Friday 30 September 2016

quantum mechanics - Application of Heisenberg's uncertainty principle on photons


While I was hobby-reading about quantum mechanics I came across Heisenberg's theory. But while I was trying to understand it I thought of this: if the speed of light (as well as the momentum of a photon) is well-defined, then if in an experiment the position of the photon was found, the researchers would know the exact position of the photon, but also its momentum which is constant. Is this an exception from the rule, or am I just ignorant?


PS, 12 grade physics education, please try to explain in simple words




quantum mechanics - How to use the Born rule to find the expected outcome of this simple Stern-Gerlach experiment



The experiment is shown below. How do I calculate the probability of observing a count in detector A, B, or C? Sakurai's text for example starts out describing how to calculate the outcome of simpler chains of S-G devices but I can't find any reference that explains how to handle only slightly-more complicated situations like this.


Experiment





Do quantum particles communicate?


So, 2 days ago I learned that quantum entanglement is just about our perception. There is not an active link between particles and they are not communicating with each other. But then, while surfing on the internet this page came up https://www.sciencealert.com/these-4-cosmic-phenomena-travel-faster-than-the-speed-of-light


It says: ""If I jiggle one electron, the other electron 'senses' this vibration instantly, faster than the speed of light. Einstein thought that this therefore disproved the quantum theory, since nothing can go faster than light," Kaku wrote." and then I read about EPR paradox and really confused. If there is not an active link between quantum particles what Kaku is talking about ? Can someone explain?




Answer



This piece is marked as syndicated from Business Insider, itself repeating Kaku himself on Big Think, so the very first step is to go there - and that quickly shows that they've omitted key qualifiers in Kaku's text:



But actually this experiment (the EPR experiment) has been done many times, and each time Einstein was wrong. Information does go faster than light, but Einstein has the last laugh. This is because the information that breaks the light barrier is random, and hence useless. (For example, let's say a friend always wears one red sock and one green sock. You don't know which leg wears which sock. If you suddenly see that one foot has a red sock, then you know instantly, faster than the speed of light, that the other sock is green. But this information is useless. You cannot send Morse code or usable information via red and green socks.)





Whether quantum particles "communicate" with each other is, if you really push it against the ropes, ultimately up to which intepretation of quantum mechanics you choose to believe. However, what's really clear is that even if they do communicate, they cannot communicate any causal information, so they cannot be used to communicate faster than light and cannot give rise to any time-travel paradoxes (which is the real worry with FTL communications). The details of how this comes about are subtle and there are plenty of questions on the subject on this site, but Kaku does a good job of explaining the limitations given the length of the piece.


However, given the inappropriate cutting they've done with the piece, I'll add this as well: as a rule of thumb, if Science Alert has content that is in direct contradiction with other media, it pays to slow down, read around, and think carefully about just how well they are representing the science. They have plenty of reasonable coverage, but they've graced these pages before in ways that didn't leave them looking very good. In this specific case, though, it just takes a bit of pulling on the thread to find the original source along with its more careful hedging of the conclusions and, moreover, pulling on that thread until you find that original source is a key bit of internet information hygiene that needs to be practiced pretty much always.


classical mechanics - Normal force in a compound pendulum (physical pundulum) system?


Consider a compound pendulum pivoted about a fixed horizontal axis, illustrated by the force diagram on the right:


image was taken from hyperphysics.phy-astr.gsu.edu/hbase/pend.html#c4


#


Okay, I can't figure out where the normal force on the pendlum should point (this force isn't indicated in the diagram).


On the one hand, I think it should point in the same direction as the tension force on the left-side diagram, in order to produce the rotation. But on the other hand, shouldn't the normal force point towards the pendulum rather than away from it? In any case, I know that normal face is perpendicuar to the contact surface, but how does "contact surface" fit into the context of this scenario?


Please help me understand the concepts involved.



Answer



If you think of the pendulum as swinging on a nail going through a smooth hole near the top, then the tension in the pendulum will pull in the direction of the pendulum's length. The normal force will be opposite to this (your first answer) and will be exerted by the nail on the point of the hole in contact with it. So if the nail has a circular cross section, the force will be exerted at the point of this circle at the "top end" of the diameter which is in the direction of the pendulum's length.



special relativity - Characteristic of photons for constant speed


What characteristic of photons causes them to travel with constant speed in all frames of reference? Till the topics I have studied, we always assume this but never got any ideas what can be in the nature of photons (the quanta of electromagnetic field) that causes this. Maybe something is not clear to me. Please explain.




kinematics - How can there be really any instantaneous velocity?


I have read about Zeno's arrow paradox that tells us there is no motion of the arrow at a particular instant of its flight. It can be inferred that there can be no velocity at any instant. Moreover we cannot calculate velocity at any instant in the real world (of course it can be done by using calculus) but how can this be possible? What is the intuition behind this concept?



Answer



At a "frozen" instant of time, the arrow may not be moving - but this is a tautology, since movement is something that requires time. However, even in that frozen instant the arrow does have a velocity (instantaneous velocity, if you will). Imagine that time is a series of huge number of discrete frames (or instead imagine that it is continuous, and that we are taking finer and finer discrete approximations). The position of the arrow jumps to the right from frame to frame. How does the arrow "know" how far to travel from one frame to the next? If the only piece of information "stored" in one frame were its position, then the arrow wouldn't be able to determine this! The necessary information, which is the instantaneous velocity of the arrow, must be as much a part of this frozen frame as all the information related to the arrow's position.



More formally, one says that the configuration space of a physical system, which is the set of all information needed to predict its future (and thus all the information associated with a point in time) includes not only the list of positions of all objects, but also their velocities.


homework and exercises - Calculating Acceleration


I'm having problems calculating acceleration for the following variables. I would have thought it would be extremely straight forward, except I am getting two different answers and do not know which one is correct.


I have the following variables:


$$\begin{align}d &= 229.75\ \mathrm{cm}\\ t &= 1.97\ \mathrm s\end{align}$$



and I need to find acceleration using these variables. I've used both of the following equations, each resulting in a different answer:


$$v=\frac{\Delta d}{\Delta t}\tag{1.1}$$


$$a=\frac{\Delta v}{\Delta t}\tag{1.2}$$


$$a=\frac{2d}{t^2}\tag2$$


Using Equations (1.1) and (1.2), I get the following:


$$\begin{align}v&=\frac{229.75\ \mathrm{cm} - 0\ \mathrm{cm}}{1.97\ \mathrm s - 0\ \mathrm s}\\ &=116.6\ \mathrm{cm/s}\end{align}$$


$$\begin{align}a&=\frac{116.6\ \mathrm{cm/s} - 0\ \mathrm{cm/s}}{1.97\ \mathrm s - 0\ \mathrm s}\\ &=59.2\ \mathrm{cm/s^2}\end{align}$$


Acceleration is $59.2\ \mathrm{cm/s^2}$, according to those equations.


Using Equation (2), I get the following:


$$\begin{align}a&=\frac{2(229.75\ \mathrm{cm})}{(1.97\ \mathrm s)^2}\\ &=118.4\ \mathrm{cm/s^2}\end{align}$$



Which is obviously a different answer than above. It is also double the answer above, which makes complete sense because if I merge Equations (1.1) and (1.2), I get:


$$a=\frac{\frac{\Delta d}{\Delta t}}{\Delta t}$$


or, essentially:


$$a=\frac{\Delta d}{\Delta t^2}$$


and the second equation is the same except it doubles displacement at the top. Therefore, the answer for my second equation is double the answer I got for my first equations.


What I don't understand is which formula I am supposed to be using, and why the formulas result in different answers. I was under the impression that I can use whatever formula I want, as long as I have enough variables to put in and am able to use algebra to solve for the variable I want.


Any ideas as to which I should use?


EDIT: Forgot to mention, acceleration is constant. Initial velocity, initial time and initial displacement are all 0. The information above was measured during a lab in which we timed a cart accelerating down a ramp from rest.



Answer



An accelerating object has a changing velocity. Obviously so since the object starts with zero velocity and the velocity increases with time according to the SUVAT equation:



$$ v = u + at $$


So your equation 1.1 is no use here. It calculates the average velocity. This could actually be used to calculate the acceleration, but the working is a bit involved so I advise not going down that path. Instead you need another SUVAT equation:


$$ s = ut + \tfrac{1}{2}at^2 $$


You know that the cart starts at rest so $u = 0$, and you know $s$ and $t$ so you can calculate $a$.


Thursday 29 September 2016

hilbert space - Getting particles from fields: normalization issue or localization issue?


There seems to be something very strange about the relationship between quantum field theory and quantum mechanics. It is bothering me; perhaps somebody can help.


