Monday 29 February 2016

quantum field theory - "Slightly off-shell"?


I'm not new to QFT, yet there are some matters which are quite puzzling to me. I often come across the statement that real particles (the ones we actually measure in experiments, not virtual ones) are "slightly off-shell". What does this actually mean? To my knowledge, something being off-shell means that it violates the relativistic energy-momentum relation. But how can this be possible for particles we actually consider to be "physical"? Please fill me in on the mathematical/experimental backing for such a statement to any degree of detail necessary.



Answer




See for instance the comments on my answer to Are W & Z bosons virtual or not?.


Basically the claim is that the observed particle represents a path internal to some Feynman diagram and accordingly there is a integral over it's momentum.


I'm not a theorist, but as far as I can tell the claim is supportable in a pedantic way, but not very useful.


quantum field theory - How are bound states handled in QFT?


QFT seems very well suited to handle scattering amplitudes between particles represented by the fields in the Lagrangian. But what if you want to know something about a bound state without including it as an extra field? For example, suppose we have electron+proton QED (ignoring the proton's structure):


$$\mathcal{L} = -\frac14 (F_{\mu\nu})^2 + \bar{\psi_e} (i\not \partial -m_e)\psi_e + \bar{\psi_p} (i\not \partial -m_p)\psi_p - e \bar{\psi_e} \not A \psi_e + e \bar{\psi_p}\not A \psi_p$$


I can use this with no problem to calculate Rutherford scattering or similar processes. But this Lagrangian should also have the hydrogen atom hidden in it somewhere. For example, I may want to use QFT to calculate the binding energy of hydrogen. Or I might want to calculate the probability of firing an electron at a proton and getting hydrogen plus photons as a result. How can this be done? Obviously this is a broad subject, so I'm just looking for an outline of how it goes.



Answer



The conventional way to handle bound states in relativistic quantum field theory is the Bethe-Salpeter equation. The hydrogen atom is in QFT usually treated in an approximation where the proton is treated as an external Coulomb field (and some recoil effects are handled perturbatively). The basics are given in Weinbergs QFT book Vol. 1 (p.560 for the Bethe-Salpeter equation and Chapter 14 for 1-electron atoms). Weinberg notes on p.560 that



the theory of relativistic effects and radiative corrections in bound states is not yet in entirely satisfactory shape.



This quote from 1995 is still valid today, 20 years later.



On the other hand, quantum chemists use routinely relativistic quantum mechanical calculations for the prediction of properties of heavy atoms. For example, the color of gold or the fluidity of mercury at room temperature can be explained only through relativistic effects. They use the Dirac-Fock approximation of QED.


newtonian mechanics - Hyper/parabolic Kepler orbits and "mean anomaly"


In an elliptical kepler orbit there is an easy recipe to describe the motion/position of a satellite at time $t$. One just follows the following steps - an important detail for me is that the numerical part has always the same error, the error doesn't increase with time, nor is it based on the timestep. (Which it would if one would solve a differential equation).


Calculate mean motion $n$ (given a semi major axis $a$, and central mass $M$, this is basically the 2 pi divided by the period): $$n = \sqrt{\frac{GM}{a^3}}$$


Now the definition of mean anomaly:


$$M(t) = M_0 + n \cdot t$$


And one can (numerically) solve the following equation, to get the eccentric anomaly $E$:


$$M = E - \varepsilon \sin(E)$$


And then simple geometry allows one to find the true anomaly $\theta$.





Now this is all fine and good working. I'm at a loss as how to to apply this to non-elliptical orbits. The main problems I have are that there's no notion of semi major axis (or when using a negative for hyperbolic orbits- it is infinite for parabolae, which would result in a mean motion of 0.).

Nor is there an meaning for the eccentric anomaly, non elliptical orbits don't have a geometric center. (The true anomaly would still be defined as the focal point).




optimization - Halve time with two timers


power outlet, 2 timers, light bulb
Two motorized 24-hour light timers are daisy chained between a power outlet and a light bulb.


For these timers, devise schedules and choose initial times that produce the following repeated lighting pattern, with the largest possible whole number $n$, beginning when the outlet's power is switched on.


     Light is on for 1/2 hour,$~$ off for $n$ hours,
      on for 1/2 hour,$~$ off for $n$ hours,
      on for 1/2 hour,$~$ off for $n$ hours,
      $~\,\vdots$



If you are unfamiliar with these timers


Each timer repeatedly cycles through its schedule of 24 intervals that last an hour each.
•$~$ A circular dial determines the current point in the schedule
•$~$ A motor rotates the dial to advance through its schedule whenever power is supplied to the timer
•$~$ You may initially set the dial to any minute of any interval
•$~$ You preset each interval to ON or OFF
•$~$ When the dial is in an interval that was set to ON, the timer acts as a direct connection for power to flow to whatever is plugged into the timer
•$~$ When the dial is in an interval that was set to OFF, the timer does not provide a power connection
•$~$ The resulting ON and OFF durations could be fractions of an hour if the timer is set to begin within an interval or if incoming power is interrupted during an interval


The first timer is plugged into the outlet.

•$~$ It runs nonstop once the outlet is switched on
•$~$ It supplies power— but only when its dial is in an ON interval — to the second timer


The second timer has the light bulb plugged into it.
•$~$ It advances through its schedule only when the first timer supplies power
•$~$ It lights the bulb, but only while powered by the first timer and when its dial is in an ON interval


Related puzzles
Odd hours with two timers
Day and night of the two timers
Third timer's a charm


(These puzzles are either directly from or related to actual botany experiments)




Answer



To show that Ivo Beckers has the longest $n$ in his answer, rewrite it like this:



Set the intervals like this where 1 is on and 0 is off:
A: 101101101101101101101101
B: 101010101010101010101010


set dial A to 30 minutes and B to 0 minutes.



That is, rotate A once to the left and set its timer to 30 instead of 90.


This works as before for a one hour delay.



Notice that A is now simply 101 repeated 8 times.


To get a 2 hour delay we'd need A to be 10010 repeated, but 5 doesn't divide 24 evenly.


Likewise, for a 3 hour delay A is set to 1000100 repeated but 7 doesn't divide 24 evenly.


etc


None of these work because:
Their length is len = n + (n + 1) for an n hour delay.
Let's call this len(A), therefore len(A) has to be odd.
The only odd number that divides 24 evenly is 3.


So 1 hour is the longest delay.


EDIT: Other patterns simply don't work.



In order to have a 1/2 hour on followed by off B can't have any sequential ones.


So B must be composed of a single one followed by one or more zeros with this pattern repeated through out B.


B must also start with a one or we can simply remove leading zeros (with no loss of generality) so that B does start with a one.


A must now start with a one and its timer set to start at 30 minutes so we begin with a 1/2 hour on.


A must then be followed by a zero so we go off after that 1/2 hour on.


And B obviously is followed by one or more zeros.


Therefore A and B must start with:


A = 10 [1]
B = 10 [2]


Now for any number of zeros that follow the initial one for B in [2], say B is now 10...0 repeated i.e a one followed by b zeros followed by a one followed by b zeros etc, then A must be 100..011..1 repeated i.e. a one followed by b zeros and then b ones followed by a one then b zeros, then b ones etc.



Therefore len(A) is 1 + 2b, so len(A) remains odd and must evenly divided 24.


So therefore len(A) can only equal 3.
and A is 101101101101101101101101
and B is 101010101010101010101010
with A starting at 30 minutes and B at 0 minutes.
Or equivalent rotations as in Ivo Beckers original answer.


electricity - Why is AC more "dangerous" than DC?


After going through several forums, I became more confused whether it is DC or AC that is more dangerous. In my text book, it is written that the peak value of AC is greater than that of DC, which is why it tends to be dangerous. Some people in other forums were saying that DC will hold you, since it doesn't have zero crossing like that of AC. Many others also say that our heart tries to beat with the frequency of ac which the heart cannot support leading to people's death. What is the actual thing that matters most?


After all, which is more dangerous? AC or DC?



Answer



The RMS (root-mean square) value of an AC voltage, which is what is represented as "110 V" or "120 V" or "240 V" is lower than the electricity's peak voltage. Alternating current has a sinusoidal voltage, that's how it alternates. So yes, it's more than it appears, but not by a terrific amount. 120 V RMS turns out to be about 170 V peak-to-ground.


I remember hearing once that it is current, not voltage, that is dangerous to the human body. This page describes it well. According to them, if more than 100 mA makes it through your body, AC or DC, you're probably dead.


One of the reasons that AC might be considered more dangerous is that it arguably has more ways of getting into your body. Since the voltage alternates, it can cause current to enter and exit your body even without a closed loop, since your body (and what ground it's attached to) has capacitance. DC cannot do that. Also, AC is quite easily stepped up to higher voltages using transformers, while with DC that requires some relatively elaborate electronics. Finally, while your skin has a fairly high resistance to protect you, and the air is also a terrific insulator as long as you're not touching any wires, sometimes the inductance of AC transformers can cause high-voltage sparks that break down the air and I imagine can get through your skin a bit as well.



Also, like you mentioned, the heart is controlled by electric pulses and repeated pulses of electricity can throw this off quite a bit and cause a heart attack. However, I don't think that this is unique to alternating current. I read once about an unfortunate young man that was learning about electricity and wanted to measure the resistance of his own body. He took a multimeter and set a lead to each thumb. By accident or by stupidity, he punctured both thumbs with the leads, and the small (I imagine it to be 9 V) battery in the multimeter caused a current in his bloodstream, and he died on the spot. So maybe ignorance is more dangerous than either AC or DC.


particle physics - What would happen if Large Hadron Collider would collide electrons?


After some reading about the Large Hadron Collider and it's very impressive instruments to detect and investigate the collision results, there is a remaining question.



What would happen if the scientists would use leptons instead of hadrons?


Especially: What would happen if they would collide electrons?


Isn't it intrinsic that all particles consist of smaller particles? With current technology, could we detect them?



Answer



First of all -- it wouldn't be called "the Large Hadron Collider", right?
Looks like one would rather call it something like "Large Electron-Positron Collider".
In that case one definitely would need another abbreviation for it. Something like "LEP" instead of "LHC"...


Now, guess what was there in the same tunnel before?




Edit: since my shenanigan got popular, I'll elaborate.





  • Yes, they actually was colliding electrons and positrons, not electrons-electrons. Mainly because of the richer physics of such collisions. (But for my theoristish point of view: positron is just an electron going back in time.)




  • Why the same tunnel? Perhaps surprisingly, tunnel is taking a substantial part of the cost of an accelerator. Digging a new one for LHC would have definitely burnt a large hole in CERN's pocket.




  • Given a fixed circular tunnel (it's radius) you actually have a bound on energy you can have for your particles. Due to synchrotron radiation -- see @emarti answer for more.





  • 27 kilometers seems to be a reasonable limit on the size of a circular tunnel. (Actually people think about 233 km, but that sounds crazy to me.) So the next accelerator most probably will be linear and it will be electron-positron.






P.S. Have you heard of a Photon Collider?


Sunday 28 February 2016

special relativity - (2+1)-d Unitary irreducible representations of the Poincare group


I am currently working through this paper on relativistic field theories in three dimensions. I have come to terms with the classification of Unitary irreducible representations (UIR's) given. The problem I am now trying to solve is the explicit realisation of the physical UIR's (corresponding to steps 4 and 5 in Binegar's program). I would be very happy with an answer/guidance for either of my questions, you need not answer both.


