Wednesday 31 August 2016

special relativity - How do I derive the expression for velocity in $S_n$ frame that has a velocity $v$ with respect to $S_{n-1}$ frame?



If there are $n$ frames and the $i$th frame has velocity $v$ with respect to $i-1$th frame. How do I derive the relation between velocity in $S_0$ and $S_n$ frame?


I found velocity in nth frame to be $u_n=\gamma^nu_0-v(\sum_{i=1}^n\gamma)$



What happens when n tends to infinity?


Here $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$



Answer



Suppose you have your base frame $S_0$ and a frame $S_1$ moving at a relative speed $v_1$. Then you have a second frame $S_2$ moving at $v_2$ relative to $S_1$. To calculate the speed of $S_2$ relative to your base frame you use the equation for the relativistic addition of velocities:


$$ v_{02} = \frac{v_1 + v_2}{1 + \frac{v_1v_2}{c^2}} \tag{1} $$


You could then use this to calculate $v_{02}$, then use it again to calculate $v_{03}$, and so on though that is rapidly going to get tedious. This is where the concept of rapidity mentioned by Ken G comes in.


Firstly let's write all our velocities as a fraction of the speed of light, $v/c$, in which case equation (1) simplifies to:


$$ v_{02} = \frac{v_1 + v_2}{1 + v_1v_2} \tag{2} $$


Now suppose we take the inverse hyperbolic tangent of this. This seems a strange thing to do, but you'll see why this simplifies things. The atanh function is:


$$ \text{atanh}(x) = \tfrac{1}{2}\ln\left(\frac{1+x}{1-x}\right) $$



If we take the atanh of equation (2) we get:


$$ \text{atanh}(v_{02}) = \tfrac{1}{2}\ln\left(\frac{1+\frac{v_1 + v_2}{1 + v_1v_2}}{1-\frac{v_1 + v_2}{1 + v_1v_2}}\right) $$


This apparently horrendous equation simplifies very easily. We just multiply everything inside the $\ln$ by $1 + v_1v_2$ and gather terms and we get:


$$\begin{align} \text{atanh}(v_{02}) &= \tfrac{1}{2}\ln\left(\frac{(1+v_1)(1 + v_2)}{(1 - v_1)(1 - v_2)}\right) \\ &= \tfrac{1}{2}\ln\left(\frac{1+v_1}{1 - v_1}\right) + \tfrac{1}{2}\ln\left(\frac{1 + v_2}{1 - v_2}\right) \\ &= \text{atanh}(v_1) + \text{atanh}(v_2) \end{align}$$


So the $\text{atanh}$ of $v_{02}$ is calculated just by adding the $\text{atanh}$s on $v_1$ and $v_2$. Calculating the relative velocity for the third frame just means adding $\text{atanh}(v_{03})$:


$$ \text{atanh}(v_{03}) = \text{atanh}(v_1) + \text{atanh}(v_2) + \text{atanh}(v_n) $$


And it should be obvious the general case is:


$$ \text{atanh}(v_{0n}) = \sum_{i = 1}^n \text{atanh}(v_i) $$


The quantity $\text{atanh}(v)$ is called the rapidity, and this is what Ken means when he says the rapidities just add together.


The reason why we get this surprising behaviour is that a frame $S_1$ moving at $v_1$ is related to our rest frame by a hyperbolic rotation of a hyperbolic angle $\theta_1 = \text{atanh}(v_1)$. A second frame $S_2$ moving at $v_2$ relative to $S_1$ is rotated relative to $S_1$ by $\theta_2 = \text{atanh}(v_2)$, and the angles of rotation just add. So relative to use it is rotated by:



$$ \theta_{02} = \theta_1 + \theta_2 = \text{atanh}(v_1) + \text{atanh}(v_2) $$


That's why the rapidities add.


wordplay - Superhero words!


Quietly walking among us are words which are actually superheroes in disguise! Just as Diana Prince spins around to become Wonder Woman, some seemingly ordinary words can spin around to reveal their super identities!


For example, STRESSED turns into DESSERTS just when you need it, and REVILED can turn around to DELIVER.



Meanwhile...
Poor little THINK is sitting all alone, thinking about how he could become the KNIGHT, if only he had a spare "G" ...


Just a mere mortal word without powers of transformation? Well, never fear! There is no shame in using an accessory (or ten) to enable your superpowers! (Respect, Batman, Green Lantern, and Iron Man!)


So now STINGIER can become the fiery REIGNITES(1), EMERGED can become the brilliant doctor DEGREE(2), and SALAMI can become fleet-footed IMPALAS(3)!
(1) with addition of an "E", sold separately
(2) after subtraction of an "M", not responsible for lost or missing pieces
(3) with addition of a "P", not included


But wait, there's more! With a well-stocked utility belt, some words can take on multiple transformations!


For example, DUAL becomes DUEL through the following transformation sequence:


 DUAL

. .
. .
. ---U-->
. .
. .
LAD
. .
. .
. <--E---
. .

. .
DEAL
. .
. .
. ---A-->
. .
. .
LED
. .
. .

. <--U---
. .
. .
DUEL


Now let's see if you can help other superhero words find their own transformation sequence!


At each step, you may add or subtract a letter as you choose. You don't need to alternate additions and subtractions.


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 conjuring up increasingly obscure words, you may be overthinking it.


The first three are just to get you warmed up, while 4-8 are serious:



1.  start:  SCAM
end: MAGES
(suggested 2 intermediate words)


2. start: TROOPER
end: TROOPED
(suggested 3 intermediate words)



3. start: XENON
end: WOK
(suggested 5 intermediate words)


4. start: SLEEPY
end: SLEEPER
(suggested 16 intermediate words)



5. start: SEAPORT
end: SPIRITS
(suggested 17 intermediate words)


6. start: KNIGHT
end: TIME
(suggested 23 intermediate words)



7. start: GURU
end: SERIES
(suggested 27 intermediate words)


8. start: BLOG
end: ABOUT
(suggested 36 intermediate words)



Super helpful hints or diabolical distractions?
guar vs. gaur
taro vs. tarot





Epilogue:


Wow, we really got a refresher on some vocabulary-building words! GARS, DIT, DEMIT, SOU, SOT, SI, STIM, STOB, NOB, TA, TAT...


1-3 were nicely handled by @Mohirl


Note on 4:




The solution provided by @PiIsNot3 is well done. If you feel that LEE is no longer as well known as it once was, take a look at the solution provided by @arbitrahj which detours around it using

SEE --> EVES --> SEVER --> REVELS --> LEVER --> REEL




Alternative solution to 5:



SEAPORT --> TROPES --> SPORT --> TOPS --> POT --> TO --> OAT --> TARO --> RAT --> TARS --> SAT --> TEAS --> SET --> TIES --> SIT --> TIPS --> SPITS --> STRIPS --> SPIRITS




Alternative solution to 6:




KNIGHT --> THINK --> KNIT --> STINK --> NITS --> SIN --> IS --> SIT --> TIES --> SET --> TEAS --> SAT --> TARS --> RAT --> TARO --> OAT --> TO --> LOT --> TOIL --> LIT --> TILE --> RELIT --> TIER --> REMIT --> TIME




Note on 7:



The solution provided by @PiIsNot3 uses GARS. Here is an alternative to the segment from RAG to PAR:

RAG --> GARB --> BAG --> GRAB --> BAR --> CRAB --> ARC --> CRAP --> PAR




Alternative solution to 8:




BLOG --> GOB --> BOGS --> SOB --> BOBS --> BOB --> BOMB --> MOB --> BOOM --> MOO --> DOOM --> MOD --> DORM --> ROD --> DOER --> RED --> DEAR --> RAD --> DART --> TAD --> AT --> A --> AD --> DAB --> BAUD --> DUB --> BUDS --> SUB --> US --> SUN --> NUBS --> BUN --> SNUB --> BUS --> STUB --> BUT --> TUBA --> ABOUT




Answer



Partial


SLEEPY -> SLEEPER (14 words):



(Based off of arbitrahj’s answer)
SLEEPY
PEELS
SEEP

PEE
WEEP
PEW
WE
EWE
EWES
SEE
EELS
LEE
REEL

LEPER
REPELS
SLEEPER



SEAPORT -> SPIRITS (11 words):



SEAPORT
TROPES
SPORT
TOPS

SOT (not sure if this is considered a “common” word or not)
TOES
SET
TIES
SIT
TIPS
SPITS
STRIPS
SPIRITS




KNIGHT -> TIME (11 words):



KNIGHT
THINK
KNIT
TIN
IT
TIP
PI
DIP

ID
DIT
TIED
DEMIT
TIME



GURU -> SERIES (25 words)



GURU
RUG

GUAR
RAG
GARS
SAG
GAPS
SPA
RAPS
PAR
RAPT
TAR

AT
TAD
DART
RAD
DEAR
RED
DEER
REEDS
SEER
FREES

SERF
FIRES
SERIFS
SIRES
SERIES



BLOG -> ABOUT (20 words):



BLOG
GOB (?)

BONG
NOB (?)
BORN
ROB
OR
ROD
DO
SOD
DUOS
SOU (?)

US
SUN
NUBS
BUN
SNUB
BUS
STUB
BUT
TUBA
ABOUT




Wigner's friend and intersubjectivity in quantum mechanics


Suppose there's a radioactive material and a 1/2 quantum probability of detecting it by a Geiger counter. This puts the system in a superposition. Also suppose you are in the same room, and the walls of the room are perfectly decoherence-proof. You observe the Geiger counter and get a definite result. Either it had gone off, or it hadn't. The Copenhagen interpretation tells you the decay or nondecay only became real when it was measured, and the result reached you. Your friend is waiting outside the room. After a very long delay, you open the door, and he observes whether or not you had seen the Geiger counter go off.



