homework and exercises - What properties do you need for building a tower?

When I was a boy I used to daydream about building a tower so tall that the top of it would project into near space.

There would perhaps be a zero gravity area in the penthouse where my friends and I could bounce around and play space versions of various earth-based games and sports in most excellent zero-g conditions.

Much to my continued disappointment and despite all the technological advances of the last thirty or so years, no one has built such a structure.

Can anyone explain the physical limitations/constraints that are preventing someone from realising my fantasy of a 'Space Tower'?

UPDATE: This Kickstarter Project seems to be pretty confident ...

Answer

First off, the limitation is a material that would not collapse under the weight - earth crust is not quite hard enough. Buckling and other instabilities, nope. Generally, forget a tower built on earth. Not a chance, no such material.

Start building from geostationary orbit and extend the "rope" both inside and outside the orbit. The outside may be just heavy counterweights, as the inside will begin to pull towards earth. Make the orbital part thicker to support extra weight, as you extend the lower part, until it reaches earth surface below.

Now the problem is the material. The only material in existence with sufficient weight-strength ratio is buckytubes. These are currently centimeters long at most, extremely expensive and you'd not only need thousands of kilometers of them... the rope to sustain its own weight would have to be about 1km thick in the thickest place (near the geostationary orbit).

Now consider:

• earth carbon supply, I don't think all coal mines combined could mine that much carbon


• construction craft fuel. This all would have to be lifted high enough. A LEO rocket takes many times more fuel than its payload weight. A geostationary orbit rocket - much more. The good news is the fuel can be hydrogen+oxygen which is water, and we have that aplenty. The bad news is you need at least as much energy to separate them as you gain from burning them, so the power consumption for fuel production would exceed whole world's power production.


• environmental impact of that much steam released into atmosphere


• account for micrometeorites that can really rain over your parade. And this thing being that big, collisions WILL happen. Also account for space junk.


• account for winds and storms once you reach the atmosphere. Also, upper atmosphere is pretty hot... not nice work environment, also nanotubes aren't extremely fire-proof.


• cost and impact on economy. Coal becomes super-expensive and we look for alternate sources of carbon.

And when you finally build it, calculate how long a lift travelling some 300km/h would take to reach 37,000km of the 0-gravity orbit...

EDIT:

I can't currently find the article that listed 1km thickness, but let us try to calculate parameters of the tower merely strong enough to sustain itself.

The nanotube tensile strength is $UTS=6422kg/mm^2$ (1)

The density is $\rho=1.4g/cm^3$

The ribbon is said to be 1m wide.

The thickness will vary. For the needed $M_0=20t=20000kg$ capacity it needs $A_1=0.31mm^2$ cross-section at the bottom. At 1000mm width that's 0.00031mm thick.

Now I'm really not in the mood to solve a differential equation of thickness - mass - tensile strength - gravity so let me try a discretization, approximating with $h=1km$ long wedges. At 35000 samples that should give us a decent approximation.

$$ V_n= {A_n+A_{n+1} \over 2}h \\ M_n=\rho V = \rho {A_n+A_{n+1} \over 2}h $$

Now we can't happily assume weight not to vary with altitude. After all, near the orbit it will be zero. It varies with distance from center of Earth. At surface, $r_0=6378km; M_{earth}=5.97 10^{24}kg; G =6.67300 × 10^{-11} {m^3 \over kg s^2}$;

So, the weight function of each segment will be

$$ Fw_n=G{M_n M_{earth} \over r_n^2} \\ r_n=r_0+n[km] $$

And the tensile strength surface $A_{n+1}$ must overcome is

$$ F_{n+1}=F_n + Fw_n \\ F_{n+1} = A_{n+1} UTS $$

We seek $A_{35000}$ which will trivially yield thickness by dividing by 1000mm.

$$ A_{n+1} UTS = F_n + Fw_n \\ A_{n+1} UTS = F_n + G{M_n M_{earth} \over r_n^2} \\ A_{n+1} UTS = F_n + G{\rho {A_n+A_{n+1} \over 2} h M_{earth} \over r_n^2} \\ A_{n+1} = F_n + (A_n+A_{n+1}){ G \rho h M_{earth} \over 2r_n^2 UTS } \\ X := { G \rho h M_{earth} \over 2r_n^2 UTS }\\ A_{n+1} - = F_n + (A_n+A_{n+1})X \\ A_{n+1} = F_n + X A_n + X A_{n+1} \\ A_{n+1} - X A_{n+1} = F_n + X A_n \\ (1-X)A_{n+1} = F_n + X A_n \\ $$ We get our two fundamental equations for numeric computation: (with helper X, which I'm really not in the mood to transform into something nicer.) $$ X = { G \rho h M_{earth} \over 2r_n^2 UTS }\\ A_{n+1} = { F_n + X A_n \over (1-X) } \\ F_{n+1} = A_{n+1} UTS $$

Now excuse me, it's 3AM and I'll finish the calculations at a different time.

quantum mechanics - Probability of an hydrogen electron staying in a certain orbital

Suppose, an election of a hydrogen atom has energy corresponding to $n=2$, where $n$ is the principal quantum number.

Is it possible to find the probability of the electron staying in $2s$, $2p_x$,$2p_y$ or $2p_z$ orbital?

Are these probabilities equal, as they are have equal energy?

Answer

The question asked in your comment is different from the question I answered so I'll put a new answer here to your question:

The atom was in ground state and it is given enough energy for n=2. What are the probabilities that now the electron will be found in 2s, 2px, 2py, or 2pz orbital? Is it possible to measure this analytically?

The answer is yes. This can be deteremined analytically. Often we have a cartoon picture in our heads where an atom "magically" absorbs a photon and moves into an excited state. This leaves out many of the important details.

The reason light interacts with atoms is because the electric field of the light pushes around the electron (and thus changes the shape of the electron wavefunction). This means that if you shine on light at a frequency which the electron will respond to you can push the electron from being in one spatial pattern (the ground state, for example) to being in another spatial pattern (an excited state, for example).

Consider the $p_x$ state. In this state the electron is kind of stretched in the $x$ direction. Well this means that you can get from the ground state to this state by shining on light which is linearly polarized along the $x$ direction. The electric field acts to stretch the atom along the $x$ direction. Similarly for the $y$ and $z$ directions.

So this means that which excited state the atom is driven to depends on the geometry of the electric field driving the atom as well as the geometry of the particular ground and excited states under consideration.

In the dipole approximation it is impossible to drive transitions from $1s$ to $2s$ because the states have the parity.

I haven't given you an analytic formula to calculated the different excited state fractions given a particular incident electric field so I haven't fully answered your question. I don't have time for that now but most atom/quantum optics textbooks will have a description of how to calculate the coupling strength between an optical field and an atomic transition, including clebsch-gordon coefficients and polarization vectors which capture the geometric effects I described earlier.

I like this textbook available online by Steck

electromagnetism - In the Pound-Rebka experiment, does light lose energy?

In the Pound–Rebka experiment the redshift / blueshift of photons is measured in small distances. This experiment one explain by the influence of gravitational field on the photon: "When the photon travels through a gravitational field, its frequency and therefore its energy will change due to the gravitational redshift."(https://en.wikipedia.org/wiki/Pound-Rebka_experiment)

The redshift of photons coming from stars far away from us is explained exclusively with the Doppler effect and the difference in the gravitational field of the source and the receiver (to a very small amount for big distances). The explanation of tired light was discarded. The frequency of photons dosn't changes during its life. Light "... consists of a finite number of energy quanta which are localized at points in space, which move without dividing, and which can only be produced and absorbed as complete units."A. Einstein Concerning an Heuristic Point of View Toward the Emission and Transformation of Light

In accordance with the second paragraph the Pound-Rebka experiment has to be interpreted by an other way. The source and the receiver are located in points with different gravitational potential and that is the reason they capable to emit and receive photons at different frequencies. Is then the statement in the thirst paragraph wrong?

particle physics - Is there a connection between the fluctuation-dissipation theorem and the Green–Kubo relations?

Is there a connection between the fluctuation-dissipation theorem and the Green–Kubo relations? I have a hard time finding out if there is a relation and what it is, because the fluctuation-dissipation theorem always seems to be stated in another way and for specific cases. Both formula seem to be a statement about how a macroscopic property is determined as an integral over Greens function'esque objects.

What is the exact relation and the hierarchy of the two concepts?

Answer

The fluctuation-dissipation theorem is a general principle that comes in many concrete forms. It expresses in each case a way how the spectrum of equilibrium fluctuations can be probed by applying weak external fields.

The Kubo relations are just one specific instance of it.

For example, Linda E. Reichl derives in her book on statistical physics the F/D theorem in Section 15D from general principles of nonequilibrium thermodynamics, and gives several quantitative applications, before she gives in Section 15H the Kubo relation as the microscopic expression for it.

Differences in measurement between quantum physics and classical physics?

