Thursday, February 28, 2019

quantum mechanics - Why can't an energy level exist containing 0.9 electron wave wavelengths? Why must it be a whole number?




So, I was reading Atom: Journey Across The Subatomic Cosmos by Isaac Asimov in order to better understand quantum mechanics when I came across this sentence:



The electron couldn't spiral into the proton because it couldn't take up an orbit with a length less than a single wave.



I was wondering why that is? Why can't an electron not have non-whole number wavelengths in atomic energy levels?



Answer



That idea came from the Bohr theory of the atom. It's good enough for the main hydrogen spectral lines. But in general and to be accurate (except for relativistic effects) you have to solve the Schrodinger equation.


For the Bohr model, Bohr hypothesized a way to get quantized orbits. He postulated a quantized angular momentum as integer multiples of h/2, which led to the quantized values of energy and the discrete orbital shells.


This was later reinterpreted by deBroglie as electrons being matter waves; the idea was relatively simple but revolutionary. Assume the electrons are wavelike, and only if they create standing waves as they orbit can they be stable. Otherwise the wave in one cycle will interfere negatively with the next cycle and so on. That gives you the idea that the orbits have to be integer multiples of the wavelength. From that you get everything else in the Bohr atom. See the Bohr model, with also the deBroglie interpretation, of the atom explained more at https://en.m.wikipedia.org/wiki/Bohr_model


Those started the idea of atomic orbits being quantized. It started with Bohr in 1913 simply hypothesizing that angular momentum came in half integer multiples of h. For deBroglie around 1924 it was that n times the wavelength is $2\pi r$, in deBroglie's model that led to the momentum (which equals h/$\lambda$) being quantized. From either one the energy levels are also quantized.



It takes the Schrodinger equation to solve for the wave functions and energy levels of stationary orbits. Other solutions are not stationary. The ones that are are so called eigenstates of the energy, which is conserved.


gravity - Place each foot on a scale: can you add the two to find your weight?


I frequent a blog from a British psychologist, and every Friday he likes to pose an interesting puzzle or riddle. The Monday after that he posts the answer. They're good fun, and IANAP but this week's answer made my it-might-not-be-quite-as-simple-as-that detector go off.


My question boils down to this: let's say I have two identical scales, and I stand on the scales with one foot on each scale. The scales read W1 and W2. Does my weight equal W1 + W2?



Answer



Yes, this is not the hardest problem ever, but here's the mechanics calculation that leads to the yes answer.


Draw a free body diagram of your body as you are standing still with one foot on each scale. You experience three forces (I will label their magnitudes): (1) The force due to gravity pulling you down, $W$ (aka your weight), (2) the normal force $N_1$ of scale 1 pushing up on one foot, and (3) the normal force $N_2$ of scale 2 pushing up on your other foot. Since your body is not accelerating, these forces balance by Newton's Second Law; $$ W = N_1+N_2 $$ Now the question is, how are these forces related to what the scales read? Well, each scale reads the force of the corresponding leg that pushes down on it. Let's call the magnitude of these forces $W_1$ and $W_2$. As it turns out, Newton's Third Law tells us that the magnitude of the force that each scale exerts on each foot (the normal force) equals the corresponding magnitude of the force that each foot exerts on the scale; $$ W_1 = N_1, \qquad W_2 = N_2 $$ It follows that $$ W = W_1 + W_2 $$ as desired.



quantum mechanics - How does one properly define the derivative of one operator-valued function?


In Quantum Mechanics we usually consider operator-valued functions: these are functions that take in real numbers and gives back operators on the Hilbert space of the quantum system.


There are several examples of these. One of them is when we work with the Heisenberg picture, where we need to consider functions $\alpha : \mathbb{R}\to \mathcal{L}(\mathcal{H})$ such that $\alpha(t)$ is the operator at time $t$.


Another example is when we deal with exponentiation of operators, like when building the time evolution operator:


$$U(t,t_0)=\exp\left(-i\frac{H}{\hbar}(t-t_0)\right),$$


Here, the $\exp$ is usually understood as being defined via the eigenvalues of $H$.


The point is, the idea of a function $\alpha : \mathbb{R}\to \mathcal{L}(\mathcal{H})$ appears quite often in Quantum Mechanics, and sometimes one needs to differentiate these. In practice we do it formally, using all the properties we would expect, but I'm curious about how one would properly define this.


If we were dealing with bounded operators, then we could use the operator norm, which is available for this kind of operator, and define the derivative as we can usually we do when there's some norm around.


The point is that in Quantum Mechanics most of the time operators are unbounded.


So in the general case, how can one define the derivative of one operator-valued function?





energy - What are the "inexact differentials" in the first law of thermodynamics?


The first law of thermodynamics states that



$$dU=\delta Q - \delta W$$




I have only just graduated high school and I am finding the above form of the equation rather difficult to understand due to the fact that I don't understand what inexact differentials are. Is it possible for anybody to please explain this to me? (I have taken an A.P course in calculus in school).



Answer



The mostly math-free explanation:


The internal energy $U$ is a function of state. It depends only on the state of the system and not how it got there. The notions of heat $Q$ and work $W$ are no such functions - they are properties of a process, not of a state of the thermodynamic system. This means that we can compute the infinitesimal change $\mathrm{d}U$ as the actual change $U$ of the function between two infinitesimally close points, but the infinitesimal changes in heat and work $\delta Q,\delta W$ depend on the way we move from one such point to the other.


More formally:


Now, you should imagine the state space of thermodynamics, and the system taking some path $\gamma$ in it. We call the infinitesimal change in internal energy $\mathrm{d}U$, which is formally a differential 1-form. It's the object that when integrated along the path gives the total change in internal energy, i.e. $U_\text{end}-U_\text{start} = \int_\gamma \mathrm{d} U$. You may think of this as completely analogous to other potentials in physics: If we have a conservative force $F = -\nabla U$, then integrating $F$ along a path taken gives the difference between the potential energies of the start and the end of the path. This is why $U$ is sometimes called a "thermodynamic potential", and this means that the $\mathrm{d}U$ is an actual differential - it is the derivative of the state function $U$.


Since $W$ and $Q$ are not state functions, there are no differentials $\mathrm{d}W$ or $\mathrm{d}Q$. However, along any given path $\gamma$, we can compute the infinitesimal change in work and heat, and also the total change $\Delta W[\gamma]$ and $\Delta Q[\gamma]$, so heat and work are functionals on paths. It turns out that, together with linearity - the work along two paths is the sum of work along each of them - this is enough to know that there are two differential 1-forms representing heat and work on the entire state space (for a formal derivation of this claim, see this excellent answer by joshphyiscs). These forms we call $\delta W$ and $\delta Q$, where we use $\delta$ instead of $\mathrm{d}$ to remind us that these are not differentials of state functions.


riddle - I can be quite bright but I'm never right. Can you guess who I am?


I can be quite bright but I'm never right.
I support you in peace and protect you in fight.


I'm seldom alone and when I am, there's no use.
Even when you select me, another you choose.


You never ask how I'm doing, even when I'm blue.
I might be unhealthy but I can't catch a flu.


Sometimes I'm in transit and sometimes home I'll stay.
Sometimes you can see a person who for me will pray.



If you don't have me at all, you're poor not having a dime
Or you might simply have fallen as a victim of crime.


I can hurt you and wound you and even make you bleed.
Regardless of your pain, I'm something you still need.


I might be unwanted but also an object of desire.
I could be damaged by earth, water and fire.


I could be expensive or cheap or even free.
When people look at you, they often see me.


You'll be both arrogant and humble if I'm put on your head.
I often bear a name that someone else before me had.



What am I?








  • We're looking for a common everyday object.







  • It's not anything abstract as "love" or "energy".






  • It's not symbolic as "the letter A" or "peace sign".







Answer



Okay, I think I've got it:



A left shoe



I can be quite bright but I'm never right. I support you in peace and protect you in fight.



Shoes can be flashy. Left shoe isn't right. Supports you, and you can kick people with it



I'm seldom alone and when I am, there's no use. Even when you select me, another you choose.




Mostly useful in pairs



You never ask how I'm doing, even when I'm blue. I might be unhealthy but I can't catch a flu.



Shoes can be blue, but you still don't inquire about their health. And a shoe can be nasty!



Sometimes I'm in transit and sometimes home I'll stay. Sometimes you can see a person who for me will pray.



You want with them or leave them behind. Some people are desperate for particular examples.




If you don't have me at all, you're poor not having a dime Or you might simply have fallen as a victim of crime.



Even very poor people have shoes.



I can hurt you and wound you and even make you bleed. Regardless of your pain, I'm something you still need.



Shoes hurt you, and can make you bleed, but you still need them.



I might be unwanted but also an object of desire. I could be damaged by earth, water and fire.




Unwanted old shoe vs designer shoe. Some are susceptible to damage from mud, water or fire.



I could be expensive or cheap or even free. When people look at you, they often see me.



Range of shoes available. Put your best foot forward. Shoes are often noticed.



You'll be both arrogant and humble if I'm put on your head. I often bear a name that someone else before me had.



Shoes sometimes have the name of their designer. A shoe on the head could be a sign of madness or a very eccentric hat such as an aristocrat might wear.




Wednesday, February 27, 2019

riddle - Who got some sun on the voyage?


We meek internationalists go on a little afternoon voyage



Who got some sun on the voyage?


Hint #1:



My question history (previous puzzles) may prove useful.



