astrophysics - neutrinos by formation of "neutron pairs"

Here :

http://www.newscientist.com/article/dn20084-neutron-star-seen-forming-exotic-new-state-of-matter.html

are news on superfluidity in a neutron star. The necessary bosons they say are pairs of neutrons. So far, so good.

But then they postulate the production of neutrinos in the formation of those neutron pairs.

When neutrons pair up to form a superfluid they release neutrinos which should pass easily through the star,...

What kind of reaction is that? n + n => neutrino + "n-n" + "XYZ" ?

Answer

The phenomenon was first predicted in this paper:
"Neutrino pair emission from finite-temperature neutron superfluid and the cooling of young neutron stars"

Flowers E. G., Ruderman M., Sutherland P. G., 1976, ApJ, 205,541
PDF here

The process they are describing is actually $n^*n^*\to\nu\overline\nu$, where $n^*$ is a quasiparticle excitation above the superfluid condensate. It is the annihilation of those quasipartilces that produces those neutrinos.

Of course, the "real" neutrons do not annihilate, but the description of the process in terms of the "real" neutrons is way too complex, because there is a whole medium made of constantly interacting neutrons.

Which is once again shows the usefulness of a concept of a quasiparticle.

mass - Does an Increase of Force affect the Speed of an object if the acceleration stays constant?

If I had a mass of $100\:\rm{kg}$ accelerating due to gravity, using $F=ma$:

$F = 100\:\rm{kg} \times 9.8\:\rm{m/s^2}$

$F = 980 \:\rm N$...

If I increased the mass to 200kg, the force would be 1960 N:

$F = 200\:\rm{kg} \times 9.8\:\rm{m/s^2}$

$F = 1960 \:\rm{N}$

Now, finally getting to my question: Does this increase in force (which is supposed to be a push/pull) mean that the object would fall faster when it weighs more?

Answer

No, the heavier object does not fall faster. Instead, they heavy and light object fall at the same acceleration (and hence the same speed if they are both simply dropped). This is an example of the equivalence principle.

The more massive object has more gravitational force on it, but it also has more inertia. Specifically, because the object is twice as massive, it has twice the inertial mass.

The force on it is doubled, so the acceleration stays the same.

If we look at

$$F = ma$$

we see that when $F$ and $m$ are both multiplied by 2, $a$ stays the same.

Check these questions for more:

Free falling of object with no air resistance

Why is heavier object more reluctant to get falling down?

Projectile motion without air resistance

general relativity - What is the difference between translation and rotation?

What is the difference between translation and rotation ?

If this were a mathematics site, the question would be at best naive.

But this is physics site, and the question must be interpreted as a question about physical theory, that is about hypotheses that can be tested, subjected to experiments and possibly disproved by experiments.

Restatement of the question
After about 3 days, 5 answers, 160 views and some more comments,

taking these contributions into account (hence the length).

First I thank all users who commented or tried to answer my ill stated question, and I apologize for not doing better. Hopefully, they did help me improve my understanding and the statement of my question. You can look at the awkward prior formulations of the question which better explain the existing answers.

I am trying to understand whether and how translation differs from rotation, and whether or why it is a necessary physical concept or possibly only a mathematical convenience.

There are two sides to the issue I am raising, one regarding space(time) symmetries, and one regarding motion. From what (very) little I understand of Noether's theorem, these cannot be unrelated as laws governing motion have to conserve charges that are derived from the space(time) symmetries. This may have been one source of my initial confusion.

One of my point is that if there are situations when rotations is not distinguishable from translation: infinitesimal angles of rotation as suggested by user namehere. Then relevant phenomena can be analyzed either as rotations or as translation, with proper accomodation, particularly to account for the existence of a radius when rotation is concerned, which changes dimensionality and the mathematical apparatus.

Of course this requires "care" when considering phenomena involving the center of rotation or phenomena indefinitely distant.

One example is torque, moment of inertia, angular momentum, vs force, mass and momentum. The possible undistinguishability of translation and rotation would seem to indicate that they are really two guises for the same set of phenomena. They relate to two distinct symmetries, but is that enough to assert that they are fundamentally different ? This is precisely what is bothering me in the last comments of user namehere attached to his answer and motivated my question initially. Actually, it was someone telling me that "angular momentum is not linear momentum going round in a circle" that started me on this issue, as I was not convinced. It may have other manifestations, but it is also that.

I am aware that the mathematical expressions, including dimensionality, are significantly different for translational and rotational concepts, and somewhat more complex for rotational concepts, as remarked by user namehere. But rotational concepts have to account for the existence of a center and a radius which may be directly involved in the phenomena being considered: this is typically the case for the moment of inertia which has to account for a body rotating on itself.

If we consider a translational phenomenon about force, mass and momentum, occuring in a plane. We can analyse it indifferently as translational, or as rotational with respect to a center of rotation sufficiently distant on a line orthogonal to the plane, so that all radiuses may be considered vectorially equal up to whatever precision you wish. Since the radiuses may be considered equal, they can be factored out of the rotational formulae to get the translational ones. That is, the rotational mathematics can be approximated arbitrarily well by its translational counterpart. This should accredit the hypothesis that it is the same phenomena being accounted for in both cases.

I am not trying to assert that rotating frames should be inertial frames. I am only asking to what extent physicists can see a difference, and, possibly (see below), whether inertial frames actually exist. When do rotational phenomena differ in substance from translational ones ? Is there some essential phenomenon that is explained by one and not by the other ? And conversely ?

Mathematics is only a scaffolding for understanding problems. They are not understanding by themselves. Mathematical differences in expression do not necessarily imply a difference in physical essence.

Then one could also question whether translation is (or has to be) a meaningful physical concept. This can be taken from the point of view of space(time) symmetries or from the point of view of motion. Why should it be needed as a physical concept, or can it (should it) be simply viewed as a mathematical convenience ? Does it have meaning independently of the shape (curvature) of space ? (I guess relativists have answers to that).

Take the 2D example of the surface of a sphere. What is translation in that space ? The usual answer is "displacement along a great circle". This works for a point, but moderately well for a 2D solid, as only a line in that solid will be able to move on a great circle. I guess we can ignore that, as any solid will be "infinitesimal" with respect to the kind of curvature radius to be considered. However there is the other problem that, very simply, every translation is a rotation, and in two different ways, with a very large but finite radius. But it probably does not matter for scale reasons.

Now, there is also the possibility that I completely missed or misunderstood an essential point. Which would it be ?

rotational dynamics - How different can the directions of angular momentum and angular velocity be?

A recent question uncovered a fact that can be very surprising to those not previously aware of it: the angular velocity $\vec\omega$ and the angular momentum $\vec L$ of a rotating body are vectors that need not point in the same direction.

In essence, this is because the two vectors are related to each other by the body's moment of inertia $I$ via $$\vec L=I\vec\omega,$$ but the moment of inertia $I$ turns out to be a matrix (often called a 'rank-2 tensor' just to break out the fancy language) instead of just a scalar. As such it can (and, generically, it will) change the direction of $\vec\omega$ to produce an angular momentum in some non-parallel direction.

One of the questions that this leaves open, however, is just how different these two directions are allowed to be. Are they allowed to be orthogonal? Are they allowed to be anti-parallel? If not, why not?

Answer

As it happens, there are indeed limits to how different the directions of $\vec L$ and $\vec \omega$ can be. This is because the moment of inertia tensor is something called positive semidefinite, which requires the inner product of any vector $\vec\omega$ with its transformation under the tensor, $I\vec\omega$ to be non-negative: $$ \vec\omega \cdot I\vec\omega \geq 0. $$ In particular, this means that the angle between $\vec L$ and $\vec \omega$ cannot be greater than 90°.

---

The moment of inertia, as a matrix, is normally defined with the components $$ I_{ij} = \sum_k \left(r_{(k)}^2\delta_{ij} -x_i^{(k)}x_j^{(k)}\right)m_k \tag 1 $$ for a cloud of particles with masses $m_k$, or as the integral $$ I_{ij} = \int\left(r^2\delta_{ij} -x_ix_j\right) \rho(\mathbf r)\mathrm d^3\mathbf r \tag 2 $$ for a continuous mass distribution. Normally you calculate the components about the principal axes of the moment of inertia, which is a frame in which the off-diagonal components like $$ I_{xy} = -\int \,xy\,\rho(\mathbf r)\mathrm d^3\mathbf r $$ vanish, and you're only left with the more usual diagonal components of the form $$ I_{xx} = \int \left(y^2+z^2\right)\rho(\mathbf r)\mathrm d^3\mathbf r. $$ In the general case, however, one needs to stick with the full matrix expressions $(1)$ and $(2)$, and it is not immediately obvious how one can prove from those expressions that the moment of inertia is positive semidefinite, so it is worth exploring that in a bit of detail.