He then has the nerve to tell you



There is only one observer, and that's me, not you. You are nothing more than a dull bound state of electrons and protons. Before the door opened, you never had the property of knowing whether the Geiger counter went off. Only when the door opened did you acquire that property.



Unnerving, but you reason, your friend is just like you. Both of you are humans and made up of a bound state of electrons and protons. What applies to you ought to apply to him and vice versa. Or are you special and inherently different from him?


You object that you remember very distinctly whether or not you heard the Geiger counter go off. He counters



Your memory of having known whether or not the Geiger counter went off only came into being when the door opened. Just because you have that memory now does not mean it was real before the door opened.



You object that applying the Copenhagen interpretation to yourself tells you you did have a definite knowledge before the door opened. He counters




It's a meaningless question whether or not you knew the state of the Geiger counter before the door opened and I measured you because there is no way in principle for me to find out. There is no inconsistency here.



Don't all observers have to agree? Both of you agree on your current memory of having had a definite state and that is observable.


Did you have a definite knowledge of the state of the Geiger counter before the door opened? Change the story a little. You haven't opened the door yet, but you know your friend is waiting outside. Do you have a definite knowledge now? You think you do, but does that make it so?


After some further reflection, you realize the present you is a different observer from your past selves. How definite was the past?




relativity - Doesn't Warp theory violate causality?




I have heard many physicists (ex:- Michio Kaku) saying "Warp speed" from Star Trek doesn't violate any known physical laws. But doesn't it violate causality? Say, we make warp drive possible and travel towards Alpha Centauri (4.22 light years away) in warp speed and reach there, say, within a year (+/- few months) and blow it up. Now using Special Relativity, we could devise a frame of reference wherein the observer would see the Star blow up before we ever left planet Earth. Wouldn't that violate causality and make warp speeds unattainable?



Answer



This is more of a meta answer, since it isn't really Physics, but it got a bit long to put in a comment. You say:



I have heard many physicists (ex:- Michio Kaku) saying "Warp speed" from Star Trek doesn't violate any known physical laws.



You need to think about precisely what this statement means. If we take the Alcubierre drive as an example it is a perfectly legitimate solution to Einstein's equation. So if General Relativity is a Law of Physics then the Alcubierre drive indeed doesn't violate any laws of physics, and it can be used to create closed time-like curves with the loss of causality that this implies.


But there are two problems with saying the Alcubierre drive doesn't violate any laws of physics:


Firstly the Alcubierre drive requires a ring of exotic matter to work, and exotic matter violates a number of energy conditions. If these count as laws of physics (and so far observation suggests they do) then the Alcubierre drive does violate laws of physics.


Secondly most of us believe that general relativity is an approximate theory that breaks down under a number of conditions (specifically in the quantum regime). It's been argued that when you take quantum effects into account a closed timelike curve causes an instability that destroys it. All this is speculative since we have no theory of quantum gravity, but if true it also means that the Alcubierre drive does violate laws of physics.



So when you see a bald statement like xxx doesn't violate laws of Physics, the statement is meaningless unless you specify the assumptions it is based on.


riddle - When does 1+1=3?



When does one plus one equal three?



Answer



The following slightly abuses the notion of "equals" but in a way that is common among non-mathematicians (more or less "And the next thing we get is...")



When a woman plus a man gives a woman plus a child plus a man.




thermodynamics - How does pressurized gas constantly push?


If a gas, such as hydrogen, is pressurized into an air tight container, a force in terms of pascals (or whatever unit you want to use) is exerted, correct? That is what pushes against every surface within the container. But what I don't understand is how the gas can constantly push against the walls without being supplied more energy. Does the force of pressure not need energy, or am I missing something? What about when the force is used to move something, such as in a hydraulics system?





no computers - E's-y Word-Square Fill-in


Not necessarily easy, but "'E's-y"!


For each word square below, the Es have been filled in for you. Your task is to complete the rest of the square using only five other distinct letters. That is, you may pick any five letters of the alphabet (other than E), and use each of those letters as many times as you wish. Each square uses a different set of five letters, although there may be overlap between squares. (e.g. if square (1) uses BFKLZ, square (2) might use ABKMR).


The solutions contain only relatively-common English words that should be known to an educated native English speaker. There are no abbreviations, slang terms, archaic words, foreign words, or anything else unusual.


$$\begin{array}{rcrc} \raise{3em}1. &\begin{array}{|c|c|c|c|c|} \hline E&&&E&\\ \hline &E&&&E\\ \hline &&E&E&\\ \hline E&&E&&\\ \hline &E&&&\\ \hline \end{array} \quad\quad\quad\quad\quad \raise{3em}2.\quad &\begin{array}{|c|c|c|c|c|} \hline &&&E&\\ \hline &&&&E\\ \hline &&&&E\\ \hline E&&&E&\\ \hline &E&E&&\\ \hline \end{array}\\ \\ \raise{3em}3. &\begin{array}{|c|c|c|c|c|} \hline &&&E&\\ \hline &&&&E\\ \hline &&&E&\\ \hline E&&E&&\\ \hline &E&&&E\\ \hline \end{array} \quad\quad\quad\quad\quad \raise{3em}4. &\begin{array}{|c|c|c|c|c|} \hline \;\;\;&&&&\\ \hline &E&&&E\\ \hline &&&E&\\ \hline &&E&&\\ \hline &E&&&E\\ \hline \end{array}\\ \end{array} $$


While I can't prevent anyone from writing a computer program to solve these, I would encourage people to try it without. Where's the fun in brute-forcing a solution?

(Feel free to use a computer to help you find word patterns for a single word, just not to solve the whole puzzle.)



Answer



Since the answers all came out piecemeal, I figured I'd post them all together in one answer here, to make it easier for people to see the solutions.



$$\require{color} \definecolor{g}{RGB}{0, 180, 0} \begin{array}{rcrc} \raise{3em}1. &NRSTV\quad\begin{array}{|c|c|c|c|c|} \hline E&\color{g}N&\color{g}T&E&\color{g}R\\ \hline \color{g}N&E&\color{g}R&\color{g}V&E\\ \hline \color{g}T&\color{g}R&E&E&\color{g}S\\ \hline E&\color{g}V&E&\color{g}N&\color{g}T\\ \hline \color{g}R&E&\color{g}S&\color{g}T&\color{g}S\\ \hline \end{array} \quad\quad\quad \raise{3em}2. &ADGSU\quad \begin{array}{|c|c|c|c|c|} \hline \color{g}S&\color{g}A&\color{g}G&E&\color{g}S\\ \hline \color{g}A&\color{g}D&\color{g}A&\color{g}G&E\\ \hline \color{g}G&\color{g}A&\color{g}U&\color{g}G&E\\ \hline E&\color{g}G&\color{g}G&E&\color{g}D\\ \hline \color{g}S&E&E&\color{g}D&\color{g}S\\ \hline \end{array}\\ \\ \raise{3em}3. &AHNSV\quad\begin{array}{|c|c|c|c|c|} \hline \color{g}A&\color{g}S&\color{g}H&E&\color{g}S\\ \hline \color{g}S&\color{g}H&\color{g}A&\color{g}V&E\\ \hline \color{g}H&\color{g}A&\color{g}V&E&\color{g}N\\ \hline E&\color{g}V&E&\color{g}N&\color{g}S\\ \hline \color{g}S&E&\color{g}N&\color{g}S&E\\ \hline \end{array} \quad\quad\quad \raise{3em}4. &ORSTU\quad\begin{array}{|c|c|c|c|c|} \hline \color{g}T&\color{g}R&\color{g}O&\color{g}U&\color{g}T\\ \hline \color{g}R&E&\color{g}U&\color{g}S&E\\ \hline \color{g}O&\color{g}U&\color{g}T&E&\color{g}R\\ \hline \color{g}U&\color{g}S&E&\color{g}R&\color{g}S\\ \hline \color{g}T&E&\color{g}R&\color{g}S&E\\ \hline \end{array}\\ \end{array} $$



homework and exercises - What the heck is negative effective mass?


I am reading this book:Solid State Electronic Devices by Ben G Streetman and Sanjay Kumar Banerjee.
I have some doubts in the article 3.2.2 Effective mass.

In this the aythors say that $E=\dfrac{1}{2}mv^2=\dfrac{1}{2}\dfrac{P^2}{m}=\dfrac{{\hbar} ^2}{2m}k^2$.




  • Electrons usually have thermal velocity of order $10^7$m/s. So Shouldn't we use $E=mc^2$ rather than $E=\frac{1}{2}mv^2$.



The author further says that electron mass is related to curvature of (E,k) relationship as $\dfrac{d^2E}{dk^2}=\dfrac{\hbar^2}{m}\tag{3.2(d)}$. then the author says that the effective mass is given by $m^{*}={{\hbar^2}/{{\dfrac{d^2E}{dk^2}}}}\tag{3.3}$




  • Is equation $3.3$ derived from equation $3.2(b)$. In equation $3.2(d)$ we have $m$ the original mass and in eqn $3.3$ we have $m^{*}$ the effective mass which is completely different from $m$.




In the end of the article the author says that the total force on the electron is given by $F_{tot}=ma$. After this the author says the external force applied on an electron is related to effective mass as: $F_{ext}=m^{*}a$.




  • From where this equation comes? Sometimes the effective mass $m^{*}$ is negative then the equation implies that if we apply certain external force on electrons in the crystal they will accelerate in the opposite direction, how this is possible? What's going on?





