logical deduction - Five Hats and Three Logicians

Three logicians were seated around a table. They were blindfolded and a colored hat was placed on each of their heads. They were told a true statement

The hats on your heads were chosen among 3 black hats, 2 white hats.

Logicians look at each other's hat, after a while, logicians look and smile at each other and somehow they all figure out what colored hats are placed on their head like at the same time without even saying a word to each other.

How and what colored hats are placed on their heads?

Answer

I claim that:

All three of them are wearing a black hat

Reasoning:

Obviously you cannot have two of them wearing a white hat, or else the third would immediately figure out that his hat was black, but the other two wouldn't immediately know their hats were white (though they would know after the first guy smiles).

If you have only one white hat, the two wearing black hats won't know at first, since both see one white hat and one black hat. But after both notice the other doesn't know at first, the two of them would each deduce they're wearing black hats (since if either wore a white hat, the other would have seen two white hats and immediately known his hat was black). The guy in the white hat however, would not have been able to smile at the same time as the other two (as he gets no new information when nobody immediately smiles; that would've happened regardless of what hat he wore).

That essentially tells me that the only possible way all three figured it out at the same time is if all three were wearing black hats. What essentially happens then is:

1. Nobody smiles immediately as everybody sees two black hats


2. After no one immediately smiles, everyone is still clueless


3. After nobody smiles immediately, and nobody is able to use that fact to deduce their hat color, all three realize nobody sees a white hat (or else as outlined before, the initial "not-knowing" step is enough for the two black hat wearing logicians to figure out their colors). Hence, all three realize they're wearing black hats, and are able to deduce their hat color at (rougly) the same time.

knowledge - Spaced Out and Obfuscated

In this puzzle I took an English phrase, spaced it differently, then replaced the resulting words with synonyms or definitions. This is an example to show you what I mean.

Spaced out and obfuscated:

that man tape player button self-esteem

Just spaced out:

he (from "that man")
rew (from "tape player button")

ego (from "self-esteem")

Answer:

he rew ego $->$ here we go

Got it? Then all you have to do is apply the same process to the following:

definite article anime first child mucus coffee container within Scot. Eng. Wal. TCP/IP reply

Answer

The answer is

There is no turning back.

Because

definite article = the

anime first child = rei

mucus = snot

coffee container = urn

within = in

Scot. Eng. Wal. = GB

TCP/IP reply = ACK

alphametic - Identify unique Verbal Arithmetic Puzzles

I am attempting to create original Verbal Arithmetic puzzles such as the famous one from this question Alphametic (Verbal Arithmetic) general strategy :

SEND+MORE=MONEY

For these problems each letter represents a different one digit number but the equation still holds true. The first letter (M or S in this case) cannot be 0 of course as that would mean they would be omitted.

My main issue with creating new ones is that most of the puzzles I develop have more than 1 solution. For example, there are 3 correct and distinct solutions to:

TWO x SIX = TWELVE

In this case, I can reduce this to one answer by making it a long multiplication problem with marked blank areas for where digits are needed but this sort of ruins the problem. My question, therefore, is what strategies can I use to develop original, unique verbal arithmetic problems that only have one possible solution?

Edit: Can anyone think of any strategies for designing the puzzles, rather than just creating them and testing for the existence of solutions and uniqueness, as suggested by Ross Millikan's answer below?

pattern - Wordplay subtraction paradox

There are words from which you can remove a "chunk", leaving a new word. Like this:

    WISHBONE

   WI SHBO NE

 WI   SHBO   NE 


WI     NE

WI            NE

   WI     NE

     WINE

There are also words that work the other way, for which inserting a "chunk" produces a new word. For example, you can insert the chunk AUTIFI into the word BEER to make BEAUTIFIER.

A "chunk" is a string of consecutive letters. It must consist of at least two letters (no single-letter chunks). It does not need to be a valid English word.

Now, what if I told you there are words into which you can insert a chunk, then remove the same consecutive string of letters, and get a different word from the original word?

What the heck am I talking about?!

I have come up with about 11 examples of this strange phenomenon.

(Too easy? Too hard? Try the counterpart addition paradox.)

---

Afterword:

Here are my 11 examples:

Start with BEECHES --> insert ES --> obtain BESEECHES --> remove ES --> obtain BESEECH

Start with BEING --> insert ING --> obtain BINGEING --> remove ING --> obtain BINGE

Start with CODER --> insert DE --> obtain DECODER --> remove DE --> obtain DECOR

Start with CODERS --> insert DE --> obtain DECODERS --> remove DE --> obtain DECORS

Start with DEDUCT --> insert ED --> obtain DEDUCTED --> remove ED --> obtain DUCTED

Start with DEFEND --> insert DE --> obtain DEFENDED --> remove DE --> obtain FENDED

Start with PALED --> insert ED --> obtain PEDALED --> remove ED --> obtain PEDAL

Start with POSES --> insert SSES --> obtain POSSESSES --> remove SSES --> obtain POSSE

Start with REDESIGN --> insert ED --> obtain REDESIGNED --> remove ED --> obtain RESIGNED

Start with PAIRED --> insert RE --> obtain REPAIRED --> remove RE --> obtain REPAID

Start with DAUNTED --> insert UN --> obtain UNDAUNTED --> remove UN --> obtain UNDATED

Answer

Any word in a pattern so that

you can remove the same chunk from two different places

works as the intermediate word. For example:

RESIGNED → R(ED)ESIGNED → REDESIGN(-ed)
A wordfinder search through Qat is here.

electromagnetic radiation - Why do X-rays go through things?

I always heard that the smaller the wavelength, the more interactions take place. The sky is blue because the blue light scatters. So why is this not true for X-rays, which go through objects so readily that we need often use lead to absorb it?

Dayan speed cube 3x3x3 all corners twisted?

I have solved 3x3x3 cubes as well as 4x4x4, 5x5x5, 6x6x6, 7x7x7 and the megaminx from some time now but this one is really bugging me. I won't consider my self an expert but I have not experienced anything like this before. One of my students messed around with the cube and suddenly the corners won't orient correctly. I tried solving it several times but without being able to solve the corners. My first thought was that she must have twisted the corners too quickly but it does not help when I twist them back either way. I am really stuck. I do hope someone can help. I have taken pictures of the cube from all sides. As you can see all corners are twisted wrong. Cheers from Peter

Answer

It’s very unsolvable. The centers are not movable, and there is no possible way to orient the corner piece in the first picture so that it would align with the centers.

Most likely, the caps on the center pieces have been swapped.

My 3D visualization isn’t perfect, but I think swapping two opposing centers (no matter which) might fix this.

This is based on the observation that each corner has a possible spot with correct neighboring center colors, but two of the colors are always the wrong way, and you would need a mirror image of the corner piece to fit. (Even taking apart the cube wouldn’t help). The most likely reason is that the centers’ order has been mirrored, which happens when any two opposing centers are swapped.

thermodynamics - heat generation due to viscosity in a 3D fluid flow

Consider an arbitrary 3D fluid flow:

$$\vec{\nu}=\vec{V}\left( \vec{x} ,t \right) \tag{1}$$

where velocity at each point $\vec{\nu}$ is a function $\vec{V}$ of position $\vec{x}$ and time $t$ (non-steady). Due to the viscosity $\mu$ there is heat being generated at each point in space. I know this heat should be a function of velocity, viscosity and div/curl of velocity:

$$ \dot{q}=F\left(\vec{\nabla},\vec{\nu},\mu \right) \tag{2}$$

But I can't just find it out.

For a 2D problem with unidirectional horizontal velocity of:

$$ \nu_x=V_x\left(x,y,t\right) \tag{3}$$

I think I can write

$$ \dot{q}=\mu \frac{\partial \nu_x}{\partial y}\nu_x \tag{4}$$

Where $\tau_x=\mu \frac{\partial \nu_x}{\partial y} $ is the shear force due to viscosity. But I'm not quite sure if that's correct.

I would appreciate if you could help me know what is the correct form for equation 2.

P.S. I'm not quite sure but I think it should be something like:

$$ \dot{q}=\mu \left( \vec{\nu} \times \vec{\nabla}\right).\vec{\nu} \tag{5}$$

mathematics - 3x3 “Magic Square” of Prime Numbers -- Part II

Glad to know the previous puzzle, which was the first puzzle I posted in Puzzling, was warmly welcomed (Thank you!), and an optimal solution was found. Inspired by the comments there, here is the Version $2$ of the puzzle.

In fact, most of the things are unchanged. We still have this $3 \times 3$ grid, which $9$ distinct prime numbers $P_1, P_2, ..., P_9$ are to be filled in. And there are $8$ sums: $3$ horizontal, $3$ vertical and $2$ diagonal, and they are named $S_1, S_2, ..., S_8$. All the requirements in the first version still hold here, which mean:

• $P_1, P_2, ..., P_9, S_1, S_2, ..., S_8$ are all distinct prime numbers (i.e. there are totally $17$ different prime numbers).