Hint #2:



interNATionalists



Hint #3:




You might get some help from a girl with a Goodly Heart



Hint #4:



You also might get help from a hot lady ogre




Answer



I think the answer is




Elliot



First clue, thanks to Smock



As Smock pointed out in the comments - "the anagram Toto, I've got a feeling we're not in Kansas anymore is contained within" the puzzle. The rest of the letters in the puzzle can be rearranged to spell "Elliot tan".



Hint 2



NAT is an anagram of tan




Hints 3 and 4



Both anagrams of Dorothy Gale - the main character from The Wizard of Oz



general relativity - Why are there negative energy particles inside a black hole?



In quantum field theory in flat spacetime, there are both positive and negative frequency solutions to the classical field equations, but upon quantization we get only positive energy particles. But in Hawking's original paper about Hawking radiation, it is stated that negative energy particles can exist inside a black hole:



Just outside the event horizon there will be virtual pairs of particles, one with negative energy and one with positive energy. The negative particle is in a region which is classically forbidden but it can tunnel through the event horizon to the region inside the black hole where the Killing vector which represents time translations is spacelike. In this region the particle can exist as a real particle with a timelike momentum vector even though its energy relative to infinity as measured by the time translation Killing vector is negative. The other particle of the pair, having a positive energy, can escape to infinity where it constitutes a part of the thermal emission



That is, Hawking is saying that the energy of a particle can be defined as $$E = p^\mu K_\mu$$ where $p^\mu$ is its four-momentum and $K^\mu$ is the time translation Killing vector. I can see how this works in Minkowski space, where $K = \partial_t$ and we get $E = p^0$ as expected.


But why is this the correct definition of energy? What kind of observer would measure $E$ to be the energy of the particle? Can this quantity be shown to be conserved? Why should we trust this equation when $K^\mu$ is not even timelike inside the black hole?



Answer



As far as I know, any observer (inertial or otherwise) with 4-velocity $U^\mu$ will measure the energy of a particle with 4-momentum $p^\mu$ to be $U^\mu p_\mu$. Since no observers have spacelike 4-velocities, no observer would measure the energy of a particle to be $K^\mu p_\mu$ inside the black hole. Trying to define the energy 'as measured at infinity' strikes me as dangerous: we can only measure local quantities in GR, so the only way to talk about an observer measuring something far away is to imagine some signal travelling between events. But this cannot happen if our particle is behind the horizon.


What I believe Hawking is trying to say is that it’s OK for the quantity $K^\mu p_\mu$ to be negative, precisely because it isn't energy. There is a brief moment, as the particle approaches the horizon, during which the particle has negative energy, which would be problematic were it not for the brevity of the moment (forgive the hand-waving). But once inside the black hole, this quantity doesn't correspond to energy, and so we no longer have a problem. Indeed, a typical observer inside the black hole would have 4-velocity something like $-\partial/\partial r$, and so would measure the particle's energy as $-p_r$, which is positive, since $p^r$ is negative.


Note: the quantity $K^\mu p_\mu$ for any Killing vector $K^\mu$ is conserved along the particle's geodesic.



I assume the mostly minus metric convention.


gravity - Which way will the pencil fall?


Let's say you had a perfect pencil, with a point which was just that one point (see this question). The pencil's mass was perfectly distributed, and there are no flaws in the craftsmanship. Let's say you had it oriented vertically (like you were going to balance it), and it was exactly straight up and down i.e. the pencil's center of mass is directly above the point where the pencil is touching the ground. In the universe where this pencil is, there are no outside forces which can affect the pencil, other than gravity.



Which way will the pencil fall, after you let go?




Is it random?




vectors - Why do we need both dot product and cross product?


I was looking for an intuitive definition for dot product and cross product. I have found two similar quesitions in SO, but I am not satisfied with the answers. Finally I found a possible answer here. It says dot product actually gives us a way to depict mathematically how parallel two lines are and on the other side cross products tells us how two lines are perpendicular to each other. So my question is why do we want both. Why cant we just have dot product?




Tuesday, February 26, 2019

particle physics - Where to find cross section data for e- + p -> p + e-?



Where to find cross section data for e- + p -> p + e-?



PDG's cross section data listing does not include it.



Answer



you can search the HEPDATA database at http://www.slac.stanford.edu/spires/hepdata/ with the query string


[reac = e- p --> e- p]

and the first result will be:


"Jefferson Lab. Measurement of the elastic electron-proton cross section in the Q*2 range from 0.4 to 5.5 GeV*2"


logical deduction - Demons' Dungeon


After winning the Fairy Logician's competition, you are reverse-summoned to a room with 5 portals at equal distances from each other. They are all locked,but fortunately, there is a logical demon in front of each portal, they are willing to unlock one portal for you. The problem is that only 1 of these portals is the exit while the others will send you to your certain death.


The demons are all knowing, but incredibly lazy. They will only allow you to ask 3 questions. They will only provide a 1-bit (yes/no) answer. And if your question requires more than 25 english words* to perfectly grammatically express they won't answer it.


Oh yeah, one more thing, though demons are supposed to lie, all but one of them are to lazy to too that, and will answer truthfully.


Can you get out alive, Logician Adventurer?





(*) words are all you can use, expressions like ∀ε ∃δ have to be expressed like this: "There is a delta for every epsilon", which is 7 words already, and you have 25.


Disclaimer: this puzzle is based heavily on the one by stack reader



Answer



Pick any demon.



1. If I asked you if one of these 3 portals (point) was the exit, would you say yes?
Because the question loops itself, it doesn't matter if the devil is a liar or not, they will be telling the truth, and I'll know which group, of either 2 or 3 portals, the exit is in. Point to one of the portals from that group. Ask any devil.
2. If I asked you if this was the exit, would you say yes?
Again, loops itself, doesn't matter if the devil's a liar. If it's the exit, we can leave. If it isn't, and this is the group of 2 portals, we know the other is the exit. If this is the group of 3 portals, we have to point at one of the remaining ones, and ask any devil.

3. If I asked you if this was the exit, would you say yes?



Now we know, and can leave.


quantum mechanics - What parameters determine the amount of Uehling potential and self-energy correction effects?


Uehling potential correction and electron self-energy correction are the two most important factors contributing to the lamb shift in the hydrogen atom, by "smearing" the wavefunction of the electron. What parameter(s) determine "how much" this smearing of the wavefunction is?




Monday, February 25, 2019

homework and exercises - Change of temperature of gas in cylinder


Assume we have a cylinder of given volume filled with gas of given temperature and pressure. The cylinder is enclosed from the top by a piston of provided mass. Now, we place a small mass on the piston. Is it possible to find how much does the temperature of gas change? We assume, for the simplicity, that no heat is exchanged with the environment and that the gas is ideal.



Answer



Let $x$ be the amount the piston moves down due to the mass $m$:


enter image description here


The answer to your question is "no, there is insufficient information provided to find the change in temperature". However, if you know the cylinder's "shape" (ie, the $h:r$ ratio above) and the gas' heat capacity (https://en.wikipedia.org/wiki/Heat_capacity), the answer is yes:



  • To support the extra mass $m$, the force on the piston must increase by $g m$. Since force is pressure times area, we have


$\Delta P (\pi r^2) = m g\to \Delta P = \frac{g m}{\pi r^2}$



Note that $\Delta P$ is independent of $x$.



  • The change in volume is simply:


$\Delta V=-\pi r^2 x$



  • By Joule's Second Law, the total energy of an ideal gas depends only on its temperature:


$E_{\text{total}}=c n R T$


where $c$ is the gas' heat capacity.



Since the gas gains the $m g x$ of potential gravitational energy that the mass loses (and $c$, $n$, and $R$ remain constant):


$\Delta E_{total} = c n R \Delta T = m g x \to \Delta T = \frac{m g x}{c n R}$



  • By the Ideal Gas Law, we know that $\frac{P V}{n R T}$ remains constant. Since $n$ and $R$ are already constant, this means $\frac{P V}{T}$ remains constant. Thus, using $P$, $V$, and $T$ as the original pressure, volume and temperature, we have:


$\frac{P V}{T}=\frac{(P+\Delta P) (v+\Delta V)}{T+\Delta T}$


Using the values of $\Delta P$, $\Delta V$, and $\Delta T$ above, we can solve for $x$:


$ x\to \frac{c g m n R T V}{\pi r^2 \left(c n R T \left(g m+\pi P r^2\right)+g m P V\right)} $


Plugging this into our formula for $\Delta T$ yields:


$ \Delta T = \frac{g^2 m^2 T V}{\pi r^2 \left(c n R T \left(g m+\pi P r^2\right)+g m P V\right)} $



We can further simplify, since $n R T\to P V$:


$\Delta T = \frac{g^2 m^2 T}{\pi P r^2 \left((c+1) g m+\pi c P r^2\right)}$


(of course, we could've also made this simplification earlier when computing $\Delta T$)


While this answer is technically correct, it appears to have an odd dependency on $r$, the radius of the cylinder, and no dependency on $V$, the initial volume.


However, if we define, $k=\frac{h}{r}$ (which measures the "shape" of the cylinder in some sense), we have $h=k r$ and thus:


$V=\pi r^2 h=\pi r^2 (k r) = \pi r^3 k \to r = \sqrt[3]{\frac{V}{\pi k}}$


Making this final substitution, we have:


$ \Delta T = \frac{g^2 m^2 T}{\sqrt[3]{\pi } P \left(\frac{V}{k}\right)^{2/3} \left((c+1) g m+\sqrt[3]{\pi } c P \left(\frac{V}{k}\right)^{2/3}\right)} $


Disclaimer: This is a purely mathematical answer. You should check that it makes sense in real life.


quantum mechanics - Normalizable wavefunction that does not vanish at infinity


I was recently reading Griffiths' Introduction to Quantum Mechanics, and I stuck upon a following sentence:



but $\Psi$ must go to zero as $x$ goes to $\pm\infty$ - otherwise the wave function would not be normalizable.



