Group Theory in General Relativity

In Special Relativity, the Lorentz Group is the set of matrices that preserve the metric, i.e. $\Lambda \eta \Lambda^T=\eta$.

Is there any equivalent in General Relativity, like: $\Lambda g \Lambda^T=g$?

(We could at least take locally $g\approx\eta$, so we recover the Lorentz group, but I don't know whether we could extend this property globally.)

Why does Group Theory have much less importance in General Relativity than in QFT and particle physics?

Answer

As you point out, the Minkowski metric $\eta = \mathrm{diag}(-1,+1, \dots, +1)$ in $d+1$ dimensions possesses a global Lorentz symmetry. A highbrow way of saying this is that the (global) isometry group of the metric is the Lorentz group. Well, translations are also isometries of Minkowski, so the full isometry group is the Poincare group.

The general notion of isometry that applies to arbitrary spacetimes is defined as follows. Let $(M,g)$ be a semi-Riemannian manifold, then any diffeomorphism $f:M\to M$ (coordinate transformation essentially) that leaves the metric invariant is called an isometry of this manifold.

A closely related notion that is often useful in relation to isometries is that of Killing vectors. Intuitively a killing vector of a metric generates an "infinitesimal" isometry of a given metric. Intuitively this means that they change very little under the action of the transformations generated by the Killing vectors.

Isometries and Killing vectors are a big reason for which group theory is relevant in GR. Killing vectors often satisfy vector field commutator relations that form a Lie algebra of some Lie Group.

Addendum (May 28, 2013). Remarks on symmetric spaces and physics.

One can show that in $D$ dimensions, a metric can posses at most $D(D+1)/2$ independent killing vectors. Any metric that has this maximum number of killing vectors is said to be maximally symmetric.

Example. Consider $4$-dimensional Minkowski space $\mathbb R^{3,1}$. The isometry group of this space, the Poincare group, has dimension $10$ since there are $4$ translations, $3$ rotations, and $3$ boosts. On the other hand, in this case we have $D=4$ so that the maximum number of independent killing vectors is $4(4+1)/2 = 10$. It turns out, in fact, that each rotation, translation, and boost gives rise to an independent Killing vector field, so that Minkowski is maximally symmetric.

One can in fact show that there are (up to isometry) precisely three distinct maximally symmetric spacetimes: $\mathrm{AdS}_{d+1}, \mathbb R^{d,1}, \mathrm{dS}_{d+1}$ called anti de-Sitter space, Minkowski space, and de-Sitter space respectively, and that these spacetimes all have constant negative, zero and positive curvature respectively. The isometry groups of these spacetimes are well-studied, and these spacetimes form the backbone of a lot of physics. In particular, the whole edifice of $\mathrm{AdS}/\mathrm{CFT}$ relies on the fact that $\mathrm{AdS}$ has a special isometry group that is related to the conformal group of Minkowski space.

newtonian mechanics - Does Newton's 3rd law of motion apply when a person is standing still on a ground?

When a person is standing still, the action of the Earth on their body is a gravitational force. What is the reaction force? Is it the person's weight acting on the Earth?

differential geometry - Difference between a vector field and a force field

In mathematics while learning about vector fields, we define a "vector field" as "a function of space whose value at each point is a vector quantity". That is, at each point in space there is a vector quantity attached to it.

Now if we talk about a force field which a three dimensional object is experiencing, the "force at a certain position of three dimensional object" cannot be interpreted as "acting at a point". Instead the force is acting over the whole three dimensional object.

Then how can it be justified that the "force field" is a "vector field" in the sense of above definition of vector field.

Answer

The force acts on every element of a 3d object.

Take for example a 3d charged object. The charges can be distributed any way at all. They could be uniform, or not. One side could be positively charged, the other negative. And so on.

That 3d object has an element of charge existing at every point. (More precisely: in every infinitesimally small volume element.) Each charge element experiences the force as determined by the vector field value at the point. The force on the object is the vector sum of all those contributions.

It is possible to find one point that has the property that the translational motion of the object acts as if the total charge on the object is concentrated at that point. One might call it "center of charge", but as you point out, one does not often hear of such a thing. That is, take that charge at that point (which turns out to be the total charge), and multiply by the value of the electric field vector at that point, and you will obtain the force on the object as a whole.

The fact that there might be one point that can be considered to be "where the force acts" is a consequence of summing, is not fundamental and has some limitations. In a sense it's an illusion. It works only if you can afford to model your object as a point particle. For example, you can't use it to calculate torque.

Imagine that your object is a cube that is uniformly positively charged on its "left" half, and is uniformly positively charged on its "right" half, and has total charge zero. In a uniform electric field there will be no force on the object. The "center of charge" will have $q=0$. The same result obtains by summing the forces on all the charge elements in the cube.

However, the cube will spin! There will be a torque on the cube.

Now, you may not care that it spins. If all you care about is the trajectory of the object, all you care about is the force. In such a case, you can afford to model the object as a point, and the "center of charge" approach will give you what you want. But if you care about rotation, you cannot afford to model the object as a point, and you cannot use the point particle model. If you are interested in total kinetic energy, you'll need to include rotational as well as translational kinetic energy. The "center of charge" approach can't give you that. The point particle model is too simple to account for rotational kinetic energy.

There's a quote attributed to Einstein, but I have my doubts about who might have said it first: "Make your models as simple as possible, but no simpler."

electromagnetism - Work done by magnetic field

I know Lorentz force don't do any work. but I want to know whether any type of magnetic field do a work or not.

dark energy - Does the cosmological constant solve the flatness of the Universe with 2 independent methods?

First. The universe is flat and not negativity curved because whatever dark energy is made from represented by the cosmological constant adds to the side of the density in the first Friedman equation:

$$\left(\frac{\dot{a}}{a}\right)^2 - \frac{8πGρ}{3} -\frac{Λc{^2}}{3} = -\frac{kc{^2}}{a^2}$$

So that the curvature $k$ is zero.

Second. The universe is not flat. It just looks flat because inflation, caused by the cosmological constant, blew up small flat patches of space.

These two explanations of why the universe is flat, have the same root (cosmological constant) but to me at least seem to contradict with each other. If they are indeed the same explanation done differently please explain.

Describe Ising model dynamics in stochastic differential equation or stochastic process

The Ising model is described by the Hamiltonian $$ H(\sigma) = - \sum_{} J_{ij} \sigma_i \sigma_j -\mu \sum_{j} h_j\sigma_j, $$ and is treated extensively by equilibrium statistical mechanics. Consider a Ising model at equilibrium is affected by a sudden change in external parameters such as the temperature and the external field and the system goes to a new equilibrium state after some time. Monte Carlo simulation can simulate this non-equilibrium process, but can this relaxation from a non-equilibrium state to a equilibrium state be described by a system of coupled differential equation or a stochastic process?

newtonian mechanics - Force felt by parachutist when opening the parachute

When falling down with air resistance, y-axis vertical (positive is up), x-axis is horizontal, we have that: $F_y=-mg-kv=ma_y$, where $k$ is the drag coefficient. Now if the man opens the parachute, then I can analize it in two ways:

1. At time $t_0$ of the deployment, I only change $k$ with $k_2>k$ and the rest stays the same.


2. I consider that the parachute will take a period of time $\delta t$ to fully open and therefore I need to consider $k(t)$.

Now If my aim is to study the force felt when opening the parachute, how should I do it?

My first idea was to work with (1), but I didn't come up with anything interesting. The second idea was to consider the derivative of the acceleration, the jerk. Hence to consider $m\frac{d^3y}{dt^3}=-k\frac{d^2y}{dt^2}$ which is the derivative of the expression above for the force. However I don't know how to minimize it. It can't be zero, otherwise, nothing would change, right? Should I then consider $k$ as in (2) and use the chain rule when differentiating? And what should I do after that if I wanted to minimize this force felt when opening the parachute?

Answer

Your model for the force is reasonable - there is a constant acceleration due to gravity, and a force that depends on velocity and size of the parachute. It is also reasonable that the parachute should take some time to deploy, and that the rate at which it opens may affect the force. One can make simplifying assumptions about this - for example, we can say that it takes the parachute a time $\tau$ from deployment to fully open, and that during that time the coefficient $k$ grows linearly - that is, $k(t) = k_0 + k_p\frac{t}{\tau}$ for $t\lt \tau$, and $k(t)=k_0+k_p$ for $t\ge \tau$ - where $k_0$ is the drag due to the body (without the parachute deployed), and $k_p$ is the drag coefficient due to the parachute.

Now we can write down the equation of motion. Let's use "down" as the positive Y direction - so the sign of gravity is positive, the velocity is positive, and the sign of the drag is negative. Then

$$\begin{align} F &= mg - k(t) v\\ m \dot v&= mg - \left(k_0+k_p\frac{t}{\tau}\right)v\\ \end{align}$$

If you want to know the derivative of the acceleration, you just divide that expression by $m$ and differentiate with respect to $t$, giving

$$\dot a = -\frac{k_p}{\tau} \left(v + \dot vt\right)$$

Substituting for the expression for $\dot v$, we get

$$\dot a = -\frac{k_p}{\tau}\left(v + \left(g - \left(\frac{k_0}{m}+\frac{k_p}{m\tau}t\right)\right)t\right)$$

Which tells us that at the time that the parachute first opens, the rate of change of acceleration is negative (you are accelerating less quickly as the drag gets bigger), but that it can actually change sign (greater acceleration in the other direction) as the parachute deploys; but the larger $\tau$, the less the jerk.

