general relativity - What is the 'apparent horizon' of a black hole?

The Wikipedia article is full of general relativistic and differential geometry jargon, and its accuracy is questionable. So, without such (or explaining said jargon without more jargon), what is the 'apparent horizon' of a black hole?

Answer

The apparent horizon is a technical term, so you're going to have to have SOME jargon to explain it--the distinctions between trapping, apparent, dynamical and event horizons does just require some knowledge of differential geometry and general relativity, even if you can explain them using common words, as below.

Freeze a moment in time in some particular coordinate system. Then, if that coordinate system contains an apparent horizon, it will be closed two-dimensional surface (think a ball, but not necessarily undistorted) where all light rays but one are forced to move into the surface when evolved in time, and the last light ray will be 'frozen' on the surface at that moment of time (think of the outgoing light ray at $r=2M$ in the Schwarzschild solution--it tries to get away, but is just stuck on the surface).

It is called apparent because it:

1. depends on the notion of constant time on that one surface


2. is therefore coordinate dependent, and in some cases, can be eliminated entirely by choice of coordinate


3. can have strange, unpredictable behaviour


4. is generally inside the proper event horizon, but



5. if the ''stack'' of apparent horizons forms a timelike surface, the surface will be two-way transversible, and therefore, doesn't correspond with what we normally think of as a 'horizon'.

atomic physics - Hydrogen Spectra

I am talking about hydrogen spectral lines such as Lyman, Balmer etc. In order to make those spectral lines series more than one electron are needed to jump from higher orbits to lower orbit. But there is only one electron in an hydrogen atom. How does hydrogen spectrum happen then?

optics - Laser beam sensed by scattering

Can a laser beam, whether it is a IR laser or of the visible to human eye spectrum, be seen/sensed by an electronic eye like a photosensor/photodetector, not human eye, if the beam is just in air? For example if a laser beam is say beamed up in the sky from the ground, can it be detected by an electronic eye like a photosensor/photodetector just because of the scattering from the air?

Answer

In principle, yes, but for clear air the scattering is very very weak, and the scattered light would probably be drowned out by other background sources, especially the blue sky. It would be easier at night.

If the air is not clear, but instead is carrying dust or water droplets or smoke, the beam would be easily visible and recordable, again much more easily at night.

There are plenty of photographs of night-time laser light shows in which the beam is easily visible.

quantum mechanics - Solving one dimensional Schrodinger equation with finite difference method

Consider the one-dimensional Schrodinger equation

$$-\frac{1}{2}D^2 \psi(x)+V(x)\psi(x)=E\psi(x)$$

where $D^2=\dfrac{d^2}{dx^2},V(x)=-\dfrac{1}{|x|}$.

I want to calculate the ground state energy(which is known to be $-0.5$) with finite difference method.

Suppose $\psi(-a)=\psi(a)=0$ for sufficiently large $a$.

Split $[-a,a]$ into $N$ sub-intervals with equal length $h$: $$-a=x_1

Now use

$$D^2\psi(x_n)\to\frac{\psi(x_{n-1})-2\psi(x_{n})+\psi(x_{n+1})}{2h^2}$$

and

$$-\frac{1}{2}D^2 \psi(x_n)+V(x_n)\psi(x_n)=E\psi(x_n)$$

we get

$$-y_{n-1}+2(1+V_n h^2)y_n-y_{n+1}=2h^2 Ey_n$$

where $y_n=\psi(x_n)$.

Recall that $y_1=y_{N+1}=0$, we get

$$H \begin{pmatrix} y_2 \\ \vdots \\ y_N \end{pmatrix} = 2h^2E \begin{pmatrix} y_2 \\ \vdots \\ y_N \end{pmatrix}$$

where

$$H=\begin{pmatrix} 2(1+V_2h^2) & -1 & & \\ -1 & 2(1+V_3h^2) & \ddots & \\ & \ddots & \ddots & -1 \\ & & -1 & 2(1+V_Nh^2) \end{pmatrix}$$

is a $N-1$ tridiagonal matrix.

If the least eigenvalue of $H$ is $\lambda$, then I expect $\dfrac{\lambda}{2h^2}$ to approximate the ground state energy $E$.

So I write a MATLAB program:

function m = onedimen1(a,M)

N = 2*M+1;


h = 2*a/N;
t = -a;

for i = 1 : N-2
    t = t+h;
    H(i,i) = 2*(1+vp(t)*h*h);
    H(i+1,i) = -1;
    H(i,i+1) = -1;
end


H(N-1,N-1) = 2*(1+vp(t+h)*h*h);

m1=eig(H)/(2*h*h);
m = m1(1:5);

  function y = vp(x)
    y = - 1/(abs(x)) ;
  end
end

but when I try

onedimen1(10,100)
onedimen1(10,1000)

the least eigenvalue of $H$ divided by $2h^2$ appears to be too small(and tends to negative infinity when $N$ increases), but the second largest divided by $2h^2$ is always near $-0.5$.

Can anyone explain why?

(P.S. If I change $V(x)$ to $V(x)=\dfrac{1}{|x|}$ then the least eigenvalue of $H$ is stable)

Answer

I'm making this an answer because its too long to be a comment but its just an expansion of the things already stated in comments:

Non-normalizable states: The Schroedinger equation has an infinity of solutions but almost all of them do not have a finite norm ($\int|\psi(x)|^2dx$ is not finite). These are not phyiscally acceptable, since there would not be a probabilistic interpretation, amongst other issues. It is possible that when you discretize your system you are getting solutions that would have infinite norm in the limit $h\rightarrow 0$ and so are not physically acceptable. I strongly suspect this is the issue. To see if this is case look at the wavefunction associated with your bad eigenvalue ( this is the eigenvector associated with the eigenvalue). If in the limit $h\rightarrow0$, your solution goes to increases faster than $\frac{1}{\sqrt{x}}$ as $x\rightarrow0$ than it is non-normalizable and unphysical.

'Softening the singularity': the issue is that you potential goes to infinity at some point and that general numerical simulations don't like it when things go to infinity. Especially having your potential go to negative infinity since the wavefunction wants to gather around the singularity. The way around this would be to replace your potential $\frac{1}{|x|}$ with say $\frac{1}{\sqrt{x^2+a^2}}$ where $a$ is some number you pick, and solve for the states. This new potential is the same as the old when $x\gg a$ but it doesn't have any singularity anymore, so it shouldn't be a problem. To get back the real answer solve for the states with smaller and smaller values of $a$ until you see that if a limit is reached. It does not necessarily reach a limit and in this case you will still get the state that diverges.

standard model - Some very basic questions on the Higgs Boson

What exactly is a boson?

Is the Higgs boson the cause of gravity or a result of it? Does the collision of particles at the LHC create a gravity field or waves or somehow interact with the gravity field of the earth?

The Higgs Boson is supposed to be quite massive and equivalent to a large number of protons. Were many particles needed to create it or only a few travelling at high speeds? Was the high energy converted into the large mass?

Why is the particle so short lived and what does it decay into?

Answer

What exactly is a boson?

A boson is a particle whose spin (= intrinsic angular momentum) is an integer number. For example, the photon (the particle that is responsible for the electromagnetic force) is a boson. Contrast this with a fermion, such as the electron, whose spin is a half integer. In everyday terms, the bosons are the microscopic particles that make up the forces: electromagnetism and gravity, as well as the weak and strong forces. Fermions are the particles that make up matter: protons, neutrons, and electrons. The fact that forces come from bosons while matter comes from fermions is a very deep observation, and is related (at least in part) to something called the 'spin-statistics' theorem, but let me not go into that.

Now let me note that all the bosons that are responsible for producing forces have spin 1 or 2. The Higgs boson is another kind of boson, which does not behave like a force. Technically, this is because its spin is 0. In fact, it is the only fundamental particle with spin 0 that we know exists: all the others that are known have spin 1/2 (fermions), 1, or 2.

Is the Higgs boson the cause of gravity or a result of it? Does the collision of particles at the LHC create a gravity field or waves or somehow interact with the gravity field of the earth?

The Higgs boson has nothing to do with gravity. It does not cause gravity and it is not the result of gravity. The particle that 'causes' gravity is called the graviton, it is a boson like all the other force-particles, and it has spin 2.

The Higgs boson does give mass some to the other fundamental particles, and this mass then interacts with gravity just like all other mass. But this is an indirect relation.

The particles at the LHC interact with the gravitational field of the earth just like all other particles. As part of the collisions, certainly some gravitons (= 'gravity particles') are also created. But these are completely negligible effects that can be ignored, because gravity is a much weaker force than all the others.

The Higgs Boson is supposed to be quite massive and equivalent to a large number of protons. Were many particles needed to create it or only a few travelling at high speeds? Was the high energy converted into the large mass?

I am not sure what you mean by 'equivalent to a large number of protons'. It is true that if you weigh one Higgs particle it will weigh as much as about 120 protons, but they are not related beyond this simple fact.

The Higgs particle, like all other particles produced at the LHC, is created by a collision between two highly energetic (= very fast) protons. Indeed, in the collisions in which a Higgs was created, the energy of the protons was converted into the mass of the Higgs particle.