(In a frame aligned with the principal axes this is easy, since in that frame you have $$ \vec\omega \cdot I\omega = \sum_j I_{jj}\omega_j^2 \geq 0 $$ as all the $I_{jj}$ are non-negative. However, the existence of such a frame is not a trivial fact - it essentially requires us to know that the matrix $I_{ij}$ is diagonalizable and to find such a diagonalization. As such, it is better to work in the general case.)

The main point about the positive semidefiniteness of $I$ is that it encodes a much simpler relationship than its components lead you to believe. To see this relationship, you start with the fundamental definition of the angular momentum of a single particle, $$ \vec L= \vec r \times \vec p = \vec r\times m\vec v, \tag 3 $$ and then you use the fundamental definition of the angular velocity vector - that any particle rotating about the origin with (vectorial) angular velocity $\vec \omega$ will have velocity $$ \vec v = \vec \omega \times\vec r, \tag 4 $$ and you put this into the expression for the angular momentum: $$ \vec L = \vec r\times m(\vec \omega\times\vec r). \tag 5 $$ This looks complicated, but the key thing to realize is that it is linear in $\vec\omega$: if you put in a linear combination like $a_1\vec\omega_1 + a_2\vec\omega_2$ in for $\vec\omega$, you will get out $a_1\vec L_1 + a_2\vec L_2$. The equation above is a linear relationship between two vectors, and all such relationships can be expressed in matrix form: $$ \vec L = I \vec \omega. $$ The moment of inertia tensor is exactly that matrix.

However, the matrix form of the relationship is not necessarily the most useful way to phrase it. As an example, let me show the proof that $I$ is positive semidefinite, directly from the relationship $(5)$ above. I want to show that $\vec\omega\cdot\vec L\geq 0$, so I start by calculating that inner product: \begin{align} \vec\omega\cdot\vec L & = \vec\omega\cdot\left(\vec r\times m\,\vec v \right) \\ & = m\vec r\cdot\left(\vec v\times \vec\omega \right) , \end{align} where I've used the cyclical property of the scalar triple product, because the cross product $\vec v\times \vec\omega$ is easy to calculate since $\vec v$ and $\vec\omega$ are orthogonal by construction. Putting in the definition $(4)$ for the velocity, we get a vector triple product, which is again easy to resolve: \begin{align} \vec\omega\cdot\vec L & = m\vec r\cdot\left((\vec \omega \times\vec r)\times \vec\omega \right) \\ & = m\vec r\cdot\left(\omega^2\vec r - (\vec\omega\cdot\vec r)\vec \omega \right) \\ & = m\left(\omega^2 r^2 - (\vec\omega\cdot\vec r)^2 \right) . \end{align} Here we're essentially done, because we know that $|\vec\omega\cdot\vec r|$ must be no larger than $r\omega$, so we get the desired $\vec\omega\cdot I\vec \omega = \vec\omega\cdot\vec L \geq 0$. Finally, if we have a cloud of particles or a continuous mass distribution, we simply sum or integrate over this inequality.

---

OK, so the above is a (long-winded) explanation of why this $I$ matrix thing obeys this positive-semidefinite property. What does this mean geometrically? Well, expanding the inner product into an angle form, it gives us $$ \vec\omega\cdot\vec L = \omega L\cos(\theta)\geq 0, $$ where $\theta$ is the angle between the angular momentum and the angular velocity; this inequality translates into $\theta \leq \pi/2=90^\circ$.

This bound is pretty much the best you can do, and it is easy to find examples where $\vec \omega$ and $\vec L$ are arbitrarily close to orthogonal. The simplest is a long, thin rod of length $L$ and radius $w$ that's rotating rapidly about an axis that is almost (but not quite) aligned with the rod:

Putting the rod along the $x$ axis, you get that the moment of inertia and angular velocity are given by $$ I= \begin{pmatrix} \frac12 Mw^2 & 0 & 0 \\ 0 & \frac13 ML^2 & 0\\ 0 & 0 & \frac13 ML^2 \end{pmatrix} \quad\text{and}\quad \vec \omega= \begin{pmatrix} \Omega \\ \delta \\ 0 \end{pmatrix}, $$ respectively, and that gives you $$ \vec L= \begin{pmatrix} \frac12 Mw^2\Omega \\ \frac13 ML^2\delta \\ 0 \end{pmatrix} $$ for the angular momentum, $$ \vec\omega\cdot \vec L = \frac12 Mw^2\Omega^2 + \frac13 ML^2\delta^2 $$ for the inner product, and $$ \cos(\theta) = \frac{\vec\omega\cdot \vec L}{L\omega} = \frac{ \frac12 w^2\Omega^2 + \frac13 L^2\delta^2 }{ \sqrt{\frac14 w^4\Omega^2 +\frac19 L^4\delta^2} \sqrt{\Omega^2+\delta^2} } \approx \frac{ 3 w^2/L^2 + 2 \delta^2/\Omega^2 }{ \sqrt{9 w^4/L^4 +4 \delta^2/\Omega^2} } $$ for the angle. Fixing the rod parameters at $w\ll L$, and then choosing an angular velocity with $\Omega\gg \delta$ and such that $$ \frac wL \ll \frac \delta\Omega $$ (i.e. the rod is much thinner than the spin axis is close to the rod axis), the cosine approximates to $$ \cos(\theta) \approx \frac{\delta}{\Omega} +\frac{3\,w^2/L^2}{2\,\delta/\Omega} , $$ which can be made arbitrarily small, with the angle $$ \theta \approx \frac \pi2 -\left(\frac{\delta}{\Omega} + \frac{3\,w^2/L^2}{2\,\delta/\Omega}\right) $$ arbitrarily close to 90°.

More geometrically, spinning a thin rod about its axis gives an angular momentum component along this axis which is very small, and any misalignment will give angular momentum describing rotation about the axis of misalignment which, if the rod is thin and long enough, can dwarf the angular momentum from the spinning itself.

---

Finally, it's important to note that the angular momentum and the angular velocity can never be fully orthogonal. This can be seen by expressing the orthogonality relation $\vec\omega \cdot \vec L=0$ in the principal frame of the body, where it takes the form $$ \vec\omega \cdot \vec L = I_{xx}\omega_x^2 + I_{yy}\omega_y^2 + I_{zz}\omega_z^2 =0, $$ i.e. the sum of three non-negative terms, all of which must be zero for the sum to vanish. This is possible by having e.g. $I_{xx}=0$ and $\omega_y=\omega_z=0$, but in that case (and all such cases) you will obtain $\vec L=0$, which doesn't really count as "orthogonal" to $\vec\omega$.

special relativity - Why light moves sideways?

Greatings! I'm trying to understand special relativity and have one question bugging me.

In almost every book or video about the subject there is a thought experiment with moving light clock. I hope I need not elaborate on the sutup and the outcome of the experiment.

So the question is this: When a light source on the MOVING light clock emits a photon of light the path of the photon is triangular.

How this can be?

I thought since light speed is constant in every reference frame the movment of the "emmiter" can't affect the movment of the light so light should shoot right up (from the point of stationary observer) and thus move backward and up from the point of moving observer.

So you see my logic is: light speed is invariant => you can't make it move UP & FORWARD only UP. i.e. you can't change it's direction once it's emited in UP direction. Why the hell light moves sideways then?

And I'm puzzeled. There is flaw in the logic but I don't see it.

Thanks for help.

Answer

There are actually more than one ways to look at this problem. I am going to explain why the light must not go straight upward in the ground frame if the light clock is moving horizontally wrt ground frame.

The fundamental postulates of relativity is:

Laws of physics are invariant in all inertial frames.