I know that there are differences in measurement due to quantisation, probability, and collapse in quantum physics, but I am having trouble explaining how these ideas relate to measurement in an understandable manner.

In what ways does measurement in quantum mechanics differ from measurement in classical mechanics?

A delve into extraordinary chess problems: Self-Stalemate 2

Following on from a slightly easier puzzle here (this series continues here).

This is a puzzle by W. A. Shinkman in 1924:

White to play and force Black to stalemate him in nine moves. As a reminder: "In a self-stalemate problem, White's objective is not to win, in the usual sense of the word, but to get himself stalemated."

Here is an interactive board for you to play on.

Here is a board editor with the puzzle set up. It may help to work a little in reverse.

Clarifications based on comments:

1. Black is playing to thwart you. So your moves (as white) must lead to a stalemate against all possible moves by black.


2. White is in stalemate at the end. Meaning black makes her 9th move, and then you, as white, will have no legal moves but will not be in check. i.e. you will be stalemated.


3. If you can achieve this feat in less than 9 moves, then please post the solution. That would be a "better" solution than the one I have. But, please, double-check that you have accounted for all possible defenses by black because I am relatively confident that the best solution will be 9 moves.


4. By "9 moves" I mean 9 moves by white. In this puzzle, that means 9 moves by black also (a more common puzzle genre involves mate, in which case it would be 8 moves by black. In this instance, black gets the last move because she is stalemating you).


5. I believe there is only one solution, and it can be solved by looking at the logic of the situation. But it is tricky!

Answer

I think I have a forced stalemate in 9 white moves,

1. Re5 e6 2. Re2 e5 3. Nc5 e4 4. Qa8 e3 5. Qg2 B? 6. Ra2+ Bxa2 7. Nb3+ Bxb3 8. Qg7+ Ka2 9. Qa1+ Kxa1
The B? in the 5th move is because it doesn't matter where the bishop goes. It would have to come back to a2 in the next move to remove the check.

Animated solution:

Method:

This is not a proof or an exhaustive check, but rather an outline of the thought process that lead to this solution.

First I tried to see what could be possible boards at the end of the game. After playing around a bit I settled on:

because:

1) It doesn't seem to involve moving the white king.

2) There is a nice way to bring the bishop to the required position: using the knight to deliver a check when the bishop is at a2.
3) The pawn's position allows us to have a clear path at the black king while keeping the pawn blocked with the rook/queen.

In view of 3 above I opted to use the rook to block the pawn because it allows us to block e5 on Black's first turn and there's no apparent advantage for using the queen here.
Now I went back to the starting position and tried playing it to get some ideas, with the aim of using the knight to force the bishop to the required position and using the rook to block the pawn. The first satisfactory result was:

There are some apparent issues here: the bishop could move to b1 instead of capturing the queen and instead of the bishop the black king could capture the rook too. But the foundation of this approach looked fairly solid so I decided to ponder this before moving on to a new one.
Now it was just trying new ideas involving the rook and the queen. After trying out some ideas (such as using checks to force bishop location and even moving the king to do a discovered check and some others which I couldn't remember precisely) I got the idea that keeping a2 attacked with the queen resolves the second issue. Using that to modify the approach, I tried to play again. After a lot of uninteresting games, and some interesting ones I had to rule out, I hit on this:

I opted for Qg2 rather than Qg8 for the attack on a2 because the black bishop has access to g8. From here I think you could find the solution if you view this with an open mind and try to utilize checks as much as possible, but I had subconsciously decided that the knight would be the last to be captured so solving this took quite some playing, but eventually I noticed the diagonal queen check and thought of sacrificing the knight before the queen which lead to this solution.

wordplay - Remove first or last letter - What is the name of this word puzzle/type of word?

When I was a child I had a puzzle book. One of the puzzles involved finding words where you can remove one letter from the beginning or end of the word, and you still have a valid word. You can continue this process with these words until there is only one letter left.

For example, "brandy" is such a word, because after each letter is removed below, we get another word:

brandy
brand
bran
 ran
  an
  a

I don't have an actual puzzle here. But instead I am looking for the name of this class of word, if it exists.

Answer

In general, these puzzles are called "Deletion Puzzles" but the usually do not include the restriction to the beginning or end letters. The most famous example is "startling" as it is the longest. It is also sometimes called "Elision". Please not that these terms also do not require it to be iterative.

"Terminal Deletion" or "Terminal Elision" refer to removing both the first and last letters to make a work.

"Decapitation" word puzzles refer to only removing the first letter to reveal a word.

You would most likely have luck calling a puzzle regarding these words an "Iterated Amputation" word puzzle. Amputation usually has the same definition as Decapitation for this context but it less common. It would be easy to draw contrast between them and describe removing either of the two "arms" of the word (right or left arm). Iterated, of course, refers to the fact you want to do this several times. Does that mean you call a word "amputatable"? I don't know.

Ultimately I don't think a single term for what you want exists and, if it did, I can't think of a context where you wouldn't have to define it for the audience anyways.

I suggest you check out "A Key to Puzzledom: Or, Complete Handbook of the Enigmatic Art" which is available on google books.

quantum mechanics - What does it mean to apply an operator to a state?

Let's say I have an operator $\hat{A}$ and a state $|\psi\rangle$. What exactly is the state $\hat{A}|\psi\rangle$? Is it just another different state that I am describing using my $\hat{A}$ and $|\psi\rangle$? For example, if

$$\hat{A} \doteqdot \text{put chocolate syrup}$$

Then is $\hat{A}$ just a tool to describe a state, like:

$$|\text{vanilla ice cream with chocolate syrup}\rangle = \hat{A}|\text{vanilla ice cream}\rangle$$

But on the other hand, we have something like

$$\hat{H} |\psi\rangle = E|\psi\rangle$$ I interpret this equation as "If you apply the (time-independent) Hamiltonian to a state, the result is proportional to your original state". But $E|\psi\rangle$ itself cannot be a new state, because it's in general not normalized. So here the operator $\hat{H}$ is just used to establish a mathematical property of $|\psi\rangle$, not to describe another state. You can't say that $\hat{H}$ is a machine that takes a state and returns another state, not in the same way $\hat{A}$ takes an ice cream and puts syrup on it. Or can you?

And what does taking a measure mean? If you measure an observable, it returns an eigenvalue and the state collapses into an eigenstate. Is this resulting eigenstate the one you obtain when you apply the operator to the state?

Answer

I think you may be misguided by the concept that we associate $\textbf{observables}$ to self-adjoint operators. They do operate on the Hilbert space, but to see them as entities that transform states or prepare them is a little bit tricky. I will describe here self-adjoint operators and preparation of states.

1) The true (physical) power of self-adjoint operators for describing observables lies in the spectral theorem, and not in its $\psi \mapsto A\psi$ action. Physically, what it means? There is a set called spectrum of an observable, and it is the set of possible outcomes on its measure for given states. For example, a spin observable $S$ on a 1/2-spin system has spectrum $\sigma(S) = \{-1/2,+1/2\}$, and decomposes as a sum of its spectral projections, $S = +1/2 P_{+} -1/2 P_{-}$. In general, there is a spectral resolution $E$, that is, a bunch of projection related to the spectrum, such that the operator can be written as $A = \int_{\sigma(A)}\lambda dE(\lambda)$.

And what are the spectral projections? Those are again (self-adjoint) operators, but the whole collection of spectral projections will give you a probability measure when coupled with a state. In the spin system example, if you take a state $\psi$, then $\langle\psi, P_+\psi\rangle$ would give you the probability of measuring a +1/2 spin, and likewise for -1/2.

Now suppose you had a 1/2 spin system with prepared state $\psi$, and you measure the spin, and get +1/2. After the measurement, your state collapses to a $|+1/2\rangle$ state.

In a more detailed formalism, suppose you have prepared a state $\psi$ and you are going to make a measurement of an observable expressed as $A = \int_{\sigma(A)} \lambda dE(\lambda)$ (where the $E$ is the spectral resolution of your operator, just think of the 1/2-spin example intuitively). Then suppose your measurement is on a subset $\Lambda \subset \sigma(A)$ (you may think of the set $\{+1/2\} \subset \{-1/2,+1/2\}$. Your state $\psi$ then collapses to the following state $\phi$:

$\psi \rightarrow \phi = \frac{E(\Lambda)\psi}{\|E(\Lambda)\psi\|}. $

(notice that $\phi$ is normalized and well defined, since $E(\Lambda)\psi=0$ then the probability of the outcome being in $\Lambda$ would be zero to start).

Summing up, you do not simply apply a self-adjoint operator on a state, since, as you have seen, it doesn't have much meaning. This is a point most introductory QM books do not stress as much as I would want. What happens with measurements and collapses and whatsoever uses, as I tried to point out, the spectral projections more than the operator itself. So, as you said about your Hamiltonian operator, it does not act like your syrup machine, which we will try to cover up next.