The author also added a footer: "A good mathematician can supply you with pathological counterexamples, but they do not arise in physics (...)".


Can anybody give such a counterexample?




Answer



Take a gaussian (or any function that decays sufficiently quickly), chop it up every unit, and turn all the pieces sideways.


enter image description here


general relativity - What is gravitational radiation?


What is gravitational radiation (in association with gravitational waves)? Is it a form of energy/mass? Or is it just another word for gravitational waves?




newtonian mechanics - What is function of force $F$ and work $W$?


I know that :


Position $$r:\mathbb{R} \to\mathbb{R^3} \ \ \ \ \ ; \overrightarrow{r}(t)$$


Velocity $$v :\mathbb{R} \to\mathbb{R^3} \ \ \ \ \ ; \overrightarrow{v}(t)$$


Acceleration $$a:\mathbb{R} \to\mathbb{R^3} \ \ \ \ \ ; \overrightarrow{a}(t)$$


Now i want to know that , what is function of force $F$ and work $W$ ?




kinematics - View from a helicopter rotor: why is the horizon distorted?


This video ("rotor panorama") was captured by a camera attached to the rotor head of a radio-controlled helicopter, with the frame rate set to the rotor's frequency. During a long segment of the video, the horizon looks distorted:


enter image description here


What causes this distortion?




gravity - How is dark matter meant to explain the faster than expected rotation of galaxies?



The stars on the outer edges of galaxies go around faster than they should be. How is dark matter meant to account for this? if you just add more mass wont that just give it a greater gravitational force? So we would still expect the outer stars of galaxies to be going slower than the ones near the center. Whats the flaw in my reasoning?




Sunday, February 24, 2019

homework and exercises - Lenz' law versus $-frac{dOmega}{dt}$


I am preparing for an exam, on one problem my answer differ from the solution.


The current $i(t)=I_0e^{-\alpha t}$ runs in a long straight conductor along $\hat z$.


Point A,B,C,D forms a triangle. A and B is at $L \hat x$ with a very small separation in $\hat z$ with a resistance R between. C and D is at $2L \hat x$ with a separation in $\hat z$ of $2L$.


The question is: What is the induced current in the triangle.



I calculated the flux $\Omega$ correctly, but my answer differ on the direction of the current.


I took $V_{ind}=-\frac{d\Omega}{dt}$, since $i(t)$ is decreasing, the magnetic flux passing through ABCDA is decreasing, and so (to my understanding) it is reasonable that $i_{ABCD}=\frac{V_{ind}}{R}$ will run clockwise (using right-hand notation).


However, in the solution to the problem they too use $V_{ind}=-\frac{d\Omega}{dt}$ but then "magically" on the next line remove the minus sign and says that according to Lenz law the current opposes the magnetic flux that caused the current and so will run counter-clockwise.


I would be thankful for an intuitive understanding for why it is opposing the magnetic flux that caused the current and not opposing the change in the magnetic flux? Thanks.



Answer



Actually, it does oppose the change in the magnetic flux causing it. If the magnetic field from the source is increasing in some direction, the magnetic field from the induced current decreases in the same direction to oppose the increase, for example. The induced current opposing the change causing it ensures that the cause must supply energy to the system to increase the induced current, so guaranteeing conservation of energy.


The magnetic field from the wire points towards $\hat y$ and is decreasing, which means the magnetic field has to increase in the same direction from the induced current. Therefore the current runs clockwise around the loop.


The negative sign in Lenz's law is needed when using the right hand rule to find the direction of the induced emf: The thumb points in the direction of the applied magnetic field, and the curled fingers point around the direction of the induced emf. In your example, the thumb points towards $\hat y$ and the fingers curl in a clockwise sense. The magnetic field decreasing cancles the minus sign so the direction is the same as the fingers - clockwise.


probability - Contagious ants on a stick


Twenty-five ants are scattered on a meter stick. At the same time, they each pick a random direction (east or west) and start marching at 1cm/second. Whenever two ants meet, they turn around. When an ant reaches the end it falls off.


The ant in the middle is infected with a cold. An infected ant will spread the disease to any ant he bumps into.




On average, by the time every ant has fallen off, how many will be sick?




Answer



The average number of sick ants will be



$13 - \frac{6}{2^{12}}$



We start with the classic ant-on-a-stick observation that




we can imagine two ants as bouncing off each other as though they passed through each other, since their identities don't matter. So, in an equivalent problem, ants don't bounce, and an ant is infected by passing through another infected ant.



Now, make an observation about how ants get infected:



The infected range spreads at most as fast as an ant walks. So any ants the start moving away from the center ant cannot be infected. Any ants that start going towards the center get infected as long as an infected ant going the other way touches the center point. Usually, that's all ants that start going towards the center ant (expected $12$), since the center ant infects all such ants in front of him, and any such ant infects all such ants behind him. But, if all ants in front of the center ant are going the same way (probability $2^{-12}$), no ants behind him are infected, decreasing the expectation by $6$. Counting the $1$ infected center ant, this gives an expectation of $1+12-2^{-12}\times 6 = 13 - \frac{6}{2^{12}}$.



enigmatic puzzle - Big game hunting on the continent


I once went on a hunting expedition for five years across Europe, searching for a wonderful trophy for my lodge. It was a wonderful experience. I started in Poland and moved to France, then went to England to meet with my 3 friends, Lee, Stuart and I forget the name of the last one. (He's a light little devil. You could throw him around without breaking a sweat) My favorite thing I found was a huge mass of fireflies, though I did manage to catch a glimpse of a tortoise. My greatest achievement was hunting three different big cats. The only issue is I have misplaced my trophies now, and I can't remember what they were. Can you tell me what they were?



Answer



I think you are referring to ...



.. World War II tanks.

The five year period probably refers to USA entering the war after the attack on Pearl Harbor in 1941.

Your friends were the M3 Stuart, M3 Lee and maybe the M22 Locust. You saw the Sherman Firefly and the Tortoise heavy assault tank. The 3 cats were probably the Panther, Tiger I and Tiger II (aka. King Tiger). There was also the Panzer VII Löwe (Lion) but according to Wikipedia it only existed on paper.



electromagnetism - Why do electrons have to fall on the nucleus in the Rutherford atomic model?


As I read on Wikipedia, the Rutherford atomic model is not correct according to classical electrodynamics, as it states that electron must radiate electromagnetic waves, lose energy and fall onto the nucleus.


I don't understand this explanation.


It is clear to me that with given acceleration directed to nucleus and proper speed, electron can move around the nucleus.


I don't understand explanation about energy, but I understand that there must be some force directed to nucleus. Also this force must not be constant because if it is, a larger speed could keep electron moving around the nucleus.


So what is that force? Why does this explanation on Wikipedia and on other resources operate with energy, not with force?



Answer



Well, I dont see a problem in any of those answers here, but, since you want in force terms... lets go.


The Lorentz force is: $$ \mathbf F = q(\mathbf E + \mathbf v\times\mathbf B) $$


Lets assume the nice and simple atom of hydrogen. A single electron is classically orbiting it. Lets say there is no magnetic field. Only electric. The electric field is a central field, meaning it is pointing only radially, meaning it will result in an orbit. And more: Its a kepler orbit (same of the planets).



$$ \mathbf F = q\mathbf E $$


But then, the electron when accelerated irradiates electromagnetic energy. Conservation of energy must apply, such that the irradiation takes away the energy of the electron. The electron loses then its energy. Energy is proportional to the momentum (kinetic energy). Thus, electron loses momentum. Changing in momentum is force. If we take Larmor Formula and make this process, we will arrive at Abraham-Lorentz force.


Now the complete force of this is: $$ \mathbf F = \frac{d\mathbf p}{dt} = m\frac{d^2\mathbf r}{dt} = q\mathbf E(\mathbf r) + \frac{\mu_0 q^2}{6\pi c}\frac{d^3\mathbf r}{dt^3} $$


Note that, for a circular orbit in xy-plane: $\mathbf r = r(\cos\omega t, \sin\omega t, 0)$, and thus: $$ \omega^2\mathbf r = -\frac{d^2\mathbf r}{dt^2} \quad\Longrightarrow\quad \omega^2\frac{d\mathbf r}{dt} = -\frac{d^3\mathbf r}{dt^3} \quad\Longrightarrow\quad \mathbf F = q\mathbf E - \frac{\mu_0 q^2}{6\pi c}\omega^2\mathbf v $$


Meaning, the third order derivative has a relationship with the speed. And not only that: Has a minus sign over there, indicating a drag force: A force always opposite to the velocity, and thus will tend to stop the motion. So, an electron orbiting a proton with no magnetic field present, will drag because this force, spiral in, and collapse into the proton.


Saturday, February 23, 2019

optics - Is fluorescence from a single atom/ion visible with the naked eye (e.g. in a strongly coupled trap or cavity)



I remember sitting in on a conference talk by a person (possibly Rainer Blatt) doing research with trapped ions (or single atoms strongly coupled to light in an optical cavity), and the person showed a photo of the trap with dots of light from the fluorescence of the single atoms/ions. I thought the person mentioned you could see this with the naked eye b/c the optical coupling to the ion in the trap was so strong, but thinking about it now I'm not sure if this can be true and I can't seem to find any (obvious) reference to this in the literature.




So my question: Is it possible to see light from a strongly coupled single atom or ion with the naked eye? If so can you point me to a reference (and hopefully an image of this as well)?