Suppose the in the ground frame, the mid-point of the lower plate of the light clock is at (0,0) at time $t=0$. The upper plate of the light clock is at ($l$,0) at the same time. The clock is moving with constant speed $v$ in wrt the ground frame. A photon leaves the lower plate at $t=0$ (in ground's clock) in such a way that it is going straight upwards in the light clock frame. Now suppose there is a switch of some bomb attached with the midpoint of the upper-plate which if hit by a photon blows the light clock. Since the photon is going straight upwards in the light clock frame it will surely hit the upper plate and blow the light clock itself. Now if the photon moves straight upward even in the ground frame then it can never hit the upper plate as the light clock is moving and the upper plate must have travelled forward and thus the light clock actually survives in the ground frame. Which just can't happen because otherwise it indicates that laws of physics are different in the light clock frame and in the ground frame which just can't be the case in accordance with the first postulate of special relativity. Further, in order to be consistent with the first postulate of relativity, light has to go precisely in the direction so that it can hit the upper plate. So rather than contradicting with the relativity, the different direction of light is a necessity for the principle of relativity to hold.

Another thing: It is not dictated by the maxwell equations that the direction of light must be the same for all the inertial frame. If it would be the case it would actually break the directional symmetry of space. Because it would define a particular direction with a preference - the direction in which the light goes. Maxwell's equations just dictate that speed of light is same in all frames where Maxwell's equations are valid. i.e. at least all inertial frames.

Why is a single-corner twist not a valid position on a Rubik's cube?

For those of you familiar with the solution to the Rubik's cube, you will know that a single-corner twist is not a valid position. It can neither be attained by scrambling, nor can it be solved if it occurs.

I understand that there's some fundamental group theory at work here; the corners on a Rubik's cube can only be rotated in groups of three. This logically implies that one can't have a single (or two, in the same direction) corners rotated.

Why, mathematically, is this the case?

Answer

Okay, I'll give a group-theoretic answer, since that was requested, which goes most of the way to a proof. I'm assuming you mean a 3x3x3 Rubik's cube. For larger cubes, the proof is just a little bit less convenient, because these cubes don't form groups (or more precisely, the action of the group of all moves on the cube on the states of the cube is not free). Basically everything still holds though for all cubes larger than 3x3x3 when talking specifically about the corner pieces, except that if the number of layers is even nothing fixes the orientation. Edge pieces and centers can get a bit more complicated though. Also, throughout, I'll use the notation $\mathbb Z_n = \mathbb Z / n \mathbb Z$ for the cyclic group of order n.

There are a number of groups one could be referring to when using the term "cube group". Here, by the "full cube group" $G$, I mean the set of all states of the cube that can be reached by disassembling it and reassembling it, keeping all the center pieces fixed in position. In order to define the product, we'll identify it as a subset of the permutation group on the stickers. Note that each corner/edge sticker is distinguishable from all 47 other stickers. This is because on the 3x3x3 cube, each cubie has a unique collection of sticker colors. Hence, each state $\Sigma$ corresponds uniquely to a permutation $\sigma_\Sigma \in S_{48}$, where $\sigma_\Sigma(s_i)$ is the sticker in the position that sticker $s_i$ would be in the state $\Sigma_0$ corresponding to the solved cube. The product structure on the full cube group is induced by this map. As a trivial consequence of this definition, the identity of the cube group is the solved cube position $\Sigma_0$. (If this multiplication looks a bit backwards to you, it's because I want my permutation group to act on the right, which is primarily since the standard notation of cube moves (e.g. R U R' U R U2 representing the Sune) is read left-to-right.)

There are $8! \; 3^8$ ways of putting in the 8 corner pieces (because each can be rotated any of 3 ways, and they can be permuted), and independently $12! \; 2^{12}$ ways of putting in the 12 edge pieces. Hence, the full cube group has order $8! \; 3^8 \; 12! \; 2^{12}$.

This group is too big for cubing purposes. When cubing, there are 6 basic moves you can make, specifically, U, D, L, R, F and B (as well as their inverses, which are typically denoted by primes e.g. U'). The cube group $H$ is the subgroup of the full cube group generated by these 6 permutations, i.e. which can be written as words in these 6 letters and their inverses. These represent the states of the cube which are reachable with a finite sequence of turns. On this subgroup, the product operation is especially easy to understand. Let $a_1 \cdots a_n$ be any word in the generators which takes you from $\Sigma_0$ to $\Sigma_1$, and $b_1 \cdots b_m$ any word taking you from $\Sigma_0$ to $\Sigma_2$. Their product $\Sigma_1 \Sigma_2$ is the state you reach from the sequence of moves $a_1 \cdots a_n b_1 \cdots b_m$. Now that we have this identification, I'll be a bit cavalier in equating moves with states, but it should always be clear in context what is meant.

We want to prove that a particular state $\Sigma'$, that with a single corner turned by $120^\circ$, is not reachable, which is to say that it is not in the cube group (though it is in the full cube group). To do this, we'll define a homomorphism $\varphi: G \rightarrow \mathbb Z_3$. We'll then show that $\varphi(a_i) = 0$ for any one of the 6 generators $a_i$. Since any state $\Sigma$ in the cube group can be written as a word in the generators, that means that $\varphi(\Sigma) = 0$. That is to say that $\varphi$ is an invariant under the cube group. However, we can compute $\varphi(\Sigma') \ne 0$, which shows that $\Sigma' \not \in H$.

---

Aside: I'll just point out that this sort of approach of defining invariants that are preserved under a set of transformations is very common in mathematics; for instance, point-set topology could be characterized broadly as the study of invariants of topological spaces under homeomorphisms. Here, we're dealing with an algebraic invariant. To be a bit fancier, we're trying to define a surjective homomorphism $\varphi$ so that the composition $H \hookrightarrow G \overset{\varphi}{\twoheadrightarrow} \mathbb Z_3$ is the trivial homomorphism. Keep in mind though that this isn't an exact sequence-I haven't required that the kernel of $\varphi$ is exactly $H$. It isn't a chain complex either, at least not in standard parlance, since $H$ and $G$ are nonabelian groups.

---

To define the invariant, label each of the corner cubie stickers in the solved state with a number 0, 1, or 2, according to this diagram of a net of the cube:

Define $\varphi(\Sigma)$ to be the total number on the stickers on the up and down faces mod 3 in the state $\Sigma$. This of course gives a (surjective) map from the full cube group to the group $\mathbb Z/3\mathbb Z$, which is not clearly a homomorphism. I won't give a full proof that it is, but I'll take you most of the way there.

You may wonder where this particular choice comes from. In fact, the only reason I chose it is because it makes it very easy to see that $\varphi(a_i) = 0$ for each generator $a_i$. For the up and down moves, these don't even change any of the numbers at all. For the other 4 moves, the numbers do change, but they change the same way for all 4. Hence, you only need to check one position, and it's easy to check $\varphi(F) = 0$. For the proof I give below, it's a bit more convenient to check that the inverses of the generators also preserve $\varphi$, which only requires checking one more state $\varphi(F')=0$.

This ambiguity of choice of edge-labels for the solved state turns out to be important for proving that $\varphi$ is a homomorphism. Specifically, you'll want to prove the following lemma: Any choice of numbering will define the same function $\varphi$ provided is satisfies two criteria:

• Each cubie contains each of the numbers 0, 1, 2, exactly once, and they are in that cyclic order when read counter-clockwise around the vertex.


• The solved state still satisfies $\varphi(\Sigma_0)=0$ for this choice of numbering.

This isn't too hard to prove, but I'll skip it here. Once you can prove that, the homomorphism property of $\varphi$ comes basically for free. For simplicity, I'll just prove that the restriction $\varphi |_H$ is a homomorphism (specifically the trivial homomorphism), which is all we need. The proof is by induction. The base case is the identity. Suppose we can show that for every word of length $n$ in the generators, $\varphi(a_1 \cdots a_n) = \varphi(a_1) \cdots \varphi(a_n) = 0$. Consider an arbitrary word $a_1 \cdots a_n a_{n+1}$. Now note that for $\Sigma = a_1 \cdots a_n$, we could define a different numbering on the cube, specifically, the one we would get by first doing $\Sigma$, then numbering the cube as in the above figure (and then of course returning it to the initial solved state). By inductive hypothesis, $\varphi(\Sigma^{-1}) = \varphi (a_n^{-1} \cdots a_1^{-1}) = 0$. But $\Sigma^{-1}$ has the same numbering in the original scheme as $\Sigma_0$ in the new one, because to get to $\Sigma_0$ from the above we just undo the moves we did before by applying $\Sigma^{-1}$. Thus, this alternate numbering pattern fulfills both requirements of the lemma (the second one is clear), so it gives a different method to compute the same function $\varphi$.