Of course this is one reason why parachutes have a little "helper chute" - not only does it provide the force needed to pull the main chute from the pack, but it also starts to slow down the parachutist - in essence, increasing $\tau$ in the equation above.

In reality, equations are not nice and linear - but this is a start.

standard model - The Makeup of the Pentaquark

Why is it that when they have the artist's rendition of the Pentaquark it shows two downs, two ups, and one anti-strange quark? Is this or is this just for show? Follow up to this question: if this configuration is just for show, what is the Pentaquark truly made of?

newtonian mechanics - Calculate impact force when time is not mentioned

I've found the following statement in ISS Wikipedia

A 7 g object (shown in centre) shot at 7 km/s (23,000 ft/s), the orbital velocity of the ISS, made a 15 cm (5.9 in) crater in a solid block of aluminium.

We can clearly see that there is no mention of time difference.I know that we can calculate Force(F) from change in momentum.

Here

initial v = 7 km/s

Final v = 0 Km/s

m = 7 g

My question is, is there any way to calculate the impact force when time is not present in description.

Answer

During an impact like this, the force acting is complicated because it varies a lot while the impact is happening. So when you talk about "impact force" you are actually referring to a mean force of the process. A simple way to calculate this mean force consists in making use of kinetic energy and work.

Assuming that we are in the reference frame of the ISS, the object (of mass $m$) is approaching with velocity $v$. Therefore its kinetic energy is given by: $$ E_k=\frac12mv^2 $$

Knowing that the alluminium plate is well fixed to the station, we can immagine that the impacting object won't modify the station's velocity, since its mass is too big with respect to the object's one. This means that all the kinetic energy owned by the object is dissipated during the impact. Energy is dissipated in many ways, through friction or heat for example: we can't really take account of all this effects, but what can do is consider that the object has penetrated the alluminium plate by a distance $h$.

So, if we assume that a constant force $F$ is acting during the impact (that mean force above), we have that the work done by this force is simply: $$ W=Fh $$

From the previous considerations, we know that all the kinetic energy was converted into this work, so it follows that $E_k=W$. Now we have an equation for $F$: $$ F=\frac{mv^2}{2h} $$

It is a simple model, but it can be a good way to estimate the order of magnitude of the force. (Using 10cm as depth, we get a force of $\sim 2\times 10^6N$.)

nuclear physics - How are computed Branching decay modes (for Higgs boson and from a general point of view)

I would like to know how are computed the branching decay diagram, like for example with Higgs boson represented below (source):

It seems there are 5 ways of decays for Higgs boson.

I suppose there are theorical calculus behind this diagram but concretely, what are the origins and the hypothesis to get these curves ?

Are cross sections implied into these calculus ?

Answer

In quantum field theory, branching ratios are calculated as the ratio between the decay width of a specific decay process and the total decay width of the particle.

For instance, for the decay of K-short meson into charged pions, $$\mathrm{BR}(K \rightarrow \pi^+ \pi^-) = \frac{\Gamma (K_S \rightarrow \pi^+ \pi^-)}{\Gamma (K_S)}$$ where the total decay width is $\Gamma (K_S)=\Gamma (K_S \rightarrow \pi^+ \pi^-)+\Gamma (K_S \rightarrow \pi^0 \pi^0)$. (Here, I have given K-short as an example because it has only two decay modes which makes it easy to talk about. I could write the complete list of the Higgs decays but it would just make the answer more crowded. See for example the Particle Data Group's booklet for all decay modes of all particles)

In order to calculate the decay width, you need to calculate the matrix element, $\mathcal{M}$, for the process in question, which is done by applying Feynman rules of the model on that process. For more details, you need to refer to a quantum field theory textbook, e.g., Griffiths or Peskin/Schröder.

The plot in your question is calculated for different Higgs masses since the ratios depend on the Higgs mass which is a free parameter of the Standard Model. Apparently plot is shown in order to say that the discovery of the boson with a mass of 125 GeV at the LHC in 2012 has the same branching ratios for the Higgs boson in the Standard Model.

schroedinger equation - What are the primary obstacles to solve the many-body problem in quantum mechanics?

(This is a simple question, with likely a rather involved answer.)

What are the primary obstacles to solve the many-body problem in quantum mechanics?

Specifically, if we have a Hamiltonian for a number of interdependent particles, why is solving for the time-independent wavefunction so hard? Is the problem essentially just mathematical, or are there physical issues too? The many-body problem of Newtonian mechanics (for example gravitational bodies) seems to be very difficult, with no solution for $n > 3$. Is the quantum mechanical case easier or more difficult, or both in some respects?

In relation to this, what sort of approximations/approaches are typically used to solve a system composed of many bodies in arbitrary states? (We do of course have perturbation theory which is sometimes useful, though not in the case of high coupling/interaction. Density functional theory, for example, applies well to solids, but what about arbitrary systems?)

Finally, is it theoretically and/or practically impossible to simulate high-order phenomena such as chemical reactions and biological functions precisely using Schrodinger's quantum mechanics, over even QFT (quantum field theory)?

(Note: this question is largely intended for seeding, though I'm curious about answers beyond what I already know too!)

Answer

First let me start by saying that the $N$-body problem in classical mechanics is not computationally difficult to approximate a solution to. It is simply that in general there is not a closed form analytic solution, which is why we must rely on numerics.

For quantum mechanics, however, the problem is much harder. This is because in quantum mechanics, the state space required to represent the system must be able to represent all possible superpositions of particles. While the number of orthogonal states is exponential in the size of the system, each has an associated phase and amplitude, which even with the most coarse grain discretization will lead to a double exponential in the number of possible states required to represent it. Thus in quantum systems you need $O(2^{2^n})$ variables to reasonable approximate any possible state of the system, versus only $O(2^n)$ required to represent an analogous classical system. Since we can represent $2^m$ states with $m$ bits, to represent the classical state space we need only $O(n)$ bits, versus $O(2^n)$ bits required to directly represent the quantum system. This is why it is believed to be impossible to simulate a quantum computer in polynomial time, but Newtonian physics can be simulated in polynomial time.

Calculating ground states is even harder than simulating the systems. Indeed, in general finding the ground state of a classical Hamiltonian is NP-complete, while finding the ground state of a quantum Hamiltonian is QMA-complete. (On the other hand, ground states are to some extent less relevant because the systems for which is is computationally hard to calculate the ground state of (at least on a QC) don't cool efficiently either.)

nuclear physics - Why are alpha particles made of 2 protons and neutrons?

When experiencing alpha decay, atoms shed alpha particles made of 2 protons and 2 neutrons. Why can't we have other types of particles made of more or less protons?

Answer

The reason why alpha particles heavily dominate as the proton-neutron mix most likely to be emitted from most (not all!) radioactive components is the extreme stability of this particular combination. That same stability is also why helium dominates after hydrogen as the most common element in the universe, and why other higher elements had to be forged in the hearts and shells of supernovas in order to come into existence at all.

Here's one way to think of it: You could in principle pop off something like helium-3 from an unstable nucleus - that's two protons and one neutron - and very likely give a net reduction in nuclear stress. But what would happen is this: The moment the trio started to depart, a neutron would come screaming in saying look how much better it would be if I joined you!! And the neutron would be correct: The total reduction in energy obtained by forming a helium-4 nucleus instead of helium-3 would in almost any instance be so superior that any self-respecting (and energy-respecting) nucleus would just have to go along with the idea.

Now all of what I just said can (and in the right circumstances should) be said far more precisely in terms of issues such as tunneling probabilities, but it would not really change the message much: Helium-4 nuclei pop off preferentially because they are so hugely stable that it just makes sense from a stability viewpoint for them to do so.

The next most likely candidates are isolated neutrons and protons, incidentally. Other mixed versions are rare until you get up into the fission range, in which case the whole nucleus is so unstable that it can rip apart in very creative ways (as aptly noted by the earlier comment).

electromagnetism - Transverse polarizations of a massless spin 1 particle

Physical polarization vectors are transverse, $p\cdot{\epsilon}=0$, where $p$ is the momentum of a photon and $\epsilon$ is a polarization vector.

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Physical polarization vectors are unchanged under a gauge transformation $\epsilon + a\cdot{p}=\epsilon$, where $a$ is some arbitrary constant.

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For a massless spin $1$ particle, in the Coulomb gauge,

one common basis for the transverse polarizations of light are

$$\epsilon_{\mu}^{1}=\frac{1}{\sqrt{2}}(0,1,i,0)\qquad \epsilon_{\mu}^{L}=\frac{1}{\sqrt{2}}(0,1,-i,0).$$

This describes circularly polarized light and are called helicity states.

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In the centre of mass frame, in the positive $z$-direction, the polarization vectors of a photon are

$$(\epsilon_{\mu}^{\pm})^{1}=\frac{1}{\sqrt{2}}(0,1,\pm i,0)\qquad (\epsilon_{\mu}^{\pm})^{L}=\frac{1}{\sqrt{2}}(0,1,\mp i,0).$$

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1. Why are physical polarization vectors transverse?


2. How is $\epsilon + a\cdot{p}=\epsilon$ a gauge transformation? The gauge transformations I know are of the form $A_{\mu}\rightarrow A_{\mu}+\partial_{\mu}\Lambda$.