First of all, I have an issue with step 5 of his program:


"Finally, from each UIR $\mathcal{D}[~~]$ of each stability subgroup $S_{\hat{O}}$ we form the induced UIR of $\pi_+^{~~\uparrow}$ (connected component of the Poincare group) given by $$U_{D}^{~\hat{O}}(a,\Lambda)\psi(p)=\exp(ip_{\mu}a^{\mu})\mathcal{D}[\Omega^{-1}(p)\Omega_{\Lambda}\Omega(\Lambda^{-1}p)]\psi(p) \tag{1}$$ where $\Omega(p)\in \text{SL}(2,\mathbb{R})$ corresponds to a Lorentz transformation which takes the momentum $p$ (on mass shell with mass m) to the stability point $\hat{p}$ (both momentum 3-vectors), $\Omega_{\Lambda}\in\text{SL}(2,\mathbb{R})$ corresponds to the Lorentz transformation $\Lambda$ and $\Omega^{-1}(p)\Omega_{\Lambda}\Omega(\Lambda^{-1}p)\in\text{SL}(2,\mathbb{R})$ is to be interpreted as its ($\Lambda$'s) projection onto the stability subgroup $S_{\hat{O}}$."


Now shouldn't $\Omega(p)$ in $\Omega^{-1}(p)\Omega_{\Lambda}\Omega(\Lambda^{-1}p)\equiv W(\Lambda,p)$ correspond to a Lorentz transformation which takes $\hat{p}$ to $p$ (not the other way around)? Otherwise I do not see how $W(\Lambda,p)$ can be an element of the stability group of $\hat{p}$. Observe that if this were the case then $$\begin{align}W(\Lambda,p)\cdot \hat{p}&=\bigg(\Omega^{-1}(p)\Omega_{\Lambda}\Omega(\Lambda^{-1}p)\bigg)[\hat{p}^{\mu}\tau_{\mu}]\bigg(\Omega^{-1}(p)\Omega_{\Lambda}\Omega(\Lambda^{-1}p)\bigg)^T\\ &=\Omega^{-1}(p)\Omega_{\Lambda}\bigg(\Omega(\Lambda^{-1}p)[\hat{p}^{\mu}\tau_{\mu}]\Omega^T(\Lambda^{-1}p)\bigg)\Omega^T_{\Lambda}\Omega^{-1T}(p)\\ &=\Omega^{-1}(p)\bigg(\Omega_{\Lambda}[(\Lambda^{-1}p)^{\mu}\tau_{\mu}]\Omega^T_{\Lambda}\bigg)\Omega^{-1T}(p)\\ &=\Omega^{-1}(p)[p^{\mu}\tau_{\mu}]\Omega^{-1T}(p)\\ &=\hat{p}^{\mu}\tau_{\mu}\\ &=\text{Id}\cdot\hat{p} \end{align}$$ Where $p^{\mu}\tau_{\mu}$ is a symmetric 2x2 matrix given in the introduction and $\cdot$ denotes the action of $\text{SL}(2,\mathbb{R})$ on $\mathbb{R}^3$. It is unlikely that he wrote it wrong because he writes it in the same way multiple times throughout the paper (not to mention that this paper has ~ 350 citations). So how should I be interpreting $W(\Lambda,p)$?


Secondly, at the end of page 3/beginning of page 4 the author derives some properties for these Wigner rotations, $W(\Lambda,p)$, for some specific orbits, viz., the (physical) massive and massless cases. For example, he claims that in the massive case for an infinitesimal rotation $R(\theta)$ we have $$W(R(\theta),p)=\Omega^{-1}(p)\Omega_{R(\theta)}\Omega(R^{-1}(\theta)p)=R(\theta)$$ and that for an infinitesimal boost $L(\vec{\theta})$ in the $\vec{\theta}$ direction, $$W(L(\vec{\theta}),p)=\Omega^{-1}(p)\Omega_{L(\vec{\theta})}\Omega(L^{-1}(\vec{\theta})p)=R\bigg(\frac{\theta_1p_2-\theta_2p_1}{E+m}\bigg).$$ Given that the author does not supply a 'standard boost', or an $\text{SL}(2,\mathbb{R})$ matrix which corresponds to one (a.k.a an explicit $\Omega(p)$) for that matter, I would assume that there is a simple way to derive these relations. However I couldn't find such a simple method, and so I constructed a standard Lorentz transformation (SLT) which does the trick and was able to show the first and almost the second of these relations (after alot of cumbersome bookkeeping). The problem is that I think I should be using an $\text{SL}(2,\mathbb{R})$ standard boost instead of a Lorentz standard boost, but it has proved extremely computationally expensive (and ugly) to obtain the $\text{SL}(2,\mathbb{R})$ matrix to which my SLT corresponds (not to mention there is that pesky $\pm$ uncertainty due to the double valued-ness of the double cover $\text{SL}(2,\mathbb{R})$). Furthermore, when trying to prove a similar relation (given just after these) for the massless case I think it is imperative to use an $\text{SL}(2,\mathbb{R})$ boost rather than a Lorentz one, so I must conclude that my method is crap. Do you know of a better approach which I can take to proves these relations?



Answer



Answer to questions:





  1. The way I interpreted it is correct; $\Omega (p)$ should be an $\text{SL}(2,\mathbb{R})$ matrix corresponding to a Lorentz transformation which takes the standard momentum $\hat{p}$ to $p$. In terms of $\text{SL}(2,\mathbb{R})$ transformations, this should be; $\Omega (p)\cdot\hat{p}=\Omega (p)\hat{p}^{\mu}\tau_{\mu}\Omega^T (p)=p^{\mu}\tau_{\mu}=\text{Id}\cdot p$.




  2. Indeed, one has to use $\text{SL}(2,\mathbb{R})$ transformations instead of Lorentz transformations since we are interested in projective (multi-valued) representations (one can go even farther and use the universal cover of the Lorentz group, which can be found here). The relations stated at the end of the OP for the massive case can be proven using the following 'standard $\text{SL}(2,\mathbb{R})$ boost':$$\Omega (p)=\frac{1}{\sqrt{2m^2+2mE(\mathbf{p})}} \begin{pmatrix}m+E(\mathbf{p})-p^1&&p^2\\p^2&&m+E(\mathbf{p})+p^1\end{pmatrix}$$$$\text{ where }E(\mathbf{p})=\sqrt{m^2+(p^1)^2+(p^2)^2}.$$ Both proof's are very tedious, the second more so than the first. I used the universal cover of the Lorentz group to prove both, but it must be possible to do it by just considering $\text{SL}(2,\mathbb{R})$ since it seems Binegar did it that way. I am still working on the proof for the massless case.




In short, don't take Binegar's notation too seriously because it is ambiguous, inconsistent and incorrect on multiple accounts throughout the paper.


homework and exercises - Which trigonometric ratio should be used to describe simple harmonic motion as a function of time?



An object is undergoing simple harmonic motion with period 1.2s and amplitude 0.6m. At $t=0$, the object is at $x=0$. How far is the object from the equilibrium position when $t=0.480$s?




I used the displacement equation :


$$x(t)=A\cos(\omega t+\phi)$$


and also found out what the angular frequency $\omega$ is (5.2rad/s). Then I found that $\phi$ is 0. I plugged my results in the equation and when I looked at the solution they used the following equation:


$$x(t)=A\sin(\omega t)$$


How was I supposed to know to use this equation instead of the cosine equation that is written on my equation sheet.



Answer



The initial phase $\phi$ should be $-\pi/2$, or in other words, the solution should be $A\sin(\omega t)$ because $\cos(\theta-\pi/2)=\sin(\theta)$. How did you find that $\phi$ is zero? You're supposed to use the fact that $x$ is zero when $t$ is zero. Which is what let's you set $\phi $ to $-\pi/2$ (modulo $\pi$).


double slit experiment - Variation of delayed choice quantum eraser


Let me base the discussion on the pictorial description of the delayed choice quantum eraser experiment on wikipedia.


quantum eraser diagram


First suppose that we do precisely the same thing with the lower parts of the blue and red beams as we do with the upper parts, i.e. we put them through a lens which ends at a new detector $E_0$.




Question. Would there be the interference pattern at $D_0$ (or - I suppose equivalently - at $E_0$)? If there woudn't be - why?



I thought there would be, but then it occured to me we could use it to communicate faster than light - by shooting the lower parts of the red and blue beams to a distant partner. Then the partner would send the mesage either by preparing the setup above (and so causing the patterns at $D_0$ to interfere), or by preparing the setup of the original delayed quantum choice eraser (and so causing the interference pattern at $D_0$ to vanish.) Actually as I've described it, it would be even a message from the future.



Question If there would be interference pattern, what's wrong in the above faster-than-light communication protocol?





classical mechanics - Is there any relation between Poisson Brackets and the Jacobian Matrix?


The Poisson brackets for $u,v$ can be written as,


$$ \frac{\partial u}{\partial q} \frac{\partial v}{\partial p} - \frac{\partial u}{\partial p}\frac{\partial v}{\partial q}. $$


We can write this as determinant of this matrix


$$ \begin{bmatrix} \frac{\partial u}{\partial q} & \frac{\partial u}{\partial p} \\\frac{\partial v}{\partial q} & \frac{\partial v}{\partial p} \end{bmatrix} $$


Which is the Jacobian matrix.


Is there any relation between them?



Answer






  1. For a 2-dimensional phase space, they are the same.




  2. More generally, for a $2n$-dimensional symplectic manifold $(M;\omega)$ with symplectic two-form $$\tag{1} \omega~=~\frac{1}{2} dz^I ~\omega_{IJ} \wedge dz^J, \qquad \omega_{IJ}~=~-\omega_{JI}, $$ the Poisson bracket is given by $$ \tag{2} \{f,g\}_{PB}~=~\frac{\partial f}{\partial z^I}\pi^{IJ} \frac{\partial g}{\partial z^J}, \qquad \pi^{IJ} ~:=~(\omega^{-1})^{IJ} . $$ See also e.g. this and this Phys.SE posts.




  3. The canonical volume form on $(M;\omega)$ $$\tag{3} \Omega~=~\omega^n~=~\rho~ dz^1\wedge \ldots \wedge dz^{2n},$$ is the $n$'th exterior power of the symplectic two-form $\omega$, with volume density $$\tag{4} \rho~=~{\rm Pf}(\omega_{IJ}) $$ given by the Pfaffian, which is a square root of the determinant. This is closely related to Liouville's theorem.





optics - Why do you need at least two rays to form an image?


Why isn't enough one light beam to form an image in your retina for example?



Answer



When a sharp image is formed, every point on the object is reproduced in the image, and all the points around that point on the object are reproduced in the same relative positions of the image.


In this first diagram the two grey rays $OXI$ and $OPI$ are the rays I used to find out where the top of the image is formed.
They are called the construction ray and these are usually the only two rays that you see on a ray diagram.


enter image description here


However in reality all rays bounded by rays $OWI$ and $OZI$ go through the lens to form the image.
The time it takes light to go from a point on the object $O$ to a corresponding point on the image $I$ is the same for all rays.


If I put an obstruction in, ie reduce the aperture of the lens, the image is still formed but with a diminished brightness because less of the light goes through to form the image.



enter image description here


So only the rays bounded by rays $OXI$ and $OYI$ will get through the optical system to form the image of the top of the object.


Having used the construction rays to find the position of the image you can draw similar diagrams to show the rays which for other parts of the image as shown below.
In that diagram if the top half of the lens was obscured ie the construction rays did not get through to form the image, the image would still be formed but with a reduced brightness.


enter image description here


geometry - What are the weights of the bangles?


John the jeweler is planning to make bangles from solid 14 karat gold. As shown below, his design will:



  • be available in 5 different diameters (D)


  • have a height (h) of 1 inch for all bangles

  • have an inside profile that is straight

  • have an outside profile with a radius of curvature (r) equal to D/2


John has fabricated a size XS prototype and found that it weighs 120 grams. John knows the density of 14 karat gold is 14 grams/cc, but is not sure how to calculate the weights of the other four bangles. He needs this information to determine the fabrication costs.


Can you help John by providing the weights of the other bangles with larger diameters?


This may look like a math textbook problem, however it is a variation of a classic geometry puzzle.