Tuesday 30 August 2016

Observing Two-Beam Interference at Home


I want to know how difficult it would be for me to observe two-beam interference at home.


I have:



  • A laser pointer.


  • A non-polarizing beam-splitter.

  • A mirror.

  • Two concave lenses.

  • An uneven shaky floor, some chairs, and tape.

  • Patience that spans an entire day.


This is the sketch of the setup that I have in my mind: enter image description here where the laser, lenses, and so on are taped to the chairs.


How close do the two paths lengths have to be to each other? In my case, the two path lengths will differ by dozens of centimeters.


The laser spot from the laser pointer is not even uniform. Will that be an issue?


Are there any other ways this could go wrong? Is there any advice on how I could observe two-beam interference?




Answer



One problem I can see with your setup is that, unless your beams are VERY long, they intersect at a steep angle. So the fringe spacing is $k\,\sin\theta$, where $\theta$ is the angle which the beams intersect at and $k$ the wavenumber. $k$ is about 12 million radians per meter or 2 million waves per meter, so even a one degree crossing angle (0.017 radians) gives you about 35 000 fringes per meter, or about a 30 micron fringe spacing. So you're going to need a microscope to see your fringes, which really is not a good idea when you are using a laser pointer - you could get quite a harmful eyeful of light. This is why one uses arrangements like the Mach-Zehnder or Michelson interferometers, so that the beams can be arranged to meet with a very small angle.


I would recommend you have a look at these instructions for a home made Michelson interferometer from the outreach people at LIGO:


LIGO Scientific Collaboration, "Build Your Own Michelson Interferometer", LIGO Document # LIGO-T1400762-v1


and there is a slightly more sophisticated version here:


LIGO Scientific Collaboration, "The Magnetic Michelson Interferometer", LIGO Document #LIGO-T0900393


It's slightly more complicated than your setup, but it will be much more controllable and there's a great many different experiments you can do with it. Be sure to post any results that show a fringe shift when you rotate the interferometer!


Another, simple setup which is easy to demonstrate is the Fizeau fringe pattern from the two surfaces of a microscope coverslip. You can get long, several centimeter by 1 centimeter coverslips of the standard thickness of about $170{\rm \mu m}$. Then, you shine the laser onto the surface of the coverslip and observe the reflexion on a screen. You get inteference between the reflexions from the two surfaces of the coverslip. See my sketch below:


Fizeau Arrangement


The great thing about this arrangement is that the amplitudes of the two reflexions are almost identical, which means you get highly visible fringes (the nulls go almost completely dark). Here's an image from a 3cm x 1cm #1 coverslip taken with a 532nm semiconductor laser. My screen was the far wall of the room, and the image you see is about a half a meter across. So the fringes are very well observable in this case. The bending you see actually comes from the distortion of the coverslip's surface: it's showing you that there is a curvature on one surface that is not present on the other.



Fringes


velocity - Seeing light travelling at the speed of light


Imagine there are two cars travelling "straight" at the speed of light*, $A$, and $B$. $B$ is following directly behind $A$.


Suddenly, $B$ switches on its headlights. Will $A$ be able to see this light?





My answer is no, since $A_v = B_v = c$ (the light will always stay stationary relative to $B$. This will probably lead to it gathering up, and intensifying.




*I realize this is impossible, but it's a question my Grade 9 [Honours] teacher asked, so we don't need to get into Relativity, $m = \frac{m_0}{\sqrt{1 – (v / c)^2}}$, cough cough. (I think.)



Answer



I can think of three ways to answer this:




  1. It can't happen.





  2. It really can't happen.




  3. See #1.




Okay, that's probably enough ;-) Since you say we don't need to consider special relativity, suppose that the universe actually obeys Galilean relativity. That's the technical term for the intuitive way to think about motion, where velocities are measured with respect to some absolute rest frame, and there's nothing special about the speed of light or any other speed. If that were the case, then yes, the light beam would never catch up to car A. The energy contained in the light would presumably pile up in the headlight where it was emitted at first, but afterwards perhaps it would spread out sideways, or would be reabsorbed by the headlight as heat. We don't really have a good answer, because that's not the way the universe works - in fact, there's a lot of physics, both experimental and theoretical, that has been done to prove that it can't work that way. No matter how you try to resolve the problem, at some point you will run into a contradiction.


The best thing you could probably do would be to draw a parallel to some sort of wave that travels with respect to some fixed reference frame, at a speed much less than that of light. Sound, for instance. Sound waves travel with a certain speed with respect to the air, which defines a single absolute reference frame, and their speed is much less than that of light, so there are no special relativistic effects to worry about. Your headlight scenario would then be roughly equivalent to an airplane traveling at the speed of sound. What happens in that case is that the airplane creates a sonic boom, a shock wave which results from the energy in the emitted sound waves piling up at the airplane and eventually being forced to spread out sideways. So one might guess that in your hypothetical situation, the headlights of car B would create a light shock wave that would spread out perpendicular to the direction of motion.


This actually can happen in certain physical situations, namely when something is traveling through a transparent material that slows down the speed of light. This means that light itself travels at a slower speed, but not that the "universal speed limit" is any different. The effect is called Cherenkov radiation and it does indeed work out much like a sonic boom would.



A Little Riddle


Here's a little riddle!



Many languages spoken in my prefix
Many stars found in my suffix
Many clauses written in my whole




What am I?



Answer



You are:



An End User License Agreement (EULA)



Many languages spoken in my prefix



EU - the European Union, which has 24 official languages.




Many stars found in my suffix



LA - Los Angeles, home to Hollywood (solved by @greysaff)



Many clauses written in my whole



My word, there really are - to the point that most people don't bother reading the whole thing before ticking the 'I Accept' box!



The word that comes after



Find a five letter word that can be placed behind each of these words:



1) Boiling


2) Check


3) Pin


4) View



Try another one:



1) Bill



2) Card


3) Side


4) Skate



And the last one:



1) Fire


2) Rain


3) Sound


4) Water





Answer



The answer for the first group is



point



Second one:



board




Last one:



proof (not sure ...)



nuclear physics - Positive/negative octupole moment of nuclei?


Does octupole moment of nuclear charge distribution show any positive/negative character, like the quadrupole moment does? Quadrupole moment has prolate and oblate types, but what about characterization of octupole moment?




condensed matter - Quantum dimension in topological entanglement entropy


In 2D the entanglement entropy of a simply connected region goes like \begin{align} S_L \to \alpha L - \gamma + \cdots, \end{align} where $\gamma$ is the topological entanglement entropy.


$\gamma$ is apparently \begin{align} \gamma = \log \mathcal{D}, \end{align} where $\mathcal{D}$ is the total quantum dimension of the medium, given by \begin{align} \mathcal{D} = \sqrt{\sum_a d_a^2}, \end{align} and $d_a$ is the quantum dimension of a particle with charge $a$.


However, I do not quite understand what this quantum dimension is, or what a topological sector (with I guess charge $a$?) is. It is usually just quoted in papers, such as $\mathcal{D} = \sqrt{q}$ for the $1/q$ Laughlin state, $\gamma = \log 2$ for the Toric code... Could someone please explain? How do I know how many topological sectors a state (? system?) has, and how do I get its quantum dimension?


In addition, I am guessing that a topologically trivial state (i.e. not topological state) has $\mathcal{D} = 1$. Would that be right? What makes a state be non-trivial topologically (i.e. have $\mathcal{D} > 1$)?



thanks.



Answer



This is a heavy question, that contains many topics in it that are worthy of their own questions, so I'm not going to give a complete answer. I am relying mainly on this excellent review paper by Nayak, Simon, Stern, Freedman and Das Sarma. The first part can be skipped by anyone already familiar with anyons.



Anyons are emergent quasiparticles in two dimensional systems that have exchange statistics which are neither fermionic nor bosonic. A system that contains anyonic quasiparticles has a ground state that is separated by a gap from the rest of the spectrum. We can move the quasiparticles around adiabatically, and as long as the energy we put in the system is lower than the gap we won't excite it and it will remain in the ground state. This is partly why we say the system is topologically protected by the gap.


The simpler case is when the system contains Abelian anyons, in which case the ground state is non-degenerate (i.e. one dimensional). When two quasiparticles are adiabatically exchanged we know the system cannot leave the ground state, so the only thing that can happen is that the ground state wavefunction is multiplied by a phase $e^{i \theta}$. If these were just fermions or bosons than we would have $\theta=\pi$ or $\theta=0$ respectively, but for anyons $\theta$ can have other values.


The more interesting case is non-Abelian anyons where the ground state is degenerate (so it is in fact a ground space). In this case the exchange of quasiparticles can have a more complicated effect on the ground space than just a phase, most generally such an exchange applies a unitary matrix $U$ on the ground space (the name 'non-Abelian' comes from the fact that these matrices do not in general commute with each other).



So we know that the ground space of a system with non-Abelian anyons is degenerate, but what can we say about its dimension? We expect that the more quasiparticles we have in the system, the larger the dimension will be. Indeed it turns out that for $M$ quasiparticles, the dimension of the ground space for large $M$ is roughly $\sim d_a^{M-2}$ where $d_a$ is a number that depends on $a$ - the type of the quasiparticles in the system. This scaling law is reminiscent of the scaling of the dimension of a tensor product of multiple Hilbert spaces of dimension $d_a$, and for this reason $d_a$ is called the quantum dimension of a quasiparticle of type $a$. You can think of it as the asymptotic degeneracy per particle. For Abelian anyons we have a one-dimensional ground space no matter how many quasiparticles are in the system, so for them $d_a=1$.