I'll consider a free Klein-Gordon field. In standard treatments (e.g. Peskin & Schroeder and Schwartz) the one-particle momentum eigenstates $| \vec{k} \rangle$ are normalized so that


$$ \langle \vec{p} | \vec{k} \rangle = 2 \omega_{\vec{p}} (2\pi)^3 \delta^{(3)}(\vec{p}-\vec{k}), \qquad 1 = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 \omega_{\vec{p}}}| \vec{p} \rangle \langle \vec{p} |. $$ Now, assuming $\langle \vec{x}' | \vec{x} \rangle = \delta^{(3)}(x-x')$ as usual, it follows that $$ \langle \vec{x} | \vec{p} \rangle = \sqrt{2 \omega_{\vec{p}}} e^{i \vec{p} \cdot \vec{x}}. $$ Now, one can compute (here in the Schrodinger picture; see Schwartz 2.76 or P&S 2.42) that $$ \langle 0 | \phi(\vec{x}) | \vec{p} \rangle = e^{i \vec{p} \cdot \vec{x}}. $$ This is supposed to mean that $\phi$ creates a particle localized at position $\vec{x}$. P&S are a little cautious about the details, but Schwartz claims that the calculation implies $$ \phi(\vec{x}) |0 \rangle = | \vec{x} \rangle. $$ But this is false because $\langle \vec{x} | \vec{p} \rangle \neq e^{i \vec{p} \cdot \vec{x}}$ with the normalization conventions used. I suppose it could be true with some weird normalization of $| \vec{x} \rangle$, but I can't see what that might be (and at the very least this is not spelled out on the text).



Even if this works out, it seems extremely strange for there to be a relative normalization between the one-particle states of field theory and the states of one-particle relativistic quantum mechanics. One ought to be able to redo the correspondence to make the normalization work out, but I don't see how. (Note that the normalizations can be easily made to agree in the non-relativistic limit $\omega \approx m$, but that's besides the point. Even if fully relativistic quantum mechanics is inconsistent [as some texts claim without reference], at the very least the perturbative corrections for $v \ll 1$ should be recoverable from field theory.)


[Edit: This seems to go beyond normalization. We can get a feel for what kind of state $\phi(\vec{x})|0\rangle$ is by computing its wavefunction as a function of $\vec{x}'$, $$ \langle \vec{x}' | \phi(\vec{x}) |0 \rangle = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2 \omega_{\vec{p}}}} e^{i\vec{p} \cdot (\vec{x}-\vec{x}')}. $$ This wavefunction is peaked (I think divergent) at $\vec{x}'=\vec{x}$, so in some sense the particle is centered at $\vec{x}$, but it seems to be quite a stretch to say that it is at $\vec{x}$ (as the books do). I would go so far as to say the claim is incorrect, since in quantum mechanics saying the particle is at particular position means the wave function is a delta-function there. I guess the same language is used in the Heisenberg picture, when two-point functions are called amplitudes for particles to propagate from one spacetime point to another. This similarly seems false by the conventional meaning of amplitude as the overlap between two localized states. Words of wisdom would be appreciated.]



Answer



Might as well collect my comments, most deleted, in this memo answer.


Essentially, QFT does not want you to go near position eigenstates of the style of QM. The eigenstate of the momentum operator, $|p\rangle$, is not the conventional QM one, nor does it have the same dimension. However, QFT distinctly does not encourage one to seek a fantastical position operator conjugate to the (P&S (2.33)) enumerating P operator it uses, and normalizes peculiarly. Angels should properly fear to tread there.


The corresponding “almost” localized conjugate state to this $|p\rangle$ I’ll call $$ \bbox[yellow]{|\tilde{x~}\rangle \equiv \phi(x)|0\rangle=\int \frac{d^3{\mathbf p}}{(2\pi)^3} \frac{e^{-i{\bf p\cdot x}}}{2\omega_p} |p\rangle} ~. $$


Schwartz unwisely labels this as $|x\rangle$, inviting one to confuse it with the standard QM state localized to x by a δ-function, which nobody uses, needs, or wants, on account of vexing paradoxes of the type you got. P&S wisely use the proportionality constant and leave things vague and evocative—but they failed to prevent your question! It is just the unique one-particle state centered at x, with this normalization property.


The momentum dimension of the QM $|x\rangle$ is 3/2, whereas that of $|\tilde{x~}\rangle$ is 1, the opposite of the QFT $|p\rangle$ we utilize in the lab.


Now, P&S (2.50-2.52) effectively normalize $|\tilde{x~}\rangle$, which I would rather rewrite as $$ \langle 0| \phi(x) \phi(y)|0\rangle=\langle \tilde{x~}|\tilde{y~}\rangle =\frac{m}{4\pi^2 r} K_1(mr), $$ with momentum dimension 2, alright, where $r\equiv|{\mathbf x} -{\mathbf y}|$, and $K_1$ is the ubiquitous modified Bessel (Basset), sharply peaked at the origin on the scale of the Compton wavelength 1/m.


Despite the mild singularity at the origin, $K_1(x)\to 1/x$ as $x\to 0$, it cuts off rapidly for large argument x, $\sqrt{\pi/2x} ~e^{-x}$. So, the states $|\tilde{x~}\rangle$ are not as fully localized at x as a δ-function QM spoils one to expect, but they lose all support outside 1-2 Compton wavelengths of the particle in question and are as good as localized. In the Figure of this equal-time wavepacket autocorrelation function, r on the abscissa is in Compton wavelength units:



K1/x in Compton wavelength units


Recall that scattering experiments effectively live in momentum space, detecting momenta and energies of classical objects -- BB pellets at this level. (The spacial info in the detectors is just a classical geometric means of determining angles of momenta.) The QM interference has all been taken care of already by QFT and Wick's theorem, at this stage of asymptotic states’ detection.


The states $|p\rangle$ are virtually classical: they do not communicate/interfere with each other, living as they do in disjoint superselection sectors of the Fock space, fully decohered. So the wavepacket $|\tilde{x~}\rangle$ is virtually classical, and its quantum nature is only apparent when operated upon with more quantum fields. In scattering experiments, one never gets to probe this slight, sub-fermi-size, nonlocality; but, who knows, in earliest big-bang cosmology, one might well contemplate to.


These wavepackets are the true (one-particle) conjugates of the momentum eigenstates (check!), $\langle \tilde{x~}|p\rangle=e^{i{\mathbf x} \cdot {\mathbf p} }$. But note this is merely a projection of a single p component out of a classical wavepacket—mere classical Fourier analysis!


Related 287759.


Quantum field theory's interpretation of double slit experiment


After reading Art Hobson's article titled, "There are no particles, there are only fields" published in The American Journal of Physics in 2013, I'm wondering what other experts think of his main thesis: The double slit experiment, in all her variations, can be completely explained through relativistic quantum physics (quantum field theory) and that the alleged particle/wave "weirdness" is only weird because people don't realize that our best physical models of the universe model that universe as composed of fields ("particles" are excitations of that field).


I get how it can explain everything in the double-slit experiment except the following- how can fields explain why, when you watch which slit the "particle" goes through, does the interference pattern disappear?




particle physics - Is there strong interaction between electrons?


I am not familiar with quantum mechanics at all. But I remember when I was at high school, we learned that strong interaction keeps protons next to each other while they repel each other because of electrostatic force ($F=\large{\frac{kq_1q_2}{r^2}}$). I saw this answer by David Z. He has written "Electron-electron collisions happen at low energy all the time". I got curious to know how is it possible? Because according to the formula above, when $r\to 0$ , $F\to \infty$. Then I saw John Rennie's comment under the other answer of the same question that was saying "collision means any close interaction causing a significant exchange of momentum" and this makes sense.


But, my questions are:




  1. Is there strong interaction between electrons?




  2. If two electrons approach to each other so much (I don't know how!), do they join together?






Answer



The strong interaction that keeps protons together is a different kind of force (the strong nuclear force) which does not affect electrons. Electrons don't feel the strong force. They only feel the electromagnetic force and the left-handed ones also feel the weak nuclear force, which converts electrons into neutrinos. As a result, even if two electrons collide at high energy they will not cling together. There is no other force that can take over as in the case of protons.


Instead, the energy exchange may be large enough to produce extra particles. In a sense this is what was done at LEP (Large Electron-Positron collider), accept that the collisions occur between electrons and anti-electrons (positrons), which would attract each other and can annihilate each other.


The strong nuclear force is described by the theory of quantum chromodynamics (QCD). However, at low energies, such as where protons attracts each other, it is rather difficult to work with this theory because it is too nonlinear.


wordplay - In which U.S. cities do the following puns apply?


It's this puzzle again, but this time I've decided to focus on cities in the United States of America.




  1. ...is everyone a young male?

  2. ...are there red sticks everywhere?


  3. ...is everybody actually from a completely unrelated Canadian province?

  4. ...does everybody know how long they've been moored?

  5. ...does the voltage never vary in their huge industrial laundromats?

  6. ...are all the small mountains lent out from other places?

  7. ...is every day the Fourth of July?

  8. ...is everyone always getting more of something?

  9. ...is the fourteenth letter always really nervous?