Why is the particle so short lived and what does it decay into?

The Higgs is short-lived essentially because it is very massive. Roughly speaking, the more massive a particle is, the faster it decays into other particles. This is not a precise statement because it really depends on what the particle can decay to. For example, the proton is massive relative to the electron, but it seems not to decay to anything due to various reasons.

The Higgs particle has different possibilities to decay to (these possibilities are called 'decay channels'). For example, it can decay to two photons, or to two quarks like bottom and anti-bottom.

angular momentum - Why does everything spin?

The origin of spin is some what a puzzle to me, everything spin from galaxies to planets to weather to electrons.

Where has all the angular momentum come from? Why is it so natural?

I was also thinking do photons spin? we always think of the wave as a standard 2d sin wave but could this rotate in 3d? What implications would this have?

And what about spacetime how does all the spinning effect?

This has always been avoided in all lectures and classes I ever went to.

quantum mechanics - Experimentally, what categorizes a measurement as corresponding to a certain observable?

I want to write a computer program.

The input to the program is:

1. A description of an experimental device (for instance a Stern-Gerlach apparatus, or a laser and a polarizer)


2. What the experimenter will do with the device (e.g., fire an electron, fire a photon)

And the output of the program is a sequence of yes/no answers to the following questions:

1. Does this experiment correspond to a position measurement?


2. Does this experiment correspond to a momentum measurement?


3. Does this experiment correspond to an angular momentum measurement?


4. Does this experiment correspond to a spin measurement?



5. Does this experiment correspond to an energy measurement?

What I want to know is:

What is the algorithm the program uses to determine the result "yes" or "no" to each of the above questions?

Background

Why am I asking such a weird question? Because I've never gotten a clear understandable answer on what constitutes a measurement type. Note, I am not asking what constitutes a measurement; I am asking what constitutes the type of measurement. The answers I get are always hand-wavy and subjective. But it can't possibly be subjective, because otherwise the experimental results would not match those computed with quantum mechanics, e.g., if I compute the eigenvalues of the wrong observable, the values aren't going to match up with those obtained from experiment.

So to give you a concrete example, suppose I shoot an electron at a flat board. I can measure the board's recoil to get the momentum of the electron, or I can measure the location the electron struck the board to get its position. But that's a hand-wavy explanation! I just "decided" that I'd call the first experiment a momentum measurement because it sounds like one, and similarly I simply decided to call the other one a position measurement. I want an exact procedure to categorize whether it is a position-determining or momentum-determining experiment.

Longest word with adjacent letters on a keyboard

What is the longest English word you can type using only letters that are adjacent on a QWERTY keyboard?

You can start anywhere and then either repeat the previous letter or move to an adjacent one.

For convenience of those who don't have a QWERTY keyboard, it looks like this.

 ┌───┬───┬───┬───┬───┬───┬───┬───┬───┬───┐
 | Q | W | E | R | T | Y | U | I | O | P |
 └─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┘

   | A | S | D | F | G | H | J | K | L |
   └─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴───┘
     | Z | X | C | V | B | N | M |
     └───┴───┴───┴───┴───┴───┴───┘

I found this puzzle on the Internet and I don't have a definitive solution.

When setting number-sequence puzzles, is there any accepted method to prevent arbitrary solutions?

When setting a number sequence puzzle, one normally gives a finite sample and asks for the rule that specifies the members.

The sample has to be finite because it's impossible to uniquely specify an infinite series without giving away the answer. (Note: Is this true?)

Of course the problem is that any finite string of numbers can have an infinite number of rules that specify it.

Example

What is the next number in this series?

2 3 4 7 9 ?

To which someone could say: These are members of the set {2 3 4 7 9 73} so the answer is 73.

or

This is the union of the sets {2, 3, 4, 7, 9} and {10, 11, 12, 13, ...} so the answer is 10.

This is taking things to an extreme but how is it possible to counter such suggestions?

Note: The real-life example that caused me to pose the question is here Can you generate the next number in this integer series and describe the rule? and the first answer proposed was some arbitrary union of sets.

Question

Are there any recognised ways, or can you suggest a way, of designing number sequence puzzles that can tie down the generating rule exactly without actually specifying the answer? If not, is the number sequence puzzle genre dead?

condensed matter - What is spontaneous symmetry breaking in QUANTUM systems?

Most descriptions of spontaneous symmetry breaking, even for spontaneous symmetry breaking in quantum systems, actually only give a classical picture. According to the classical picture, spontaneous symmetry breaking can only happen for non-linear systems. Classical linear systems, such as Harmonic oscillators, can never have spontaneous symmetry breaking. (Here "linear" means that the equation of motion is linear.)

But the real QUANTUM systems are always linear since the Schrodinger equation is alway linear. So how can a linear quantum system have spontaneous symmetry breaking? Do we have a simple intuitive understanding for spontaneous symmetry breaking WITHIN QUANTUM mechanics? (without using the classical picture, such a Mexican hat -- the logo of physics.stackexchange)

The Mexican hat does give us an intuitive and pictorial understanding of spontaneous symmetry breaking in classical systems. Do we have an intuitive and pictorial understanding of spontaneous symmetry breaking in quantum systems.

Answer

Bei Zeng and I wrote a paper http://arxiv.org/abs/1406.5090 , which addresses this question:

A symmetry breaking phase for finite group G is a gLU equivalent class formed by symmetric many-body states that have GHZ entanglement.

In other words, a symmetry breaking phase is a set of

1. symmetric states $U_g \Psi = \Psi, g \in G$, and


2. those symmetric states have the same GHZ entanglement $\Psi = \sum_\alpha \Psi_\alpha ,\ \ \alpha \in G/H,\ \ H\ \subset G$, where $\Psi_\alpha$'s are locally distinguishable.

We say those symmetric states are equivalent. The set of equivalent symmetric states is a symmetry breaking phase.

So symmetry breaking = GHZ entanglement which are classified by pairs $(G , H),\ H \in G$.

More precisely:

1) A symmetric many-body state has spontaneous symmetry breaking implies that the state has a GHZ entanglement.

2) One can detect spontaneous symmetry breaking in a symmetric many-body state even without knowing the group and/or order parameter of the symmetry. One can detect spontaneous symmetry breaking in a symmetric many-body state using only probes that respect the symmetry.

3) The symmetric exact ground state of a generic symmetric Hamiltonian has spontaneous symmetry breaking iff it has GHZ entanglement.

cosmology - Results of WMAP mission

The NASA WMAP mission mapped the cosmic microwave background (CMB), the the residual high temp photons from the condensed space 380000 years after big bang that expanded into the much high volume we have now into a lower energy level in the microwave spectrum with KT about 2°.

Also, over the past 30 years, multiple Hubble constant measurements - cephid variables, type 1a supernovae intensity models, parallax on short range sources - have generated an estimate that the universe is about 70% dark energy 25% dark matter and 5% baryonic matter, ie quarks, or proton and neutrons. This 5% is all we can "see" or directly measure apart from its influence on things we can measure like gravity. The WMAP data was also used to verify these estimates.

I am trying to understand how the the WMAP data is used in this verification.

Is this extracted by the change in Energy density due to the expansion of universe volume since 380000 years ABB, perhaps using some assume multimillion degree temperature back at that time?

Answer

The WMAP, and later the Planck, spacecrafts have measured the cosmic microwave background (CMB) radiation, specifically its anisotropy. As you say, the temperature of the CMB is around 2.7 K, but this is only the average, and although the variation from direction to direction on the sky is extremely small (roughly $1\!\!:\!\!10^5$), these small fluctuations are extremely important.

Most basically, to extract information out of the CMB requires a cosmological model for how these anisotropies look. This model should be consistent with other observables (e.g. observations of supernovae, gravitational lenses, galaxy two-point correlations, and what not), and in addition it should then predict how the anisotropies look, not only qualitatively but also quantitatively.

To quantify the anisotropies, one calculates the power spectrum of the CMB. The power spectrum is a measure of how large variation you have on the different angular scales — that is, how much does the temperature vary between two random points separated by, say, 90º, by 10º, by 10 arcmin, and so on. It looks like this:

^{CMB power spectrum from Planck Collaboration et al. (2016) with my own approximate corresponding angles annotated in red.}

From measuring the exact positions of the peaks in the power spectrum, the heights, the ratio between heights, etc. one can then distinguish between various cosmological models, in this way extracting a plethora of information. Here's an incomplete list:

• The position of the peaks are sensitive to the curvature of the Universe, and the first peaks shows (or is at least consistent with) a flat Universe.


• The shapes of the peaks are given by the density of dark and baryonic matter (but only very little by dark energy).


• The second peak tells something about the baryon acoustic oscillations (baryons enhance every other peak, and the second peak is suppressed relative to the first and third).


• The third peak shows the density of dark matter.


• The higher peaks shows that radiation dominated the Universe at early times.

Additionally, the polarization of the CMB can tell you about inflation at the Big Bang, as well as when (roughly) the Universe was reionized.

A more complete list of all the information you can extract can be found on Wayne Hu's webpage.

newtonian mechanics - How deep can a whirlpool descend?

