age        friendship   shipyard           WRECK---age

                                   shipshape    wreckage
                                   shipwreck    yardage


                                         For real
                          the           ----------
                         __|__
                   he---|_____|                                  chic
                         __|__                                   __|__
               super    |__4__|                       handle    |_____|

               __|__     __|__     _____     _____     __|__     __|__
        for---|_____|---|__4__|---|__2__|---|_____|---|_____|---|__4__|
                 |  \                          |
                 |   \_____                    |        put
                 |   |_____|                 __|__     __|__
               __|__    \                   |_____|---|_____|
        con---|_____|---ally                 __|__
                                      rep---|_____|
                                             __|__
                                    stall---|_____|

This puzzle follows the lead of Word connections and Word connection squares!.

Answer

                          the
                         __|__
                   he---|_re__|                                  chic
                         __|__                                   __|__

               super    |name_|                       handle    |_hi__|
               __|__     __|__     _____     _____     __|__     __|__
        for---|_ego_|---|less_|---|_on__|---|side_|---|_bar_|---|ring_|
                 |  \                          |
                 |   \_____                    |        put
                 |   |maniac                 __|__     __|__
               __|__    \                   |_real|---|ties_|
        con---|centric--ally                 __|__
                                      rep---|_lot_|
                                             __|

                                    stall---|_ion_|

Notes:

I got LOT from Mike Q's answer. (There are other commonalities but they were independent discoveries.) NAME could also be FORM, as in BaSzAt's answer, in which case we could also replace RE with REIN as in his answer. RING could also be KING. CENTRIC and CENTRICALLY are slightly obscure. CHICHI, too. I originally had TIE instead of TIES, which yielded two (valid but) terribly obscure words, but Khale_Kitha pointed out in chat that pluralizing fixes everything.

lagrangian formalism - Variation of gamma and Christoffel

I am trying to experiment with Lagrangian densities and came across a term similar to $$\gamma^i\Gamma^j_{ik}A^k$$ where the $\gamma$ are the gamma matrices, $\Gamma$ are the Christoffel Symbols and $A$ are the field components.

In the variation of the term my question is somewhat twofold: is there any "nice" way to compute $\dfrac{\delta}{\delta g^{lm}}\gamma^a$, knowing the relation $\{\gamma^i,\gamma^j\}=2g_{ij}\mathbf{I}$?

---

---

Secondly, from what I have seen $\delta\Gamma^i_{jk}=\Gamma_{jkl}\delta g^{il}+g^{il}\delta\Gamma_{jkl}$, commonly rewritten in a different form. The question Variation of the Christoffel symbols with respect to $g^{\mu\nu}$ does not seem to have a satisfactory answer for my purpose; it seems like the variation terms will stay in the field equation so it does not make sense to leave variations in the expression. Is it valid to naively assume that $\dfrac{\delta}{\delta g^{lm}}\Gamma^j_{ik}=\Gamma_{jkl}$?

newtonian mechanics - Component of Component of a vector

NOTE :

By perpendicular component of $\vec{F}$, I mean a vector which is a component of $\vec{F}$, but perpendicular to it.

In the image above, the red vectors are a possible set of rectangular components of $\vec{F}$, and blue vector is the sine component of cos component of $\vec{F}$, ie, perpendicular component of $\vec{F}$.

Now, logically, the perpendicular component of $\vec{F}$ should be zero, since its projection is zero.

But if we consider the image above, the Perpendicular component, $|\vec{F}|\cos\theta\sin\theta$, is not zero.

What is the cause of this discrepancy, and how to take components properly?

Answer

If you take again the components of the vector whose magnitude is $Fsin\theta$, you will see one component of it will cancel the vector represented by the blue arrow both of which will have a magnitude of $Fcos\theta sin\theta$:

The other components of the components of $\vec F$ will be vectors whose magnitudes are $Fcos^2\theta$ and $Fsin^2\theta$ whose sum will equal to $F$.

rebus - The Lord of the Puzzles

In 1904 (at the age of 12) J.R.R. Tolkien sent a "puzzle letter" (pictured below) to a family friend; Father Francis Morgan.

What does it say?

^{_{Obviously it would be trivially simple to google the answer, so I want you to promise me that you didn't :-)}}

Answer

The sender's address (top right) was:

Woodside House, Rednal (wood, S, eye, D=500, E, house, {N, A, L=50} in red)

The letter reads:

My dear wise owl Fr. Francis, (M=1000, eye, deer, Y's, owl, France with 'Fr.' on it, hiss)

You are too bad (U, R, 2, bee, A, D=500)

not to come in (knot, 2, sea, O, M=1000, E, inn)

spite of Fr. Dennis. (spit, E, OF, Fr., D=500, hen, eye, S)

I am so sorry you (Hyam, sew, saw, rye, yew)

did not like the (D=500, eye, D, knot, L=50, bike-B, tea, HE)

word 'piano' in my l... (words, pea, note-T, inn, M, eye, L)

...ast letter. So I... (mast-M, letter, sow, eye)

Thanks to Pete and d'alar'cop and LeppyR64 and JNF for help with some of the clues (see comments below).

cosmology - Is the universe flat?

There are more than one way to view the description of the universe as flat. There is the description of an open, flat or closed universe in terms of it's fate, expansion forever away from gravity, or a big crunch.

Then there is the description of our observable universe being flat in it's geometry, or is it that there is no way to measure the tiny curve of space in our part of the sphere.

Another probably wrong idea is that the entire universe is flat. Are these things confused, even by the experts?

enigmatic puzzle - Mixing business and pleasure

I asked my manager if it was okay for my friends to come hang around at the shop. Here's his answer:

As I've said, they're always very welcome. But can I just remind that we'll keep working? Clients come first, your friends must naturally respect that. Did I mention things move quickly 'round the warehouse? Every hand is needed in there this time. Forget not our owner's order: "remember shop's slogan".