2) Now what you describe as "tools", in your example, the putting syrup, is not a measurement per se, it is a preparation of states, which would grab a state without syrup and put syrup in it. The modeling of such procedure is usually ignored, at least to my knowledge.

One choice would be just saying "my state now is syrup", end of discussion.

Other option is using unitary operators ($U$ such that $UU^* = U^*U = 1$). Those transform state vectors in state vectors.

If you would like more sophisticated examples, it starts to get tricky, and I will shut up before I say something very wrong about it. But rest assured this is not easy at all, and your question is really nice. Hope to see some other inspiring aswers.

special relativity - reflection at speed of light when both mirror and viewer is travelling at the speed of light

consider me sitting on the top of a train which is travelling close to the speed of light, will I be able to see my image on a mirror which I'm holding in my hand??

Linear polarized 3D glasses and the physical shape of light waves

Looking into how linear polarized 3D glasses work, I keep getting explanations that boil down to this:

However, I always assumed that a light wave was depicted in diagrams like this...

...to more easily plot on paper its properties like frequency and amplitude. I never thought that light waves would physically actually look like that. I always imagined light waves being emitted in the form of a sphere centered at its source just like sound:

If the latter is correct, then this explanation for polarized filters doesn't really make sense; or do I have some major misconceptions about the nature of waves?

mathematics - Finding Doctor No

James Bond is invited to a party with altogether $128$ participants (including Bond himself, the host, and the hostess). At the beginning of the party the host takes James Bond aside and asks him to identify the mysterious Doctor No. It is known that Doctor No knows all the other people at the party, but none of the others does know Doctor No.

James Bond starts asking questions of the following type: he asks some person A, whether A knows some other person B.

What is the minimal number of questions which James Bond needs to ask (in the worst case) to identify the mysterious Doctor No?

Answer

When James Bond asks A if he knows B:

• 
  

  
  

  If the answer is yes, this eliminates B as a candidate, since no one is supposed to know Doctor No.

  
  

  
  


• 
  

  
  

  If the answer is no, this eliminates A as a candidate, since Doctor No is supposed to know everyone.

  
  

  
  

James Bond just keeps asking an arbitrary non-eliminated person about another arbitrary non-eliminated person. Every question eliminates one person. At the beginning there are 125 candidates: 128 participants minus James Bond (whom the host knows), minus the host (whom James Bond knows), minus the hostess (whom the host knows).

So the answer is...

...124 questions in the worst case.

astronomy - Are there websites or programmes that permit a simulation of the night sky in the past and the future on an ordinary computer?

Are there websites or programs that permit a simulation of the night sky in the past and the future on an ordinary computer?

(For the past, I would be content with objects visible to the naked eye.)

Answer

The two main free destop programs that I know about, Stellarium and Celestia, do not include the proper motion of the stars when they move forward and backward in time. At least according to documentation that I've seen.

These programs claim to do it but I have no experience with them:

However, the Distant World Star Mapper at the Astoronomy Nexus will compute the proper motion when rendering sky maps. It's not a full simulation but can provide some information.

Setting of renormalization scale in field theory calculations

In dimensional regularization an arbitrary mass parameter $\mu$ must be introduced in going to $4-\epsilon$ dimensions. I am trying to understand to what extent this parameter can be eliminated from physical observables.

Since $\mu$ is arbitrary, physical quantities such as pole masses and scattering amplitudes must be independent of it. Nevertheless at any fixed order in perturbation theory these quantities contain residual $\mu$-dependence. One expects this dependence to decrease at higher orders in perturbation theory.

For concreteness, consider dimensional regularization with minimal subtraction of $\phi^4$ theory, which has bare Lagrangian

$\mathcal{L}_B = \frac{1}{2}(\partial \phi_B)^2 - \frac{1}{2}m_B^2 \phi_B^2 - \frac{\lambda}{4!}\phi_B^4$

Here are the 1-loop expressions for the physical mass $m_P$ and 4-point coupling $\lambda_P \equiv (\sqrt{Z})^4 \Gamma^{(4)}$ after minimal subtraction of poles and taking $\epsilon \to 0$:

\begin{equation} m_P^2 = m_R^2 \left\{1 + \frac{\lambda_R}{2(4\pi)^2}\left[\log\left(\frac{m_R^2}{4\pi\mu^2}\right)\right] + \gamma - 1 \right\} \end{equation}

$$\lambda_P = \lambda_R + \frac{3\lambda_R^2}{2(4\pi)^2}\left[\log\left(\frac{m_R^2}{4\pi\mu^2}\right) + \gamma - 2+\frac{1}{3}A\left(\frac{m_R^2}{s_E},\frac{m_R^2}{t_E},\frac{m_R^2}{u_E}\right) \right] $$

where $A\left(\frac{m_R^2}{s_E},\frac{m_R^2}{t_E},\frac{m_R^2}{u_E}\right) = \sum_{z_E = s_E,t_E,u_E} A\left(\frac{m_R^2}{z_E}\right)$ and $A(x) \equiv \sqrt{1+4x}\log\left(\frac{\sqrt{1+4x}+1}{\sqrt{1+4x}-1}\right) $.

Both of these quantities ($m_P$ and $\lambda_P$) are physically observable.

Suppose we conduct an experiment at a reference momentum $p_{E0}\equiv(s_{E0},t_{E0},u_{E0})$ and make measurements of the pole mass and 4-point coupling with the result $\lambda_{P0}, m_{P0}$. We now have a system of two equations in the three unknowns ($\lambda_R,m_R,\mu$). This means that in principle I can solve for $\lambda_R = \lambda_R(\mu)$ and $m_R = m_R(\mu)$.

I would now like to make a prediction for the 4-point amplitude at a different momentum $p_{E}' \neq p_{E0}$. Since I have two equations in three unknowns I need to guess a suitable value for $\mu$ (say $\mu' = \sqrt{s_{E}'}$) which allows me to fix $\lambda_R$ and $m_R$ and then calculate $\lambda_P (p_E')$. This procedure seems quite ad hoc to me because a different (arbitrary) choice of $\mu$ (e.g. $\mu'/2$ or $2\mu'$) will lead a different physical answer (albeit only logarithmically different).

From what I can gather from the literature, the problem of determining the renormalization scale I described above is a genuine problem in actual calculations of QCD (e.g. http://arxiv.org/abs/1302.0599) which leads theorists to introduce so-called "systematic uncertainties".

What concerns me is that I haven't been able to find any mention of this problem in any textbook on quantum field theory that deals with QED or QCD (anyone know of a reference?). Since this problem appears already in arguably the simplest QFT of $\lambda\phi^4$, I would expect it also to occur in e.g. 1-loop calculations of Bhabha scattering but I haven't been able to find any mention of it in this context.

Does anyone know how this problem is dealt with in real loop calculations (e.g. at LEP or LHC?)

Also, I would be interested to know if there is any analogue of this problem in condensed matter theory.

mathematics - Ernie and the Island of Stability

Unfortunately, it appears that I may have misled you a little in this puzzle (as you probably know - my memory of events isn't always perfect). When I was writing it, I re-checked the Kzijekistanian postal rules on Gurgle-translate, but somehow stumbled on an old cached web-page with out of date data. Notes 1, 2, and 3, were unchanged but Note 4 was misleading. As a result, it appeared that even with the best packing, the pinot noir would have had to be so execrable that the wine store would have paid Ernie to take it away!!! I have edited the story below to show the correct postal rules for four or more items (and have retained the old note in strikeout form for posterity). J Richard Snape did all the hard work for the first part of the puzzle - which was meant to be the hard part - so I am accepting his answer as correct.

"Are you familiar with the Island of Stability?", Ernie asked, as we sat drinking coffee and perusing some light literature at the local cafe. I had to admit that I wasn't, and so Ernie took it as an opportunity to reduce my ignorance. He read the following from his magazine Nuclear Analysis Monthly "In nuclear physics, the island of stability is a set of predicted, but as-yet undiscovered, heavier isotopes of trans-uranic elements which are theorized to be much more stable than some of those closer in atomic number to uranium." and pointed at a diagram showing the predicted island. "The current belief is that the right isotophe of Flerovium may have a half life measuring in millions of years!", he continued emphatically.

I murmured a note of vague interest and returned my attention to my much more interesting reading - The Journal of Trans-Chatcal Affairs. I read on for a while and then felt a mild note of deja-vu. "What was that element you were talking about before?", I asked. "Flerovium", Ernie replied, "But chances are there is no really stable isotophe or it would have been synthesised by now, if only in picogramme quantities". I paused to savour the moment - it wasn't often that I was in the position to tell him something that he didn't know. "So I guess you wouldn't give much credence to this report", I replied, and read out a short section of the article I was reading. It has been reported by the People's Council of Mining and Metallurgy, in the break-away republic of Kzijekistan, that a large quantity of elementally pure Flerovium-298 has been found in a prehistoric meteoric impact crater - situated in a remote region south-west of the Kyshtut river. The Fleovium occurs in crystalline form as perfect regular tetrahedrons - all exactly identical in size. As the republic has no suitable research facilities of their own, in the spirit of international brotherhood, they are offering uncut crystals, charging only for extraction costs and postage, to researchers around the globe.