Bangle



Answer



They are exacty the same so also 120 gram.



By making the diameter bigger you also make the ring narrower. And these cancel eachother out.


A better explanation here


special relativity - Why is there a universal speed limit?


I am looking for an answer that does not rely on Special or General Relativity -and without recourse to the fact that the speed of light is frame invariant.


Is there another way of showing this universal speed limit to be necessary -one that it would have been possible to find before Einstein made his theories?


For myself I think there should be a universal speed limit because there is a law of diminishing returns when we look for ways to accelerate an object (we have to mine further and further regions of the universe which means that even an infinite universe would only have a finite amount of accessible energy)


However, following my reasoning it does not follow that this limit would be the same as the speed of light in vacuum. (I don't deny that it is identical -just that my "method" does not show this)



Answer



Backing up what zeldredge said, what you asked about is known as "relativity without light". According to the intro of this paper (arXiv link) for instance, the original argument was given as early as 1910 by Ignatowski, and has been rediscovered several times. There is a modern version due to David Mermin, in "Relativity without light", Am. J. Phys. 52, 119-124 (1984), but a pretty accessible presentation may also be found in Sec.2 of this paper by Shan Gao: "Relativity without light: A further suggestion" (academia.edu link). The basic idea is that the existence of an invariant speed follows directly from the homogeneity and isotropy of space and time, and the principle of relativity. No reference to a speed limit is needed, but it does follow that the invariant speed acts as a speed limit. The only alternative is a universe without a speed limit (infinite invariant speed), where kinematics is governed by the Galilei transformations. Why it is that our universe has a finite invariant speed, and not an infinite one, remains an open question. Gao's "further suggestion" is that the invariant speed is related to the discreteness of space and time at the Plank scale, which is an intriguing thought in its simplicity, but then it remains just a "thought" so far.


electrostatics - What will happen to the charges if we dig a hole in a hollow charged conductor(example a hollow conducting sphere)?


I am confused about how will the charges on outer surface of a hollow charged conductor would distribute themselves to make electric field zero inside after we dig a hole in the hollow conductor going from outer surface till the inner surface(imagine a thick hollow sphere having a hole piercing completely the thick layer of the sphere).my question is that are the charges now going to distribute themselves on the outer as well as the inner surface, as the inner surface is no longer inner because its connected to outer surface now because of the hole we dug?





Saturday 27 February 2016

newtonian mechanics - Force required to lift a block




Let us consider block of mass $m$ hanging by rope with force $T$. When no other horizontal and vertical forces are acting on a block, the block is at equilibrium if $$T=mg.$$ And if the rope is pulled up a distance $h$, $$\text{work done}=mgh$$


Don't we need more force to move the block upward? If we just use $T$, the body will be in equilibrium. Don't we need more force to lift the block?




energy - A question on the redshift of photons due to cosmic expansion


Given that the universe is expanding over time, in the sense that the (spatial) metric is changing over time, corresponding to the physical distance between objects increasing, naïve intuition leads me to the conclusion that the wavelength of a photon travelling from a distant galaxy (receding us) will be stretched, and consequently its frequency will decrease, leading to the energy of the photon decreasing (with this energy simply being lost, since time translation symmetry is broken due to the big-bang). The problem with this is, relative to another observer, wouldn't the wavelength be stretched by a different amount and hence the redshift the photon will be different, corresponding to a different amount of energy being lost by the photon.


This all leaves me feeling confused on the subject. Does the photon actually get redshifted and lose energy due to cosmic expansion, or is it simply an observer effect (akin to the standard Doppler effect). It does seem a little counter intuitive that a photon would lose energy simply due to its propagation through space?!


Is the whole point of this that it is an observer dependent phenomenon and the energy of an object is an observer dependent quantity (energy is not conserved one moves between two different frames of reference)?!



Answer



You need to concentrate on the observer making the measurements. Photons carry energy, but they don't lose energy just because they travel. The "loss" of energy is not the cause of the redshift, only if the photon scatters off something will it lose energy.


However, not all observers will agree that photon has the same amount of energy. Assume you are in a frame in which the photon is observed as green. An observer in a different frame moving relative to yours measures the photon as red.


Because measurements are made in different reference frames, the conservation of energy principle is not violated. Ultimately the energy of a photon is frequency dependent and different observers measure different frequencies.


Analogously, if you toss a coin whilst being driven, the coin has a different velocity to you, as to that measured by a bystander. Keep in mind that energy is conserved within each reference frame. The law of energy states that, in any given reference frame, the amount of energy doesn't change, but it does not apply to the way in which the energy in one frame is related to the energy in another frame.



A good read on this is Preposterous Universe.


newtonian mechanics - If an object in uniform motion with zero net force, can its mass change?



According to newton laws if something is in uniform motion and equivalent force is equal to zero then its mass cannot change.



Our book says an object stays in state of rest or motion unless compelled by external force. Now if mass is changing and velocity is not changing there will be non-zero equivalent force, but if equivalent force is equal to zero then mass cannot change. I deduced this, so I am asking for confirmation


I am just a school kid. I just entered physics so please explain kindly and simply.




What is the meaning of internal Energy?


What is the meaning of internal energy. Is it the amount of energy that a body contains?



Answer



Standard Definition : The internal energy of a system is identified with the random, disordered motion of molecules; the total (internal) energy in a system includes potential and kinetic energy.


Since this molecular motion is primarily a function of temperature, the internal energy is sometimes called 'thermal energy'.


Total energy of the system $(E)$ = U (Thermal energy) + $E_{Kinetic\space of\space system}$ + $E_{Potential\space of\space system}$ ....



Here, the two terms viz, $E_{Kinetic\space of\space system}$ and $E_{Potential\space of\space system}$, are for the system and not the particles inside the system. For the particles the $KE$ and $PE$ is included in the internal energy.


In most of the cases, kinetic energy, potential energy, chemical energy of the system is constant, thus we conclude,


$\Delta{E}=\Delta{U}$


The proper form of the first law of thermodynamics is given by,


$\Delta E = q + W$,


but for the reasons stated above,


$\Delta U = q + W$.


Therefore, summing it up, Internal energy includes :


1). Translational KE


2). Vibrational and rotational kinetic energy



3). Potential energy from intermolecular forces


Friday 26 February 2016

optics - Would obstacles cause light diffraction if it has the same refractive index as the surrounding material?


Restatement of title:


Would an obstacle still cause diffraction of light if it has the same refractive index as the surrounding material?



Answer



Generally speaking we use the term diffraction when we have some apparatus that blocks part of the light. So for example a diffraction grating absorbs light except at the slits in it. When no light is being absorbed we normally use the term refraction. They're both the same physics, but the distinction is often convenient. Your question implies the obstacle doesn't absorb light, so your question is about refraction not diffraction.


Anyhow, refraction relies on different regions of the light encountering media with different refractive indices. The differing refractive index causes a path dependant phase change in the light that leads to refraction. In the case you describe where the refractive index is everywhere constant there will be no refraction.



This is used in the measurement of refractive indices of powders. If you do first year crystallography at university you are likely to do a practical where you measure the refractive index of a powdered mineral by placing the powder in liquids of different refractive indices and measuring the light absorption. When the refractive index of the powder is the same as the refractive index of the liquid the powder becomes virtually invisible and the light absorption falls to almost zero.


particle physics - Overall symmetry of pion wave function


Mesons are bosons, therefore their wavefunction must be symmetric under particle exchange. Overall, the meson wave function ($\text{WF}$) has the following contributions:


$$\text{WF} = \lvert \text{flavor}\rangle \lvert \text{spin}\rangle \lvert \text{radial}\rangle \lvert\text{color}\rangle.$$


Mesons are a color singlet, their color wavefunction looks like: $$\frac{1}{\sqrt{3}}(\lvert r\bar{r}\rangle + \lvert b\bar{b}\rangle + \lvert g\bar{g}\rangle)$$ which is symmetric.


For the pseudoscalar mesons the spin part is antisymmetric because the spins are a spin singlet.


The radial part is symmetric for the pseudoscalar/vector mesons, because the angular momentum $\ell$ is zero.


When it comes to the flavor part of the wavefunction for the pseudoscalar mesons, it is symmetric as well: for example $$\pi^+: \frac{1}{\sqrt{2}}(\lvert u\bar{d}\rangle + \lvert \bar{d}u\rangle).$$



This gives us an overall antisymmetric wavefunction because of the antisymmetric spins. But the wavefunction has to be symmetric! The same goes for the vector mesons: there we just have an antisymmetric flavor part times symmetric spin. So it is again overall antisymmetric.


What's wrong with this reasoning?




Do differences in physical properties of different substances correspond directly with differences in the energy which composes those substances?


A follow up question to In $E=mc^2,$ does it not matter what constitutes the mass?


Do the different physical properties of chemical elements and compounds corresponds with the different sources of energy which compose of the mass of those elements and compounds?


In answers to my previous question it was clarified that different forms of energy make up the total energy in a gram of sugar vs. a gram of water vs. a gram of lead, but ultimately a 1 gram mass of any of those objects has the same total energy. That has me thinking: the differences in properties between these objects must be accounted for in the differences in forms of energy composing each object. For example, the forms of energy which compose a gram of water (hydrogen bonding, covalent bonding, and more) are different than the forms of energy which compose a gram of salt (ionic bonding and other forces).


Aside from the extreme complexity of the various forms of energy interacting to compose water from elementary particles, is the relationship between energy and mass/matter really this simple? Or are differences in the properties of objects dependant on more than the forms of energy of which they are made of, in which case, what other influences on properties of physical objects are there at play?




Answer



You are confusing things some. Chemical energy and nuclear energy, and other kinds of energy are indeed different forms of energy, due to different forces, or equivalently different kinds of interactions (In physics there are four forms of interactions, the so called 4 forces). They will form different looking pieces of matter, and have different effects on on other matter when say they are mixed, but if the total matter for each case is the same, they have the same total energy.


But this is the important thing. If you somehow take two equal amounts of energy and are able to convert them each totally into matter, their mass will be identical. Exactly given by E = m$c^2$. You have to be careful that you don't add other uncounted energy in. For instance, all that matter in each case has to be at rest, or else you have to add their kinetic energies in. Also, the matter 'things' in each case need to be one clump, or you'd have to account for gravitational and other forces between clumps. So you have to be careful and make sure you didn't miss something (such as neutrinos which are hard to detect).


There is more. The strong equivalence principle says that that mass, from whatever energy, in each clump, say 1 gram of sugar and 1 gram of sand, will exert exactly the same gravitational force. Of course their inertial masses are also identical.


An interesting way to see that the specifics of the forces makes no difference regarding total mass, consider two black holes at rest at some very large distance at rest with masses m and m (to make it easy). They will be attracted to each other (or equivalently in general relativity, GR, moves in geodesics in the spacetime towards each other) and start moving towards each other, accelerating as they get closer. They'll get very close, and if their trajectories are not perfectly in line, rotate around each other a number of orbits, and merge with each other into one larger black hole. The mass of the new black hole will be M, with


$M = 2m - \mu$, and amazingly gravitational waves will be radiated and its total energy at infinity (or far enough away that you can consider it as no longer interacting with the black hole left behind), will be e, and the kinetic energy of the resultant black hole KE, with


$\mu c^2$ = e + KE


The changes in mass and energy were multiple, but the translations between them always follow Einstein's mass energy equation. The mass of the remaining black hole is exactly what some body (small enough to call it a test body, i.e. Ignore its own gravitational field), some astronomical distance away will conclude, from the black hole's gravitational attraction, is the mass of that black hole. The energy that escaped in the gravitational wave and the kinetic energy of the final black hole, make it balance out. Notice that you don't need to know the black hole's internal binding energy (which does not exist, but it would in any other body) or its rotational energy (which exists because their trajectories were not exactly aligned and so there needed to be some angular momentum) as it all goes to define its total mass M.