However, then the state $a_1 \cdots a_n a_{n+1} = \Sigma a_{n+1}$, by construction in the new edge-numbering, will have the same numbering as if you started with the original edge-numbering in the figure above and did the single turn $a_{n+1}$, which we know will still give $\varphi(a_1 \cdots a_{n+1}) = 0$. Hence, for any reachable state $\Sigma \in H$, we'll have $\varphi(\Sigma) = 0$, so $\varphi |_{H} = 0$.

That's all you need, but I'll just say that proving that $\varphi$ is a homomorphism on $G$ isn't too much harder. You have to choose a different generating set though, since the generators for the cube group don't generate the full cube group. The generators you want to choose are basically "swap corners i and i+1" and "twist corner 1 by $120 ^\circ$". Those generate all possible configurations of the corner pieces (you should prove this). Of course, you also need to include things which generate the edge permutations, though these aren't important at all since we only care about the corners. The proof will proceed much the same as above, with a bit of special consideration taken for the "twist corner 1" generator.

Now, once you know $\varphi(\Sigma) = 0$ whenever $\Sigma \in H$, you just have to compute $\varphi(\Sigma')$ for the state with a single corner turned. This is easily seen to be either 1 or 2 depending on which direction you twist the single corner. So $\Sigma' \not \in H$, and we're done with the proof. As you can see, it's not really that hard, though it is very detailed and long.

---

I'll take a moment to mention some nice group theoretical stuff that we're actually not far off from proving. Note that as I said before, the sequence of homomorphisms $H \hookrightarrow G \twoheadrightarrow \mathbb Z_3$ is not exact. However, we can basically completely understand the group structure of the cube group. It is a normal subgroup of the full cube group, and the quotient is given by $H \hookrightarrow G \twoheadrightarrow \mathbb Z_3 \times \mathbb Z_2 \times \mathbb Z_2$ (which is exact). The latter factors of $\mathbb Z_2$ are also describable simply in terms of cube positions. The first represents edge orientation; a state with a singe edge flipped is not solvable. The second represents odd permutations. A state with 2 corner pieces (or 2 edge pieces, but not both) swapped, but with all other pieces in the right position, is not solvable. It's impossible to permute a single pair of pieces, though it is possible to permute 2 pairs simultaneously. For any cube state, when solving it, you'll either eventually come to a fully solved cube, or you'll reach something which has a single corner twisted and/or a single edge flipped and/or a single pair of either corners or edges swapped. In any of these latter cases, the cube is not solvable and will need to be disassembled to return it to the solved state.

From the above, the cube group $H$ is an index 12 subgroup of the full cube group $G$, which means that 1/12 of the positions will be solvable. The cube group $H$ thus has order $8! \; 3^7 \; 12! \; 2^{10}$. Furthermore, if you know a bit of group theory, it's pretty easy to see that the full cube group is isomorphic to $\mathbb Z_3 \wr S_8 \times \mathbb Z_2 \wr S_{12}$. The edge and corner orientations are just the diagonal homomorphisms of their respective wreath products (which incidentally is the easiest way to prove the above lemma by showing that this gives you the same homomorphism), while the permutation parity is a bit more complicated to understand. Once you've understood these fully, you can compute that the cube group is isomorphic to $(\mathbb Z_3^7 \times \mathbb Z_2^{11}) \rtimes ((A_8 \times A_{12}) \rtimes \mathbb Z_2)$.

Finally, since it's relevant to my interests, I'll briefly note that much of what I said here still works for higher dimensional cubes. See, for instance, this MO question.

lagrangian formalism - Derrick’s theorem

Consider a theory in $D$ spatial dimensions involving one or more scalar fields $\phi_a$, with a Lagrangian density of the form $$L= \frac{1}{2} G_{ab}(\phi) \partial_\mu \phi_a \partial^\mu \phi_b- V(\phi)$$ where the eigenvalues of $G$ are all positive definite for any value of $\phi$, and $V = 0$ at its minima. Any finite energy static solution of the field equations is a stationary point of the potential energy $$E = I_K + I_V ,$$ where $$I_K[\phi]= \frac{1}{2} \int d^Dx G_{ab}(\phi) \partial_j \phi_a \partial_j\phi_b$$ and $$I_V = \int d^Dx V(\phi)$$ are both positive. Since the solution is a stationary point among all configurations, it must, a fortiori, also be a stationary point among any subset of these configurations to which it belongs. Therefore, given a solution $\phi(x)$, consider the one-parameter family of configurations, $$f_\lambda(x)= \bar{\phi}(\lambda x)$$ that are obtained from the solution by rescaling lengths. The potential energy of these configurations is given by \begin{align} E_\lambda &= I_K(f_\lambda) + I_V(f_\lambda)\\ &=\lambda^{2-D} I_K[\bar\phi]+\lambda^{-D} I_V[\bar\phi]\tag{1} \end{align}

$\lambda = 1$ must be a stationary point of $E(\lambda)$, which implies that $$0=(D-2) I_K[\bar\phi]+D I_V[\bar\phi] \tag{2}$$

My problem is, how they got the equation(2) from (1)?

Answer

I didn't go through all of your equations. However, if you take (1), differentiate it w.r.t $\lambda$ and set $\lambda = 1$, then since $\lambda=1$ is the stationary point $E'(\lambda)|_{\lambda=1} = 0$. This is equation (2)

soft question - Can we tell when an established theory is wrong?

I was reading the following answer from this question:

In physics, you cannot ask / answer why without ambiguity. Now, we observe that the speed of light is finite and that it seems to be the highest speed for the energy.

Effective theories have been built around this limitation and they are consistent since they depend of measuring devices which are based on technology / sciences that all have c built in. In modern sciences, one doesn't care of what is happenning, but of what the devices measure.

I think this raises a good point, can we get a false positive of a theory being right just because the instruments doing the measuring have that theory built in? For example, if it were the case that there is a particle which travels faster than light, could we even tell that's the case since some of our methods of measuring distance involve relativity (which assumes the speed of light as an upper bound)?

Answer

can we get a false positive of a theory being right just because the instruments doing the measuring have that theory built in?

This sounds dangerously close to a contradiction-in-terms, so let me carefully read you as saying that the instruments doing the measuring "are interpreted according to that theory," possibly by calculations that sit between their transducers and the screen which we read numbers off of.

And the answer then is, "sort of: yes that's a real thing that can happen, yes it can throw you for a loop (especially if it means that you're not measuring what you think are measuring), but no, the theory is not really in trouble in those cases."

For example, when the OPERA neutrino debacle was unfolding, there were lots of guesses about what was going wrong, and many of them were guesses that "you're expecting the GPS satellite system works according to this theory; maybe it works according to that one instead." It turned out that the theory was correct but the timing-signal cables were faulty; but that was a very real possibility that a lot of people took very seriously.

Why, however, would the theory not be in trouble? This is due to a pervasive fiction that we teach kids in high school, because we've got a bad textbook selection process that creates echo-chambers and no real strong impetus for good scientists to take "shifts" teaching high-school science or reviewing textbooks -- and even then, not much communication between the philosophers of science and the scientists. So I forgive you if you've never heard any of this!

The fairy tale is that scientists first observe the facts of the world, look for patterns in those facts, propose those patterns as a "hypothesis", collect more facts, and see if those new facts confirm or deny that hypothesis. If a hypothesis has no denying-facts, then we call it a "scientific theory". One reason that you should find this suspect is that it's really hard to track down who we should ascribe this to: it comes from some time after Francis Bacon and Isaac Newton, but I haven't been able to really track down whose model it really was. In any case I regard this as a "fairy tale" because in my professional career in science I have never seen someone say "okay, we have to stop what we're doing and start observing facts and see what some patterns are, so that we can form a hypothesis." To the extent that this stuff happens it's totally subconscious; it's not a "method" by any normal sense of that term.

One of the first big advances was due to Karl Popper, who wanted to discriminate science from pseudoscience by suggesting that science was very different from other fields of endeavor in a particular way: good science "sticks its necks out," or perhaps "has a stake", in what it says about the world. He observed that pseudoscience usually could be compatible with any statement of fact, whereas proper science has some sort of experiments which you can run which would hypothetically prove that idea wrong. This isn't really embodied in the above "scientific method" but it's not too hard to add it. A lot of people take it a step further than maybe Popper would have, using it to say that scientific beliefs can be split into only two essential categories: "falsified" and "not yet falsified", rather than "false" and "true". Everything gets reconceptualized as a ruthless survival-of-the-fittest system.