"And the article references a web-site where they claim you can order the stuff", I said. Our first thoughts were that this must be an internet scam of some sort, but Ernie ran a quick search and found several academic pre-prints already showing preliminary data from their Fleovium samples. Ernie entered the link into his tablet. The site was written in Kzijekistanese, but Gurgle-translate dealt with that and soon we were reading the following:

Fleovium crystals are all precisely 5.29 cm along each edge. Any number may be purchased at extraction price (plus postage) which can be calculated using the following algorithm:

1) Calculate the precise volume of Fleovium required (in cubic cm), then round down to the nearest cubic cm.
2) Multiply by the density of Fleovium (#%@##**^ grams per cm), then round up to the nearest gram.
3) Multiply by the cost price (6.02 cents per gram), then round up to the nearest cent (all prices in $US).

Gurgle-translate seemed to have problems converting the density to standard Arabic numerals but Ernie said he was familiar with Kzijekistanian digits, so he would calculate the extraction price if I was willing to deal with postage. I loaded the post office site onto my tablet to find:

The Kzijekistan Postal Department charges all post by standardized volume, not weight. The postal charge is calculated by the following algorithm:

1) Determine the standard volume for the item or items (in cubic cm) and round off to the nearest cubic cm.
2) Multiply by postal price (63.28 cents per cubic cm), then round off to the nearest cent (all prices in $US).
Note 1: The standard volume for a single item is that of the smallest x*x*x cube into which the item can be packed.

Calculating the optimal packing of a tetrahedron into a cube wasn't too hard, but I made a couple of small models from folded paper to confirm I had got it right. Nonetheless, when I told Ernie the post charge, he was very unhappy. "But that is more than the extraction cost!", he exclaimed, "I have no interest in subsidizing the profit of the Kzijekistan Postal Department". "Well, I could always use two crystals, maybe we can pack them more efficiently", Ernie muttered. I scrolled further down the KPD's web page and read the following:

Note 2: The standard volume for two items is that of the smallest x*x*y rectangular prism into which both items can be packed simultaneously. 

Once again, I calculated the post cost (I needed to make a second tetrahedron model, and it took a little longer to find the optimal packing), but Ernie was still disappointed: "The post cost is still higher than the materials cost. How about if I buy three?". There was another note further down the page:

Note 3: The standard volume for three items is that of the smallest x*y*z rectangular prism into which all three items can be packed simultaneously. 

It took a lot of work to confirm the best packing this time. By the time I was finished the table was littered with little model tetrahedra and boxes. I decided Ernie had chosen the much simpler task. "Well, at least this time the postal charge is less than the extraction cost", I said when Ernie told me his number. "True", he replied, but now the total cost is 10c more than what I have in my PayPall account! How about if I buy four of them?". This seemed like a fools errand to me - if three were too expensive what cope could there be in buying four crystals? There was a final note on the KPD site:

Note 4: The standard volume for four or more items is that of the smallest hexahedron constructed from six quadrilateral faces into which all the items can be packed simultaneously.

Note 4: The standard volume for four or more items is that of the smallest closed volume (constructed from planar faces) with a maximum of twelve  edges into which all the items can be packed simultaneously. 

I sighed and reached for the scissors to build another tetrahedron. It could take all afternoon to work out how the most efficient packing, but then Ernie gave a chuckle and told me he had already worked out the postal cost and it was much improved. "I can afford four crystals and still have money left over!", he exclaimed. He was so pleased that after he had filled out the online purchase form, he spent all the money left in his account on a nice bottle of Pinot Noir.

---

Here is the point where you may be able to help me. I believe there is sufficient information to work out the density of Fleovium crystals (and the error limit in that value). Also, as it is my turn to buy the wine next week it would be good to know approximately how much Ernie spent on the celebratory wine.

---

Hint 1: For 1, 2, and 3 crystals the answer is all in the rounding.

Hint 2: For n = 1, 2, 3 crystals, the exact un-rounded mass and exact un-rounded volume scale exactly with n.

Hint 3: For n = 4, to find the packing density you need to "Think outside the box". Note that this is different from "Think outside the box", if that helps...

Answer

Well, here are some thoughts - they rely a bit on the hints and a maths result that I can't quite show for myself for the very last bit. Don't know whether it's right, but it seems to work out...

I realised that the floored volume of the crystals themselves (as needed for the material calculation) will go up as 17, 34 and then 52 (not 51!).

You then work out the rounded packed volume by realising that the smallest cube you need is the one with the tetrahedron side as a diagonal on the cube face, so you get a cube of side x == 3.74059.... so the volume is 52.338590.... or rounded off to 52 and then working out the progression to 2 and 3 tetrahedra.

You then set up the inequalities of material cost (left) vs postage cost (right) as follows:

ceil(ceil(17 * d) * 6.02) < round(63.28 * 52)
ceil(ceil(34 * d) * 6.02) < round(63.28 * 103)
ceil(ceil(52 * d) * 6.02) > round(63.28 * 157)

For all three to be true, d must lie in the range

31.73077 and 32.11764 grams / cm³.

This further yields a total cost for the third scenario of between

$198.75 and $199.95

Thus, if we go conservative on the density, let's say he has in his Paypal:

$198.65

Next, we must work out the cost for the 4 crystals. We already have an estimate for d - so we can work out the material cost (taking the same lower bound on density as we did for the Paypal amount):

ceil(ceil(69*d) * 6.02) == $131.84

Now - we know how much he has left at most for paying for postage - to give us a clue on the search space for the volume:

$66.81 Thus we know that the packed volume must be somewhat less than 105.578 cm³ so that all criteria are satisfied. This is plausible as the actual volume of the crystals is 69.78479.... cm³

So, all that remains is to work out what the optimal hexahedron is to contain the 4 tetrahedral crystals.

Ernie had obviously read an article about the best density of packing for 4n tetrahedra and had realised that we could get them into a volume of

4671 / 4000 * 69.78479 == 89.4912 cm³

Thus costing us in total

$183.41

and leaving

$15.24

For wine. Looks like he could just about manage a Pinot noir :)

EDIT:

Well, I've been thinking about this a little more and built a model. I think there is a hexahedron that might get a slightly better volume than the one above, but I haven't been able to calculate that volume as yet. Perhaps the OP will give us some clues as to whether this angle (excuse the pun) is the right one... I built a model, in the hidden picture below, along with some translucent shapes overlaid of what I think the overlaid hexahedron might look like (ish)...

I'm somewhat doubtful, as I don't think even Ernie could calculate the volume of this shape without pen and paper...

My proposed arrangement of tetrahedra:

Overlaid faces of hexahedron (top and bottom view)

Note the faces that look like parallelograms here are not coplanar, so I don't think the volume calculation is simple...

If gravitational waves exist are they technically just another form of light/electromagnetic wave?

I would imagine a gravitational wave would have very similar characteristics to electromagnetic wave, what kind of differences are there?

Answer

If gravitational waves exist are they technically just another form of light/electromagnetic wave?

No.

Electromagnetic waves are (classically) disturbances in the electromagnetic field that propagate with speed $c$.

Gravitational waves are disturbances in the geometry of spacetime that propagate with speed $c$.

I would imagine a gravitational wave would have very similar characteristics to electromagnetic wave

Electromagnetism is (classically) linear. However, the gravitational field equations are non-linear (which are approximately linear in the 'small-signal' approximation).

While there is dipole electromagnetic radiation, the lowest order gravitational radiation is quadrupole. This is related to the fact that the gravitational field is a rank 2 tensor field as opposed to a vector field.

So, other than the fact that they are waves and propagate at $c$, they aren't similar at all. In a comment, Hassan has provided a relevant link for further reading.

mathematics - Dave switched the calculator keys

My calculator has the usual number keys:

0123456789

Dave swapped two number keys around, and won't give it back until I work out which keys he's swapped.

I can give him one sum, using one or more of the basic operators +−÷×, and he will tell me the result on the screen after he presses the equals key.

What sum should I give, so that I will know what keys were swapped when he tells me the result? Or is this simply not possible?

Answer

Perhaps the most obviously correct way to do this is

123456789 + 123456789

because

You can halve the result, producing the number that the key sequence 123456789 corresponds to (as the key swap will affect both halves equally), and then you'll know which digit each of the keys from 1 to 9 corresponds to (and can determine 0 by elimination). This actually works for any permutation of the number keys, not just swaps.

Is linear monemtum conserved with in angular momentum?