Although we used the analogy to a tensor product of Hilbert spaces, note that in that case the dimension of each Hilbert space is an integer, while the quantum dimension is in general not an integer. This is an important property of non-Abelian anyons that differentiates them from just a set of particles with local Hilbert spaces - the ground space of non-Abelian anyons is highly nonlocal.



More details on anyons and the quantum dimension can be found in the review paper cited above. The quantum dimension can be generalized to other systems with topological properties, maintaining the same intuitive meaning of asymptotic degeneracy per particle. It is in general very hard to calculate the quantum dimension, and there is only a handful of papers that do (most of them cited in the paper by Kitaev and Preskill that inspired this question).



I can also try and give a handwaving argument for why the quantum dimension would be related to entanglement. First of all, the fact that the entanglement entropy of a bounded region depends only on the length of the boundary $L$ and not on the area of the region is very clearly explained in this paper by Srednicki, which is also cited by Kitaev and Preskill. Basically it says that the entanglement entropy can be calculated by tracing out the bounded region, or by tracing out everything outside the bounded region, and the two approaches will yield the same result. This means the entanglement has to depend only on features that both regions have in common, and this rules out the area of the regions and leaves only the boundary between them.


Now for a system with no topological order the entanglement would go to zero when the size of bounded region goes to zero. However for a topological system there is intrinsic entanglement in the ground space which yields the constant term $-\gamma$ in the entanglement. The maximal entanglement entropy a system with dimension $D$ has with its environment is $\log D$, so in an analogous manner the topological entanglement is $\gamma=\log D$ where $D$ is the quantum dimension. Again this last argument relies heavily on handwaving so if anyone can improve it please do.


I hope this answers at least the main concerns in the question, and I welcome any criticism.


Monday 29 August 2016

logical deduction - What is the answer to the following questions?



Q1. Which is the first question where c) is the correct answer?


a) Q3
b) Q4
c) Q1
d) Q2


Q2. Which is the first question where a) is the correct answer?


a) Q4
b) Q2
c) Q3
d) Q1



Q3. Which is the first question where d) is the correct answer?


a) Q1
b) Q2
c) Q4
d) Q3


Q4. Which is the first question where b) is the correct answer?


a) Q2
b) Q4
c) Q3
d) Q1




Answer



Q1. Which is the first question where c) is the correct answer


a) Q3 b) Q4 c) Q1 d) Q2



D



Q2. Which is the first question where a) is the correct answer


a) Q4 b) Q2 c) Q3 d) Q1



C




Q3. Which is the first question where d) is the correct answer


a) Q1 b) Q2 c) Q4 d) Q3



A



Q4. Which is the first question where b) is the correct answer


a) Q2 b) Q4 c) Q3 d) Q1



B




newtonian mechanics - Why do heavier objects fall faster in air?


We all know that in an idealised world all objects accelerate at the same rate when dropped regardless of their mass. We also know that in reality (or more accurately, in air) a lead feather falls much faster than a duck's feather with exactly the same dimensions/structure etc. A loose explanation is that the increased mass of the lead feather somehow defeats the air resistance more effectively than the duck's feather.


Is there a more formal mathematical explanation for why one falls faster than the other?



Answer




We also know that in reality a lead feather falls much faster than a duck's feather with exactly the same dimensions/structure etc




No, not in reality, in air. In a vacuum, say, on the surface of the moon (as demonstrated here), they fall at the same rate.



Is there a more formal mathematical explanation for why one falls faster than the other?



If the two objects have the same shape, the drag force on the each object, as a function of speed $v$, is the same.


The total force accelerating the object downwards is the difference between the force of gravity and the drag force:


$$F_{net} = mg - f_d(v)$$


The acceleration of each object is thus


$$a = \frac{F_{net}}{m} = g - \frac{f_d(v)}{m}$$



Note that in the absence of drag, the acceleration is $g$. With drag, however, the acceleration, at a given speed, is reduced by


$$\frac{f_d(v)}{m}$$


For the much more massive lead feather, this term is much smaller than for the duck's feather.


Sunday 28 August 2016

quantum mechanics - Finding the energy levels of an electron in a plane perpendicular to a uniform magnetic field


Suppose we have an electron, mass $m$, charge $-e$, moving in a plane perpendicular to a uniform magnetic field $\vec{B}=(0,0,B)$. Let $\vec{x}=(x_1,x_2,0)$ be its position and $P_i,X_i$ be the position and momentum operators.


The electron has Hamiltonian


$H=\frac{1}{2m}((P_1-\frac{1}{2}eBX_2)^2+(P_2+\frac{1}{2}eBX_1)^2)$


How can I show that this is analogous to the one dimensional harmonic oscillator and then use this fact to describe its energy levels?


I have attempted to expand out the Hamiltonian and found:


$(\frac{P^2_1}{2m}+ \frac{1}{2} m (\frac{eB}{2m}))^2X^2_1+(\frac{P^2_2}{2m}+\frac{1}{2}m(\frac{eB}{2m})^2)X^2_2+\frac{eB}{2m}(X_1P_2-P_1X_2)$


This looks very similar to the 2D harmonic oscillator, if anyone can help/point out where I am wrong I'd much appreciate it!



Answer



Starting from the canonical momenta:



$ \Pi_1= P_1 -\frac{1}{2}eBX_2$ and $ \Pi_2= P_2 +\frac{1}{2}eBX_1$,


We get


$[\Pi_1, \Pi_2] = ieB\hbar$


Thus the operators:


$a = \frac{1}{\sqrt{2eB\hbar}}(\Pi_1+i\Pi_2)$ and


$a^{\dagger} = \frac{1}{\sqrt{2eB\hbar}}(\Pi_1-i\Pi_2)$


satisfy the canonical commutation relation $[a, a^{\dagger}] = 1$


Making the substitution in the Hamiltonian we obtain:


$H =\hbar \omega (a a^{\dagger}+\frac{1}{2})$, with $\omega=\frac{eB}{m}$.


The difference from the Harmonic oscillator lies in the fact that now the energy levels are infinitely degenerate for example, the ground states equation:



$a\Psi(X_1,X_2) = 0$


has an infinite number of solutions and any function of the form:


$\Psi = f(X_1+iX_2) \mathrm{exp}(-eB(X_1^2+X_2^2)/4\hbar)$


is a ground state. These are the lowest Landau levels.


geography - T-Rex goes cooking


This puzzle is part 3 of Gladys' journey across the globe. If you're new to the series, feel free to start at the beginning: "Introducing Gladys".







Dear Puzzling,


Today I visited an important historic site from the Civil War era. This is the first of many museums I expect to be seeing on my trip. The puzzle looks a little like a tyrannosaurus wearing a cook's hat, don't you think? Have fun! I'll write you again soon from my next destination.


Wish you were here!
Love, Gladys.





enter image description here



Across
1. Cry
4. Loathing
7. Infuriate
8. Fred's frequent partner
9. Hoover or Poe
10. Evil glances
11. Doctrine
15. Old cloth or Joplin piece
16. Glide across ice

18. Bourbon ingredient
19. The Crocodile Hunter
20. Naturally illuminated at night
21. Football legend


Down
1. Waller's piano style
2. Cointreau flavour
3. Panhandler
4. Male pronoun
5. L.A. player

6. Grown-up fawns
11. Renounced
12. Elderly women
13. Mousepad or yoga cushion
14. Become older
17. Whale food
18. Cut a photograph
19. Woman turned into a cow






Gladys will return in "Devoted to high culture".



Answer



She is at



Tredegar Iron Works



Filled in crossword grid



enter image description here




And the path (thanks to w l in the comments)



You need to get each snake to eat all the apples in their grid
enter image description here



grid deduction - Double feature: Health shakes


This puzzle is part 10 of the Double feature series (first part here). The series will continue in "Double feature: Computer problems".





enter image description here


Rules of Slitherlink1



  • Draw a single continuous loop in the grid, following cell borders.


  • The loop never branches off in multiple directions or crosses itself.

  • The numbers indicate how many lines belonging to the loop surround that cell. If no number is indicated, the cell may be surrounded by any number of lines.


Across
2. Red camp ground lacking space (7)
5. Surge of water is eventually concealed (4)
7. Following a drain without delay (4)
8. In Spain, the primary official language (3)
9. Irish author's delight: finishing parts of Celtic tale (5)
13. Brew a lime peel (3)

14. A feline male chromosome (4)
15. Water of the first degree (5)


Down
1. A gel spread almost too smoothly (6)
2. Leading lawyer follows California gangster's right-hand man… (4)
3. …as well as Capone's organization's head (4)
4. Hearing to take note of Mr. Chagall (4)
6. Father of 007 in a cast (3)
9. Giants share their home with black bears, at last (4)
10. Some mystery, enigmatic note from Tokyo (3)

11. Make an impression in ketchup (4)
12. Apple computer's physical address (3)
13. Dismiss a gender-neutral pronoun (3)


1 Paraphrased from the original rules on Nikoli.



Solve both puzzles to answer the question: What are health shakes?