My question is based around comparing the physics of a tornado against, what I imagine/assume to be the marine equivalent of a tornado, that is a whirlpool located either at sea, in a tidal region or in an estuary.

My assumptions are:

• 
  

  They both have the same basic cause, opposing streams of fluid, (air versus water) meet and an "inward" velocity vector arises, creating the inital spinning motion, which then, temporarily, becomes self perpetuating, dragging in more fluid and increasing the angular momentum of the system.

  
  


• 
  

  I assume, from video footage of tornados that I have seen, that tornados generally seem to be created at the height of the cloudbase, which can, it seems to me, range from 100 to 500 metres in vertical height.

  
  


• 
  

  I would assume that, due to the much greater density of water relative to air, a whirlpool could not descend very far, as the forces required to maintain it are far greater that it's atmospheric equivalent. However, I don't know enough about fluid dynamics to estimate how far down it could actually descend before it dissipates. My (pretty obvious, I admit) guess would be on the order of metres, i.e. far less than tornado height.

  
  

In other words, how good is my analogy of picturing a whirlpool as the marine equivalent of a atmospheric tornado and where does it break down?

Answer

According to Heimholz second theorem it goes all the way. It even can't end in the fluid. You of course mean just the air, but it's merely a matter of how the situation is developing. If the flow conditions goes over Froude number 1 you will allways have the connection. It's explained here; Air core Vortex; Physical explanation of the "air Entrainment Hook" at $F_{co}=0.7$ -on my answer.

And according to experience gathered at hydropowerplants it really goes "all the way." Here is a nice study about the issue see ie. picture in page one.

Note that the pressure in intake is not the same, and "all the way" is the connection between two pressures.

electromagnetism - Coincidence, purposeful definition, or something else in formulas for energy

In the small amount of physics that I have learned thus far, there seems to be a (possibly superficial pattern) that I have been wondering about.

The formula for the kinetic energy of a moving particle is $\frac{1}{2}mv^2$.

The formula for kinetic rotational energy is $\frac{1}{2}I\omega^2$.

The formula for energy stored in a capacitor is $\frac{1}{2}C \Delta V^2$.

The formula for energy delivered to an inductor is $\frac{1}{2}LI^2$.

Finally, everyone is aware of Einstein's famous formula $e=mc^2$.

I realize there are other energy formulas (gravitational potential energy, for example) that do not take this form, but is there some underlying reason why the formulas above take a similar form? Is it a coincidence? Or is there a motivation for physicists and textbook authors to present these formulas the way they do?

Answer

Usually, linear equations are very common in physics. Something like $\text{Quantity}=\text{constant}\times\text{variable}$

So, you have $p=mv,L=I\omega,Q=CV$, etc.

Now, it just turns out that when you multiply these linear equations with a small increment of the variable, you get an expression for energy. Why this happens isn't so easy to figure out. The easiest explanation is that we usually define our variables such that force and similar quantities (e.g, net charge, etc) are linear.

Now, if you sum up a linear equation multiplied by a small change in the linear part, you get something like $\int cx dx$, which is a quadratinc term of the form $\frac12 cx dx$

quantum mechanics - What is meant by the term "single particle state"

In a lot of quantum mechanics lecture notes I've read the author introduces the notion of a so-called single-particle state when discussing non-interacting (or weakly interacting) particles, but none that I have read so far give an explicit explanation as to what is exactly meant by this term.

Is it meant that, in principle, each individual state constituting a multi-particle system can be occupied by a single particle (contrary to an entangled state, where is impossible to "separate" the particles), such that the state as a whole can be de-constructed into a set of sub-states containing only one particle each, and each being described by its own Hamiltonian?

Sorry to ramble, I'm a bit confused on the subject, in particular as I know that more than one particle can occupy a single-particle state (is the point here that the particles can still be attributed their own individual wave functions, and it just so happens that these individual wave functions describe the same state?).

Answer

Many particle wavefunctions are generally appallingly complicated objects. One way to get a handle on them is to break them down into simpler parts, understand those parts and then put them back together again. We do this by constructing the space of many particle wavefunctions as either a tensor product space or a Fock space.

An obvious way break down a many particle system is to try to consider what each particle is doing individually. Obviously there will be emergent effects in the many body system due to entanglement that were not present when only considering one particle and for strongly interacting systems this breakdown may not be possible, but often it is the only method we have.

So the single particle states are those states which on their own describe a single particle and from which we construct the full space as a tensor product space (i.e. the tensor product of single particle states and linear combinations thereof) or Fock space.

acceleration - Calculate speed from accelerometer

I am trying to use my accelerometer on my mobile device (smart watch to be specific) to calculate a persons arm swing speed.

The data returned from the accelerometer is in $m/s^2$.

Since the acceleration of a persons arm is not constant I cannot use the equation v0 + at to calculate the velocity.

I have never been good at physics so how do I calculate speed with varied acceleration?

Answer

You need to integrate acceleration to get the velocity.

$$v(t) = \int_{t=0}^{t} a. dt$$

There are a number of ways of doing this numerically.

I assume that you get these readings regularly with a spacing of $\delta t$, for example $\delta t = 100 ms$ or something like that.

About the simplest way to do it is

$$v(t) = v(0) + \sum a \times \delta t$$

where $v(t)$ is the velocity at time $t$. but there are more sophisticated ways of doing it - I will not repeat them here, but you might want to look at using Simpson's rule, which is described here.

The problem is complicated by velocity being three dimensional - so you need to integrate in each of the three dimensions x, y and z separated.

It depends how the phone gives you the information about the acceleration, but if you get $a_x$, $a_y$ and $a_z$ at regular intervals then you can do the following...

vx += ax * dt;
vy += ay * dt;
vz += az * dt;

if you get accleration as a raw number and angle then you will have to convert from I guess polar coordinates to xyz components to be able to add them up.

Total speed, $|v|$ is, of course, given by $|v|=\sqrt{v_x^2 + v_y^2 + v_z^2}$

I would, of course, try to start at $v=0$

Curious One, raises a really interesting point about $g$ - the best way to test this is to code it and try it - shake the phone and see if the velocity returns to zero when it is at rest after shaking it or moving it....

... can you post your results if you do this and try it out?

Another issue is twisting the phone and twisting the accelerometer - this would require you to think about angular acceleration etc., but the basic principles outlined here would be the same if you needed to think about angles.

geometry - Continue the Pattern of Circles

The following shapes are sequenced from left to right. Describe or create the next five shapes. Bonus points to those who describe how the pattern works.

• The pattern could continue on forever.


• The line thicknesses you see here are 1, 3, and 5; and the next five shapes require only these.

Spoilers: I presented this as a programming question and folks made some very cool animations.

Answer

Here is your sequence:

And the explanation is:

These are Roman numerals. Read the circles from inside out. Thickness 1 is I, thickness 3 is V, thickness 5 is X.

Calculating the size of the "lensing sphere" of a black hole

Given a black hole of some size, say $10^8$ solar masses, how can the size of its sphere of influence of light be calculated?

To clarify, ultimately I'd like to be able to calculate the apparent angle of a black hole's lensing effects, as viewed from some distance.

(I'm no physicist, so I might be making incorrect assumptions about gravitational lensing. If this is the case, sorry!)

Answer

The calculation of light bending isn't hard if you make the assumption that you're well away from the event horizon, but it's probably a bit much for a non-physicist so I'll just quote the result. If you stay well away from the event horizon the approximate equation for the angle that the light is bent is:

$$\Delta \phi = \frac{4GM}{c^2r}$$

where $M$ is the mass of your object and $r$ is the distance of closest approach. So for example for a light ray just grazing the Sun, mass $1.99 \times 10^{30}$kg at a distance of $6.96 \times 10^8$m the equation tells us the light is bent by about $8.5 \times 10^{-6}$ radians or just under 2 seconds of arc. This was the bending that Eddington measured in his famous (if now disputed) 1919 experiment to test general relativity.

You can use this to work out how much light is bent by your $10^8$ solar mass black hole. However note that there is no cut-off i.e. no distance at which light stops being bent. It just gets bent less and less as you get farther away. You would have to decide for yourself what angle you consider negligable and work out the corresponding distance.

Fit the most chess pieces on a board with these rules

Fit the most pieces on a chess board.

• You are only allowed to place Rooks, Knights and Bishops.



• The number of rooks, knights and bishops must be approximately equal (specifically, the number of rooks, knights and bishops must all be either $k$ or $(k+1)$ for some $k$).


• Every piece has to be attacked by exactly one other piece.

I don't know what the answer to this problem is. Highest legit answer wins, internet bonus points if you can prove that you found the optimum.

Answer

POST MORTEM EDIT (well, post-tick, anyway) There was way too much flexibility left in the earlier solution, I just had to keep wiggling everything until I got a bishop on the other long diagonal too. Here it is:

9 knights, 9 bishops and 8 rooks, for a total of 26 pieces.

Please do double check it; I checked it several times already, but wishful thinking often makes it impossible to see one's own mistakes.

