Can neutrinos travel faster than the speed of light?

Possible Duplicate:
Superluminal neutrinos

What would be the immediate effects if light does not go at the maximum speed possible?

This is a hot topic right now, so I thought we should get a question going on it and hopefully keep it up to date with the latest evidence for or against this discovery.

Sources: [1] (Associated Press), [2] (Guardian.co.uk), [3] (Original Publication - Cornell University)

Scientists around the world reacted with cautious shock on Friday to results from an Italian laboratory that seemed to show that certain subatomic particles can travel faster than light.

The journey would take a beam of light around 2.4 milliseconds to complete, but after running the Opera experiment for three years and timing the arrival of 15,000 neutrinos, the scientists have calculated that the particles arrived at Gran Sasso 60 billionths of a second earlier, with an error margin of plus or minus 10 billionths of a second. The speed of light in a vacuum is 299,792,458 metres per second, so the neutrinos were apparently travelling at 299,798,454 metres per second.

Ignoring the boilerplate media hype about the possibilities of time travel and alternate dimensions - I'm looking for academic sources that might suggest how this could be true, or alternatively, how this discrepancy could be accounted for.

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I read the article published with their findings. It looks like they took an insane amount of care with their measurement of distance and time.

One of the most common skepticism of people who know nothing about the experiment is stuff like:

You might worry about[...] have they correctly accounted for the time delay of actually reading out the signals? Whatever you are using as a timing signal, that has to travel down the cables to your computer and when you are talking about nanoseconds, you have to know exactly how quickly the current travels, and it is not instantaneous. [2]

This experiment doesn't use that sort of 'stopwatch' timing mechanism though. There is no 'T=0', and no single firing of neutrinos. What is detected is watermark patterns in the steady stream of particles. The streams at the input and output are time stamped using the same satellites and any position along each stream has a precise time associated with it. By identifying identical patterns at input and output streams, they can identify how long it took particles to travel between the points. [1]

As for distance, they use GPS readings to get the east, north, and altitude position along the path travelled to great precision. So much so that they even detect slow earth crust migration and millimetres of changes in distance between source and destination when something like an earthquake occurs. When your particles are travelling on the scale (730534.61 ± 0.20) metres, this is more than enough precision:

It's going to take a lot more than grassroots skepticism to think of what could have caused this discrepancy. I've seen suggestions such as the gravity of the earth being different along the path of the neutrinos, which warps space/time unevenly. The neutrino might not actually be travelling as far as they think if space/time is contracted at one or more points along the path where gravity varies.

Anyways, I'll be interested in seeing how it pans out. Like most scientists, my guess is an unaccounted for systematic error (because they definitely have statistical significance and precision and on their side) that has yet to be pointed out, but it probably won't take too long with all the theoretical physicists that will be pouring through this experiment.

mathematics - A new PSE member with square reputation

A new Puzzling Stack Exchange member (reputation = 1) asked a good question.
He got a square number upvotes, and a square number downvotes.
(more Upvotes than Downvotes)
So, His reputation is a square number below 500.

Then he answered a question, and got good responses.
Again, he got a square number upvotes, and a square number downvotes.
(more Upvotes than Downvotes)
Again, his reputation now is a square number, but still not more than 500.

How many upvote and downvote he get after asking?
Then, How many upvote and downvote he get after answering?

Bonus Question : How if he answer first before asking?

Answer

In the beginning, when he asks he gets

64 upvotes and 16 downvotes ending up with 289 rep = 17$^2$ rep

and then after he answers the question, he gets

16 upvotes and 4 downvotes ending up with 441 rep = 21$^2$ rep

so we are done!! :)

---

For the Bonus:

After answering he gets

49 upvotes and 25 downvotes ending up with 441 rep = 21$^2$ rep

and then after asking he gets

9 upvotes and 1 downvote ending up with 484 rep = 22$^2$ rep

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EDIT: HOW I FIGURED IT OUT

The riddle never said anything about having to use pencil and paper so I made an assumption and assumed that both numbers were less than 200. Then I wrote a short C++ program and tested all possibilities, ending up with both my answers. Total time taken: 4 min and 30 sec.

cipher - What is MOehm trying to tell us?

MOehm has an interesting (updated) profile picture:

It is a lower case and italic 'm' (supposedly for 'M'Oehm) surrounded by what appears to be braille.

The braille says (in UEB, may mean different in other languages): (* used to indicate braille that has no apparent meaning)

ka;ak*k l*,*

a,*b,la* ,;ak

Some of the unknown symbols could be rotated or flipped symbols. However the rest doesn't make sense.

Note: I don't know the answer. If it turns out rude or inappropriate (Which I doubt) don't blame me. Puzzles like this have been done before, and are therefore on-topic.

Answer

If you

XOR / OR (doesn't matter since the input is never 11) each braille code with the one on the opposite side of it.

you get

MISFORMARTIN,

which means

a wordplay of MISFORMATTING (his first name is Martin).

M IS FOR MARTIN (Thanks to Deusovi for enlightening me).

Giving the real name

Martin Oehm

knowledge - Gladys visits a campus

_{This puzzle is part 9 of Gladys' journey across the globe. Each part can be solved independently. Nevertheless, if you are new to the series, feel free to start at the beginning: Introducing Gladys.}

---

Dear Puzzling,

Today I wanted to see a local educational institution. It may not be the most exciting tourist destination, but I enjoyed it a lot. Museums and churches are great, of course, but it's nice to get to talk with young people for a change.

Wish you were here!
Love, Gladys.

P.S. One rule for today's puzzle: A number is worth two letters!

Across
2. Destination of an emigrant trail
3. Wanted to tear down a wall

4. Program used in AA
6. Green vegetable
7. 25%
8. Historical military ruler
9. Segregated urban area
11. Musician behind Live Aid
13. President's daughter
15. Waiter or computer
16. Object with magical powers
17. James Bond

18. Burnt colour
19. Existing being

Down
1. Used for dating by archaeologists
3. Greatly enjoy
5. With sign, light or fighter
6. Sydney Harbour or George Washington
10. Tropical bird
12. Median lethal dose
14. Japanese sword

15. David Fincher thriller

---

_{Gladys will return in "The hearts of gold".}

Answer

Gladys is at

Oran 1 University in Oran, Algeria

The gimmick with this crossword is that

every square has to be occupied by either two letters or one numeral. For example, 4A is 12 STEP, so you put the 1 and 2 in the first and second squares, then ST and EP in the last two.

Completed gridI:

(Credit to @Rubio for this much better and cleaner grid! If you want to see the original (terrible) handwritten grid, click here.)

On the shape of magnetic and electric fields in an electromagnetic wave

Electromagnetic waves are generally depicted like this:

Where the electric fields and magnetic fields exist in the planes perpendicular to the direction of propagation. I also realize that as the electric field changes while the wave is propagating, a magnetic field is induced and vice versa (by faraday's and maxwell's laws of induction). But, those laws predict that the fields will be circular. So, won't the electric and magnetic fields look different? Won't they be circles along arrows that are drawn in the figure? I haven't seen anything written about this anywhere.

mathematics - Make numbers 33-100 using only digits 2,0,1,8

Use the numbers 2, 0, 1, and 8 (ALL of them...only ONE time each) to make every integer from 33 to 100.

• Allowed operations: +, -, x, ÷, ! (factorial), exponentiation, square root and Parentheses


• No specific order is needed


• The modulus operator is not allowed


• Rounding is not allowed (e.g. 201/8=25)

newtonian mechanics - Vertical uniform circular motion.. can it really be uniform?

This is the picture in my mind:

For centripetal force, I learned that: $T-mg\cos\theta= \frac{mv^2}{r}$

In vertical circular motion, the velocity is naturally going to decrease as kinetic energy is converted to potential energy as the particle moves up the circle- and thus resulting into a non-uniform circular motion.

However, I was told that if tension changes accordingly, the velocity will remain constant(as from the equation). What I do not get about this part of the explanation is that which force will balance mgsinθ (tangential force in the diagram )?

Answer

In the situation at hand, you'll never be able to achieve uniform circular motion.

$\frac{d\vec{v}}{dt} = \frac{1}{m}\sum \vec{F}_{\text{ext}}$

This is a vectorial equation. If you look at the picture you've drawn, you have forces on the radial as well as the tangential direction.