  10. ...do all the people like to make scenes of themselves in their state's Capitol building, or at least sorta sound like they do?


Note that all the cities I'm thinking of have a population greater than 50,000.




Answer



...is everyone a young male?



Boise, ID
Young males = Boys = Bois (Thanks, WBT)



...are there red sticks everywhere?



Baton Rouge, LA
Baton = stick, Rouge = red




...is everybody actually from a completely unrelated Canadian province?



New Brunswick, NJ or Ontario, CA
Also a Canadian province



...does everybody know how long they've been moored?



Anchorage, AK
moored = anchored, how long = age




...does the voltage never vary in their huge industrial laundromats?



Washington, D.C.
voltage never vary = DC, huge = ton, laundromats = washing



...are all the small mountains lent out from other places?



Hillsboro, OR
small mountains = hills, lent = borrow (borough)




...is every day the Fourth of July?



Independence, MO
July 4th = American Independence Day



...is everyone always getting more of something?



Gainesville, FL
more = gains




...is the fourteenth letter always really nervous?



Cheyenne, WY (Thanks, chrylis!)
Cheyenne = shy N



...do all the people like to make scenes of themselves in their state's Capitol building, or at least sorta sound like they do?



Austin, TX
Austin, Texas sounds like ostentatious which means to make a scene of oneself




Wednesday 28 September 2016

wordplay - Holiday cookies word attrition [humans only]


I am inspired by the story of the boy who came upon an assortment of holiday cookies beautifully arranged on a platter. The boy quietly helped himself to a cookie and artistically rearranged the remaining cookies to make it appear as if none were missing. Then, he helped himself to another and rearranged again. And another. And another. Until all the cookies were gone. (And that's when his mother noticed...)


It's very simple:




  • I will give you a starting word.

  • You will remove a letter of your choosing, and then rearrange the remaining letters to form a new word.

  • Then, remove another letter and rearrange the remaining letters to form a new word.

  • Then, remove another letter and rearrange the remaining letters to form a new word.

  • ...

  • And so on until all the letters are gone.



For example:


Starting word is ALTERCATIONS


-------------------------------------------

Solution:


ALTERCATIONS
(remove "T" and rearrange)
LACERATIONS
(remove "C" and rearrange)

SENATORIAL
(remove "A" and rearrange)
RELATIONS
(remove "O" and rearrange)
ENTRAILS
(remove "N" and rearrange)
REALIST
(remove "L" and rearrange)
SATIRE
(remove "T" and rearrange)

RAISE
(remove "I" and rearrange)
EARS
(remove "R" and rearrange)
SEA
(remove "E" and rearrange)
AS
(remove "S" and rearrange)
A
(remove "A" and you're done)



You can do as much or as little rearrangement as you like. E.g., you are permitted to remove the "R" from BRAKE and leave BAKE, or even just snap the "-S" off the end of a word and leave all the remaining letters as they are.


As always, I construct my puzzles in such a way that they can be solved using only well-known words, so if you find yourself going down a path of increasingly obscure words, you may be overthinking it.


Multiple solution paths may exist, so don't worry if your path varies slightly from your neighbor's path, especially at the tail ends.


Below are 12 starting words. See if you can reduce each one to nothing through attrition.


1.  DISSEMINATORS

2. ENCRUSTATIONS


3. DECENTRALIZED

4. IMPERSONATING

5. SHAREHOLDINGS

6. INVESTIGATORS

7. INDOCTRINATED


8. REGISTRATIONS

9. DETERIORATING

10. INDETERMINATE

11. REINSTALLATION

12. RESUSCITATIONS




Nice work, @hexomino and @Randal'Thor !


I'm giving @hexomino the answer for solving three of them, and an upvote to @Randal'Thor for solving two of them.


FYI, this puzzle may have been a little too tough to do manually, so I've opened up the remaining unsolved words to computers. See here.


Alternative solution for #4:



If you're not happy with ORATING, you might prefer this solution:

IMPERSONATING
IMPREGNATION
GERMINATION
EMIGRATION

ORIGINATE
RIGATONI
RIOTING
ORIGIN
GROIN
IRON
ION
IN
I




Alternative solution for #7:



If you're not comfortable with CITRATE, you might prefer the solution:

INDOCTRINATED
INDOCTRINATE
INTERACTION
RECITATION
INTRICATE
INTERACT
CERTAIN
RETAIN

INERT
TINE
TIE
IT
I




Answer



Partial


4.




IMPERSONATING
GERMINATIONS
GERMINATION
EMIGRATION
MIGRATION
RIGATONI
ORATING
RATING
GRAIN
RAIN

RAN
AN
A

My general tactic here was to alternate between using -ATE, -TION and -ING endings to proceed. The only tricky part was RIGATONI. Note: MIGRATION may be equally replaced by ORIGINATE.



7.



INDOCTRINATED
INDOCTRINATE
INTERACTION
RECITATION

INTRICATE
INTERACT
CITRATE
ATTIRE
TREAT
TEAR
TEA
AT
A

This one is tricky. As before I tried to use -TION and -ATE endings together with experimenting with INTER- at the beginning. The temptation was to go with ITERATION after RECITATION (or try to 2-step to CITATION) but I couldn't figure out how to proceed in either case so back-tracked to get INTRICATE instead. Had to check that CITRATE is a word, but we could also replace it with CATTIER.




9.



DETERIORATING
INTERROGATED
INTERROGATE
RETREATING
INTEGRATE
TREATING
TEARING
EATING

TINGE
TINE
TIN
IN
I

Similar tactic to 4, trying to keep -ING and -ATE(D) endings as long as possible.



newtonian mechanics - When the oscillator is a system with an angle can we define the angular frequency to be the radians per unit time covered by the system itself?


I read on stackechange that in springs or any one dimensional oscillator the angular frequency is just describing a rate of angle change in the associated circle on which it's projected. Something like this : enter image description here


My question is: suppose you have a pendulum as an oscillator. Would it be correct to say that the omega / angular frequency. Is a measure of radians per unit time that the pendulum itself going through. Or there is still some other hidden circle for which this is defined?


Edit: is the following definition possible? : we take the whole edge on which the pendulum is passing through and circulate it. Meaning, we make a closed circle out of it . Would then the radians in that circle can be thought of as the angular frequency? If it's correct, would that be correct for the line on which a linear spring is oscillating?




statistical mechanics - A die versus a quantum experiment


Let suppose you roll a die, and it falls into a hidden place, for example under furniture.
Then although the experiment has already been made (the die already has a number to show), that value can not be known, so the experiment was not fully realized.

Then till you see the die's top side, the probability remain p = 1/6.
I see no difference between this and the wave function collapse, at least as an analogy.
Can someone explain a deeper difference?


Thanks



Answer



You're absolutely right: the probabilities predicted by quantum mechanics are conceptually fully analogous to probabilities predicted by classical statistical mechanics, or statistical mechanics with a somewhat undetermined initial state - just like your metaphor with dice indicates. In particular, the predicted probability is a "state of our knowledge" about the system and no object has to "collapse" in any physical way in order to explain the measurements.


There are two main differences between the classical and quantum probabilities which are related to one another:




  1. In classical physics - i.e. in the case of dice assuming that it follows classical mechanics - one may imagine that the dice already has a particular value before we look. This assumption isn't useful to predict anything but we may assume that the "sharp reality" exists prior to and independently of any observations. In quantum mechanics, one is not allowed to assume this "realism". Assuming it inevitably leads to wrong predictions.





  2. The quantum probabilities are calculated as $|c|^2$ where $c$ are complex numbers, the so-called probability amplitudes, which may interfere with other contributions to these amplitudes. So the probabilities of outcomes, whenever some histories may overlap, are not given as the sum over probabilities but the squared absolute value of the sum of the complex probability amplitudes: in quantum mechanics, we first sum the complex numbers, and then we square the result to get the total probability. On the other hand, there is no interference in classical physics; in classical physics, we would surely calculate the probabilities of individual histories, by any tools, and then we would sum the probabilities.




Of course, there is a whole tower of differences related to the fact that the observable (quantities) in quantum mechanics are given by operators that don't commute with each other: this leads to new logical relationships between statements and their probabilities that would be impossible in classical physics.


A closely related question to yours:



Why is quantum entanglement considered to be an active link between particles?




The reason why people often misunderstand the analogy between the odds for dice and the quantum wave function is that they imagine that the wave function is a classical wave that may be measured in a single repetition of the situation. In reality, the quantum wave function is not a classical wave and it cannot be measured in a single case of the situation, not even in principle: we may only measure the values of the quantities that the wave function describes, and the result is inevitably random and dictated by a probability distribution extracted from the wave function.


mathematics - Non-Pythagorean coins (Part 1)


To make payments, the Non-Pythagoreans use coins in three denominations of 999, 1000, and 1001 Oboloi. What is the largest integer amount of Oboloi that can not be represented by using these three types of coins?