Note: The best I can find is the image below from the Blatt research group taken with a CCD (details here). However it is not at all obvious that this would be visible to the naked eye, or if the exposure was just set very high on the camera.


enter image description here



Answer



Experiments with trapped ions generally use fluorescence for detecting the ions. This means that they use a strongish pump to take the ion from its ground state to a dipole-allowed excited state and wait for the ion to decay by emitting a photon in a random direction, and then re-run the cycle over and over. This means that each ion essentially emits one photon per natural lifetime of the line, which can be quite often (in atomic terms) for dipole-allowed lines.


One example, taken from this paper, is an $S$-$P$ line in $\mathrm{Ca}^+$ with a linewidth of 21 MHz, at 397 nm. This means that each individual ion will emit about 20 million photons per second or so, evenly spread about $4\pi$ solid angle. This is in general just below the limit of human sensitivity but if you had a high enough numerical aperture and a dark enough background you could in principle do it. I asked the authors and they said that they can't do it because of lack of sufficient optical access (they have huge magnets in the way) and they don't know anyone who does but in principle it's just about doable.


conservation laws - Separability of Hamilton Jacobi Equation


When we talk about integrability of classical systems in terms of Hamiltonian or Lagrangian mechanics, it's all to do with counting independent conserved quantities.


Then when we move to the Hamilton-Jacobi formalism, suddenly everything is about separability of the Hamilton-Jacobi equation and Staeckel conditions. How do these two concepts relate to one-another? Does the existence of a certain number of conserved quantities imply separability of the Hamilton-Jacobi equation in some coordinate system?



Answer



The answer to your question is yes, the existence of $n$ conserved quantities with $n$ degrees of freedom implies separability of HJ.


The massless HJ equation is $$g^{MN}\frac{\partial S}{\partial x^M}\frac{\partial S}{\partial x^N}=E.$$ It separates if there exists a new set of coordinates $Y^M$ such that $$ S(Y_1,...,Y_n)=\sum_{i=1}^n S_i(Y_i),$$ which implies existence of $n$ conserved quantities, because each term in the HJ equation depends on its own variable. The same procedure is used when we solve PDE. For example in $2D$ $$S=S_x(x)+S_y(y), \quad f(x)(\partial_x S_x(x))^2+f(y)(\partial_y S_y(y))^2=E.$$ The latter means that both terms in LHS are constants separately. So we have two independent conserved quantities.



wordplay - A Rhyming Six-Words Puzzle


In the spirit of the original Four-words puzzle by Prem, and all those that followed.



My first completely overshadows anything you see.
My second's used extensively to mean extensively.
My third are pithy aphorisms commonly dispersed.
My fourth's what you'll be feeling if Deusovi solves this first.

My fifth's predominantly white, though legend says it's golden.
My sixth is what will keep the world consistently unfolding.


My whole's a square that's six by six with letters in each cell.
There is no difference in the words read flat or vertical.




Answer



How about



D W A R F S
W I D E L Y

A D A G E S
R E G R E T
F L E E C E
S Y S T E M



EDIT: Mooooooooooom, do I have to explain every line? FINE.


My first completely overshadows anything you see.



aye, something that dwarfs everything could be said to overshadow it




My second's used extensively to mean extensively.



widely, even. Judge I object to having to explain this one, res ipsa loquitur



My third are pithy aphorisms commonly dispersed.



adages is another word for aphorisms



My fourth's what you'll be feeling if Deusovi solves this first.




well, yes, I would be regretful if the god of false-egg-plurals had beaten me to it, but they did not



My fifth's predominantly white, though legend says it's golden.



fleece is white, but there's that whole Jason thing



My sixth is what will keep the world consistently unfolding.



systems are the logic by which things can be said to run




thermodynamics - Ultra-relativistic gas


What is the physical significance of the relation $E=3NkT$ for classical ultra-relativistic gas? Why is it greater than ideal gas for which $E=(3/2)NkT$?



Answer



Nice question. Sometimes we get used to a certain fact, such as equipartition with $(1/2)kT$ per degree of freedom, that we forget that it's not always true, or what assumptions are required in order to make it true. I had to refresh my memory on how equipartition works.


Basically the $(1/2)kT$ form of the equipartition theorem is a special case that only works if the energy consists of terms that are proportional to the squares of the coordinates and momenta. The 1/2 comes from the exponent in these squares.


The WP article on equipartition has a discussion of this. There is a general equipartition theorem that says that


$$\langle x \frac{\partial E}{\partial x} \rangle = kT,$$


where $x$ could be either a coordinate or a conjugate momentum. If $E$ has a term proportional to $x^m$, the partial derivative has a factor of $m$ in it. In the ultrarelativistic case, where $E\propto\sqrt{p_x^2+p_y^2+p_z^2}$, you don't actually have a dependence on the momenta (momentum components) that breaks down into terms proportional to a power of each momentum. However, I think it's pretty easy to see why we end up with the result we do, because in one dimension, we have $|\textbf{p}|=|p_x|$, which does have the right form, with an exponent of 1.


Friday, February 22, 2019

optics - For the derivation of the form of a paraboloidal reflector, there is much more in need as just Fermats principle?


I am currently learning about Optics using an MIT OpenCourseWare Course. I have in particular a question about the application of Fermat's principle in Lecture 3, i.e. deriving the shapes of parabolic reflectors and ellipsoids of refractive material for a given focal point. The lecture notes could be found here, and the derivations in question are on the first slides.


For example, consider the derivation of the paraboloidal reflector (3rd slide), he does this derivation at around minute 6 in the video.


In short, he says that all rays getting reflected and hitting the focus point F should be the same length by Fermats principle that the optical path length (or the time taken by the photons/wave fronts) should be minimal.


But I doubt that reasoning and have certain objections about his arguments.


1) First you cannot measure distance for points in infinity (as $\infty + c = \infty$), but if we replace infinity with some unspecified far away point, then we can put some plane behind F and measure distance from this plane onwards (as the rays before hitting that plane have all the same distance). So this is easily fixed.


2) In my opinion Fermats principle is just applied for the ray that goes directly through F and is reflected in the exact opposite direction (i.e. the ray on the optical axis), and that it must have this path is easily seen with Fermats principle. But the reasoning that all the other paths must have the same length (or must take the same time) is not Fermats priniple as I see it, as I see it an additional argument must be in order here, I can image that we might require that the photons that start at the same time (or the wavefronts thereof) should all meet in F, hence should all take the same time (or path length).


Am I right? Is the author a little bit sloppy here and are additional arguments in order? I have the feeling that Fermats principle is not enough, or said more provocative that it is not the crucial part of the argument (mainly that all paths should have the same length), but it just gives the magnitude of the common path (i.e. $2f$ in the lectures).


PS: A student seems to ask a similar question at around minute 11, but the answer does not seem to address it properly in my opinion (and btw. also thread infinity as a point from which we can measure distance).




puzzle identification - Kruskal's hard to refold envelope


In Wikipedia's bibliography of Martin Kruskal it is said:



Martin, who had a great love of games, puzzles, and word play of all kinds, also invented several quite unusual origami models including an envelope for sending secret messages (anyone who unfolded the envelope to read the message would have great difficulty refolding it to conceal the deed).



Unfortunately, this is the only information in Wikipedia about Kruskal's secret message envelope.


Does anyone know the history of this envelope, or where I can find more information about it?


This envelope seems like a puzzle, since there may have been a clever trick to refold it.



Answer



It is really hard to find something on the Web about Martin Kruskal in relation to origami.



There is a link missing in Wikipedia, so maybe this information is just gossip.


If you want to see how to fold an envelope for sending secret messages, you can check out:




  1. This method for a hard-to-open envelope.




  2. This method and this method for kind-of-hard-to-fold-back envelopes.





Is Liouville's equation an axiom of classical statistical mechanics?


Suppose we have a classical statistical problem with canonical coordinates $\vec{q} = (q_1, q_2, \dots, q_n)$ and $\vec{p} = (p_1, p_2, \dots, p_n)$ such that they fulfill the usual Poisson brackets: \begin{align} \{ q_i, p_j \} & = \delta_{i,j} \\ \{ q_i, q_j \} & = 0 \\ \{ p_i, p_j \} & = 0 \\ \end{align} One can show that the time evolution of every dynamical quantity $f(\vec{q}, \vec{p}, t)$ is given by $$ \frac{d f}{d t} = \frac{\partial f}{\partial t} + \{ f, H \} = \frac{\partial f}{\partial t} + \sum_{i=1}^n \frac{\partial f}{\partial q_i} \dot{q_i} + \frac{\partial f}{\partial p_i} \dot{p_i} = \frac{\partial f}{\partial t} + (\nabla f) \cdot \vec{v} = \frac{\partial f}{\partial t} + \nabla (f \cdot \vec{v}) $$ with $\nabla = (\frac{\partial}{\partial q_1}, \dots, \frac{\partial}{\partial q_n}, \frac{\partial}{\partial p_1}, \dots, \frac{\partial}{\partial p_n})$, $\vec{v} = (\dot{q_1}, \dots, \dot{q_n}, \dot{p_1}, \dots, \dot{p_n})$ and $H$ the Hamiltonian of the system.


Liouville's theorem states that: $$ \int d^n p ~ ~ d^n q = \int d^n p' ~ ~ d^n q' $$ if $(\vec{q}~', \vec{p}~')$ and $(\vec{q}, \vec{p})$ are both canonical coordinates, e.g. related by a canonical transformation. So the phase space volume is a constant between systems which are described by canonical coordinates.