On the radial direction, there is the tension force, and $mg\cos\theta$. The equation you have given is the outcome of a calculation which tells you that in order to have the object moving at velocity $v = |\vec{v}|$ on a circle of radius $R$, you have to apply to it a net perpendicular force of equal to $\frac{mv^2}{R}$. That's all it says.

However, there is the tangential part still.

$\frac{d}{dt}v_{tan} = mg\sin{\theta} \neq 0 \text{ in general}$

This means that the speed along the circle is always subject to a gravitational force. Thus, the speed along the circle varies. Since the tension force is orthogonal to it, it can not act on the magnitude of this speed, it can only act on its direction, so it can do nothing to keep it constant. The examples given in the reply of User58220 are misleading because they have added mechanisms which contribute to give forces such, that the tangential part of the gravitational force is cancelled.

In short : no, tension won't make it go at a constant speed.

special relativity - Is measure relative velocity the same for both observer n particle

A particle is moving at velocity v. A stationary observer tries to measure its velocity. From the observer reference frame, he will measure a shorter distance travel as compared to what the particle will measure due to length contraction. The observer will also measure a greater time compared to what the particle will measure. Won't this mean that the observer measured velocity of the particle is different from what the particle measure?

Edit: I think the problem here is that the particle can't measure its own velocity as it must be wrt something. But I thought about what if it is moving from point x to point y. Surely it will be able to measure its velocity through that.

Name the Game #3

_{This is part of a series of puzzles where you must identify the name of a video game
Previous puzzle is here: Name the Game #2}

---

Finding my cubes is key in this place
I could never exist in three-dimensional space
Some might call me the opposite of a room
But my exit is right there, get to it soon!

Name the video game

Answer

Is it

Anti-chamber?

For the clues:

Finding my cubes is key in this place

Progress involves finding different colored cubes that allow you further progress into the game

I could never exist in three-dimensional space

A lot of the puzzles involve impossible spaces and using perspective to shift/connect isolated blocks

Some might call me the opposite of a room

The opposite of a chamber is an anti-chamber

But my exit is right there, get to it soon!

From what I remember, you start next to the exit but have to complete everything else before you're able to leave. And going fast is always good

gravity - How do we determine the mass of a black hole?

Since by definition we cannot observe black holes directly, how do astronomers determine the mass of a black hole?

What observational techniques are there that would allow us to determine a black hole's mass?

waves - How is the relationship between the end correction and the pipe diameter related to acoustic impedance?

I know that Levine and Schwinger calculated the exact value of the end correction by doing something with the acoustic impedance but I don't understand their calculations.

I've looked at paisanco's answer to a similar question, but several points confused me.

• 
  

  How can a wave have mechanical impedance? I thought impedance was a property of the medium.

  
  



• 
  

  Why does the imaginary part of the "radiation impedance of the pipe end" lead to the end correction?

  
  

I'm looking for a conceptual explanation rather than a mathematics based one, but any help would be appreciated.

Answer

The reason the end correction impedance is imaginary (not real) is because it represents energy dissipation from the pipe, by radiation into the infinite region of fluid (air) outside the pipe.

Since it dissipates energy, you can think of it physically as a type of damping. All types of damping is represented by imaginary terms in the transfer function - see any textbook or web site on damped harmonic motion for examples.

The impedance is a property of the medium, not of "the wave". The relevant properties include the geometry of the region - i.e. the diameter of the pipe, but not the properties of the wave itself. For example the end correction is independent of the wavelength of sound waves in the pipe.

In a first course (high school level) on sound waves in pipes, you were probably told that at the "open end" of a pipe the pressure is zero and the velocity is non-zero. In fact that is only approximately correct. If it was exactly true, there would be no work done on the air outside the pipe, and you would not hear any sound produced by the pipe. (Work = $\int P\, dV$, which is $0$ if $P = 0$).

Levine and Schwinger produced a series approximation which is a better approximation to the true end condition, but it is still not exact in real world situations. For example, it ignores the viscosity of the air, which creates a boundary layer in the flow inside the pipe and therefore means that the axial flow velocity over a cross-section through the pipe is not constant.

If the wavelength of the wave in the pipe short, with the same order of magnitude as L&S's end correction formula, the approximation breaks down and the L&S formula becomes "completely wrong," but that situation only arises for a very short pipe where the length is similar to the diameter, or for a high harmonic of the oscillation in a longer pipe.

Physically, you can think of the end correction as representing a cylindrical "lump" of air outside the end of the pipe (with length = the end correction, diameter = the pipe diameter) which is forced to oscillate by the vibrating air inside the pipe, and then radiates all its sound energy away into free space.

Note, the math in the L&S derivation is quite "advanced", and their overall approach to solving the problem predates the use of numerical methods (in 1948, computers had recently been invented, but had not yet become commonplace or powerful enough to attack problems like this) - so don't worry about the fact that you don't understand it, unless you are currently studying "theoretical physics" at graduate level! If you set up a computer model to calculate the 3-D flow pattern in and around the pipe using current methods (e.g. the Boundary Element method for acoustic problems) you will get the "end correction" included automatically because of the "no energy reflection" boundary conditions you specify for the flow field infinitely far from the pipe, without having to do anything "special" to include it in the model.

You might find this paper interesting and perhaps more accessible than the original L&S paper: https://arxiv.org/pdf/0811.3625.pdf

general relativity - Growth of black holes and the no-hair theorem

Let's assume there is a black hole, and there are two starships (A and B), hovering above the black hole at the same distance from the event horizon, but on the opposite sides of the black hole. According to the "no-hair" theorem, both starships should measure exactly the same gravitational force from the black hole. The situation is depicted in the picture.

Now let's throw a very massive object C from the direction where the starship A is positioned to the black hole. As the object C is falling, the starship A should measure a greater gravitational force, because it is closer to the massive object C. Now the situtation looks like in this picture:

According to the theory of general relativity, an outside observer should never see an object cross the event horizon - instead the object will freeze outside the event horizon. I know that the object will become more and more redshifted, and finally it will fade to black. But it should still exert a gravitational force. If it is so, the spaceship A will continue to detect a greater gravitational force, even after it stops receiving any light from the object C.

If this is right, how can a black hole ever grow and how can it consume anything? And if it somehow consumes the falling object C, how can the "no-hair" theorem be true, as you can still measure presence of C? And if my reasoning is not correct, where have I made a mistake?

I know there are similar questions to mine, but as far as I could find none answers exactly my question.

Answer

The mistake seems to be that intuition developed in the context of one approximate model is being extrapolated to a situation that breaks that approximation.

The "approximate model" that I'm talking about is the model in which the spacetime metric is fixed, as it would be for an eternal black hole in an otherwise empty universe, and none of the other objects being considered have any effect on the metric. This can be a useful approximation with those other objects have negligible mass compared to the mass of the black hole. The statement "the object will freeze outside the event horizon" would be appropriate (asymptotically) in this approximation.

However, the question describes a situation in which the mass of object C is not negligible. To determine what the spacetime metric would be in this situation, in principle, we would need to solve the Einstein field equation in a dynamic scenario with two masses, the one labelled BH and the one labelled C. This equation is non-linear; there is no generally useful sense in which two single-mass solutions can be "added" to obtain a two-mass solution. (I wrote "added" in scare quotes because I don't even know what that would mean here.) So the picture of object C approaching the original (say, spherical) event horizon of BH arbitrarily closely is not correct, because the event horizon will change while that is happening.

Solving the Einstein field equation in such a two-mass scenario is very difficult, with numerical calculations on powerful computers being the leading approach today. However, for the sake of illustrating the dynamic nature of the event horizon, a crude back-of-the-envelope exercise might be sufficient.

Consider an eternal Schwarzschild black hole with mass $M$, let $r$ denote the usual radial coordinate, and let $R=2GM$ denote the Schwarzschild radius (in units with $c=1$). The innermost circular orbit is at $r=3R/2$, and something orbiting at that radius must be moving at the speed of light. Anything that is initially moving in a tangential direction inside that radius will necessarily spiral into the black hole. Now, here's the crude back-of-the-envelope exercise. Suppose that objects BH and C both have mass $M$, so that they would both have event horizons at $R=2GM$ if they existed in isolation — but suppose that they are both present, with "coordinate distance" $< R$ between their would-be event horizons. (For the sake of this crude exercise, we can pretend that these statements are well-defined.) Then the midpoint between them has "coordinate distance" $ from both of the two would-be individual event horizons, which means that a flash of light emitted from that point would not be able to escape along any direction at all, because it is closer than the innermost circular orbit with respect to both of the individual "black holes", even though it is not inside the naive event horizon of either one of the individual "black holes." The conclusion of this crude exercise is that the real event horizon is not a pair of spherical surfaces centered on the two masses, because the real event horizon also encloses the midpoint between the two masses before their naive event horizons touch each other.