What is our slogan?

Answer

Always keep your things in order?

Reason:

Could be a red herring, but the words in every sentence are ordered alphabetically except for those six.

classical mechanics - Idea of integrable systems

I do not quite understand the idea of an integrable dynamical system. Does it mean that the EOMs are analytically and exactly solvable? What are the necessary and sufficient conditions such that a system is integrable?

Answer

There has been extensive discussion on this here. Infact the question was posed by Gil Kalai, a well known mathematician. I hope it helps.

general relativity - Difference tensor between two connections

I am using these supergravity lecture notes by Gary W. Gibbons. On page 18, the author claims that geodesics and autoparallels coincide for a theory with totally antisymmetric torsion, and proves it by using the following identity (which he states without proof):

$$\Gamma_{\mu\;\;\nu}^{\;\;\rho}=\{_{\mu\;\;\nu}^{\;\;\rho}\}+K_{\mu\;\;\nu}^{\;\;\rho},$$

where

$$K_{\mu\rho\nu}=-\frac{1}{2}(T_{\mu\rho\nu}+T_{\rho\mu\nu}+T_{\rho\nu\mu}).$$

Here, $\Gamma_{\mu\;\;\nu}^{\;\;\rho}$ are the coefficients of an arbitrary connection with totally antisymmetric torsion and $\{_{\mu\;\;\nu}^{\;\;\rho}\}$ are the coefficients for the Levi-Civita connection, while $T_{\rho\nu\mu}$ is the torsion. Is there a proof of this identity somewhere?

Answer

OP's question is probably spurred by the fact that Ref. 1 forgets to mention that:

1. 
   

   The other connection $\nabla:\Gamma(TM)\times \Gamma(TM)\to \Gamma(TM) $ with lower Christoffel symbols $$\Gamma_{ij,k}~:=~ \Gamma_{ij}^{\ell}~g_{\ell k}\tag{1}$$ is still assumed to be compatible with the metric $$\nabla g = 0\qquad\Leftrightarrow\qquad \partial_kg_{ij}=\Gamma_{ki,j}+\Gamma_{kj,i}.\tag{2}$$

   
   



2. 
   

   The contorsion tensor $K~\in~\Gamma\left(T^{\ast}M\otimes \bigwedge^2(T^{\ast}M)\right)$ is defined$^1$ as the difference between the metric-compatible $\nabla$ and the Levi-Civita connection $\nabla^{LC}$, i.e. $$K_{i,jk}~:=~\Gamma_{ij,k}-\Gamma^{LC}_{ij,k} ~\stackrel{(2)}{=}~-K_{i,kj}.\tag{3} $$

   
   


3. 
   

   The torsion tensor $T~\in~\Gamma\left(\bigwedge^2(T^{\ast}M)\otimes T^{\ast}M \right)$ is defined$^1$ as $$ T(X,Y)~:=~\nabla_XY-\nabla_YX-[X,Y] $$ $$\quad\Leftrightarrow\quad T_{ij,k}~:=~\Gamma_{ij,k}-\Gamma_{ji,k}~\stackrel{(3)}{=}~K_{i,jk}-K_{j,ik}.\tag{4} $$

   
   


4. 
   

   One may show that the inverse relation to (4) is $$ K_{i,jk}~\stackrel{(4)}{=}~ \frac{1}{2}\left(T_{ij,k}-T_{jk,i}+T_{ki,j}\right),\tag{5}$$ and vice-versa.

   
   


5. 
   
   

   Let us decompose the contorsion tensor $$K_{i,jk} ~=~ \frac{1}{2}(K^+_{i,jk}+K^-_{i,jk}), \tag{7}$$ in components $$ K^{\pm}_{i,jk}~:=~K_{i,jk}\pm K_{j,ik}, $$ $$ K^+_{i,jk}~=~T_{ki,j}+T_{kj,i}, \qquad K^-_{i,jk}~=~T_{ij,k}, \tag{8} $$ that are symmetric/antisymmetric wrt. the first two indices $i\leftrightarrow j$.

   
   


6. 
   

   The geodesic equation reads $$0~=~\nabla^{LC}_{\dot{\gamma}}\dot{\gamma} \qquad\Leftrightarrow\qquad \ddot{\gamma}^{\ell}~=~-\Gamma^{LC,\ell}_{ij}\dot{\gamma}^i\dot{\gamma}^j$$ $$\qquad\Leftrightarrow\qquad -g_{k\ell}\ddot{\gamma}^{\ell}~=~\Gamma^{LC}_{ij,k}\dot{\gamma}^i\dot{\gamma}^j.\tag{9} $$

   
   


7. 
   

   In contrast, the auto-parallel equation $$0~=~\nabla_{\dot{\gamma}}\dot{\gamma} \qquad\Leftrightarrow\qquad \ddot{\gamma}^{\ell}~=~-\Gamma^{\ell}_{ij}\dot{\gamma}^i\dot{\gamma}^j$$ $$\qquad\Leftrightarrow\qquad -g_{k\ell}\ddot{\gamma}^{\ell}~=~\Gamma_{ij,k}\dot{\gamma}^i\dot{\gamma}^j~=~\left(\Gamma^{LC}_{ij,k}+\frac{1}{2}K^+_{i,jk}\right)\dot{\gamma}^i\dot{\gamma}^j\tag{10} $$ can detect the symmetric part $K^+_{i,jk}$ of the contorsion.

   
   


8. 
   