Now consider the phase space density $\varrho(\vec{q}, \vec{p}, t)$ which is the density of dynamically allowed trajectories at a given point $(\vec{q}, \vec{p})$ in phase space for a given instance of time $t$.


Liouville's equation reads: $$\frac{d \varrho}{d t} = 0$$ which (together with the equation for $f$) says that $\varrho$ is a (locally) conserved density in phase space. Because $\varrho \ge 0$ one can conclude that there are no sources of trajectories in phase space: trajectories do not start, end or cross.


Usually the Liouville equation is derived from Liouville's theorem. However I haven't seen such a derivation for which at some point it wasn't assumed that $\varrho$ is a (locally) conserved density. Hence, that reasoning is circular.


Do you know a non-circular derivation of Liouville's equation or is it indeed an axiom of classical statistical mechanics?



Answer




Why There is a Need for a Further Axiom


To derive Liouville's equation, you indeed need another axiom further to your assumptions. Something like: "there is no nett creation or destruction of any particle of any species throughout the particle system state evolution". The easiest way to understand the need for this axiom is to cite a system wherefor Liouville's equation cannot hold, even though particles undergo dynamical evolution described by Hamilton's equations throughout their lifetimes: a system of particles undergoing a far-from-equilibrium chemical reaction. In such a system, reactant particle species are consumed by the reaction, and disappear from phase space. Reaction product particles appear in phase space in their place. Moreover, chemical energy is converted to kinetic energy (or contrariwise), so that a product species will "suddenly" appear at a different point in phase space from the one where the correspondingly consumed reactant species particles vanished. Liouville's equations would conceptually be replaced by a coupled system of equations, one for each species $j$, of the form:


$$\frac{\partial\,\rho_j(X,\,t)}{\partial\,t} = \{H,\,\rho_j\} + \sum\limits_k \int_\mathcal{P} M_{j\,k}(X,\,X^\prime)\,\rho_k(X^\prime,\,t)\,\mathrm{d}\Gamma^\prime$$


where the integral is over all phase space $\mathcal{P}$, $\Gamma$ is the measure defined by the volume form and the kernel $M_{j\,k}$ expresses detailed stochimetric balance between the chemically reacting species as well as other physical principles such as conservation of energy, momentum and strict increase with time of entropy. Note that I said "nett" creation or destruction: Liouville's equation would work if the reaction were at equilibrium.


Complete Axioms


The following axioms (1. and 2. are equivalent to yours) will get you Liouville's equation:



  1. Axiom 1: Phase space is a $2\,N$ dimensional $C^2$ manifold $\mathcal{P}$;

  2. Axiom 2: Points in phase space always and only evolve with a flow parameter $t$ through Hamilton's equations defined by a $C^2$ Hamiltonian $H:\mathcal{P}\times \mathbb{R}\to\mathbb{R}$, the latter possibly time varying (hence the domain $\mathcal{P}\times\mathbb{R}$);

  3. Axiom 3: The full states of particles are points in $\mathcal{P}$ evolving according to axiom 2 and there is no nett creation or destruction of any particle of any species throughout the particle system state evolution.



From Complete Axioms to Liouville's Equation


From these axioms, the chain of inference you need is as follows:



  1. Inference 1: From axioms 1. and 2., deduce that any $X\in T_p\,\mathcal{P};\;\forall p\in \mathcal{P}$ expressed in canonical co-ordinates (i.e. ones for which the Hamilton equations hold) that is Lie-dragged by the Hamiltonian flow evolves according to $\dot{X} = A(t)\,X$ where $A(t)\in\mathfrak{sp}(N,\,\mathbb{R})$, thus the symplectic 2-form $\omega(X,\,Y)\stackrel{def}{=} X^T\,\Omega\,Y$ where, for the special case of canonical co-ordinates, $\Omega =\left(\begin{array}{cc}0&-1_N\\1_N&0\end{array}\right)\;\forall p\in\mathcal{P}$ is conserved under the mapping $\mathcal{P}\mapsto \Phi(H,\,t)\,\mathcal{P},\,\forall t\in\mathbb{R}$ induced by the Hamiltonian flow. (Indeed, at any given point $p\in\mathcal{P}$ find $N$ different $C^2$ Hamiltonians such that the tangents to their flows span $T_p\,\mathcal{P}$ to deduce that the Lie derivative of $\omega$ in any direction is nought, thus $\mathrm{d}\omega=0$ from Cartan's formula relating Lie and Exterior derivatives, but this information is further to our immediate needs). Take heed that inference 1 holds whether or not the Hamiltonian be time-varying. In the latter case, the Hamiltonian is not constant along the flow, but the flow still conserves the symplectic form.

  2. Inference 2: From inference 1, we have immediately that the volume form $\Gamma = \omega^N$ ($N^{th}$ exterior power) is conserved under Hamiltonian flows. Thus deduce Liouville's theorem (as opposed to equation). Alternatively, the conservation of the symplectic form shown in Inference 1 implies that the Jacobi matrix of the transformation $\mathcal{P}\mapsto \Phi(H,\,t)\mathcal{P}$ is a symplectic matrix (member of $\mathrm{Sp}(N,\,\mathbb{R})$), which always has a unit determinant. Thus the volume form is conserved.

  3. Inference 3: But the volume form is also the Jacobian of the transformation $\mathcal{P}\mapsto \Phi(H,\,t)\,\mathcal{P}$ and $J(p,\,\Phi(H,\,0))=J(p,\,\mathrm{id})=1$. Since the volume form is conserved, the Jacobian $J(p,\,\Phi(H,\,t))=1,\forall p\in\mathcal{P},\forall t\in \mathbb{R}$. Thus $\Phi$ is everywhere a local bijection (inverse function theorem). Alternatively, we can make the same deduction this straight from Inference 1, which implies that the Jacobi matrix of the transformation $\mathcal{P}\mapsto \Phi(H,\,t)\mathcal{P}$ is a symplectic matrix (member of $SP(N,\,\mathbb{R})$), which is never singular and indeed always has a unit determinant.

  4. Inference 4: From Axioms 2 and 1, deduce that the distance function defined in canonical co-ordinates by $d(p_1,\,p_2) = (p_1-p_2)^T\,(p_1-p_2)$, zero iff $p_1=p_2$, between any pair of points $p_1,\,p_2\in\mathcal{P}$ must be a continuous function of the flow parameter $t$ (continuous with respect to the topology with basis of open balls defined by this distance function);

  5. Inference 5: From Inference 3, at any $p\in\mathcal{P}$ and $t\in \mathbb{R}$, there is an open set $\mathcal{U}_p$ small enough such that $\Phi(H,\,t):\mathcal{U}_p\to\Phi(H,\,t)(\mathcal{U}_p)$ is a bijection. The question now arises as to whether $\Phi(H,\,t)$ can map any point outside $\mathcal{U}_p$ into $\Phi(H,\,t)(\mathcal{U}_p)$ (which situation would make $\Phi(H,\,t)$ a local bijection but globally many to one for some $t\in\mathbb{R}$). However, if two or more points are mapped to one point $\tilde{p}\in\Phi(H,\,t)(\mathcal{U}_p)$, from inference 4. deduce that $\exists t$ small enough that the two chosen preimages of $\tilde{p}$ both lie in $\mathcal{U}_p$, thus contradicting local bijectivity. (Informally, from inference 4, multiple points of a function can only arise from mappings along connected "forked" flow lines, so zoom in near enough in on the fork point and thus contradict local bijectivity, showing that forks are impossible). Repeating the reasoning for $-t$ lets us deduce that multiple points are impossible and $\Phi(H,\,t):\mathcal{P}\to\mathcal{P}$ is a global bijection (indeed a symplectomophism in the light of inference 1, but, again, this information is further to our needs);

  6. Inference 6: From inference 5 and axiom 3, deduce that if there is some number $M$ of particles in any subset $\mathcal{V}\subseteq\mathcal{P}$, then there are precisely $M$ particles in $\Phi(H,\,t)(\mathcal{V})$. From inference 2. deduce that $\mathcal{V}$ and $\Phi(H,\,t)(\mathcal{V})$ have the same volumes. Therefore infer that the average particle density in any subset $\mathcal{V}\subseteq\mathcal{P}$ is constant if the particle states and subsets evolve by Hamiltonian flows;


  7. Inference 7: Apply inference 6 to a small open set that is shrunken according to an appropriate limiting process to deduce that the density function $\rho(p,\,t)$ at point $p$ and at time $t$ must be the same as the density at point $\Phi(H,\,-\mathrm{d}t)\,p$ at time $t-\mathrm{d}t$. Putting these words into symbols: $\mathcal{L}_{-X}\rho=\frac{\partial\,\rho}{\partial\,t}$, where $X$ is the vector field tangent to the Hamiltonian flow $\Phi$. This is, of course, $\{H,\,\rho\}=\frac{\partial\,\rho}{\partial\,t}$, or Liouville's equation.


Circularity of Other Proofs


Ultimately, I don't believe that proofs of Liouville's equation grounded on the divergence theorem are different from the above: I think that they are tacitly introducing Axiom 3 as "obvious" (even though I hope I have shown at the beginning of my answer that it doesn't always hold) and then the continuity equation and incompressible flows are simply an expression of this tacitly assumed axiom. So I don't think that these "proofs" are circular, just somewhat badly written in making use of tacit assumptions.