Unfortunately, then the boat was rocked even more by Thomas Kuhn, who pointed out something which you also see in the OPERA experiment above: theories can be "established". That's very different from Popper's model, which says that we're eagerly looking to cut down all of our theories. In fact in the OPERA statement most physicists pushed back against the report of superluminal neutrinos: "no, you must be wrong, there is something wrong with your timing circuitry or your GPS models or something." In fact Kuhn went a step further: he said that true theories are unfalsifiable, which is like knifing Popper in the back!

Kuhn pointed out that there are two levels: "theory" and "model". A scientific theory is a paradigm -- a way of structuring how you think about the world, how you ask questions, and how you decide what questions are worth asking. You need to already have some theory in order to offer explanations in the first place, because theories define the "ground rules" of how their models work, so that you can model a phenomenon. It's like the theory is the paper, paint, and brush whereas the model is the painting you make with them. When something is different between the picture and what you see in the world, you can shelve your model as a limited approximation, then tear off a new sheet of theory-paper and mix new theory-paint, and try a different painting -- maybe one which is similar in the broad strokes but changes when you start to paint it with a finer brush. Kuhn regarded this model-selection process as Popperian.

But every once in a while, Kuhn said, we see these scientific revolutions where the art supplies themselves get redesigned. He didn't have a great model for how exactly this happens, but definitely thought that it was crucial to scientific progress. His idea was basically, "everyone gets together, agrees that they're painting with too broad a brush, comes up with a better brush-shape and better pigments and paper and what have you, lots of ideas are floated, people keep whichever is aesthetically pleasing while serving the purpose, and then we settle back down into our normal scientific work again." Of course, people didn't like this because it seemed to open the door back to pseudoscience: might astrology now claim that it is a "scientific theory" but that many "astrological models" could be proposed and scrapped underneath? When Kuhn tried to defend himself, he didn't have much of an answer for this. He basically said that scientists have a lot of aesthetic values that they want out of their theories, and those aesthetic values banned astrology due to it not being very simple or parsimonious with how we think about the rest of the cosmos -- a principle sometimes called Ockham's Razor. Kuhn's work was really important, but it was clear that there was no good answer to this question.

I will go one step farther than I think he did, and say that theories tend to be Turing-complete for some domain. We now know that if you include the centrifugal and Coriolis forces, you can put the Earth at the center of the solar system. Geocentrism is 100% as valid as heliocentrism. "Was this not the subject of a huge scientific revolution? How awful, that in Newton's laws we can prove that it makes no difference!" Well, yes -- but what was at stake was not "which theory is right?" but rather "which theory is better?"

To my mind this tension was best resolved by Imre Lakatos, whose works take a little effort to understand. You have to understand that theory choice is actually driven by lazy grad students. You can regard a scientific theory as the "genes" for its scientific models, and those genes make it easy or hard for that model to predict surprising new observations which can be confirmed by experiment -- which spurs an interesting publication which can excite other researchers into making more publications with that theory. The theories that simplify the models will thus "reproduce;" the ones that don't will get ignored by these grad students who don't want to do all of that work if they can easily avoid it. So, heliocentrism won because it did not need the "epicycles" that the geocentrists used. Those epicycles were totally valid -- it's the Fourier series of some periodic motions, after all -- but they were complicated and hard to use. Putting the Sun at a fixed point in your coordinates meant that you didn't need to calculate them. Lazy grad students therefore chose heliocentric coordinates. Theory choice is therefore by natural selection. A more recent example: we now have a deterministic quantum mechanics via "pilot waves"; why is nobody using it? Because it's complicated! Nobody really likes the Copenhagen interpretation philosophically, but it's dead simple to use and predicts all of the same outcomes.

So that's why the theory is not at stake in such circumstances: only the model is at stake, until you get to larger concerns about "it's too hard to model this in that way." Such concerns will seldom be because the instruments assume some other theory -- more likely, it's because some new instrument comes out that forces new measurements which complicate our entire understanding of an existing theory. Then someone comes along with something like quantum mechanics, "hey, we can do this all a lot better if we just predict averages rather than exact values, and if we use complex numbers to form our additive probability space." This changes the questions you ask; you gloss over "how did this photon get here rather than somewhere else?" for questions of "what are the HOMO and LUMO for this molecule?"

homework and exercises - Index notation matrix calculation (Intro to Relativity)

Consider the next matrix:

$$M_{ab} = \left(\begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array}\right)$$ and $$N_{ab} = \left(\begin{array}{cccc} 0 & 0 & \frac{-1}{2} & 0 \\ 0 & 1 & 0 & 0 \\ \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array}\right)$$

Calculate: $ I_M = M_{ab}\Delta x^a\Delta x^b $ and $ I_N = N_{ab}\Delta x^a\Delta x^b $ for the next subtractions:

• 
  
  

  $a) \Delta x^a = (1,0,1,0) $ and

  
  


• 
  

  $ b) \Delta x^a = (1,0,0,0) $

  
  

The problem is I don't really understand the notation nor, given $M_{ab}$ and $N_{ab}$ (as well as $\Delta x^a$), how to operate the matrix.

Any help on this particular example or a general procedure to operate this will be really appreciated.

Answer

Let's look at $M$ and $N$ first. Here, $a$ refers to the row, and $b$ to the column. So, $M_{11}$ corresponds to the upper left corner, and has a value of $-1$. $N_{13}$ corresponds to row $1$, column $3$, with a value of $-\frac{1}{2}$.

Now, lets look at $\Delta x$. This is very similar, but there's only one dimension to work with. So, for b), $\Delta x^{1} = 1$, while for $a = 2, 3, 4$ the value is $0$.

Finally, let's put everything together. The basic concept is that, when an index occurs in both the top and bottom of an equation, there is an implied summation over all values of that index. See below:

$ X_{\gamma} = \begin{bmatrix}1 \\ 2 \\3 \end{bmatrix}$, $Y^{\gamma} = \begin{bmatrix}4 & 5 & 6 \end{bmatrix}$

Then,

$$X_{\gamma}Y^{\gamma} = \sum_{\gamma = 1}^{3} X_{\gamma}Y^{\gamma} = X_{1}Y^{1} + X_{2}Y^{2} + X_{3}Y^{3} = (1*4) + (2*5) + (3*6)$$

Your problem is a bit more complex, as it includes a double summation.

$$M_{ab}x^{a}x^{b} = \sum_{a = 1}^{4} \sum_{b = 1}^{4} M_{ab}x^{a}x^{b}$$

So, now you're out of Einstein notation and in the somewhat more familiar realm of double summations. In case you need a refresher, first let $a=1$, and let $b$ cycle through $1 \to 4$. Now let $a = 2$, and repeat. Add up all these terms, and you have your answer. Alternatively, there are plenty examples of double summations online, the "Einstein notation" part is just understanding how to write them as such.

visual - Don't clear your terminal history 2

Previously, on terminal use...

The door accepted your input and slid open! Unfortunately, so did a trap door under your feet. You slide down and end up in another room. The only apparent exit is a ladder leading up to a locked hatch. Across the hatch in large, friendly letters are the words Don't Panic Gab Hero. Next to the ladder, you see another terminal that reads:

> emmrme arrate
Access granted. Door opened.
> higLss geeLeg
Access granted. Door opened.

> gnninn iiinaa
Access granted. Door opened.
> suMsma uuouoo
Access granted. Door opened.
> pfreTT mfpOlA
Access granted. Door opened.
ERROR: ENTRY MISSING! CONTINUING TO NEXT ENTRY
>

Looks like it is waiting for your input again...

Hint 1:

On the wall across from the terminal exists the following poem:
Wonder, wonder where they are? For wondrous things, you'll travel far.
Where, oh where could they be at? Their position, you know exact.

Hint 2:

In an attempt to gather more information from the terminal, you enter the following:
> sudo help

As if in response, the wall next to you slides up, revealing the following image:

Answer

I think I know the method.

Since the picture is

Statue of Zeus at Olympia

and that is

one of the ancient wonders of the world

all of which together

have locations that spell gab hero:
Giza
Alexandria
Babylon
Halicarnassus

Ephesus
Rhodes
Olympia

And I'm pretty sure the ones that have been entered already correspond to

the first six in order, except Rhodes is skipped

Finally, the passwords are six letters followed by a space, followed by six more letters.

I wanted this to have to do with longitude and latitude, since Hint 1 tells us we know the "position exact". But I can't make those numbers correspond with the letters in the title of the Wonder, in order to make a password. I think I can now, although it makes little sense.