Yes, mass and energy are numerically equivalent, with the $c^2$ being the translation factor. In advanced physics one uses natural units where c is set to equal 1, so mass and energy are indistinguishable in terms of their total values. We just call it mass energy, sometimes use either term for the other.


Note that the mass or energy may have different forms, and do different things, even when the total value is always the same. So a gram of sand or sugar may have different amounts of nuclear mass and chemical or electromagnetic energy, that will always depends on differences between the two substances, but total masss for both are the same, and if you somehow converted each into total energy, of whatever kind, the total for each would be the same. And their gravitational effects on other matter are exactly equal (even though 1 gram is a small gravitational force).



So for total mass or energy the constituent parts contributions to total mass, or total gravitational field created (assuming they both are point masses as an idealization) by each, are the same. Regardless of internal details. Stil a gram of sugar you may be able to eat ok, you wouldn't a gram ((or say a kg) of sand. Same mass, but Lots and of other differences.


cipher - The Twenty Doors! (ROOM 2)


This is part of The Twenty Doors series.
The previous one is The Twenty Doors! (ROOM 1)
The next one is The Twenty Doors! (ROOM 3)




ROOM 2:


After entering the first door, you find yourself in a bigger room. You look at the keypad.



[+] [-] [X]
[÷] [∞] [±]
[ENTER]

You look around the room. Then, you see the paper.



KDKGEOVB



And, as before, there is a hint carved into the wall - A to Z Alphabet


This time, the answer seems to be a maths symbol. Or symbols. Still, you've got this far, so you may as well try.



HINT 1:



There is only one symbol involved.



HINT 2:



A to Z was not required. It could have been A to B, I to C or anything. Each one changes the encrypted text. Now, which cipher is that?



HINT 3:




It is an eight-letter word.



HINT 4:



AAAABABAAAAABABABAAAAAABB I love bacon.



MEGA HINT:



The cipher used above is called Baconian.




The next door will be added when the current door is solved!



Answer



We should press



INFINITY



Deciphering AAAABABAAAAABABABAAAAAABB gives:



BIFID using the Baconian Cipher




Deciphering KDKGEOVB:



Using Bifid cipher with Alphabet as a key and Translate A to Z gives INFINITY



quantum mechanics - Consequences of the new theorem in QM?



It seems there is a new theorem that changes the rules of the game in the interpretational debate on QM:


http://www.nature.com/news/quantum-theorem-shakes-foundations-1.9392


Does this only leave Bohm,Everett and GRW as possible candidates?



Answer



The paper does not go into details about what interpretations would be disproved by their results. There's a good reason for this: There are no interpretations that would be disproved by their results. They are disproving a straw-man. Here is the central result proved by the paper, phrased in a less obscure way:


"If a system is in the pure state $|+_Z\rangle$ then it is definitely not in some other different pure state $|+_X\rangle$ or whatever."


If this seems obvious and uncontroversial, it is! Admittedly, in the conclusions section, they claim they are saying things that are not obvious...but they're wrong.


Let's start from the beginning. They define the debate by saying there are two pure quantum states, $|\phi_0\rangle$ and $|\phi_1\rangle$. There is one procedure to prepare $|\phi_0\rangle$ and a different procedure to prepare $|\phi_1\rangle$. They say there are two schools of thought. The first school of thought (the correct one) is that "the quantum state is a physical property of the system", so that "the quantum state is uniquely determined by [the physical situation]". That's the one that they will prove is correct. They say the alternative (the incorrect one) is that "the quantum state is statistical in nature", by which they mean "a full specification of [the physical situation] need not determine the quantum state uniquely".


Let's say you have a spin-1/2 system, in state $|+_Z\rangle$. Then...HOLD ON A MINUTE! I just committed myself to the first school of thought! I said the system was really in a certain quantum state!


In fact, everyone doing quantum mechanics is always in the first school of thought, because we say that a system has a quantum state and we do calculations on how the state evolves, etc., if the system is in a pure state. (Not necessarily true for mixed states, as discussed below.)



What would be the second school of thought? You would say, "I went through a procedure which supposedly prepares the system into the pure state $|+_Z\rangle$. But really the system doesn't just have one unique state. It has some probability somehow associated with it. This exact same procedure might well have prepared the state $|+_X\rangle$ or whatever.


Real physicists have a way to deal with this possibility: Mixed states, and the density matrix formalism. If you try to prepare a pure state but don't do a very good job, then you get a mixed state, for example the mixed state which has a 70% chance of being $|+_Z\rangle$ and a 30% chance of being $|+_X\rangle$.


So again, as I said at the start, they have proven the obvious fact: "If a system is in the pure state $|+_Z\rangle$ then it is definitely not in some other different pure state $|+_X\rangle$ or whatever."


With such an obvious and uncontroversial premise, how do they purport to conclude anything that is not totally obvious? Let's go to the conclusions section. They conclude that the "quantum process of instantaneous wave function collapse [is different from] the (entirely nonmysterious) classical procedure of updating a probability distribution when new information is acquired." Indeed, if a spin is in the state $|+_Z\rangle$, then acquiring new information will never put it in the state $|+_X\rangle$. You have to actually do something to the system to change a spin from $|+_Z\rangle$ to $|+_X\rangle$!! For example, you could measure it, apply a magnetic field, etc.


Let's take a more interesting example, an EPR pair in the state $(|++\rangle+|--\rangle)/\sqrt{2}$. After preparing the state, it really truly is in this specific quantum state. If we carefully manipulate it while it's isolated, we can coherently change it into other states, etc. Now we separate the pair. Someone who wants to describe the first spin as completely as possible, but has no access to the second spin, would take a partial trace like usual to get a density matrix. He then gets an email that the second spin is in state +. He modifies his density matrix to the pure state +. You will notice that their example does not show that this so-called collapse violates any laws of quantum mechanics. Their disproof is specific to pure states, and would not work in this mixed state example. Therefore, they cannot conclude that "the quantum collapse must correspond to a real physical process" in the EPR case.


One more example: A spin in state $|+_Z\rangle$, and you measure it in the X-direction. The Schrodinger equation, interpreted with decoherence theory, says that the wavefunction of the universe will coherently evolve into a superposition of (macroscopic measurement of +) and (macroscopic measurement of -). In the paper, they say this in a different way: "each macroscopically different component has a direct counterpart in reality". This is just saying the same thing, but sounds more profound. I should hope that anyone who understands decoherence theory will agree that both of the macroscopic measurements are part of the universe's wavefunction, and that the universe really does have a unitarily-evolving wavefunction even if we cannot see most of it. We rarely care, however, about the wavefunction of the universe; we care only about the branch of the wavefunction that we find ourselves in. And in that branch it is quite reasonable for us to collapse our wavefunctions and to say that the other branches are "not reality" (in a more narrow sense).


-- UPDATE --


I tried to reread the paper in the most charitable way that I can. Now, I think I was a bit too harsh above. Here is what the paper proves:


CENTRAL CLAIM: Say you have a hidden-variables theory, so when you "prepare a pure state $|\psi\rangle$", you actually prepare the state $\{|\psi\rangle,A\}$, where A is the hidden variable which randomly varies each time you prepare the state. It is impossible to have $\{|\psi_0\rangle,A\}=\{|\psi_1\rangle,A'\}$, if $|\psi_0\rangle\neq|\psi_1\rangle$. In other words, the hidden-variable-ensembles of different pure states do not overlap.


They are disproving a straw-man because there is no interpretation of quantum mechanics that asserts that the hidden-variable-ensembles of different pure states must overlap with each other. Even hidden-variable theories do not assert this. There is a so-called "statistical interpretation" in the literature (advocated by L. Ballentine), which also does not assert this.



So this is a straw-man, because nobody ever argued that hidden-variable ensembles of different states ought to overlap. But, it's not a manifestly ridiculous straw-man. At least, I can't think of any much simpler way to prove that claim. (Admittedly, I do not waste my time thinking about hidden-variables theories.) I can imagine that someone who was constructing a new nonlocal-hidden-variables quantum theory might like to know that the hidden-variable-ensembles should not overlap.


logical deduction - The coolest checkerboard magic trick. Version 2


Version 1: The coolest checkerboard magic trick


You and your friend are imprisoned. Your jailer offers a challenge. If you complete the challenge you are both free to go. The rules are



The jailer will take you into a private cell. In the cell will be a triangular chessboard and a jar containing 100 coins.


The jailer will take the coins, one-by-one, and place a coin on each triangle on the board. He will place the coins randomly on the board. Some coins will be heads, and some tails (or maybe they will be all heads, or all tails; you have no idea. It's all at the jailers whim. He may elect to look and choose to make a pattern himself, he may toss them placing them the way they land, he might look at them as he places them, he might not …). If you attempt to interfere with the placing of the coins, it is instant death for you. If you attempt to coerce, suggest, or persuade the jailer in any way, instant death. All you can do it watch.


Once all the coins have been laid out, the jailer will point to one of the triangles on the board and say: “This one!” He is indicating the magic triangle. This triangle is the key to your freedom.


The jailer will then allow you to turn over one coin on the board. Just one. A single coin, but it can be any coin, you have full choice. If the coin you select is a head, it will flip to a tail. If it is a tail it will flip to a head. This is the only change you are allowed to make to the jailers initial layout. You will then be lead out of the room. If you attempt to leave other messages behind, or clues for your friend … yes, you guessed it, instant death!


The jailer will then bring your friend into the room. Your friend will look at the board (no touching allowed), then examine the board of coins and decide which location he thinks is the magic triangle. He gets one chance only (no feedback). Based on the configuration of the coins he will point to one triangle and say: “This one!” If he guesses correctly, you are both pardoned, and instantly set free. If he guesses incorrectly, you are both executed.


The jailer explains all these rules, to both you and your friend, beforehand and then gives you time to confer with each other to devise a strategy for which coin to flip.


What strategy would you use to escape?


enter image description here




soft question - Is it Possible to Determine Radiation Levels Using Satelites?


Given recent events in Japan, this got me wondering. Is it possible to determine radiation levels reliably not having Geiger counters near the possible radiation contaminated zone? According to wikipedia the Chernobyl disaster was first (other than Soviet Union naturally) detected by Swedish via radioactive particles found on clothes of nuclear plant workers. Surely more efficient ways should have been developed by now.



Answer




Short answer: no.


Longer answer: No, excepting neutrinos none of the products of radioactive decay has the penetrating power to pass through the atmosphere, and neutrino detection is not something we can do from satellites.


To elaborate, the immediate products of radioactive decay are (some set of, depending on the decay in question) fission fragments, electrons, positrons, alphas, neutrons, photons (gamma rays) and neutrinos. Plus the remnant nucleus. The only secondary product which might be interesting is Cerenkov light.


The electrons and positrons will travel a number of cm in air (at ground level). The gamma might go a few meters. The heavy stuff has no penetrating power at all.


Even if lofted to the top of the troposphere, there is just too much air in the way.


Cerenkov light will, of course, go through a lot of atmosphere, but you'd be looking for a pale blue glow against the general light background. For dispersed radionucleides (i.e. contamination), the intensity will be awfully low.




N.B. I too have seen various TV show and movie where some character from some agency says "We can track the radiation with satellites!". I believe this to be misinformed babbling of desperate script writers.


Thursday 25 February 2016

homework and exercises - Why are triangles drawn like so when working with gravity on an inclined plane?