Answer



Health shakes are



MEAL REPLACEMENTS




Solved cryptic crossword (with explanations) + Slitherlink:



health_shakes
Across
2. Red camp ground lacking space (7) = (RED CAMP)*
5. Surge of water is eventually concealed (4) = _RISE_
7. Following a drain without delay (4) = A + SAP
8. In Spain, the primary official language (3) = LA + O_
9. Irish author's delight: finishing parts of Celtic tale (5) = JOY + _C + _E

13. Brew a lime peel (3) = A + L_E
14. A feline male chromosome (4) = MAN + X
15. Water - of the first degree (5) = TONIC (ddef)

Down
1. A gel spread almost too smoothly (6) = (A GEL)* + TO_
2. Leading lawyer follows California gangster's right-hand man… (4) = CA + _R + L_
3. …as well as Capone's organization's head (4) = ALS + O_
4. Hearing to take note of Mr. Chagall (4) = MARC (homophone of MARK)
6. Father of 007 in a cast (3) = (IN A)*
9. Giants share their home with black bears, at last (4) = JET + _S
10. Some mystery, enigmatic note from Tokyo (3) = _YEN_

11. Make an impression in ketchup (4) = _ETCH_
12. Apple computer's - physical address (3) = MAC (ddef)
13. Dismiss a gender-neutral pronoun (3) = A + XE



To extract the phrase,



overlap the Slitherlink path on the crossword grid, then take all the letters not inside the loop, excluding the ones outside the 9x9 grid, and read from left to right, top to bottom:
health_shakes_sol



cipher - Room 2 of the Maze


Still excited about your quick triumph over Room 1, you head straight for the suspended note and read it:



Congratulations on making it through the warm-up to Room 2. From here on out, getting the answer from Key 2 will now require the use of the answer from the previous room in addition to Key 1. You will be unable to get the correct answer without the use of these two values (though their roles in the solution process will differ between rooms). In case you forget, we will always include the answer from the previous room in our letter.
This room will be more challenging than the first, Key 2 has been encrypted, and Key 1 seems to be of no help. You'll need to use the answer from the previous room up front. Good Luck!!


Key 1: Key 2 is much more than the sum of its parts.
Key 2: UAST OIGT UQH VBPU PMBNQPXFAGBHT. UMIG PVTHPX UIQM PI PTH. PUTH FIID PI DTX IHT



For those who want to start from the beginning, search for Room 1 of the Maze. The answer for the previous room, for those who forgot or who want to join in starting in this room, is below.



You notice that the keypad in this room has only the digits 0-9 as well as the usual Enter button. What do you enter to move on to the next room in the Maze?


Answer from Previous Room:



ANKARATURKEY



EDIT: Halfway through your calculations, a small portal opens in the center of the room and a note materializes from it. You hurriedly grasp it and read its contents:



We hear that your kind is quite fond of the number 10; something about the number of digits on your extremities. We've heard that some of your kind have learned to use other numbers in it's place and translate between them; hopefully you are one of these. If it is any consolation, we have 349 fblthps on each of our kjaposenjfq, so you can imagine what math is like on our planets!




This room has now been solved. Here is Room 3



Answer



Solving Key 2 as a keyed Caesar cipher with the key being the answer from room 1, you get:



FAVE SOME FUN WITF TRIBUTYLAMINE. FROM TWENTY FOUR TO TEN. TFEN LOOK TO KEY ONE



The H's in the string are incorrectly coded to F's - they should be Y's in the coded version but they are presented as U's.


H being coded to F in the string could mean we should also code it to F in the chemical (C12F27N instead of C12H27N)


EDITS:




Based on the new info, we need to convert C12F27N from base 24 to base 10, which gives 2302070591.



Then, the sum of its parts gives the answer as:



29



riddle - See me once, see me twice #3


You guys seem to like these riddles, so here you have another one:



See me once, the winged creatures visiting.
See me twice, small children's eyes glistening.

See me four times, are you even listening?



Here are the previous riddles in this series (the solutions there have nothing to do with this one).
#1
#2



Answer



You are



LA




See me once, the winged creatures visiting.



LA stands for the city Los Angeles, "the angels".



See me twice, small children's eyes glistening.



LALA land is used to mean a fantasy world, often related to kids.



See my four times, are you even listening?




LALALALA while plugging your ears.



quantum mechanics - Non-locality and Bell's theory


Non-Locality – (just ) one more question?


I have read comments that Bell’s theory proves quantum mechanics is non-local, and also comments that it does not. I have read a comment by a very eminent person who stated what it means is that a measurement of a particle at one location "influences the state of the other particle". So my question is this: Two particles in the singlet state (as always) go their separate ways. At time one each passes through a SG apparatus with orientation in direction $z$. At time two particle 1 passes through a second SG apparatus but with orientation direction $x$; nothing happens to particle two. At time three the spin of both particles is measured in direction $z$. What will be the outcome of the spin measurements?




solid state physics - Are they the same thing: Wigner distribution in quantum Boltzmann equation and Wigner function in quantum optics?


We know that quantum Boltzmann equation (QBE) is an equation of motion for the interacting Green's function $G^<(\vec{x}_1,t_1;\vec{x}_2,t_2)\equiv\mathrm{i}\langle \psi^\dagger(x_2)\psi(x_1)\rangle$ when transformed to center of mass and relative coordinates, also coined as Wigner distribution. Besides, in quantum optics and quantum information, we have another Wigner function for phase space description, defined as $W(x,p) = \frac{1}{2\pi\hbar} \int_{-\infty}^{\infty}{\mathrm{d} \xi \mathrm{e}^{\mathrm{i} p \xi/\hbar} \langle x-\frac{\xi}{2} \vert\psi\rangle \langle \psi\lvert x+\frac{\xi}{2}\rangle} $. (Certainly, we have more complete definition for more modes and density matrices.)


Unluckily, I happen to be that kind of person who cares about nomenclature. Both of these two Wigner distribution functions are quasiprobability distributions, i.e., not necessarily nonnegative. And I even remember that QBE shares some similar form with certain equation for Wigner function in quantum optics. However, these are no more than faint traces. I guess they are different things, but there must be some plausible relation. Can anyone shed light on this?


Update For anyone who are interested in this question, besides the answer underneath, it may also be helpful to refer to Chap. 4 & 16 of Schieve and Horwitz's book Quantum Statistical Mechanics.




Answer



I would say they are not entirely the same, but it depends on the context. First the definitions:




  • the Wigner transform of an operator $\hat{A}$ is defined as $$\tilde{W}\left[\hat{A}\right]=\int dz\left[e^{\mathbf{i}pz/\hbar}\left\langle x-z/2\right|\hat{A}\left|x+z/2\right\rangle \right]$$ and this is a strange function. You see that on the left, the operator is projected onto a real-space representation, then Fourier transformed. You may find more details (especially the link with the Weyl transform) on the wonderful review by Hillery, M., O’Connel, R. F., Scully, M. O. & Wigner, E. P. Distribution functions in physics: Fundamentals, Phys. Rep. 106, 121–167 (1984) which is unfortunately beyond a paywall.




  • the Wigner transform of the density operator $\hat{\rho}=\left|\Psi\right\rangle \left\langle \Psi\right|$ is then naturally defined as the Wigner transform $$W\left(p,x\right)=\int dz\left[e^{\mathbf{i}pz/\hbar}\left\langle x-z/2\right|\hat{\rho}\left|x+z/2\right\rangle \right]$$ and it is coined Wigner function in that context.





  • the Green function is not an operator, it is a correlation function, defined as $G\left(x_{1},x_{2}\right)=\left\langle \hat{T}\left[\hat{a}\left(x_{1}\right)\hat{a}^{\dagger}\left(x_{2}\right)\right]\right\rangle $ where $\hat{T}$ is the time-ordering operator, $\hat{a}$ is the (fermionic or bosonic) destruction operator, and $\left\langle \cdots\right\rangle $ represents the averaging process: it could be $\left\langle \cdots\right\rangle =\left\langle N\right|\cdots\left|N\right\rangle $ if you're working with number states $\left|N\right\rangle$, or $\left\langle \cdots\right\rangle =\text{Tr}\left\{ e^{-\beta H}\cdots\right\} /\text{Tr}\left\{ e^{-\beta H}\right\} $ if you're working with thermal averaging ($\beta=\left(k_{B}T\right)^{-1}$ is an inverse temperature in that case), ... Note there are other conventions for the Green functions, but it does not matter here. The Fourier transform of the Green function reads $$G\left(p,x\right)=\int dz\left[e^{\mathbf{i}pz/\hbar}G\left(x-z/2,x+z/2\right)\right]$$ and it looks like a Wigner transform of the Green function, but it should be more appropriate to call it a Fourier transform of the Green function when you choose $x_{1,2}=x\mp z/2$ for the components. In condensed matter theory, $G\left(p,x\right)$ is often called a mixed-Fourier Green function (the full Fourier transform would have given $G\left(p_{1},p_{2}\right)$ instead) or a quasi-classical Green function for the reason to come.




In the limit $\hbar/\tilde{p}\tilde{x}\ll1$ (called quasi-classical limit), with $\tilde{p}\tilde{x}$ the phase-space exploration of the system, the equation of motion of the quasi-classical Green function is the Boltzmann's (transport) equation. The quasi-classical Green functions are not normalised, so they can not be interpreted (whatever it means) as quasi-probability distribution.


As far as I remember, the quasi-classical equation of motion for the Wigner function is not the Boltzmann's one, but the Liouville's one: the collision term is absent, since there is no self-energy method associated with the density matrix. One needs to work with the Lindblad equation for the density matrix, whereas the self-energy method is sufficient when you work with the quasi-classical Green function. Other method to deal with open systems when working with the density matrix is the so-called stochastic method, see e.g. Walls, D. F. & Milburn, G. J. Quantum optics (Springer-Verlag, 1994).


To conclude, note I've put the time under the carpet in the above explanation. That's for a good reason: time is always more complicated to deal with in the Wigner-Weyl transform, especially in the quasi-classical limit and with the Green functions method. The use of the Wigner function is not a big problem when time is taken into account. Of course dealing with the Lindblad equation is not a simple issue... but that's an other story :-)


cosmology - How would we estimate, ahead of time, "the chances" of LIGO spotting black holes colliding in the period that it has been operating?