Here's an example:

#1 is the Great Pyramid of Giza
Its longitude and latitude is 29°58′45.03″N 31°08′03.69″E
If I translate this to decimal instead, but ignore the decimal part of the seconds, and then round, the longitude and latitude are 29.9792 N and 31.1342 E
2, 9, 9, 7, 9, 2 would be wrong, but add 1 to all of them (I guess we are supposed to skip the starting letter)
(That is 3, 10, 10, 8, 10, 3)
The 3rd letter of "Great Pyramid of Giza" is e

The 10th letter is m
The 8th letter is r
So 3, 10, 10, 8, 10, 3 --> e, m, m, r, m, e
and you get emmrme (the first half of the password)
same with 3, 1, 1, 3, 4, 2 -- add 1 and you get arrate (the second half of the password)

So for the one you actually need:

Statue of Zeus at Olympia longitude/latitude is 37°38′16.3″N 21°37′48″E
Decimalized as described above is 37.6378 N, 21.63 E

Per comment from OP, we need 13 letters, so I'll pad out with zeroes
I'm going to guess it's tfotfZ atotSS

So I'm (no longer) stuck.

water - Why are snowflakes symmetrical?

The title says it all. Why are snowflakes symmetrical in shape and not a mush of ice?

Is it a property of water freezing or what? Does anyone care to explain it to me? I'm intrigued by this and couldn't find an explanation.

quantum mechanics - Eigenstates in QFT and amplitude of a field operator

I've seen in different posts (such as here) that given a field $\hat{\phi}(x)$, its eigenstates $|\phi\rangle$ are of the form:

$$|\phi\rangle\ = e^{\int dx\phi(x)\hat{\phi}(x)}|0\rangle\tag1$$

I don't understand where this came from.

Also, I have another question. If I had a one particle state $|\vec{p}\rangle$, and I want to know the amplitude of some shape of the field, let's call it $f(x)$, I guess I have to compute

$${\rm amplitude} =\ \langle f(x)|\vec{p}\rangle \ =\ \langle f(x)|a_{\vec{p}}^\dagger\ |0\rangle \tag2$$

Where $|0\rangle$ is the vacuum state. But, could I write equation (2) as:

$${\rm amplitude} = 4\sqrt{\pi}f(x)\exp\{-\frac{1}{2}\int\ d^3d^3yf(x)\Omega_{x, y}f(y)\}\tag3$$

In analogy with harmonic ($4\sqrt{\pi}f(x) =$ norm of state in Harmonic oscillator times the 1-degree Hermite's polynomial) and with $\Omega_{x, y}$ is the kernel that you can see in Eq. (3) of Interaction term in QFT

Answer

Your best bet might be Itzykson & Zuber's overdetailed QFT text, insert in 3.1.2. They explain that your overly naive expression (1) yields eigenstates of $\hat \phi (x)_-$, not $\hat \phi (x)$. The answer you are citing distinctly warns you this is not a field operator eigenstate. (By the way, your $|f(x)\rangle$ should really be $|f\rangle$, as it covers all xs. The argument is the function, not its value at some x.)

Rather than perish in a nightmare of sequential Fourier transforms, rampant indexing and normalizations to enforce Lorentz invariance and Hermiticity, I'll give you an easy hint.

I'll eliminate the entire infinite array of oscillators of QFT, retaining only one, and remind you of the basic coherent state maneuvers, adumbrating the trail map for your proof involving an infinity of oscillators.

So, just keep a single oscillator, and go to here ; here; and here for technical details. $\hat \phi$ here corresponds to $\hat x$, $\hat \phi_+$ to $a^\dagger$, but $\hat p$ to the canonical conjugate field $\pi$; distinctly not your oscillator label, fluff, for your $|\vec p\rangle$, which has reduced to a single value here (along with your x fluff label).

Recall $[a , a^\dagger ]=1$. Then,taking x=f for your classical configuration, here, location, $$ |f\rangle = N(f) ~ \exp\left ( -\frac{(a^\dagger -f\sqrt{2})^2} {2}  \right )|0\rangle   \Longrightarrow  \hat{~x}~|f\rangle=f|f\rangle,\\ |p\rangle= N(p) ~ \exp \left( \frac{(a^\dagger +ip\sqrt{2})^2}{2}   \right ) |0\rangle  \Longrightarrow  \hat{~p}~|p\rangle=p|p\rangle ~. $$ You may try to fix $N(x)=e^{x^2/2}/\pi^{1/4}$ from $\langle 0|x\rangle$, the Schrödinger ground state of the oscillator. (Indeed, N is the inverse of the Gaussian!) It is then shown that $$ \langle p |x\rangle= e^{ixp}/ \sqrt{2\pi}. $$ The basic maneuver is that a acts like a derivative operator on the $a^\dagger$s in the exponentials.

The analog of the conjugate of your amplitude (2) here is $$\langle 0| a |f\rangle = \langle 0|f\sqrt{2}-a^\dagger|f\rangle= f\sqrt{2} ~ \langle 0 |f\rangle =f\sqrt{2}~e^{-f^2/2}/\pi^{1/4}.$$ I am unsure what you would learn from this for an isolated oscillator, and, a fortiori, in scalar field theory, but there it is.

Note that if you had only kept the cross terms in the exponent, i.e., if you had chucked away the quadratic term in the exponent, you would have gotten the very same answer for (2), as the operative shift function of the a is the very same for coherent states in QM!

Of course, the real McCoy here is but the state itself, $$\frac{ a +a^\dagger}{\sqrt{2}} |f\rangle = f |f\rangle . $$ This might help illuminate the role of the quadratic term in the exponent: dropping that quadratic term in the exponent would yield an eigenstate of merely a.

cipher - Can you help me find out what this guy's name is?

I'm a publisher. Yesterday I received an email from someone who wanted me to publish his book. However, he refused to send it to me unless I figured out his name. I sent back an email with the answer 'Rumplestiltskin', but he said no. Instead, he sent me this:

bar.gjb.gjb.gjb.gjb/gjb.gjb.gjb.gjb.bar|bar.bar.bar.bar.bar|bar.bar.bar.bar.bar ^
bar.gjb.gjb.gjb.gjb|bar.bar.gjb.gjb.gjb/bar.bar.bar.bar.bar ^
bar.bar.bar.bar.bar|bar.gjb.gjb.gjb.gjb/bar.bar.gjb.gjb.gjb ^
bar.bar.bar.bar.bar|bar.gjb.gjb.gjb.gjb/bar.bar.gjb.gjb.gjb <>
bar.bar.bar.bar.bar|bar.gjb.gjb.gjb.gjb/bar.bar.bar.gjb.gjb ^
bar.bar.bar.bar.bar|bar.bar.bar.bar.bar <>
bar.gjb.gjb.gjb.gjb|gjb.gjb.gjb.gjb.bar|bar.bar.gjb.gjb.gjb/bar.bar.bar.bar.bar ^
bar.bar.bar.bar.bar|bar.gjb.gjb.gjb.gjb/gjb.gjb.gjb.gjb.bar ^

bar.gjb.gjb.gjb.gjb|bar.bar.gjb.gjb.gjb/bar.bar.bar.bar.bar|bar.bar.bar.gjb.gjb|gjb.gjb.gjb.bar.bar ^
bar.bar.bar.bar.bar|bar.gjb.gjb.gjb.gjb/bar.bar.bar.gjb.gjb ^
bar.gjb.gjb.gjb.gjb|bar.bar.gjb.gjb.gjb/bar.bar.bar.bar.bar ^
bar.bar.bar.bar.bar|bar.bar.bar.bar.bar ^
bar.bar.bar.bar.bar|bar.gjb.gjb.gjb.gjb/bar.bar.gjb.gjb.gjb

Can you help me figure out what Mr. Bar Gjb's real name is?

Hint 1:

13. Decaying.

Hint 2:

What's an anagram of here come dots?

Answer

The first hint...

... alludes to rot13 encoding, which converts bar and gjb to one and two. _{Confirmed by OP}

The second hint...

... refers to Morse code. _{Confirmed by OP}

The bar/gjb sequences are...