Comment 1: valid representations use non-negative numbers of coins


Comment 2: Pythagorean coins



Answer



I think the answer is



498500
because $498500=498\times999+998$ and in modular arithmetic we have $1000\equiv1\pmod{999}$ and $1001\equiv2\pmod{999}$, so the remainder of $998$ cannot be made by fewer than $499$ additions of positive numbers less than or equal to $2$.




Tuesday 27 September 2016

mazes - Reprogramming the ColorBot


As the strange professor leads you to the next room, he says "And think, that was my most functional bot. Take a look at this one. Its right yellow number is jammed to 3 and its red dials don't work at all. If it lands on a red square, it just fails. The kicker being, it still works. There is still a way to get it from the start (upper left, pointing down) to the shutdown panel (in the bottom right) without hitting a wall (black). Can you find it?"


enter image description here


"To remind you, these dials set how he moves. He's got a camera on the bottom. When turned on, he checks the color of the space he's on and moves according to the dials of that color. First he either turns 90 degrees [L]eft, 90 degrees [R]ight, or keeps going [F]orwards, based on the left dial. Then he moves the number of spaces shown on the right dial; so one, two or three spaces. Once he moves, he checks the new space and starts again. The matching dials glitch is in this model too; if two sets of dials match exactly, he won't move at all.



Answer



Set Purple to



F2




Set Yellow to



L3



Set Green to



R3



Set Cyan to




Anything not used



Set Orange to



Anything not used



The final path is



A9, A7, D7, D4, G4, J4, J1

enter image description here




wordplay - Community Metapuzzle - The Broken Time Machine


(This is the community metapuzzle, discussed in this meta thread. It requires answers from nine other puzzles linked below.)


You find a mysterious device sitting on the ground. It's labelled in a language you don't recognize and doesn't seem to be like anything you've ever seen before. Picking it up, you notice no form of input or output except the button that was face down... which you now seem to have released by lifting it off the ground. The device starts to make a whirring noise. As it vibrates, you suddenly find yourself transported to all sorts of locations, time periods, and situations with seemingly nothing in common...



As the whirring slows down, you hear a lot of yelling... and notice several spears pointed directly at you.


Welp, this is bad.


You're gonna have to find a way to calm these people down, and fast. You desperately try to think of things to show them to convince them that you're worth keeping around, at least out of curiosity. Mathematics? No, they don't have algebraic notation, so you'd have to somehow convey a concept without words. Maybe there's something on your phone? But there aren't any internet connections in ancient Rome (or at least, you think that's where you are). While fumbling around in a desperate panic, you notice that the back of the device has come loose, and inside there appears to be some sort of diagram. Maybe this was placed in there intentionally? Could this give you the clue you need?



enter image description here



Answer



Metapuzzle Answers



Gordon K: Silicon
Beastly Gerbil: Bigfoot
Hugh Myers: Ditto
Rand Al'Thor: Simplistic
Alconja: Violin
GentlePurpleRain: Incognito

Julian Rosen: Idiotic
'Q': Digit
Dcyfj: Windowshopping



I noticed:



The only vowels are I and O, and an I starts every such string, treat these as binary.



Plugging these into the using this decoding blatantly stolen from ffao


Results in




Gate A: BeastlyGerbil + Gordon K = 4 + 6 = 10
Gate B: Alconja ^ Hugh Myers = 5 ^ 2 = 25
Gate C: Hugh Myers * GentlePurpleRain = 2 * 10 = 20
Gate D: Julian Rosen - Rand Al'Thor = 13 - 7 = 6
Gate E: Dcyfj / 'Q' = 9 / 3 = 3



And the next step:



Gate F: D - Ground = 6 - 0 = 6

Gate G: B * C = 25 * 20 = 500
Gate H: A ^ E = 10 ^ 3 = 1000
Gate I: D / D = 6 / 6 = 1
Gate J: E + D = 3 + 6 = 9



And now, to get letters



We're in Rome (maybe), and these look like Roman Numerals so...
VIVID MIDI MIX




So I pull out my phone



Open the music app, and play this sick video game music jam that was apparently downloaded onto it by whoever drew this diagram. The centurions are impressed by this wondrous sound, and don't even stab me a little bit.



Monday 26 September 2016

nuclear physics - About mass defect


Here's how my book explains mass defect:



Particles inside the nucleus interact with each other - they feel attraction. The potential energy $U$ of such attraction is negative, because in absence of these forces we consider the potential energy to be zero. So we can write the total energy as: $$E=E_{rest}+U$$ Dividing $E$ by $c^2$ we obtain the mass, and because $U<0$ the mass of the nucleus is less than the sum of individual nucleons.



Now, I have problem with the $U$ term. We know that we can choose the zero level for PE arbitrarily. Thus, $E$ can't be defined well (up to constant). However, real measurements "obey" the standard convention of zero PE at infinity. So how can I solve the contradiction? (Obviously, I'm wrong, but I fail to understand why).


This question leads me to a more general question regarding the $E=mc^2$ relation. It follows that $m$ has no certain value when we're dealing with potential energies. Only the change in mass matters, because only the change in potential energy has physical meaning (and can be defined precisely). But mass is a quantity which we measure everyday very precisely, and there's no ambiguity in its value, despite the fact that the systems we measure include quite often some potential energy.





electromagnetism - Force between two point charges moving parallel to each other


When we observe two point charges moving parallel to each other we can see two forces acting on each of the charges:



  • the Coulomb force

  • the magnetic force ($\mathbf{F}=q\mathbf{v}×\mathbf{B}$) (similar to the force between two parallel wires with the same current)


However, if we change to the charges' frame of reference, there will only be one force - the Coulomb one, and moreover, the amplitude of this force should be the same as the previous one's (the distance between the charges is constant and the force is independent of velocity). The resulting force should be independent of the frame of reference, yet it isn't and I can't find the missing piece. I know that a magnetic field is just an electric field viewed from a different frame of reference, but it doesn't help in this specific case.


I found this question to be similar: Two electron beams exert different forces on each other depending on frame of reference?, but in one charge scenario the charge density has no meaning.





Potential energy in Special Relativity



In Special Relativity, the energy of a free particle is $E^2=p^2c^2+m^2c^4$.


But what would be the energy when there is potential energy?


If it's something like $E=\sqrt{p^2c^2+m^2c^4}+U$, what does it mean if a particle has zero or less energy?


Addendum 2013/09/26


The potential momentum is used only in gauge theories (like EM). But could it be used in SR+Newton's gravity, without introducing the concept of curvature (GR).



Answer



Let's start with Newtonian mechanics. Of the fundamental forces of nature, the only one that can be handled at all by Newtonian mechanics is gravity. Newtonian mechanics can't handle electromagnetism. Electromagnetism is inherently relativistic (i.e., Maxwell's equations only make sense in the context of SR, not Galilean relativity).


Now let's pass from the Newtonian approximation to SR. We lose the ability to model gravity, since that would require GR. We gain the ability to model electromagnetism. In electromagnetism, we don't really have a useful concept of a scalar potential energy $q\Phi$, where $\Phi$ is the electric potential. The reason for this is that although the charge $q$ is a relativistic scalar, the electrical potential $\Phi$ is not a relativistic scalar, it's the timelike component of a four-vector. The conserved energy in Maxwell's equations is not really the energy of a point particle in some external field, it's the energy of the electromagnetic field itself, which depends on energy densities proportional to $E^2$ and $B^2$.


particle physics - Why are neutrino flavour eigenstates expressed in terms of the elements of the complex conjugate of the PMNS matrix?


If we have


$$ \begin{pmatrix} \nu_e\\ \nu_{\mu} \\ \nu_{\tau} \end{pmatrix} =\begin{pmatrix} U_{e1} & U_{e2} & U_{e3} \\ U_{\mu1} & U_{\mu2} & U_{\mu3} \\ U_{\tau1} & U_{\tau2} & U_{\tau3} \end{pmatrix} \begin{pmatrix} \nu_1 \\ \nu_2 \\ \nu_3 \end{pmatrix},\tag{1} $$


then why are flavour eigenstates expressed using the complex conjugated elements of the PMNS matrix:


$$ |\nu_{\alpha}\rangle=\sum_{i=1}^3U_{\alpha i}^*|\nu_i\rangle, $$ (where $\alpha = e,\ \mu,\ \tau$) rather than $$ |\nu_{\alpha}\rangle=\sum_{i=1}^3U_{\alpha i}|\nu_i\rangle, $$ as suggested by $(1)$. Wouldn't it make more sense to define the PMNS matrix as its complex conjugate in the first place?





Sunday 25 September 2016

kinematics - Physics of simple collisions


I'm building a physics simulator for a graphics course, and so far I have it implementing gravitational and Coulomb forces. I want to add collisions next, but I'm not exactly sure how to go about doing it.