POST POST MORTEM EDIT: Here’s a completely different approach with the same result (now with much more symmetry)

9 knights, 9 bishops, 8 rooks = 26 pieces.

This time much easier to confirm: the rooks attack everything except the very corners, nothing else attacks anything, except the corner bishops that attack each other.

The original accepted answer is below the line.

---

Remorselessly stealing Daniel Duque's incredible solution, which had just enough wiggle room to add another knight:

25 pieces (9 knights, 8 bishops, 8 rooks)

---

Here's another with the same number of pieces, but with a lot more room for improvement, it seems (I hope there aren't any mistakes, I'm getting pretty cross-eyed here..)

9 knights, 8 bishops, 8 rooks = 25 pieces
(There are lots of promising lookin variations going on in here, please feel free to use this one in your own answer)

---

Here's my first (original) attempt with

20 pieces: 7 bishops, 7 knights and 6 rooks. (with some wiggle room left in the bottom middle)

Like so:

As you can probably tell, squeezed everything together by starting with a greedy approach with the difficult pieces on the top side.

newtonian mechanics - Why are orbits elliptical?

Almost all of the orbits of planets and other celestial bodies are elliptical, not circular.

Is this due to gravitational pull by other nearby massive bodies? If this was the case a two body system should always have a circular orbit.

Is that true?

Answer

No, any ellipse is a stable orbit, as shown by Johannes Kepler.

A circle happens to be one kind of ellipse, and it's not any more likely or preferable than any other ellipse. And since there are so many more non-circular ellipses (infinitely many), it's simply highly unlikely for two bodies to orbit each other in a perfect circle.

riddle - I am big and small?

Find the answer :

• My first is in my third


• My second used to walk fast



• My third is you + me

Who am I?

Answer

Are you:

Uranus?

Starting with the last clue first...

My third is you + me

You + me = us

My first is in my third

"U" is in "us"

My second used to walk fast

Ran, past tense (used to) of run (walk fast)

Put it all together:

U ran us

The title: "I am big and small?"

The planet Uranus is obviously very big. It is has the fourth-largest planetary mass in our solar system and is termed an ice giant. But as the seventh planet from the sun it is very far away and so appears very small to us.

optics - Are infinite reflections between two mirrors really infinite?

As written in the title it's a somewhat rhetoric question, so let me be more clear.

Let's say I have two mirrors facing each other. They are perfectly aligned and perfectly reflective. I'm looking through one of them (using the one-sided mirror effect) and see what seems to be infinite reflections.

If I look closely at the vanishing point, I obviously can't see very well what goes on there, because the reflections are too small. My question is, are there other limitations?

a) If I look close enough to the vanishing point, is there a point at which the light would have to travel too much distance to reach me that no photon would actually make it?

b) If I start looking immediately after turning on the light, it should take a tiny amount of time until all the reflections are created, i.e., it takes time for light to do all the back and forth trips it takes to create the very small images. How good a microscope (telescope?) would I need to be able to see the image being created? That is, to look at a spot and see nothing, and then some epsilon of time later see an image appear.

wordplay - TV Title Mondegreens

Inspired by the original Twisted TV Titles and its follow-up, I came up with a similar puzzle. This one builds on the idea of Mondegreens - misheard lyrics (like “Excuse me while I kiss this guy” instead of “Excuse me while I kiss the sky”). This is misheard TV Show Titles.

I’ll give a description of what the show could be, and you can give the misheard title, along with the real show title. For example:

Clue: Linda never realized how much she relied on her android until she lost it

Answer: Missed Her Robot - Real title: Mister Robot

1. A story about herbed black tea - Clue: This one is way simpler than you think. If you just name a black tea, you'll likely get it right!


2. The creator of X-Files and Breaking Bad opens an optometry clinic


3. Briefly looking at clones


4. From The Darjeeling Limited to Rushmore, we learn all about this director’s universe


5. Due to the bartender - Clue: This refers to a well-known bartender on a show that is currently airing (as of March, 2018) - Clue 2: It refers to a very animated bartender


6. Will definitely stay closed without a key



7. Broken Buildings


8. Wander around


9. A man who never drives during the day

Answer

Answers so far:

1) A story about herbed black tea

The Chai => The Chi BruceWayne

2) The creator of X-Files and Breaking Bad opens an optometry clinic

Vince Gilligan opens an optometry clinic => Gilligan's Eye Land => Gilligan's Island - PuzzledPig

3) Briefly looking at clones

Twin peeks => Twin Peaks - Alconja

4) From The Darjeeling Limited to Rushmore, we learn all about this director’s universe

Wes Anderson's universe => Wes' world => Westworld - Somebody

5) Due to the bartender

`Cuz Mo => Cosmos - Cgz

6) Will definitely stay closed without a key

Sure-locked => Sherlock - Xenocacia

7) Broken Buildings

Faulty towers => Fawlty Towers - Kobi

8) Wander around

Roam => Rome - PuzzledPig

9) A man who never drives during the day

Night rider => Knight Rider - Alconja

newtonian mechanics - Relation between torque and moment of inertia

I have seen 2 formulas: $\tau = I \alpha$ and $\tau = I \dot\omega + \omega \times I \omega$. Which one shall I use in which case please? Also, am I right to think that $\alpha$ from the first formula is $\dot\omega$ from the second one?

Answer

Yes, $\alpha = \dot{\omega}$, being the angular acceleration. The first equation is special case of the second equation.

For a general object the moment of inertia is not just a scalar (a single value) but a tensor, in that case you have to use your second equation. $\tau$ and $\omega$ are then vectors and $I$ is a 3x3 matrix.

But when you spin an object around one of its high symmetry axes (one of the eigenvectors of the inertia matrix $I$), the equation simplifies to your first equation. Proof:

If $\vec{\omega}$ is an eigenvector of $\hat{I}$ it holds that: $\hat{I} \vec{\omega} = \lambda \space \vec{\omega} = \vec{\omega} \space \lambda$ Therefore your second equation becomes: $\vec{\tau} = \hat{I}\dot{\omega} + \vec{\omega} \times \vec{\omega} \space \lambda$ and $\vec{\tau} = \hat{I}\dot{\omega} + 0$ since the crossproduct of a vector with itself is 0.

newtonian mechanics - Why is it effectively impossible to balance this rectangle block on a triangle block? (See photo)

I've always wondered why it's basically impossible to balance a rectangle block on top of a triangle one like this. This is my nine-year-old giving it a try.

I have never in my life gotten it to giving even a hint of balancing, not even for a fraction of a second.

Why is this so difficult?

(Unsure about the question-tags...)

Answer

It's a general consequence of Newton's Laws of Motion that an object is stationary when no net forces or momenta act on it. Avoiding net forces or momenta is precisely what makes perfectly balancing a rectangular block on a triangular one so difficult.

Above is a rectangular block with mass $m$, perfectly balanced on a triangular block. I've added the diagonal cross lines which intersects at the centre of gravity of the rectangle and an $x$-axis. The centre of gravity here corresponds with the $0$ point of the $x$-axis.

This block is perfectly balanced because the weight force vector $mg$ runs precisely through the top point $P$ of the triangle. The triangle provides an equal and upward reaction force $-mg$ and there are no net forces or momenta acting on the rectangular block: it is perfectly stationary.

But that is so easily changed. Suppose we move the bloc along the $x$-axis to the right, by a tiny amount that I'll call $\Delta x$. Now the weight vector $mg$ no longer passes through the point $P$. The product $mg\Delta x$ now provides a small moment ('torque', if you prefer) which causes the rectangle to start rotating clockwise around the point $P$ and eventually the bloc falls off the triangle.

Note that exactly the same would happen if we had moved the bloc a little to the left ($-\Delta x$). The bloc would then have started rotating counter-clockwise around $P$.

So getting the position of the centre of gravity right with respect to $P$ is crucial here for precise balancing.

Edit (on request of the OP) - some factors that affect ease of balancing:

1. Sharpness/bluntness/hardness:

In the schematic above the previously sharply defined pivot point P has been (exaggeratedly) replaced by two points P and P’. On this ‘blunt’ triangular bloc’s top edge there’s now a narrow flat ridge of width PP’. As long as the rectangular bloc’s weight vector $mg$ passes through the interval PP’ there is no moment that can cause the rectangular bloc to start moving. The bloc has now a little freedom in the $x$ direction. So it’s easier to balance on a blunt pivot than on a sharp one.

As a consequence it’s also easier to balance on a softer pivot than a harder one: the softer one will 'sag' slightly under the weight of the rectangular bloc, causing a geometry that’s similar to the case of the blunter pivot point. For example, high precision balancing scales use very hard and very sharp anvils for maximum precision.

2. Aspect ratio:

For the rectangular block we can define the aspect ratio $AR$ as:

$AR=\frac{W}{D}$ with $W$ the widest horizontal dimension and $D$ the largest vertical dimension. For the rectangular bloc used in the schematics, clearly $AR > 1$.

Now imagine an object with a very large AR like a 30 cm ruler: the $AR$ is in the order of $\frac{30cm}{0.5cm}=60$. Such a ruler is fairly easy to balance horizontally on the pivot.