Can anyone summarize calculations that have been done about the theoretical probability of a detectable black hole collision happening in the observable universe within the time that LIGO has been operating?


I mean, given what we know about processes and parameters in the universe - about densities of black holes, rate of explosion of stars etc - how often would we expect an observable event?



This question is looking for an analysis independent of observations of actual GW events. It's the scientific "what does theory predict" question.


If we just think for a moment about black hole mergers (as opposed to other GW events), the types of thing that obviously need to be taken into account is "what is the density of black holes throughout the lifetime of the universe?", to determine what is the chance that they merge. This is a complicated question in itself, because the relevant density comes from earlier and earlier in the universe's life the further away the event is (obviously?).


But interestingly, there's also the problem if the planar nature of GW wave fronts. What is the chance that Earth happens to be in the plane of the GW event at the time - it seems to me that this aspect alone must dramatically reduce the number of probable observations...


It appears to be an obvious question, but googling the exact question, and creative variants I could think of, does not uncover anyone else asking it or addressing it. Similarly, ligo.org does not appear to have any material talking about this topic.


The consideration is alluded to in this question and answer, but not given any direct elaboration that I could find.


Interestingly, the answers in this similar question are massively different. One says "possibly daily" and the other says "should see one by 2020". Neither offer any justification.




Saturday 27 August 2016

quantum mechanics - Is there such a thing as "Action at a distance"?


What ever happened to "action at a distance" in entangled quantum states, i.e. the Einstein-Rosen-Podolsky (EPR) paradox? I thought they argued that in principle one could communicate faster than speed of light with entangled, separated states when one wave function gets collapsed. I imagine this paradox has been resolved, but I don't know the reference.



Answer



It's not possible to communicate faster than light using entangled states. All you get out of entanglement is a correlation between the values of two measurements.; the entanglement doesn't allow you to influence the value measured at another location in a non-causal way. In other words, the correlation only becomes evident after combining the results from the measurements afterwards, for which you need classical information transfer.


For example, consider the thought experiment described on the Wikipedia page for the EPR paradox: a neutral pion decays into an electron and a positron, emitting them in opposite directions and with opposite spins. However, the actual value of the spin is undetermined, so with respect to a spin measurement along a chosen axis, the electron and positron are in the state


$$\frac{1}{\sqrt{2}}\left(|e^+ \uparrow\rangle|e^- \downarrow\rangle + |e^+ \downarrow\rangle|e^- \uparrow\rangle\right)$$


Suppose you measure the spin of the positron along this chosen axis. If you measure $\uparrow$, then the state will collapse to $|e^+ \uparrow\rangle|e^- \downarrow\rangle$, which determines that the spin of the electron must be $\downarrow$; and vice versa. So if you and the other person (who is measuring the electron spin) get together and compare measurements afterwards, you'll always find that you've made opposite measurements for the spins. But there is no way to control which value you measure for the spin of the positron, which is what you'd need to do to send information. As long as the other person doesn't know what the result of your measurement is, he can't attach any informational value to either result for his measurement.


general relativity - How much faster would a Clock without gravity run?


Pardon the misleading title.


It is to my understanding that moving/heavy clocks run slow. The Earth itself is under gravitational influence from many sources, and is moving. Is there a way to know how much 'faster' a clock would run if those influences were removed? I don't need a precise number, an order of magnitude is fine, just to get the idea.


Any insight is appreciated.



Answer



Neither effect can really be calculated in a meaningful way.


Kinematic time dilation describes the time dilation of one frame of reference relative to another. There is no preferred frame of reference, so there is no way to say what it would mean to remove the effect of kinematic time dilation.



Gravitational time dilation is a concept that makes sense in a static spacetime, and in that case the time dilation between two different points is given in terms of the difference $\Delta\Phi$ in gravitational potential as $e^{\Delta\Phi}$. Here again you have the problem of what to compare to. Do you want to compare to interplanetary space? Interstellar space? Space outside our local cluster of galaxies? As you continue this process, you reach cosmological distances, at which point you run into the problem that cosmological spacetimes aren't static, and the whole thing becomes meaningless.


This is what relativity is all about. There's no best measure of time. It's all relative.


quantum field theory - Must vacuum states in QFT have non zero energy?


Vacuum states of different quantum fields corresponds to different energy levels. Some sources say they are mostly zero, some other says that it is a limit imposed by heisenberg uncertainty principle, while some other also say that it is an instrinsic energy level for some fields which is not related to heisenberg uncertainty principle at all. In a sense of absolute energy level which of these claims is correct and where did they come from?



Answer





Vacuum states of different quantum fields corresponds to different energy levels.



It is not entirely clear what this statement is supposed to mean in the actual formalism, but it is almost certainly incorrect. The vacuum energy - or zero-point energy - of a quantum field theory is not observable in non-gravitational theories (in which it corresponds instead to the cosmological constant), and is in fact a renormalization parameter in such theories that can be set to an arbitrary value.


It has no physical content, and, loosely spoken, all other energy levels are ordinarily measured relative to the energy of the vacuum. You also cannot meaningfully compare the energy levels of different quantum field theories because their states live in different Hilbert space with different Hamiltonian, i.e. the very definition of what energy means differs between theories.


(In fact, the above is not strictly true. When your theory has more than one perturbative vacuum, such as in the case of instantons, it may become meaningful to talk about their energy or at least their energy differences. But such phenomena are specific to certain types of quantum field theories and there is in my opinion little use in trying to make any sort of generic statement about these cases.)


Anyways, the uncertainty principle is entirely unrelated to the vacuum energy. And actually, it is often misapplied, especially in popular presentation, when it comes to energies. See this excellent answer by joshphysics for the proper meaning of an uncertainty principle that involves energy.


Friday 26 August 2016

wordplay - Another seven letter word


This is the same principle as here.



The 1st, 2nd and 3rd letters spell a body part.


The 2nd, 3rd and 4th spell a mean of expression.


You use the 4th, 5th and 6th everyday.


And the 5th, 6th and 7th is an animal.



What is the word we are looking for?





The next one is here.



Answer



I think the answer is:



Earthen



Explanation:


The 1st, 2nd and 3rd letters spell a body part.




Ear.



The 2nd, 3rd and 4th spell a mean of expression.



Art is a means of expression.



You use the 4th, 5th and 6th everyday.



The word 'the' is used commonly in speech.




And the 5th, 6th and 7th is an animal.



Hen.



semiconductor physics - Change in the width of depletion layer with doping



I have previously learnt that increasing the doping will decrease the width of the depletion layer and vice-versa. However, I am unable to understand this. Does it have some relation with force of repulsion?



Answer




The depletion region forms due to the equilibrium between drift (field driven) and diffusion ( concentration gradient driven) currents. If you have very low doping, the depletion region will be large because a large volume of depleted semiconductor is needed to generate enough electric field to balance the diffusion current. On the other hand, if you have very high doping a much smaller region is required to balance the diffusion of carriers.


I was actually doing this calculation last week, here are some results from my simulation of the Poisson-Boltzmann equation for a Ge/GaAs pn-junction for different doping levels. I have indicated the approximate width of the depletion layer with the blue background. The blue and greens lines are the conduction and valance bands, the red and light blue lines are the intrinsic Fermi-level and the Fermi-level, respectively. The top plot is for light doping $10^{16}\text{cm}^{-3}$, the bottom is for higher doping $10^{17}\text{cm}^{-3}$.


Ge GaAs pn junction at different doping levels


This width of the depletion with applied bias was discussed here In an avalanche breakdown, where are the electrons that break free from?, so that might also be interesting.


optics - Why can't we reach the ends of rainbow?


Rainbows are spectacular things. But I imagine why we can't reach to the ends of rainbows. Do rainbows have no ends ? If so, why can't we reach at them? If you go closer,they will go further. Why?



Answer




Why can't we reach the ends of rainbow?



The WIkipedia article you linked to contains the explanation




The rainbow is not located at a specific distance, but comes from an optical illusion caused by any water droplets viewed from a certain angle relative to a light source. Thus, a rainbow is not an object and cannot be physically approached. Indeed, it is impossible for an observer to see a rainbow from water droplets at any angle other than the customary one of 42 degrees from the direction opposite the light source.






If you go closer,they will go further. Why?



The primary rainbow is refracted at an angle of 42 degrees. If you are at an angle of say 52 degrees from some droplets you won't see any light refracted by those particular droplets.


enter image description here


As you walk 1 meter towards the droplets that are refracting the rainbow you initially see, you move outside the the "beam" from those specific droplets and move into the "beam" refracted by droplets 1 m further back.


Thursday 25 August 2016

irreversible - Particle sources and particle detectors in quantum field theory



I am looking for a resource that clearly exposes the concepts of a particle source and a particle detector in the context of Quantum Field theory. I want to understand Irreversibility in this context.



Answer



Typically one thinks of the sources as being at infinite past, and the detection at infinite future; then a reversible S-matrix description applies. For photons, a corresponding treatment of sources and detectors can be found in Mandel & Wolf's treatise on quantum optics. But their treatment doesn't give any hint on irreversibility.


[Edit] Detection is always irreversible; nothing counts as detected unless there is an irreversible record of it. There is no really good account from first principles how an irreversible detection event is achieved. From the 1999 article ''Some problems in statistical mechanics that I would like to see solved'' by Elliot Lieb http://www.sciencedirect.com/science/article/pii/S0378437198005172:


    The measurement process in quantum mechanics is not totally understood, even after three quarters of a century of thought by the deepest thinkers. At some level, the problems of quantum mechanical measurement are related, distantly perhaps, to the problems of non-equilibrium statistical mechanics. Several models (e.g. the laser) indicate this, but the connection, if any, is unclear and I would like to see more light on the subject.