A quick summary of how this is working so far is: All objects are spheres of a set radius, mass and charge. The mass's and charges of the spheres are treated as point charge/mass for the calculation. Every step in time (about 1/50th of a second) the forces acting on each object are calculated in a nice big nested for loop that figures out the coloumb and gravitational force between 2 objects, for every set of 2 objects, and then they are summed together. I use this net force to determine the acceleration, and the rest is fairly obvious from there.


What I want to add in is collisions. I can deteremine pretty easily if a collision is happening (if the distance between them <= radius of one + radius of other), what I am not so certain of is how I should add in the collision force (and would need to do it component wise). I want elastic collisions, though for now I'd just be happy with getting conservation of momentum


The information I have easily available are: the velocity of each object (and I mean velocity, speed and direction), the position of each object, the mass/charge (charge obviously not needed here) and the Net force calculated so far for each object for the next step in time. I dont need exact formulas (would prefer if they arent exact), just need a nudge in the right direction.




thermodynamics - Difference between sound and heat at particle level


If heat (or thermal energy) are vibrations of particles and sound is a wave that is propagated through medium e.g vibration of air particles, what indicates if vibration of particles will be perceived as sound or heat (what is the difference between these vibrations) ?




quantum mechanics - Where is locality used in CHSH/Bell's inequality?


A very similar question is asked here, but I'm still confused :(


From Bell, in a hidden variable model, $A = A(\lambda, a)=\pm 1$ is the observed spin of the first particle around axis $a$, and $B = B(\lambda, b)=\pm 1$ is the same for the 2nd particle around $b$. The CHSH proof is then $E(AB)+E(A'B)+E(AB')-E(A'B')=E(AB+A'B+AB'-A'B')\leq 2$, since $|AB+A'B+AB'-A'B'|=|A(B+B')+A'(B-B')|\leq 2$.



But we could do the same trick if $A$ depends on $b$ too, so where is locality used? The link says that $E(AB)+E(A'B)+E(AB')-E(A'B')=E(AB+A'B+AB'-A'B')$ is unjustified, but aren't expectations always linear?



Answer



For better clarity I will here be using the notation $A_0$ and $A_1$, instead of $A$ and $A'$, to denote the outcomes for different measurement setups, and same with $B$.


The expectation values are defined as $$ E(A_xB_y)\equiv\int d\lambda\, q(\lambda)\, E_\lambda(A_xB_y) =\int d\lambda\, q(\lambda)\sum_{ab} ab \,p(ab|xy,\lambda), $$ where $q(\lambda)$ is the probability of the hidden variable having the value $\lambda$, and $p(ab|xy,\lambda)$ is the (joint) probability of getting the outcomes $a$ and $b$ given the measurement settings $x$ and $y$ and hidden variable $\lambda$.


The locality assumption (plus assumption of independence of measurement choices) is embedded in the following factorization for $p$: $$p(ab|xy,\lambda)=p(a|x,\lambda)p(b|y,\lambda).\tag A$$


This relation is needed to have $E_\lambda(A_x B_y)=E_\lambda(A_x)E_\lambda(B_y)$, so that \begin{aligned} E(A_0B_0)+E(A_0B_1) &=\int d\lambda \,q(\lambda)[E_\lambda(A_0)E_\lambda(B_0)+E_\lambda(A_0)E_\lambda(B_1)] \\ &= \int d\lambda \,q(\lambda)E_\lambda(A_0)[E_\lambda(B_0)+E_\lambda(B_1)], \end{aligned} where $$E_\lambda(A_x)\equiv\sum_a a \,p(a|x,\lambda),\quad E_\lambda(B_y)\equiv\sum_b b \,p(b|y,\lambda).$$ Without locality assumption, the above would not hold.


An analogous argument leads to $$E_\lambda(A_1B_0)-E_\lambda(A_1B_1) = E_\lambda(A_1)[E_\lambda(B_0)-E_\lambda(B_1)].$$


The conclusion is now straightforward from here. Define $$S_\lambda\equiv E_\lambda(A_0B_0)+E_\lambda(A_0B_1)+E_\lambda(A_1B_0)-E_\lambda(A_1B_1).$$ Then, using the above equalities, we have $$S_\lambda= E_\lambda(A_0)[E_\lambda(B_0)+E_\lambda(B_1)] + E_\lambda(A_1)[E_\lambda(B_0)-E_\lambda(B_1)].$$ The triangle inequality, together with the fact that, by definition of the numbers we are attaching to the possible outputs, we have $0\le E_\lambda(A_i)\le 1$, now gives $$\lvert S_\lambda\rvert\le \lvert E_\lambda(B_0)+E_\lambda(B_1)\rvert + \lvert E_\lambda(B_0) - E_\lambda(B_1)\rvert = 2 \max(E_\lambda(B_1),E_\lambda(B_2)).$$ Using again that by definition $0\le E_\lambda(B_i)\le 1$, we conclude that $\lvert S_\lambda\rvert \le 2$.


The full $S$ is now defined by averaging $S_\lambda$ over the hidden variable $\lambda$, and because a convex mixture of numbers in $[-2,2]$ remains in $[-2,2]$, we reach the conclusion: $$\lvert S\rvert\le 2.$$




But we could do the same trick if A depends on b too, so where is locality used?



No, you could not.


If the outcome of $A$ directly depends on $B$, then (A) does not hold, thus the probabilities do not factorize ($E(AB)\neq E(A)E(B)$), thus $E(A_0B_0)+E(A_0B_1)\neq E(A_0(B_0+B_1))$, thus the CHSH argument cannot be applied.



Aren't expectations always linear?



Indeed they are. The locality assumption is needed not for the linearity but to factorize the probabilities/expectation values, in order to put them into the form $E_\lambda(A_0)[E_\lambda(B_0)+E_\lambda(B_1)]$, at which point the CHSH argument applies.


electromagnetism - Force on matter in inhomogeneous magnetic field (diamagnetism and paramagnetism)


I found on this site a formula (4.101). It describes wich force acts on matter in an inhomogeneous magnetic field. $ F_z = m_z * \frac{\partial B_z (z0)}{\partial z}$ What does the fraction consist of? What does the $z$ mean?



Answer



What you call "the fraction" is the derivative of the magnetic induction in respect position along the z axis. This is the factor that measures the inhomogeneity of the magnetic field. It is actually the component of the gradient of the magnetic field along the z axis. z is an index for one of the direction or axes in a Cartesian coordinate system. The other two directions are usually labeled x and y. So the formula shows how the component of the magnetic force along the z axis is related to the magnetic moment along the same axis and the gradient of the magnetic field (again, along same axis).


story - The Knights and the Hounds


(With thanks and congratulations to Lee Leon)





We rejoin our heroes several days after the events of The Knights' Gift, at the head of a pack of enormous staghounds, 1000 strong. Such a pack had never before been seen, and the barking and howling carried for many leagues.


It was a mighty pack, full of hounds of every color and pattern – 540 were solid, 2 for every hound that was mottled. 330 were brown but not spotted, and 120 were spotted but not brown.


Among the white dogs, twice as many were solid as were mottled, and half again were mottled as were spotted. And among the 380 grey animals, half as many were spotted as were mottled.




Said one knight, “Did you really have to ask for 1000 enormous staghounds? We shall surely go broke feeding them, let alone finding a suitable place for them to live!


“Fear not!”, the other replied. “For I am in possession of a large quantity of land. We can divide it into 100 plots – 10 on each side – and there will be ample space for 10 staghounds per plot, even enormous as they may be. We shall divide the 100 plots into 9 rectangular pens, so that the animals may be sorted according to color and pattern.”


“Well, enough space is one thing, but how shall we arrange it? After all, not all the hounds get along – we must avoid letting any of the spotted dogs share fences with any of the mottled dogs. Well… except for the white spotted dogs, which seem to get along with the mottled dogs just fine.”


“Yes, that’s an excellent plan! And we must keep their special needs in mind, as well. The solid brown hounds should have at least 2 fence sections along the Southern edge, so that they can feel the sun on their fur, and the solid white hounds need at least 2 fence sections along the Eastern edge, so that they may feel the wind at their backs. And lastly, the grey solid and grey mottled animals are very friendly with each other and should share at least one fence section.”


“Very well, but with so many constraints, are you sure there’s a way to arrange them all?”


The knights stopped for a few moments, considering, then one opened a piece of parchment and began to sketch…





Hint #1



Solid brown has exactly 4 fence sections along the southern border. Solid white has exactly 2 fence sections along the eastern border. Grey solid/grey mottled share exactly 6 fence sections.




Answer



With the help of Hint 1, here's the full answer.


First, the number of each kind of hound.