Summary


User Image sums all this up nicely (I was perhaps too brainfried to make the last step):



For Axiom 3 however, you showed that Axiom 3 $\Rightarrow \frac{d \varrho}{d t} = 0$. The other direction $\frac{d \varrho}{d t} = 0 \Rightarrow$ Axiom 3 is readily discussed in any textbook (trajectories do not start, end or cross etc.). So in fact we have Axiom 3 $\Leftrightarrow \frac{d \varrho}{d t} = 0$ when we are in the context of Axiom 1+2, e.g. classical mechanics. Hence, Liouville's equation is an axiom.



and indeed, in the presence of the other two, my axiom 3 is logically equivalent to Liouville's equation. My version is perhaps more physically transparent, but open to interpretation, and so the assertion of Liouville's equation as an axiom is perhaps more succinct and precise. So the answer to the title question is that Liouville's Equation must indeed be added as an axiom, and, in the presence of Axioms 1 and 2, it has the meaning that particle number of all species is conserved.



mass - Massive excitations in Conformal Quantum Field Theory


Single particle states in quantum field theory appear as discrete components in the spectrum of the Poincare group's action on the state space (i.e. in the decomposition of the Hilbert space of quantum states into irreducible representations of the Poincare group). Classification of irreducible unitary representations of the Poincare group leads to the notions of mass and spin.


Now, suppose we have a conformal QFT and are doing the same trick with the conformal group. Which irreducible representations do we have?


We still have the massless particles (at least I'm pretty sure we do although I don't immediately see the action of special conformal transformations). However, all representations for a given spin $s$ and any mass $m > 0$ combine into a single irreducible representation.




  • What sort of physical object corresponds to this representation?

  • Is it possible to construct a scattering theory for such objects?

  • Is it possible to define unstable objects of this sort?




mathematics - The Calculator with Misbehaving `+` and `*`


Well, that calculator looked like an ordinary one when I saw it in the shop, but it behaves most extraordinarily! I tried using the + and * keys on small integers, but the results I got were totally wrong and usually decidedly non-integer. However, there does seem to be some method in its madness. Its results are never negative and always fairly small. Provided that it doesn't display Error -- which it does in many cases, unfortunately.


Whatever these operations are that the + and * keys do, they behave in some ways like $+$ and $*$. If I denote them by $\oplus$ and $\otimes$, then, for any $a, b, c$:


$a\oplus b=b\oplus a \\ (a\oplus b)\oplus c=a\oplus(b\oplus c) \\ a\otimes b=b\otimes a \\ (a\otimes b)\otimes c=a\otimes(b\otimes c) \\ a\otimes (b\oplus c)=(a\otimes b)\oplus(a\otimes c)$


That is, $\oplus$ and $\otimes$ are commutative and associative, and $\otimes$ is distributive over $\oplus$.



$10$ has this simple property: for any $a$, $10\otimes a=a$.


Hint 1:



Each other number $a$, as entered on the keypad or displayed, has an internal equivalent $f(a)$. Whenever $f(a)+f(b)=f(c)$, $a\oplus b=c$, and whenever $f(a)*f(b)=f(c)$, $a\otimes b=c$. The image or range of $f$ is a field.



I also found this curious relationship between $\oplus$ and $\otimes$:


$a\otimes b=s\oplus d$


where $s$ and $d$ are the sum and difference of $a$ and $b$, found, I hasten to add, by using ordinary arithmetic, not by using the + key!


Occasionally it produces an integer. What integers are these values? In each case, give the lowest possible.




  1. $13\oplus 7$

  2. $z$ where, for all $a$, $a\oplus z=a$ and $z\otimes a=z$

  3. $24\otimes 14\otimes 7\otimes 6$

  4. $14\otimes 6\otimes 4\otimes 3$

  5. $13\otimes 12\otimes 11$

  6. $\sqrt{10\oplus 10\oplus 10}$


And what's the hint in them taken collectively?



Answer



I have an answer that works. I haven't yet proved it to be unique. We can take




$f(x) = 2 \cos (\frac{\pi}{30}x)$.



Then the mysterious sum-and-difference identity turns into



a standard trigonometrical identity.



The magic number



30 is there to make $f(10)=1$ as required.




The answers to the requested calculations are, in order,



3, 15, 19, 9, 14, 5



which are the alphabet-positions of the letters of



cosine.



I think the statement in hint 1 is not quite correct, by the way:




we have f(x)=f(-x) for any x and f(x+60)=f(x), so f doesn't have a well-defined inverse, which means that it can't possibly be true that e.g. "whenever $f(a)+f(b)=f(c)$, $a\oplus b=c$" -- because whenever the stated condition holds for a,b,c it also holds for a,b,-c and a,b,c+60 and so on, and $a\oplus b=c$ can't be true for more than one choice of c.



Thursday, February 21, 2019

special relativity - Does a constant 4-acceleration implies constant acceleration?


Before asking the question I explain here my line of thoughts:


I'm almost sure that a constant acceleration $\bf{a}$ implies constant 4-acceleration $w^\mu$. So I assume constant acceleration, then, in the rest frame of the particle where its velocity is zero we have $w^\mu=(0,\bf{a})$, thus, if $\bf a$ is constant, the quantity $w^\mu w_\mu$ will be constant aswell and Lorentz-invariant. If there's a mistake you're free to point it out.


Anyway, assuming what I've said true I ask the opposite question: does constant four acceleration implies constant acceleration?


I didn't come up with a satisfying yes, I tried to think in analogy of what I've written above, but I only obtained that just in the comoving reference frame i have constant acceleration, but then in another reference frame where $w^\mu$ is function both of the velocity and the acceleration of the particle, how can I know they don't modify themselves in order to keep the initial assumption of constant 4-acceleration valid?



Answer




The norm of the four-acceleration (i.e. the proper acceleration) defined by:


$$ A^2 = g_{\alpha\beta} a^\alpha a^\beta $$


is a Lorentz scalar and therefore the same for all observers. However there is nothing to stop me from choosing some bizarre coordinates in which the four acceleration is time dependent but this is cancelled out by the time dependence in the metric to give a constant proper acceleration. So in this sense a constant proper acceleration does not imply a constant four acceleration or vice versa.


What is a charge?



If mass is the measure of the inertia of a body, then what is a charge?


Everyone tells that it is a fundamental property but i believe it must have a unique definition. Anyone know in this regard?




quantum mechanics - Assumptions in Heisenberg's 1925 paper


I am not entirely clear as to what were the bases for Heisenberg's assumptions in his 1925 paper. He claims that one cannot consider relations between quantities that are unobservable "in principle", like the position and period of revolution of an electron.


To quote some text : "These rules (the abovementioned relations) lack an evident physical foundation, unless one still wants to retain the hope that the hitherto unobservable quantities may later come within the realm of experimental determination.



This hope might be justified if such rules were internally consistent and applicable to a clearly defined range of quantum mechanical problems."


My first query is why does he claim the position and period of an electron to be unobservable "in principle"? There was theoretically no reason (at THAT time) to doubt that these quantities could be measured, though certainly they were indeterminate practically.


Secondly, just because a theory dealing with those quantities is inconsistent, or not general enough, why does it imply that we cannot define or measure quantities that that theory deals with? We may be able to measure some quantities perfectly, but still formulate an incorrect theory around them.


Finally, is there any ad-hoc basis to decide what these "uncertain" quantities are? More specifically, how could Heisenberg pinpoint position of an electron as an uncertain parameter and not any other quantity (like some electric field, etc.)?


Thanks in advance. (by the way I'm studying the original paper solely to look more closely at the motivation for assumptions underlying the theory)



Answer




My first query is why does he claim the position and period of an electron to be unobservable "in principle"? There was theoretically no reason (at THAT time) to doubt that these quantities could be measured, though certainly they were indeterminate practically.



Werner Heisenberg obviously disagreed with this assumption of yours and it just happened that his ability to disagree made him a founder of quantum mechanics.



He has spent several years by trying to develop "quantized planetary" models of the helium atom etc. before he understood that this failing project is failing for fundamental reasons. Such a helium with well-defined positions would be described by a chaotic 3-body problem and there would be no way how it could be consistent with the known regular behavior of the helium atom (and other atoms and other coherent systems), including the sharp spectral lines.


So Heisenberg was able to see in 1925 something that you can't see now: that the electrons can't be going along any particular trajectories while they're in the atoms. Instead, what is observed is that they have a totally sharp energy from a possible list, the spectrum – something we can really observe via the photons that atoms emit or absorb. To conclude that electrons can't be going along particular classical trajectories in the atoms, he didn't have to wait for measuring apparatuses that would be sufficiently accurate. He was able to make this conclusion out of the available data by "pure thought", and he was right.



Secondly, just because a theory dealing with those quantities is inconsistent, or not general enough, why does it imply that we cannot define or measure quantities that that theory deals with? We may be able to measure some quantities perfectly, but still formulate an incorrect theory around them.



Many combinations of options would be possible in a generic hypothetical world and you're right that the combination of options you mentioned would be logically possible in another world but Heisenberg was talking about our world. He learned his message from special relativity that one shouldn't talk about things that can't be operationally defined – such as the simultaneity of events (which is observer-dependent) and tried to maximally apply this positivist mode of reasoning to the world of atoms. His analysis dictated that he may assume that the electron in the atom has a particular energy for a long time but it can't have a well-defined position or velocity. So he reformulated physics around the notion of the energy which is measurable and found out the first formulation of quantum mechanics in the energy eigenstate basis Heisenberg picture.



Finally, is there any ad-hoc basis to decide what these "uncertain" quantities are? More specifically, how could Heisenberg pinpoint position of an electron as an uncertain parameter and not any other quantity (like some electric field, etc.)?



You are mixing apples with oranges here. Heisenberg's paper wasn't discussing the electromagnetic field. It was discussing the general logical framework underlying physics and the examples he took were those from mechanics – rigid rotator and anharmonic oscillator – that were meant to be later generalized to a theory of atoms in particular just by a new choice of the potential energy formula.