This is my first year as a physics student, and I've never learned about vectors past a basic level, so this is confusing me. When we have gravity on an inclined plane, we separate it into two components, which I understand. However, consider the image below, and there's a box at point A. When separating the gravity components, you draw a triangle AGC (sorry, the G and D are on top of each other and difficult to distinguish). AG becomes the force of gravity in the y-direction, and GC becomes the force of gravity in the x-direction. Then you do the trig functions from there. However, when I tried this myself, I drew triangle ADF instead and tried the trig functions from there. It didn't work. I'm having trouble understanding why you can't compute the trig functions from AGF. The only partial solution I came up with was that force of gravity in the y-direction can't be the hypotenuse, as the force of gravity in the y-direction is always less than the force of gravity. But I think I'm missing something more. enter image description here



Answer



You can decompose the forces along AD and DF rather than AG and GC, but they won't be the relevant forces you're looking for. What makes this confusing is we often work entirely in magnitudes, whereas fundamentally we are manipulating vectors, and we only get away with this with good choices of decompositions.


Suppose the downward force of gravity is $\vec{f}$ with magnitude $f$ and direction along AC. The "right" method will say there is a normal force $\vec{f}_{\!\perp}$ in the direction of AG and a parallel force $\vec{f}_{\!\Vert}$ in the direction of GC, with $\vec{f} = \vec{f}_{\!\perp} + \vec{f}_{\!\Vert}$. The formulas for the magnitudes are $$ f_{\!\perp} = f \cos\beta, \qquad f_{\!\Vert} = f \sin\beta. $$ Moreover, because AG and GC are orthogonal, we know $\vec{f}_{\!\Vert}$ cannot have any effect on pushing into the inclined plane; all such effects are captured by $\vec{f}_{\!\perp}$.


Now consider the "wrong" decomposition of $\vec{f} = \vec{f}_1 + \vec{f}_2$, with $\vec{f}_1$ along AD and $\vec{f}_2$ along DF. We can get these magnitudes too: $$ f_1 = f \sec\beta, \qquad f_2 = f \tan\beta. $$ The problem is, $\vec{f}_1$ doesn't fully capture the normal force, because $\vec{f}_2$ contributes to this as well.


For an extreme example, imagine a mass sitting on horizontal ground with weight $\vec{f}$ directed downward. We could write $\vec{f} = \vec{f}_1 + \vec{f}_2$ with both $\vec{f}_1$ and $\vec{f}_2$ also pointing downward. We cannot just look at $\vec{f}_1$ and neglect $\vec{f}_2$ when considering the weight of the mass on the ground.


Another way of looking at things is that we are silently taking dot products. The real, unambiguous definition of the normal force of the mass on the block is the dot product of its weight vector with the unit normal vector to the surface, $\vec{f}_{\!\perp} = (\vec{f} \cdot \hat{n}) \hat{n}$ (give or take a sign). When computing dot products, you can only ignore components of $\vec{f}$ that are orthogonal to $\hat{n}$; if you do a decomposition where neither component is orthogonal, you have to include both terms. In equations, this is the difference between $$ \vec{f} \cdot \hat{n} = \vec{f}_{\!\perp} \cdot \hat{n} + \vec{f}_{\!\Vert} \cdot \hat{n} = (f_{\!\perp}) (1) \cos0^\circ + (f_{\!\Vert}) (1) \cos90^\circ = f_{\!\perp}. $$ and $$ \vec{f} \cdot \hat{n} = \vec{f}_1 \cdot \hat{n} + \vec{f}_2 \cdot \hat{n} = (f_1) (1) \cos0^\circ + (f_2) (1) \cos69.91^\circ. $$


wordplay - Some four word puzzles (and possible bounty)


A four word puzzle (see this and this) is basically a 4x4 grid containing 4 4-letter words, and reads the same top to bottom or left to right. For example,


A B C D
B E F G
C F H I

D G I J

Here are some puzzles with hints (not necessarily in the same order):


Sample puzzle


a shape
an identity
a valley
an audio track


Answer


S O N G

O V A L
N A M E
G L E N

Puzzle 1


a flower
a plant
a stage in life
a fraction of a whole


Puzzle 2



a medium of writing
an examination
a mark joining two points
a narrow cut


Puzzle 3


a kind of footwear
a kind of sensation
obedience
squashing things together


Puzzle 4



stack
volume
termination
a metal


Possible bounty


I know it's quite unlikely, but if a single person manages to be the first one to solve 3 of the 4 puzzles, you will get a bounty of 100 rep (plus up-votes from many people).



Answer



Puzzle 1



PART - a fraction of a whole

ALOE - a plant
ROSE - a flower
TEEN - a stage in life



Puzzle 3



MASH - squashing things together
ACHE - a kind of sensation
SHOE - a kind of footwear
HEED - obedience




Puzzle 4



PILE - stack
IRON - a metal
LOUD - volume
ENDS - termination



Wednesday 24 February 2016

mathematics - Can you find this abundant dad and his six prime sons?


We are six little prime brothers. All our ages add up to the age of our abundant and old dad who is obviously not in his prime.


However, if you pair him up with either of the prime neighbors, they can have a prime party.


Additionally, any five of our ages add up to a prime.


Deductive logic and small calculations might be enough to find us. Who are we?



Based on some initial comments, I will add few more comments: none of the brothers were born at the same time, and their dad is abundant and has seen all the Super Bowls since their inception.



Answer



The sons are



5, 7, 11, 19, 29, and 37. (All primes.)



The dad is



5+7+11+19+29+37 = 108 (an abundant non-prime number).
Any five of the sons' ages sum to a prime (71, 79, 89, 97, 101, and 103).

His neighbors are both primes (107 and 109).
The "parties" (concatenations of the father's age with a neighbor)
    are 108107 and 108109 — also primes.
Our virile father then sired his youngest at the venerable age of 103. Impressive!

To address the added comments, the sons' ages are all unique, the father's age is an abundant number, and the father is old enough to have seen the first Super Bowl in 1967.



mathematics - Reassembling the Marquetry


You decided to visit a woodworker's home. This woodworker is very famous for their skills in marquetry - that is, setting pieces of wood to create patterns - and had a piece of a very geometrical nature assembled on top of their work bench, not yet glued together.


Unfortunately, you are a rambunctious young whippersnapper with no patience for these old things, so you decide to play an impromptu game of "the floor is lava" while in the workshop. In the course of this game, you accidentally knock the whole workbench over, spilling the contents all over the floor and causing the assembled design to scatter in many pieces.


You notice that the pieces are of three varieties as pictured and described below: enter image description here




  • There are $24$ square pieces. Let us say they have side length $1$.





  • There are $40$ rhombuses. Each rhombus has two angles measuring $45^{\circ}$ and two measuring $135^{\circ}$. All their side lengths are $1$ as well.




  • There are $8$ isosceles right triangles. They have two sides of length $1$ and one side of length $\sqrt{2}$.




You decide that, in order to avoid trouble, you should reassemble the pattern based on the brief glimpse you got of the assembled piece before you carelessly careened into it. You recall three things: Firstly, the pieces fit together to form a regular octagon, leaving no empty space inside the octagon nor any overlap of multiple tiles. Secondly, you recall that squares did not border* other squares, nor did triangles border triangles, or squares border triangles. Finally, you recall that if two rhombuses bordered each other, then they were in rotated differently - that is, two bordering rhombi were never mere translations of each other.


You figure that as long as you reassemble a pattern satisfying all that, the woodworker will probably not notice any difference. How can these tiles be put together to form a suitable octagon?


(*"Border" here means "share an edge with in the completed tiling")



Answer





Reassembled marquetry



(All credit for the neatness of the diagram goes to 2012rcampion -- the diagram I posted was a mess and he, entirely unprompted, replaced it with the rather nice one now seen above. Thanks, 2012rcampion!)


This was largely trial and error, beginning with the observations that



the side length has to be $2+\sqrt{2}$ and therefore we need one triangle-hypotenuse on each side



and (to save effort)




looking only for highly symmetrical solutions.



The first thing I tried was to



put a triangle in the middle of each edge and a rhombus on each side, but that is readily seen not to work. Next was the pattern above, starting at the edges and maintaining rotational symmetry; doing what felt most natural at each stage produced an octagonal ring with a smaller internal octagon to be completed; then there was only one halfway plausible way to place the squares, after which everything else was forced.



wavefunction - Modern explanation of the Young experiment with Quantum Field Theory?


In the Young double slit experiment it is possible to detect the arrival of individual photons as well as an interference pattern.


It doesn't makes much sense to me that something could be either a particle either a wave, neither the use of the wave function collapse to explain individual photon observation.


I have therefore been reading answers to this question on stack exchange, and quite reasuringly, wave-particle duality is considered wrong. Quantum field theory is refered as the correct answer, however it is never explained in any details.



What is the (advanced) layman explanation of Young experiment with QFT?


Thinking about this paradox, I found an explanation that I believe is convincing:



  • there is no individual particle travelling, only waves

  • waves only probabilistically interact with matter (like it is the case with neutrinos)

  • waves have a certain probability to interact with matter which depend on the energy at a given physical location (which depend on waves superposition) and on the material used to build the detector (should not be transparent at the wavelength used by the coherent source)


Is this explanation consistent with QFT?




Tuesday 23 February 2016

electromagnetism - what the explicit formula magnetic field a electron that moving on the curve path?


,hi,we suppose ,we have a single electron that moving on a curve path,the velocity is v (it is variable),the path moving is a curve not direct path.i saw maxwell equation my question is ,is there a explicit( not implicit ) equation for magnetic filed a single electron that moving on curve path.


thanks for reply



Answer



Maxwell alone doesn't specifically attribute an electric or magnetic field due to any specific charge. But an example of a solution to Maxwell can be provided if both the electric and magnetic field are computed as the electric and magnetic parts of the electromagnetic field given by Jefimenko's equations:


$$\vec E(\vec r,t)=\frac{1}{4\pi\epsilon_0}\int\left[\frac{\rho(\vec r',t_r)}{|\vec r -\vec r'|}+\frac{\partial \rho(\vec r',t_r)}{c\partial t}\right]\frac{\vec r -\vec r'}{|\vec r -\vec r'|^2}\; \mathrm{d}^3\vec{r}' -\frac{1}{4\pi\epsilon_0c^2}\int\frac{1}{|\vec r-\vec r'|}\frac{\partial \vec J(\vec r',t_r)}{\partial t}\mathbb{d}^3\vec r'$$ and $$\vec B(\vec r,t)=\frac{\mu_0}{4\pi}\int\left[\frac{\vec J(\vec r',t_r)}{|\vec r -\vec r'|^3}+\frac{1}{|\vec r -\vec r'|^2}\frac{\partial \vec J(\vec r',t_r)}{c\partial t}\right]\times(\vec r -\vec r')\mathbb{d}^3\vec r'$$ where $t_r$ is actually a function of $\vec r'$, specifically $t_r=t-\frac{|\vec r-\vec r'|}{c}.$


These reduce to Coulomb and Biot-Savart only when those time derivatives are exactly zero, which is statics. So Jefimenko is an example of proper time dependent laws for the electromagnetic field.



If you have a single charge you can consider the situation where the charge is a Delta function and the current is a Delta function times the velocity.


You will find that the term that depends on the change in current requires knowing the acceleration of the charge. And you'll notice that every point in space requires knowing the location and the velocity and acceleration of the charge at different (and earlier) times in the past. And none of them depend on the charge right now.


Simply put. No one can magically see where the charge is now. And no fields are affected by that. Instead, for every place $\vec r$ where the charge isn't located you make a backwards lightcone, and see where, $\vec w,$ it intersects the worldline of the charge. Then you take the acceleration, $\vec a,$ velocity, $\vec v,$ and position $\vec w$ of the charge at that time and compute the electric field and magnetic field at that place $\vec r$ based on those three vectors. To compute the fields you can use $\vec d=\vec r -\vec w$ and $\vec u=c\hat d-\vec v$ and get (adapted from Griffiths' Introduction to electrodynamics):


$$\vec E(\vec r,t)=\frac{q}{4\pi\epsilon_0}\frac{d}{(\vec d\cdot \vec u)^3}\left[(c^2-v^2)\vec u+\vec d\times(\vec u\times \vec a)\right]$$


and $$\vec B(\vec r,t)=\frac{1}{c}\hat d\times \vec E(\vec r,t).$$


Recall that the $t$ is not the time when the charge had the position $\vec w,$ velocity $\vec v,$ and acceleration $\vec a.$ Those are at an earlier time, the time back when the charge broadcasted its position $\vec w,$ velocity $\vec v,$ and acceleration $\vec a$ just in time to arrive at $\vec r$ at time $t.$


condensed matter - Yet another question on the Lindhard function


Here's another question concerning the Lindhard function as used in the physical description of metals.