   Observation: $$T_{ij,k} \text{ is totally antisymmetric} \qquad\Leftrightarrow\qquad K_{i,jk} \text{ is totally antisymmetric} $$ $$\qquad\Leftrightarrow\qquad K^+_{i,jk}~=~0.\tag{11}$$ In the affirmative case, we have $T_{ij,k}=2K_{i,jk}$, and the geodesic & auto-parallel equations (9) & (10) are the same.

   
   
   

--

$^1$ Pertinent applications of the musical isomorphism are implicitly implied from now on.

quantum mechanics - Is the Moyal-Liouville equation $frac{partial rho}{partial t}= frac{1}{ihbar} [Hstackrel{star}{,}rho]$ used in applications?

This answer by Qmechanic shows that the classical Liouville equation can be extended to quantum mechanics by the use of Moyal star products, where it takes the form $$ \frac{\partial \rho}{\partial t}~=~ \frac{1}{i\hbar} [H\stackrel{\star}{,}\rho]. $$ I've seen this mentioned as a possible classical interpretation of quantum mechanics, but I've never seen this used in anger, so I want to ask: is this representation actually useful in understanding dynamics? In what contexts does it provide insights about physical processes that are harder to obtain through alternative routes?

I'm not really looking for an exhaustive list of places where this representation turns up, but I would like to see a few representative examples to get a better feel of how it looks like in applications.

Answer

"Used in anger" or "killer ap"? To my knowledge, no problem has been solved in the phase-space quantization language that was not solvable in the other two formulations/pictures (Hilbert space or path integrals). This is in contrast to, e.g., path integrals (whose gauge fixing, Faddeev-Popov, and Fujikawa anomaly applications to gauge theories are virtually irreplaceable; as well as the lattice gauge theory simulations that have revolutionized strong-coupling QCD).

In this sense, the best days of the formulation are still ahead, alas....

But since it is best suited to exploring the quantum-classical interface, given the identical variables it uses for both regimes, uniquely!, study of decoherence truly relies on it for visualization, and even results! A classic book: Decoherence and the Appearance of a Classical World in Quantum Theory by Joos, Zeh,‎ Kiefer, Giulini,‎ Kusch, and Stamatescu goes to town in Ch 3.2.3. It convincingly uses the formulation far, far beyond jee-whizzy novelty and "lies to children" which is sometimes the mode in such forays.

As a corollary, one uses it credibly in quantum computing, quantum carpets—don’t skip the movies, and importantly “quantum chaos” (whether it exists or not): I’ll let you find your least stressful reference.

So much for deep irreplaceability. When it comes to salutary intuition, however, it is peerless. (Personally, I cannot resist running to this picture to get a visual impression of any classical limit, in semi-quantitative terms.)

We detail in our booklet (A Concise Treatise on Quantum Mechanics in Phase Space, T L Curtright, D B Fairlie, C K Zachos, World Scientific (2014) ISBN-13:978-9814520430) how the SHO rigid rotation helps bypass umpteen confusions; e.g., by inspection, automatically manifest which states of the oscillator annihilate the anticommutator of $\hat x$ and $\hat p$, in this formulation just $2xp$, a plain phase-space hyperbola: behold Fig 6.

The quantum oscillator in Hilbert space appears to possess quasi-magical wave-packet periodic reconstitution features―until its evident rigid rotation in phase space is appreciated, leading to the less intuitive x or p projections, as above. The logic of coherent states (yellow) is also apparent. Thence, the logic of the interaction representation of Dirac, where the states rotate rigidly on a SHO turntable, while further wiggling and morphing according to the interaction.

The meaning of phase-integrated states yielding stationary palisades, morphing into cookie-cutters, in the large n, small ℏ limits are also useful (Fig. 4), a quick appreciation of what stationary states are in classical mechanics, where they are hardly useful.

Many students groove on the classical limit of the two-slit interference structure (Fig. 1) as the distance between the slits gradually increases.

More seriously, it personally paved the way for us to dispel the persistent myth that Nambu brackets were somehow unquantizable… it demonstrated that nature had already quantized them while the sophisticates were not looking… (Admittedly, we subsequently had to transcribe the message to Hilbert space to convince the masses, but lacking the phase-space tool would automatically place the beholder at a conceptual disadvantage…) Specifically, it indicated missing quantum conjugate variables in odd dimensions, and the odd-even phase space dichotomy barely known to mathematicians but not physicists. But I’m ranting…

---

Response comment to comment/question : Moyal mechanics, the evolution equation you wrote (actually written down by Wigner in his 1932 paper), is the von Neumann equation in this language, and describes the evolution of the Wigner function ρ. It is completely equivalent to the corresponding *-equations for the Husimi, Glauber-Sudarshan, etc. reps, as we detail in the later chapters of our booklet: each has its proper *, ρ and H, systematically connected to the "Moyal" one, (actually discovered by Groenewold ). Perhaps by "quantum" you mean using a and $a^*$ instead of x and p? They are all c-number variables, though... Choosing any is fully a matter of convenience.

For example, as we exemplify, the oscillator *-genvalue equation is much simpler in the Husimi, and some people, unwisely, are impressed by this one because its distribution is positive semidefinite. (Unwisely because they ignore the *-measure in computing expectation values, only absent in the Moyal/Wigner language --that's why some call the Wigner the "Cartesian coordinate system" of the formulation. So Husimi distributions are far less intuitive vis-a-vis the classical limit, not more, as people cannot convolve quantities in their mind! They then go on to write wrong papers...). But playing favorites among these is as unreasonable as playing favorites between polar and Cartesian coordinates. Usually the problem chooses the handiest.

cosmology - What is the best explanation for the fine tuning of the Universe?