There's no observable concept of "electric fields" in the description of an atom or anharmonic oscilator at all. Even in classical physics, one deals with functions of positions and momenta. He figured out that not all functions are equally observable: energy (a particular function of positions and momenta) is much more observable and stable.


The underlying logic he has developed was later (soon) applied to other systems in mechanics such as atoms and molecules as well as field theory such as electromagnetism. But the essence isn't in describing which degrees of freedom are there (they're kept as close to those in the corresponding classical theory as possible); the essence of quantum mechanics is in the totally new set of postulates and methods to make predictions.


He realized that the right goal wasn't just to find another classical theory just with some new degrees of freedom, which is the intrinsic, fundamental, and completely flawed assumption of your whole question from the beginning to the end. He realized that the new insights force physicists to formulate a completely new theory – and he (and others) has (have) already used the completely new term "quantum theory" for it – and he just did so, discovering some of the new explicit quantum formulae for nontrivial predictions (beyond the spectrum of the Hydrogen atom that was "explained" by Bohr's toy model).


You may repeat many times that a complete conceptual revolution in physics (switching from the classical to the quantum) wasn't needed and one should have only discussed new classical models with new variables (paying no attention to whether or not they may be actually observed) except that Heisenberg knew that it was needed and the months (and a few years) that followed his discovery made his assumption unquestionable.


conservation laws - What is the Noether charge associated with the the color $SU(3)$ symmetry of QCD?



A version of the Noether's theorem applies to local gauge symmetries. What is the Noether's charge associated with a non-abelian gauge symmetry such as the color $SU(3)$ and how is that derived? I want an expression for the color charge operator like we have an expression for the electric charge operator. Please see Eq. (9) and (11) of the answer here.



Answer



The $\mathrm{SU}(3)$ gauge symmetry is a local symmetry, and therefore it is not Noether's first, but Noether's second theorem that applies to it, which does not yield conserved quantities.


For $\mathrm{U}(1)$ gauge symmetries like the electromagnetic symmetry, there is also a global $\mathrm{U}(1)$ symmetry, and hence a conserved quantity. But the global symmetry associated to a non-Abelian gauge symmetry is just the center of the gauge group, which is discrete for $\mathrm{SU}(3)$, and hence there is no conserved quantity associated to it. This center symmetry has physical significance e.g. in models of confinement, see this question and its answer.


fluid statics - Surface Tension-- Sessile drop


For background :- Surface tension and capillarity I was reading the above answer. Particularly the sessile drop thing. I understand that as water is being pulled into the solid(assuming it is hydrophilic).. correspondingly the force Ysl acts as shown on the sessile drop's surface.


But am confused regarding why the solid-gas surface tension is acting on liquid surface as shown. I understand that, for example if gas is trying to move away from solid.. A surface tension will be there on the gas' surface (of magnitude Ysg).. but am unable to understand why the liquid's surface will be pulled with Ysg as shown in the question above.


Please help in clarifying. I would be grateful. I am just a high school student. Thank you for your help.


Also, does this imply that surface tension of solid-liquid interface will also act on the gas??



Also.. If you are interested and capable please answer this too.. I would be very happy :- Pressure and surface tension



Answer



It is not about the forces on the liquid. The boundary of the liquid; in better words, the outline of the shape that the sessile drop traces on the solid, is the interface between three media: solid, liquid, and gas. For a particle at this interface to be stable, net force acting on it must be zero. This is what gives Young's equation. It's not the force on the liquid. It's the force on the molecule at the triple interface. And since it's a triple interface, there'll be three forces.


This should clear it up.


Detecting if resistances are parallel or series in complex circuits



I know how to detect when resistors are arranged in parallel or series arrangement and
I can also find their equivalent resistance in simple circuits or when resistances are connected in form of
triangle but what happens when the arrangement is complex like this :
Resistors


Which resistors are parallel and which are in series ? How can I find the equivalent resistance in such cases ? Is there rule or method for figuring this out ?




Wednesday, February 20, 2019

strategy - What is the lowest you can get in 2048?



What is the lowest score you can get in 2048? Creating high numbers is obviously bad, but what about not matching numbers? What is the chance of getting this lowest score?



Answer



If you were unlucky enough and were not trying...You could in theory end up with a square full of alternating 2s and 4s. It should even be possible to do this without matching, which means a score of 0.


board games - Unfinished Snakes and Ladders


An entry for Fortnightly Topic Challenge #40, have fun with non-Chess puzzle! ;)




Three young children: Red, Green, and Blue; were playing a classic game of Snakes and Ladders, when suddenly their mom called them to have a dinner.


When they came back to their room, they forgot whose turn it was even they were not sure either about the order of the play!


This was the last condition of the board, including the die. Could you tell them whose turn it is and the order of the play?


Unfinished Snakes and Ladders



Some notes and clarifications:




  • They put their markers at tile 1 at the beginning.

  • They will win if their markers land on exactly at tile 100. They are using bouncing back version (roll 3 will move the marker from tile 99 to 98).

  • As long as they roll 6, they will get an additional extra roll after moving the marker.

  • The ladders should be used to go to higher number only, and the snakes are used to go to lower number. For the "interrupted" ones, both are immediately used (e.g. landing on 39 will bring you to 25, same as landing on 20 will bring you to 25 too.)

  • If someone land on a square where there is already a player, it doesn't sent him back to square 1.




Answer



The other guys already got the answer. (Maybe not with completely airtight arguments, but nevertheless.) The explanations would be a lot easier to follow if they had pictures, so here's one:




enter image description here (errata: 25 should be red)



Legend:



  • Red square: after turn 1, you are in one of these.

  • Green square: after turn 2, you are in one of these, or in some red square (except square 2).

  • Red circle with R: if Red was the last player to move, Red's turn started at one of these.


From there, the solution is pretty simple:




1. Green has had either one or two turns.
2. Therefore, both Red and Blue have had at most three turns.
3. Therefore, because no green square overlaps with a letter R, Red was not the last to move.
3.1 There is no way to get to Red's current position within two turns, so Red has taken at least three turns. Combined with point 2, Red has taken exactly three turns.
3.2 Since Red has already taken three turns, but wasn't the last player to play, someone else must have also played three turns. It can only be Blue.
4. Because Red has played three turns, Green has taken 2 turns, rolling first 1, then 2.



So the only remaining things to check are that



5. Red really can get to its current square in 3 turns (yes, starting the third turn at 66, 62, 60 or 56.)

6. Blue can get to its current spot in 3 moves, ending with a 4 (yes, starting the third turn at 58, 60, 65, or indeed 77)



So finally, the answer is



It's Green's turn, and after that it's time for the fourth round, with Red playing next.



general relativity - Equation of motion of a photon in a given metric


I have this metric: $$ds^2=-dt^2+e^tdx^2$$ and I want to find the equation of motion (of x). for that i thought I have two options:




  1. using E.L. with the Lagrangian: $L=-\dot t ^2+e^t\dot x ^2 $.





  2. using the fact that for a photon $ds^2=0$ to get: $0=-dt^2+e^tdx^2$ and then: $dt=\pm e^{t/2} dx$.




The problem is that (1) gives me $x=ae^{-t}+b$ and (2) gives me $x=ae^{-t/2} +b$.




letter sequence - Scrambled Songs


I've used quite a simple algorithm to create the appearance of a jumbled up bunch of letters. These letters represent a song. What is the song title?






(Everyone should know this song, unless you live in a non-English speaking country.)



Answer



The answer is



"Twinkle Twinkle Little Star", each letter corresponds to the first letter of each word in the song.



newtonian mechanics - Connection between moment/torque and centre of gravity?


So I understand how moments work with regards to basic examples like pushing a door, in that the further you are away from the hinges of the door, the greater the moment, which is like a turning force. I also know, though perhaps do not fully understand, the formula for calculating torque, $\tau=\mathbf r\times \mathbf F$ where $\mathbf F$ is the force applied, $\mathbf r$ is distance from centre, and $\times$ denotes cross product.


However, I don't really understand how it is connected to the centre of gravity (sometimes called centre of mass). For instance, a problem in my textbook addresses the issue of the centre of gravity of a triangular-shaped piece of thin card, with equal mass distribution. How do moments help me find the centre of gravity? What forces are being applied to the piece of card?



Answer




  • When pushing down on a lever to lift something heavy, your push is a force. You apply it at some point.


Now consider the exact same situation, just with gravity being the force instead of your push:



  • gravity pulls down on the lever. It applies this pull at some point.



This point is the center of gravity. Let's call it CoM. That is all. You talk about the CoM in connection to where gravity pulls.



How do moments help me find the centre of gravity?



Only moments caused by gravity can help you, since only then does the moment have anything to do with the CoM.


Gravity will pull in the CoM as a force applied at a point. So as long as this CoM is not straight under the rotation point, gravity will try to rotate the object (it will cause a torque, since the distance $r$ is not zero in your formula, unless the CoM is exactly under the rotation point).


So when your object hangs still and doesn't rotate anymore, you know that the CoM must be somewhere directly below the rotation point. On the vertical line below.


If you now hang the object in another point so that you have a new rotation point, then when the object hangs still again, you again know that the CoM must be directly below somewhere on the vertical line under the new rotation point. These two lines have only one point in common. So this point must be the CoM, because it must be the same for both situations (since you didn't redistribute the mass).




What forces are being applied to the piece of card?



If it hangs freely in a hook for example, so it can rotate freely, then only two forces act on the card. Gravity from the CoM and the hook's normal force holding it.