3. How do you define transverse polarization of light? For example, why are $\epsilon_{\mu}^{1}=\frac{1}{\sqrt{2}}(0,1,i,0)$ and $\epsilon_{\mu}^{L}=\frac{1}{\sqrt{2}}(0,1,-i,0)$ the transverse polarizations of light?


4. Why do the polarization vectors $\epsilon_{\mu}^{1}=\frac{1}{\sqrt{2}}(0,1,i,0)$ and $\epsilon_{\mu}^{L}=\frac{1}{\sqrt{2}}(0,1,-i,0)$ correspond to transverse polarizations of light?


5. Why are these polarization vectors $\epsilon_{\mu}^{1}=\frac{1}{\sqrt{2}}(0,1,i,0)$ and $\epsilon_{\mu}^{L}=\frac{1}{\sqrt{2}}(0,1,-i,0)$ called helicity states?



6. In the centre of mass frame, in the positive $z$-direction, why are the polarization vectors of a photon given by $(\epsilon_{\mu}^{\pm})^{1}=\frac{1}{\sqrt{2}}(0,1,\pm i,0)$ and $(\epsilon_{\mu}^{\pm})^{L}=\frac{1}{\sqrt{2}}(0,1,\mp i,0)?$

Answer

I will answer the first two questions and leave the rest to others.

First question - We describe the photon with $A_{\mu}$ but this has 4 degrees of freedom.But we ultimately need just 2. To reduce the number we use the transverse condition and say the momentum is perpendicular to the polarization vector. So any polarization vector will be transverse to the direction of motion. This gives us 3 degrees of freedom. To get to 2, we set up an equivalence class of polarization vectors. This leads to second question

Second question - $ A_{\mu} \rightarrow A_{\mu }+\partial_{\mu} \lambda $ in momentum space becomes $ \epsilon \rightarrow \epsilon + \alpha p $ . This is gauge transformation because notice $ \epsilon \cdot p \rightarrow e \cdot p + \alpha p \cdot p $ but $ p \cdot p = 0 $ so in fact $ \epsilon\text{ and } \epsilon + \alpha p $ represent the same physical state in the same way that $ A_{\mu} $ and $A_{\mu }+\partial_{\mu} \lambda $ represent the same physical state.

pattern - Fill The grid - 3.0

Can you fill this grid ? Version ($\pi$)

In text:

?14516
243748

120581
17186?
01?857

HINT 1

It's a big spiral!!!

HINT 2

It starts from 0 at the center!

HINT 3

Colors included to identify

Answer

The final grid is

014516

243748
120581
171869
010857

And the pattern is

A clockwise spiral out from the center of the following function, displayed such that each digit of a multidigit number occupies one grid square
f(1)=0, f(n+1)=f(n)+O(n+(-1)^n) where O(n) is the nth odd number, or 2n-1
example: f(2) = f(1)+O(3) = 0+5 = 5, F(3) = f(2)+O(2) = 5+3 = 8...

homework and exercises - Guidance needed in finding scattering amplitude

If I have the Lagrangian $$\mathcal{L}=\bar{\psi}(i\gamma ^\mu \partial_\mu - m)\psi -g\bar{\psi}i\gamma^5\phi\psi,$$ where $g$ is a coupling constant.

How to find the scattering amplitude for $$ \phi \psi \to \phi \psi $$ What I only learned in class was electron-electron scattering and electron-proton scattering and can't seem to relate this case above to any of them. I ask only for guidance. Please and thank you!

combinatorics - Possible pawn combinations

This may seem simple, but I have a problem calculating it. It may be because it's Monday morning.
How may possible valid combinations of one color pawn (white or black, your choice) positions are there on a regular chess board?
To explain it better....
Start a normal chess game.

1. So we have the first pawn combination: All white on row 2 and/or all black on row 7.


2. (Almost) Every pawn (of the color you choose) move means a new combination .


3. Every other piece move does not increase the number of pawn combinations.


4. Pawns on the last row (8 for white, 1 for black are not pawns anymore, they are considered as promoted)


5. (exception for 2.) In the example below, capturing the white pawn on c4 results in the same combination, not 2 different ones. Example is provided for black pawns, but it applies if you choose white also.

I will settle for partial responses, strategies or anything.

Answer

I'm going to be working from the assumption that you still want pawns from both sides for this answer. If you only care about 1 side, the numbers get much more manageable (but slightly less interesting).

First let's establish a massive upper bound and see if we can work it down from there:

Each of the 48 positions is either black, white, or unoccupied.

3^48 = 7.97666e+22

Working down from that, you can assume that at most 16 spaces from the 48 can be occupied.

The sum of all n from 0 to 16 for 48 choose n gives us 4.1243046e+12 possible choices for pawn locations.

To account for pawn color, you multiply each term from the above equation by 2^n before adding it together. After accounting for piece color, that gives us 1.9341983e+17 states.

We can possibly reduce this down a bit further as the above number allows for anywhere from 0 through 16 pawns of each color, but we can limit it to 8 of each color.

To do that, we can choose 0 through 8 pieces for one color and then for each have 48 minus that number choose the second number for all combinations of 0 to 8 pieces on each side.

This gives us 4.90955e+13 possibilities. This number does still encompass a few impossible arrangements, but it should be a pretty close upper bound.

Now let's work up from what the pawns can do.

To establish a strong lower bound for this number, we assume that all pawns may only move forward and do not capture other pieces. The white pawn can move from 0 to 5 spaces and then the black pawn can move from 0 to 5-i spaces. This is the set of possible configuration states for the column as a whole. The configuration of the whole board can then be expressed as:

or 2.56289e+9. This is a strict lower bound of the answer.

If we assume that any pawns can be safely removed from the board by a combination of player cooperation and a knight being able to reach any square on the board, we can add in the empty state for no pawns, six states for the white pawn and six states for the black pawn giving:

or 3.77801e+11. Again, assuming any pawn can be surgically extracted by a knight, we have an even better strict lower bound

So far we have 3.77801e+11 < states < 4.90955e+13

It might be possible to extend this line of thinking to include cases with 2 pawns of one color and 1 of another where a pawn joined his buddy's file. And 3 and so on. In order to still keep this as a lower bound, we just have to prove that all the positions covered by the equation are reachable but don't necessarily cover all the possible behaviors of the pawns. This is where things start to get sticky. The ability to move sideways on a board is a limited resource, so if we have a state that involves more than 7 file displacements, it's unreachable without causing side effects to the opposing pawns. More on that in the next section.

Now on to capturing

Pawns can only change their file by capturing and no pawn may capture more than 5 times (since it puts you forward a space). Each type of pawn also opens up different amounts of possible positions for itself with some number of captures. A single capture let's a rook's pawn open up all but 1 of the positions in the adjacent knight's file, adding 5 potential position whereas a knight's pawn with a single capture gains access to the rook's file or the bishop's file, adding 10 possible positions. For 2, 3, 4, and 5 captures, each piece gains a different number of possible positions.

I'll be naming the pawns for their file name.

Positions with 1 capture:

• Rook 12


• Knight 17


• Bishop 17


• King 17

Positions with 2 captures:

• Rook 16


• Knight 21


• Bishop 25



• King 25

Positions with 3 captures

• Rook 19


• Knight 24


• Bishop 28


• King 31

Positions with 4 captures

• Rook 21


• Knight 26


• Bishop 30


• King 33

Positions with 5 captures

• Rook 22


• Knight 27



• Bishop 31


• King 33

So let's make another upper bound, this time based on per-piece positions and assuming as many captures as we want:

Since we have 4 of each type of pawn on the board, we get:

33^4 * 31^4 * 27^4 * 22^4 = 1.3634786e+23 possible configurations.

This bound is way higher than the previous one as this method produces a lot of "duplicate" board states such as a knight and rook pawn trading files as well as not handling co-occupied spaces between our same color pieces or opposite ones, so this is a looser upper bound.

There are 7 "free" captures per side on the board that don't affect other pawns. These include capturing Rooks, Knights, Bishops, and the Queen. Since they are irrelevant to our pawn configurations, they're "free" ways to capture.

When you add in opposing pawns, there are 15 captures per side in total, so the total pawn file displacement must be less than or equal to 15. Each displacement beyond the 7th, however, produces side effects. Since we removed an opposing pawn to shift over, this restricts one of the other side's pawns to one option (non-existence).

Working from just the 7 "free" captures per side, let's assume each pawn captures strategically to add the most possible moves to its move pool.

The highest utility each piece gets from a move is +10 for knight/bishop/king pawns on the first capture, so lets' use 6 of them for that. The next hightest is +8 on a bishop/king pawn on its second capture, so let's use the final one on one of them.

4x 0 cap Rook(7 moves) + 10x 1 cap Knight/Bishop/King (17 moves) + 2x 2 cap Bishop/King (25 moves)

7^4 * 17^10 * 25^2 = 3.0252508e+18

So our 3.0252508e+18 number is the maximum number of non-destructive board states that can be made in a single game before removing co-occupying pieces. We still aren't accounting for co-occupancy, so this can't be a strict lower bound. We also aren't accounting for all the possible destructive captures either, so this isn't an upper bound either.

If we want to refine our non-destructive capture number to include all possible games, we'll need to distribute those 7 "free" captures to each of 2 rook pawns, 2 knight pawns, 2 bishop pawns, and 2 king pawns. For each side, we have to use a partition operator on 7 and square the result.