But this time, one more additional requirement:

• The grand total $P_1 + P_2 + ... + P_9 + S_1 + S_2 + ... + S_8$ also has to be a prime number.

The challenge: To minimize the grand total.

With the additional requirement, some solutions satisfying the previous puzzle do not satisfy this version. And, the optimal solution will be different.

Below is one possible solution I come up with, which has a grand total of $601$, but it is not the optimal solution:

Feel free to have a try!

Answer

As our friend @KSab said, the best solution ever is 541. There are exactly 16 possible solutions which are shown in the images belown:

mathematics - What is the "linguistically hardest" number less than $10^9$?

The linguistic hardness ($LH$) of a natural number is the ratio of the amount of letters in the writing of this number in English to the amount of its digits.

For example, $LH(1234) = 7.75$, as:

$$\frac{\mbox{Number of letters}}{\mbox{Number of digits}} = \frac{\mbox{N(one thousand two hundred thirty four)}}{N(1234)}= 31/4 = 7.75$$

What the is $N: 0

P.S. This puzzle appears to be much more interesting in Russian, but this is not Russian site to ask it here.

Answer

For any given digit range the numbers with the longest spelling have the highest LH.

7-only-combinations are always among those numbers. Here's a table with their corresponding LH:

 number from here | number segment | letters from here | LH
------------------+----------------+-------------------+-------
        777777777 | seven hundred  | 87                | 9 2/3 = 9 14/21
         77777777 | seventy        | 75                | 9 3/8
          7777777 | seven million  | 68                | 9 5/7 = 9 15/21
           777777 | seven hundred  | 56                | 9 1/3
            77777 | seventy        | 44                | 8 4/5
             7777 | seven thousand | 37                | 9 1/4

              777 | seven hundred  | 24                | 8
               77 | seventy        | 12                | 6
                7 | seven          | 5                 | 5

Looks like 7777777 beat my previous suggestion. It worries me that LH(8878878) is the same (alongside others like 3878373).

Given entanglement, why is it permissible to consider the quantum state of subsystems?

Quantum entanglement is the norm, is it not? All that exists in reality is the wave function of the whole universe, true? So how come we can blithely talk about the quantum state of subsystems if everything is entangled? How is it even possible to consider subsystems in isolation? Anything less than the quantum state of the whole universe at once. Enlighten me.

cosmology - Can we have bubbles of the tenth dimension in our universe?

If the gap between the branes in the Big Bounce theory is the 10th dimension, and branes have wrinkles as they collide, can our universe have 'bubbles' in the 10th dimension? This is in reference to the Ekpyrotic theory or Big Bounce theory formulated by Dr Neil Turok and Dr Paul Steinhardt.

electromagnetism - How does reflection work?

In Newton's model of light as being composed of particles, it's easy to imagine reflection as being the rebounding of individual corpuscles off a surface. However, since light can also behave like a wave, it poses a challenge in visualizing reflection.

How does a wave reflect off of a surface, whether it be specular reflection or diffuse reflection? Must the wave be first absorbed, and then re-emitted? Or is there a different mechanism?

Answer

I'll just go ahead and write this down even though it's already been covered in the other thread.. but I didn't post there so :)

First, if you think of light as a (scalar) wave (which is really a semi-classical way of thinking but might be enough to answer your question) you can invoke the Huygen-Fresnel principle which in this case boils down to considering every point on the reflecting surface as an origin of a re-emitted spherical wave with a starting phase directly related to the phase the point got from the incident wavefront.

The superposition of these wavefronts, after you let them destructively interfere with each other, will amount to a new combined wavefront which propagates according to Snells law (angle of incidence = angle of reflectance). See this image for the corresponding illustration of refraction (which is very similar, I couldn't quickly find a good image of reflection):

Now, light really doesn't behave "sometimes as a particle, sometimes as a wave". It is always detected as quanta (particles) but the probability amplitudes (phases) propagate in a wavelike fashion. One way of expressing the propagation is to say that a photon is sort of split and takes every possible path between A and B (or, in the case of a reflector, from A to any point on the reflector and then from there to point B by any means). Every path gets a phase contribution, and all the indistinguishable paths are summed. Most paths are simply cancelling each other but some constructively interfere, creating a large contribution (in case you don't know QM, the probability amplitude squared is the probability of the described event so a large contribution means this result will most probably occur). There is a VERY good image and description of this process in Feynman - QED The Strange Theory of Light and Matter (as I wrote in the comment above).

In the case of the reflector, the large contribution occurs at the classical angle of reflectance (Snell's law again). Notice the similarity between this formulation (called the path integral approach) and the semi-classical principle outlined above; this is not a coincidence of course.

Also to briefly digress on your implied question on the per-atom non-zero "reflectance" time - saying that an electron-orbit absorbs the photon energy for a while the re-emits it a non-zero time later is of course also a slight simplification. In reality the electron interacts with the a photon, changes its momentum a bit, the re-emits (interacts) with the new photon and changes its momentum again. This scattering process occurs at all allowed momenta and intermediate times, which are then all superpositioned like above and hence I'm not sure it's meaningful to talk about any appreciable time of reflectance. Notice that this scattering is in practice very different from scattering which can excite the electron to another orbit.

general relativity - Dimensional analysis of metric tensor

In the geometry of GR, the metric tensor $g$ can determine the Riemannian connection and curvature tensor by combining the spatial derivatives (w.r.t. the 4d coordinate system) properly.

I am curious about the dimensional analysis of the metric tensor.

According to the geometric picture of GR, the connection as the potential is related with energy and the curvature tensor is related with force strength (with the mass to connect the connection/curvature with energy/force).

Then what's the dimension of the metric tensor? Intuitively it should be dimensionless, but how its first/secondary spatial derivative is related with energy/mass and force/mass=acceleration respectively?

Another observation is from the representation of Lorentz group. Where the rotation/boost is related with $SU(2)$ and $SL(2)$ transformations. If we take the $SU(2)$ or $SL(2)$ as transformations $U$ on quantum states, then they are dimensionless. So the acceleration (boost/time) can be regarded as $$dU/dt=H/\hbar=1/t$$ so we get boost is dimensionless, so time=length and energy=mass (these are normal conclusions since we usually take c=1). The reason that I check the Lorentz group representation is that the general metric tensor is generated from the Minkowski metric by dimensionless operation $GL(4)$, so this seems to confirm that the metric tensor should be dimensionless.

But if we go back to the former analysis, where the spatial derivative of the dimensionless metric tensor gives energy/mass, which is then also dimensionless since energy=mass. So we get the spatial derivative of a dimensionless value is still dimensionless.

There must be something wrong with my deduction. Can anybody help to clarify this?

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?

logical deduction - The round table

Three couples are dining together at a round table and they've all just ordered their desserts. Two waiters arrive at the table, both of them holding three plates. One is bringing the desserts for the three people on the left side of the table, the other is bringing the desserts for the three people on the right side of the table.

Based on the following (and previous) information, can you answer the two questions below?

1. None of the couples are sitting next to each other.



2. Ted is sitting in seat 6, in between two ladies.


3. Hugh's wife ordered the Danish pastry.


4. Faye is sitting exactly opposite to the person who ordered the jellyroll.


5. The upside-down cake was ordered by the person in seat 1, and is brought by the same waiter that brings the dishes for the people in seats 2 and 3.


6. The chocolate mousse was ordered by a person whose name comes later alphabetically than the name of the person who ordered the rice pudding. Both these dishes are brought by the same waiter, however.


7. Benjamin and Vanessa, who are sitting next to each other, are served by different waiters.


8. Yvonne is not seated exactly opposite to her husband.


9. One of the six guests ordered nut brittle.

Question 1: If the seats are numbered clockwise, who's sitting where, what are the three couples and what did everybody order? (Please supply the steps taken to arrive at the answer)

Question 2: What's the color of the table: red, green, blue, yellow, white, black or brown?

N.B.: The answer to the first question can be found using logic alone; for the second question a little lateral thinking is needed.

Hints for question 2:

Look at initials only. Seats 1-3 are on the right side, seats 4-6 are on the left side. Their dishes are all coming up right behind them...

quantum mechanics - Probability amplitude in Layman's Terms

What I understood is that probability amplitude is the square root of the probability of finding an electron around a nucleus, but the square root of the probability does not mean anything in the physical sense.

Can any please explain the physical significance of the probability amplitude in quantum mechanics?

I read the Wikipedia article on probability amplitude many times over. What are those dumbbell shaped images representing?

spacetime - In Special Relativity, is it allowed to ask 'How much time has elapsed in a second inertial frame at a particular moment in the first inertial frame'?