But see
http://arxiv.org/pdf/quant-ph/0702135
http://arxiv.org/pdf/1107.2138

A field theoretic discussion of irreversibility necessitates a statistical mechanics treatment. This more detailed modeling is done in practice in a hydrodynamic or kinetic approximation. They treat sources as generators of beams with an extended distribution in space or phase space, respectively. The dynamics of both descriptions is irreversible, and may be computed in terms of $k$-particle irreducible ($k$PI) Feynman diagrams for $k=1$ and $k=2$, respectively.


The kinetic description is based on the Kadanoff-Baym equations in the 2PI Schwinger-Keldysh (CTP) formalism. The Kadanoff-Baym equations are dynamical equations for the 2-particle Wightman functions and their ordered analoga, and are used in practice to model high energy heavy ion collision experiments. See, e.g.,

http://arxiv.org/abs/hep-th/9605024
and the discussions in
Good reading on the Keldysh formalism
What is known about quantum electrodynamics at finite times?


The hydrodynamic description is based on the simpler 1PI approach, but it is (to my knowledge) used mainly theoretically; see, e.g.,
Reviews of Modern Physics 49, 435 (1977)
and the papers
http://arxiv.org/pdf/hep-ph/9910334
http://arxiv.org/pdf/hep-ph/0101178
http://arxiv.org/abs/gr-qc/9805074



optics - Advantage of using a polygonal mirror with larger number of faces in Michelson method of measuring the speed of light and its value


The following image is from Concepts of Physics by Dr. H.C.Verma, from the chapter "Speed of Light", page 447, topic "Michelson Method":



enter image description here


For higher image resolution click here.



The following text is from the "Science Hero" article - Michelson’s Method for Determining Velocity of Light, under the topic "Disadvantages of Michelson’s method":



At high speeds [angular speed of the rotating mirror], the rotating mirror may break. But speed can be reduced by increasing the number of faces of the mirror.




I can understand that when the number of faces in the rotating mirror is increased, the clear image of the source could be seen at lower angular speeds since the angle by which the mirror needs to rotate for the next face to take the position of the adjacent face is decreased.


The speed of light as measured by this method is given by


$$c=\frac{D\omega N}{2\pi}$$


where $D$ is the distance travelled by the light between reflections from the polygonal mirror, $\omega$ is the minimum possible angular speed of rotation of the mirror when the image becomes steady and $N$ is the number of faces in the polygonal mirror.


As $c$ is a constant, the product $\omega N$ is also constant. So, it can be seen that when we increase the number of faces the rotating mirror, the clear image could be obtained at lesser angular speeds. Now as $N$ gradually approaches infinity, i.e., the polygonal mirror becomes a cylindrical mirror, the angular speed $\omega$ tends to zero. So I think there must be a highest possible value for $N$ which gives the most benefit. What is its value, and what is the reason for this choice? Are there any other advantages of using a larger number of faces in the rotating mirror besides the one discussed in the question?




Related question asked by me: Number of reflecting surfaces in the rotating mirror in the Michelson method of determination of speed of light


I think Michelson method of determining the speed of light is different from the Michelson Morley experiment. So, I had to use the query michelson speed of light -morley as my initial results were populated with the second experiment which has a similar name.


This method of determination of speed of light is briefly discussed here and here.



Answer




The purpose of having a relatively large number N of mirrors on the polygon is to increase the switching rate for a given rotational speed. This allows the distance to the retroreflecting mirror to be short enough to be practical. An important factor is that the beam size needs to be large enough to ensure that a significant fraction of the beam will reach the distant retroreflector. As you probably know, a light beam spreads faster when its waist diameter is small. This means the mirrors need to be relatively large, depending on how far away the retroreflector is. In turn, this means that N must be relatively small for a fixed-size wheel.


general relativity - How would it be to look at the sky if the earth were near the edge of the universe?


By looking at this picture:


http://earthspacecircle.blogspot.com/2013/01/earths-location-in-universe.html


The earth is near the center of the universe. I've read that the universe look the same no matter where the observer is located. It is the same distance everywhere.


So I understand that for general relativity the universe need to be homogeneous and isotropic, so it will look the same no matter where I am.


But what if I'm on one of the planets near the right or left of the image, then if I draw the same picture of the universe, but from my perspective, then I would be also located in the center? If that's not the case (I'm actually near an edge), then part of my sky would be completely dark, and all the sky that way won't be isotropic?




Answer



The universe has no edge so to speak. It is, however, finite in age, so light can only have traveled a given distance to get to us. Call this distance $R$, the "radius" of the universe. Any observer, anywhere, will see out to a distance $R$ in all directions from their location.


Now two different observers will have different origins for their respective observable universes, and so will see slightly (or vastly) different patches of the "full universe." (Be careful when talking about things outside your observable patch by the way - it is very easy to end up talking about impossible scenarios that produce nonsensical results.) This can happen even if the universe is closed (read: finite), so long as its size is bigger than something like $R$.


So no, no one is at the "edge" of the universe.


By the way, general relativity in no way requires homogeneity and isotropy. These are simply assumptions cosmologists make in order to take an utterly intractable problem (evolving the whole universe) and make it absurdly simple (see the FRW metric, which, although it may look complicated at first, is pretty much the most trivial thing you can do with general relativity). The homogeneous/isotropic assumptions, by the way, turn out to be justified on cosmological scales, though this was discovered only after the early days of GR-based cosmology, once we had very deep galaxy surveys.


electromagnetism - Why is the Hodge dual so essential?



It seems unnatural to me that it is so often worthwhile to replace physical objects with their Hodge duals. For instance, if the magnetic field is properly thought of as a 2-form and the electric field as a 1-form, then why do they show up in Ampere's and Gauss' as laws as their duals, i.e.


$$ \int_{\partial M} \star \mathcal B^2 = \int \int_M \left( 4\pi j^2 + \frac{\partial \star \mathcal E^1}{\partial t} \right)$$


$$ \int \int_{\partial U} \star \mathcal E^1 = 4 \pi Q_{\mathrm{enc}} $$


Similarly, angular momentum is considered nearly everywhere as a pseudo-vector instead of as a 2-form. Do these laws have formulations that do not use Hodge duals? Is this just for the sake of simplicity, since tensors are less familiar to the physics community than vectors?



Answer



In the language of differential forms in spacetime, the field strength $2$-form $F = E\wedge\mathrm{d}\sigma + B$ gives Gauss's law for magnetism and Faraday induction: $$\mathrm{d}F = 0\text{.}$$ Meanwhile, the electromagnetic excitation $2$-form $H = -\mathcal{H}\wedge\mathrm{d}\sigma + \mathcal{D}$ provides a natural formulation of Gauss's law and Ampère's circuital law: $$\mathrm{d}H = J\text{,}$$ where $J$ is the electric current 3-form.



For instance, if the magnetic field is properly thought of as a 2-form and the electric field as a 1-form, then why do they show up in Ampere's and Gauss' as laws as their duals, i.e. ...



Because it's a qualitatively different role: $F$ being a closed form is a necessary property to ensure conservation of magnetic flux and that the existence of a potential $1$-form $A$ for which $F = \mathrm{d}A$. But $H$, instead of conservation of magnetic flux, expresses the conservation of charge, with $H$ acting as a "potential" for the electric current $J$.



Of course, if you know that $H\propto\star F$, then you can eliminate the $(\mathcal{D},\mathcal{H})$ excitation fields put everything in terms of $(E,B)$ only. Or the reverse, if you wished. This naturally introduces at least an implicit Hodge dual into the equations, as you have above. But doing so obscures the fundamentally metric-free character of Maxwell's equations: the only place the metric appears is in the Hodge dual. So instead, one can think of the Hodge dual as providing a simple constitutive relation for free space, with vacuum having its own meaningful $\mathbf{D}$ and $\mathbf{H}$ fields.


In that kind of presentation, the appearance of the Hodge dual is natural and necessary to turn electromagnetism into a fully predictive theory--the metric must make an appearance eventually, but Maxwell's equations themselves are metric-free!


There are other possible relations between $H$ and $F$ independent of Maxwell's equations per se, leading to alternative theories of electromagnetism, such as Born-Infeld theory and Heisenberg-Euler vacuum polarization, etc. Generally, the requirements of the relation being local and linear gives $36$ independent components, which $15$ are dissipative and don't contribute to Lagrangian ("skewon") and $1$ that contributes to Lagrangian but doesn't affect light propagation or electromagnetic stress energy (a ghostlike "axion").


For the differential form presentation of electromagnetism that emphasizes the logically independent roles of $F$ and $H$, a good place to start is Hehl and Obukhov's arXiv:physics/0005084, since it works exclusively in $1+3$ decomposition and hence much more clearly corresponds to the more usual presentation of electromagnetism in terms of $(\mathbf{E},\mathbf{B})$ and $(\mathbf{D},\mathbf{H})$. They have also the book on this: Foundations of Classical Electrodynamics, though it's more demanding.


Additionally, MTW's Gravitation has many nice illustrations of what would be $F$ and $H$, although in MTW's presentation they correspond to the "Faraday tensor" and the "Maxwell tensor", respectively, and differ by a conversion factor.


Wednesday 24 August 2016

mathematics - Dinosaur egg drop


Archeologists discover two dinosaur eggs, and you are given the chance to test the durability of these eggs (bad move on their part). Suppose that these eggs will absorb a specific amount of force with no cumulative damage. In other words, if they don't crack, it is as if they never fell.