There are 15 ways to partition 7 into addition, but some of those (7+0 and 6+1) involve more than 5 captures for a single pawn, so they can be discarded, leaving 13. Each unique partition generates its own choose function that gets added on to deal with the distribution of 0s, 1s, 2s, etc across the 8 possible slots.

Theoretically, each of those distributions would create a set of states for a single game, each depending on how the captures were distributed. The number will be massive but there should be a lot of collision, which means we can theoretically drop a fair number of them.

Captures beyond the 7 "free" ones should always yield fewer moves that the non-destructive game. For instance, sacrificing the Rook pawns to give the Bishop/King pawns more moves gives:

4x dead Rook(1 option) + 10x 1 cap Knight/Bishop/King (17 options) + 4x 2 cap Bishop/King (25 options) for 18 captures, taking all 4 Rook pawns off the table

1^4 * 17^10 * 25^4 = 7.8749762e+17

Thus, the additional states provided by each of these should be less than each of the non-destructive games.

Note, we still haven't taken out doubly occupied positions.

Self-overlapping can be reduced out by determining the amount of overlap each pawn has with another per amount of captures it makes. For instance, a 1 cap Rook's pawn has a range that covers 5 of the 7 locations of the adjacent 0 cap Knight's pawn, meaning we should treat the move contributions of the 0 cap Knight's pawn as 2 moves instead of the usual 7. We would need to create a mapping table for each type of piece's interference against each other type and use it for each of the partition calculations mentioned earlier. This will remove a lot of duplicate or erroneous board states.

Note: We still havn't dealt with pawns being blocked by others and all sorts of other things.

I'll add more later if I get some more time.

newtonian mechanics - Goalkeeper does the work when he stop the ball

When a goalkeeper stops the ball he does the work but how he only stops the ball but dont displaces the ball please explain

Answer

Does the goalkeeper do Work while stopping the ball?

The amount of Work done by the goalkeeper’s mass equals the amount of kinetic energy lost by the ball.

Computing the velocity of the mass-plus-ball system after the goalkeeper captures the ball gives us a quantitative answer as to the amount of Work done by the goalkeeper’s mass.

To solve this problem, assume the goalkeeper captures the ball in a frictionless process, (e.g., the ball sticks to the goalkeeper, and the goalkeeper begins to slide on a frictionless surface). The goalkeeper gains kinetic energy, and the ball loses kinetic energy. The total amount of kinetic energy in the ball before the capture is the same as the kinetic energy of the ball-goalkeeper system after the capture.

The big picture: the goalkeeper’s mass exerts an inertial force on the ball, causing the ball to decelerate. Likewise, the ball exerts an inertial force on the goalkeeper, causing him/her to accelerate. The amount of kinetic energy lost by the ball is the amount of Work done by the inertial force exerted by the ball on the mass of the goalkeeper.

The calculation is performed by equating the kinetic energy of the ball before the collision, with the kinetic energy of the goalkeeper-plus-ball after the collision, and then solving for the velocity.

If we choose the goalkeeper at rest as the $(0,0)$ origin of the frame of our system, in this frame, the goalkeeper has zero kinetic energy. The ball has velocity $v_i$ and mass $m_{b}$, and therefore $KE=\frac{1}{2}m_{b}v_i^2$.

After the goalkeeper catches the ball, the goalie plus ball system will have velocity $v_f$. The system of the ball-goalkeeper will have the same total kinetic energy as was present in the ball before the collision. The transfer of energy (and Work done) is obvious in hockey when the goalkeeper is on frictionless ice. After catching the ball, both goalkeeper and puck move backward with a lower velocity than the puck. The work done by the mass of the goalkeeper on the puck/ball is equal to the change of kinetic energy of the puck/ball. $$W=\Delta KE = \frac {1}{2} m_b (v_i -v_f) ^2$$

The key to understanding this concept is realizing that mass exerts an inertial force on any body attempting to accelerate that mass. Thus, the total Work done in the process can be calculated by integrating the force acting over each increment of distance moved. $W=\int F dx$. Obviously, this is a difficult way to calculate the Work done by the ball on the goalkeeper because the force exerted by the ball changes each moment over a very short period of time. We would need equipment to measure the force vs. time function to do the calculation. The easier method is to simply measure the speed of the ball before capture, and speed of the goalkeeper-ball system afterward, and calculate the change in kinetic energy of the ball.

In the process of the collision (catching the ball), the goalkeeper accelerates, and the ball decelerates, which means the ball exerts an inertial force on the goalkeeper, and the goalkeeper exerts an inertial force on the ball.

The inertial force exerted by a mass is the force the mass exerts against a force attempting to accelerate that mass. Note: the inertial force exerted by a mass will always be opposite and equal to the force applied to the mass. For example: The goalkeeper feels the inertial force of the ball when it hits his/her hands. As the ball contacts the goalkeeper’s hands, the force applied on his/her hands is opposite and equal to the force applied by his/her hands on the ball. But, because of the conservation of energy and momentum (and a lot more discussion of what that means), even though the forces are opposite and equal, the ball decelerates, and the mass of the ball-goalkeeper system accelerates.

In short, the inertial force exerted by the ball on the goalkeeper is the force referred to in Newton's second law, $F_{inertial}=ma$.

Work, in its most general sense, is defined as the amount of energy transferred into or out of a system. If energy leaves a system, we typically refer to that transaction as the system “doing work,” and vice versa. In the case of the goalkeeper, the ball did work on the goalkeeper by accelerating him/her. The perspective of "doing Work on," or "having Work done to" a system, changes depending on the frame chosen and system boundaries.

In this question, we see the goalkeeper is accelerated, and the ball is decelerated, and so energy is transferred, and hence Work is done.

electricity - Can lightning be used to solve NP-complete problems?

I'm a MS/BS computer science guy who is wondering about why lightning can't (or can?) be used to solve NP complete problems efficiently, but I don't understand the physics behind lightning, so I'm posting here.

What seems peculiar to me about lightning is that it seems to follow the "easiest" path to a given destination, as opposed to just the shortest path. This seems analogous (identical?) to the weights in the traveling salesman problem.

Since the lightning process ionizes the air and seems to pick the easiest route quickly, I have 2 questions:

1) Does lightning use a method similar to Dijkstra's approach (such that a "good enough" path is found in polynomial time, but not necessarily the "best of all options")? Is it basically an NP "solver"?

If the answer to the above question is "no" and it is indeed the best pick:

2) Could the selection process of lightning be used in a computer algorithm to solve a traveling salesman-type problem in polynomial time, and

3) If not, could an arbitrary series of weighted nodes be physically set up somewhere and have electricity passed through it and have the results fed back into some kind of computer mechanism for every time a user wants to know a shortest path?

If you could use this approach to solve NP-hard problems quickly, you would have your name on the news and maybe in the history books. So why hasn't anyone used lightning to solve this problem?

UPDATE: Another aspect I find interesting: when lightning strikes it ionizes the air to find the shortest path. If the world were totally flat (and the clouds were also) this would ionize all the air equally. Which leads me to believe the route "picked" by the lightning incorporates all points, which seems to be a requisite for finding the best of all solutions.

Answer

1) yes, it basically will find a non-optimal solution. At every point, the top of the ray looks for the bigger potential gradient, the charge in the surrounding volume grows, polarizing surrounding material (air, in this case) until a bigger gradient shows up and the ray continues over that direction. This is why the lightining path looks like a jigsaw; its basically a Monte Carlo search.

Not to say, that you can't apply this to find efficient solutions of the travelsman problem. This would require making a programmable configuration with weights related to capacitance so the electric arc finds the best path that also solves your particular adjacency matrix problem. This would probably be only hampered by the fact that the arc, in order to do any traversing, must burn its way across the capacitance barrier, so unless you find a cheap, repleaceable material that can also be configurable (the weight needs to adjust to the adjacency matrix weights) it will be cheaper to do this computation in a standard computer.

quantum mechanics - What are independent parameters in Hellmann–Feynman theorem?

A typical example in textbooks about the application of Hellmann–Feynman theorem is calculating $\left\langle\frac{1}{r^2}\right\rangle$ in hydrogen-like atoms. Wikipedia has a nice demonstration of this. At some point in the Wikipedia derivation is used that

$$\frac{\partial n}{\partial \ell}~=~1. \tag{1}$$

But why is eq. (1) true? I know that $$n=n_r+\ell+1,$$

but $n_r$ is just another variable with different physical meaning, so why is $n_r$ independent from $\ell$, whereas $n$ is not? The Wikipedia proof for Hellmann–Feynman theorem does not address the problem of independence of different parameters. What variables are kept fixed during the differentiation $(1)$ and why?

The Wikipedia page seems to have only a vague notion of $\frac{\partial\hat{H}}{\partial\lambda}$ and $\frac{\partial E}{\partial\lambda}$, unlike in, e.g., thermodynamics, where all partial derivatives are typically written like $$\left( \frac{\partial U}{\partial V} \right)_S \qquad ,\qquad \left( \frac{\partial U}{\partial V} \right)_p \qquad ,\qquad \ldots $$ so that it is clear which variables are kept fixed during the differentiations.

Answer

1. 
   

   The application of Hellmann–Feynman theorem to calculate the expectation value $$ \langle n\ell m | \hat{r}^{-2} | n\ell m \rangle \tag{1}$$ of a radial operator e.g. $\hat{r}^{-2}$ does only depend on the radial wave function $R_{n\ell}(r)$ and not the spherical harmonics $Y^m_\ell (\theta, \phi)$.