Or is it a meaningless question?

For example, A and his friend B are the same age initially. B travels relative to A at a very high speed. A keeps observing B from his frame. At one moment, A observes that B has aged only half as himself (say A is aged 60 when he makes the observation and observes B to be 30).

The question is 'How old is B from his own (B's) frame of reference at the moment A made the above observation?' If it has an answer, what is it? If the question is meaningless, why?

Answer

'How old is B from his own (B's) frame of reference at the moment A made the above observation?'

Your problem lies in the very moment you are talking about. In SR there isn't an unique moment for universe. A moment will be meaningful only if you specify your refrence frame of that moment.

i.e you should have asked something like this:

1.'How old is B from his own (B's) frame of reference at the moment $t$ in B's frame where A made the above observation?'

Or

2.'How old is B from his own (B's) frame of reference at the moment $t'$ in A's frame where A made the above observation?'

The answer to these questions is quite different! Unfortunetly i can't go on without math to explain differences. If you understood what i am saying then you need not to continue reading, unless you are interested in math as well.

To use lorentz transformations, we need to define events in spacetime. Let's say that A and B clocks are synchronized at $t=t'=0$. B arranges his birthday party when he become 30y/o. So we can assign an event in spacetime for this party $E_1=(ct,x)=(c(30y),0)$, (B is at orgin of his frame so $x=0$). If we were to use Galilean transformation to analyze this event in A's frame we would get $E_1'=(c(30y),-v(30y))$, in other words, according to Newton himself A will be 30y/o when he observes B's birthday party. (One might argue that light travel time between observers is not considered here. well, it's simple. Because observers are fully aware of their distance, they can tell how much time it took for the light to travel. So A can tell you guys when B's birthday actually happened though A will recieve singal much later in reality). With Lorentz transformation however, we will get $E_1'=(c\gamma(30y),\gamma(-vt))$. i.e according to Einstein, A is much older than B, when he observers B's birthday. Now back to the question, here i can show you why it's meaningless to ask

'How old is B from his own (B's) frame of reference at the moment A made the above observation?'

to answer your question From analogy above, one might jump to conclusion that of course B would be 30y/o, because after all we assumed so!(just check how did i define $E_1$) on the other hand, one might use inverse Lorentz transformation for event $E_1'$ above to see that $E_1=(c\gamma^2(30y),\gamma(vt))$ (as it was done by @RogerJBarlow) and conclude that B will be 120y/o. How is it possible?! well its because the moment you are talking about is different for every observers. in fact, if you were to ask the first question, the first answer would be your solution. on the other hand if you were to ask second question, the second answer would be right.

Update: Note that inverse Lorentz transformation is $t=\gamma (t'+vx'/c^2)$. if you pick $x'$ from event $E_1'=(c\gamma(30y),\gamma(-vt))$ you will arrive at $t=t$ which is obvious because using two Lorentz transformation at the same time does not change anything. However, if you assume another event ("observing itself") in A's frame such that $E_1'=(ct',x')=(c\gamma(30y),0)$ ($x'=0$ because A is at his orgin) then you will get the second answer.

TL;DR

In B's point of view, the moment he become 30y/o is not simultaneous with the moment that A observes him. And no, It's not because of light travel time. So he conclude that when A observes him, he is 120y/o while what A observes is B 30y/o!

Why is charge of the electron negative?

How did scientists figure out that the charge of the electron was indeed negative? I know how the cathode ray tube experiment works, but how did Thompson know that the plate that the cathode ray beam was attracted to was positive, meaning the cathode ray was negative? What is the history behind positive and negative charges leading up to the cathode ray experiment. Ben Franklin postulated that a body with excess electricity was positive as in a surplus of electricity, and that surplus flows from positive to negative. But how and when was it discovered that electricity flows from negative to positive?

homework and exercises - Gravitation and equilibrium (...and similar problems)

One problem that I'm having trouble with (as opposed to the other):

The Messenger is a probe that orbits Mercury $700 \rm km$ from the surface. What is the tangential velocity it should be rotating at so that it doesn't precipitate towards the planet, in $\rm m/s$?

Data: Mercury's mass $3,3 \times 10 ^{23} \rm kg$ Diameter: $4870 \rm km $ . Gravitational constant: $G=6,67 \times 10^{-11} \rm m^{3}kg^{-1}s^{-2}$

I assume I need to use the following $$a_{c}=\frac {V^2} r$$

$$F=G\frac{m_1 m_2}{d^2}$$

EDIT:

Solving for $V$ in $$G\frac{{{m_1}{m_2}}}{{{d^2}}} = {m_1}\frac{{{V^2}}}{d}$$ did it.

electricity - Electron-Current Relationship

I have a rather simple conceptual question that I cannot seem to find a straightforward answer to online. My question is "Does increasing the amount of electrons or increasing the speed of the electrons increase the current?

general relativity - What's the difference between the equivalence principle and curvature of spacetime?

Calculating using the equivalence principle only accounts for half the deflection of light, whereas the other half is from curvature of space-time.

But isn't the equivalence principle the same thing as curvature of spacetime?

Deflection of light has two parts: The first part is falling people measure the deflection. Why is it significant for falling people to do this and how does this follow principle of equivalence? Falling people means gravity and curvature of spacetime doesn’t play a role. This means we’ll know how much deflection is caused by the motion of the photon and general relativity? The second part is the curvature of space. For example, if you have a rod, the rod itself would bend around the Sun. These two parts are like two halves, each of it is one half the deflection.

Answer

There's a very intuitive, visual explanation of how the equivalence principles relates to the bending of light here, which includes an explanation as to how the geometry of space enters the picture in terms of link up up infinitesimal patches of space together.

Note that it the freely-falling frame is important here because according to the principle, physics should follow special relativity (over a small enough patch). That's what allows one to conclude that the light beam should be straight in that frame... and therefore must bent in the other.

But answer your question directly, no, it is simply not the case that the equivalence principle is "the same thing as" the curvature of spacetime. Rather, the equivalence principle establishes that gravity and the geometry of spacetime are inextricably linked, but it does not by itself prescribe a particular geometry that spacetime must follow.

Einstein's version of the equivalence principle, from which the deflection of light follows, is:

The outcome of any local non-gravitational experiment in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime.

In any spacetime geometry, a small enough patch looks flat, so in a free-falling frame spacetime looks just like special relativity (to first order). But this is true in any theory of gravity in which gravity manifests as spacetime geometry with gravitational free-fall following a geodesic. It does not have to be the geometry predicted by general relativity.

quantum field theory - Renormalized mass

I am reading Schwarz QFT and I reached the mass renormalization part. So he introduces, after renormalization, a physical mass, defined as the pole of the renormalized propagator, and a renormalized mass which can be the physical mass, but can also take other values, depending on the subtraction schemes used. Are these masses, other than the physical one, observable in any way experimentally, or are they just ways to do the math easier (using minimal subtraction instead of on-shell scheme, for example)?

Also, in the case of charge renormalization, the explanation was that due to the vacuum polarization, the closer you are to the charge, the more you see of it, so the charge of the particle increases with the momentum you are testing the particle with. However, I am not sure I understand, from a physical point of view, why do you need to renormalize the mass. Is this physical mass (defined as the pole of the renormalized propagator) the same no matter how close you get, or it also changes with energy? And if it changes, what is shielding it at big distances?

riddle - A reprieve from the brain-stretching - Clue Twenty Three

_{<<---First clue
<--Previous clue}

---

You push the door open. You look around, expecting another hard puzzle like the previous one.

You look at the floor. It says:

In recognition of you work, we would like to award you a Nob... sorry, wrong speech.

In recognition of your work, we are giving you an easy puzzle. Just a plain, simple, undisguised riddle. Hope you're happy. The next one's impossible.

The first part of me:
Apparently, I’m very lovely… but I’m also dangerous.
If you try to hug me, you may be pricked.
Don’t make me blush – I’m red enough.

But I’m also as pale as a ghost.
And yellow as the sun.

The second part of me:
Bloody? Me? Well, yes.
But that was a long time ago.
At least I wasn’t like my father!
I am going to have mutton for dinner.
This livestock is really getting on my nerves.

Add us together, and what do you see?
This is the answer to Clue 23!

---

_{Next clue--->}

Answer

I see

ROSEMARY

Apparently, I’m very lovely… but I’m also dangerous.
If you try to hug me, you may be pricked.

Lots of people like roses, but their thorns are painful to touch.

Don’t make me blush – I’m red enough.
But I’m also as pale as a ghost.
And yellow as the sun.

Roses come in red, white and yellow.

Bloody? Me? Well, yes. But that was a long time ago.

Mary I of England is known as "Bloody Mary".