Now imagine we rotate the ruler by 90 degrees (so it stands upright on one of its edges). Its AR is now $\frac{0.5cm}{30cm}=0.0167$. In the upright position, the ruler is far harder to balance.

In general, the larger the AR of the object, the easier it is to balance it.

enigmatic puzzle - Rainbow Road Rabbit Hole

Question: Who is in first place?

---

It's difficult to add hints in the puzzle itself given its nature so I'll try to add some notes here as text instead.

Red: [Solved]

Orange: [Solved]

Yellow:

K1

Answer

This puzzle is rainbow (road)-themed, and consists of a series of colored images. Each image contains the URL of the next image, encoded or obfuscated in some fashion.

Steve Eckert started us off by solving the "red" and "orange" images.

The "red" image (the image in the OP) used a traditional steganographic technique:

the URL was hidden in plain text, in a color one bit different from the background.

The "orange" image uses a classic encoding method:

The URL is encoded as Morse code.

f'' was next to step in, solving the "yellow" image:

The URL was enciphered using the same Vigenere cipher as the famous Kryptos sculpture whose coordinates were in the image.

This revealed the "green" image:

http://i.stack.imgur.com/b8qwn.png

This image contains a number of barcodes. Each of the barcodes on the first row encodes a single character from the URL:

The first is an Aztec code. Using Mathematica to decode it produced m. (I verified its contents by hand, since I've been working a lot with 2d barcode formats in preparation for my next puzzle.)

The second is a Code 128 barcode, containing the letter L. I used Mathematica to decode this one as well.

The third and fifth are Code 39 barcodes, containing +C and W. The former is actually using the full ASCII extension to code a lowercase c. I decoded these by hand using the picture on the Wikipedia page.

The last large barcode is a MaxiCode, used mainly by UPS. This was by far the most difficult part of the puzzle, since it's hard to find information on the format, and there aren't any online MaxiCode readers (since nobody except UPS needs to read the barcodes, only generate them). I eventually installed a trial version of some computer vision software to reveal that the barcode contained only .png. (It was obvious in retrospect that the repeated pattern in the upper half of the barcode is just padding in the outer data area, since the barcode is a fixed size.)

This gives us all but one of the characters in the URL. Brute-forcing the remaining character by programmatically trying all 62 possible characters leads us to the "blue" image:

http://i.stack.imgur.com/mLc1W.png

We can see the letters p, n, and g on the right, so it's apparent that we must rearrange the letters using the values on the left in order to retrieve the next URL.

I'm too impatient to figure out what $B'_L$ and $\beta$ are, so I programmatically searched all 120 permutations of the right-hand side that ended in .png, leading to the "magneta" image:

http://i.stack.imgur.com/xfrqT.png

The solution seems to indicate that we were supposed to order the left-hand side by value, so we know that $\beta < B'_L < \sqrt{2}$. It could be that $B'_L$ is supposed to be Legendre's constant, $1$, in which case $\beta < 1$ (but I still don't know what $\beta$ is; oh well).

The next URL is encoded with a simple

Caesar cipher, which I figured out by noticing the difference between c and a was the same as the difference between p and n (what I would expect to be in the plaintext there). (As Zandar points out, this particular Caesar cipher is rot13, with to "rot5" for the numbers.)

Using a known-plaintext attack on the extension (.png), we get:

http://i.stack.imgur.com/T*gf*.png

I wasn't sure how the cipher was supposed to affect the digits, so I once again brute-force searched all 100 possible digit assignments, which turns up (along with two false positives) the "purple" image:

http://i.stack.imgur.com/T1gf2.png

You can tell that we're dealing with a transposition cipher, since all the characters from the expected URL format (http://i.stack.imgur.com/*.png) are here. Playing around some reveals that it's a simple

transposition cipher with 12 columns.

Note that in order to decode it correctly, some of the spaces in the text have to be doubled or trebled. With the correct spacing, I get:

Y       s i a t . i o 9
o a t H   n g t s m m .
u l h e t a e p t g / j
  m e r h l : : a u 5 p
a o r e e     / c r Y g
r s e     i   / k . 8
e t . i f m h i . c u 

or:

You are almost there. Here is the final image: http://i.stack.imgur.com/5Y8u9.jpg

Yielding the final part of the puzzle:

http://i.stack.imgur.com/5Y8u9.jpg

In the spirit of the challenge, I'll present the answer as an image of my own:

cosmology - Is the Cosmological constant part of or predicted by or supported by the Dirac Equation?

By Cosmological constant here I mean the constant as would be predicted by the vacuum. The one and the same that if compared to the actual expansion of the universe it is off by a factor of $10^{120}$. I am curious since Dirac's remarkable equation does predict a considerable number of elementary particles I am wondering if it says anything about space time or it's expansion. They may be disparate concepts in this context.

Answer

No: the cosmological constant is not predicted by any known physical model. In the context of particle physics, $\Lambda$ is one of the multiple free parameters of the Standard Model. As this model does not include gravitation, it is customary to set $\Lambda=0$, because the origin of energies is arbitrary anyway (in the absence of gravity). The condition $\Lambda=0$ is in fact a renormalisation condition, and it effectively eliminates all vacuum diagrams.

homework and exercises - What are the general solutions to a hard sphere collision?

Surely someone has found the solutions to the hard sphere collisions (in $n$ dimensions) of two bodies of mass $m_1$ and $m_2$, respectively--that is the resultant velocities (or momenta) of the two bodies after a hard sphere collision given $\vec{p}_1$ and $\vec{p}_2$. I have a solution (both for elastic and inelastic collisions), but I wonder whether or not it's correct.

FYI, here are my solutions for elastic, hard-sphere collisions:

$$\vec{p}_{1f} = \vec{p}_1 + 2\frac{\left(m_1\vec{p}_2 - m_2\vec{p}_1\right)\circ \hat{r}}{m_1 + m_2}\hat{r}, \\ \vec{p}_{2f} = \vec{p}_2 - 2\frac{\left(m_1\vec{p}_2 - m_2\vec{p}_1\right)\circ \hat{r}}{m_1 + m_2}\hat{r}.$$

wher $\hat{r}$ is in the direction of the "hard sphere" (i.e. the unit vector connecting the two centers of the two spheres). You can see that my formulation trivially conserved momentum--the question is whether or not it conserves energy as well.

If it helps, here is my full explanation (this is quite long and the link will expire in 30 days from 09/23/2014). For these purposes, you should skip to the section on n-dimensional cases (page 6). I wrote this as a reference to myself for doing physics simulations since I could not find any reference on hard sphere collisions (other than references setting up the problem which give no solutions).

Here is an abbreviated work on finding the final momenta from the elastic case:

Just as in the $1$D case, we add and subtract the same momentum from each particle--except now we are adding and subtracting a vector: $\vec{p} = p\hat{r}$, which means that $|\vec{p}| = p$ (such that $\vec{p}_{1f} = \vec{p}_1 + p\hat{r}$ and $\vec{p}_{2f} = \vec{p}_2 - p\hat{r}$). This gives the final energy as:

$$\begin{align} E_f = \frac{|\vec{p}_1 + p\hat{r}|^2}{2m_1} + \frac{|\vec{p_2} - p\hat{r}|^2} {2m_2} \end{align}$$

We now write the final energy in terms of the magnitude of $p$ and the dot product between $\hat{r}$ and $\vec{p}_1$ and likewise for $\vec{p}_2$. Note that if you choose the direction for $\vec{p}$ to act a priori (like in the hard-sphere case), the dot products $\vec{p}_1\cdot \hat{r}$ and $\vec{p}_2\cdot\hat{r}$ will be known:

$$\begin{align*} E_f = \frac{p_1^2 + p^2 + 2p\hat{r}\circ\vec{p}_1}{2m_1} + \frac{p_2^2 + p^2 - 2p\hat{r}\circ\vec{p}_2}{2m_2} \end{align*}$$

As usual the elastic case greatly simplifies:

$$\begin{align*} \require{cancel} \cancel{\frac{p_1^2}{2m_1}} + \cancel{\frac{p_2^2}{2m_1}}= \frac{\cancel{p_1^2} + p^2 + 2p\hat{r}\circ\vec{p}_1}{2m_1} + \frac{\cancel{p_2^2} + p^2 - 2p\hat{r}\circ\vec{p}_2}{2m_2} \\ \frac{m_1 + m_2}{2m_1m_2}p^2 - 2\frac{m_1 \vec{p}_2\circ\hat{r} - m_2\vec{p}_1\circ\hat{r}}{2m_1m_2}p = 0 \end{align*}$$

This leads to a trivial solution of $p = 0$ (no collision occurs) and:

$$\begin{align} \frac{m_1 + m_2}{2m_1m_2}p - 2\frac{m_1 \vec{p}_2\circ\hat{r} - m_2\vec{p}_1\circ\hat{r}}{2m_1m_2} = 0 \\ p = 2\frac{m_1 \vec{p}_2\circ\hat{r} - m_2\vec{p}_1\circ\hat{r}}{m_1 + m_2} \end{align}$$

Hence my solutions of:

$$\vec{p}_{1f} = \vec{p}_1 + 2\frac{\left(m_1\vec{p}_2 - m_2\vec{p}_1\right)\circ \hat{r}}{m_1 + m_2}\hat{r}, \\ \vec{p}_{2f} = \vec{p}_2 - 2\frac{\left(m_1\vec{p}_2 - m_2\vec{p}_1\right)\circ \hat{r}}{m_1 + m_2}\hat{r}.$$

Answer

Finding the missing equations

Coordinate transforms just complicate the issue. The heart of the matter is that in n dimensions you have n degrees of freedom for the velocities of the COM of each sphere, and you only have n momentum conservation equations plus one energy conservation equation. That means you need an additional n-1 equations to solve the problem.