I am trying to understand conservation linear momentum with in angular momentum

Herewith i just explain what i am looking for. Imagine an object of mass(M), velocity(V), is undergoing circular motion
with radius(R) Angular momentum L = RMV (R-radius M-mass- V- tangential velocity)

Mass is constant here

Since L is conserved so when R is decreased tangential velocity(V) has to be increased to keep angular momentum constant.

Does this mean linear momentum not conserved when radius of rotation changes?

When linear momentum is not conserved then what is the force that increases the velocity of object when R decreases.

Could anyone clarify this. or could any one say is my basic understanding about momentum is wrong. for easy understanding i am only considered magnitude of momentum's not cross product of vectors

Answer

Two ways of answering your question.

Consider the mass as the system which is acted on by a gravitational force - a central force.

The central force does not apply a torque on the mass and this means that the angular momentum of the mass is constant.
However even if the radius stays the same the linear momentum of the mass changes (in direction) because there is a net force on the mass.
However the speed and the kinetic energy of the mass does not change because the displacement of the mass is at right angle to the line of action of the force and so the force does no work on the mass.

If the radius of the orbit decreases then to conserve angular momentum the mass starts to move faster and so there is an increase in the kinetic energy and the magnitude of the linear momentum of the mass.
This comes about because as the radius of the orbit is decreased work is done on the mass by the central force because there is a displacement of the mass along the line of action of that force.
That work done increases the kinetic energy of the mass and the magnitude of the linear momentum of the mass.

Now consider as the system your mass and another mass which is providing the central force on your mass.
N3L tells you that there must also be a force on the other mass due to your mass which is equal and opposite.

Again angular momentum is conserved because there are no external torques acting on the two masses but this time the angular momentum is carried by both masses because they rotate about their common centre of mass.
The linear momentum of the two masses also does not change because there are no external force acting on the two masses.

Any change in linear momentum of one mass is compensated for by an equal and opposite change in the linear momentum of the other mass.
The equal and opposite forces on the two masses act for exactly the same tine on the two masses.
If the separation of the two masses decreases then both masses have more kinetic energy that energy coming from the decrease in the gravitational potential energy of the two masses.

It is often the case that the mass of one object is much greater than the other and the assumption is made that the orbit of the less massive object is centre at the centre of mass of the more massive object. This can simplify calculations without much loss os accuracy.

classical mechanics - Distance a curveball travels?

I've seen some discussions regarding the movement of a spinning object, say a curveball. However, all have been largely qualitative. I was wondering if anyone has seen or worked through a calculation of how far a curveball moves laterally on its way from the mound to homeplate - even in an order-of-magnitude sense.

geometry - Create a 3 inch measurement

How will you create a 3 inch (within say plus or minus 0.05 inch) side on a standard piece of paper ( 8.5 x 11 inches) merely by Folding? No marking of any kind allowed. Only one paper available. Please explain why. Please try to do with minimum number of foldings. Of course no other measurement tool available

Answer

I can do it in

2

folds, by

folding one corner down to the opposite side, then folding the top of the paper down to the bottom:

The diagonal line splits the right side into one line of 8.5" and another of 2.5", and the middle line splits the right side into two lines of 5.5". Thus the red line is 5.5" - 2.5" = 3" long.

rotational dynamics - Is work done in rolling friction?

I am confused by rolling friction.

Suppose you have a cylinder rolling which starts at rest at the top of an incline plane and begins to roll down the plane without slipping. Is work done by the incline on the cylinder?

I know from doing some problems that the total kinetic energy (translational and rotational) is $mgh$, which is true only if the only work done on the cylinder is by gravity. But also, the cylinder must have nonzero angular acceleration, so friction must be exerting torque on the cylinder, so work must be done by friction. One of these statements is wrong.

quantum mechanics - Equivalence between QFT and many-particle QM

My understanding from my QFT class (and books such as Brown), is that many-particle QM is equivalent to field quantization. If this is true, why is it not an extremely surprising coincidence? The interpretation of particles being quanta of a field is -- at least superficially -- completely different from the quantum mechanical description of N point particles.

calculation puzzle - Magic: the Gathering - Challenge #7: In Your Dreams

_{Previous Challenge}

Background

Normal MtG rules apply here. You may not assume that your opponent will cooperate with you (e.g. choose not to block any of your creatures when they can). Assume that your opponent will block the largest power creatures that they legally can with as many creatures as they legally can, regardless of whether or not it will kill their blockers (in the event of tied power, assume they choose to block the first of the tied attackers alphabetically). If you would draw a card, you may assume that you draw any card remaining in your library, regardless of shuffling.

Puzzle Setup

It's your main phase 1, you haven't played a land for the turn. You're being demolished by your opponent and they have you dead next turn: reverse the tides and win this turn! You've just cast Three Dreams and are about to resolve it. Good luck!

Your hand:
Nothing

Your board:
Nothing

Your graveyard:
Aura Shards
Auramancer
Eternal Witness

Your library:
Abduction
Animate Dead
Battle Mastery
Daybreak Coronet
Dragon Fangs
Dual Casting
Empyrial Armor
Enslave
Fallen Ideal

Field of Reality
Flight of Fancy
Galvanic Arc
Hammerhand
Infiltrator's Magemark
Mirror Mockery
Nature's Chosen
Necromancer's Magemark
Righteous Authority
Sisay's Ingenuity

Spirit Loop
Splinter Twin
Triclopean Sight
Vow of Malice
Vow of Wildness

Your life:
1

Your mana:
: 41
: 11

: 8
: 3
: 7
: 1

Opponent's hand:
Nothing

Opponent's board: (all untapped, summoning sick)
Aether Membrane
Arashi, the Sky Asunder
Drift of Phantasms

Pillar of War

Opponent's graveyard:
Nothing

Opponent's library:
60 Swamp

Opponent's life:
60

Opponent's mana:
Nothing

Answer

I feel like I found an answer far easier than intended. EDIT: Here's a solution that deals 78 damage.

Resolve Three Dreams (hand: 3 cards)
Cast Animate dead, returning Eternal Witness, getting back Three Dreams
Cast Three Dreams (hand: 5 cards)
32 C
9 W
8 U
2 B
7 R
1 G

Cast Hammerhand on Eternal Witness (trigger targets Drift of Phantasms). Cast Splinter Twin on Eternal Witness, activate it, get back Three Dreams
Cast Dual Casting on the Eternal Witness token, then cast Dual Cast Three Dreams (hand: 8)
25 C
8 W
8 U
2 B
2 R
1 G
Cast Triclopean Sight on the Twin Witness to untap it
Activate Twin for a token to get back Auramancer (hand: 7 + Aurmancer)

24 C
7 W
8 U
2 B
2 R
1 G
Play Fallen Ideal on the new Witness token.
Play Abduction on Twin Witness, untapping it.
Activate Twin, then with the trigger on the stack, sacrifice the Twin Witness to Fallen Ideal.
Abduction triggers, bringing back Eternal Witness, bringing back Abduction.

Splinter Twin resolves, you get a token, bringing back Three Dreams.
Board: Fallen Token (4/2), Dualcaster Token, Plain Token, Original Witness
Hand: 5 + Auramancer, Three Dreams, Abduction
20 C
7 W
6 U
1 B
2 R
1 G
Play Nature's Chosen on the Dualcaster, untap it.

Dualcast Three Dreams (hand: 10 + Auramancer, Abduction)
16 C
6 W
6 U
1 B
1 R
Sacrifice Original Witness to Fallen Ideal.
Cast Auramancer for Animate Dead.
Cast Animate Dead for Witness, for Three Dreams (hand: 10 + Abduction, Three Dreams)
Board: Fallen Token (6/3), Dualcaster Token, Plain Token, Original Witness, Auramancer

13 C
5 W
6 U
1 R
Cast Three Dreams (hand: 13 + Abduction)
Cast Battle Mastery, Righteous Authority, Empyrial Armor, Mirror Mockery on Fallen Token (hand: 9 + Abduction)
2 C
4 U
1 R
Cast Abduction on Auramancer

Sacrifice Auramancer (8/4), abduction triggers, bringing back Auramancer, getting back Hammerhand
Cast Hammerhand on Fallen Token, trigger targets Aether Membrane (hand: 9)
Attack with Fallen token (opp has no flying blockers because of Hammerhand triggers), Mirror Mockery triggers, token gets back Three Dreams (hand:10)
Sacrifice all five creatures, attacker is 2 base + 16 Ideal + 1 Hammerhand + 10 Armor + 10 Authority. 78 damage with Double Strike.

@NeedAName mentioned that this solution is easier than intended because Arashi doesn't have flying, you can adapt this solution to beat a flying Arashi by Witnessing back Abduction instead of Three Dreams and Abducting the opponent's Arashi (you lose 8 damage, but the solution can then be optimized to save a bit of mana so that you can cast Flight of Fancy for the draw 2, so you only lose 4 damage).

special relativity - Why does the minus sign in the Minkowski metric mean that nothing can move backwards in time?