At least I wasn’t like my father!

Henry VIII ordered the execution of many people.

I am going to have mutton for dinner.
This livestock is really getting on my nerves.

Reference to the poem Mary had a little lamb.

Can we see a spaceship falling into a black hole and entering the event horizon?

Or it pauses in time because the spaceship reaches the speed of light, c?

Answer

There is a powerful time dilation element to what is seen by a distant observer, and the falling object appears to grow every slower (as well as ever more red shifted) as it approaches the event horizon. Asymptotically so, such that (if you are observing on a long enough wavelength) the falling object never quite vanishes.

soft question - Why "Dark Energy" is called energy instead of force?

The overly simplified explanation I'm giving myself right now is dark energy causes the opposite of what gravity does, that's why the universe is expanding. Now where gravity is a force, why dark energy is "energy"? Why it's not called "Dark Force" instead? I think I must be missing something here.

Answer

Because as far as we understand general relativity, it's not doing "the opposite of what gravity does." Gravity can be locally attractive or repulsive, depending on whether the stress-energy content satisfies or violates the strong energy condition. For ordinary matter, the stress-energy is dominated by the mass, the SEC holds, and its gravity is attractive. But this needn't be the case in general.

Since gravity depends on more than the (mass-)energy content, in cosmological models, the universe accelerates or decelerates proportionally to $\rho+3p$, where $\rho$ is the energy density and $p$ is the pressure. The factor of $3$ for the pressure comes from the fact that there are three spatial dimensions, but only one temporal dimension. In particular, the cosmological constant corresponds to the case of a perfect fluid with $p = -\rho$ (which can be generalized to other "equations of state"), and so a positive "dark energy" density has a repulsive effect, not because it's "the opposite of what gravity does", but rather because its negative pressure gravitates in addition to its energy.

word - Have You Heard My New Riddle?

A stirrup I own that no stallion can wear.
I am not a smith though an anvil I bear.
My hammer? No Middle Earth dwarf gives a care.

With diamonds or pearls adorned may I be.
The heart of my heart holds the truth secretly.

In truth rearranged are my letters, you see.

The answer is a single English word.

Answer

The answer is:

ear

A stirrup I own that no stallion can wear.

A stirrup is the name of a bone in the ear. Stapes.

I am not a smith though an anvil I bear.

Anvil is another bone. Incus.

My hammer? No Middle Earth dwarf gives a care.

The other bone. Malleus.

With diamonds or pearls adorned may I be.

Earrings/studs etc

The heart of my heart holds the truth secretly.

The middle (heart) of the word heart is ear.

In truth rearranged are my letters, you see.

Are is an anagram of ear.

If the dimension of a space is prime, are quantum states in it guaranteed to be entangled?

A rather obvious question perhaps but if I have a Hilbert space of dimension $d$ and $d$ is prime, I cannot possibly write my state as $$\rho = \sum_i p_i\rho_a\otimes\rho_b$$ simply because the dimensions don't add up correctly, yes?

In general, is the number of factors of $d$ useful if one wants to check if a state is entangled or not?

Answer

If the dimension is prime, there is no point in talking of two subsystems (which would be entangled), and thus no point in talking about entanglement (which is a property between two subsystems). Since, if subsystem 1 has dimension $d_1$ and subsystem 2 has dimension $d_2$, the total Hilbert space dimension would be $d_1d_2$, which is not prime.

Why are materials that are better at conducting electricity also proportionately better at conducting heat?

It seems like among the electrical conductors there's a relationship between the ability to conduct heat as well as electricity. Eg: Copper is better than aluminum at conducting both electricity and heat, and silver is better yet at both. Is the reason for this known? Are there materials that are good at conducting electricity, but lousy at conducting heat?

Answer

See http://en.wikipedia.org/wiki/Thermal_conductivity In metals, I think it generally has to do with the higher valence band electron mobility, but it gets more interesting elsewhere.

In metals, thermal conductivity approximately tracks electrical conductivity according to the Wiedemann-Franz law, as freely moving valence electrons transfer not only electric current but also heat energy. However, the general correlation between electrical and thermal conductance does not hold for other materials, due to the increased importance of phonon carriers for heat in non-metals. As shown in the table below, highly electrically conductive silver is less thermally conductive than diamond, which is an electrical insulator.

special relativity - Faster than light possibility?

According to general relativity, speed is relative, so for example if you are running at 20 km/h and a car passes you at 30 km/h, the runner is actually moving at 50 km/h relative to the car.

Now imagine that 2 spaceships are traveling both at 99% the speed of light, and they pass each other in a linear fashion. Could it be argued that from 1 spaceships perspective, you are traveling at 198% the speed of light relative to the other spaceship, according to special relativity, yet violating the constant speed of light?.

electromagnetic radiation - Are the Properties of the EM Spectrum Fluid?

To put it simply. We group our EM waves into groups such as X-rays, microwaves, visible light etc. I was wondering, if the properties of say, x-rays, slowly change into the properties of say, gamma rays, or as soon as the wavelength is larger than 10 picometers it's properties become that of an x-ray. If it's the former, how do we calulate the standards for identifying a wave.

Answer

The EM waves form a continuous spectrum, thus your first description of a gradual change in wave behaviour is correct. However, because of this; there is also no precise way of determining exactly at which frequency a category of EM wave starts or finishes at.

In fact, sometimes the classification of waves can be so hand wavy that the typical description of an EM wave: $$c=f\lambda$$ is treated as a secondary determinant for which category an EM wave belongs to! One such example of this is with X rays and $\gamma$ rays. Their frequencies overlap one another at approximately $10^{18}\:\mathrm{Hz}$ so it is difficult to quantitatively differentiate higher frequency X rays and lower frequency $\gamma$ rays. Instead, we let the origin of the waves decide, $\gamma$ rays are emitted from the nucleus of the atom while X rays are emitted from outside the nucleus; by excited electrons.

To conclude, the EM spectrum can in no way be easily divided into discrete categories and as such there is always some overlap between adjacent categories of wave.

general relativity - Alcubierre Drive - Clarification on relativistic effects

On the Wikipedia article on the Alcubierre drive, it says:

Since the ship is not moving within this bubble, but carried along as the region itself moves, conventional relativistic effects such as time dilation do not apply in the way they would in the case of a ship moving at high velocity through flat spacetime relative to other objects.

And...

Also, this method of travel does not actually involve moving faster than light in a local sense, since a light beam within the bubble would still always move faster than the ship; it is only "faster than light" in the sense that, thanks to the contraction of the space in front of it, the ship could reach its destination faster than a light beam restricted to travelling outside the warp bubble.

I'm confused about the statement "conventional relativistic effects such as time dilation do not apply".

Say Bob lives on Earth, and Jill lives on a planet in Andromeda, and we'll say for the sake of argument that they're stationary. If I were to travel from Bob to Jill using an Alcubierre drive such that the journey would take me, say, 1 week from my reference frame... how long would Jill have to wait from her reference frame? Do the time dilation effects cancel out altogether? Would she only wait 1 week?

Answer

Spacetime can dynamically evolve in a way which apparently violates special relativity. A good example is how galaxies move out with a velocity v = Hd, the Hubble rule, where v = c = Hr_h at the de Sitter horizon (approximately) and the red shift is z = 1. For z > 1 galaxies are frame dragged outwards at a speed greater than light. Similarly an observer entering a black hole passes through the horizon and proceeds inwards at v > c by the frame dragging by radial Killing vectors.

The Alcubierre warp drive is a little spacetime gadget which compresses distances between points of space in a region ahead of the direction of motion and correspondingly expands the distance between points in a leeward region. If distances between points in a forwards region are compressed by a factor of 10 this serves as a “warp factor” which as I remember is $w~=~1~+~ln(c)$, so a compression of 10 is a warp factor 3.3. The effect of this compression is to reduce the effective distance traveled on a frame which is commoved with the so called warp bubble. This compression of space is given by $g_{tt}$ $=~1~-~vf(r)$.

Of course as it turns out this requires exotic matter with $T^{00}~<~0$, which makes it problematic. Universe is also a sort of warp drive, but this is not due to a violation of the weak energy condition $T^{00}~\ge~0$. Inflationary pressure is due to positive energy. The gravity field is due to the quantum vacuum, and this defines an effective stress-energy tensor $T^{ab}$ with components $T^{00}~=~const*\rho$, for $\rho$ energy density, and $T^{ij}~=~const*pu^iu^j$, for $i$ and $j$ running over spatial coordinates $u^i$ velocity and $p$ pressure density. For the de Sitter spacetime the energy density and pressure satisfies a state $p~=~w*\rho$ where $w~=~-1$. So the pressure in effect is what is stretching out space and frame dragging galaxies with it. There is no need for a negative energy density or exotic matter.