First we define the general Lindhard function in the Random Phase approximation as $\chi(q,\omega)=\frac{\chi_{0}}{1-\frac{4\pi e^2}{q^2}\chi_{0}}$ where $\chi_{0}$ denotes the "bare" Lindhard function" excluding any feedback effects of electrons. A reference is Kittel's book "Quantum Theory of solids" Chapter 6, however he is using the dielectric function instead of the Lindhard function.


By writing $\chi_{0}(q,\omega)=\chi_{01}+i\chi_{02}$ as a complex number and using the relation $\frac{1}{z-i\eta} = P(1/z) + i\pi\delta(z)$ in the limit that $\eta$ goes to zero from above and P(..) denotes the principal value it is straight forward to see that the imaginary part of the lindhard function is zero except of the case when we have electron-hole excitations of the form $\hbar\omega=\epsilon_{k+q}-\epsilon_{k}$.


Now let's assume we study plasma excitations using the actual Lindhard function (not the bare one). These plasma excitations appear for small q outside of the electron hole continuum. Therefor we have a vanishing imaginary part of $\chi_{0}$ in this case and consequently since also a vanishing part of $\chi$ itself, since its a function of $\chi_{0}$ which is "real" except of the "$\chi_{0}$"-part.



However when deriving the plasma excitations using a small q expansion we obtain a none vanishing imaginary part of $\chi$ as given in equation (3.48) of the following document: http://www.itp.phys.ethz.ch/education/lectures_fs10/Solid/Notes.pdf


I don't see why this is consistent.


I'd be more than happy if you could help me.


Thanks in advance.




homework and exercises - Why is electric field zero inside a hollow metal sphere ?



The figure below shows a hollow metal sphere with a positive point charge $Q$ sitting outside it. What is the electric field at the center of sphere ? The answer is zero (look at here at the beginning of page 4), but I do not understand why?


enter image description here




Monday 22 February 2016

thermodynamics - Why is $PV^gamma$ constant in an adiabatic process?


In non isolated systems where there is no adiabatic process, $PV$ is constant. But the graph gets steeper in adiabatic process because of the $\gamma$ over the $V$. Why is it there in adiabatic processes and why only over the $V$?





dimensional analysis - Question about Radian as a unit


I'm having a hard time trying to understand the units between angular velocity and basic velocity of a circle. For angular velocity the units are Radian(s) per second(s) or degree(s) per second(s). The speed or velocity of the circles circumference is the angular velocity times the radius, but the units for this is meter(s) per second(s). So where did the radian go? It counts as a unit for angular velocity but why it doesn't count for the speed?



Answer



Actually the radian does not have units as it is defined as the ratio arclength/radius. Since the arclength is a distance and thus has units of meters, and the radius is also in meters, the ratio is dimensionless.


In particular for angular opening $\theta$ the arclength is $r\theta$ for a circle of radium $r$, and going around the circle in full once give a ratio $2\pi r/r=2\pi$ rad, where the arclength is the full circumference in this case.



Are all instances of quantum non-locality problem artifacts of the use of classical concepts in quantum physics?


Consider experiments involving entangled spins, say two-spin 1/2 particles in the singlet state:


$$\left|\psi\right> =\frac{1}{\sqrt{2}}\left[\left|\uparrow\downarrow\right> - \left|\downarrow\uparrow\right>\right]$$


where Alice is going to measure one spin in the $z$-direction and Bob the other spin in the same $z$-direction. Alice and Bob are space-like separated at the time they perform their measurements. There then seems to be a (benign) locality paradox. Suppose that in some frame, Alice performs her measurement first. She then knows what Bob will find, but how can the information of Bob's result already exist before Bob makes his measurement?


The problem here is that we know that there are no local hidden variables underlying quantum mechanics. John Bell demonstrated this by showing that a local hidden variables theory gives rise to certain inequalities (now know as the Bell inequalities) for certain correlations of spin measurements of entangled spins, while quantum mechanics implies that these inequalities can be violated in certain cases (the same singlet state we're considering here, can show correlations that do not satisfy the Bell inequalities). later the violations of these inequalities was verified in an experiment conducted by Alan Aspect.


So, the non-existence of local hidden variables means that the information that Alice has about what Bob will find cannot also be present locally at Bob's place. Now, in principle, there is no problem in principle with information about this popping up non-locally at Alice's place. One cannot exploit this to allow for faster than light communications. Also we can note that non-local hidden variable theories can be compatible with quantum mechanics. But this does go against the spirit of locality. It then seems that there is an inherent benign non-local aspect to quantum theory.


However, in the MWI, there is no non-locality at all. After Alice measures her spin, there are two copies of her and both copies are equally for for Bob, so it's not true that in the frame where Alice has made her measurement and Bob hasn't, that the information about what Bob will find already exists in Alice's place.


Now, the whole point of the MWI is to treat observers and measurement devices in a quantum mechanical way too instead of in an ad-hoc classical way and thereby to get rid of the collapse of the wavefunction. But if we do treat observers in a classical way and stick to a single universe theory, then we do get a non-locality problem, albeit a benign one.



Then having resolved the source of the apparent non-locality in this case as a classical spanner thrown into the quantum world, we can ask if this is always the case. E.g. we can read here that Lev Vaidman claims that this is also true for the famous Aharonov-Bohm effect, but also that this is in dispute. Vaidman's argument is rather technical, but he argues that ultimately the whole non-locality issue there is artifact of using classical potentials.


Is there an example of a non-local aspect of quantum mechanics that cannot explained away as an artifact of a classical concept?




conformal field theory - Correlator of a single vertex operator


In any textbook on CFT vertex operators $V_\alpha(z,\bar{z})=:e^{i\alpha\phi(z,\bar{z})}:$ are introduced for the free boson field $\phi(z,\bar{z})$ and their correlation function is computed $\left\langle V_{\alpha_1}(z_1,\bar{z_1})\dots \right\rangle=\prod_{i. Also, this equation holds only if $\sum_i{\alpha_i}=0$, otherwise the correlator is zero.


Consider now the correlator of a single vertex operator $\left\langle V_{\alpha}(z,\bar{z}) \right\rangle$. On the one hand, it should vanish as failing the neutrality condition. On the other hand, its expansion is $\left\langle V_{\alpha}(z,\bar{z}) \right\rangle=\left\langle 1 \right\rangle+\sum_{n>0}\frac{(i\alpha)^n}{n!}\left\langle :\phi(z,\bar{z})^n: \right\rangle$. My understanding is that all $n>0$ terms vanish by definiton of normal ordering, but why does the $n=0$ term, which is the identity, also give zero?




Answer



I don't think it's the case that all $n>0$ terms vanish, because the mode expansion of $\phi$ has a zero mode $\phi_0$. Its expansion is


\begin{equation}\phi \left(z,\bar{z}\right) = \phi_0 - i\pi_0 \log\left(z\bar{z}\right) +i \sum_{n\neq 0} \frac{1}{n} \left(a_n z^{-n} + \bar{a}_n \bar{z}^{-n}\right)\end{equation}


Computing $\langle:\phi^n:\rangle$ for $n>0$, the only term that contributes when we take the vacuum expectation value is $\phi_0^n$. This is because $a_n$ and $\bar{a}_n$ annihilate the vacuum for $n>0$, and $\pi_0|0\rangle=0$ as well. Any cross-terms involving $a_n$ and $a_{-m}$ will be zero due to the normal ordering, as will any terms involving $\phi_0$ and $\pi_0$ (as $\pi_0$ is placed to the right).


As a result, we just get \begin{equation} \langle V_\alpha \left(z\right) \rangle =\langle \sum_{n} \frac{\left(i\alpha \phi_0\right)^n}{n!} \rangle= \langle e^{i\alpha \phi_0} \rangle. \end{equation} Because of the commutation relations between $\pi_0$ and $\phi_0$, $e^{i\beta \phi_0} |\alpha\rangle = |\alpha+\beta\rangle$, so the vacuum expectation value is $\langle e^{i\alpha \phi_0}\rangle = \delta_{\alpha,0}$; this is just the charge neutrality condition.


It's easier to obtain this result by using the definition of normal ordering [see e.g. Di Francesco]; \begin{equation}V_\alpha = \exp\left(i\alpha \phi_0 + \alpha \sum_{n>0} \frac{1}{n}\left(a_{-n}z^n + \bar{a}_{-n} \bar{z}^n\right)\right) \exp \left(\alpha \pi_0 \log\left(z\bar{z}\right) - \alpha \sum_{n>0}\frac{1}{n} \left(a_{n}z^{-n} + \bar{a}_{n} \bar{z}^{-n}\right)\right).\end{equation} The last exponential acts trivially on $|0\rangle$, and the $a_{-n},\bar{a}_{-n}$ with $n>0$ map $|0\rangle$ on to its descendants, which are orthogonal to $|0\rangle$. So when taking the vacuum expectation value, the operator is just $e^{i\alpha \phi_0}$ as before.


Alternatively, one can use the Ward identities; the Ward identity for translational invariance $\partial_z \langle V_{\alpha} \left(z\right)\rangle = 0$ means the correlator is constant. The Ward identity $ \left(z\partial_z + h_{\alpha}\right) \langle V_{\alpha}\left(z\right)\rangle =0$ then implies that $h_\alpha \langle V_{\alpha} =0 \rangle$: since $h_\alpha = \alpha^2/2$ is non-zero for $\alpha \neq 0$, the correlator must be zero. If $\alpha=0$, $V_{\alpha} = 1$ and the correlator is just 1.


causality - The transactional interpretation of quantum mechanics


John Cramer’s transactional interpretation of quantum mechanics (TIQM) is billed as resolving the fuzzy agnosticism of the Copenhagen interpretation while avoiding the alleged ontological excesses of the Many Worlds Interpretation. Yet it has a low profile.



Is this because no-one care anymore about ontology in physics, or is there something about TIQM which undermines belief in it?



Answer



Nobody has explained to me how Shor's quantum factorization algorithm works under the transactional interpretation, and I expect this is because the transactional interpretation cannot actually explain this algorithm. If it can't, then chances are the transactional interpretation doesn't actually work. (I have looked at some of the papers that purport to explain the transactional interpretation, and have found them exceedingly vague about the details of this interpretation, but assuming this interpretation is actually valid, maybe somebody else with more determination could figure these details out.)


experimental physics - Energy for burning things!


I have a cylindrical lens & a surface that can be burned.( for example wood , sheet ,...) In a sunny day , I wanna burn the surface but I don't know the "Energy for burning surfaces" If the $Power$ of sun is $1400 W/m^2$ , could you tell me how can I burn them?




Sunday 21 February 2016

pattern - What is a Southpaw Word™?



This is in the spirit of the What is a Word/Phrase™ series started by [JLee][2] with a special brand of [Phrase™][3] and [Word™][4] puzzles.




If a word conforms to a special rule, I call it a Southpaw Word™.


Use the following examples below to find the rule.


Southpaw Words


And, if you want to analyze, here is a CSV version:


SOUTHPAW WORD™, NOT SOUTHPAW WORD™
FAST, QUICK
SCREW, NAIL
EVADERS, DEFLATERS

SWEET, SOUR
TREATS, TRICKS
AVERTED, WELCOMED
REAR, FRONT
CREED, MANTRA
@45266, COMMUNITY

Answer



A Southpaw word



is typed with only the left hand on a keyboard.




units - What are the proposed realizations in the New SI for the kilogram, ampere, kelvin and mole?