I just watched this video:

http://www.youtube.com/watch?v=GkCWywO93b8#t=27

and there Mr. Cox states that because of the minus sign in the Minkowski metric nothing can move backwards in time. It's the first time I hear this argument and does anyone know how to show this?

To be precise: I'm looking for an explicit computation that shows why the minus sign in the Minkowski metric means that we can't rotate into a frame where $t \rightarrow -t$

thermodynamics - Is everything 100% efficient at heating?

Everyone always talks about the efficiency of their appliances. I was wondering if everything was 100% efficient at heating its surroundings ?

quantum mechanics - Does a magnetic field disentangle an EPR pair?

The spins of the both electrons of an EPR pair are undetermined before one conducts a measurement on one of them. Does a magnetic field determine (set up or down) the spin of one partner of an EPR pair if it moves through it and hence disentangle the pair? Is there a potential measurement process involved like emission of a photon from this electron? (I can't imagine how this could happen because the electron was in an indefinite energy state too) But if so, can the EPR pair stay entangled when making the process unobservable (by example closing the field in a box)?

quantum mechanics - Is half-life a statistical average of variable decay times?

Is the half life of a material only accurate as long as you are still in a macroscopic regime? If I had 8 particles in a box would I observe a fluctuation in half lives, and what would occur within the 4th half life?

electromagnetism - Eddy currents Vs. Inducing EMF in opposing the change?

In the following circuit there is a power supply applying a voltage(+$V$) to a circuit with resistance ($R$), current($I$) is now flowing in the circuit, and there is a movable part like so:

The white part is able to move freely. Now I've introduced a magnetic field($B$) to that movable part:

Due to the Lorentz force($F_L$) that movable part will accelerate:

(Ignoring self-inductance) Due to that Lorentz force, there is a change in magnetic flux($\phi$) and by Faraday & Lenz law, there must be a phenomenon that resist's that change. My confusion is with the following, intuitively, there is an induced EMF that oppose the applied voltage(that is causing the motion) like so:

However, would that mean... that there is also Eddy currents that resist the change? By creating a drag force like so:

Are both phenomenons coexistent in this case to resist the cause of change? Eddy current's and induced EMF? I think that Eddy current's is a form of opposition to the change the induces it, like dropping a magnet in a copper bar, or moving a piece of copper slab into a magnetic field, but when the change is caused like the circuit above... the form of "opposition" would purely be a negative induced EMF.

Also to note, this is quite similar to the basics of an electric motor, which I found having no eddy current losses(besides iron-core related). Only back/counter EMF that resist the change that is caused from motion that is caused by the applied power source.

Answer

Attempting an answer:

Eddy currents are induced where there is a change in the field. The way you have drawn the situation, there is no place where the field changes while the white rectangle moves: $\frac{dB}{dt}=0$ everywhere in the conductor. So while there is a current flowing around the loop, there is no eddy current induced (that I can see).

The same is not true if the B field doesn't have such a nice uniform shape / boundary...

I must admit that without doing a proper simulation, I am not sure that my analysis is correct - but right now I can't see anything wrong with it.

mathematics - Find this Unique UVC Palindrome ( ignoring signs and decimal) from Given Fractional Relationship

Given:

U, V, C are three distinct digits ( 0 to 9 ).

UVVVV and CVVVV.U are concatenated numbers.

Dot “.” Stands for decimal.

Relation:

$UVVVV/C= CVVVV.U$

Find U, V , C

Answer

Finding $C$

1. 
   

   
   

   If dividing an integer by $C$ gives a fraction with exactly one digit after the decimal point (note that $U=0$ doesn't work), then $C$ must be non-coprime with $10$, i.e. it must be one of $2,4,5,6,8$.

   
   

   
   


2. 
   

   
   
   

   If $C\geq45$, then the right-hand side is more than $40,000$, and after multiplying by $C$ it won't be a 5-digit number any more. So we must have $C=2$.

   
   

   
   

Finding $U$ and $V$

1. 
   

   
   

   Since $C=2$, the division by $C$ must give $U=5$.

   
   

   
   
   


2. 
   

   
   

   Since $UVVVV$ divided by $2$ is not an integer, $V$ must be odd. Trying the possibilities in turn shows that $V=9$ is the only one which works.

   
   

   
   

Summary

$U=5,V=9,C=2$. The equation is $59999/2=29999.5$.

newtonian mechanics - Optimal speed for the water wheel

The hydroelectricity plants extract the potential energy of highly deployed massive object (water) as it falls down. Without turbine, all that energy would be converted into speed (kinetic energy) at the bottom of the waterfall and further into heat. The turbine produces energy by slowing water down.

The efficiency of turbine, how much energy is extracted by turbine, can be charactarized by the exhaust speed: the faster is the output stream, the less efficient our turbine is since not all speed/energy is extracted. So, slower the turbine spins, the higher is its the efficiency. The extraction is 100% when turbine does not spin and no electricity is produced at all. So, there must be a trade-off between the efficiency and amount of the output, the trade-off determined by the turbine spinning speed (exhaust speed). How is it decided?

I read that large modern water turbines operate at mechanical efficiencies greater than 90%. Since couple of percent losses are inevitable whatever you do, it seems that they say that theoretical efficiency is 100%. Identical efficiency is provided by switching power supply converters, which are 100% efficient in theory. I understand the secret exploited by SMPS. My question is how similar, 100% energy extraction, is achieved through the turbines, which seem to operate linearly (spinning at the same pace) rather than switching mode pumping. What is the water release speed when 100% energy extraction is achieved?

This question is actually is not limited to water wheels. Today wind turbines are becoming more popular and I am curious how do you extract all power from the wind flow. If turbine spins quckly, the air is realased at high speed, which means that you do not slow down the flow, which means that it makes no work. On the other hand, if if you stop the flow completely, your turbine stops and no power is extracted either. What is the optimal turbine speed?

gravity - Photons and Relativity

Consider a Photon from Sun and travels with a velocity $c$. Now think we are that photon. For us, it looks like Sun is moving away from us with a velocity $c$. So, why don't we get attracted back towards Sun, because the mass of Sun would be infinite for us since it moves away from us with a velocity $c$.

Answer

You have completely mixed the modern and classical concepts of relativity. If you're talking about mass increment, you shouldn't calculate speed of Sun based on absolute time & space notion.

For you as a photon, space will be contracted to zero and time will be dilated to infinity. So, you can't calculate a speed (which is a time-like spacetime event) of Sun.

While its a nice satisfactory explanation, its not the real one.
Real Answer:
Relativistic physics doesn't allow you to take position of a photon. In other words, relativistic physics doesn't allow photons to be an observer. Its because a photon can see itself stationary which breaks the framework of relativistic physics. Relativistic physics doesn't allow photons to be at rest in any reference frame.

electromagnetism - Vector Potential for Magnetic field when the field is not in simply-connected region

According to Poincare's Lemma, if $U\subset \mathbb{R}^n$ is a star-shaped set and if $\omega$ is a $k$-form defined in $U$ that is closed, then $\omega$ is exact, meaning that there's some $(k-1)$-form, say $\eta$ with $\omega = d\eta$. Now, translating to vector fields, if we consider $U$ a star-shaped set in $\mathbb{R}^3$ and if $B$ is a vector field inside $U$ such that $\nabla\cdot B= 0$, then there's some vector field $A$ defined in $U$ such that $B = \nabla \times A$.

I've heard that Poincare's Lemma turns out to be true even if $U$ is not star-shaped, but rather, just contractible. Now, in the hypothesis of Poincare's Lemma, the fact that the magnetic field satisfies $\nabla\cdot B = 0$ implies the existance of the vector potential $A$, with $B = \nabla \times A$. But now, what happens if the magnetic field is defined in some region of space that is not simply connected? In that case, the vector potential could not exist according to Poincare's Lemma (it doesn't says that it doesn't exists, but it doesn't guarantees the existance).

So, if the region where the field is defined has holes in it, what happens? Is really a chance that the vector potential doesn't exists? In that case, what are the physical consequences of that? Since I was always told that the vector potential was just a mathematical tool introduced to make life easier, I think that there wouldn't be so great impact on the point of view of conceptual explanation of the situation, however, I'm not sure.

Answer

You ask

if the region where the field is defined has holes in it, what happens?

Well, in that case you can define the vector potential on simply-connected sub-regions $R_i$ whose intersection is the whole non-simply connected region $R$ and such that they differ by only by a gauge transformation on the regions of overlap. This is a physically well-motivated thing to do, because it means that up to gauge transformation, the vector potential can be defined on $R$.