... organised in groups of five. In Morse code, the numerals all have five signals. Taking bar/one as dit and gjb/two as dah, the message reads:

1 / 9 | 5 | 5 ^
1 | 2 / 5 ^
5 | 1 / 2 ^
5 | 1 / 2 <>

5 | 1 / 3 ^
5 | 5 <>
1 | 9 | 2 / 5 ^
5 | 1 / 9 ^
1 | 2 / 5 | 3 | 8 ^
5 | 1 / 3 ^
1 | 2 / 5 ^
5 | 5 ^
5 | 1 / 2
_{Confirmed by OP}

Each line...

... represents a letter. The caret denotes the end of a letter, the diamond denotes the end of a word. _{Confirmed by OP}
As @yuzuki observed in a comment, the slash probably means that two adjacent digits are joined into a larger number. After joining the numbers and removing the end markers of the letters, we get:

19 | 5 | 5
1 | 25
5 | 12
5 | 12
5 | 13
5 | 5

1 | 9 | 25
5 | 19
1 | 25 | 3 | 8
5 | 13
1 | 25
5 | 5
5 | 12

The next step is...

... to convert the numbers to letters. The number represents the position of the letter in the alphabet, so 1 is A, 2 is B and so on:

see ay el el / em ee / aiy es aych em ay ee el

These are the letters as they are spelt when read out loud. (Although I'd have used ELL and AITCH.)

The sentence reads:

CALL ME ISHMAEL

which is the famous opening line from Herman Melville's Moby-Dick. So the author's name is Ishmael. (The name could also be Herman Melville, but nobody will know the famous quote if the book hasn't been published.)

Footnote:

I had found the answer without figuring out the second step, where series of numbers are treated as letters. It was enough to treat each individual sequence of numbers, say 5|12, as a letter in a monoalphabetic substitution cipher. Just finding that the caret and diamond are letter and word separators was enough to attack the monoalphabetic cipher.

electromagnetic radiation - Against what force are we doing work when we accelerate an electron?

In vol. I, chapter 32, of The Feynman Lectures, Feynman says:

If we take a charged body and accelerate it up and down it radiates energy; if it were not charged it would not radiate energy. It is one thing to calculate from the conservation of energy that energy is lost, but another thing to answer the question, against what force are we doing the work?

Then he says:

this problem has never been solved.

Has this problem been solved since?

electromagnetism - Why is the magnetic flux proportional to the current (Inductance)?

I'm trying to understand the reasoning behing this formula, or how to get there, if possible, using Maxwell's equations:

$$\Phi_i = L_i I_i + M_{ij} I_j$$

From the wikipedia page of inductance I got this formula which is basically faraday's law:

$$v(t) = - \frac{d\Phi(t)}{dt}$$

But then they somehow get to this one which is what I don't understand:

$$v(t) = L \frac{d i(t)}{dt}$$

I've tried doing some things with the Maxwell equations but I couldn't really get to anything interesting.

riddle - Four Directions #2 - Where and what am I?

To the North is a place you could wear on your feet
To the South is a place that could mean alone
To the West they play a ba' game in the street

To the East is where Hallvard sits on his throne

I've been a tall and proud trio for three decades and a half.
Where and what am I?

Four Directions #1

Answer

Partial: I think you are somewhere in :

Norway or North sea

Explaination:

To the West they play a ba' game in the street - BA game is a version of medieval football played in Scotland, primarily in Orkney and the Scottish Borders, around Christmas and New Year.
To the East is where Hallvard sits on his throne - Hallvard Kirke Norway
To the South is a place that could mean alone - sola - Sola is a municipality in Rogaland county, Norway. it mean Alone in Spanish
Sandal is also a place in Norway, just to the north from Sola.

The part between these places is

Norway or North sea

Damn I found it:

I've been a tall and proud trio for three decades and a half.
It's in Stavanger: Sverd i fjell - The monument was created by sculptor Fritz Røed from Bryne and was unveiled by King Olav V of Norway in 1983. The three bronze swords stand 10 metres (33 ft) tall and are planted into the rock of a small hill next to the fjord.

riddle - Mysterious Murder Mystery 2

"Take her away!" commands the police officer, as two other officers handcuff the blonde woman. "Good work, Detective! I can see why you were named the number one detective. Keep up the great work!". You stand there trying to be mature about it, but deep down you are overwhelmed by the comment, which makes you very happy.

You then remember that after every crime you must write a detailed report and send it to your superiors, so you head back to your office. As you arrive at your building, you park in the designated parking spot, get out of your car and lock it. As you head to the building's front door, you look up to see that there are dark clouds approaching. "Seems like we're gonna have a storm tonight," you say as you enter through the front door.

You head to your office, sit down on your chair, and start writing a report of what happened. You reach into one of your drawers to get a stapler, when you notice a file you have not seen. You open the file to see that this file belonged to a murder case from years ago. You see that the victim's name was Ling O, and she was brutally murdered. They caught the murderer; her name is Bella B. She tortured Ling for several days before police found out. The murder case was solved by a detective named Randy White.

"That name sounds familiar," you think to yourself, and then it hits you: Randy is a detective in the same department as you. "He might have misplaced this file by accident; I will just place it back in the vault with the others," you think to yourself.

As you finish your report, you realise that it is almost time for you to head home. Since this was your first crime solved in Chicago, you decide to celebrate by going to the new bakery not too far from your house. You look out your office window to see that the dark clouds have taken over the earlier sunny day, but you don't think much of it.

You start driving to the new bakery shop, but all of a sudden you get violently cut off. You swerve out of control, but manage to stop safely and don't crash. You look up to see that the other car is also stopped right in front of you. Through the driver's window you see a man with shaggy brown hair, a small nose, and big ears. You give a big sigh. "Not again!"

You see the man get out of his car and walk towards you. You get out of your car as well. Just as you're about to ask if he is OK, the man starts yelling at you: "LEARN HOW TO DRI...." he pauses. "Hey, I know you! you're the one who put my sister in jail! She's innocent!"

The man slowly approaches you; you can tell that he is getting more and more angry. "You put the wrong person in jail! Do you know what you have done to my family? DO YOU?" yells the man. As he is about to get physical, two police officers get between the two of you and separate you. The man is arguing with one of the officers, and does not seem to be calming down.

"Good thing we were patrolling around the neighbourhood; it could have ended badly," says the officer next to you. As you are walking back to your car, the man calls out your name. "YOU HAVEN'T SEEN THE LAST OF ME! I WILL GET REVENGE FOR MY SISTER, YOU HEAR ME?".

You take a deep breath and get back in your car. You finally arrive at the bakery and sit down at one of the tables. "Welcome to Niki's Bakery, how may I help you?" says the waitress as she approaches you. You take a look at the menu and choose a couple of snacks. "Ok, I'll be right back with your food," she says. You notice that she is wearing the bakery's uniform, has blonde hair, is very beautiful, and has a name tag that says "Heather".

As you wait, you look around the bakery. In one corner you see a TV, and the news is on. "BREAKING NEWS! An unknown person has left a note at the front door of the police station. Its contents seem to be addressed to the new head detective of Chicago. The letter says, and I quote, 'To the Chicago detective department: I have killed someone, and the body is located somewhere in RuffNear Park. I challenge the head detective to find me. For everyday that you do not find me, I will kill someone else. I wish you the best of luck, because you will never find me! I am invisible like objects in a December storm; you will not see me coming! My story will go down in history! Better watch out for the children, because they will remember this story forever!' End quote. The police are warning citizens to report any suspicious activities, and...."

The TV goes blank, and all you see is the word "Reconnecting...", and then you hear a loud noise: BOOM! You look outside the bakery's window to see that it is raining heavily outside, with thunder and lightning striking every so often. "It seems like the weather has affected the reception," you think to yourself.

"Here's your food," says the waitress. You ask her if you cam take it to go instead. You call the police department, and one of the police officers picks up. "Good afternoon, don't worry about the message and the body; we have already dispatched a group of search teams to search for the body. We will contact you again when we find it." The officer ends the call.

You decide to head home to get some rest. The next morning, you receive a call: RING RING RING RING. You pick up the phone. "Good morning. I would like to let you know that we have found the body. Please come to RuffNear Park by the washroom ASAP." The call ends. You quickly drive over to the location given by the officer. When you arrive, the forensics team is already at work taking pictures and discussing the matter. You take a look at the body. There do not seem to be any kind of knife or gunshot wounds. You notice that the body is lying down as if asleep. The victim has blonde hair, blue eyes, and is skinny. He still had everything with him: his wallet, his cell phone, and his keys; nothing seems to be missing. Checking his wallet shows that the victim is around 5' 1" in height, and his name is Alex T.