The metrology world is currently in the middle of overhauling the definitions of the SI units to reflect the recent technological advances that enable us to get much more precise values for the fundamental constants of nature than were possible when the SI was drawn up. This has already happened to the second and the meter, which are defined in terms of a caesium transition and the speed of light, and it is being extended to the other units. Thus, in the new system, known as the 'new SI',



four of the SI base units, namely the kilogram, the ampere, the kelvin and the mole, will be redefined in terms of invariants of nature; the new definitions will be based on fixed numerical values of the Planck constant ($h$), the elementary charge ($e$), the Boltzmann constant ($k$), and the Avogadro constant ($N_A$), respectively.



The proposed draft of the SI brochure gives more details, but it stops short of describing the recommended mises en pratique. For example, for the kilogram, the definition goes




The SI unit of mass, the kilogram



  • The kilogram, symbol kg, is the SI unit of mass; its magnitude is set by fixing the numerical value of the Planck constant to be exactly 6.626 069 57 10−34 when it is expressed in the SI unit for action J s = kg m2 s-1.


Thus we have the exact relation h = 6.626 069 57 10−34 kg m2 s-1 = 6.626 069 57 10−34 J s. Inverting this equation gives an exact expression for the kilogram in terms of the three defining constants $h$, $\Delta \nu$(133Cs)hfs and $c$: $$ \mathrm{kg} =\left(\frac{h}{6.626 069 57\times10^{−34}}\right)\mathrm{m}^{2}\:\mathrm s^{-1} =1.475521\ldots\times 10^{40}\frac{h \Delta \nu\left(^{133}\mathrm{Cs}\right)_\mathrm{hfs}}{c^2} $$ The Planck constant is a constant of nature, whose value may be expressed as the product of a number and the unit joule second, where J s = kg m2 s-1. The effect of this definition is to define the unit kg m2 s-1 (the unit of both the physical quantities action and angular momentum), and thus together with the definitions of the second and the metre this leads to a definition of the unit of mass expressed in terms of the value of the Planck constant $h$.


Note that macroscopic masses can be measured in terms of $h$, using the Josephson and quantum-Hall effects together with the watt balance apparatus, or in terms of the mass of a silicon atom, which is accurately known in terms of $h$ using the x-ray crystal density approach.



However, the brochure is pretty scant as to what the specific realizations through watt balances actually imply in terms of a route from measured physical quantities to values of fundamental constants or to inferred masses. For the specific case of the watt balance, for example, the physical constants at play are much more naturally the Josephson and von Klitzing constants, $K_J=2e/h$ and $R_K=h/e^2$, if I understand correctly, so there is some re-shuffling of experimental results to be done.


The SI brochure is similarly vague for the other three base unit / fundamental constant pairs.


This brings me, then, to my specific questions. For each of these four base unit / fundamental constant pairs,





  • what are the proposed experimental realizations, what are the basics of their operation, and what physical effects do they rely on?




  • what other fundamental constants are used to go from experimentally measured values to inferred parameters? (i.e. the meter depends on the second. Does the kilogram depend on the value of the electric charge?)




  • what specific natural constants are measured by the experiment, and how are they reshuffled to obtain final results?





Additionally, what is the dependency tree between the different definitions of the base units? What units depend on what others, either directly or indirectly?



Answer



So the BIPM has now released drafts for the mises en pratique of the new SI units, and it's rather more clear what the deal is. The drafts are in the New SI page at the BIPM, under the draft documents tab. These are drafts and they are liable to change until the new definitions are finalized at some point in 2018. At the present stage the mises en pratique have only recently cleared consultative committee stage, and the SI brochure draft does not yet include any of that information.


The first thing to note is that the dependency graph is substantially altered from what it was in the old SI, with significantly more connections. A short summary of the dependency graph, both new and old, is below.


$\ \ $


In the following I will explore the new definitions, unit by unit, and the dependency graph will fill itself as we go along.


The second


The second will remain unchanged in its essence, but it is likely that the specific reference transition will get changed from the microwave to the optical domain. The current definition of the second reads




The second, symbol $\mathrm{s}$, is the SI unit of time. It is defined by taking the fixed numerical value of the caesium frequency $\Delta\nu_\mathrm{Cs}$, the unperturbed ground-state hyperfine splitting frequency of the caesium 133 atom, to be $9\,192\,631\,770\:\mathrm{Hz}$, where the SI unit $\mathrm{Hz}$ is equal to $\mathrm{s}^{–1}$ for periodic phenomena.



That is, the second is actually implemented as a frequency standard: we use the resonance frequency of a stream of caesium atoms to calibrate microwave oscillators, and then to measure time we use electronics to count cycles at that frequency.


In the new SI, as I understand it the second will not change, but on a slightly longer timescale it will change from a microwave transition to an optical one, with the precise transition yet to be decided. The reason for the change is that optical clocks work at higher frequencies and therefore require less time for comparable accuracies, as explained here, and they are becoming so much more stable than microwave clocks that the fundamental limitation to using them to measure frequencies is the uncertainty in the standard itself, as explained here.


In terms of practical use, the second will change slightly, because now the frequency standard is in the optical regime, whilst most of the clocks we use tend to want electronics that operate at microwave or radio frequencies which are easier to control, so you want a way to compare your clock's MHz oscillator with the ~500 THz standard. This is done using a frequency comb: a stable source of sharp, periodic laser pulses, whose spectrum is a series of sharp lines at precise spacings that can be recovered from interferometric measurements at the repetition frequency. One then calibrates the frequency comb to the optical frequency standard, and the clock oscillator against the interferometric measurements. For more details see e.g. NIST or RP photonics.


The meter


The meter will be left completely unchanged, at its old definition:



The metre, symbol $\mathrm{m}$, is the SI unit of length. It is defined by taking the fixed numerical value of the speed of light in vacuum $c$ to be $299\,792\,458\:\mathrm{m/s}$.




The meter therefore depends on the second, and cannot be implemented without access to a frequency standard.


It's important to note here that the meter was originally defined independently, through the international prototype meter, until 1960, and it was to this standard that the speed of light of ${\sim}299\,792\,458 \:\mathrm{m/s}$ was measured. In 1983, when laser ranging and similar light-based technologies became the most precise ways of measuring distances, the speed of light was fixed to make the standard more accurate and easier to implement, and it was fixed to the old value to maintain consistency with previous measurements. It would have been tempting, for example, to fix the speed of light at a round $300\,000\,000 \:\mathrm{m/s}$, a mere 0.07% faster and much more convenient, but this would have the effect of making all previous measurements that depend on the meter incompatible with newer instruments beyond their fourth significant figure.


This process - replacing an old standard by fixing a constant at its current value - is precisely what is happening to the rest of the SI, and any concerns about that process can be directly mapped to the redefinition of the meter (which, I might add, went rather well).


The ampere


The ampere is getting a complete re-working, and it will be defined (essentially) by fixing the electron charge $e$ at (roughly) $1.602\,176\,620\times 10^{–19}\:\mathrm C$, so right off the cuff the ampere depends on the second and nothing else.


The current definition is couched on the magnetic forces between parallel wires: more specifically, two infinite wires separated by $1\:\mathrm m$ carrying $1\:\mathrm{A}$ each will attract each other (by definition) by $2\times10^{-7}\:\mathrm{N}$ per meter of length, which corresponds to fixing the value of the vacuum permeability at $\mu_0=4\pi\times 10^{-7}\mathrm{N/A^2}$; the old standard depends on all three MKS dynamical standards, with the meter and kilogram dropped in the new scheme. The new definition also shifts back to a charge-based standard, but for some reason (probably to not shake things up too much, but also because current measurements are much more useful for applications) the BIPM has decided to keep the ampere as the base unit.


The BIPM mise en pratique proposals are a varied range. One of them implements the definition directly, by using a single-electron tunnelling device and simply counting the electrons that go through. However, this is unlikely to work beyond very small currents, and to go to higher currents one needs to involve some new physics.


In particular, the proposed standards at reasonable currents also make use of the fact that the Planck constant $h$ will also have a fixed value of (roughly) $6.626\,069\times 10^{−34}\:\mathrm{kg\:m^2\:s^{-1}}$, and this fixes the value of two important constants.





  • One is the Josephson constant $K_J=2e/h=483\,597.890\,893\:\mathrm{GHz/V}$, which is the inverse of the magnetic flux quantum $\Phi_0$. This constant is crucial for Josephson junctions, which are thin links between superconductors that, among other things, when subjected to an AC voltage of frequency $\nu$ will produce discrete jumps (called Shapiro steps) at the voltages $V_n=n\, \nu/K_J$ in the DC current-voltage characteristic: that is, as one sweeps a DC voltage $V_\mathrm{DC}$ past $V_n$, the resulting current $I_\mathrm{DC}$ has a discrete jump. (For further reading see here, here or here.)


    Moreover, this constant gives way directly to a voltage standard that depends only on a frequency standard, as opposed to a dependence on the four MKSA standards as in the old SI. This is a standard feature of the new SI, with the dependency graph completely shaken for the entire set of base plus derived units, with some links added but some removed. The current mise en pratique proposals include stabs at most derived units, like the farad, henry, and so on.




  • The second constant is the von Klitzing constant $R_K = h/e^2= 25\,812. 807\,557 \:\Omega$, which comes up in the quantum Hall effect: at low temperatures, an electron gas confined to a surface in a strong magnetic field, the system's conductance becomes quantized, and it must come as integer (or possibly fractional) multiples of the conductance quantum $G_0=1/R_K$. A system in the quantum Hall regime therefore provides a natural resistance standard (and, with some work and a frequency standard, inductance and capacitance standards).




These two constants can be combined to give $e=K_J/2R_K$, or in more practical terms one can implement voltage and resistance standards and then take the ampere as the current that will flow across a $1\:\Omega$ resistor when subjected to a potential difference of $1\:\mathrm V$. In more wordy language, this current is produced at the first Shapiro voltage step of a Josephson junction driven at frequency $483.597\,890\,893\:\mathrm{THz}$, when it is applied to a resistor of conductance $G=25\,812. 807\,557\,G_0$. (The numbers here are unrealistic, of course - that frequency is in the visible range, at $620\:\mathrm{nm}$ - so you need to rescale some things, but it's the essentials that matter.


It's important to note that, while this is a bit of a roundabout way to define a current standard, it does not depend on any additional standards beyond the second. It looks like it depends on the Planck constant $h$, but as long as the Josephson and von Klitzing constants are varied accordingly then this definition of the current does not actually depend on $h$.


Finally, it is also important to remark that as far as precision metrology goes, the redefinition will change relatively little, and in fact it represents a conceptual simplification of how accurate standards are currently implemented. For example, NPL is quite upfront in stating that, in the current metrological chain,




All electrical measurements below 10 MHz at NPL are traceable to two quantum standards: the quantum Hall effect (QHE) resistance standard and the Josephson voltage standard (JVS).



That is, modern practical electrical metrology has essentially been implementing conventional electrical units all along - units based on fixed 'conventional' values of $K_J$ and $R_K$ that were set in 1990, denoted as $K_{J\text{-}90}$ and $R_{K\text{-}90}$ and which have the fixed values $K_{J\text{-}90} = 483.597\,9\:\mathrm{THz/V}$ and $R_{K\text{-}90} = 25\,812.807\:\Omega$. The new SI will actually heal this rift, by providing a sounder conceptual foundation to the pragmatic metrological approach that is already in use.


The kilogram


The kilogram is also getting a complete re-working. The current kilogram - the mass $M_\mathrm{IPK}$of the international prototype kilogram - has been drifting slightly for some time, for a variety of reasons. A physical-constant-based definition (as opposed to an artefact-based definition) has been desired for some time, but only now does technology really permit a constant-based definition to work as an accurate standard.