Hound table | brown | white | grey | total solid | 280 | 120 | 140 | 540 mottled | 50 | 60 | 160 | 270 spotted | 70 | 40 | 80 | 190 total | 400 | 220 | 380 | 1000

We start by finding pattern totals. solid = 540, mottled = solid/2 and spotted is what's left. 120 are spotted not brown so there are 190-120 = 70 that are spotted and brown. There are 70+330 = 400 total brown, and since grey = 380, white = 220. Then (white solid) = 2*(white mottled), (white mottled) = 1.5*(white spotted), so white = 220 = 2*(white mottled) + (white mottled) + 2/3*(white mottled) = 11/3*(white mottled) therefore, white mottled = 60. Following white spotted = 40 and white solid = 120. grey spotted = 190-70-40 = 80, then doubled for grey mottled = 160. The rest can be figured out be subtracting the other two from the total.



Possible pen dimensions



Now that we have the totals we can find all the possible pen dimensions. The longest side must be 10 or lower to fit on the land. Here are the possibilities.

solid brown = 4x7
solid white = 2x6, 3x4
solid grey = 2x7
mottled brown = 1x5
mottled white = 1x6, 2x3
mottled grey = 2x8, 4x4

spotted brown = 1x7
spotted white = 1x4, 2x2
spotted grey = 1x8, 2x4

Taking Hint 1 into account we can further reduce the following:

solid white = 2x6
mottled grey = 2x8



Pen arrangement



I found the arrangement with simple trial and error and a few good guesses.

solid brown = A
solid white = B
solid grey = C

mottled brown = D
mottled white = E
mottled grey = F
spotted brown = G
spotted white = H
spotted grey = I
HHIIIIIIII HHCCBBBBBB FFCCBBBBBB FFCCDAAAAG FFCCDAAAAG FFCCDAAAAG FFCCDAAAAG FFCCDAAAAG FFEEEAAAAG FFEEEAAAAG



Saturday 24 September 2016

quantum field theory - Confining a particle into a region shorter than its Compton wavelength


In Coleman's lecture on quantum field theory he says that when a particle is confined in in a region shorter than its Compton wavelength, very many particles can be produced. My question is whether this will happen even in vacuum and, if so, where does the energy required for their production come from?


People explain these using uncertainty principle in the form of the energy-time uncertainty relation. However, in quantum field theory, the only uncertainty principle is the uncertainty between a field $\phi(x)$ and the conjugate momentum $\pi(x)$.


EDIT: Page 16 of this note by Coleman.



Answer



Yes, this will happen. But you cannot confine particle in the vacuum. To confine a particle, you must have some potential. The energy to produce pairs must come exactly from this binding potential. For example, you can confine electron using a very strong electric field. To confine an electron in a region smaller than its Compton wavelength you need a field with enough energy to create electron position pairs. Particle in a vacuum will never be confined.


Calculating Energy of a Wave



So my physics examinations are coming up and I was going through my notes on waves, but I realized that there were some discrepancies.


In my notes, the energy of a wave is directly proportional to the square of the amplitude, ie. $E \propto A^2$


However, I recalled that, in one of my physics lessons, our physics teacher told us that the energy of a wave can be calculated using $E=hf$, where $h$ is the Planck constant and $f$ the frequency.


Hence I was rather confused and tried searching google for answers but couldn't find any suitable ones. To the best extent of that research, what I found out was (apparently) (for visible light), the frequency of the wave could be used to calculate the energy of the wave, while the amplitude was used to determine the intensity of the wave.


So I was wondering, firstly, whether the above statement was correct, and secondly, in the event it is correct, whether it would be applicable to all kinds of waves, (ie. Sound waves, water waves, other EM waves, etc.), and thirdly, back to the original question, how do we calculate the energy of a wave?


Thanks. :)



Answer



Both the equations you cite are correct.


The energy carried by a wave is indeed proportional to the amplitude squared. for what it's worth, you don't even need a propagating wave, any harmonic oscillator (e.g. a pendulum) will follow that rule. The validity of this rule remains unaffected even in quantum mechanics (actually, since in QM everything can be described by a wave function, it is even more fundamental there).


The second formula expresses the energy of a single photon. A photon is the smallest quantity of radiation that can exist at that frequency. This is completely unrelated to the total energy of the wave! For instance even a small light bulb will emit something like $10^{20}$ photons each second. Each carries an energy of $hf$. Together they sum up to the total power of the beam.



hilbert space - What happens to the global $U(1)$ symmetry in alternative formulations of Quantum Mechanics?


The global $U(1)$ symmetry in Quantum Mechanics corresponds to the freedom to shift the phase of the wave function $$ \Psi \to e^{i\varphi} \Psi \, $$ and can be used to understand the conservation of electric charge.



Where, if at all, does this symmetry show up in alternative formulations of Quantum Mechanics, such as



  • the path integral formulation,

  • the phase space formulation,

  • the pilot wave formulation?




atomic physics - Necessary photon energy for electronic excitation


What happens if a photon incident on an atom has greater energy than the difference of energies between successive energy levels corresponding to electronic excitation(say between ground state and next energy level) but smaller for multi energy-level hop? Will the photon be absorbed or not?



Answer



If the photon energy coincides with the energy difference between two levels, and the initial level of that pair is populated, then that photon will excite the transition. If it doesn't, it won't.


This includes some amount of 'smudging':




  • The photon energy will typically span some nonzero bandwidth $\Delta\omega$, which is limited by the Heisenberg Uncertainty Principle (HUP) to be no smaller than $2\pi/T$, for $T$ the total duration of the experiment.





  • The electronic transitions will also have some nonzero width, which can come from multi-atom aspects ('inhomogeneous broadening' effects, like thermal Doppler shifts, which move the transition energy of each atom in the sample slightly up or down, by different amounts) and from intrinsic effects ('homogeneous broadening', also known as the 'natural linewidth', a HUP effect coming from the fact that transitions between non-ground-state effects cannot be held for unlimited times, since all excited states decay to the ground state after enough time).




If there is an overlap between these two bandwidths, the transition will happen. If there is no meaningful overlap, then no population transfers will take place.


electromagnetism - Legal values of spin-1 field can take: $mathbb{R}$, $mathbb{C}$, $mathbb{H}$, ..?


For the spin-1/ boson field $A_\mu$, we may choose it to be a vector which needs to be




  • real $\mathbb{R}$ usually for photon field. The field strength $F= dA$ is also real. Same for the nonabelian case $F= dA+A^2$ is also real.




  • but can $A$ be complex $\mathbb{C}$?





  • but can $A$ be quaternion $\mathbb{H}$?




What are the legal values of spin-1 field can take? real $\mathbb{R}$, complex $\mathbb{C}$, quaternion $\mathbb{H}$ , ..? And what may be the QFT like in these cases?




In the double slit experiment, using only one photon, will it create a diffraction pattern on an ultra sensitive screen?


Given the double slit experiment, send one photon through the open slits, then stop. With just that one photon traveling through both slits it will appear as a point on the screen. If the screen was ultra sensitive would we see the typical diffraction pattern in addition to the single point?





Friday 23 September 2016

optics - 3D glasses giving the opposite effect to that expected


I have just finished watching the new Star Wars movie (The Force Awakens), and during the end credits, text is shown upon a background of stars. Wearing the 3D glasses, I noticed that the text appears in the foreground, and the stars appear in the background. On removing the glasses, I then noticed that the text is crisp, clear and drawn only once on the screen, whereas the stars are all drawn twice, with a constant distance (from left to right) between the two instances of each star.


However, from my understand of how 3D glasses work, objects whose two instances are further apart should appear closer to the eyes, and objects whose two instances are close together (or at the same location, as with the text), should appear further away from the eyes. So why did I experience the opposite effect when I watched the credits?



Answer



The others have already provided good explanations, but since it sounded like an interesting question and I already sketched up a diagram, I thought I would show it, too. As already mentioned, if you have an object that is to be shown as the exact same distance as the distance between you and the screen, it's very easy to represent that: It's just a single object on the screen that looks the same to both eyes. If, on the other hand, you want to show an object which is far away, then you need to 'trick' your eyes by showing two separate images on the screen, one for the left eye and another for the right eye. That's indicated by the two hollow green dots on the screen on the diagram below. And if you want to show an object which is closer to you than the actual screen distance, then to trick your eyes two images at the locations of the hollow blue dots need to be presented on screen. Note that for objects that are to appear closer than the screen distance that the placement of the images is reversed: The image for the left eye is to the right of the image for the right eye, and the image for the right eye is to the left of the image for the left eye.


It's interesting to note that the apparent distances of the objects are all scaled to the actual distance of the observer to the screen, so audience members will have somewhat different impressions of the distances to the on-screen objects depending on how far from the screen they are seated. Perhaps something to take into account the next time you go to see a 3D movie.



(P.S.: On the diagram the actual observer-to-screen distance is assumed to be 30 feet. Hence, an object to be presented at an apparent distance of 30 feet is represented by a single dot on the screen.)


Diagram


mass - Why are particles different sizes?


Is it correct in saying that a particles size is it's rest energy, and that particles don't actually have size (in the way you get different size objects)?


What defines what sizes a particle can be? Why do particles have discrete sizes, and there's not a continuous spectrum of particles varying in size?