Negative energy density or negative mass fields have serious pathologies. Principally since they are due to quantum mechanics the negative eigen-energy states have no lower bound. This then means the vacuum for these fields is unstable and would descend to ever lower energy levels and produce a vast amount of quanta or radiation. I don’t believe this happens. The Alcubierre warp drive then has a serious departure between local laws of physics and global ones, which is not apparent in the universe or de Sitter spacetime. The Alcubierre warp drive is then important as a gadget, along with wormholes as related things, to understand how nature prevents closed timelike curves and related processes.

Addendum:

The question was asked about the redshift factor and the cosmological horizon. This requires a bit more than a comment post. On a stationary coordinate region of the de Sitter spacetime $g_{tt}~=~1~-~\Lambda r^2/3$. This metric term is zero for $r~=~\sqrt{3/\Lambda}$, which is the distance to the cosmological horizon.

The red shift factor can be considered as the expansion of a local volume of space, where photons that enter and leave this “box” can be thought of as a standing wave of photons. The expansion factor is then given by the scale factor for the expansion of the box $$ z~=~\frac{a(t_0)}{a(t)}~-~1 $$ The dynamics for the scale factor is given by the FLRW metric $$ \Big(\frac{\dot a}{a}\Big)^2~=~\frac{8\pi G\rho}{3} $$ for $k~=~0$. The left hand side is the Hubble factor, which is constant in space but not time. Writing the $\Lambda g_{ab}~=~8\pi GT_{ab}$ as a vacuum energy and $\rho~=~T_{00}$ we get $$ \Big(\frac{\dot a}{a}\Big)^2~=~H^2~=~\frac{\Lambda}{3} $$ the evolution of the scale factor with time is then $$ a(t)~=~\sqrt{3/\Lambda}e^{\sqrt{\Lambda /3}t}. $$ Hence the ratio is $a(t)/a(t_0)~=~ e^{\sqrt{\Lambda /3}(t-t_0)}$.The expansion is this exponential function, which is Taylor expanded to give to first order the ratio above $$ a(t)/a(t_0)~\simeq~1~+~H(t_0)(t_0-t)~=~1~+~H(t_0)(d-d_0)/c $$ which gives the Hubble rule. $z~=~a(t)/a(t_0)~-~1$. It is clear that from the general expression that $a(t)$ can grow to an arbitrarily large value, and so can $z$. On the cosmological horizon for $d~-~d_0~=~r_h~=~\sqrt{3/\Lambda}$ we have $z~=~1$.

Looking beyond the cosmological horizon $r_h~\simeq~10^{10}$ly is similar to an observer in a black hole looking outside to the exterior world outside the black hole horizon. People get confused into thinking the cosmological horizon is a black membrane similar to that on a black hole. Anything which we do observe beyond the horizon we can never send a signal to, just as a person in a black hole can see the exterior world and can never send a message out.

classical mechanics - What is the difference between configuration space and phase space?

1. 
   

   What is the difference between configuration space and phase space?

   
   
   


2. 
   

   In particular, I notices that Lagrangians are defined over configuration space and Hamiltonians over phase space. Liouville's theorem is defined for phase spaces, so is there an equivalent conservation law for the configuration space?

   
   


3. 
   

   In the process of modeling a physical system, when it is appropriate to use one instead of the other?

   
   

Fluid Mechanics explanation of an object levitated next to an air jet

I just saw the following video:

Can anyone tell me how this works? It seems to be through Bernoulli's principle, where the air pressure inside the tape roll decreases due to lower speed of the air through it. Or is it Coanda effect?

Answer

Coanda. Essentially two things are going on here:

• 
  

  Coanda effect: The jet of air tend to follow the surface: so it's thrown to the right and down in this video, which (Newton's third law) means the object is pushed up and to the left; but if it goes too much to the left, it'll be pushed back by the direct "hit" of the jet. That's what makes it stable.

  
  


• 
  

  Dynamic pressure, which helps keeping the object afloat against the pull of gravity: the jet has upwards momentum, is to a great extent driven sideways as the object blocks its path and, in doing so, exerts force on it. It also keeps the cylinder next to the jet, instead of directly at it, as can happen at lower speeds.

  
  
  

That's very well explained in this video from the Veritasium YouTube channel: Hydrodynamic Levitation!

quantum mechanics - Hubbard-Stratonovich transformation and decoupling channels

I'm studying an example of the Hubbard-Stratonovich transformation in Altland and Simons' Condensed Matter Field Theory (2nd ed.), pp. 246-247.

In it they say that...

one is frequently confronted with situations where more than one Hubbard-Stratonovich field is needed to capture the full physics of the problem. To appreciate this point, consider the Coulomb interaction in momentum space. $$S_{int}[\bar{\psi},\psi] = \tfrac{1}{2} \sum_{p_1,...,p_4} \bar{\psi}_{\sigma, p_1} \bar{\psi}_{\sigma', p_3} V(\textbf{p}_1-\textbf{p}_2) \psi_{\sigma', p_4} \psi_{\sigma, p_2} \delta_{p_1-p_2+p_3-p_4}.$$ In principle, we can decouple this interaction in any of the three channels...

discussed in the previous page. If one chooses to decouple in all three channels then the action becomes ...

$$ S_{int}[\bar{\psi},\psi] \simeq \tfrac{1}{2} \sum_{p,p',q} ( \bar{\psi}_{\sigma, p} \psi_{\sigma,p+q} V(\textbf{q}) \bar{\psi}_{\sigma', p'} \psi_{\sigma',p'-q} - \bar{\psi}_{\sigma, p} \psi_{\sigma',p+q} V(\textbf{p'}-\textbf{p}) \bar{\psi}_{\sigma', p'} \psi_{\sigma,p'} - \bar{\psi}_{\sigma, p} \bar{\psi}_{\sigma', -p+q} V(\textbf{p'}-\textbf{p}) \psi_{\sigma,p'} \psi_{\sigma',-p'+q} )$$

where the first term is decoupled via the

direct channel $\rho_{d,q} \sim \sum_{p} \bar{\psi}_{\sigma,p} \psi_{\sigma,p+q}$, second in the exchange channel $\rho_{x,\sigma\sigma',q} \sim \sum_{p} \bar{\psi}_{\sigma,p} \psi_{\sigma',p+q}$, and third in the Cooper channel $\rho_{c,\sigma\sigma',q} \sim \sum_{p} \bar{\psi}_{\sigma,p} \bar{\psi}_{\sigma',-p+q}$.

It's generally a good strategy to decouple in all available channels when one is in doubt, then let the mean-field analysis sort out the relevant fields.

My question is, if we choose to decouple the quartic term via 3 different channels (for example) is it necessary to multiply the resulting terms by a factor of $\tfrac{1}{3}$? This isn't discussed in the textbook and I'm confused by the liberal use of $\sim$ and $\simeq$ in the examples.

Answer

No. You should not add a factor of $1/3$. As you can see in page 244 of Altland and Simons, the HS transformation is done by multiplying by a unity expressed as a functional integral over an auxiliary field. In this case, they just choose to introduce 3 different fields - 1 for each term.

Does it matter in which direction I travel in relativity theory?

I'm not a physicist, but I'm still very interested in the relativity theory, especially in how the twin paradox is explained. Actually, it does not make sense to me and I hope you can answer my following question to help me understand how it works:

Does it matter in which direction i travel in the relativity theory? When I first read about the theory, I thought, that it does not matter if A moves away from B, or B moves away from A, because every party will see the effect of "time going slower" for the other pary, meaning A sees B aging slower and B sees A aging slower.

But regarding the twin paradox, when A leaves the earth with insane speed, and returns years later, A has aged less than the people on earth. If this is actually true, then my previous assumption is wrong and it does matter if A moves away from B or the other way round, because the effect could be turned around, so that the earth ages less than A, if the earth moved away from A instead.

If this is true, then this brings up a next question: How can we know if B moves away from A or A moves away from B? Since the earth, our solar system, and even our galaxy is already traveling in the universe at some speed in a specific direction, we could actually stop moving by flying in the exact opposite direction at the same speed. but would this be considered as the earth moving away or as we moving away?

I really hope you can help me with that :)

Update for @Alfred Centauri's answer

the stay-at-home twin does not change direction while the travelling twin does.

When I look at it from only the travelers perspective, then yes. But when I look at it from the earth's perspective, then the earth indeed changes its direction, while the traveler does not. Why is that changing direction thing that important?

As I could see in Will's answer, the change of speed in universe seems to be the critical difference. In the time where the traveler is still on the earth, both the traveler and the earth move at the same speed in the universe, which means they age exactly the same. As soon as the traveler starts to accelerate or decelerate (move faster or slower than the earth) he slows his own time, no matter if he accelerates or decelerates.