The additional equations come from the fact that if you assume a frictionless impact*, then the impulse/force imparted to each object must be normal to the surface. This gives a constraint on the direction of the impulse. Along with the magnitude, this direction would give us n degrees of freedom. So without it, it gives us n-1: exactly the number we need.

Solving for final velocities

Since we have spheres, the direction normal to the impact surface will be the vector that points from one COM to the other. Let's call this normal vector $\hat{j}$.

Now the velocities of balls A and B can be given as $V_{a1}$ and $V_{b1}$ before the collision and $V_{a2}$ and $V_{b2}$ after the collision.

Let's call the impulse that is imparted on ball A $J$. Then we have:

$$J \bullet\hat{j} =|J|$$ $$V_{a1} + \frac{J}{M_a}=V_{a2}$$ $$V_{b1} - \frac{J}{M_b}=V_{b2}$$

Then we can write the conservation of energy (multiplied by 2) as:

$$M_a V_{a1} \bullet V_{a1} + M_b V_{b1} \bullet V_{b1} = M_a V_{a2} \bullet V_{a2} + M_b V_{b2} \bullet V_{b2}$$

Plugging in for our velocities after to collision gives:

$$M_a V_{a1} \bullet V_{a1} + M_b V_{b1} \bullet V_{b1} = M_a \left( V_{a1} + \frac{J}{M_a}\right) \bullet \left( V_{a1} + \frac{J}{M_a}\right) + M_b \left( V_{b1} - \frac{J}{M_b}\right) \bullet \left( V_{b1} - \frac{J}{M_b}\right)$$

$$M_a V_{a1} \bullet V_{a1} + M_b V_{b1} \bullet V_{b1} = M_a V_{a1}\bullet V_{a1} + 2 J\bullet V_{a1} + \frac{J\bullet J}{M_a} + M_b V_{b1}\bullet V_{b1} - 2 J\bullet V_{b1} + \frac{J\bullet J}{M_b}$$

$$0 = 2 J\bullet V_{a1} + \frac{J\bullet J}{M_a} - 2 J\bullet V_{b1} + \frac{J\bullet J}{M_b}$$

dividing by the magnitude of $J$ yields:

$$0 = 2 \hat{j} \bullet (V_{a1}-V_{b1}) + |J|\left(\frac1{M_a} + \frac1{M_b}\right)$$

$$|J| = 2 \frac{\hat{j} \bullet (V_{a1}-V_{b1})}{\frac1{M_a} + \frac1{M_b}}$$

Final Velocities

$$V_{a2}=V_{a1}+2 \frac{\hat{j} \bullet (V_{a1}-V_{b1})}{M_a\left(\frac1{M_a} + \frac1{M_b}\right)}$$ $$V_{b2}=V_{b1}+2 \frac{\hat{j} \bullet (V_{b1}-V_{a1})}{M_b\left(\frac1{M_a} + \frac1{M_b}\right)}$$

Finding Direction of Impulse

Let's call the vector pointing from the center of ball B to the center of ball A $R$. We know that before the impact $R=R_0+R't$ where $R_0$ is the difference in the balls' initial positions $P_a-P_b$ and $R'$ is the difference in the balls' initial velocities $V_{a1}-V_{b1}$.

At the moment of impact:

$$\hat{j}=\frac{R}{|R|}$$ and $$|R|=r_a+r_b$$ Where $r_a$ and $r_b$ are the radii of the balls.

We can transform this second equation to find t:

$$R\bullet R=(r_a+r_b)^2$$ $$R'\bullet R't^2+2R_0\bullet R't+R_0\bullet R_0-(r_a+r_b)^2=0$$ $$t=-\frac{R_0\bullet R'+\sqrt{(R_0\bullet R')^2-R'\bullet R'(R_0\bullet R_0-(r_a+r_b)^2)}}{R'\bullet R'}$$

If this time is negative or imaginary, then the collision will not occur. Otherwise $R$ at the moment of impact can be found by plugging in t:

$$R=R_0-R'\frac{R_0\bullet R'+\sqrt{(R_0\bullet R')^2-R'\bullet R'(R_0\bullet R_0-(r_a+r_b)^2)}}{R'\bullet R'}$$

Wrap Up

So given initial positions $P_a$ and $P_b$, initial velocities $V_{a1}$ and $V_{b1}$, and ball radii $r_a$ and $r_b$ the final velocities $V_{a2}$ and $V_{b2}$ can be found be following the following procedure:

Calculate the relative positions and velocities:

$$R_0=P_a-P_b$$ $$R'=V_{a1}-V_{b1}$$

Calculate impact time:

$$t=-\frac{R_0\bullet R'+\sqrt{(R_0\bullet R')^2-R'\bullet R'(R_0\bullet R_0-(r_a+r_b)^2)}}{R'\bullet R'}$$ If t is negative or imaginary, then no collision occurs, and the initial velocities are the final velocities. Otherwise:

Calculate the impulse direction:

$$\hat{j}=\frac{R_0+R't}{|R_0+R't|}$$ Calculate the impulse: $$J=2 \hat{j}\frac{\hat{j} \bullet R'}{\frac1{M_a} + \frac1{M_b}}$$ Calculate the final velocities: $$V_{a2} = V_{a1} + \frac{J}{M_a}$$ $$V_{b2} = V_{b1} - \frac{J}{M_b}$$

*The collision must be frictionless if we're not considering moments of inertia as any friction would apply torque to the ball which would transfer energy and rotational inertia to the ball. The problem becomes slightly more complicated in that case as one has to assume how the friction will cause the balls to interact, which is not trivial as many of the simple assumptions conflict with energy conservation.

homework and exercises - Wave on a guitar string, differential equation

The mechanical energy of an element of the string is :

$$dE(x,t) = \frac{1}{2}\left[T_0\left(\frac{\partial y}{\partial x}\right)^2 + \mu\left(\frac{\partial y}{\partial t}\right)^2\right]dx$$

Where $T_0$ is the tension at rest and $\mu$ is the linear mass density.

I am to find a differential equation between the linear energy density $e(x,t) = \frac{dE}{dx}$ and the power $P_{1\rightarrow 2} (x,t)$ received by the right part (2) of the string from the left part (1) .

How am I supposed to do that ?

Answer

D'Alembert equation reads, for the considered case,

$$\mu\frac{\partial^2 y}{\partial t^2} = T_0 \frac{\partial^2 y}{\partial x^2}\:.\tag{1}$$ It is nothing but $F=ma$ along the vertical direction ($y$). Here $y$ denotes the small deformation of the string along the vertical direction from the stationary (horizontal) configuration. We disregard horizontal deformations. $T_0 \frac{\partial^2 y}{\partial x^2}$ is the $x$-derivative of vertical component of force acting on a infinitesimal segment of string $dx$, when the deformation is small, assuming that the absolute value of the tension $T_0$ is constant. In other words $$f_y = T_0 \frac{\partial y}{\partial x}\tag{2}$$
is the $y$ component of either the tension or the force acting at the endpoints.

From (1) and the defintion of $e(t,x)$

$$e(t,x) = \frac{1}{2}\left[T_0\left(\frac{\partial y}{\partial x}\right)^2 + \mu\left(\frac{\partial y}{\partial t}\right)^2\right]$$ one easily sees (just computing derivatives) that $$\frac{\partial e}{\partial t} = \frac{\partial}{\partial x} \left(\frac{\partial y}{\partial t} T_0\frac{\partial y}{\partial x}\right)\:.\tag{3}$$ Integrating this identity along the string, from the point (maybe endpoint) $x_1$ to the point (maybe endpoint) endpoint $x_2$, we have $$\frac{d}{dt}\int_{x_1}^{x_2} e(t,x) dx = \left.\left(\frac{\partial y}{\partial t}T_0\frac{\partial y}{\partial x}\right)\right|_{x_2} - \left.\left(\frac{\partial y}{\partial t}T_0\frac{\partial y}{\partial x}\right)\right|_{x_1}\tag{4}$$ As $\frac{\partial y}{\partial t}$ is the vertical velocity of the endpoint, taking (2) into account, we conclude that $\frac{\partial y}{\partial t}T_0\frac{\partial y}{\partial x}$ is the power of the force acting on the endpoint (or at a generic point of the string).

Equations (3) and (4) establish relations, respectively local and global, between the density of energy and the power of the forces acting along the string.

quantum field theory - What is meant by a local Lagrangian density?

1. 
   

   What is meant by a local Lagrangian density?