Here's a simple example. Let $\ell=\{(x,y,z)\,|\, x=0, y=0\}$ denote the $z$-axis, then the region $R=\mathbb R^2\setminus\ell$ is not simply connected. To see this, simply consider a closed loop enclosing the axis; there is no way to continuously shrink it down to a point while staying in $R$. Because of this, the there is no $\mathbf A$ defined on all of $R$. However, let $\ell_+$ denote the positive $z$-axis, and let $\ell_-$ denote the negative $z$-axis, then the regions $R_- = \mathbb R^3\setminus \ell_+$ and $R_+ = \mathbb R^3\setminus \ell_-$ have the property that they are each simply connected and $R = R_+\cap R_-$. Moreover, we can define a vector potential $\mathbf A_+$ on $R_+$ and $\mathbf A_-$ on $R_-$ such that there exists a scalar function $\Lambda$ for which \begin{align} \mathbf A_+(\mathbf x) - \mathbf A_-(\mathbf x) = \nabla\Lambda(\mathbf x), \qquad \text{for all $\mathbf x\in R$} \end{align} In fact, here are the explicit expressions in spherical coordinates $(r,\theta,\phi)$: \begin{align} \mathbf A_{\pm} &= -g\frac{\cos\theta\mp 1}{r\sin\theta}\hat{\boldsymbol \phi} \end{align} I'll leave it to you to determine $\Lambda$; it's a fun exercise.

what are the physical consequences of that?

Well, in the context of quantum mechanics, these sorts of topological issues are physically relevant (I am unsure if there are examples in which they are relevant at the classical level, but I don't think so). I won't go into the details here (unless perhaps there is some demand), but the very vector potentials I wrote down in the example above come up when discussing magnetic monopoles and the quantization of electric charge (see Dirac quantization).

These topological issues also become significant in discussing the famous Aharonov-Bohm effect.

thermodynamics - Why doesn't soda go flat immediately after opening?

So, soda is under pressure and has gas dissolved in it. But, when you open it, the gas is still dissolved in it. But, if we wait a few hours, the gas has escaped into the atmosphere.

What factors determine the rate at which gas escapes the soda-gas solution?

thermodynamics - Is there a limit to how hot an object can get?

If heat is the measure of how fast the atoms are moving in an object, than isn't there a limit to how hot that object can get as nothing can go as faster than the speed of light. So because the atoms can't vibrate that fast, will there be a limit to how hot the object can get?

Answer

Wikipedia says at above $1.416785\times 10^{32}~\rm{K},$ all theories breakdown. So, that is theoretical limit. In actuality, 7.2 trillion deg F is the highest known temperature, and that is in Large Hadron Collider (LHC) when they smash gold particles together.

In terms of motion of atoms, the limit would be much lower because, the atoms will fly away as gas. Higher temperatures can be achieved by containing the atoms from flying by compressing them at high pressures. At some point, the compressor will also blast, or evaporate.

One way it can reach very high temperatures is where the heated matter also provides its own compression. That can happen when gravity itself creates compression, so that there is no problem of blast, or evaporation. May be temperatures at the time of big bang, or that of a singularity.

However, the main problem would be that of measuring such temperatures, so, the temperature would be limited by range of the measuring mechanism.

special relativity - Trying to understand one of Einstein's thought experiments

I try to understand Einstein's Relativity: The Special and the General Theory, chapter IX., "The Relativity of Simultaneity". Here's an online version: http://www.bartleby.com/173/9.html.

             |----M'----v->
####-----A--------M--------B-----####

Einstein considers two lightning strokes happening at two points A and B along a railway embankment. At this moment, two observers M and M' meet each other at the mid-point between A and B. M resides on the embankment, M' is riding on the train moving with velocity v towards B.

Then, paragraph 3 reads:

"... he [the observer at M'] is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A."

As far as I understood SR until now, this is actually the non-relativistic point of view. Bearing relativity in mind, the moving observer M' should experience both light beams from A and from B approaching with light speed as well, the beam from B being blue-shifted and the one from A red-shifted. In respect to M' they have different energies, but they should arrive at the same time.

The longer I think about it, the more I doubt whether Einstein found a very fortunate thought experiment here. What do you think about it, could that be true?

explosions - Why do meteors explode?

A report on the Chelyabinsk meteor event earlier this year states

Russian meteor blast injures at least 1,000 people, authorities say

My question is

• Why do meteors explode?


• Do all meteors explode?

Answer

Meteoroids come in a very large range of sizes, from specks of dust to many-kilometer-wide boulders. Explosions like that of the Chelyabinsk meteor are only found meteors that are larger than a few meters in size but smaller than a kilometer.

Though the details are argued endlessly by those who study such phenomena (it is very hard to get good data when you don't know when/where the next meteor will occur), the following qualitative description gets much of the important ideas across.

The basic idea is that the enormous entry velocity into the atmosphere (on the order of $15\ \mathrm{km/s}$) places the object under quite a lot of stress. The headwind places a very large pressure in front of it, with comparatively little pressure behind or to the sides. If the pressure builds up too much, the meteor will fragment, with pieces distributing themselves laterally. This is known as the "pancake effect."

As a result, the collection of smaller pieces has a larger front-facing surface area, causing even more stresses to build up. In very short order, a runaway fragmentation cascade disintegrates the meteor, depositing much of its kinetic energy into the air all at once.

This is discussed in [1] in relation to the Tunguska event. That paper also gives some important equations governing this process. In particular, the drag force has magnitude $$ F_\mathrm{drag} = \frac{1}{2} C_\mathrm{D} \rho_\mathrm{air} A v^2, $$ where $C_\mathrm{D} \sim 1$ is the geometric drag coefficient, $\rho_\mathrm{air}$ is the density of air, $A$ is the meteor's cross-sectional area, and $v$ is its velocity. Also, the change in mass due to ablation is $$ \dot{m}_\text{ablation} = -\frac{1}{2Q} C_\mathrm{H} \rho_\mathrm{air} A v^3, $$ where $Q$ is the heat of ablation (similar to the heat of vaporization) of the material and $C_\mathrm{H}$ is the heat transfer coefficient. Since the mass-loss rate scales as $A \sim m^{2/3}$, sublinearly with mass, smaller objects will entirely ablate faster, setting a lower limit on the size of a meteor that can undergo catastrophic fragmentation before being calmly ablated.

Meteors that are too big, on the other hand, will cross the depth of the atmosphere and crash into the ground before a pressure wave (traveling at the speed of sound in the solid) can even get from the front to the back of the object. There simply isn't time for pressure-induced fragmentation of the entire object to occur, meaning the kinetic energy isn't dissipated until the entire body slams into Earth.

[1] Chyba et al. 1993. "The 1908 Tunguska explosion: atmospheric disruption of a stony asteroid." (link, PDF)

How does the Physics work for the Quantum Suicide thought experiment?

On page 5 of this paper written by Max Tegmark, Tegmark discusses a thought experiment called 'Quantum Suicide'. As far as I understand it, this experiment was created to show the experimental difference between the Copenhagen interpretation and the Many Worlds Interpretation of Quantum Mechanics.

I'm really not sure how the physics works out here. As far as I understand it, the Copenhagen Interpretation assures us that after each measurement, the wave-function collapses for the particle, and asserts that the experimenter will either be dead or alive after each measurement. So the experimenter, as long as she isn't incredibly lucky, should be dead within a couple of measurements. This all works out in my head very nicely, and makes a lot of sense. However, somehow when we are dealing with the Many Worlds Interpretation, "the observer will hear click with 100% certainty". Somehow, the experimenter is guaranteed to survive. At first, I thought this concept is just a parallel of the Anthropic Principle, but on further consideration that doesn't seem to be the case. The Wikipedia Article for this thought experiment states:

However, if the many-worlds interpretation is true, a superposition of the live experimenter necessarily exists, regardless of how many iterations or how improbable the outcome. Barring life after death, it is not possible for the experimenter to experience having been killed, thus the only possible experience is one of having survived every iteration.

Sure, there is a possibility that the experimenter survives. But isn't that the same with the Copenhagen interpretation? Basically, I am very confused, and any help to understand this thought experiment, and its implications (or the maths, because I don't understand equation 4), would be great.

cosmology - The status of the BICEP2 'discovery' after Planck 2014

The tumultous period after the original announcement that the BICEP2 experiment had supposedly detected strong evidence of cosmological inflation in the form of B-mode polarization in the cosmic microwave background brought a lot of excitement and speculation. It seems that now, more than half a year later, the dust is finally starting to settle (har-har-har).

In all seriousness though: There have been a lot of rumor going around about the possibility that BICEP2's signal could be purely due to dust. At my university, one of the founding fathers of inflationary theory has been publicly proclaiming that the BICEP2 experimenters are 'crap', and that the claimed discovery is 'nonsense'. With the new data release by Planck, it may be time for a new, possibly decisive, analysis of the claimed discovery. Was it real? Was it dust? Is there still room for speculation?