If this is true, this means that not only the relative speed of the objects influences the time. Also the change of speed in the universe does.

But this opens another question: What if 2 travelers move to the exact same position in the universe, but traveler A goes there at twice the speed? Are the two travelers the same age when they meet at the destination? Or is one traveler older because he moved faster?

Answer

If this is actually true, then my previous assumption is wrong and it does matter if A moves away from B or the other way round,

In the context of the Twin "Paradox", it doesn't matter. It isn't the direction that makes the difference, it is the path through spacetime.

The key to understanding the twin paradox without the frightening mathematics of GR is this: from any IRF (inertial reference frame), the stay-at-home twin does not change direction while the travelling twin does.

For example, chose as your reference frame, instead of the one above, the frame of the travelling twin on the outbound leg (we've switched which twin is "moving away").

In this frame, it is the stay-at-home twin that ages slowly, during the outbound phase, compared to the travelling twin. However, and crucially, the travelling twin must change direction in order to return to the other twin.

So, at the halfway point, the travelling twin changes direction and is now travelling faster than the stay-at-home twin.

Now, it is the travelling twin that is ageing more slowly than the stay-at-home twin. Moreover, due to the non-linear time dilation factor, the travelling twin is ageing slowly enough on the inbound leg such that the total ageing along the outbound and inbound paths is less than the total ageing along the straight path of the stay-at-home twin.

The bottom line is that the situation isn't symmetric. The stay at home twin never changes direction while the travelling twin does.

But when I look at it from the earth's perspective, then the earth indeed changes its direction

No, from any inertial (non-accelerated) reference frame, the stay-at-home twin's path through spacetime is straight.

You're not taking into account that the travelling twin changes from one IRF (the outbound leg) to another (the inbound leg), i.e., the travelling twin boosts from one frame to another while the stay-at-home twin does not. The travelling twin's frame of reference is non-inertial during the turnaround.

This is the crucial difference between the two paths through spacetime; one has a "kink" (the turnaround), and the other does not.

Why is that changing direction thing that important?

Because a straight spacetime path connecting two events is different from a curved or kinked spacetime path connecting the same two events.

Intuitively, in Euclidean geometry, the shortest path between two points is a straight line.

Counter-intuitively, in the Minkowski geometry of spacetime, the straight path between two events is the "longest" in the sense that the elapsed time (proper time) along a straight path is larger than for any other path.

Since the stay-at-home twin's path is straight, the ageing along that path is greater than for any other path.

puzzle creation - Help me to find a small but hard and clever maze

For a competition, I need to choose a maze puzzle, only one. There are tons of mazes, but most of them are not clever, i.e. they are just solved by trial and error.

Few days ago I saw A blue, white and red maze. It's a clever puzzle: when you approach it with trial and error, you can easily fail and conclude that it doesn't have a solution, but it does. I liked it a lot. A problem with it? It is too simple, as I think most of the people can solve it eventually.

So, I need a maze, which:

• has a solution;


• is clever, i.e. not a usual maze where you just walk in all possible ways until you find an exit. In this puzzle, you should be required first to find an idea how to approach it, only then to try different paths;


• hard, i.e. the idea should not be trivial and most of the people who try to solve the maze should feel like there is no solution.


• relatively small, with rules, which are easy to understand. Let's say it should fit to a piece of squared paper of size about 15 by 15 cells;

Could you help me to find such a puzzle?

P.S. Looking at Near-impossible puzzle for Christmas, I suppose this is not off-topic.

mathematics - Arranging a long line of cards

Annie has a very long line of cards, numbered from 1 to $n$, such that every odd card is face-up and every even card is face-down. The only way she can modify the line is by taking any segment of consecutive cards that has an even number of face-up cards and reversing the order of the cards in this segment (note that none of the cards are flipped).

As a neat freak, she wants to get all the face-up cards to be next to each other. Tell her how to achieve this or show that it is impossible if:

1. $n = 2014$


2. $n = 2016$

Answer

For 2014, first execute the following operations:

$(1, 4)$ $(2, 8)$ $(3, 12)$ $(4, 16)$ ...

($(a, b)$ is reversing the segment starting with the $a$th card and ending with the $b$th one)

Then:

Arrange the blocks of 2 face-up cards together with the lone one and it's done.

For 2016,

I think it's impossible. Notice that the operation is invertible. The final state for 2016 cards is a contiguous block of 1008 face-up cards. However, the operation cannot make a block of odd length if there wasn't one to begin with, because the reversed segment should start and end with face-up blocks of the same parity (because in its middle can only be blocks of even length). And we are supposed to end up with a lot of separate face-up cards.

logical deduction - Left coin, right coin, last coin?

In this game, all players start sitting at a round table, each with a coin in their hand. At every turn, all players toss their coins together and put them on the table in front of them. Then each player looks at the coins tossed by his left and right neighbours. If the neighbours tossed different faces of the coin (one obverse and one reverse), the player in the middle survives for the next turn. Otherwise, if the neighbours both tossed the same face of the coin (both obverse or both reverse), the game is lost for the player sitting in between them.

After everybody knows who's continuing to the next turn, the losers pick up their coins and leave the table.

The game is won when there is exactly one player left sitting at the table, who will be the winner and grab the lucky $10 bonus.

If all players are eliminated at the same time, the game is a tie and nobody gets the bonus. If there are only two players left, since left and right neighbours will be the same opponent, the players will necessarily eliminate each other at the next turn.

You are offered a chance to play this game and grab the lucky $10 bonus. You can choose to play in a game starting with 10, 11, or 12 players including yourself.

What would you choose and why?

If the lucky $10 bonus sounds boring, replace it with 10 rep points or whatever you find worth playing for.

Answer

It doesn't matter which option you choose, because

it's impossible to win.

Your probability of survival if you're one of n players left is as follows:

n=1: 100% (obviously!)
n=2: 0%
n=3: 0%

n=4: 0%
n=5: 0%
n=6: 0%
n=7: 0%
n=8: 0%
n=9: 0%
n=10: 0%
n=11: 0%
n=12: 0%

Informal proof
It was established in the question that if there are only 2 players left, they must annihilate each other. So if there are 3 left, your only hope of survival is to annihilate both your opponents, which can only happen if your result (WLOG, heads) is the same as both of theirs, which means their results must be the same, which means you're annihilated too.

The same argument goes on. If there are $n$ players left, where $n>1$, your only hope is to annihilate them all in one go (since all lower cases have been checked), which means annihilating yourself too. The proof is slightly different for odd and even $n$.

Formal proof
Let $S_n$ be the statement that if there are n players left, you can't possibly win. We've already proved $S_2$ and $S_3$, so by strong induction it will suffice to prove that

if $S_k$ is true for $1

If $S_k$ is true for $1all your opponents in one go. (Otherwise, at the next stage you'll be one of $k$ players remaining where $1

If $n$ is odd, you're already done at this stage because by proceeding in steps of 2 you cover all the other players before coming back to yourself (whom you don't want to eliminate!), so everyone gets heads and you get eliminated too.

If $n$ is even, you now know that you need every second player, starting from yourself, to get heads. In order to eliminate C, you need B and D to get the same (doesn't matter whether it's heads or tails); then to eliminate E, you need D and F to get the same, and so on. Taking this all the way round, you find that B, D, F, H, ..., V, X, Z (ending with Z since you know $n$ is even) to get the same result. But B and Z getting the same result means you get eliminated too.

So you can't win when there are $n$ players remaining. QED.

classical mechanics - Motion of a bouncing sphere with a spring attached inside

Imagine a sphere with inside a spring attached (between opposite sides). You let it fall from a certain height, after which it bounces from a flat surface. The sphere is rigid.

Will the following motion be chaotic or can we put a nice function upon its subsequent motion? For simplicity, we’ll let friction out of the game.

Answer

I assume that the spring’s mass is distributed homogeneously across the spring.

In this case, you have to use a partial differential equation (PDE) to perfectly describe the spring, whose solution would for example be a function which describes how each infinitesimal segment of the spring is shifted from its resting position (when no external forces and gravity are applied). Under benign boundary conditions the solution to these PDE would be longitudinal waves propagating on the spring.

However, the boundary conditions of the PDE for this problem are described by the sphere’s motion but also feed back into the sphere’s motion: The spring perceives a “kick” (starting a new wave) whenever the sphere bounces, and the sphere’s motion is affected by waves reaching the end of the spring. Thus I strongly doubt that there is a nice analytic solution to this problem – except perhaps in some benign cases, such as when the spring’s natural frequency matches the frequency of the bouncing.

As for chaoticity, you have infinitely many degrees of freedom (the position and momentum of the sphere as well as that of every point in the spring). You also have at least one clear non-linear component introduced by the sphere’s bouncing off the surface (you can have more if the spring is non-linear or due to the PDE’s boundary conditions). So, all the necessary ingredients for chaos are present.