In my view, this is one of the most profound unsolved mysteries of Science. How to explain the fine tuning of the Universe without recourse to saying that a God/Designer did it.

Examples of fine tuning:

We exist because of a crucial one part in a billion difference that favors matter over antimatter.

If the ratio of the nuclear strong force to the electromagnetic force had differed by 1 part in 10 to the 16th power, no stars would have formed.

If the strong force constant were 5% weaker, there would be no stable stars

If the strong force constant were 2% stronger, there would be no stable hydrogen, no long-lived stars, no hydrogen containing compounds

If the expansion rate of the Universe were larger, no galaxies would form

If the expansion rate of the universe were smaller, the universe would collapse even before starts formed

If the mass density of the Universe were larger, overabundance of deuterium from the big bang would cause stars to burn rapidly, too rapidly for life to form

If the mass density were smaller, insufficient helium from the big bang would result in a shortage of heavy elements

I could go on, but lets cut to the chase.

How do you account for this:

(A) Inflation (B) Multiverse (C) Top Down Cosmology (D) God/Designer (E) Some new innovation you're working on or have that can win a Nobel Prize in Physics

Let's be clear on the options. Either we have some law that forces the fine tuning, which does not exist unless you know otherwise, or we put it to chance, which is obvious nonsense given the math, or it's a designer. Unless you have a fourth option?

mathematics - Breaking Balance (Part C)

For a starting number of otherwise identical coins there are among them TWO IDENTICAL counterfeit coins which are either heavier or lighter than the rest. Using a three-pan balance (described in detail below), and three weighings, what is the most number of starting coins (including the counterfeits) you can have while being able to always determine the counterfeits and whether they are both heavier or both lighter?

This should not be confused with the balance detailed in Find 2 heavy coins among 27 with a 3-pan balance as it has a bunch of different rules.

So there isn't confusion, I want to explicitly state how this thing works. It has three equal-length arms supporting three pans (120deg apart), and is balancing on a point. It works like other balances in that the center-of-gravity is below the balance point and thus will tilt to counteract imbalance in the pans.

There are exactly 7 outcomes the balance will give you. The first is the "balanced" outcome, where all pans stay at the same height. The other six are variations on the following:

One "lone" pan will go up or down, and the other two pans will be equal to each other but move opposite the lone pan. Given that there are TWO IDENTICAL counterfeit coins, this may be less informative than you might otherwise hope for. My shorthand for the 7 outcomes is as follows:

• A = B = C



• A > B = C


• A < B = C


• B > A = C


• B < A = C


• C > A = B


• C < A = B

For sake of convention, I consider the coins to be numbered 1 through whatever, and the pans labeled A, B and C. The 4th location, D, is used to denote a coin not being the scale.

Warning: I do not believe there is a "cute" solution and it took me weeks to solve it... and even then it isn't quite a "proof". I really enjoyed going through it, but seriously... this isn't for everybody. And for those that do get it... there's a bunch of much harder variants awaiting you!

Answer

@A.P. Provided a great analysis! I am picking up from were he/she stopped:

Given $18$ coins, I found (with the help from my computer :-) a second weighing that would split potential counterfeit coins into 7 different groups with each having either 7 or 6 potential counterfeit pairs. I thought the problem is solved, but unfortunately some of the groups with 6 coins were impossible to crack with just one remaining weighing. My findings are described bellow. Using this second weighing that I found, I also show how to solve the problem with total $17$ coins.

So $17$ coins is the best total number of coins when I could still find counterfeits and their type with just 3 weighings. I do not know if it is doable with $18$ - it might be. At some point I will modify my program to do the exhaustive search :-)

Reasoning:

As @A.P. pointed out, we have to only analyze the case $A > B = C$ with 4 coins on each pan. We will number the coins from 1 to 18. We will use notation $AN$ to refer to the content of pan in weighing $N$. In the analysys of the results weighing $N$ we will use scenario $XN$ if a counterfeit coin is heavier than normal and $YN$ if a counterfeit coin is lighter. Let's (following @A.P.) make the first weighing:

$A1:$ 1, 2, 3, 4;
$B1:$ 5, 6, 7, 8;
$C1:$ 9, 10, 11, 12;
$D1:$ 13, 14, 15, 16, 17, 18;

We have 2 possible scenarios for the distribution of counterfeit coins:

$X1:$ Heavy case: at least one is in $A1$, the other is in $A1$ or $D1$;

$Y1:$ Light case: $B1$ has one and $C1$ has one.

Let's proceed with weighing #2 in the following way

$A2:$ 1, 5, 13;
$B2:$ 2, 6, 14;
$C2:$ 3, 7, 15;
$D2:$ 4, 8, 9, 10, 11, 12, 16, 17, 18;

Let's look at the 7 cases one by one.

Case 1, $A2 = B2 = C2:$
$X2:$ 4 and one of 16, 17, 18;
$Y2:$ 8 and one of 9, 10, 11, 12;
Weighing #3:
$A3:$ 9, 16
$B3:$ 10, 17
$C3:$ 11, 18

We have at most one counterfeit coin on the scale. If only one pan is down, we have a heavy case and the down pan uniquely determines the second heavy coin. If only one pan is up we have a light case and the up pan uniquely defines the second light coin. If $A3 = B3 = C3$ then coins 8, 12 are light counterfeits.

Case 2, $A2 < B2 = C2:$
$X2:$ counterfeit is one of the pairs {2, 3} {2, 15}, {3, 14}
$Y2:$ 5 and one of 9, 10, 11, 12;
Weighing #3:
$A3:$ 9, 1
$B3:$ 10, 14
$C3:$ 11, 15
Notice that again - we have at most one counterfeit coin on the scale and potentially lighter coin is paired with a potentially heavier coin. This solved in the same way as case 1. If $A3 = B3 = C3$ then coins {2, 3} are heavy counterfeits.