   
   


2. 
   
   

   The angular part of the hydrogen-like Hamiltonian $$\hat{H} ~:=~ \frac{1}{2\mu r^2}\left\{ - \hbar^2 \frac{\partial}{\partial r}r^2\frac{\partial}{\partial r} +\hat{L}^2\right\} - \frac{Z e^2}{r},\qquad e^2 ~:=~ \frac{e_0^2}{4\pi\varepsilon_0} , \tag{2}$$ depends on the angular momentum operator $\hat{L}^2$. We now replace $\hat{L}^2$ with its eigenvalue $\hbar^2 \ell(\ell+1)$. The resulting Hamiltonian $$\hat{H}_{\ell} ~:=~ \frac{ \hbar^2}{2\mu r^2}\left\{- \frac{\partial}{\partial r}r^2\frac{\partial}{\partial r} +\ell(\ell+1)\right\} - \frac{Z e^2}{r}\tag{3}$$ depends on the radial variable $r$ but not the angular variables $(\theta, \phi)$.

   
   


3. 
   

   Thus we can formally think of space $$\mathbb{R}^3 ~=~ [0,\infty[ ~\times~ S^2 \tag{4}$$ as just a halfline $[0,\infty[$, where the radial variable $r$ lives, as the angular variables $(\theta, \phi)$ have become irrelevant for the problem.

   
   


4. 
   

   When we eliminate the two-sphere $S^2$, we eliminate spherical symmetry $SO(3)$. Recall that the number $\ell$ had to be an integer to have finite-dimensional unitary representations of $SO(3)$. But in the radial half-line picture, the number $\ell$ has lost its geometric meaning, and we can formally proceed with a continuous $\ell\in[0,\infty[$. This is needed to apply Hellmann–Feynman variational method.

   
   


5. 
   

   But we still have to solve the radial time-independent Schrödinger equation (TISE) $$\hat{H}_{\ell}R_{n\ell}(r) ~=~E_n R_{n\ell}(r) \tag{5}$$ in this new situation. The upshot is that for real $\ell\in[0,\infty[$, we still derive a quantization condition, namely, that the bound state energy levels $E_n$ are still discrete, and that the variable $$ n_r ~:=~ n-\ell -1 ~\in\mathbb{N}_0 \tag{6}$$ should be a non-negative integer. Here the 'principal number' $n\in[0,\infty[$ is defined to make the standard energy formula for the hydrogen-like bound state energy spectrum $$E_n~=~ -Z^2\alpha^2\frac{\mu c^2}{2n^2}\tag{7}$$ still hold with the caveat that $n$ might not be an integer! In other words, eq. (7) is a definition of $n$ in terms of the bound state energy $E_n$.

   
   
   


6. 
   

   Thus if we vary $\ell$, we must also vary $n$ by the same amount to keep $n_r$ an integer.

   
   

differential geometry - What is a Killing vector field?

I recently read a post in physics.stackexchange that used the term "Killing vector". What is a Killing vector/Killing vector field?

Answer

I think https://en.wikipedia.org/wiki/Killing_vector_field answers your question pretty good:

"Killing fields are the infinitesimal generators of isometries; that is, flows generated by Killing fields are continuous isometries of the manifold. More simply, the flow generates a symmetry, in the sense that moving each point on an object the same distance in the direction of the Killing vector field will not distort distances on the object."

A killing vectorfield $X$ fulfills $L_X g=0$ where $L$ is the Lie derivative or more explicit $ \nabla_\mu X_\nu + \nabla_\nu X_\mu =0$.

So in a layback manner: When you move the metric $g$ a little bit by $X$ and $g$ doesn't change, X is a killing vectorfield.

For example the Schwarzschildmetric https://en.wikipedia.org/wiki/Schwarzschild_metric has two obvious Killing vectorfields $\partial_t$ and $\partial_\phi$ since $g$ is independent of $t$ and $\phi$.

Edit: On recommndation I add a nice link to a discussion of how to use Killing vector fields: See the answer of Willie Wong at Killing vector fields

mathematical physics - Subtlety in the proof of 2-to-1 homomorphism between $SU(2)$ and $SO(3)$

In physics, it's common to use the relations $$\textbf{r}^\prime=\mathscr{R}\textbf{r};~~\text{and}~~\textbf{r}^\prime\cdot\boldsymbol{\sigma} =\mathscr{U}(\textbf{r}\cdot\boldsymbol{\sigma}) \mathscr{U}^{\dagger}\tag{1}$$ to establish a two-to-one homomorphism between ${\rm SU(2)}$ and ${\rm SO(3)}$ where $\textbf{r}\in \mathbb{R}^3$, $\mathscr{R}\in {\rm SO(3)}$, $\mathscr{U}\in {\rm SU(2)}$ and $\boldsymbol{\sigma}=(\sigma_1,\sigma_2,\sigma_3)$ are three Pauli matrices. Both the relations of Eq.(1) represent rotation of coordinates in real three-dimensioanl space because both of them satisfy $|\textbf{r}^\prime|^2=|\textbf{r}|^2$. It's easy to see from (1) that corresponding to every $3\times 3$ matrix $\mathscr{R}\in {\rm SO(3)}$ there exist two $2\times 2$ matrices $\pm \mathscr{U}\in {\rm SU(2)}$ that represent the same rotation.

Question Note that the above proof of 2-to-1 homomorphism is based on fundamental representations of $SO(3)$ and $SU(2)$. But for any odd-dimensional representation of $SU(2)$, if $\mathscr{U}$ has determinant $+1$, $-\mathscr{U}$ is not a representation of $SU(2)$ since it has determinant $-1$. Hence, if $\mathscr{U}$ is a member of an odd-dimensional representation of $SU(2)$. $\mathscr{U}$ is not. Does it mean that 2-to-1 homomorphism between $SU(2)$ and $SO(3)$ is not true in general?

Mass in terms of energy

Given the relationships between mass and energy in relativity, and given that particles with mass can be created given energy over the threshold energy, and vice-versa, can we say that mass is simply an extremely dense form of energy? Or is there a deceptive parallel between the two?

cosmology - Doesn't dating the universe violate the concept of spacetime's inseparability?

It would seem that measuring an age of the universe from the big bang requires separating spacetime into a 3D coordinate system and a time track. I fail to understand why it is appropriate to take the reference of a comoving observer as being stationary to the cosmic microwave background when this seems to be a choice based upon our subsequent detection of that entity.

quantum field theory - Dark matter: degrees of freedom

I'm afraid this question could sound a little too vague. I don't even know if dark matter (DM) can be genuinely described by quantum field theory, or if quantum field theory should be somehow "modified" in order to include dark matter.

Assuming that ordinary QFT describes DM, what can be said (or what is known) about the number of degrees of freedom dark matter should have?

Answer

It is trivial to design a dark matter candidate that is compatible with quantum field theory: massive sterile neutrinos are a moderately popular possibility already.

But that doesn't prove anything, because it is just a dark matter candidate. Indeed the question is rather speculative until we know something about what the dark matter is rather than just things about what it isn't.

cosmology - Will acceleration rate of expansion of space become faster than speed of light?

From watching cosmology lectures, it seems that the space between galaxies is expanding at an accelerating rate, my question is since it is the space that is (acceleratingly expanding), the special relativity does not apply in this case? in other words since it is not anything that is accelerating with relative distance, would at some point galaxies that are far enough from each other separate faster than speed of light?

Please note, I am not asking if the galaxies would move faster than speed of light, but whether the rate of expansion of space be faster than speed of light so in effect each galaxy would be as if it is behind an event horizon where even light can not escape?

If yes, is there a name for such an event horizon that is cause by expanding space outside rather than the conventional black hole event horizon caused by stretching of space within the event horizon?

How would one be able to differentiate between a black hole even horizon and a galactic even horizon caused by accelerating expansion of space outside? (not sure if the concepts of inside and outside are still meaningful in such cases)

Answer

Yes indeed, in the circumstances you describe a horizon does form, and it's called a cosmological event horizon. Googling for this term will lots of articles on the subject, though for once Wikipedia has let me down and does not have a good article on the subject. However each galaxy wouldn't be behind it's own horizon as groups of galaxies tend to be gravitationally bound together. For example the Milky Way would stay bound to Andromeda and a dozen or so smaller galaxies.

Your question suggests you're think of this horizon as a sort of shell that would stop outsiders looking in on the galaxy behind the horizon, but it's really the other way round. The cosmological event horizon is like a shell that stops us looking out. In this respect it's the opposite of a black hole that stops us looking in.

A quick footnote: I had another Google and found http://www.mso.anu.edu.au/~charley/papers/DavisLineweaver04.pdf, which seems interesting reading on the subject.

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?

electromagnetism - Formal Connection Between Symmetry and Gauss's Law

In the standard undergraduate treatment of E&M, Gauss's Law is loosely stated as "the electric flux through a closed surface is proportional to the enclosed charge". Equivalently, in differential form, and in terms of the potential (in the static case):

$$\nabla^2 \phi = -\frac{\rho}{\epsilon_0}$$

Now, when using the integral form, one typically uses the symmetries of a known charge distribution to deduce related symmetries in the electric field, allowing the magnitude of the field to be factored out of the integral. To do so, one usually relies on intuitive, heuristic arguments about how the field in question "ought to" behave$^1$.