I thus would expect chaos for at least some parameter choices. While I cannot prove this system without actually simulating it, I can appeal to intuition: If the parameters of the system are such, that the reflection of a spring wave at the spring’s end can be seen as a singular event, it can make a considerable difference whether this reflection happens before or after the sphere bouncing. Here a tiny difference in the state can eventually considerably affect the dynamics and thus you have the butterfly effect. Moreover the system bears a certain resemblance to the kicked oscillator, which exhibits chaos and is arguably less complicated.

elasticity - Elastic band around a cylinder

An elastic band is stretched using a known force and then placed around a cylinder. How are the forces or tensions distributed? I assume there will be two components: firstly, a tangential or circumferential component, and secondly, a radial or centripetal component. How are these components calculated?

astronomy - How is it the Voyagers are a few seconds closer to Earth than earlier?

The Voyager 2 tweet of March 01, 2013 put it's distance at 14 hrs 04 mins 23 secs of light-travel time from Earth. A more recent (earlier today) tweet says it is 14 hrs 04 mins 22 secs of light-travel time from Earth. An older tweet put it still further out at 14 hrs 04 mins 25 secs of light-travel time.

The Voyager 2 tweets on Voyager 1's proximity to Earth too has reduced similarly to 17 hrs 07 mins 07 secs of light-travel time earlier today from 17 hrs 07 mins 11 secs of light-travel time reported on Feb 28 2013

It could quite simply be a typo ... If not a typo then my curiosity is piqued; I was under the impression the Voyagers are outbound from the Solar System now without any major in-system activities now.

How is it the Voyagers are closer to Earth than before?

Answer

"Note: Because Earth moves around the sun faster than Voyager 1 is traveling from Earth, the distance between Earth and the spacecraft actually decreases at certain times of the year. " - quoted from http://voyager.jpl.nasa.gov/where/index.html

harmonic oscillator - Why can waves (audio, eletromagnetic, etc.) be represented by a circle?

I found an intuitive reason for the sine function, but want one that can be used for all kinds of waves. I've found a good explanation summarized in the following figure:

It is logical because a circle is symmetric and starts to change in height fast near the x axis, but gets slow near the y axis. But intuitively, why is it just a circle?

Answer

After much research, I found a posted answer that does not depend on such arbitrary facts as assuming Hooke's law is true.

Suppose Energy(E) as a continuous function of displacement (y)

$$\color{blue}{E = f(y)}$$

Using Taylor Series for a continuous function we have (where $\color{blue}{'}$ represent a derivative)

$$\color{blue}{E(y) = E(0) + E\,'(0)y + \frac{E\,''(0)y^2}{2\,!} + \frac{E\,'''(0)y^3}{3\,!} + ...}$$

$\color{blue}{E(0)}$ is a constant which depends on reference. It can be considered as $\color{blue}{0}$ and selected as, for instance, the position of the tip of a stretched spring, when the movement changes direction. This is a characteristic of any type of oscillatory movement.

$\color{blue}{E\,'(0)}$ is $\color{blue}{0}$. Reversing the direction of motion in a continuous function means that the first derivative is null in the reference point. Non-oscillatory movements are not included in that reasoning.

$\color{blue}{E\,''(0)}$ as a constant $\color{blue}{k}$, since it is the value of a function at a given point.

$\color{blue}{E\,'''(0)}$ and so on can be dismissed in simpler models.

So

$$\color{blue}{E(y) = \frac{k\,y^2}{2}}$$

This is the formula of the potential energy of a spring, when someone pull a spring and hold it. Now we can derive it in relation to x (Energy = Force * Distance):

$$\color{blue}{E\,'(y) = F = -ky}$$

As the force acts for something to return to a stable position, the constant $k$ shoud be preceded by a - signal.

The above expression can be expressed as a function of the acceleration, 2nd. derivative of displacement function, when expressed as a function of time t (m is mass).

$$\color{blue}{ m\,y\,''(t) = -k\,y(t)}$$

The most general solution of the above differential equation where the second derivative is the function itself, with the changed signal corresponds to:

$$\color{blue}{y(t) = A\,sin(\omega t + \Phi)}$$ $$\mathsf{or}$$ $$\color{blue}{y(t) = A\,cos(\omega t + \Phi)}$$ $$\mathsf{or}$$ $$\color{blue}{y(t) = A\,sin(\omega t) +B\,cos(\omega t)}$$

Where amplitude is maximum value ($\color{blue}{A}$ in the first two, and $\color{blue}{\sqrt{A^2+B^2}}$ in the third ) , $\color{blue}{\Phi}$ is phase and it is shown easily that $\color{blue}{\omega = 2\pi/T}$ ($\color{blue}{T}$ is the period, time for a full lap).

There are 2 different constants because are 2 freedom degrees in double derivation.

For instance, if one derives the first solution twice, one gets

$$\color{blue}{y''(t) = -A\,\omega^2 cos(\omega t)}$$

That can be rewritten as

$$\color{blue}{y''(t) = -\omega^2 y(t)}$$

So

$$\color{blue}{\omega = \sqrt { k/m}}$$

Remembering that it is indifferent to use sine or cosine because

$$\color{blue}{sin(y+\pi /2) = cos(y)}$$

Let's forget the other 2 solutions and let's focus on the first solution ($\color{blue}{\sin}$)

To visualize better, if we imagine this function as expressing the vertical oscillation of an longitudinal oscillatory motion. Let's consider $\color{blue}{\Phi = 0}$ so $\color{blue}{y(t) = A\,sin(\omega t)}$.

It's possible interpret the $\color{blue}{y}$ value as $\color{blue}{\sin}$ value in a uniform circular motion in a circle of radius $\color{blue}{A}$, see in a profile view, as the eye in the below figure:

Any above expression represents simple harmonic motion (SHM), that can be displayed as circular movement with constant angular speed. Over time, the amplitude draws a sine graphics.

Most oscillating system will behave like a vibrating spring, so long as the oscillations are small enough. For this reason, the vibrating spring, or simple harmonic oscillator (SHO) as it is called is very important in Physics.

aerodynamics - NASCAR drafting at Daytona

This year, the racetrack at Daytona has been repaved. The track was always faster than other tracks NASCAR raced at and several cars in a "train" were faster than single cars or smaller trains. This year is different, however, because two car groups are by far the faster than any other groupings. When a third car joins a tandem, all three slow down noticeably. Why is this? Why wouldn't it still be faster to have more cars pushing?

geometry - Drawing something using one pen stroke

Can you determine if it's possible to draw a geometric figure (made up from shapes like rectangles, triangles, and other regular shapes) with one pen stroke and not drawing the same line twice.

I am thinking about something related to graph theory or something like that.

For example, something like this:

Answer

Yes! This is a classic graph theory problem. Anywhere where lines meet is called a vertex, and the degree of a vertex is the number of lines that meet there. Euler proved that as long as a graph has either 0 or 2 vertices of odd degree, and the graph is connected (consists of a single piece), then it can be drawn as you specified. Furthermore, if there are any odd vertices, then any successful Eulerian path (as such a pen stroke is called) must start at one, and end at the other.

The main reason this is true is as follows: every time you enter and leave a vertex, you kill two of its edges. So, if a vertex has an odd number of edges, not all of its edges can be taken care of by passing through it, so the pen must start or end there. There are only two start and end points, so there can be at most two odd degree vertices.

In your picture, there is a square with horizontal sides, all of whose vertices have 5 edges coming out of them. This means that you have $4$ odd-degree vertices, which is $2$ too many, so your graph can't be drawn with a single non-overlapping pen stroke. But, if you erased one of the diagonals of this middle square, then you'd be in business.

This result applies to any shape with various curves on a page which meet at various places, not just straight line figures.

electromagnetism - What is the minimum wavelength of electromagnetic radiation?

As a first approximation, I don't see how a wavelength of less than 2 Planck distances could exist. The question is: Are there any other limits that would come into play before that?

For example:

• Would the energy density cause the photon to turn into a black hole or something like that?


• Would the energy of the photon exceed the total mass+energy of the universe?

How does matter transform into energy and vice versa?

In what ways can energy transform into matter and vice versa? Annihilation is one way to tranform matter to energy. Fission is another (when splitting and atom, what happens to its two parts?)

Are quantum fluctuations one way to transform energy to matter?

Answer

In what ways can energy transform into matter and vice versa?

Energy and matter are connected according to special relativity and this has been experimentally demonstrated . It is the famous formula:

$E=mc^2$ , where $m$ is the relativistic mass and $c$ the velocity of light. or

$E^2=m_0^2c^4 +p^2c^2$ , for a particle with rest mass $m_0$ moving with momentum $p$.