Case 3, $A2 > B2 = C2:$
$X2:$ counterfeit is one of: {1, 13} {1, 16} {1, 17} {1, 18} {1, 4} {4, 13}
$Y2:$ impossible, we know that counterfeit coins are heavy

I was not able to find one weighing to deferenciate all 6 cases. But if we were dealing with only 17 coins (no coin #18), the following weighing #3 would solve this case:

$A3:$ 2, 4
$B3:$ 3, 13
$C3:$ 5, 6

$A3 = B3 = C3$ ==> $X3:$ {1, 17}
$A3 > B3 = C3$ ==> $X3:$ {1, 4}
$B3 > A3 = C3$ ==> $X3:$ {1, 13}
$C3 > A3 = B3$ ==> $X3:$ {1, 16}
$C3 < A3 = B3$ ==> $X3:$ {4, 13}

Case 4, $B2 < A2 = C2:$
$X2:$ counterfeit is one of: {1, 15} {1, 3} {3, 13}
$Y2:$ counterfeit is 6 and one of the 9, 10, 11, 12

This can be solved the same way as case 2.

Case 5, $B2 > A2 = C2:$
$X2:$ counterfeit is one of: {2, 14} {2, 16} {2, 17} {2, 18} {2, 4} {4, 14}

$Y2:$ impossible, we know that counterfeit coins are heavy

The same as case 3 - could not solve in one weighing, but can be solved by dropping coin 18

Case 6, $C2 < A2 = B2:$
$X2:$ counterfeit is one of: {1, 14} {1, 2} {2, 13}
$Y2:$ counterfeit is 7 and one of the 9, 10, 11, 12
This can be solved the same way as case 2.

Case 7, $C2 > A2 = B2:$
$X2:$ counterfeit is one of: {3, 15} {3, 16} {3, 17} {3, 18} {3, 4} {4, 15}
$Y2:$ impossible, we know that counterfeit coins are heavy

The same as case 3 - could not solve in one weighing, but can be solved by dropping coin 18

classical mechanics - How can earthquakes shift the earth's axis?

One often comes across news articles that claim that an earthquake shifted the earth's axis.

http://news.google.com/?q=earthquake%20shifted%20OR%20shifts%20earth%27s%20axis

If you ignore the influence of other celestial bodies, an internal event like an earthquake surely can't change the direction of the angular momentum of the Earth (unless stuff is ejected out of Earth), since angular momentum has to be conserved in the absence of an external torque. So the axis has to remain fixed.

Am I missing something? Or are geologists trying to say that the resulting movement of tectonic plates causes a change in the point of intersection of the axis (which remains the same) and the plates that include the poles, so that it seems as if the axis has shifted?

EDIT Some articles mention the value of the shift in the axis and also the change in the length of the day. If, as Ted Bunn's answer indicates below, the shift in the axis isn't actually real but is because of the movement of tectonic plates with respect to the axis, shouldn't the shift be different at the north and south poles? How are the shifts and the change in day-length calculated?

Answer

Angular momentum doesn't change, but the angular velocity vector does. This is effectively due to a shift in the body's moment of inertia tensor.

special relativity - Confusion about Length Contraction (ex in Muon decay)

I am a bit confused about the implications of length contractions;

For example, in the muon decay problem, we assume that the distance between the muon and the earth is contracted only in the frame of reference of the muon (I've heard justification saying this is because it is Earth's atmosphere) but suppose there were no atmosphere then in the reference frame of Earth shouldn't the length between them also be contracted?

What I don't get is if one object is moving at a speed relative to another object, shouldn't this movement affect the distance between them in the reference frames of both of them, since their movement is merely relative?

Answer

Don't confuse time dilation and length contraction,(even if length contraction is a consequence of time dilation) because there is no length contraction for the distance between two reference frames. The length (or distance) must be found in one reference frame, and the observer must be in another reference frame, he may not be part of the reference frame of the distance.

By consequence, it is not the distance between muon and Earth which is contracted. It is the distance between the starting point A of the muon and Earth (if we suppose that the muon is traveling from A to Earth).

From the Earth frame the distance A-Earth is at its largest. For the muon frame, the distance A-Earth is contracting according to its relative velocity.

The example of David Z is the opposite example: There is a distance in the muons' frame (between two muons which are belonging to the same frame) which is observed from the Earth frame.

electricity - Formula for Electrical Arc Length

I was playing with some High-Voltage the other day, when a question popped into my head. Can you calculate length of an electrical arc? It probably would be proportional to :-
1. Voltage of the source
2. Spark gap
3. Relative Humidity (in air only)
4. Resistivity of Medium
5. Shape of the Electrodes

Can anyone come up with a dimensional formula from this...?

astrophysics - Relationships between optical depth, $tau$, and probability of being scattered

The optical depth is given by many as:

$$ \tau=n_e\sigma_{cs}L $$ Where $n_e$ is the number density of the medium, $\sigma_{cs}$ is the scattering cross section and L is the linear extent of the medium.

I was wondering, what is the relationship between the optial depth and the probability of scattering? I can't seem to find it amongst online lecture notes and other sources. A really concise answer would be appeciated.

fluid dynamics - Reynolds number and inertial force

The Reynolds number is defined as the ratio of ´inertial´ forces to viscous forces.

$$ Re = \frac{\text{Inertial Forces}}{\text{Viscous Forces}}$$

Now, viscous forces make sense to me. They are frictional shear forces that come about due to the relative motion of the different layers in a flowing fluid, resulting in different amount of friction, hence, different viscosity values.