I'm looking for an in-depth exposition of the credibility of the BICEP2 'discovery' of B-mode polarizations, in light of the latest data (if anyone feels other data than Planck latest dataset is relevant here, feel free to discuss it!). How much, if any, faith should we still put in the claimed discovery? Can (part of) the signal be dismissed as coming from galactic dust? Have we seen inflation?

Answer

OK, I found a recent link:

Planck versus BICEP2

Despite the new data, the collaboration did not give any insights into the recent controversy surrounding the possible detection of primordial "B-mode" polarization of the CMB by astronomers working on the BICEP2 telescope. If verified, the BICEP2 observation would be "smoking-gun" evidence for the rapid "inflation" of the early universe – the extremely rapid expansion that cosmologists believe the universe underwent a mere 10^–35 s after the Big Bang. A new analysis of polarized dust emission in our galaxy, carried out by Planck earlier in September, showed that the part of the sky observed by BICEP2 has much more dust than originally anticipated, and while this did not completely rule out BICEP2's original claim, it established that the dust emission is nearly as big as the entire BICEP2 signal. Both Planck and BICEP2 have since been working together on joint analysis of their data, but a result is still forthcoming.

This article was updated on 5 December 2014

Now as the BICEP2 paper had used published Planck data available at that time on the dust in their window of the sky, maybe this means that Planck had underestimated dust at that time? The forthcoming paper should surely clear this up.

logical deduction - A Dozen Golden Eggs

Twelve golden eggs are arranged on the shelf according to shell thickness with weight tags in ounces. You've been told that two of the eggs were swapped by some culprit! You must put them back again on the right positions.

• You may use a balance to test the relative weight of a single egg or group of eggs against a different single egg or group of eggs.


• You may use a digital scale to determine the weight of a single egg or group of eggs.


• You may use the balance twice only


• You may use the digital scale twice only

How can you sort out the two swapped eggs?

Answer

Firstly, use the balance scale once to compare the weights of

$A_1:=\{1,2,3,10,11,12\}$ and $B_1:=\{4,5,6,7,8,9\}$.

1. 
   

   If $A_1$ and $B_1$ both weigh the same, then both swapped eggs are in the same one of these two sets. In that case, use the balance scale again to compare the weights of

   
   

   
   

   $A_2:=\{1,4,5,6,11,12\}$ and $B_2:=\{2,3,7,8,9,10\}$.

   
   

   
   
   
   
   • If $A_2$ and $B_2$ both weigh the same, then the two swapped eggs must both be in the same one of $A_1,B_1$ and also both in the same one of $A_2,B_2$. So they must be either two of $\{1,11,12\}$ or two of $\{2,3,10\}$ or two of $\{4,5,6\}$ or two of $\{7,8,9\}$. To find out which, use the digital scale to weigh
   
   
   
   
   

   
   

   $\{1,10,5,8\}$.

   
   

   
   

   If this weighs more than 1 different from what it should, then we know exactly which pair was swapped. If it weighs 1 more than what it should, then the pair swapped is either $5,6$ or $8,9$, and it only takes one more use of the digital scale to find out which. Similarly if it weighs 1 less than what it should. If it weighs exactly what it should, then the pair swapped must be either $11,12$ or $2,3$ or $4,6$ or $7,9$; to find out which, use the digital scale to weigh

   
   

   
   

   $\{2,4,9,12\}$.

   
   

   
   
   
   
   • If $A_2$ is lighter than $B_2$, then the two swapped eggs must be both in the same one of $A_1,B_1$, but the lighter one (i.e. the one with the higher value on the label) must be in $A_2$ and the heavier one in $B_2$. So they must be (one of $11,12$) and (one of $2,3,10$). To find out which, use the digital scale to weigh
   
   
   
   
   

   
   

   $\{10,12\}$.

   
   

   
   

   If it weighs exactly what it should, then 2 and 3 were swapped. If it weighs one less than what it should, then 11 and 12 were swapped. If it weighs more than what it should, we can tell which of 2,3 was swapped with 10.

   
   
   
   
   • If $A_2$ is heavier than $B_2$, then the two swapped eggs must be be both in the same one of $A_1,B_1$, but the heavier one (i.e. the one with the lower value on the label) must be in $A_2$ and the lighter one in $B_2$. So the swapped pair is either (one of $4,5,6$) and (one of $7,8,9$) or (one of $2,3,10$) and $1$. To find out which, use the digital scale to weigh
   
   
   
   

   
   

   $\{1,4,7\}$.

   
   
   

   
   

   If it weighs one, two, or nine more than it should, then we know what 1 was swapped with. If it weighs four or five more than it should, then we know what 4 was swapped with. If it weighs less than it should, then we know what 7 was swapped with. If it weighs exactly what it should, then either $4,7$ were swapped or (one of $5,6$) was swapped with (one of $8,9$). To find out which, use the digital scale to weigh

   
   

   
   

   $\{4,6,9\}$.

   
   

   
   


2. 
   

   If $A_1$ is lighter than $B_1$, then the two swapped eggs must be (one of $10,11,12$) and (one of $4,5,6,7,8,9$). To find out more, use the digital scale to weigh

   
   

   
   

   $\{10,11,12\}$.

   
   
   

   
   

   How much less than it should this set weighs will reduce the number of possibilities to at most three: for instance, if it weighs 28, then the swapped pair must be $10,5$ or $11,6$ or $12,7$. To distinguish between these three possibilities, use the digital scale to weigh

   
   

   
   

   11 together with the egg which might have been swapped with 10 (5, in the example above).

   
   

   
   


3. 
   

   If $A_1$ is heavier than $B_1$, then the two swapped eggs must be (one of $1,2,3$) and (one of $4,5,6,7,8,9$), and we can proceed exactly as in point 2 above to find the solution.

   
   

singularities - Was the singularity at Big Bang perfectly uniform and if so, why did the universe lose its uniformity?

Am I right in understanding that current theory states that Big Bang originated from a single point of singularity?

If so, would this mean that this was a uniform point?

If so, as the universe expanded, what factors contributed to transforming its expansion in a non-uniform manner, to the point of what we can now observe? E.g., non uniform densities of matter and non-uniform cosmic microwave background noise, the latter we are told being the imprint of the very early universe.

I'm making assumptions here but I would be interested in understanding how, if it started uniformly, what made the universe become diverse.

cosmology - Does (it make sense to say that ) the universe has a center?

I was reading this page:

http://www.guardian.co.uk/science/2011/oct/23/brian-cox-jeff-forshaw-answers

and I found this sentence by Brian Cox:

That seems to imply that everything is flying away from us and we're therefore somehow in a privileged position; that isn't true. The way it's often described is if you imagine some bread with raisins in it that you're baking in the oven and as you heat it, it expands. On any particular raisin, if you look, you can see all the other raisins receding from it. So it's space that stretching, it's not that everything's flying away.

I already heard this raisins analogy, but it never persuaded me:

I understand that the "big bang" is more like a "big stretch", and I see how every 2 observers in the universe are being distanced farther and farther away (regardless of their position)

Yet one of the Big Bang ideas is that the universe isn't anymore considered infinite and completely homogeneous

But the fact that the universe is finite, while inflating to me implicates that it should have some kind of bounds (not that we can reach these "bounds", since our distance to them is getting bigger, but they should still exist)

(And the fact that it's spreading inhomogeneous mass and energy over big distances, is thus making it more homogeneous, but this doesn't probably matter)

So: the very idea of a big bang seems to me in contradiction to the assertion that there's no such thing as a "center of the universe":

If it has a finite mass and some kind of bounds, then it should also have a barycenter.

And if we consider the bread with raisins analogy: the bread has a center from which it's expanding

Surely, the universe isn't homogeneous (like the distribution of the raisins), and so, in its hypothetical center, there may not be actually anything... but I think (even if it's really unlikely) it should still be theoretically possible to have a raisin in the exact centre of the bread

Answer

The question of the center of the universe is a question of whether the universe is the same at all points. The easiest way to see that the universe now does not have a center is to use the Newtonian big bang. In such a description, everything is flying away from everything else with a velocity vector proportional to the position vector, where we are at the origin:

$$ v= a r $$

Suppose you are on one of the objects at position r. Then, from your point of view, everything is shifted in $r$, because of your new center $r\rightarrow r-r_0$, but everything is also shifted in $v$, because your velocity is not zero relative to us, but you will describe yourself as stationary. So $v\rightarrow v-ar_0$. The result is that you describe the objects as flying away from you with a speed proportional to their position vector.

The Newtonian big-bang is homogenous--- everyone feels that they are at the center. It is exactly analogous to the relativistic big-bang, which is also homogenous. But the Newtonian big-bang is infinite, while the relativistic big-bang is finite, in that there is no horizon in Newton.

The horizon in relativity occurs where the objects fly away at the speed of light, or equivalently, where the light-rays that reach you emerge straight from the big-bang (since looking further out is looking back in time). The horizon makes the space bounded, but it does not pick out a center, because every point has a horizon symmetric around itself.