   
   
   


2. 
   

   How will a non-local Lagrangian look like?

   
   


3. 
   

   What is the problem that we do not consider such Lagrangian densities?

   
   

Answer

What is a local Lagrangian density?

A classical field theory on Minkowski space $\mathbb R^{d,1}$ is specified by a space $\mathcal C$ of field configurations $\phi:\mathbb R^{d,1}\to T$, and an action functional $S:\mathcal C\to\mathbb R$. The set $T$ is called the target space of the theory, and is often a vector space. If there exists a function $L:\mathcal C\times\mathbb R\to \mathbb R$ for which

$$\begin{align} S[\phi] = \int_{\mathbb R} dt\, L[\phi](t), \end{align}$$ then we call $L$ a lagrangian for the theory. If, further, there exists a function $\tilde L$ such that $$\begin{align} L[\phi](t) = \int_{\mathbb R^d} d^d\mathbf x \,\tilde L[\phi](t, \mathbf x) \end{align}$$ then we call $\tilde L$ a langrangian density for the theory. Finally, if there exists a positive integer $n$ and a function $\mathscr L$ such that $$\begin{align} \tilde L[\phi](t, \mathbf x) = \mathscr L(t,\mathbf x,\phi(t, \mathbf x), \partial\phi(t,\mathbf x), \dots, \partial^n\phi(t,\mathbf x)) \end{align}$$ then we say that the lagrangian density is local. In other words, the lagrangian density is local provided its value at a given spacetime point depends only on that point, the value of the field at that point, and a finite number of its derivatives at that same point.

An example of a non-local Lagrangian density.

Consider $T = \mathbb R$, namely a theory of a single real scalar field. Let $\mathbf a\in\mathbb R^d$ be given, and define a Lagrangian density by

$$\begin{align} \tilde L[\phi](t,\mathbf x) = \phi(t,\mathbf x) + \phi(t, \mathbf x+\mathbf a). \end{align}$$ This Lagrangian density is not local because the value of the Lagrangian at a given point $(t,\mathbf x)$ depends on the value of the field at that point and on the value of the field at the point $(t,\mathbf x+\mathbf a)$. If we were to Taylor expand the second term $\phi(t,\mathbf a)$ about $\mathbf x$, then we would see that the Lagrangian density depends on an infinite number of derivatives of the field, thus violating the definition of a local Lagrangian density.

What's the issue with theories with non-local Lagrangian densities?

I'm no expert on this, so I'll divert to another user. I will say, however, that people do study theories with non-local Lagrangian densities in practice, so there's nothing a priori "wrong" with them, but they might generically exhibit some pathology that you might prefer not to have.

Perhaps most relevant, though, if you're taking QFT from a high energy theorist, for example, is that the Lagrangian density of the Standard Model is local, so there's no need to consider non-local beasts if one is studying the Standard Model.

electrostatics - What are the units or dimensions of the Dirac delta function?

In three dimensions, the Dirac delta function $\delta^3 (\textbf{r}) = \delta(x) \delta(y) \delta(z)$ is defined by the volume integral:

$$\int_{\text{all space}} \delta^3 (\textbf{r}) \, dV = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \delta(x) \delta(y) \delta(z) \, dx \, dy \, dz = 1$$

where

$$\delta(x) = 0 \text{ if } x \neq 0$$

and

$$\delta(x) = \infty \text{ if } x = 0$$

and similarly for $\delta(y)$ and $\delta(z)$.

Does this mean that $\delta^3 (\textbf{r})$ has dimensions of reciprocal volume?

As an example, a textbook that I am reading states:

For a collection of $N$ point charges we can define a charge density

$$\rho(\textbf{r}) = \sum_{i=1}^N q_i \delta(\textbf{r} - \textbf{r}_i)$$

where $\textbf{r}_i$ and $q_i$ are the position and charge of particle $i$, respectively.

Typically, I would think of charge density as having units of charge per volume in three dimensions: $(\text{volume})^{-1}$. For example, I would think that units of $\frac{\text{C}}{\text{m}^3}$ might be possible SI units of charge density. If my assumption is true, then $\delta^3 (\textbf{r})$ must have units of $(\text{volume})^{-1}$, like $\text{m}^{-3}$ for example. Is this correct?

Answer

Yes. The Dirac delta always has the inverse dimension of its argument. You can read this from its definition, your first equation. So in one dimension $\delta(x)$ has dimensions of inverse length, in three spatial dimensions $\delta^{(3)}(\vec x)$ (sometimes simply written $\delta(\vec x)$) has dimension of inverse volume, and in $n$ dimensions of momentum $\delta^{(n)}(\vec p)$ has dimensions of inverse momentum to the power of $n$.

dirac equation - Components of the Weyl spinor field

In the Weyl basis we can separate the spinor field into 2 components: the right-chiral spinor and the left-chiral spinor. Each of these fields has again 2 components which are coupled. What is the physical interpretation of these 2 components that make up the left-chiral (or right-chiral) field?

In the Dirac basis the interpretation of the 4 components is:
1. Electron spin-up
2. Electron spin-down
3. Positron spin-up
4. Positron spin-down

So my question is what is the corresponding interpretation in the Weyl basis (in the massless case). Is it like this?
1. Left-chiral electron $\psi_{4}$
2. Left-chiral positron $\psi_{3}$
3. Right-chiral electron $\psi_{2}$
4. Righ-chiral positron $\psi_{1}$

If this is the case than I don't understand why the left-chiral electron $\psi_{4}$ couples to left-chiral positron $\psi_{3}$ as can be seen in the equations:

$$\partial_{t} \psi_{4} + \partial_{x} \psi_{4} - i\partial_{y} \psi_{4} + \partial_{z} \psi_{3} = 0$$ $$\partial_{t} \psi_{3} + \partial_{x} \psi_{3} + i\partial_{y} \psi_{3} - \partial_{z} \psi_{4} = 0$$ $$\partial_{t} \psi_{2} - \partial_{x} \psi_{2} + i\partial_{y} \psi_{2} - \partial_{z} \psi_{1} = 0$$ $$\partial_{t} \psi_{1} - \partial_{x} \psi_{1} - i\partial_{y} \psi_{1} + \partial_{z} \psi_{2} = 0$$

Answer

The meaning of different components in the chiral representation are

1. Left-handed spin up,


2. Left-handed spin down,


3. Right-handed spin up,


4. Right-handed spin down.

Spin up and down are with respect to some arbitrary axis, which we often set to the $z$-axis .

Electron and positron(or negative-energy electron) states are identified from the solutions of the Dirac equation. It turns out that in the massless limit (for which the chiral representation is most convenient), the left- and right-handed sectors decouple, and electron(positron) states in the left-handed sector are left(right)-handed, viz. the spin is anti-parallel(parallel) to the momentum, and similarly for the right-handed sector.

newtonian mechanics - What happens to a satellite's orbit when you speed it up?

If a satellite is initially orbiting the earth with a constant velocity $v$ on a circular orbit.

What would happen, if the velocity is suddenly increased by 5%?

Would the orbit still be circular?

I guess it will be elliptic, but how to prove this and find the eccentricity of the ellipse?

logical deduction - Can you solve this Bruce Wayne logic puzzle?

This variation of the classical Farmer Goat puzzle was in a puzzle contest at my college two years ago.
Here it goes.

Mr. Bruce Wayne is very fond of parties. Well, he’s more fond of girls, but, let’s just say that he’s fond of parties.
So, he plans a party at an island off the coast of Gotham Bay. All his guests have already reached. And we all know, Mr. Wayne likes to make an entrance. So, he decides to go in the chopper along with two supermodels. But unfortunately for Mr. Wayne, the chopper can hold only two people.

To add insult to the injury, the Mayor arrives with his two assistants to get a ride in the chopper. And there’s Mr. Fox, but with a secret case. Never mind, let’s just go to the party. But it seems someone doesn't want Mr. Wayne to enjoy tonight because there is no pilot to fly the chopper. And only Mr. Wayne, The Mayor and Mr. Fox know how to fly the chopper.

Oh no, wait! That’s not it. Mayor’s assistants won't stay with Mr. Wayne without their Mayor (I don’t know Bruce Wayne freaks them out); the supermodels won’t stay with the Mayor without Bruce (they think the Mayor is a pervert but Bruce isn't!) and Mr. Fox would never leave the case.

Let’s revise the situation.

The passenger list is:

1. Bruce Wayne


2. Bruce's 2 supermodels



3. Mayor


4. His 2 assistants


5. Mr. Fox


6. The secret case

And the constraints are:

• The capacity of the chopper is two persons.
  


• The secret case counts as a person (what, it's important!).


• Only Bruce, Mayor or Mr. Fox can fly the chopper.


• The Supermodels won’t stay with the Mayor without Bruce; Assistants won’t stay with Bruce without their Mayor and Mr. Fox won’t leave the case alone.

So much for making an entrance, Mr. Wayne! All things said and done, we still have a party to attend to. Now that you’ve understood the situation, help Mr. Wayne and his friends (and the case, too) reach the party.

I think I was able to solve this one, can you solve this puzzle?