However, I am not really sure how to think about the 'inertial force'. This, to me, is somewhat of a dynamic effect since large Re numbers indicate turbulence in most cases, where there is a lot of motion, vortices and eddies. But what exactly is the inertial force and how can it be explained physically?

Answer

Inertial force, as the name implies is the force due to the momentum of the fluid. This is usually expressed in the momentum equation by the term $(\rho v)v$. So, the denser a fluid is, and the higher its velocity, the more momentum (inertia) it has. As in classical mechanics, a force that can counteract or counterbalance this inertial force is the force of friction (shear stress). In the case of fluid flow, this is represented by Newtons law, $\tau_x = \mu \frac{dv}{dy}$. This is only dependent on the viscosity and gradient of velocity. Then, $Re = \frac{\rho v L}{\mu}$, is a measure of which force dominates for a particular flow condition.

The inertial forces are what gives rise to the dynamic pressure. Another way to look at the Reynolds Number is by the ratio of dynamic pressure $\rho u^2$ and shearing stress $μ v/ L$ and can be expressed as $$Re =\frac{\rho u^2} {μ v/ L} = \frac{ u L} {\nu} $$

At very high Reynolds numbers, the motion of the fluid causes eddies to form and give rise to the phenomena of turbulence.

quantum mechanics - Von Neumann entropy in terms of the mutual overlap?

I have $N$ pure, but nonorthogonal, states $|\psi_n\rangle$ with density matrix $\rho_n=|\psi_n\rangle\langle\psi_n|$.

Say we call the the total density matrix $\rho=\frac{1}{N}\sum_n \rho_n$.

Are there any formulas to calculate $S_{vn}[\rho]=-\mathrm{Tr}[\rho\log{\rho}]$ solely from the overlaps $\langle\psi_n|\psi_m\rangle$ (and the fact that all individual states are pure)? Gram-Schmidt orthogonalization might be a possibility but perhaps there is an easier way/existing result?

Intuitively it would seem at first that knowledge of the $\mathrm{Tr}[\rho_n\rho_m]=|\langle\psi_n|\psi_m\rangle|^2$ would be sufficient because the mixing between the states is noncoherent, but is this true?

Answer

Let us denote the overlap matrix by $O$, this is, $O_{nm}=\langle\psi_n\vert\psi_m\rangle$. Then, $$ S_{\mathrm{vN}}(\rho)=S_\mathrm{vN}(O)\ . $$ More generally, $O$ has the same non-zero eigenvalues as $\rho$, so any function of the eigenvalues (which is insensitive to zero eigenvalues) can be evaluated on $O$ instead of $\rho$.

This can be seen by defining a matrix $X$ whose columns are the states $\vert\psi_n\rangle$, i.e. $X_{kn}=\langle k\vert\psi_n\rangle$. Then, $\rho=XX^\dagger$, while $O=X^\dagger X$. However, two matrices $AB$ and $BA$ always have the same non-zero eigenvalues, and thus, the non-zero eigenvalues of $\rho$ and $O$ are the same.

newtonian mechanics - Deformation of an elastic bar

We know that if we fix a bar at one of its ends, then the other one will descend by $s = A \cdot F l^3, A = const.$ (we can assume that $F$ is the gravitational force.

||
---------

||
^
solidly fixed here

We can assume that we know the descend of the end in this case and mark it by $x$

If we put it freely, like this

----------
||      ||
^       ^
held here 

Then I heard we can treat it like two bars solidly fixed in the middle (not sure why)

But what if we fix it at the both ends, like this:

||      ||
----------
||      ||
^       ^
solidly fixed here

How low will the lowest point of the bar descend?

general relativity - Interpretation of a singular metric

I'm interested to find out if we can say anything useful about spacetime at the singularity in the FLRW metric that occurs at $t = 0$.

If I understand correctly, the FLRW spacetime is a combination of the manifold $\mathbb R^{3,1}$ and the FLRW metric $g$. The metric $g$ has a singularity at $t = 0$ because at that point the proper distance between every pair of spacetime points is zero. Presumably though, however the metric behaves, the manifold remains $\mathbb R^{3,1}$ so we still have a collection (a set?) of points in the manifold. It's just that we can no longer calculate distance between them. Is this a fair comment, or am I being overly optimistic in thinking we can say anything about the spacetime?

I vaguely recall reading that the singularity is considered to be not part of the manifold, so the points with $t = 0$ simply don't exist, though I think this was said about the singularity in the Schwarzschild metric and whether it applies to all singular metrics I don't know.

To try to focus my rather vague question a bit, I'm thinking about a comment made to my question Did the Big Bang happen at a point?. The comment was roughly: yes the Big Bang did happen at a point because at $t = 0$ all points were just one point. If my musings above are correct the comment is untrue because even when the metric is singular we still have the manifold $\mathbb R^{3,1}$ with an infinite number of distinct points. If my memory is correct that the points with $t = 0$ are not part of the manifold then we cannot say the Big Bang happened at a point, or the opposite, because we cannot say anything about the Big Bang at all.

From the purely mathematical perspective I'd be interested to know what if anything can be said about the spacetime at $t = 0$ even if it has no physical relevance.

mathematics - How many Strobogrammatic numbers are there from 0 to 99999

0,1,2,5,8,11,69,96 are Strobogrammatic numbers.

We call a Strobogrammatic numbers if:

When it is typed on a calculator, and the calculator is spun 180 degrees, the number visually looks the same.

How many Strobogrammatic numbers are there from 0 to 99999?

This is a no-computer puzzle; only the first right answer with explanation will be accepted.