Again, this was a very crude back-of-the-envelope exercise, but it's sufficient for showing that an event horizon is generally not a static thing: it changes shape during the process. This is confirmed by numerical calculations and is illustrated in figures 1 and 15 from "Revisiting Event Horizon Finders," https://arxiv.org/abs/0809.2628, which are reproduced here for convenience:

After two black holes merge, the system quickly settles into a nominal configuration with a traditionally-shaped event horizon, producing gravitational waves in the process. The No-Hair "Theorem" only says that the final result should be a nominal black hole with a traditionally-shaped event horizon.

The original question isn't about a merger of two black holes; it's about what happens when an object C with significant mass (not necessarily a black hole itself) falls into a black hole BH. But the reasoning and the conclusion is qualitatively similar: when object C is close enough to the original BH, that little bit of extra gravity will extend the region from which light cannot escape, which defines the event horizon. The event horizon will change shape during the process, finally settling back down to a nominal shape, emitting gravitational radiation in the process. Working this out explicitly is exceedingly difficult, but this reasoning is enough to resolve the paradoxes posed in the OP. The No-Hair "Theorem" applies only to the final state, not to the process; and the picture of the infalling object freezing outside the original event horizon is not correct, because the event horizon will change dynamically, becoming momentarily non-spherical, for precisely the reason stated in the OP: the infalling object adds to the overall gravitational effect, which affects not only the observer at A, but also any light that is being emitted from the object.

fluid dynamics - Why does the sound of my tea stirring go up in tone the faster I stir?

I've noticed that when stirring my tea, as I stir faster the tone generated by the stirring goes up. Why is that? Is it something to do with the Doppler Effect?

EDIT --------------------------

Thanks for the responses guys - Victor's and Carl's seem like good answers to the question. Unfortunately I realise I may have described the situation wrong - I think I may actually be hearing the pitch go up as I stir at a generally constant speed. It may have something to do with some accumulating increase in the velocity of the tea. But as that's my stupid mistake I'll tick Victor's response to this particular question.

Answer

I would go for this: Imagine the bottom of the cup as a saw. The noise or chattering of the spoon jumping on the sawteeth is higher the faster spoon moves. Those "sawteeth" on the cup bottom are very small, but the principle is the same. Therefore the faster stirring the higher pitch.

mathematical physics - Diffeomorphisms, Isometries And General Relativity

Apologies if this question is too naive, but it strikes at the heart of something that's been bothering me for a while.

Under a diffeomorphism $\phi$ we can push forward an arbitrary tensor field $F$ to $\phi_{*}F$. Is the following statement correct?

If $p$ is a point of the manifold then $F$ at $p$ is equal to $\phi_* F$ at $\phi(p)$, since they are related by the tensor transformation law, and tensors are independent of coordinate choice. ()

I have a feeling that I'm missing something crucial here, because this would seem to suggest that diffeomorphisms were isometries in general (which I know is false). (*)

However if the statement isn't true then it menas that physical observables like the electromagnetic tensor $F^{\mu \nu}$ wouldn't be invariant under diffeomorphisms (which they must be because diffeomorphisms are a gauge symmetry of our theory). In fact the proper time $\tau$ won't even be invariant unless we have an isometry!

What am I missing here? Surely it's isometries and not diffeomorphisms that are the gauge symmetries?! Many thanks in advance.

Answer

If p is a point of the manifold then F at p is equal to ϕ∗F at ϕ(p), since they are related by the tensor transformation law, and tensors are independent of coordinate choice.

This is roughly true. Initially, there is no meaning when one says that tensors at different tangent spaces are equal. However, the diffeomorphism induces an isomorphism between $T_p M$ and $T_{\phi (p)} M$ (the isomorphism is nothing but the vector push forward). The two tensors are equal with respect to this isomorphism.

I have a feeling that I'm missing something crucial here, because this would seem to suggest that diffeomorphisms were isometries in general...

This is actually true in a sense that is relevant. If $(M,g)$ is a spacetime and $\phi \in \text{diff}(M)$, then while there is no reason to think that $\phi$ is an isometry between $(M,g)$ and itself, $\phi$ is always an isometry between $(M,g)$ and $(M,\phi_{\star}g)$.

This last point saves your concern about proper time. If $\gamma$ is a normalized timelike path between two events $a$ and $b$, we can always consider $\phi \circ \gamma$ as a timelike path in $(M,\phi_{\star}g)$. You can check that the new path is normalized with respect to the new metric $\phi_{\star}g$. The domains of the two paths are exactly the same so the proper time between $\phi(a)$ and $\phi(b)$ is the same as the original path's proper time.

electrical engineering - Could we run an electric car on electric eels?

Could we run an electric car on a tank full of electric eels?

I've heard electric eels are around 400 to 500 volts and can keep shocking for up to an hour. They also have a 10 volt system to sense with, which might be good for the headlights and cd player.

Normalization of the Chern-Simons level in $SO(N)$ gauge theory

In a 3d SU(N) gauge theory with action $\frac{k}{4\pi} \int \mathrm{Tr} (A \wedge dA + \frac{2}{3} A \wedge A \wedge A)$, where the generators are normalized to $\mathrm{Tr}(T^a T^b)=\frac{1}{2}\delta^{ab}$, it is well known the Chern-Simons level $k$ is quantized to integer values, i.e. $k \in \mathbb{Z}$.

My question is about the analogous quantization in $SO(N)$ gauge theories (A more standard normalization in this case would be $\mathrm{Tr}(T^a T^b)=2\delta^{ab}$ ). Some related subtleties are discussed in a (rather difficult) paper by Dijkgraaf and Witten Topological Gauge Theories and Group Cohomology, but I am not sure about the bottom line.

Does anyone know how to properly normalize the Chern-Simons term in $SO(N)$ gauge theories, or know a reference where this is explained?

Answer

Let me normalize the action as $$S=\frac{k}{4\pi}\int\langle A\wedge dA + \frac{1}{3} A\wedge[A\wedge A]\rangle$$ for $\langle,\rangle$ being the Killing form. This coincides with your normalization for $SU(N)$.

Variation of the Chern-Simons action under a gauge transformation $g:M\rightarrow G$ is given by $$S\rightarrow S + \frac{k}{24\pi}\int_{g_*[M]} \langle\theta\wedge[\theta\wedge\theta]\rangle,$$ where $\theta\in\Omega^1(G;\mathfrak{g})$ is the Maurer-Cartan form (Proposition 2.3 in http://arxiv.org/abs/hep-th/9206021). The last term is also called the Wess-Zumino term. Therefore, $\exp(iS)$ is invariant if $$\frac{k}{24\pi}\int_{[C]} \langle\theta\wedge[\theta\wedge\theta]\rangle\in2\pi\mathbf{Z}$$ for $[C]$ the generator of $H_3(G;\mathbf{Z})$.

For $G=SO(N)$, the homology is generated by $SO(3)\subset SO(N)$, and that term can be computed as follows. As you say, $$\frac{1}{24\pi}\int_{SU(2)} \langle\theta\wedge[\theta\wedge\theta]\rangle=2\pi,$$ but $SU(2)\rightarrow SO(3)$ is a 2:1 local diffeomorphism, so $$\frac{1}{24\pi}\int_{SO(3)} \langle\theta\wedge[\theta\wedge\theta]\rangle=\pi.$$

Therefore, the level $k$ in this case has to be even. See also appendix 15.A in the conformal field theory book by Di Francesco, Mathieu and Senechal.

quantum mechanics - Density operator as a function of time

Given the density operator $\rho = \sum_iw_i | \alpha^{i} \rangle \langle \alpha^{i}|$, how does the density operator change with time. Apparently I should get $$i \hbar \frac{\partial \rho}{\partial t} = \sum_{i}w_i(H| \alpha^{i}(t) \rangle \langle \alpha^{i}(t)| - | \alpha^{i}(t) \rangle \langle \alpha^{i}(t)|H).$$ I am having difficulty getting this, it seems that I have to use Shrodingers equation on the $(| \alpha \rangle \langle \alpha|)$ since the intial population $w_i$ is constant in time, but I'm not sure how to differentiate this since $\langle \alpha|$ as I understand is a bra which is a functional in a sense, how does the product rule for differentiation apply then in this case?