Since the normal force by the hook works in the rotation point, this gives no torque (the distance $r$ is zero). Therefore only gravity is left to do a torque to make it rotate.


And if the CoM is vertically below the rotation point, then also gravity gives no torque as explained above. Therefore the above method to find the CoM will only work if no more forces than the gravity can cause a torque.


electromagnetism - What is the difference between gravitation and magnetism?


If you compress a large mass, on the order of a star or the Earth, into a very small space, you get a black hole. Even for very large masses, it is possible in principle for it to occupy a very small size, like that of a golf ball.



I started to think, how would matter react around this golf ball sized Earth? If I let go of a coffee mug next to it, it would go tumbling down toward the "golf ball". Isn't that exactly how magnets work, with paperclips for example?


Magnets are cool because they seem to defy the laws of gravity, on a scale that we can casually see. Clearly, the force carrier particles that produce electromagnetic attraction are stronger than gravity on this scale (or are at least on par: gravity plays some role in the paperclips path, but so does electromagnetism).


My question is, why do we try to consider gravity as anything different than magnetism? Perhaps "great mass" equates to a positively (or negatively) charged object. Pull so much matter in close and somewhere you've crossed the line between what we call electromagnetic force and gravity force. They are one in the same, no?



Answer



There are several qualitative and quantitative differences between gravity and magnetism.




  1. When you attract 'neutral' bits of metal with a magnet, or attach it to something like a plate of metal, what's happening is that individual atoms of the metal react to the magnetic force. In a ferromagnetic metal, one with a similar electronic structure to Iron or Nickel, the individual atoms work like nanoscopic magnets; but they are very weak, and they are not lined up with one another, so that their fields cancel one another out over any macroscopic distance. But if you bring a "large" magnet (such as a fridge magnet) up to them, the field of the large magnet causes them to align with the field, so that they are pulled towards the magnet — and the magnet is pulled towards them. This is why some metal objects are attracted to magnets.


    Other metals, such as aluminum or silver, also react to magnets, but much more weakly (and in some cases repulsively): the way that they react to magnetic fields is described as paramagnetism (for materials which align very weakly with magnetic fields) and diamagnetism (for materials which align very weakly against magnetic fields).


    The very fact that different materials react differently to magnetic fields is something that sets magnetism apart from gravity. Gravitation works equally with masses of any sort, and is always attractive (as noted by Nic); magnetism can both attract and repel, and do so with different degrees of force, as between ferromagnetic, paramagnetic, and diamagnetic materials. But of course, quite famously, even a single object can be both attracted and repelled by magnetic forces: the north poles of two magnets repel each other, as do the south poles; only opposite poles attract each other. (This, of course, is the basis on which compasses work.)





  2. The way that these forces operate over distance also varies. Gravity very famously (but only approximately) obeys an inverse-square law; the field far from a bar magnet, however, decreases like the inverse of the cube of the distance from the magnet.




  3. Finally, moving electric charges produce magnetic forces; whereas they don't cause any gravitational forces which could not be accounted for just by the fact that the charged particles have mass (whether moving or at rest).




So, on both the macroscopic level and on the level of individual atoms, the forces of gravity and magnetism act quite differently.


Tuesday, February 19, 2019

classical mechanics - How do waves have momentum?


A question on a practice test I'm taking is as follows:




By shaking one end of a stretched string, a single pulse is generated. The traveling pulse carries:
A. mass
B. energy
C. momentum
D. energy and momentum
E. mass, energy and momentum



How would one describe the momentum of a wave?




capacitance - In an RLC series circuit on resonance, how can the voltages over the capacitor and the inductor be larger than the source voltage?


Consider an RLC circuit in series, of the form



If the source drives the circuit in AC at the resonance frequency $\omega =1/\sqrt{LC}$, the peak-to-peak voltages on the capacitor and the inductor, $$ V_C=\left|\frac{Z_C}{Z_\mathrm{tot}}\right|V_S=\frac{\frac{1}{\omega C}}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}V_S \quad \text{and}\quad V_L=\left|\frac{Z_L}{Z_\mathrm{tot}}\right|V_S=\frac{\omega L}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}V_S ,$$ can both be larger than the peak-to-peak voltage $V_S$ of the source.


The math might say one thing, but this is till terribly counterintuitive. How can this be?




special relativity - Trying to understand relativistic action of a massive point particle



I got badly lost in derivation of relativistic formulas for energy and momentum.


I stumbled upon relativistic action as follows (which should explain relativistic motion of a classical particle):


$$ S = \int Cds=C\int_{t_i}^{t_f}\sqrt{c^2-(x')^2}dt $$


Where $C$ is some constant (depends on what kind of physics we put in equation) and $s$ is relativistic interval. Later on Lagrangian $$ L(x')\equiv C\sqrt{c^2-(x')^2} $$ is used in derving relativistic energy and momentum.


I am familiar with Lagrangians and symmetry rules which connect energy and momentum to Lagrangian formalism. What I do not understand is this action - weren't action sum over time? Why all of a sudden it is sum over relativistic interval?



Answer



The action is commonly written in terms of $ds$ because it is a Lorentz scalar. $dt$ is not a Lorentz scalar, but $dt \sqrt{1 - v^2} = dt / \gamma = ds$ is, so you can write the action as an integral over time if you want: $$ S = - m \int ds = - m \int dt \sqrt{1 - v^2} $$ We can check this in the non-relativistic limit: $$ S = - m \int dt \sqrt{1 - v^2} \approx \int dt \left(-m + \frac{1}{2} m v^2 \right) $$ The constant $-m$ does not affect the equations of motion, so it can be removed, and we are left with $L = m v^2 / 2$, as expected.


The speed of light $c=1$ above.


special relativity - Derivation of Lorentz boosts


I was deriving the matrix form of Lorentz boosts and I came up with a doubt. I don't think I quite understand hyperbolic rotations.


The stadard basis of the Minkowski space is given by $\{e_0, e_1,e_2,e_3\}$, where the square norm of any vector $x=x_0 e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3$ is $|x|^2 = x_0 ^2 - x_1 ^2 - x_2 ^2 -x_3 ^2$. Note: $x_0$ is the time component ($e_0$ indicates time axis), $x_1, x_2, x_3$ is the spatial component ($e_1 , e_2, e_3$ indicate the $x, y,$ and $z$ axis).


Let $B_i$ be a Lorentz boost in the ith direction. This boost will only modify the time component and the $ith$ component, and like any other lorentz transformation, it will preserve the norm of any vector. Consider $B_i e_0 = a e_0 +b e_i = e_0 '$. Then,


$(B_i e_0)^2 = e_0 ^2 = 1$


$a^2 - b^2 = 1$


The solutions of this lie on a hyperbola along the $e_0$ axis. We can parametrize it and obtain:


$a=cosh \theta$ and $b=sinh\theta$


I do the same procedure for $B_i e_i = a e_0 +b e_i = e_i '$. $a$ and $b$ satisfy the equation: $b^2 -a^2 = 1$ which is a hyperbola along the $e_i$ axis.



My intuition: I imagine the $e_0$ as the horizontal axis in the x-y plane and the $e_i$ as the vertical axis. If I shift $e_0$ by and angle $\theta$ (counterclockwise) over the hyperbola, the resulting vector $e_0'$ will fall on the first quadrant. So $e_0 ' = cosh \theta e_0 + sinh \theta e_i $. Now I need to shift $e_i$ by the same angle (counterclockwise) over the corresponding hyperbola so that I can keep $e_i$ and $e_0$ orthogonal to each other. The resulting vector $e_i'$ will fall over the 2nd quadrant. So $e_i ' = -sinh \theta e_0 + cosh \theta e_i $


For $i=1$, I end up with a boost $B_i$ of the form: $$B_i= \left( \begin{matrix} \cosh \theta & -sinh \theta & 0 & 0\\ sinh \theta & cosh \theta & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{matrix}\right) $$ when I should end up with:


$$B_i= \left( \begin{matrix} \cosh \theta & sinh \theta & 0 & 0\\ sinh \theta & cosh \theta & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{matrix}\right) $$ Edit: The columns of this matrix are correct because the matrix satisfies Lorentz transformation condition $\eta = B_1 ^T \eta B_1$ (where $\eta$ is the minkowski metric tensor) while my boost matrix doesn't.
Question : Since taking the square might make me lose/add a negative sign, is there a better way to obtain $e_0'$ and $e_1'$ with all the necessary negative signs in its components? Right now I would just be using trial an error, until the boost matrix I obtain satisfies $\eta = B_1 ^T \eta B_1$ I appreciate any insights, thanks!



Answer



In your attempted construction, you first use the future unit-hyperbola for the tips of your unit-timelike vectors. Note that this hyperbola has spacelike tangents. In fact, following Minkowski's own construction from "Space and Time"



We decompose any vector, such as that from O to x, y, z, t into four components x, y, z, t. If the directions of two vectors are, respectively, that of a radius vector OR from O to one of the surfaces ∓F = 1, and that of a tangent RS at the point R on the same surface, the vectors are called normal to each other. Accordingly, $$c^2tt_1 − xx_1 − yy_1 − zz_1 = 0$$ is the condition for the vectors with components x, y, z, t and $x_1$, $y_1$, $z_1$, $t_1$ to be normal to each other.



If R is the tip of the unit-timelike vector OR from O, the tangent to the hyperbola at R is [Minkowski-]orthogonal to OR. The "intuition" to have is that the tangent to the "circle" in that geometry is orthogonal to the radius vector.



(You don't need to use the other hyperbola [with timelike tangents].)


classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?

I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...