Answer

assumption: we can not usually have leading zeroes on a calculator, they are usually stripped.

the spinnable digits are 0, 1, 2, 5, (6/9), and 8

For 1 digit

Pos 1: we can have any of the five non-paired options (0 is not stripped in this one instance, as 0 is displayed on the calculator)

Total: 5

For 2 digits:

Pos 1: we can have any of the six non-zero spinnable digits

Pos 2: we can only have the inverse digit of pos 1

Total 6x1 = 6

For 3 digits:

Pos 1: we can have any of the six non-zero spinnable digits
Pos 2: we can have any of the five self-referential spinnable digits
Pos 3: we can only have the inverse digit of pos 1

Total 6x5x1 = 30

For 4 digits:

Pos 1: we can have any of the six non-zero spinnable digits
Pos 2: we can have any of the seven spinnable digits
Pos 3: we can only have the inverse digit of pos 2
Pos 4: we can only have the inverse digit of pos 1

Total: 6x7x1x1 = 42

For 5 digits

Pos 1: we can have any of the six non-zero spinnable digits
Pos 2: we can have any of the seven spinnable digits
Pos 3: we can have any of the five self-referential spinnable digits
Pos 4: we can only have the inverse digit of pos 2
Pos 5: we can only have the inverse digit of pos 1

Total: 6x7x5x1x1 = 210

Thus all spinnable numbers from 0 to 99999 is the sum:
5+6+30+42+210 = 293

So:

There are 293 spinnable numbers between 0 and 99999

homework and exercises - Current Density and the Dipole Moment $int _v vec{J} dV = d vec{p}/dt$

Show that: $$\int _v \vec{J} dV = \frac{d \vec{p}}{dt}$$

Here's my attempt: $$\vec{p} = \int _v \rho \vec{r} dV \rightarrow \frac{d \vec{p}}{dt}= \frac{d}{dt} \int _v \rho \vec{r} dV = \int _v \frac{\partial \rho}{\partial t}\vec{r}$$

Because of the fact that $\nabla \cdot \vec{J} = - \frac{\partial \rho}{dt}$ we have $$\frac{d \vec{p}}{dt}= -\int _v (\nabla \cdot \vec{J})\vec{r}dV$$

I'm not exactly sure how to turn this divergence of the current density per unit volume into the original $\int _v \vec{J} dV$

The last expression is very close to equaling the current, maybe we could use that?

Answer

You have the right idea. Let's use index notation for convenience. We have \begin{align} p^i(t) = \int_{\mathbb R^3} d^3x \,x^i\rho(t,\mathbf x) \end{align} and as you note, we have the continuity equation; \begin{align} \partial_jJ^j = -\frac{\partial\rho}{\partial t} \end{align} so we get \begin{align} \dot p^i = -\int_{\mathbb R^3} d^3 x \,x^i\partial_j J^j \end{align} where I'm using the summation convention for the dot product. This is the point to which you've basically gotten. The trick is now to essentialy perform an integration by parts. In other words, we use the product rule for differentiation to note that \begin{align} \partial_j(x^i J^j) &= (\partial_j x^i)J^j + x^i \partial_jJ^j \\ &= \delta^i_jJ^j +x^i\partial_jJ^j \\ &= J^i + x^i\partial_jJ^j \end{align} and plugging this into the above expression for $\dot p^i$ gives \begin{align} \dot p^i = -\int_{\mathbb R^3} d^3 x \partial_j(x^iJ^j) + \int_{\mathbb R^3} d^3x J^i \end{align} the first term is the volume integral of the divergence of $x^iJ^j$. Since we are integrating over all space, this turns into a boundary term at infinity which vanishes for any finite (or sufficiently rapidly decaying) charge density. The result you want then follows \begin{align} \dot p^i = \int_{\mathbb R^3} d^3 x J^i \end{align}

optics - Why do rainbows have distinct colors?

When I searched on the Internet for the reason of formation of rainbows, I got many explanations like this one & this. All the explanations consider only one spherical water droplet (like this one).

Why don't the refracted waves coming from the adjacent droplets interfere with each other? For example, why doesn't a red ray coincide with a violet ray coming from another droplet at a suitable position?

How are we able to see rainbows with distinct colors? Since there are thousands of raindrops in the sky when it is raining, shouldn't it be a mess due to overlapping of different wavelengths?

Please explain.

Diffusion vs Gravity in water: does a dissolved ion tend to "sink"?

Im a french student in geochemistry.

My question might be silly, but I became really too confused to answer it myself. Does gravity affect the diffusion of ions in water ?

Lets imagine a vertical volume of water. There is no temperature gradient. If some NaCl salt is placed (without disturbing the water) at the bottom, it will be dissolved quickely. But then ? : -Na+ and Cl- will start to diffuse as to get their concentrations homogeneous everywhere. There will be an upward flux . - Does gravity want them to remain at the bottom because both of them are heavier than water molecule ? Then it would be a downward flux.. (?)

If these two process occur together, how could I evaluate which is dominant, or what would be the equilibrium ?

I thought I could use something like a Boltzmann distribution ? Like Concentration = $\mathrm{exp}(mgz/kT)$ ? Im not sure though of how to consider an ion (Cl or Na) sinking between water molecules because of gravity

Thanks for help

astronomy - What made us think that Earth moves around the Sun?

Trying to observe the night sky for a few weeks, the motion of the Sun and the stars pretty much fits into the Geocentric Theory i.e. All of them move around the Earth.

What then, which particular observation, made us think that it could be the other way around, that all the planets move around the Sun?

Answer

which particular observation, made us think that it could be the other way around

Retrograde motion must be a prime candidate.