Thanks.

Answer

Start with \begin{align} i\hbar \frac{d}{dt}\vert{\alpha}\rangle &=H\vert\alpha\rangle \end{align} and take the adjoint $$ -i\hbar \frac{d}{dt}\langle {\alpha}\vert =\langle \alpha\vert H $$ where $H^\dagger=H$ has been used. Then simply use the product rule: \begin{align} i\hbar\frac{d}{dt} \left[\vert \alpha\rangle\langle\alpha\vert \right]&= \left[i\hbar\frac{d}{dt}\vert \alpha\rangle\right]\langle\alpha\vert + \vert\alpha\rangle\left[i\hbar\frac{d}{dt}\langle \alpha\vert\right]\, ,\\ &=\left[H\vert\alpha\rangle\right]\langle\alpha\vert - \vert\alpha\rangle \left[ \langle\alpha\vert H\right]\, . \end{align}

electromagnetism - How to make a small tokamak?

$\require{mhchem}$I made a fusor once, like the easy science project: deuterium-deuterium ones, but they're really inefficient. I was wondering if it would be possible to make a small tokamak; not one that will make a ton of power, but I just want to make one for the sake of making one, like a tokamak the size of a pizza box. What materials would I need and how would I go about assembling it? Assume it is deuterium-deuterium or $\ce{p-^11B}$ fusion, maybe $\ce{^3He-^3He}$.

cosmology - Could the missing antimatter lie outside the observable universe?

While I was reading a similar question asking if other galaxy could be made of antimatter, to which the answer was: if they were, we should detect the radiation from matter interacting with antimatter on that sort of scale. But what if the missing antimatter lies outside the observable universe. Wouldn't that result in the matter dominated observable universe we live in, without any of the radiation from the antimatter lying outside it being detected?

wavefunction collapse - How can two electrons repel if it's impossible for free electrons to absorb or emit energy?

There is no acceptable/viable mechanism for a free electron to absorb or emit energy, without violating energy or momentum conservation. So its wavefunction cannot collapse into becoming a particle, right? How do 2 free electrons repel each other then?

Answer

It is true that the reactions $$e + \gamma \to e, \quad e \to e + \gamma$$ cannot occur without violating energy or momentum conservation. But that doesn't mean that electrons can't interact with anything! For example, scattering $$e + \gamma \to e + \gamma$$ is perfectly allowed. And a classical electromagnetic field is built out of many photons, so the interaction of an electron with such a field can be thought of as an interaction with many photons at once. There are plenty of ways a free electron can interact without violating energy or momentum conservation, so there's no problem here.

quantum mechanics - Why does the wave description say that probability oscillates, while the phase interpretation says constant amplitude?

The wave description of a particle illustrates an oscillating probability of the particle being found in any point in space.

When a particle travels, it carries along with it a phase that oscillates in the complex plane. When the particle travels, the amplitude never changes although the phase does, which doesn't matter. The only property that matters is the amplitude.

Why does the wave description say that probability oscillates and the phase interpretation says constant amplitude? What am I missing?

special relativity - How does matter turn to energy at the atomic level?

When matter is converted to energy by means of $E=mc^2$, it produces quite a lot of "energy". What I am having trouble understanding is exactly how the matter is transformed to energy at the atomic level. Do the atoms gain something or lose something in their internal structure? Do they just vibrate at different frequencies when the conversion occurs?
Edit: This is a duplicate of the question, pardon me. Sorry.

Answer

Let us get down to basics, to convert matter to energy the special relativity algebra has to be used. This describes elementary and complex particles by a four vector, whose "length" is the invariant mass of the system described, invariant to Lorenz transformations.

The length of this 4-vector is the rest energy of the particle. The invariance is associated with the fact that the rest mass is the same in any inertial frame of reference.

The $M$ in the famous $E=Mc^2$ coincides with the invariant mass only in the rest frame of the particle/system, because this $M$ is a function of velocity and is called the relativistic mass and has nothing to do with the energy budget of particles, except at the rest frame of the system.

The fact that energy can be extracted from particles and systems with an invariant mass depends on the quantum mechanical nature of atoms. Atoms are composed out of electrons trapped in the electric potential well of the nucleus, in stable orbitals, , but the energies stored are of order of keV, not really exploitable, also because the orbitals are stable.

A lot of energy can exist in a nucleus , order of MeV, where neutrons, protons, are bound by the strong force in potential wells, and also where there also exist instabilities that can be exploited, by forcing changes in nuclear structure, i.e. the type and number of nucleons.

This is the binding energy curve

for the nuclei. It gives for each known nucleus the average binding energy per nucleon, in the collective strong potential well. The fact that one can extract energy from transitions is based on this curve.

Mass defect is defined as the difference between the mass of a nucleus, and the sum of the masses of the nucleons of which it is composed. The mass defect is determined by calculating three quantities. These are: the actual mass of the nucleus, the composition of the nucleus (number of protons and of neutrons), and the masses of a proton and of a neutron. This is then followed by converting the mass defect into energy. This quantity is the nuclear binding energy, however it must be expressed as energy per mole of atoms or as energy per nucleon.

Nuclei where the nucleons, protons and neutrons, are less bound , if fused will give off energy order of MeV. Heavy nuclei, like uranium, broken into pieces with less binding energy will give off energy again order of MeV.

One more link for fusion.

particle physics - Is baryon wavefunction antisymmetry always true?

The baryon wavefunction is comprised of the direct product of contributions forming different Hilbert spaces such that: $$|\Psi \rangle = |\text{spin} \rangle \otimes |\text{flavour}\rangle \otimes | \text{colour}\rangle \otimes |\text{space}\rangle. $$ The necessity for a colour degree of freedom is usually motivated in the literature from the delta(++) spin 3/2 containing quark content $uuu$. It is flavour symmetric by inspection, is symmetric in the spin quantum numbers and for lowest lying states, has symmetric space state. The state thus contains identical fermions but is overall symmetric under interchange of any of the quarks. This is in violation of the PEP – the resolution was of course the addition of the colour degree of freedom which is necessarily antisymmetric so as to conform to the principle.

It’s then said that the generic wavefunction for a baryon is overall antisymmetric. Is this only the case where the baryon wavefunction is of the form $$\epsilon_{ijk} \psi^{(1)}_i \psi^{(1)}_j \psi^{(1)}_k$$ or $$\epsilon_{ijk} \psi^{(1)}_i \psi^{(1)}_j \psi^{(2)}_k.$$

Where in the former case we have all three $qqq$ the same and in the latter only two are identical? I say this because if we consider one of the baryons with no flavour symmetry e.g $|uds \rangle$ this is a state with no identical fermions so does the the PEP (i.e total wavefunction antisymmetry) have to hold for this case? (I guess analogously to the fact that the mesons have no requirement of antisymmetry because the content of the valence quarks is $q \bar q$ and this is never two identical quarks)

Answer

The valence quark content of a baryonic state is $qqq$, that is to say three identical quarks make up the content of the bound state amongst a sea of partons. Each quark therein carries a series of quantum numbers which serve to denote the different possible states of the quark and, collectively, must be such that a simultaneous permutations of the degrees of freedom yields an overall antisymmetric state, in accordance with the fact the quantum state must obey Fermi-Dirac statistics.

For non excited states (the so called lowest lying states), the spatial component of the direct product is always symmetric ( $S$-wave orbital angular momentum) and colour is antisymmetric so the combination $|\text{spin} \rangle \otimes |\text{flavour} \rangle$ must be symmetric. The observable states transform under irreducible multiplets of (approximate) $SU(3)$ flavour symmetry and so the '$|uds \rangle$' state is actually a flavour symmetric combination in the $10$ decuplet, a flavour antisymmetric combination in the $1$, and appears with mixed flavour symmetries in the remaining two octets in agreement with the group theoretic decomposition $3 \otimes 3 \otimes 3 = 1 \oplus 8 \oplus 8 \oplus 10$.