I'm wondering how one goes about formalizing this notion in precise mathematical terms. In particular, it seems that there ought to be an equivalent statement for Gauss's law in differential form, along the lines of "symmetries in $\rho$ induce related symmetries in $\phi$". Is there a way to formally state this claim? In particular:

1. How would one formulate a proof of the conditions (necessary and sufficient) under which a symmetry of $\rho$ induces a symmetry in $\phi$?


2. When it exists, how does one explicitly state the induced symmetry in terms of the known symmetry?


3. Can such a result be generalized for arbitrary linear PDEs whose source terms exhibit some symmetry?

It seems to me like there must exist a concise, elegant, and general way to state and prove the above, but I can't quite seem to connect all the dots right now.

---

$^1$ See, for example, in Griffiths, Example 2.3, p. 72: "Suppose, say, that it points due east, at the 'equator.' But the orientation of the equator is perfectly arbitrary—nothing is spinning here, so there is no natural "north-south" axis—any argument purporting to show that $\mathbf{E}$ points east could just as well be used to show it points west, or north, or any other direction. The only unique direction on a sphere is radial."

Answer

Let $\mathcal{D}$ be the operator corresponding to your equation ($-\epsilon_0 \nabla^2$ in this case). Let $U$ be some operator corresponding to the symmetry. It might be a rotation or parity transformation, etc.

If $U f = f$, we say the function $f$ is symmetric.

If $U\mathcal D=\mathcal D U$ as operators, then we say your equation is symmetric.

Let, $\mathcal {D} f= g$. If $\mathcal{D}$ is symmetric and $g$ is symmetric, then we can easily show $\mathcal{D}f = \mathcal{D}\mathcal U f$. If we can take an inverse of $\mathcal{D}$ we've proven $f$ is symmetric.

Taking an inverse of $\mathcal{D}$ is the the same thing as being able to solve the equation uniquely. In your particular case we can solve the equation uniquely if we restrict our function space to have some boundary conditions, say vanishing at infinity.

So this is the example you had in mind and this a formalization of the argument that $\phi$ must be symmetric.

Now if we can't solve the equation uniquely then there may be a loophole in the argument. A particular case I have in mind is a magnetic monopole which is rotationally symmetric, but the vector potential solution has a Dirac string and is not. But any two solutions $f$ and $Uf$ in this case are connected by a gauge transformation.

logical deduction - Don't put yourself out, dear

I asked my mother the other day, "What do you want for dinner? I'll get you anything your little heart desires."
My mother, true to form, radiated the stereotypical motherly vibe: "Don't worry about me. Just make whatever you want, dear."
So I said, "You want a nice cold-cut sandwich?" to which she replied, "I dunno, maybe something warm."
So I racked my brain. "How about some mac and cheese? You love my mac and cheese!"
She said "Maybe something meatier."
I said, "A steak? How about a nice, juicy steak?"
"Too much work, with the knife and the fork and the cutting and the chewing."
Hmm. "Spaghetti and meatballs? Meatballs are meaty!"
"Too round. Maybe something flatter?"

What did my mother want to eat?

Answer

Perhaps

she wanted a hamburger. No cutting, meat, flat, warm. But be sure to make the meat patties square.

quantum mechanics - Why is the singlet state for two spin 1/2 particles anti-symmetric?

For two spin 1/2 particles I understand that the triplet states ($S = 1$) are: $\newcommand\ket[1]{\left|{#1}\right>} \newcommand\up\uparrow \newcommand\dn\downarrow \newcommand\lf\leftarrow \newcommand\rt\rightarrow $

\begin{align} \ket{1,1} &= \ket{\up\up} \\ \ket{1,0} &= \frac{\ket{\up\dn} + \ket{\dn\up}}{\sqrt2} \\ \ket{1,-1} &= \ket{\dn\dn} \end{align}

And that the singlet state ($S = 0$) is:

$$ \ket{0,0} = \frac{\ket{\up\dn} - \ket{\dn\up}}{\sqrt2} $$

What I'm not too sure about is why the singlet state cannot be $(\ket{↑↓} + \ket{↑↓})/\sqrt2$ while one of the triplet states can then be $(\ket{↑↓} - \ket{↑↓})/\sqrt2$. I know they must be orthogonal, but why are they defined the way they are?

Answer

Let's temporarily forget that the two $m=0$ states exist, and consider just the two completely aligned triplet states, $\newcommand\ket[1]{\left|{#1}\right>} \newcommand\up\uparrow \newcommand\dn\downarrow \newcommand\lf\leftarrow \newcommand\rt\rightarrow $ $\ket{\up\up}$ and $\ket{\dn\dn}$. There's not any physical difference between these: you can "transform" your state from one to the other by changing your coordinate system, or by standing on your head. So any physical observable between them must also be the same.

Either of the single-particle states are eigenstates of the spin operator on the $z$-axis, $$\sigma_z = \frac\hbar 2\left(\begin{array}{cc}1&\\&-1\end{array}\right),$$ and "standing on your head," or reversing the $z$-axis, is just the same as disagreeing about the sign of this operator.

But let's suppose that, on your way to reversing the $z$-axis, you get interrupted midway. Now I have a system which I think has two spins along the $z$-axis, but you are lying on your side and think that my spins are aligned along the $x$-axis. The $x$-axis spin operator is usually $$\sigma_x = \frac\hbar 2\left(\begin{array}{cc}&1\\1\end{array}\right).$$

Where I see my single-particle spins are the eigenstates of $\sigma_z$, $$\ket\up = {1\choose0} \quad\text{and}\quad \ket\dn = {0\choose1},$$ you see those single-particle states as eigenstates of $\sigma_x$, \begin{align} \ket\rt &= \frac1{\sqrt2}{1\choose1} = \frac{\ket\up + \ket\dn}{\sqrt2} \\ \ket\lf &= \frac1{\sqrt2}{1\choose-1} \end{align}

If you are a $z$-axis chauvinist and insist on analyzing my carefully prepared $\ket{\rt\rt}$ state in your $\up\dn$ basis, you'll find this mess:

\begin{align} \ket{\rt\rt} = \ket\rt \otimes \ket\rt &= \frac{\ket\up + \ket\dn}{\sqrt2} \otimes \frac{\ket\up + \ket\dn}{\sqrt2} \\ &= \frac{\ket{\up\up}}2 + \frac{\ket{\up\dn} + \ket{\dn\up}}2 + \frac{\ket{\dn\dn}}2 \end{align}

This state, which has a clearly defined $m=1$ in my coordinate system, does not have a well-defined $m$ in your coordinate system: by turning your head and disagreeing about which way is up, you've introduced both $\ket{\up\up}$ and $\ket{\dn\dn}$ into your model. You've also introduced the symmetric combination $\ket{\up\dn} + \ket{\dn\up}$.

And this is where the symmetry argument comes in. The triplet and singlet states are distinguishable because they have different energies. If you propose that the symmetric combination $\ket{\up\dn} + \ket{\dn\up}$ is the singlet state, then you and I will predict different energies for the system based only on how we have chosen to tilt our heads. Any model that says the energy of a system should depend on how I tilt my head when I look at it is wrong. So the $m=0$ projection of the triplet state must be symmetric, in order to have the same symmetry under exchange as the $m=\pm1$ projections.

torque - Calculating the angle of tipping given force applied

This is my first time here, and I have a question regarding the calculation of the angle that an object would tip at, when a certain force is applied to its center of gravity.

(This question seems similar to this one, but that doesn't really answer my question; because its looking for a threshold height, but I'm looking for the angle of tipping.)

So, let's suppose we have a cylinder with a circular base of radius $r$, whose center of gravity is a perpendicular distance $h$ above its circular base. I am interested in looking at the angle that it would tip at, if we apply a horizontal force through its center of gravity.

The cylinder is placed on a horizontal surface, offering as much friction as is necessary to prevent the object from sliding (the friction acting is equal to the force applied).

This is the free body diagram I came up with.

Here, if the torque caused by the weight force is greater than the torque by the applied force, it would not tip. However, if we apply a bigger force, we can assume that the object tips, and we have the following situation.

Notice that $ER=h$ and $RB=r$.

The angle of tipping must be one that cancels out the torques of the weight and the applied force.

With some geometry, we have the following results.

$\angle EQR = \theta$. Hence $EQ=\dfrac{h}{\sin\theta}$, and $RQ=\dfrac{h\cos\theta}{\sin\theta}$.

$QB=RB-RQ=r-\dfrac{h\cos\theta}{\sin\theta}$.

Also, $\angle PQB=\theta$. Therefore, $PB=QB\sin\theta$, and $QP=QB\cos\theta$.

$PB=\left(r-\dfrac{h\cos\theta}{\sin\theta}\right)\sin\theta=r\sin\theta-h\cos\theta$.

The torque $\tau_W$ by the weight force is $mg\times PB$.

$$ \tau_W=mg(r\sin\theta-h\cos\theta)\qquad\qquad\qquad(1) $$

$QP=\left(r-\dfrac{h\cos\theta}{\sin\theta}\right)\cos\theta=r\cos\theta-\dfrac{h\cos^2\theta}{\sin\theta}$.

$EP=EQ+QP$

$\qquad=\dfrac{h}{\sin\theta}+r\cos\theta-\dfrac{h\cos^2\theta}{\sin\theta}$

$\qquad=r\cos\theta+\dfrac{h}{\sin\theta}(1-\cos^2\theta)=r\cos\theta+h\sin\theta$.