I ask because I was told that particles mass depends on its size, as the bigger a particle is the more it interacts with the Higgs boson and so the more mass it has. So why are there so few particles with specific masses/sizes?




mathematics - Simple Math Problem #1


Here is a simple math problem. Can you solve it?



Examples:



  • 678+57-24*2 = 7

  • 265+34-12*3 = 8

  • 328+58-22*1 = 12


Problem:



541+13-21*0 = ?




HINT 1:


Will see if this hint makes any sense or will edit it in another way, sorry making hints are hard for this(atleast for me):P



678+ 57- 24* 2 = 7



HINT 2:



(678+)(57-)(24*)(2) = 7




Answer




I think I found if, but it is a bit complicated.



The hint(added spaces) seems to imply that the symbols affect the previous number so.
678+ 57- 24* 2 = 7
21(6+7+8) 2(7-5) 8(2*4) 2
21 + 2 - 8 * 2 = 7 (Then I find a chain of formula with the resulting numbers that gives to proper answer for all 3 examples)

265+34-12*3 = 8
13 1 2 3
13 + 1 - 2 * 3 = 8

328+58-22*1 = 12
13 3 4 1
13 + 3 - 4 * 1 = 12

Now if I apply that chain of equation(+-*) to the problem, I get

541+13-21*0 = ?
10 2 2 0
10 + 2 - 2 * 0 = 12

Note that the final chain of equations is the same as the initial one, so basically we need to apply each equations 2 times, once within the number itself and then between each numbers.



soft question - Where did the concept of energy come from?


Energy seems to me to be a very abstract thing, and while it clearly works out very nicely, I don't understand how anyone would have thought to come up with it. Where does the concept of energy find it's roots, and how was it settled down on as a 'useful' quantity as opposed to something else?



Answer




energy (n.)

1590s, "force of expression," from Middle French énergie (16c.), from Late Latin energia, from Greek energeia "activity, operation," from energos "active, working," from en "at" (see en- (2)) + ergon "work, that which is wrought; business; action" (see urge (v.)).


Used by Aristotle with a sense of "force of expression;" broader meaning of "power" is first recorded in English 1660s. Scientific use is from 1807. Energy crisis first attested 1970.



http://www.etymonline.com/index.php?term=energy





Huygens (1650's) was the first to develop the terminology, stating that:



  • energy is not like matter

  • energy does not have size, shape or occupy space


  • energy does not have inertia


Instead, it was defined that energy is a measure of the ability of a physical system to perform work



http://abyss.uoregon.edu/~js/ast122/lectures/lec03.html


electromagnetism - What non-metal is attracted by a magnet?


Are there any non-metal objects that are attracted by magnets?



Answer



Oxygen, for one. In its gaseous state it moves too fast to be affected, but liquid oxygen can be trapped between the poles of a magnet:


enter link description here



Materials can be broadly classified into three sets:


Diamagnetism: All materials are diamagnetic, but their diamagnetic propoerties are easily masked by paramagnetic/ferromagnetic nature. Diamagnetism is the property of an object to be weakly repelled by all magnetic fields. doesn't matter if its near a north or south pole. It will always be repelled. With stronger magnets, the "weakly" becomes less weak, and we get levitating frogs:



Yup, that's a live frog, but more importantly(except to the frog I guess), he's diamagnetic. And he floats in the magnetic field--poor chap must be confounded.


Paramagnetism This is basically the opposite of diamagnetism. Paramagnetism is the property of a material to be attracted towards a magnetic field--again, it doesn't matter north or south. The strength of the attraction varies widely, but its always greater than the diamagnetic repulsion, and generally much less than ferromagnetic attraction. Paramagnetism is only observed in materials with unpaired electrons. Oxygen is paramagnetic (so is diatomic boron), so its attracted by the magnetic field. Note that not all metals are paramagnetic--in fact many are just plain diamagnetic materials (not sure of this)


Ferromagnetism: This is the property of a material to get permanently magnetised. Only a few elements are ferromagnetic (iron, cobalt,nickel, neodymium, and a few others). These are generally strongly attracted to a magnetic field.


Thursday 22 September 2016

electrostatics - What is the electric field in a parallel plate capacitor?


When we find the electric field between the plates of a parallel plate capacitor we assume that the electric field from both plates is $${\bf E}=\frac{\sigma}{2\epsilon_0}\hat{n.}$$ The factor of two in the denominator comes from the fact that there is a surface charge density on both sides of the (very thin) plates. This result can be obtained easily for each plate. Therefore when we put them together the net field between the plates is $${\bf E}=\frac{\sigma}{\epsilon_0}\hat{n}$$ and zero everywhere else. Here, $\sigma$ is the surface charge density on a single side of the plate, or $Q/2A$, since half the charge will be on each side.



But in a real capacitor the plates are conducting, and the surface charge density will change on each plate when the other plate is brought closer to it. That is, in the limit that the two plates get brought closer together, all of the charge of each plate must be on a single side. If we let $d$ denote the distance between the plates, then we must have $$\lim_{d \rightarrow 0}{\bf E}=\frac{2\sigma}{\epsilon_0}\hat{n}$$ which disagrees with the above equation. Where is the mistake in this reasoning?


Or more likely, do our textbook authors commonly assume that we are in this limit, and that this is why the conductor behaves like a perfectly thin charged sheet?



Answer



When discussing an ideal parallel-plate capacitor, $\sigma$ usually denotes the area charge density of the plate as a whole - that is, the total charge on the plate divided by the area of the plate. There is not one $\sigma$ for the inside surface and a separate $\sigma$ for the outside surface. Or rather, there is, but the $\sigma$ used in textbooks takes into account all the charge on both these surfaces, so it is the sum of the two charge densities.


$$\sigma = \frac{Q}{A} = \sigma_\text{inside} + \sigma_\text{outside}$$


With this definition, the equation we get from Gauss's law is


$$E_\text{inside} + E_\text{outside} = \frac{\sigma}{\epsilon_0}$$


where "inside" and "outside" designate the regions on opposite sides of the plate. For an isolated plate, $E_\text{inside} = E_\text{outside}$ and thus the electric field is everywhere $\frac{\sigma}{2\epsilon_0}$.


Now, if another, oppositely charge plate is brought nearby to form a parallel plate capacitor, the electric field in the outside region (A in the images below) will fall to essentially zero, and that means


$$E_\text{inside} = \frac{\sigma}{\epsilon_0}$$



There are two ways to explain this:




  • The simple explanation is that in the outside region, the electric fields from the two plates cancel out. This explanation, which is often presented in introductory textbooks, assumes that the internal structure of the plates can be ignored (i.e. infinitely thin plates) and exploits the principle of superposition.


    electric fields in superposition




  • The more realistic explanation is that essentially all of the charge on each plate migrates to the inside surface. This charge, of area density $\sigma$, is producing an electric field in only one direction, which will accordingly have strength $\frac{\sigma}{\epsilon_0}$. But when using this explanation, you do not also superpose the electric field produced by charge on the inside surface of the other plate. Those other charges are the terminators for the same electric field lines produced by the charges on this plate; they're not producing a separate contribution to the electric field of their own.


    electric field from one plate to the other





Either way, it's not true that $\lim_{d\to 0} E = \frac{2\sigma}{\epsilon_0}$.


general relativity - Relativistic reference frames - What is the independent variable?


Currently reading the following document - Moyer (1971). Excellent read if anyone is interested.


I note that in Eqs. $(12)-(20)$, Moyer uses the coordinate time as the affine parameter in the Euler - Lagrange equations of motion. I am terribly confused about this. Normally, for a timelike geodesic the affine parameter is the proper time. Later in the document the independent variable will be the geocentric coordinate time or the Barycentric coordinate time. Does anyone have a simple explanation why they don't use the proper time?


The notation is the same in the IERS technical note for Eq. $(10.12)$.



Answer



The reason is simple: The goal in that technical report is to establish a relativistically-correct framework via which an N-body problem can be solved numerically, where N is smallish, but is larger than two. In particular, the goal is to describe the solar system.


General relativistic effects in the solar system are small, considerably smaller than are the perturbations that result from the Newtonian gravitational interactions amongst planets. This means it makes much more sense to use coordinate time rather than proper time. (This raises the question: Proper time of what? This is an N-body problem, after all.)


quantum mechanics - Detecting coherence


Is there a way to know if a particle is acting as a wave or a particle? Alternatively, if an entangled particle was already measured?


A - Yes


So any experiment over an entangled particle that let you know if they are coherent or decoherent, so would let we know that the distant particle is same "state" (coherence or decoherence), so independiently of any quantum state measured, having detected that "the wavicle has been measured", let you send information coded as decoherence time intervals.


B - No


So, does still have sense to talk about a coherence/decoherence "state" if there is no way to detect it?