So. e.g for the totally flavour symmetric combination, the spin state of the wavefunction must be in one of the symmetric spin $3/2$ states.

electricity - Conceptual Doubt Regarding Electric Circuits

There are a few questions which have always bugged me. Unfortunately, due to time constraints, I end up memorizing an example, hoping to imitate that knowledge in an exam question. Consider the following situation, given a circuit with a battery (negligible internal resistance) and a light bulb that is glowing. Let $V$ denote the emf and let $I$ denote the current flowing in the circuit.

1) By increasing which variable will we be able to increase the Brightness of the Bulb?

2) This is a more specific question: Suppose the Bulb is glowing normally (given the initial conditions [$V,I$]) then if consider this new circuit (illustrated below). $R$ is a resistor with Resistance equal to $R$. The question is: What will be the resulting emf?

quantum mechanics - What is the difference between metastable states and excited states?

In the book Mathematical concepts of quantum mechanics ,Stephen J. Gustafson Israel,Michael Sigal, they say

The notion of a resonance is a key notion in quantum physics. It refers to a metastable state – i.e., to a state which behaves like a stationary (bound) state for a long time interval, but which eventually breaks up. In other words, the resonances are states of the essential spectrum (i.e. scattering states), which for a long time behave as if they were bound states. In fact, the notion of a bound state is an idealization: most of the states which are (taken to be) bound states in certain models, turn out to be resonance states in a more realistic description of the system

What is the difference between metastable states and excited states?

word - Fastest way from 8 to 7

Inspired by this great question by Uvc (actually, it's the same logic but with different words).

How fast one can get from EIGHT to SEVEN?

The rules are the same:

• You are only allowed to change one letter at a time.



• You have to keep the word length the same.


• At each step of the way, you must have a legitimate English word. Thus, something like going from "EIGHT" to "EIGHN" is not a valid step, but from "FIGHT" to "NIGHT" would be.

Here is an example 11-step way (but with very obscure words involved): eight-sight-sighs-sikhs-sikas-simas-simar-sizar-sizer-siver-sever-seven.

Answer

Here's my 8-step solution:

EIGHT SIGHT SIGHS SINHS SINES SENES SEMES SEMEN SEVEN

(I don't know if SINHS - plural for SINH ("hyperbolic sine function") is an accepted word)

riddle - Bathed in violence... Say my name!

In the darkness, I stay hidden, then with the cleansing sweep,
I step out from the shadows, as I watch the whole world weep.

I was torn apart, my body bent, forever segregated.
Broken, I at first give life. Then as a weapon, take it.

Elusive, you won't trap me, nor the hoard kept at my feet,

Sometimes haunted by my twin, though cursed to never meet.

Riding on your gallows, bathed in violence, I start.
Say my name, mere mortal, before I again depart.

Answer

You are a:

RAINBOW

In the darkness, I stay hidden, then with the cleansing sweep,

The "cleansing sweep" refers to the rain. "In the darkness" means that it cannot be seen (unless under some certain conditions).

I step out from the shadows, as I watch the whole world weep.

The rainbow emerges from nowhere. The "world weeping" again refers to the rain.

I was torn apart, my body bent, forever segregated.

"Body bent" refers to the shape of the rainbow.

Broken, I at first give life. Then as a weapon, take it.

"Broken" refers perhaps to the division of the colours of the rainbow. "At first give life" refers to RAIN, "then as a weapon take it" refers to BOW.

Elusive, you won't trap me, nor the hoard kept at my feet,

You cannot contain a rainbow. There is said to be a pot of gold at the end (or "feet") of the rainbow.

Sometimes haunted by my twin, though cursed to never meet.

Sometimes there is a second rainbow on a wider arc.

Riding on your gallows, bathed in violence, I start.

ROYGBIV...

Say my name, mere mortal, before I again depart.

I said already dammit! RAINBOW!

As a personal comment: This is a world class riddle.

newtonian mechanics - Why do you have less control over a bicycle with a broader steer?

I recently had an argument with someone on why a bicycle with a broader steer gives you more control. We both had different answers to this question and neither could convince the other.

First a schematic drawing of the situation:

Both claims assume that the only force on Body is gravity, denoted $F_g(Body)$. This force is carried on through the bicycle to where the wheel touches the floor. The bicycle is assumed to have no mass.

Now the two claims:

1. 
   

   Your body and both hands form a triangle. This means that the force $F_g(Body)$ is split up in horizontal component vectors and vertical component vectors. The claim is that this by this, not all of $F_g(Body)$ is directed downwards anymore. So the force acting on the floor through the wheel would be less than $F_g(Body)$. If the steer is broader, the horizontal components will increase in size and thus the vertical components will decrease in size. Meaning that you wouldn't press down on the floor as hard with a broader steer resulting in less friction=less control.

   
   

   The steer would provide a counter force for the horizontal components.

   
   


2. 
   

   When you're steering, the bicycle is leaning. So the point where the wheel is touching the floor could then be considered a fulcrum. A broader steer would mean a greater distance to the fulcrum, thus a bigger moment. This would be harder to compensate=less control.

   
   
   

My question now is: "Why do you have less control over you bicycle when the steer is broader. I would like to know the correct physical explanation and what is wrong with the claims (or perhaps somethings wrong with our assumptions)."

N.B. I am not a physicist myself (I'm a mathematician), so forgive if I made some obvious mistakes.

Answer

The mechanics of steering a bicycle are more complicated than you think - as evidenced by the fact that it's quite easy to ride a bike without touching the handlebars, for example.

The key to the stability of a bicycle is the angle of the fork - both the angle of the main pivot, and the offset of the wheel with respect to that pivot - and the point where that line meets the road (green dashed line in the below):

When the bike starts to "fall over", the wheel will start to turn because the green line intersects the road in front of the point of support of the wheel. And when the dashed line is in front of the wheel, it will turn the wheel in the right direction to "counter the lean" - in other words, it is a stable situation. If the fork is bent so the green line ends up behind the support, the bike becomes impossible to control.

Now assuming that the fork is correctly sized and aligned for the bike, the question becomes "what is the role of the handlebars?". These allow you to either amplify, or counter, the force that the wheel is exerting by itself. Wider handlebars do three things:

1. because they provide a longer lever, you will feel less force when the wheel is trying to turn: you therefore "feel" the natural movement of the bike less well


2. The longer lever means that you need to move your hands more to make a small adjustment - this makes it feel like your actions are not doing anything


3. At the same time, once you get used to the longer handlebars, you would in principle be able to make more precise adjustments


4. Really long (and heavy) handlebars could in principle change the inertia of the system (you notice this when you hang heavy shopping bags on your handlebars), but it's not likely to be a problem with normally-sized bars

I think that points (1) and (2) in particular will give you the sense that you have less control over the bike, although (3) could, in experienced hands, provide the opposite sensation. But by FAR the most important factor in the control of your bike is the correct alignment of the fork.

oscillators - Linearized equations

1. 
   

   What is $V_{\alpha\beta}$?

   
   
   


2. 
   

   And what is a symmetric, positive definite potential energy matrix?

   
   


3. 
   

   And why is there a linearized equation like this?

   
   

homework and exercises - Velocity of a body in a pulley problem involving a spring fixed to the ground

So the picture shows the whole situation. My question is, how do you find the velocity of the object after it had descended 10 cm (0.1 m) ? Pulley's radius 0.2 m, moment of inertia 0.2 kg/m

I have a problem because the pulley is supposed to have friction if it's rotating,right? Because the torques on either side shouldn't be equal. My teacher used $$ \frac {Iw^2}{2} + \frac {mv^2}{2} + \frac {kx^2}{2} = mgh$$ Where $$ v=rw $$ and $$h=0.1m $$ as given. So the answer came as 0.5 m/s. My problem is, if there is friction why aren't we taking the energy loss due to that into account?

$mgh $ would have to be spent on that as well. Plus where do the individual tensions on each side come into play? I understand that $mgh $ kind of accounts for those tensions, but not exactly how.

If mgh is doing work on the pulley, does that include work done by/against tension as well as the energy loss due to friction?

Plus, tension on one side is stretching the spring, right? So why shouldn't we consider that separately?

If somebody could clarify I'd be grateful. Thanks so much. No working out needed, I just want to know exactly what happens to the energy

newtonian gravity - Pluto's gravitational pull on a person on the Earth's surface?

My physics teacher stated that Pluto has a gravitational pull on objects on Earth, namely humans. Is this true? What is the free-fall acceleration of Pluto with respect to being on the Earth's surface (i.e. the Earth's free-fall acceleration is $9.8$ m/s$^{2}$)?