The torque by the applied force is $\tau_F$ is $EP\times F$.

$$ \tau_F=F(r\cos\theta+h\sin\theta)\qquad\qquad\qquad(2). $$

We need $(1)=(2)$ for equilibrium, so we have

\begin{align} mg(r\sin\theta-h\cos\theta)&=F(r\cos\theta+h\sin\theta)\\ \\ \sin\theta(mgr-Fh)&=\cos\theta(Fr+mgh)\\ \\ \tan\theta&=\dfrac{Fr+mgh}{mgr-Fh}\\ \\ \theta&=\arctan\left(\dfrac{Fr+mgh}{mgr-Fh}\right). \end{align}

However, this is clearly wrong, because as $F$ increases, $\theta$ increases, when it must in fact, decrease.

(The following is a variation of $\theta$ against $F$).

What is wrong here?

---

I also approached this problem in a different way, and got a more realistic answer.

We know that the friction force and normal reaction have a resultant, inclined at some angle to the horizontal. For the object to be in equilibrium, it must tip at some angle so that this resultant passes through its center of gravity.

Suppose $\vec{R}$ is the resultant of the friction force and normal reaction.

We have that $f=F$ and $N=mg$, because the object is in equilibrium.

The angle $\angle EBS$ that $R$ makes with the horizontal is $\arctan \dfrac{N}{f}=\arctan \dfrac{mg}{F}$.

$\angle EBR=\arctan \dfrac{ER}{RB}=\arctan\dfrac{h}{r}$.

If we measure the angle $\angle RBS$ as the angle of tipping in this case, we have the following.

\begin{align} \angle RBS&=\angle EBS-\angle EBR\\ \\ &=\arctan\dfrac{mg}{F} -\arctan\dfrac{h}{r}\\ \\ &=\arctan \left(\dfrac{\dfrac{mg}{F}-\dfrac{h}{r}}{1-\dfrac{mgh}{Fr}}\right)\\ \\ &=\arctan\left(\dfrac{mgr-Fh}{Fr-mgh}\right). \end{align}

Now this makes sense, because as $F$ increases, $\angle RBS$ increases, as seen in the plot below.

So which way is actually correct? Why do the two approaches give different results?

(Please do answer in as simple terms as possible, because I'm still a beginner :) )

Thank you for taking your time.

Answer

Approach 1: ... because as F increases, θ increases, when it must in fact, decrease.

The $F$ in your problem is the external force required to keep the object in equilibrium at inclination $\theta$. And it makes sense that $\theta$ decreases as $F$ decreases: If I had two positions/orientations of cylinders (one at $\theta_1$ and another at $\theta_2$, where $\theta_2>\theta_1$) and I asked you to exert a force $F$ to maintain equilibrium, I think you would agree that less force is required to maintain the equilibrium for $\theta_1$ as compared to $\theta_2$.

In fact, $F$ is zero when the centre of gravity is directly above the point of contact with the floor (point B in your figure): This means that no external force is required to maintain equilibrium. This is the critical angle $\theta_c = \tan^{-1}(\frac{h}{r})$ after which $F$ changes sign.

Approach 2: We know that the friction force and normal reaction have a resultant, inclined at some angle to the horizontal. For the object to be in equilibrium, it must tip at some angle so that this resultant passes through its center of gravity.

They do yield consistent results. Check with the identity below. $$\tan^{-1}x-\tan^{-1}y=\tan^{-1}(\frac{x-y}{1+xy})$$ Also, you've mistakenly identified ($90^o-\theta$) as $\theta$ in approach 2: $\angle RBS = 90^o - \angle EQR$.

quantum mechanics - Are dark energy and zero-point energy the same thing?

According to Quantum Mechanics is it possible that the famous "dark energy" and "zero-point energy" are the same thing that drives the accelerated expansion of the universe or maybe related to each other?

pattern - What is a Close-knit Word™?

_{My first contribution to the What is a Word/Phrase™ puzzle series started by JLee over a year ago.}

If a word conforms to a special rule, I call it a Close-Knit Word™.

Use the list of examples below to find out what this rule is.

$$\begin{array}{|c|c|} \hline \textbf{Close-Knit Words™}&\textbf{Not Close-Knit Words™}\\ \hline \text{EDUCATION}&\text{LEARNING}\\ \hline \text{FOGHORN}&\text{KLAXON}\\ \hline \text{ARTISTIC}&\text{IMAGINATIVE}\\ \hline \text{BOOTSTRAPS}&\text{SHOELACES}\\ \hline \text{ELIMINATE}&\text{EXTERMINATE}\\ \hline \text{PONYTAIL}&\text{PLAIT}\\ \hline \text{HIKING}&\text{WALKING}\\ \hline \text{REMUNERATION}&\text{RETURN}\\ \hline \text{FAVOURITE}&\text{BOOKMARK}\\ \hline \text{BACKDROP}&\text{SCENERY}\\ \hline \text{FENCED}&\text{PRIVATE}\\ \hline \text{JAYWALKING}&\text{BURGLARY}\\ \hline \text{INQUIRIES}&\text{INVESTIGATION}\\ \hline \text{DISGUSTING}&\text{REVOLTING}\\ \hline \end{array} $$

Text-only version for analysis:

EDUCATION     LEARNING

FOGHORN       KLAXON
ARTISTIC      IMAGINATIVE
BOOTSTRAPS    SHOELACES
ELIMINATE     EXTERMINATE
PONYTAIL      PLAIT
HIKING        WALKING
REMUNERATION  RETURN
FAVOURITE     BOOKMARK
BACKDROP      SCENERY
FENCED        PRIVATE

JAYWALKING    BURGLARY
INQUIRIES     INVESTIGATION
DISGUSTING    REVOLTING

Answer

I think "close-knit words" are

ones with three consecutive letters of the alphabet in them.

Thus:

CDE EDUCATION
FGH FOGHORN
RST ARTISTIC
RST BOOTSTRAPS
LMN ELIMINATE
NOP PONYTAIL
GHI HIKING
MNO REMUNERATION
TUV FAVOURITE
ABC BACKDROP

DEF FENCED
JKL JAYWALKING
QRS INQUIRIES
STU DISGUSTING

supersymmetry - Are there SUSY Lagrangian terms that are not D-term nor F-term?

I've read that a way to construct supersymmetric invariant lagrangian could be either to integrate a superfield in the whole superspace, i.e. in all anticommuting coordinates (D-term), or in half of them (F-term).

Obviously I call F-term a lagrangian term that can't be written as D-term, because all D-term could be written trivially as integrals in half the superspace.

But now I can't understand why couldn't be supersymmetric invariant lagrangian terms that are not even F-term, but they are however invariant.

EDIT

I thought that the answer could be that given an ordinary lagrangian $F(x)$ term (dependent only on space-time coordinates) I can make it a part of a chiral superfields, as a coefficients of $\theta\theta$ in $y-\theta$ expansion

$$ \Phi(y,\theta) = \phi(y) + \sqrt{2} \theta \psi(y) - \theta\theta F(y)$$ $$ y^\mu = x^\mu + i\theta\sigma^\mu\bar{\theta}$$

choosing arbitrarily the $\phi$ and $\psi$ functions. The question now become: does it work?

gravity - Are there models/simulations of antigravitational antimatter-galaxies?

In the comments to another question's answer, I started wondering:

Assuming antimatter possessed negative gravitational mass^§ (which is not proven impossible to date, though deemed unlikely), basically entire galaxies consisting of antimatter could form. Due to the repulsion of ordinary matter, they would rarely collide with usual ("pro-matter-")galaxies^% and actually drift apart, maybe even partly contributing to inflation. But I don't want to speculate too much, so my questions is

Have entire galaxies consisting of negative-gravitational-mass antimatter been considered in any sound theory, model or simulation?

---

^§ I.e. while such antimatter gravitationally repulsed conventional matter, it would still attract other antimatter.

^% Unless there exists e.g. a (currently unknown) force that attracts antiparticles to their exact counterparts

Answer

Short answer: that depends on your definition of sound theory. For instance, it is possible to find peer-reviewed papers considering such possibilities.

The idea that antimatter can be gravitationally repulsed from ordinary matter is definitely not the most popular one. Nevertheless, some people do try to apply it in astrophysical context. Let us have a look at the works of M. Villata:

Villata, M. "CPT symmetry and antimatter gravity in general relativity." EPL (Europhysics Letters) 94.2 (2011): 20001. arxiv:1103.4937.

This one attempts to establish 'why' we can expect such repulsion. Subsequent works develop the cosmological model resulting from such repulsion. The latest one is:

Villata, M. "On the nature of dark energy: the lattice Universe." Astrophysics and Space Science (2013): 1-9. arxiv:1302.3515.

From the abstract of that one:

... Here we start from the recent theoretical results that come from the extension of general relativity to antimatter, through CPT symmetry. This theory predicts a mutual gravitational repulsion between matter and antimatter. Our basic assumption is that the Universe contains equal amounts of matter and antimatter, with antimatter possibly located in cosmic voids, as discussed in previous works. From this scenario we develop a simple cosmological model, from whose equations we derive the first results. While the existence of the elusive dark energy is completely replaced by gravitational repulsion, the presence of dark matter is not excluded, but not strictly required, as most of the related phenomena can also be ascribed to repulsive-gravity effects. With a matter energy density ranging from ∼5 % (baryonic matter alone, and as much antimatter) to ∼25 % of the so-called critical density, the present age of the Universe varies between about 13 and 15 Gyr. The SN Ia test is successfully passed, with residuals comparable with those of the ΛCDM model in the observed redshift range, but with a clear prediction for fainter SNe at higher z.