^{As seen from Earth against star background, Mars occasionally slows down and goes backwards. Our moon doesn't.}

It probably became clear to people constructing orreries that heliocentric models were enormously simpler and more convincing. They also tied in with simple inverse square laws of gravitation and planetary motion.

---

The discovery by Galileo Galilei of Jupiter's moons also provided firm evidence of the existence of heavenly objects that, perversely, did not orbit the Earth.

^{Photo: Thomas Bresson (Galileo probably didn't have a Nikon / mobile phone handy)}

^{Luckily, he had available a corner of a napkin, a goose and some soot (or equivalents)}

special relativity - Why can't we make measurements in a photon's rest frame when loop diagrams make measurements possible?

It is one of the axioms of special relativity that the photon has no rest frame; light travels at speed c when measured in any inertial frame of reference. As a corollary, it is often said that if one were in a photon's rest frame, infinite time-dilation and length-contraction would make the universe appear (from a photon's perspective) to be unchanging, with zero length in its direction of motion. In other words a measurement would not be possible in the reference frame of a photon, because there would be no time or space in which one could conduct it.

My question is: how does this accord with the fact that parallel-moving photons can interact with each other via loop diagrams[*]? Photons can therefore be used to perform measurements in the photon's rest frame. I suppose an answer may be "the act of measurement changes the photons momentum vector and therefore its co-moving frame is non-inertial" however this can be said of any measurement.

[*] Or can they? I can't find an explicit amplitude to check this (for, say, the photon-photon "box" diagram), however as far as I can tell from the Euler-Heisenberg Lagrangian nonlinear effects should be present for parallel photons.

logical deduction - Introducing: Compoundoku!

Rules of Compoundoku:

• Solve both left and right Sudokus.


• In addition, the board below them is the Compound Board of both Sudokus.



• Each number on the Compound Board should tell either: (1) the number on the left Sudoku, or (2) the sum of both numbers on the left and right Sudokus; in the respective position.

Answer

I think this is the answer

Partial Reasoning

For starters the 1 must appear on the left and the 6 must be the sum of both.
Once we make a choice of how to split the 6, the rest of the grid is filled in very quickly (4x4 Sudokus are very restrictive). If we put 3 on the left and 3 on the right or 2 on the left and 4 on the right, we can quickly reach a contradiction. However, putting 4 on the left and 2 on the right allows us to complete the grid in a unique way using the logic. A key step is realising that splitting the bottom 4 will eventually produce a contradiction so it must be that a 4 appears in that position on the left.

How to build pattern search puzzles

This is a question about puzzle-building. I very much enjoyed (albeit not solved) the puzzle Find the missing elephants by GOTO 0. And, as it is often the case with intriguing puzzles, it sparked some ideas of my own. (No need to worry, I'm not planning to put tons of clones online anytime soon...)

However, I then asked myself: How does one best build and verify such puzzles? Simple kernel-convolution puzzles I would have a good idea how to build & test, but anything with a complex conditional like in this particular puzzle, I don't know.

What tools/methods are you using/would you use to build such puzzles?

Answer

I cannot give any general recommendations on how to create great puzzles, but I'm glad to explain how the particular puzzle mentioned in the question was made.

The elephant puzzle was built (you guessed it) with a computer program: I tried out a few different algorithms before coming up with one that seemed good enough to make a puzzle.

Random numbers are produced with a seeded random number generator: this makes sure that we can always replay an interesting arrangement after we find one.

The board is created in three successive steps. First, all directions are set up randomly. Next, each row is scanned and cells are partitioned in subsets of color-matching elephants, ehm, elements. Each of these subsets is assigned a different color, chosen randomly from a palette of four colors. The last step is the rendering. This is done in Javascript/HTML. Just grab a screenshot and use GIMP to export the board.

The source code of the generator can be found here: https://github.com/fasttime/Elephant-Puzzle-Maker

general relativity - Are the gravitational wave mass estimates redshift dependent?

In answers to previous questions about (see, especially, this one) the LIGO detections it is stated that LIGO is unable to determine redshift, only luminosity distance. This is made clear in the papers. They do not provide a similar caveat to the mass/energy estimates, though. Schematically, because I don't know the equations, black hole mass determines orbital period and inspiral rate. Thus, the mass determines the frequency of the wave produced. The waves get redshifted in transit to Earth. So how is it that their mass estimates are not, in some way, redshift dependent? Are they some kind of "comoving mass" (mass multiplied by some factors of $1+z$), or are they true rest frame masses?

Put another way: what are the details of how they go from waveform to binary parameters and luminosity distance, with particular attention to where the redshift dependence cancels?

Answer

The final waveform visible in the detector depends on, 1) Intrinsic parameters: 2 Masses, 6 Spins 2) Extrinsic Parameters: Declination, Right Ascension, Luminosity distance, Orbital inclination, polarization and coalescence phase.

There are 3 degeneracies i.e. two different values will produce the same waveform 1) Luminosity distance - observed mass: A source mass Ms will be seen as Ms*(1+z) in the detectors. With detector noise in the observed waveform you cant tell what is the right combination ? Off-course the more you go away in z from true value you will also go away in waveform amplitude. 2) Inclination angle-luminosity distance: You cant tell if binary is poorly aligned and clode or nicely aligned(face-on) and far away. The waveform is the same 3) Partial degeneracy in spin-mass ratio: There is a partial degeneracy because a no-spin+high mass ratio (low_mass/high_mass components) will produce the same waveform as as high-spin+low mass ratio.

Now you create this mammoth parameter-space covered by these 9 parameters and see where does you match of generated waveform matches best with the observed waveform.

With 2-detectors its a real pain in the neck. With 3-detectors at least you can localize and get good initial numbers for sky-location (Ra, dec).