Answer

This is true. Newton's law of universal gravitation says everything attracts everything. To get the free-fall acceleration of some object on Earth towards Pluto, take Newton's law and divide by the object's mass to get $$a=\frac{F}{m}=\frac{GM}{r^2}.$$ Subbing in reasonable values - $M=0.002$ Earth masses, and $r$ between 29 and 49 AU you get something like $10^{-14}\textrm{ m s}^{-2}$.

nuclear physics - What decides fission products?

I am learning about nuclear fissions and learned about the fission fragment distribution. It was interesting to see that the fission fragments have unequal masses.

I was wondering as to what governs which fission products are generated in each fission?

What parameters decide which fission products will have higher yield?

special relativity - Can something travel faster than light if it has always been travelling faster than light?

I know there are zillions of questions about faster than light travel, but please hear me out. According to special relativity, it is impossible to accelerate something to the speed of light. However, we can still have objects (like photons) going at speed $c$ if they never had to accelerate in the first place, i.e., if they always go at the speed of light.

What I was wondering is, would it be possible for us to discover some particle that travels faster than the speed of light? Again, I don't mean something that can be accelerated beyond $c$, but rather something that always goes faster than light. Does that contradict relativity?

Answer

Yes, if a particle would be travelling faster than light, it would always travel faster than light. This is what's called a tachyon, and they have in some sense imaginary mass.

The three regimes, time-like, light-like and space-like (i.e. subluminal, luminal and superluminal space-time distances) are invariant under Lorentz transformation. Therefore anything on a super-luminal 'mass-shell' would always stay there and could not be decelerated to light/ or sub-light speed.

The problem is not that it would violate relativity, but rather causality, since with faster than light information propagation one could 'travel back in time', therefore leading to paradoxes.

For an introduction check out Wikipedia

particle physics - What the heck is the sigma (f0) 600?

At one point, I decided to make friends with the low-lying spectrum of QCD. By this I do not mean the symmetry numbers (the "quark content"), but the actual dynamics, some insight.

The pions are the sloshing of the up-down condensate, and the other pseudoscalars by extending to strangeness. Their couplings are by soft-particle theorems. The eta-prime is their frustrated friend, weighed down by the instanton fluid. The rho and omega are the gauge fields for flavor SU(2), and A1(1260) gauges the axial SU(2), and they have KaluzaKlein-like echoes at higher energies, these can decay into the appropriate "charged" hadrons with couplings that depend on the flavor symmetry multiplet. The proton and the neutron are the topological defects. That accounts for everything up to 1300 but a few scalars and the b1.

There are scalars starting at around 1300 MeV which are probably some combination of glue-condensate sloshing around and quark-condensate sloshing around, some kind of sound in the vacuum glue. Their mass is large, their lifetime is not that big, they have sharp decay properties.

On the other hand, there is nothing in AdS/QCD which should correspond to the sigma/f0(600), or (what seems to be) its strange counterpart f0(980). While looking around, I found this discussion: http://www.physicsforums.com/showthread.php?t=241073. The literature that it pointed to suggests that the sigma is a very unstable bound state of pions (or, if you like, tetraquarks).

This paper gives strong evidence for an actual pole; another gives a more cursory review. The location of the pole is far away from the real axis, the width is larger than the mass by 20% or so, and the mass is about 400MeV. The authors though are confident that it is real because they tell me that the interpolation the interactions of pions is safe in this region because their goldstone properties dominate the interactions. I want to believe it, but how can you be sure?

I know this particle was controversial. I want to understand what kind of picture this is giving. The dispersion subtraction process is hard for me to visualize in terms of effective fields, and the result is saying that there is an unstable bound state.

Is there a physical picture of the sigma which is more field theoretical, perhaps even just an effective potential for pions? Did anyone who convinced himself of the reality of the sigma have a way of understanding the bound state properties? Is there an analog unstable bound state for other goldstone bosons? Any insight would be welcome.

optics - How the amplitude of light increases its brightness?

Amplitude is the maximum value reached by a wave.

In case of the light wave how the bigger amplitude gives more brightness/intensity ?

word - Definitely not suspicious

This unsuspicious ... post - has just a bunch of random letters.

No secret words, no silly ciphers, nothing.

Nothing to see here. Goodbye.

Urgent secret message: Go to another question

ahem, what are you still doing here?

aeioubdfhjstvngclmrykpqwz (What? Don't look!)

Haha! You can't see it.

Can everyone apart somewhere? Another riddle?

No no no! Don't look at the second one!

1. suufbudyeliley. fyezeled, sumihv, guzhogfrcgeri 2. jiccuazug. -tvinin

general relativity - How (or why) equivalence principle led to Einstein field equations?

If equivalence principle was origin of general relativity what was the process that this principle led Einstein to developed his theory of general relativity?

mathematics - Three mathematicians are forever in Prison

I'm excited to share the following riddle. It was given to me more than two years ago and I finally solved it last summer (after not thinking about it for a long time). In my desperation, I tried to find a solution online, but couldn't even find the riddle anywhere else. I'm excited to see if somebody knows the riddle or if not, how you approach the solution.

Three mathematicians are in prison. Each of them is in a single cell and they are not able to communicate in any way. They are imprisoned for an arbitrary number of days.

Each cell has a single light bulb that is either on or off on a given day. The warden tells the mathematicians that the light system of the prison has three modes:

• neutral mode, where each lightbulb is independent of the others


• bright mode, where two bulbs turn on every day and the other turns off


• dark mode, where two bulbs turn off every day and the other turns on

(All distributions are not necessarily uniform.)

The prison starts in neutral mode. After an unknown but finite number of days, the warden will select either bright mode or dark mode, which is locked in permanently.

After countably infinitely many days have passed, the mathematicians are asked which one the warden picked. They may discuss strategy before going into the cells, but there will be no communication afterwards. They have unlimited capacities to communicate and remember strategies that they come up with. Two of the three need to guess correctly to escape; how can they ensure this? You may assume that the axiom of choice holds.

Answer

The night before their first day in prison, the three mathematicians choose a non-principal ultrafilter on the set of days they are in prison. A non-principal ultrafilter is a rule for classifying some sets of days as large, and the rest as small, subject to the following conditions:

1. Every set of days containing a large set is large,


2. If the set of all days is partitioned into finitely many sets, exactly one set is large, and


3. No finite set of days is large.

Constructing a non-principal ultrafilter requires the axiom of choice, and communicating an ultrafilter from one mathematician to another requires an infinite amount of information. These mathematicians have unlimited mental capacity though, so maybe this is possible.

After the countably infinite number of days has passed, each mathematician guesses dark mode if the set of days when their light was off is large, and guesses bright mode if the set of days their light was on is large (property 2 above guarantees that exactly one of these conditions is met).

Suppose warden eventually selects dark mode. Consider the following four sets of days:

• neutral mode days,


• dark mode days when the first mathematician's light is on,



• dark mode days when the second mathematician's light is on,


• dark mode days when the third mathematician's light is on.

Again, property 2 of the ultrafilter says that exactly one of these sets of days is large. By property 3, it cannot be the first, because that set is finite. This means exactly one of the mathematicians saw a light for a large set of days, so that mathematician guesses bright mode and the other two guess dark mode. Success!

The argument in the case the warden selects bright mode is identical.

cipher - Lepidopterology needed!

This puzzle belongs to the puzzle series: hyper-modern art

---

The two friends in the gallery of Hyper-modern art are still discussing.

"You know what, to certain extent I start enjoying this Hyper-modern art, but I still think it lacks something important."

"And what should that be, my friend?"

"Beauty. Aesthetics. It is all well and good that the observer has to fully engage for deeper understanding, but I still would like to, well, like an image. It should be pleasing to watch."

"Hmm, yes, I get your point. But maybe the next example is more to your liking."

They move on to the forth room of the gallery, where a gigantic painting of 11x16 metre fills the whole side of the room.

"See what we have here: The painting is called 'Nature tells a story', and I think it is rather pleasing, don't you?"

"Well, it for sure is impressive... And yes, it is also nice. But what is the story nature wants to tell us here?"

"See, now we need our HUD again..."

---

Full resolution image, 1.3Mb, 6582 x 4539 pixels

---

The goal of the puzzle is to find the message, this painting is telling us. This message is written in English and consists of five words. The puzzle will likely require some patience and work, and it does require the full-resolution image provided by the link.