You have a 100 story building, and you are allowed only 20 trials.


Questions:




  1. Is there an algorithm such that you can determine the highest floor in which one of these dinosaur eggs can be dropped and not break?

  2. If so, what would be the highest building for which this algorithm would be able to work?

  3. Supposing the number of trials were still 20, but you had as many dinosaur eggs as you needed. At what point would extra eggs not increase the highest floor for which you can test?


Bonus question:



  1. What would be the highest building you could test for with 3 eggs at your disposal?



Answer







  1. Take the first egg. Drop it from floors 10, 20, 30, ..., 100 until it breaks. This will take at most 10 of the 20 trials. When it breaks, go 9 floors down, and test every floor with the remaining egg. E.g. if the first egg breaks at 70, drop the second egg from 61, 62, 62, ..., 69. That's at most 19 trials in total.






  2. We can do a bit better though. If the first egg breaks on the first trial, we've got 19 trials left, so we can actually drop it from the 20th floor right away, because there are enough remaining trials to test each floor below. If it doesn't break, we can go 19 floors up for the second trial (with the same egg). Continuing like this, we might be lucky to be using the first egg for all 20 trials without risking not having enough trials to figure out the exact floor. This means we can go up to 20 + 19 + 18 + ... + 1 = 210 floors, using






This:



$$ \sum_{n=0}^{19} n+1 = \sum_{n=1}^{20} n = \frac{20(20+1)}{2} = 210. $$



Just to break down the first version of this:



There are 20 possibilities for when the first egg breaks. For each case $n$ is the number of trials left. And for each such case we can test $n$ floors with the second egg and $1$ floor with the egg we potentially broke. Hence the summand of $n+1$.







  1. With more eggs, you can can basically apply a binary search, halving the potential range of floors each time. That lets you check 220 = 1048576 floors. The first egg is dropped from floor 219, then you go up or down 218 floors depending on whether the egg broke on the first trial.






  2. With three eggs, we can test considerably more than with two eggs. Consider the first egg breaking on the first trial. Then we've got 19 trials and two eggs left. Applying algorithm 2, we can still test 190 floors with those. So we can drop the first egg from floor 191. If the first egg breaks on the second trial, we've still got 18 trials and two eggs left, with which we can test 171 floors, so we can make the second trial already from floor 373, and so on. This holds as long as there are at least two trials left, so that we can make use of both eggs. If we're still dropping the first egg on trial number 19, then we've only got one trial left, so we can only go up two floors for the 19th trial. I think this coincides with the previous formula, but I'm not sure it would if we generalise this beyond 3 eggs. Basically, for the last two trials, this reduces to the above formula.






So we have with 3 eggs:



$$ \sum_{m=1}^{18}\left(1+\sum_{n=0}^m n+1\right) + \sum_{n=0}^{1} n+1 = 1350 $$



lagrangian formalism - What do the derivatives in these Hamilton equations mean?



I have a Hamiltonian:


$$H=\dot qp - L = \frac 1 2 m\dot q^2+kq^2\frac 1 2 - aq$$


In a system with one coordinate $q$ (where $L$ is the Lagrangian). One of the Hamilton equations is:


$$\dot q =-\frac {\partial H} {\partial p}$$


But when I try to derive $H$ with respect to $p$, I get very confused. What is the derivative of $q$ with respect to $p=m\dot q$, for instance? When I boil it right down, my confusion stems from the fact that I realize I don't know what that partial derivative means. A partial derivative of a multi-variable function should be taken with respect to an index (you just specify which variable, thought of as a "slot" in the function, you're deriving with respect to). I suppose I'm not clear on what multi-variable function $H$ represents (I mean, $q$ and $\dot q$ are functions of $t$, so you could say it's a one variable function...), or how I should interpret $p$ as a variable.


I have similar difficulties with the equation $\dot p=\frac {\partial H} {\partial q}$, although I think I can understand $\frac {\partial H} {\partial t} = \frac {dL} {dt}$. The left hand side should give $m\dot q\ddot q + kq\dot q - a\dot q$, right?



Answer



Let's restrict the discussion to one spatial dimension for simplicity.


What's going on with partial derivatives?


The lagrangian is a function of two real variables. We commonly label these variables $q, \dot q$ because of their physical significance. For example, the lagrangian for a one-dimensional simple harmonic oscillator is \begin{align} L(q, \dot q) = \frac{1}{2}m \dot q^2 -\frac{1}{2}k q^2 \end{align} Notice that we could just have easily written \begin{align} L(\heartsuit, \clubsuit) = \frac{1}{2}m\clubsuit^2 - \frac{1}{2}k\heartsuit^2, \end{align} because we are just using labels for the two slots that can be any symbols we choose. However, it is convenient to stick to a particular labeling, because then, we can use some convenient notation for the partial derivatives. For example, if we use the $q, \dot q$ labeling, then the expressions \begin{align} \frac{\partial L}{\partial q}, \qquad \frac{\partial L}{\partial \dot q} \end{align} mean the derivatives of $L$ with respect to its first and second arguments ("slots") respectively. But notice that if we have used the second labeling above, we just as easily could have written \begin{align} \frac{\partial L}{\partial \heartsuit}, \qquad \frac{\partial L}{\partial \clubsuit} \end{align} for the same derivatives.



What exactly is the hamiltonian...really?


Now, the Hamiltonian is also a function of two real variables, and we conventionally call them $q$ and $p$, but how is this function generated from a given Lagrangian $L$? Well we need to be careful here because this is where physicists tend to really abuse notation.


What we do, is we first define a function $\bar p$ (the canonical momentum conjugate to $q$) as a certain derivative of the Lagrangian: \begin{align} \bar p = \frac{\partial L}{\partial \dot q}, \end{align} where I am using the conventional labeling for the arguments of the lagrangian. I put a bar on $p$ here to avoid the common abuses of notation you'll see in physics, and to emphasize the actual mathematics of what's going on. Notice, in particular, that in this conventional labeling, $\bar p$ is a function of two real variables $q$ and $\dot q$.


Next, we write the relation \begin{align} p = \bar p(q, \dot q), \end{align} and we invert it to write $\dot q$ in terms of $q$ and $p$, so now we have \begin{align} \dot q = \text{some expression in terms of $q$, and $p$} = f(q,p), \end{align} Finally, we define \begin{align} H(q,p) = p f(q,p) - L(q, f(q,p)). \end{align} Notice, again, that we just as easily could have labeled the arguments of $H$ with whatever labels we wanted, but once we do this, we typically stick to that labeling in which case the same remarks that we made above for the derivatives of the Lagrangian can be made here.


Note that intuitively what's happening here is that the Hamiltonian is defined "as a function of $q$ and $p$; you should never be writing it "as a function of $q$ and $\dot q$."


Example. Consider, again, the one-dimensional simple harmonic oscillator. We have \begin{align} \bar p (q, \dot q) = \frac{\partial L}{\partial \dot q}(q, \dot q) = m \dot q \end{align} So now the relation $\bar p (q, \dot q) = p$ is \begin{align} m \dot q = p \end{align} and therefore inversion to write $\dot q$ in terms of $q$ and $p$ is super easy in this case; \begin{align} \dot q = \frac{p}{m} = f(q,p). \end{align} It follows that \begin{align} H(q,p) &= p f(q,p) - L(q, f(q,p)) \\ &= p(p/m) - \frac{1}{2}m(p/m)^2 +\frac{1}{2}kq^2 \\ &= \frac{p^2}{2m} + \frac{1}{2}kq^2 \end{align} Now, taking derivatives with respect to $q$ and $p$ simply means taking derivatives with respect to the first and second arguments of this function of two real variables.


What about Hamilton's equations etc.?


Now, that we know what the hamiltonian is and how it's computed, let's address equations like Hamilton's equations: \begin{align} \dot q = \frac{\partial H}{\partial p}, \qquad \dot p = -\frac{\partial H}{\partial q}, \end{align} Again, your confusion is not surprising because physicists are notorious for abusing notation in these instances and not pointing that out.


To interpret this properly, we note that in the Hamiltonian formulation, the state of the system at any given time $t$ consists of a pair $(q(t),p(t))$ giving the value of the position of the system and of its canonical momentum at that time $t$. Actually, in order to avoid perpetuating common confusions, let's use a different notation and write $(\gamma_q(t), \gamma_p(t))$ for the state of the system at time $t$ and reserve $q$ and $p$ for labels of the argument of $H$.


Then Hamilton's equations are really saying that if the pair $(\gamma_q(t), \gamma_p(t))$ is a physical motion realized by the system, then \begin{align} \dot \gamma_q(t) = \frac{\partial H}{\partial p}(\gamma_q(t), \gamma_p(t)), \qquad \dot \gamma_p(t) = -\frac{\partial H}{\partial q}(\gamma_q(t), \gamma_p(t)). \end{align} for all $t$. In other words, we get a system of coupled, first order ODEs for the functions $\gamma_q(t), \gamma_p(t)$. You can see, because of this notation, that for example, $\dot q$ in Hamilton's equations is a different beast than $\dot q$ as used to label the arguments of the Lagrangian. In the former case, it is a function, in the latter case, it is just a label.



You can always stave-off this ambiguity by using different notations for these animals in the different contexts as I have done here. However, once you know what you're doing, you can happily once again revert back to overloading the symbols you're using, and you probably won't make a mistake either procedurally, or conceptually. In fact, in practice almost everyone who knows what he's doing does this because it's faster.


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...