The rules of transformation follow Quantum Mechanical solutions of kinematic and potential problem equations .

Annihilation is one way to transform matter to energy.

Yes

Fission is another (when splitting and atom, what happens to its two parts?)

In the quantum mechanical description of nuclei they are represented by potential wells with energy levels, some filled. The number of baryons ( protons and neutrons) bound in this potential well characterize the nucleus. Nucleus A that is struck by a neutron ( for example) becomes a nucleus B higher up in baryons by absorbing it into an energy level of this potential well. In fission this higher up nucleus is unstable and falls into a lower energy state, giving up part of its mass in energy according to the relativistic formulae, and breaking into smaller nuclei and free neutrons which go on to sustain the fission on another original nucleus. Generally a form of fission happens if a nucleus is unstable.

There is also fusion, two deuterium nuclei adhering at a lower energy level and giving up energy. The binding energy curve shows whether nucleons can fuse or fission and give up as energy a part of their mass.

Are quantum fluctuations one way to transform energy to matter?

No, quantum fluctuations are virtual . If you mean tunneling, yes.

quantum field theory - Contact terms in Dyson-Schwinger equation can be ignored?

According to this text here

http://www.physics.indiana.edu/~dermisek/QFT_09/qft-II-4-4p.pdf

contact terms do not affect the scattering amplitude. But These contact Terms are there; the question is: When contact Terms are relevant for scattering Amplitude computation?

My idea:

By starting with the connected Partition function $G[J]:=\log Z[J]$ where $Z[J]$ is ordinary Partition function corresponding to the Action

$$S = S_\mathrm{theory} + \int d^4x\ J\phi$$

for some fields $\phi$ and the source $J$ one can derive cumulants belonging to $S_\mathrm{theory}$ by multiple Derivation of $G[J]$ by $J$ and Setting $J=0$. Only the equation for quadratic cumulants $\langle0|\phi(x) \phi(y)|0\rangle$ will contain an equation with the contact term $\delta(x-y)$. More precisely

$$\mathcal{H} \langle0|\phi(x) \phi(y)|0\rangle = f(\text{others})+\delta(x-y)$$

for an Operator $\mathcal{H}$ that I assume to be linear and nonlinear corrections $f(\text{others})$.

Neglecting nonlinearities I see that $\langle0|\phi(x) \phi(y)|0\rangle$ is exactly the Green function generated by $\mathcal{H}$. This Green function $\Delta(x-y)$ vanishes if the Observation time $t$ is set to $\infty$. And infinitely Observation times are assumed in the LSZ formula for scattering amplitudes.

Will contact Terms be relevant for finite Observation times? Why on scattering Amplitude/ cross section computation infinitely Long Observation times are assumed?

No real process has infinitely Long Observation times. But maybe uncertainty in energy is cancelled if $\Delta t \mapsto \infty$ is assumed.

Help would be greatly appreciated.

Does gravity keep nearby galaxies from flying apart in space expansion?

On a large scale, the universe expansion pulls galaxies apart while gravity keeps galaxies from expanding. So there seems to be a certain scale, at which the expansion and gravity roughly cancel each other. Within a galaxy gravity wins, but remote galaxies fly apart.

What happens if galaxies are close to each other? I know Andromeda is not a good example, as it is already on a collision course with Milky Way, but generally, do neighboring galaxies of the same group tend to be kept together by gravity or do they fly apart with the expansion of the universe? At what scale does the expansion finally overcome gravity?

Answer

Yes, galaxy clusters can remain bound together despite the expansion of the universe.

Analysing the balance between expansion and gravitational attraction produces an estimate where test particles remain bound in circular orbit around a larger mass if $2GM_{obj} \geq \beta^* H_0^2 r_0^3$ where $\beta^*\approx 5.3$, producing the criterion $$\frac{M_{obj}}{10^{12} M_\odot} > 3h_{70}^2 \left( \frac{r_0}{1 \mathrm{Mpc}} \right)^3.$$ The Milky Way has a sphere of influence is 0.7 Mpc, while a typical star of $0.5 M_\odot$ (with no other competition) has a sphere of influence 55 pc across. A less restrictive bound uses 1.18 rather than 3.

See this paper for a comparison with actual supercluster data. It also looks at non-circular orbits, where things are more complicated.

riddle - What do they have in common?

What is the link between an astronotical fish, an owner of crown jewels, Chelsea, an honoured Akira, and a synthesizer?

Hint 1:

The link is a name

Hint 2:

A wilde name

Answer

The answer is

Oscar

for the following reasons:

astronotical fish -> astronotus ocellatus = oscar fish
an owner of crown jewels -> Oscar of Sweden (King of Sweden and Norway, 1799-1859)
Chelsea -> Oscar dos Santos Emboaba Junior (plays for Chelsea)
an honoured Akira -> Oscar academy award for Akira Kurosawa
a synthesizer -> OSCar synthesizer (by the Oxford Synthesiser Company)

quantum mechanics - Second order perturbation of a degenerate system with no first order correction

Consider the following Hamiltonian, in arbitrary units:

$$ H = \begin{bmatrix} 0 & 0 & g\\ 0 & 0 & g\\ g & g & 1 \end{bmatrix}$$

where $g<<1$. It is relatively straightforward to find the spectrum of this Hamiltonian analytically. Simply solve the characteristic equation and find that the three eigenvalues of this matrix are:

$$\begin{aligned} \epsilon_1 &= \frac{1}{2}\left(1 - \sqrt{1+8g^2}\right) \approx-4g^2 \\ \epsilon_2 &= 0 \\ \epsilon_3 &= \frac{1}{2}\left(1 + \sqrt{1+8g^2}\right) \approx1+4g^2 \end{aligned}$$

I am asked to calculate the spectrum of this Hamiltonian to second order by pertubative methods. So we split the Hamiltonian up into:

$$ H = H_0 + g \tilde H$$

But of course we realise immediately that $H_0$ is degenerate.

How do we perform perturbation theory on this Hamiltonian? References are welcome, but will not be considered answers, as at the moment I cannot access them.

Note, I am at home, and the only reference I have on the subject is Alastair Rae's Quantum Mechanics, 5th edition. Here there is a section on perturbation theory for degenerate levels, but this approach does not work here. That's because it relies on finding first order corrections to the energy levels, which we know don't exist, thanks to our analytical solution.

I have also partially cheated, by recognising that the vector:

$$ v_2 = \frac{1}{\sqrt{2}} \begin{bmatrix} ~~1~ \\ -1~ \\ ~~0~ \end{bmatrix}$$

is both the eigenvector of both $H$ and $H_0$ for the eigenvalue $0$, but it hasn't helped me much.

Finally, I am aware that there is a question about the same Hamiltonian, but there is no answer for it.

Answer

Indeed, the first order firmly establishes the vanishing of the energy corrections but fails to completely specify the wavefunction corrections, and you must keep going, to 2nd and 3rd order to specify your unknowns in underdetermined systems. Courant and Hilbert (cited here ) describe the procedure, but maybe you don't want to go there... Anyway, the other two eigenvectors are $$ v_1 \propto \frac{1}{\sqrt{2}} \begin{bmatrix} ~~1~ \\ 1~ \\ ~~c~ \end{bmatrix} \qquad c=\left (1-\sqrt{1+8g^2}\right )/2g \approx -2g, \\ v_3 \propto \begin{bmatrix} ~~g~ \\ g~ \\ ~~b~ \end{bmatrix} \qquad b=\left (1+\sqrt{1+8g^2}\right )/2 \approx 1+2g^2, \\ \Longrightarrow v_3 \propto \begin{bmatrix} ~~0~ \\ 0~ \\ ~~1~ \end{bmatrix} + \sqrt{2} g v_1 +O(g^2),...$$ All 3 eigenvectors are mutually orthogonal.

The first order specifies the $v_3$ correction, but one knows nothing about 1-2 mixing, at this order, and so you need to go to the next level, to get those, etc.

Instead, your limited cheating was just enough, and all you need is to project out the $v_2$ subspace and end up with a trivial non-degenerate 2×2 system. Consider the "Foldy-Wouhuysen" transformation, $$ U = \begin{bmatrix} 1/\sqrt{2} & 1/\sqrt{2} & 0\\ -1/\sqrt{2} & 1/\sqrt{2} & 0\\ 0 & 0 & 1 \end{bmatrix}$$ yielding $$ U^\dagger H ~U = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & \sqrt{2}g\\ 0 &\sqrt{2} g & 1 \end{bmatrix},$$ an equivalent system to the original, but now with the first subspace, your cheating eigenvector, projected out, and non degenerate to boot, so the second order energy shift is $E^{(2)}=\pm 2g^2$ as per the standard dull formula. The pros call this a 3-scale problem: 0, $g^2$, and 1, as you saw.