The kilogram, as mentioned in the question, is defined so that the Planck constant $h$ has a fixed value of (roughly) $6.626\,069\times 10^{−34}\:\mathrm{kg\:m^2\:s^{-1}}$, so as such the SI kilogram will depend on the second and the meter, and will require standards for both to make a mass standard. (In practice, since the meter depends directly on the second, one only needs a time standard, such as a laser whose wavelength is known, to make this calibration.)


The current proposed mise en pratique for the kilogram contemplates two possible implementations of this standard, of which the main is via a watt balance. This is a device which uses magnetic forces to hold up the weight to be calibrated, and then measures the electrical power it's using to determine the weight. For an interesting implementation, see this LEGO watt balance built by NIST.


To see how these devices can work, consider the following sketch, with the "weighing mode" on the right.




Image source: arXiv:1412.1699. Good place to advertise their facebook page.


Here the weight is attached to a circular coil of wire of length $L$ that is immersed in a magnetic field of uniform magnitude $B$ that points radially outwards, with a current $I$ flowing through the wire, so at equilibrium $$mg=F_g=F_e=BLI.$$ This gives us the weight in terms of an electrical measurement of $I$ - except that we need an accurate value of $B$. This can be measured by removing the weight and running the balance on "velocity mode", shown on the left of the figure, by moving the plate at velocity $v$ and measuring the voltage $V=BLv$ that this movement induces. The product $BL$ can then be cancelled out, giving the weight as $$mg=\frac{IV}{v},$$ purely in terms of electrical and dynamical measurements. (This requires a measurement of the local value of $g$, but that is easy to measure locally using length and time standards.)


So, on one level, it's great that we've got this nifty non-artefact balance that can measure arbitrary weights, but how come it depends on electrical quantities, when the new SI kilogram is meant to only depend on the kinematic standards for length and time? As noted in the question, this requires a bit of reshuffling in the same spirit as for the ampere. In particular, the Josephson effect gives a natural voltage standard and the quantum Hall effect gives a natural resistance standard, and these can be combined to give a power standard, something like



the power dissipated over a resistor of conductance $G=25\,812. 807\,557G_0$ by a voltage that will produce AC current of frequency $483.597\,890\,893\:\mathrm{THz}$ when it is applied to a Josephson junction



(with the same caveats on the actual numbers as before) and as before this power will actually be independent of the chosen value of $e$ as long as $K_J$ and $R_K$ are changed appropriately.


Going back shortly to our NIST-style watt balance, we're faced with measuring a voltage $V$ and a current $I$. The current $I$ is most easily measured by passing it through some reference resistor $R_0$ and measuring the voltage $V_2=IR_0$ it creates; the voltages will then produce frequencies $f=K_JV$ and $f_2=K_JV_2$ when passed over Josephson junctions, and the reference resistor can be compared to a quantum Hall standard to give $R_0=rR_K$, in which case $$ m =\frac{1}{rR_KK_J^{2}}\frac{ff_2}{gv} =\frac{h}{4}\frac{ff_2}{rgv}, $$ i.e. a measurement of the mass in terms of Planck's constant, kinematic measurements, and a resistance ratio, with the measurements including two "artefacts" - a Josephson junction and a quantum Hall resistor - which are universally realizable.


The Mole


The mole is has always seemed a bit of an odd one to me as a base unit, and the redefined SI makes it somewhat weirder. The old definition reads




The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in $12\:\mathrm{g}$ of carbon 12



with the caveat that



when the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.



The mole is definitely a useful unit in chemistry, or in any activity where you measure macroscopic quantities (such as energy released in a reaction) and you want to relate them to the molecular (or other) species you're using, in the abstract, and to do that, you need to know how many moles you were using.


To a first approximation, to get the number of moles in a sample of, say, benzene ($\mathrm{ {}^{12}C_6H_6}$) you would weigh the sample in grams and divide by $12\times 6+6=78$. However, this fails because the mass of each hydrogen atom is bigger than $1/12$ of the carbon atoms by about 0.7%, mostly because of the mass defect of carbon. This would make amount-of-substance measurements inaccurate beyond their third significant figure, and it would taint all measurements based on those.


To fix that, you invoke the molecular mass of the species you're using, which is in turn calculated from the relative atomic mass of its components, and that includes both isotopic effects and mass defect effects. The question, though, is how does one measure these masses, and how accurately can one do so?



To determine that the relative atomic mass of ${}^{16}\mathrm O$ is $15.994\,914\, 619\,56 \:\mathrm{Da}$, for example, one needs to get a hold of one mole of oxygen as given by the definition above, i.e. as many oxygen atoms as there are carbon atoms in $12\:\mathrm g$ of carbon. This one is relatively easy: burn the carbon in an isotopically pure oxygen atmosphere, separate the uncombusted oxygen, and weigh the resulting carbon dioxide. However, doing this to thirteen significant figures is absolutely heroic, and going beyond this to populate the entire periodic table is obviously going to be a long exercise in bleeding accuracy to long chemical metrology traceability chains.


Now, as it happens, there can in fact be more accurate ways to do this, and they are all to do with the Avogadro project: the creation of a shiny sphere of silicon with a precisely determined number of ${}^{28}\mathrm{Si}$ atoms. This is done by finding the volume (by measuring the diameter, and making sure that the sphere is really round via optical interferometry), and by finding out the spacing between individual atoms in the crystal. The cool part happens in that last bit, because the spacing is found via x-ray diffraction measurements, and those measure naturally not the spacing but instead the constant $$\frac{h}{m({}^{28}\mathrm{Si})}$$ where $h$ is Planck's constant. And to top this up, the $h/m(X)$ combination can be measured directly, for example by measuring the recoil shift in atomic spectroscopy experiments (as reported e.g. here).


This then lets you count the number of silicon atoms in the sphere without weighing it, or alternatively it lets you measure the mass of the sphere directly in terms of $h$ (which is itself measured via the prototype kilogram). This gives a mise en pratique of the new SI kilogram (where the measured value of $h$ is replaced by its new, fixed value) but that one seems rather impractical to me.


More importantly, though, this gives you a good determination of the Avogadro constant: the number $N_A$ of elementary entities in a mole. And this is what enables you to redefine the mole directly as $N_A$ elementary entities, with a fixed value for $N_A$, while keeping a connection to the old standard: by weighing the silicon sphere you can measure the relative atomic mass of silicon, and this connects you back to the old chemical-metrological chain of weighing different species as they react with each other.


In addition to that, a fixed value of $N_A$ enables a bunch of ways to measure the amount of substance by coupling it with the newly-fixated values of other constants, which are detailed in the proposed mises en pratique.



  • For example, you can couple it with $e$ to get the exactly-known value of the electrical charge of one mole of electrons, $eN_A$, and then do electrolysis experiments against a current standard to get accurate counts on electrons and therefore on the aggregated ions.

  • Alternatively, you can phrase the ideal gas law as $pV=nRT=n(N_Ak_B)T$ and use the newly-fixed value of the Boltzmann constant (see below) and a temperature measurement to get a fix on the number of moles in the chamber.

  • More directly, the number $n$ of moles of a substance $\mathrm X$ in a high-purity sample of mass $m$ can still be determined via $$n=\frac{m}{Ar(\mathrm X)M_u}$$ where $Ar(\mathrm X)$ is the relative mass of the species (determined as before, by chemical means, but unaffected because it's a mass ratio) and $$M_u=m_uN_A$$ is the molar mass constant, which ceases to be fixed and obtains the same uncertainty as $m_u$, equal to $1/12$ of the mass of $N_A$ carbon-12 atoms.



As to the dependency of the standards, it's clear that the mole depends only on the chosen value of $N_A$. However, to actually implement it one needs a bunch of additional technology, which brings in a whole host of metrological issues and dependence on additional standards, but which ones come in depends exactly on which way you want to measure things.


Finally, in terms of why the mole is retained as a dimensional base unit - I'm personally even more lost than before. Under the new definition, saying "one mole of X" is exactly equivalent to saying "about 602,214,085 quadrillion entities of X", saying "one joule per mole" is the same as "one joule per 602,214,085 quadrillion particles", and so on, so to me it feels like the radian and the steradian: a useful unit, worth its salt and worthy of SIness, but still commensurate with unity. But BIPM probably have their reasons.


The kelvin


Continuing with the radical do-overs, the kelvin gets completely redefined. Originally defined, in the current SI, as $1/273.16$ of the thermodynamic temperature $T_\mathrm{TPW}$ of the triple point of water, in the new SI the kelvin will be defined by fixing the value of the Boltzmann constant to (roughly) $k_B=1.380\,6\times 10^{-23}\mathrm{J/K}$.


In practice, the shift will be mostly semantic in many areas. At reasonable temperatures near $T_\mathrm{TPW}$, for example, the proposed mises en pratique state that



The CCT is not aware of any thermometry technology likely to provide a significantly improved uncertainty on $T_\mathrm{TPW}$. Consequently, there is unlikely to be any change in the value of $T_\mathrm{TPW}$ in the foreseeable future. On the other hand, the reproducibility of $T_\mathrm{TPW}$, realised in water triple point cells with isotopic corrections applied, is better than $50\:\mathrm{µK}$. Experiments requiring ultimate accuracy at or close to $T_\mathrm{TPW}$ will continue to rely on the reproducibility of the triple point of water.



In other words, nothing much changes, except a shift in the uncertainty from the determination of $k_B$ to the determination of $T_\mathrm{TPW}$. It seems that this currently the case across the board of temperature ranges, and the move seems to be to future-proof against the emergence of accurate primary thermometers, defined as follows:




Primary thermometry is performed using a thermometer based on a well-understood physical system, for which the equation of state describing the relation between thermodynamic temperature $T$ and other independent quantities, such as the ideal gas law or Planck's equation, can be written down explicitly without unknown or significantly temperature-dependent constants.



Some examples of this are



  • acoustic gas thermometry, where the speed of sound $u$ in a gas is related to the average mass $m$ and the heat capacity ratio $\gamma$ as $u^2=\gamma k_BT/m$, so characterizing the gas and measuring the speed of sound yields the thermodynamic temperature, or

  • radiometric thermometry, using optical pyrometers and Planck's law to deduce the temperature of a body from its blackbody radiation.


Both of these are direct measurements of $k_BT$, and therefore yield directly the temperature in the new kelvin. However, the latter is the only standard in use in ITS-90, so it seems that the only direct effect of the shift is that pyrometers no longer need to be calibrated against temperature sources.


Since the definition depends on the joule, the new kelvin obviously depends on the full dynamical MKS triplet. Metrologically, of course, matters are much more complicated - thermometry is by far the hardest branch of metrology, and it leans on a huge range of technologies and systems, and on a bunch of empirical models which are not entirely understood.


The candela



Thankfully, the candela remains completely untouched. Given that it depends on the radiated power of the standard candle, it depends on the full dynamical MKS triplet. I do want to take this opportunity, however, to remark that the candela has full rights to be an SI base unit, as I've explained before. The definition looks very innocuous:



The candela, symbol $\mathrm{cd}$, is the SI unit of luminous intensity in a given direction. It is defined by taking the fixed numerical value of the luminous efficacy $K_\mathrm{cd}$ of monochromatic radiation of vacuum wavelength $555\:\mathrm{nm}$ to be $K_\mathrm{cd}=683\:\mathrm{cd/(W\:sr^{-1})}$.



However, the thing that slips past most people is that luminous intensity is as perceived by a (standardized) human eye, ditto for luminous efficacy, and more generally that photometry and radiometry are very different beasts. Photometric quantities require access to a human eye, in the same way that dynamical quantities like force, energy and power are inaccessible to kinematical measurements that only implement the meter and the second.




Further reading



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...