While the full resolution is needed to see all details, the puzzle does not require any digital information. You can print the image and still solve it from that.

Bonus: An additional message is hidden in the image. This message is a clue for the actual puzzle. What is the message?

---

Hints

Hint 1:

You need a hint? Count!

Hint 2:

You think the flying direction of butterflies is random? Think again!

Hint 3:

A single butterfly (in one direction). But two?

Hint 4:

Have you taken a close look at hint 3 above?

Answer

The solution is

Darwin was right no creationism

1. Finding the hint

The butterflies have to be arranged according to their orientation. There are 16 different orientations (17 if you count the grey ancestor-butterfly in the center, but it can be ignored for this), evenly distributed with 22.5° difference each. Based on JTL's arrangement, slightly altered here, the first hint becomes visible (reading counter-clock-wise like mathematicians do and unlike cuckoo clocks): Hint: Baconian by shape

2. Preparing

To get the correct order, the butterflies have to be ordered: Starting with the grey butterfly in the middle the next butterfly-generation has exactly one property changed: either the shape (wings or antennae) or the color or pattern. With this rule a family-tree can be created like this (based on Alconja's work)

3. Solving

With the hint on Bacon's cipher the solution is hidden as a binary code with A=00000, B=00001, etc. In this riddle a butterfly is considered as 1 if its shape is different (wing-shape or antennae) and is considered as 0 if the shape is the same (and the color or pattern is different). Starting on the second image, the first branch becomes
000110000010001101100100001101
split into chunks of 5 yields
00011 = D, 00000 = A, 10001 = R, 10110 = W, 01000 = I, 01101= N.
Each new branch of the butterflies forms a new word, resulting in the solution: Darwin was right no creationism

A big thank you for this beautiful and challenging riddle! And of course to Alconja and JTL who did all the hard work. I'm new here, so if I can split the bounty, I'm happy to do so.

special relativity - Help understanding Bell's spaceship paradox

The problem statement of Bell's Spaceship paradox is this:

Two spaceships float in space and are at rest relative to each other. They are connected by a string. The string is strong, but it cannot withstand an arbitrary amount of stretching. At a given instant, the spaceships simultaneously (with respect to their initial inertial frame) start accelerating (along the direction of the line between them) with the same acceleration. (Assume they bought identical engines from the same store, and they put them on the same setting.) Will the string eventually break?

And the solution is here: http://www.physics.harvard.edu/uploads/files/undergrad/probweek/sol11.pdf

The very first statement made in the solution to this problem is "To an observer in the original rest frame, the spaceships stay the same distance, d, apart.". But why do they stay the same distance apart to an observer in the original rest frame? Shouldn't the distance between the spaceships undergo length contraction, as they are connected by a rope? I asked a similar question here, and the answer that I got was that

Length contraction only applies to situations where you have a system with two "ends" that are moving at the same velocity, and you know the distance L between these ends in the frame S where they are at rest, and want to know the distance L' between them at any given instant in some other frame S' where they are moving at velocity v along the axis joining the two ends.

Well the two ends of the rope are moving at the same velocity, I know the distance between them at frame $S$ when they are at rest, and I do want to know the distance between them at any given instant when they are moving at velocity $v$ along the axis joining the two ends. How then can I make the statement that for an observer in the rest frame that the spaceships stay the same distance, $d$, apart?

Answer

"To an observer in the original rest frame, the spaceships stay the same distance, d, apart.". But why do they stay the same distance apart to an observer in the original rest frame?

The spaceships move with constant mutual distance in the original rest frame, since their corresponding parts have the same velocity function of time. The description of the situation in the original question directly implies this.

Shouldn't the distance between the spaceships undergo length contraction, as they are connected by a rope?

No, this would be contrary to the specified situation.

Software Riddle

 This riddle is a computer program/language. Enjoy!

I used to love the Sun
Now I pray to my oracle
I'm said to virtualize all
I'm not picky, working anywhere...

I love my classes-all of them
I love my matter-known as objects
I love running parallel-too lazy to wait
Fans worldwide build and break with me!

My syntax-derived from a letter
many programmers found me familiar

I have nothing included-very greedy
And I must import all!

I want every line torn apart, as this is an easy one :)

Answer

You are

Java

I used to love the Sun
Now I pray to my oracle

Sun Microsystems used to be the standards keeper for the JRE and JDK; now Oracle Ltd. has inherited that role.

I'm said to virtualize all
I'm not picky, working anywhere...

All methods in Java classes are what C++ calls virtual by default, referring to the fact that they are dynamically bound. Java code is deployable across numerous operating systems and platforms, in numerous "flavours" (standard, micro, and enterprise). A great deal of firmware also runs on Java.

I love my classes-all of them
I love my matter-known as objects

Java OOP structure primitives are called classes, and all Java classes are a subclass of Object

I love running parallel-too lazy to wait
Fans worldwide build and break with me!

Java 5 and later come with a significant concurrency (i.e. parallel execution) API. Java enjoys a broad adoption in numerous platforms worldwide.

My syntax-derived from a letter
many programmers found me familiar

Java inherits much of its syntax from the programming language C.

I have nothing included-very greedy
And I must import all!

By default, without imports, a Java file only has access to other classes in its package and classes in java.lang.*, which is an incredibly sparse collection of objects such as boxing primitives, System interface methods, and some common exceptions.

Why does a particle-antiparticle collision produce $2$ photons instead of $1$?

In one of the questions that I did, it asked: What is produced when an electron and positron collides with each other? The answer is $2$ photons. Why doesn't it merge into one? After all, the photon with that amount of energy can decay back into a particle and antiparticle.

Is my understanding correct?

PS. I'm just a beginner learning about the standard model. Would be nice if the answer is simplified to a lower level. Thanks.

Answer

Photons have momentum $p=\frac{E}{c}$ (despite having no mass) Thus to conserve linear momentum, multiple photons are formed, moving in different directions. Also, photons are typically formed for low energy collisions. High energy collisions can result in exotic heavy particles forming. Furthermore, the annihilation (or decay) of an electron-positron pair into a single photon can occur in the presence of a third charged particle to which the excess momentum can be transferred by a virtual photon from the electron or positron.

Trouble connecting stress and force in continuum mechanics with my concept of force from point mechanics

I'm not very familiar with continuum mechanics and have a hard time combining my knowledge of forces from simple mechanics with what I read about continuum mechanics.

Let's suppose we have a metal rod of a certain length and a quadratic cross-sectional area $A$ which is put under stress $\sigma$.

1) What exactly is the physical significance of the force $F = \sigma A$? My current understanding of forces is that they need a point of application - where would this point be? Is it a single point? All points of the cross-sectional area at once? The latter seems to conflict with the notion that force gets smaller if I consider only a part of the area:

If I partition the cross-sectional area $A$ into a number of smaller areas $A_i$, I can calculate forces $F_i = \sigma A_i$. Since $A = \sum_i A_i$, we also have $F = \sum_i F_i$. This makes it seem that $F$ is some kind of cumulative quantitiy and raises the question: what is the physical significance of the $F_i$?

Trying to generalize this further, we can also admit different values of stress $\sigma_i$ for the different areas $A_i$ and even make the areas infinitesimal so that we get a stress distribution $\sigma(x,y)$ (in order to simplify things let's suppose that we chose the distribution such that there's no net torque on the rod). What is the physical significance of the "force" $F = \int_A \sigma(x,y) dA$?

2) What's going on mathematically? Does the notion of force as a vector with a point of application need to be replaced by some kind of vector field in continuum mechanics (considering only stress in a single direction and ignoring additional complications related to the tensor nature of stress)? If yes, how can these "area forces" (I've also read the term "surface force") be combined with forces which have a point of application (for example with the weight of a point mass or a rigid body where the force can be described as acting on its center of mass)? In order to be combined, they need to be described by the same mathematical structure.

3) Does the notion of force somehow lose its meaning along the path which I sketched in 1) above? Is the integral quantity $F = \int_A \sigma(x,y) dA$ a force in one of the senses covered above or is it a quantity with units of force but without physical significance as a force?

Beyond QKD, which types of quantum entanglement experiments will QUESS likely facilitate?

In the recent Xinhua press release China launches first-ever quantum communication satellite the last two sentences (which are two paragraphs) outline several distinct experiments planned for the Quantum Experiments at Space Scale (QUESS) satellite in conjunction with either one or two ground stations.