EDIT


The wave has his meta-existence as a probability of a detectable particle interaction, but.. if there is no way to know about the system coherence or decoherence -before- it have "expressed" his behaviour (it's done always as a particle) then we could say that coherent state is never detectable?, I mean if you have measured a photon by a detector, you already know coherence is lost, but with engangled particles A, B, if you decoherence A, then what happend to B? only we could know that there is no "random"(umpredictable) value anymore.. then could we know if a particle is entangled with other? or explicitly could we know if a particle was measured or decoherenced from another entangled partner ?



Fantastic way to say it: "Suposse you are employed to discover if our entire world is being spied from other distant world via entangled particles and the random we see (and its subsecuent evolution) was already known via an engangled model of our world, so.. there is a way to know if someone had got access from a distant engangled particle?, or there is no way to know if any measured value was already determined or is the current experiment where it's first known and fixed (and last years as a lab note in a yellow paper..) or was the experiment just redundant, because the value was already known in an entangled distant experiment (and by a distant world, with beings laughing at our surprise) it doesn't matter if the value is random, anyway is a known value, and they saw it first"



Answer



the answer is No, at least up to the moment after both particles in the entangled pair are measured. In fact, each particle in the world is always behaving both as a wave and as a particle. One of these two behavior may turn out to be more important - or exclusively important - for the description of a particular experiment. But which one it was can only be determined retroactively.


This is demonstrated by the "delayed choice quantum eraser" experiments, see some comments here:


http://motls.blogspot.com/2010/11/delayed-choice-quantum-eraser.html


It's an experiment in which some sloppy interpreters could think that the particle is already "doomed" and any possible interference pattern that the particle could have created has been damaged. However, the experiment may continue in such a way that the interference pattern is "revived" without any problem. The lesson is that one can never try to determine the "results" of anything before things are actually measured. All intermediate properties - and all histories, using Feynman's approach to quantum mechanics - can contribute and always do contribute to the quantity that matters, namely the probability amplitude whose only goal is to predict the probabilities of outcome at the very end of the experiment. When the measurement is completed, the outcome becomes an objective fact. But there are no objective facts about any system before it is measured.


Just like you can't say whether a particle is coming through slit 1 or slit 2 before you measure it - and if you don't interrupt the particle, both possible histories actually contribute to the interference pattern which is what you measure after many repetitions of the double slit experiment - you are also unable to say that a particle behaves "only as a particle" or "only as a wave" before anything is measured. Once again, both possible intermediate histories must be allowed to contribute to what's going on. The wave function has to be carefully evolved in time. And even though some portions of the wave function may later become irrelevant because they describe histories that were not realized, you can never make such a reduction prematurely. And you can't even imagine that in principle, one of the answers was already correct.


So be sure that you can't transmit any information by entangled particles faster than light. It's not possible by the measurements of their usual quantum numbers and it's not possible by your seemingly more sophisticated procedure which is actually not more sophisticated but it is the same thing with different types of measurements.


Finally, you wrote that because I answered No, it doesn't make sense to talk about a coherent state. Well, a coherent state is a particular state that exists in a Hilbert space. (More precisely, a "coherent state" means something else than you say - it's a term denoting states of a shifted harmonic oscillator ground state - but you mean "some state that exhibits coherence.)


And if you make infinitely many repetitions of many experiments to test the character of a state, you may prove that it is a particular state, e.g. your "coherent state", up to an overall phase. So the coherent state surely exists both as a mathematical object as well as an actual situation that leads to very particular predictions that differ from the predictions of any other state. So it's just incorrect to say that it makes no sense to talk about this state - or any other state. It makes a perfect sense.



What isn't right is to claim that a particular particle - or any other quantum object - is determined to have one property or another before it is measured. Indeed, whenever the wave function admits that both properties (up or down spin; or coherent or incoherent behavior) may be true, you must patiently admit that both possilibities are true until the very moment when all the measurements are completed. Then you may perhaps do some retroactive interpretations of what happened. But it's important that there can't be any interpretation before the measurements are over, and not even God knows whether two particles are coherent or incoherent with each other before all the manipulations with the particles are over and the particles are actually measured.


That's how quantum mechanics works. States in the Hilbert space exist and maybe realized, but they can only be tested by statistically comparing their probabilistic predictions with repetitions of the same experiments. If you want to discuss a much more specific experiment and notion of "coherence", you will have to clarify your question a bit.


Best wishes Lubos


measurements - Comparing scales of atomic level objects to scales of everyday size objects


I am trying to come up with everyday size objects comparisions of atomic scales items, e.g. if a proton probability cloud was of size basketball how far would the next atoms to it be?


reason being is to give an idea of the space in between the atoms and giving student some idea of relative atomic object sizes.


Looking for examples of how does dnsity of materials can be reflected in this comparsion scheme, for example comapring a dense material to a relatively far less dense material and much closer the basketball sized atomic cloulds would relatively be closer to each other.


PS : No models of everyday objects as analogy to atomic structure will be used, only relative distances of the structures is the main objective.



Answer




It's easy enough to look up the sizes of a proton and a typical atom and determine their ratio, then you can use that to compare to real-world objects. For example, hydrogen has a nucleus consisting of a single proton. The radius of the proton is about $1\text{ fm}$, and that of the electron cloud in a hydrogen atom is about $25\text{ pm}$ (though there are some subtleties in the definition of "radius," which I won't go into here). Thus the hydrogen atom is about 25000 times the size of the proton. So if you're using a basketball, with a $24\text{ cm}$ diameter, to represent the proton, the electron cloud would be represented by something 25000 times larger, or about $6\text{ km}$ across - probably the size of a small town.


One comparison that I've often heard is a kernel of corn to a football field. A kernel of corn is typically about $5-7\text{ mm}$ across (by my estimation), whereas a football field at 100 yards is about 13000 times larger. That's not actually the right ratio for a hydrogen atom, but it does give a sense of the huge scale difference involved.


Wednesday 21 September 2016

Why does a black hole have a finite mass?


I mean besides the obvious "it has to have finite mass or it would suck up the universe." A singularity is a dimensionless point in space with infinite density, if I'm not mistaken. If something is infinitely dense, must it not also be infinitely massive? How does a black hole grow if everything that falls into it merges into the same singularity, which is already infinitely dense?



Answer




If something is infinitely dense, must it not also be infinitely massive?



Nope. The singularity is a point where volume goes to zero, not where mass goes to infinity.


It is a point with zero volume, but which still holds mass, due to the extreme stretching of space by gravity. The density is $\frac{mass}{volume}$, so we say that in the limit $volume\rightarrow 0$, the density goes to infinity, but that doesn't mean mass goes to infinity.



The reason that the volume is zero rather than the mass is infinite is easy to see in an intuitive sense from the creation of a black hole. You might think of a volume of space with some mass which is compressed due to gravity. Normal matter is no longer compressible at a certain point due to Coulomb repulsion between atoms, but if the gravity is strong enough, you might get past that. You can continue compressing it infinitely (though you'll probably have to overcome some other force barriers along the way) - until it has zero volume. But it still contains mass! The mass can't just disappear through this process. The density is infinite, but the mass is still finite.


gravity - What forces are at work causing sand to migrate to the centre of a spinning bucket of water?




A bucket is filled with water and a handful of sand. The water is then spun. Why and what forces are in play which cause the sand particles to congregate in the centre of the bucket?




quantum field theory - Reading list in topological QFT



I'm interested in learning about topological QFT including Chern Simons theory, Jones polynomial, Donaldson theory and Floer homology - basically the kind of things Witten worked on in the 80s. I'm looking for pedagogical reviews rather than original articles. Though these things sit at the interface of mathematics and physics I'm interested in them more as a physics student. I remember someone asking for a suggested reading list for topological QFT in mathoverflow. The suggested papers were almost uniformly rigorous mathematics written by mathematicians. I am not looking for something like that.




enigmatic puzzle - Five-Minute Comics: Part 2


Part of the Fortnightly Topic Challenge #35: Restricted Title 1




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Didn't actually follow the five minute rule this time -- though i guess i should have, because i nearly ran out of time! There's only 10 minutes left in the day for me. Also these aren't comics, and while we're at it, this is actually the 13th part in my series of puzzles.


Visited some ideas that I never ended up getting to make with smaller minipuzzles.



Answer





Each of these can have a three-letter RNA codon (sequence of A, G, C, and U) to make a word:

AUGMENTATION

GUACAMOLE
COLLEAGUES
MORTGAGORS
EVACUATING
???

When these are converted to DNA (change U to T), their abbreviations spell MYSELF.





The clue answers are FED, MUK, I'M JK, LOKI, I, HUN

The used keys on a QWERTY keyboard for each answer are all near each other. Interpreting them as Braille spells HOT GAS.






enter image description here
The text spells COURSE, taking the large border to be a C.





Each word can have its fifth letter changed to spell another word; the new letters spell IRONIC.






...The answer is EARWAX.





The answers are:


MYSELF
??????
HOTGAS
COURSE

IRONIC
EARWAX


The diagonals spell MATRIX and FIGURE.

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...