For the criticism of this approach from the theoretical grounds see:

Cross, Daniel J. "Response to" CPT symmetry and antimatter gravity in general relativity"." arXiv:1108.5117.

which states that

... This repulsion or anti-gravity is derived by applying the CPT theorem to general relativity. We show that this proposal cannot work for two reasons: 1) it incorrectly predicts the behavior of photons and 2) the CPT transformation itself is not consistently applied.

Another one rebuttal:

Cabbolet, Marcoen JTF. "Comment to a paper of M. Villata on antigravity." Astrophysics and Space Science 337.1 (2012): 5-7. arxiv:1108.4543

For experimental constraints on such antimatter hypothesis see

Ting, Yuan-Sen. "Experimental constraints on anti-gravity and antimatter, in the context of dark energy." arXiv:1310.6089.

Its abstract:

In a paper by Villata (2011), the possibility of a repulsive gravitational interaction between antimatter and ordinary matter was discussed. The author argued that this anti-gravity can be regarded as a prediction of general relativity, under the assumption of CPT symmetry. Stringent experimental constraints have been established against such a suggestion. The measurement of free-fall accelerations of various nuclei by the Eot-Wash group and searches for equivalence principle violation through the gravitational splitting in kaon physics consistently establish null results on any difference between the gravitational behaviour of antimatter and ordinary matter. The original arguments against antigravity were questioned by Nieto & Goldman (1991). In the light of new experiments as well as theoretical developments in the past 20 years, some of Nieto & Goldman's concerns have been addressed. While a precise measurement of the free-fall acceleration of antihydrogen will eventually lay this issue to rest, the purpose of this short letter is to argue that the substitution of dark energy with anti-gravity by antimatter, as suggested by Villata, is highly unlikely.

quantum mechanics - If two photons collide, does the resulting particle have zero velocity?

If two photons traveling in opposite directions along the same line collide, will the resulting particle have a velocity of zero relative to the rest of time space in the instant of the collision?

Answer

If two photons traveling in opposite directions along the same line collide will the resulting particle have a velocity of zero relative to the rest of time space in the instant of the collision?

Photons are quantum mechanical particles. In the microscopic dimensions where quantum mechanical particles interact there are Nature's rules that dominate these dimensions, though they are usually insignificant in macroscopic dimensions.

One of these rules is the Heisenberg uncertainty principle, HUP,: one cannot define the location of a particle and the momentum of a particle with accuracy better than:

where $\hbar =6.62606957(29)×10^{−34}$ Joule second a very small number that is why it is effectively zero in macroscopic dimensions.

Thus two photons even with the same energy will not collide at a point.

Going into the mathematics of it, photon-photon interactions are very very weak, since there is no first-order interaction between two photons, but they have to go through a particle loop. In addition, momentum conservation requires two particles out.

A Feynman diagram (box diagram) for photon-photon scattering, one photon scatters from the transient vacuum charge fluctuations of the other

Feynman diagrams have one to one correspondence with calculable integrals that will give the probability for a given interaction.

A photon carries energy, two photons have an invariant mass. In their center of mass system, depending on the energy available from each, the output can be again two photons, or if there exists energy enough to generate massive particles, there will exist a quantum mechanical probability for the interaction. They are proposing high energy photon colliders, gamma gamma colliders.

quantum electrodynamics - Using photons to explain electrostatic force

I am trying to understand the idea of a force carrier with the following example.

Let's say there are two charges $A$ and $B$ that are a fixed distance from each other. What is causing the force on $B$ by $A$?

Classically charge $A$ has an associated electric field which causes a force on $B$. From the standard model, photons are the force carrier for the electromagnetic force. With this view does it mean that $A$ is constantly emitting photons but in a way that the magnetic component cancels out? If that is the case then doesn't that mean that charge $A$ is constantly losing energy?

Thermodynamics of evaporation

If water is introduced in a container maintained at 20 °C in vacuum conditions, a gaseous phase will appear and the pressure will stabilize at the vapour pressure for the given temperature inside the container. So far so good.

Now imagine the experiment is repeated but instead of vacuum conditions, the water is presurized with nitrogen at 1 atm. According to the phase diagram of water, liquid is the stable form of water in these conditions. Yet it is commonly observed that the water molecules with the highest kinetic energy will escape and form a gaseous phase. The partial pressure of gaseous water will be equal to the saturation pressure at this temperature.

Why do we say that this is a liquid/vapour equilibrium in this case, since the liquid phase is not at pressure required for this equilibrium to appear?

Edit: I just realized that my question seems to be answered by Raoult's law.

Answer

Single species

At the surface of a liquid, molecules are constantly being ejected from the liquid surface and gas molecules rejoining it. When those two rates are equal there's no net change and the system is in equilibrium.

The rate at which liquid water will eject a molecule is roughly independent of pressure and is based solely on temperature. The temperature determines how much kinetic energy the molecules have while the pressure just determines how tightly packed they are and they're already very tightly packed so the density doesn't change much. There are always water molecules right on the surface because of this high density.

The rate at which gaseous water molecules rejoin the water depends on how often one hits the surface which depends on the density. By the ideal gas law, the density is related to pressure and temperature, so if we've already set the temperature, it only depends on pressure. Thus the higher the pressure, the higher the density, the more gaseous water molecules there are to hit the surface and join it.

Locally, the density and pressure of the gaseous water molecules will increase until the rate of rejoining is equal to the rate of evaporation.

Multi gas species

Other gasses can feel free to bombard the surface of the water, compressing it a little bit, and increasing the pressure significantly. However, this doesn't increase the rate at which water molecules escape, so the vapor will still equilibrate at the same partial pressure.

Multi liquid species

This is where Raoult's law comes in. On the liquid surface there are now two species of molecules but the rate at which molecules are ejected is still the same. Now those ejections must be split between species, and according to Raoult's law, it's based on the concentration. If 30% of the liquid's molecules are water then 30% of the ejections will be water, so the rate of ejection is reduced to 30%. So in equilibrium the rate of rejoining must also be 30% so there must be 30% as many collisions thus 30% the density and thus 30% the vapor pressure.

Non Equilibrium

Examining the gas directly above the surface would show a partial pressure of water that approaches the vapor pressure of water for that temperature very quickly. These water molecules would then diffuse through the nitrogen which would slowly lower the local partial pressure if it was not being replenished by additional evaporation from the surface. This evaporation requires energy to overcome the latent heat and as such it will lower the local temperature of the water. Water is a much better heat conductor than air so this heat is drawn from the water.

In some evaporation processes the slow diffusion of the gas dominates, in others, the temperature drops until the diffusion process dominates. This is why liquid nitrogen gets and stays so cold, and why wet bulb temperatures are lower than dry bulb temperatures.

If you'd like to learn more about the steady state solution I would recommend reading about relative/absolute humidity, condensation etc.

If you'd like to learn more about the rate limiting effect look into gas diffusion.

If you'd like to learn more about the evaporation condensation process learning about statistical thermodynamics would help.

thermodynamics - Heat of vaporization of water - dependence on relative humidity?

Does the heat of vaporization of water depend strongly on the relative humidity of the gas into which it evaporates?

Some context: If we want to calculate the dew point of water, we find the temperature at which the partial pressure of the water lies on the liquid/vapor boundary of the water phase diagram. This is why water can evaporate from our bodies even though we do not heat it to anywhere near its boiling point.

The heat of vaporization should be pressure dependent (in addition to temperature dependent). Yet, when specifying the heat of vaporization, most references only specify the total ambient temperature, usually 1 atmosphere. Why is the total pressure used in this case instead of the partial pressure? And if the partial pressure is what matters after all, then wouldn't relative humidity be important when calculating heats of vaporization? Of course, relative humidity governs the rate and the total amount of evaporation, which is why we can't cool ourselves by sweating in humid weather, but that's not what my question is about.

Answer

Heat of vaporization is related to enthalpy change, while dew point is related to free energy change, i.e. enthalpy plus entropy. That's why they are very different concerning relative humidity.

The enthalpy of a gas is more-or-less independent of pressure or partial pressure, because gas molecules don't really interact with each other. At insanely-high pressures there would be some effect on enthalpy of course, but the effect at everyday pressures is very low. Pressure mainly affects a gas via entropy not enthalpy.

The enthalpy of a liquid is somewhat dependent on total pressure: A high pressure will push the molecules closer together and therefore change their interaction energies. But obviously the enthalpy of the liquid doesn't depend on what the gas partial pressures are, it can only depend on the liquid's own total internal pressure.

So the answer is: Heat of vaporization, being related to enthalpy not entropy, has essentially no dependence on relative humidity. (given a constant total air pressure)

-- UPDATE --

Oops, whenever I wrote "enthalpy" I should have said "enthalpy per molecule" or "enthalpy per mole" ["molar enthalpy"]. You can check for yourself that the enthalpy per molecule of an ideal gas is independent of pressure or partial pressure. For a real-world gas, it's approximately independent. The "per mole" quantities are what matter for dew point etc.