With the help of the new satellite, scientists will be able to test quantum key distribution between the satellite and ground stations, and conduct secure quantum communications between Beijing and Xinjiang's Urumqi.

QUESS, as planned, will also beam entangled photons to two earth stations, 1,200 kilometers apart, in a move to test quantum entanglement over a greater distance, as well as test quantum teleportation between a ground station in Ali, Tibet, and itself.

If I try to make an outline, I get the following.

1. QKD between the satellite and ground stations


2. secure quantum communications between Beijing and... Urumqi


3. beam entangled photons to two earth stations (1,200km apart)


4. test quantum teleportation between a ground station... and itself

Here is my current understanding of those:

1. QUESS produces a pair of entangled photons, sends one photon to a ground station and simultaneously measures the other internally.


2. Don't understand - QKD by #3, or something different? Uses second public path for the encrypted data? Since Beijing to the Xinjiang Observatory is over 2400km, I'm thinking Store and Repeat?


3. QUESS produces a pair of entangled photons, sends both photons - each to different ground stations for simultaneous measurement.


4. Not sure - is this more of a "science experiment" related to entanglement but not necessarily for secure information transfer?

I'm looking for an answer which helps me understand the most likely interpretation of the experiments mentioned in items 2 and 4 - a clarification of what the experiment is likely to be intended to do, not an explanation of the underlying quantum mechanics of entanglement.

Because of the mixture of publicly conducted science and government interests, the answer may need to contain a bit of speculation or 'most likely experimental scenario' from someone familliar with the field.

This helpful answer and the Nature article cited there has helpful information, and also goes on to mention some confusion on geographic sites in a different (New York Times) article. But I don't find an answer to my question there.

Answer

It's very hard to answer this question, because all details are missing - and the detailed description of the QKD protocol is necessary.

But here is my guess: The main goal of the satellite seems to be quantum key distribution between two ground stations. That's the obvious goal - that's where real world impact happens. Everything else is academic.

But how do you do that? Well, a first step is clearly establishing QKD between one ground station and the satellite. To do this, there are several possibilities, but all of them rely on entangled photons. Clearly, in order to do QKD between two stations, you'll therefore need:

beam entangled photons to two earth stations (1,200km apart)

We start with one station and the satellite. Now we want to do QKD. What this means is that you need to check whether the states are actually nearly maximally entangled. This can be done in several ways - one example would be a Bell test. If that doesn't work, you won't have QKD. In any case the second step after entangling photons is

QKD between the satellite and ground stations

is not precise enough to know what they attempt to do - we need the name of the protocol (I couldn't find it) or a description of what's going on.

Now we have QKD between the satellite and a ground station. If we want to extend this to QKD between two ground stations, there are also several possibilities. You could do QKD between ground station to satellite and then QKD between satellite and ground station again and do decoding and reencoding on the satellite. But that's not really good, because then the satellite actually has the decoded message, which you don't want, so what you really want is to use the satellite only as an intermediary to create entanglement between the two ground stations directly, so that they can generate the secret key directly.

In order to do this, the satellite will need to be a quantum repeater, which works by teleporting the qubits. So in order to implement a quantum repeater to implement QKD between two ground stations, we need to do teleportation. We can start by

test quantum teleportation between a ground station... and itself

Doing this with itself means that we can just work with one lab and not two for starters. But we'll need to extend it later.

So in the end, all of these are probably just toy experiments to implement certain steps of the real thing:

secure quantum communications between Beijing and... Urumqi

Saying "QKD between two ground stations via a satellite" comprises everything else discussed. The other "experiments" are just there to drop a few buzzwords and to show that the mission can still be a success if the main goal fails.

buoyancy - Energy Conservation in Kinetic Power Plants

Quite recently the company Rosch has developed a new kind of power plant that supposedly utilizes the buoyancy effect to generate electricity. The apparatus consists of a vertical conveyor belt with buckets attached to it. The whole construct sits inside a cylinder of water. Through a compressor, air is blown into the upturned buckets which decreases their density and allows them to rise. At the top the air is released and the buckets go back to the bottom of the cylinder. This turns the conveyor belt and a generator that produces electricity. A schematic is shown below.

Now this technology is very controversial and many claim that it is a fraud but the company has had a test facility running for quite a while and generated enough electricity to power both the compressor and a number of household appliances.

I have spent a lot of time trying to figure out problems with the system because it seems that it does not conserve energy. Sure it uses gravity but still there is no loss of energy elsewhere, like the loss of potential energy in a hydroelectric facility for example. I simply do not understand where the energy is coming from. Would this not count as a perpetual motion machine?

(If this is more related to engineering feel free to migrate the question)

quantum mechanics - How can a Photon have a "frequency"?

I picture light ray as a composition of photons with an energy equal to the frequency of the light ray according to $E=hf$. Is this the good way to picture this? Although I can solve elementary problems with the formulas, I've never really been comfortable with the idea of an object having or being related to a "frequency". Do I need to learn quantum field theory to really understand this?

Answer

All you need is quantum mechanics, i.e. that nature in the microcosm is dual,sometimes it can manifest wave properties and sometimes particle properties.

It depends on the measurement/experiment if the wave or the particle nature will manifest itself. Electrons manifest this duality: in the two slit experiment their wave nature appears governed by the de Broglie wavelength. Photons do the same too, displaying the wavelength/frequency associated with the collective classical electromagnetic wave.

The classical electromagnetic wave is built out of photons in a consistent way, and you could study this link if you are interested in this more complicated problem.

Do Sidebands mean the frequency of an AM radio wave is not constant?

I'm studying for A-level now. I have read some other posts explaining how the sidebands generated after the carrier wave is modulated. So, if there is sideband frequency, the frequency isn't constant, right? However, they have written that there is constant frequency for amplitude modulated wave on the book, so which one is right?

Also, there is a question (text below) asking for the frequency of carrier wave according to the graph showing the amplitude modulated radio wave carrying a signal, and the question can be solved by counting the number of modulated wave in certain time. Does it mean that the frequency of AM wave is constant and same as the frequency of carrier wave or we can find out the frequency because the sidebands cancel each other out?

If a radio station carries music, the wave transmitted by the radio station will differ from the wave shown in Figure 33.3. There is only one signal frequency present in the signal in Figure 33.3. Music consists of many, changing frequencies superimposed so that it has a more complex wave pattern. The amplitude of the carrier wave will change as the music pattern changes. The carrier wave frequency does not change but the amplitude of the trace will change with time.

In amplitude modulation (AM), the frequency of the modulated wave is constant. The amplitude of the modulated wave is proportional to, and in phase with, the signal.

logical deduction - What is the solution of the last operation?

Each symbol represents a number between 0 and 9. You need found the number which is the solution to the last operation. The two triangles in the first operation and the triangle and the rectangle of the third operation represent a number of two digits, not a multiplication operation.

So, which is the number that corresponds to the last operation?

Answer

Let G, R, Y, B denote the digits corresponding to the four shapes in the order they appear (green star, red octagon, yellow triangle, blue rectangle).

In the first equation,

two 1-digit numbers sum to YY, so Y=1 (22 is too large).

In the second equation we learn B=R+1.

Now the third equation is

1B-R = 1(R+1)-R = 11,

so the answer is

11, or two yellow triangles.

quantum field theory - How to replace $T$-product with retarded commutator in LSZ formula?

I am reading Itzykson and Zuber's Quantum Field Theory book, and am unable to understand a step that is made on page 246:

Here, they consider the elastic scattering of particle $A$ off particle $B$:

$$A(q_1) + B(p_1) ~\rightarrow~ A(q_2) + B(p_2)$$

and proceed to write down the $S$-matrix element using the LSZ formula, with the $A$ particles reduced:

$$S_{fi}=-\int d^4x\, d^4y e^{i(q_2.y-q_1.x)}(\square_y+m_a^2)(\square_x+m_a^2)\langle p_2|T \varphi^\dagger(y) \varphi(x)|p_1 \rangle \tag{5-169}$$

Then they say that because $q_1$ and $q_2$ are in the forward light cone, the time-ordered product can be replaced by a retarded commutator:

$$T \varphi^\dagger(y) \varphi(x) ~\rightarrow~ \theta(y^0-x^0)[\varphi^\dagger(y),\,\varphi(x)]\,.$$

This justification for this replacement completely eludes me. What is the mathematical reason for this?