Answer

Let me give it a try: Let's call the models B1 and B2 since they belong to B and the assistants M1 and M2 and the case F1.

B and M

B back

B and B1
B and M back

B and B2
B back

M and B
M back

F and F1
B back

B and M

M back

M and M1
B and M back

M and M2
M back

finally M and B

astronomy - What are good books for graduates/undergraduates in Astrophysics?

There are no book recommendations for Astrophysics here. I will write my own answer, but I am also interested in what are others' views on the question (I will NOT mark my own answer as the best one).

thermodynamics - Reversible and Quasi-static processes

Do we have any proof that reversible processes are always quasi-static or is it just a fact that hasn't been violated till date? If there is a proof then please provide a link.

general relativity - Is Moon really rotating about its axis?

Wikipedia says yes but on Newtonian arguments. From general relativistic point of view Moon is not rotating but moving along geodesic trajectory. And like parallel transport of a vector (pointing to the direction of motion) on a surface of a sphere along equator, from outside it seems like rotation. So, would Moon still be rotating if we remove the Earth? Would Foucault pendulum on Moon detect its rotation?

acoustics - Does the human body have a resonant frequency? If so, how strong is it?

Inspired by this question on Music beta SE, I'm wondering if the human body has a strong resonant frequency. I guess the fact that it's largely a bag of jelly would add a lot of damping to the system, but is that enough to dampen it entirely?

What models for resonance might be used to model the human body? (E.g. weight-on-a-spring, with legs as springs?) What about individual, semi-independent body parts, like legs, or lung cavity (acoustic resonance?).

Answer

There seem to be a lot of human body mechanical models, such as this one:

As for applications, I have heard that sub-audio frequency vibrations have been considered as nonlethal weapons for riot control.

thermodynamics - Lowest temperature possible in the universe?

The third law of thermodynamics states that nothing can reach to absolute zero temperature. What is the lowest possible temperature that can be in the universe? Has any experiment reached to a billionth of a Kelvin? Is there any restriction on how low it can be? Is $10^{-1000}\;\rm K$ possible? Or is there a lowest quantum of temperature?

general relativity - Non-local gravitational energy tensor

The well-known derivation of the Landau-Lifshitz gravitational energy pseudotensor, relies on several requirements:

• 
  

  1) that it be constructed entirely from the metric tensor

  
  



• 
  

  2) that it be index symmetric so that it conserves angular momentum

  
  

• 3) when added to the stress–energy tensor of matter, $T^{\mu \nu}$, its total 4-divergence vanishes (this is required of any conserved current) so that we have a conserved expression for the total stress–energy–momentum.

• 4) that it vanish locally in an inertial frame of reference (which requires that it only contains first and not second or higher derivatives of the metric). This is because the equivalence principle requires that the gravitational force field, the Christoffel symbols, vanish locally in some frame. If gravitational energy is a function of its force field, as is usual for other forces, then the associated gravitational pseudotensor should also vanish locally.

The only quantity that satisfies all the above turns out not to be a tensor. My question is:

Is there a consistent derivation that doesn't assume requirement 3) but that still manages to be a well-behaved tensor?

I suspect that removing this requirement leaves too many possible degrees of freedom

electromagnetism - General relation between $E$ and $H$ field magnitudes

If $\eta=\sqrt{\mu/\epsilon}$ is the wave impedance of a homogeneous infinite medium in which a free-space electromagnetic wave propagates (i.e. plane wave), then we know that the magnitude of the electric and magnetic fields are related by

$$|E|=\eta |H|$$

Will this simple equation generally hold also for any field at resonance (e.g. say for an electromagnetic field resonating inside some cavity of arbitrary shape) or is it only known to be true for plane waves?

quantum field theory - Why does the electromagnetic and weak coupling strength do not meet at the electroweak scale?

The running of the coupling strengths is usually visualized on a logarithmic scale like here

What surprises me is that the weak and the electromagnetic coupling strength do not meet before the GUT scale. Why is this the case?

A common argument in Grand Unified Theories is that all elementary forces meet at some energy scale. Above this threshold we have only one interaction, describe by a gauge group $G$ and correspondingly only one coupling strength. The symmetry gets broken spontaneously to the standard model gauge group $G \rightarrow SU(3) \times SU(2) \times U(1)$ at lower energies, the coupling strength split and the new gauge bosons and possibly exotic fermions get a mass comparable to the GUT scale (this is called survival hypothesis).

Now, this is speculative beyond the standard model stuff, but in the standard model something very similar happens. The standard model gauge group $SU(3) \times SU(2) \times U(1)$ gets broken at energies below the electroweak scale.

$$SU(3) \times SU(2) \times U(1) \rightarrow SU(3) \times U(1)$$

Most books and papers talk about a unified electroweak interaction. Shouldn't this mean that the electromagnetic and weak coupling strength get unified?

And bonus: Shouldn't all fermions and bosons get a mass comparable to the Electroweak scale? Even without the neutrino the mass difference between the lightest (electron) $\approx 0,5 \cdot 10^{-3}$ GeV and heaviest (top) $\approx 170$ GeV is six orders of magnitude.

Answer

Answer to the main question:

It is a well regarded fact that the terminology unified electroweak interaction is a bit of an abuse of terminology. What the term means is that both Quantum Field Theories, the Hypercharge ($U(1)_Y$) and Weak ($SU(2)_L$), are unified in a common framework, which predicts the low energy electromagnetism ($U(1)_{em}$) through the Higgs mechanism

$$U(1)_Y \times SU(2)_L \to U(1)_{em}$$ It does not refer to an unification scenario as in the Grand Unified Theory (GUT) setup, it's meant to refer to an unification in the sense Weak decays and Electromagnetism are understood as remnants of a higher-energy theory, the Standard Model.

In comparison with GUTs the terminology can be applied if you think that regular GUT setups predict unification into a gauge group which is composed of solely one (semi) simple Lie group, e.g. $SU(5)$, $SO(10)$ being 2 of the most popular. In this sense the couplings do unify. The electroweak unification can be regarded as a unification into a group with 2 (semi) simple Lie group factors, the $SU(2)$ and $U(1)$. It is in this way of thinking about it that people refer to as unification. Notice that in the later case each factor can has its own coupling, and so the couplings are not equal, i.e. do not unify.

Answer to the bonus question:

What you asked is a big open question. Fermionic masses come from Lagrangian terms called Yukawa couplings, for example for the electron

$$y H L e+\mbox{h.c.}$$ for example, and the masses are then $$m \sim v y$$ where $v$ is the vacuum expectation value of the Higgs field, $H$, and $y$ are the Yukawa couplings are not specified in the Standard Model and one should expect them to be of order 1. But this only happens to the top quark, while all the other fermions have a lot smaller (in some cases many orders of magnitude smaller) than 1. Why this is like this is still an open question in Physics.

classical mechanics - What's the name of the symmetry $L to L + frac{d Lambda}{dt}$?

In the Lagrangian formulation of Classical Mechanics, we have the freedom to add a total time derivative of an arbitrary function $\Lambda$ to the Lagrangian:

$$L \to L + \frac{d \Lambda}{dt} .$$

Does this symmetry of the Lagrangian have any particular name?

Answer

Some authors call that a gauge transformation of the Lagrangian function. Others, don't give any specific name and may object to the previous one.

For a reference for the former denomination, see for instance F. Scheck, Mechanics, Springer, 2010.

electromagnetism - Why do like charges repel and opposite charges attract?

Why do like charges repel and opposite charges attract?

newtonian mechanics - Will centre of gravity coincide with centre of mass if density of object is non-uniform?

I read that for bodies of very large dimensions, but having non-uniform density, the centre of gravity does not coincide with centre of mass. I can understand that with large dimensions the strength of gravity can vary and hence affect the centre of gravity. But how does having a non- uniform density makes the centre of gravity not to coincide with the centre of mass?

Answer

The centre of gravity will always be the same as the centre of mass in a uniform gravitational field (constant in magnitude and direction). This applies for bodies with non-uniform density as well as those with uniform density. The Earth's gravitational field can be considered uniform if the dimensions of the object are much smaller than its distance from the centre of the Earth.

The reason for this is, when calculating CM constituent elements of mass are 'weighted' by their mass $m_i$ and vector distance $\vec{r_i}$ from an arbitrary fixed point :

$$\vec{r_{CM}} \times \sum m_i = \sum m_i \vec{r_i} \tag{1}$$

whereas when calculating CG the same elements are weighted by their gravitational weight $m_i g(\vec{r_i})$ which might vary with position $\vec{r_i}$ :

$$\vec{r_{CG}} \times \sum m_i g(\vec{r_i}) = \sum m_i g(\vec{r_i})\vec{r_i} \tag{2}$$

If $g(\vec{r_i})=g$ is the same for all elements (ie the gravitational field is uniform) then the calculation gives the same centre (CM=CG). If $g(\vec{r_i})$ is not constant (ie the gravitational field is not uniform) then the positions of CM and CG are likely to be different.

However, it is still possible for the CM and CG to coincide in special cases. For example, the effects of non-uniform field and non-uniform density could cancel out, making the CM and CG coincide.