You do not see any small yellow cones with numbering around, so you ask one of the officers.

"Nothing seemed to be left behind; the only clue we have is this cloth on top of the victim's face." The officer shows you a ziplock bag with the cloth inside. You take a closer look at the body and find a small puncture hole in the victim's right arm. "It rained pretty heavily last night, it might have washed away some clues and evidence. We are going to bring the body and the cloth back to the forensics lab to more analysis," one of the detectives says. With that, you make your way back to your office.

Soon after, you receive a call from the forensic scientist telling you to head over to the forensics lab for more details on the victim. When you arrive, he says, "Glad you could make it so soon. The victim did not have any foreign fingerprints on him, nor any foreign hair on his clothing. The small hole in his right arm seems to have come from something sharp, like a needle. The cause of death is a very toxic and deadly poison."

You stand there and ponder things for a minute. You call the police station asking if they have found any suspects yet, but none have came up. You know that time is running out before the day ends and the next victim is killed.

With no leads, you head back to RuffNear Park where the body was found. As you arrive, you try to search for any clues that might have been missed by the other police and detectives. You search for hours but find nothing.

You look at the time on your wrist watch; it says 8:32 p.m. It has been more 24 hours since the arrival of the note received by the police, and there have not been any reports of another dead body, so you decide to head home to get some rest. You arrive home, change out of your clothes, go to bed, and close your eyes.

You wake up to a loud sound: "RING RING RING RING RING!" You look at your alarm clock, which reads 6:15 a.m. You pick up the phone.

"Sorry to disturb you so early, but we just got a report of another body found." You ask for the location, and the officer directs you to it.

You arrive at the scene, and, like every other case, you see police and detectives everywhere and a yellow police line around the area, but no yellow cones. You walk up to the body, and, just as you feared, you notice no signs of physical injury on the body; no gunshots, no knife wound, nothing. You check the victim's right arm to find that there is a small puncture hole. "It must be him," you mutter to yourself. You check the victim's wallet, her name is Patricia G., 5' 2" in height. You stand there pondering, trying to find a relation between the first victim and the second victim, but you can't think of any.

The murder case goes on for another four days, and for each day another person is murdered. On the third day, it is a 34-year-old male, Peter L., 5' 2" in height. On the fourth day, a 12-year-old male named Leo E., 4' 9". On the fifth day, a 68-year-old female, Elsa D., 5' 2". The pattern seems to continue, until the sixth day. On the sixth day, another person has been murdered, but this time there is no wallet found on the victim's body, and the same for the seventh day. Forensic scientists confirm that all of the victims died from the same deadly poison in the right arm.

Just when you thought the case couldn't get any weirder, the murders stop after the seventh day. For the next 2 days, there are no reports of any dead bodies, no murders, no notes, nothing. On the tenth day, you receive a call. "We have found some suspects; come down to the police department pronto." The person on the other end of the phone hangs up.

You rush over to the police department. When you arrive you notice that they have a lineup of four suspects. "I hope we can get any new information soon; we are out of ideas," says one of the officers. You noticed that he looks to be a bit out of breath, his eyes looking here and there. You look at the ID badge pinned to the left side of his chest; it says "ID: 7777777 Officer: George Snoe". You ask him if everything is OK. "Yeah I'm OK; I'm just not in the best of shape like I used to be," he replies.

You don't think much of it after that, so you start interrogating each of the suspects. The first suspect is a 25-year-old female, beautiful, with a slim body, blonde hair, and blue eyes. "I didn't commit any murders. I was working every day from 8:00 a.m. to 4:30 p.m. I teach health and phys. ed.," says Patricia F.

The second suspect is a 34-year-old male, tall, about 6' 9", with black hair and brown eyes. "I couldn't have murdered anyone. I work as a investment banker at the Chicago bank, and after work I have night school. I don't have time to murder anyone," he says. Before you can ask for his name, he receives a phone call. He says it is an emergency, and you let him take the call.

The third suspect is a 41-year-old male, 6' 7" in height, very groomed, with nicely-styled hair and brown eyes. "I am a doctor at the Chicago Hospital. I did surgery on two of the days all day, and on three of the days I was working my usual hours. On the weekend I was with my family, so it couldn't have been me." You ask for his name, and he replies, "My name is Dr. David N. Edle".

The fourth and last suspect is a 24-year-old male. He is wearing a bandana, jeans hanging low, and nice white shoes. Looking at some papers the officers gave you beforehand, you read that he was caught doing all sorts of drugs. He is an addict and also a drug dealer. You asked him where he was during the week. "Nowhere man, just been chillin' with my homies. I didn't do shit, man. You pigs always arresting my brothers and shit man. Been working at the local fast food chain Rocky Ranchy's Burger Joint, just trying to get by. Don't need you poppin' me for doing nothin' wrong man."

You pondered everything for several minutes, trying to put together all the clues and suspects and other information, trying to find the murderer. Then it hit you. "How could I have been so blind?"

Who is the murderer?

newtonian mechanics - Given Newton's Third Law, how can an inanimate object create a force?

Newton's Third Law states that

Whenever any force is exerted by a body#1 on any other body#2, another force which is equal in magnitude and opposite in direction is exerted on body#1 by body#2.

In theory, suppose a boy is pushing a rock heavier than him on ice. Due to the relative lack of friction, the boy would probably be pushed back more than the rock was pushed forward. How is this possible? How can an inanimate object create a force?

pressure - Hydrostatic paradox

I'm aware of the hydrostatic paradox that water will exert same pressure from same height irrespective of the shape of container (mass of liquid). But is it true water in different shapes of container with upto same height and same base will weigh the same on weighing scale?

special relativity - Understanding the "$pi$" of a rotating disk

Let us say you are in an inertial reference frame with a circular planar disk. If you take your meter measuring rods (or perhaps tape measure) you can find the diameter and circumference of the disk. If you divide the circumference by the diameter, you will get exactly $\pi$. Now you start rotating the disk. A book I have claims that now the ratio will be different (lets call it $\pi_\circ$ to avoid ambiguity.) This has caused to reevaluate my understanding of special relativity (this is a set up to general relativity, but the problem itself only requires the special theory.) My problem is that, sure, the measuring rods (or tape) would contract on the circumference, but wouldn't the circumference contract a similar amount, therefore giving a measurement of $\pi$ in every reference frame according to your own measuring rods. Of course, this reasoning goes on to make me question how I understood all of special relativity up this point.

classical mechanics - Why certain rotations are unstable? (Euler Equations)

We have the Euler equations for a rotating body as follows

$$I_1\dot\omega_1+\omega_2\omega_3(I_3-I_2)=0\\ I_2\dot\omega_2+\omega_1\omega_3(I_1-I_3)=0\\ I_3\dot\omega_3+\omega_2\omega_1(I_2-I_1)=0$$ Where $I_i$ are the moments of inertia about the $x_i$ axis, and $\omega_i$ is the angular velocity about this axis.

It can be shown (*) that if $I_1>I_2>I_3$, then objects with angular velocity very close to $\vec\omega=(0,1,0)$ are unstable. Why is this and how can I try to picture it?

I tried to picture this using a ball, but realised this is probably not a good way to visualise it, since a ball is spherically symmetric, so the moments of inertia are not distinct. Is there any visualisation or animation that could allow me to see this rotation, and possibly understand why it is unstable?

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(*) In response to @SRS's comment:

I am not sure about any references, but I know how to do it: Let $\omega_1=\eta_1,\omega_3=\eta_3$ where $\eta$ is a small perturbation, and suppose $\omega_2=1+\eta_2$. Then the Euler eqns become$$I_1\dot\eta_1=(I_2-I_3)\eta_3+O(\eta^2)\tag1$$$$I_2\dot\eta_2=O(\eta^2)\tag2$$$$I_3\dot\eta_3=(I_1-I_2)\eta_1+O(\eta^2)\tag3$$Differentiate $(1)$ and sub in $(3)$ to the resulting expression$$\ddot\eta_1=\frac{(I_2-I_3)(I_1-I_2)}{I_3I_1}\eta_1$$If $I_1>I_2>I_3$, then the constant on the right hand side is positive, so the solution to this equation is an exponential (if it was any other order, then the solution would be a $\sin/\cos$). Therefore it is unstable.

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Edit:

To clarify, I posted this question to see other more visual ways of understanding this effect rather than solving the equations as I did above, and to see how this effect comes into play in real life. So I don't think it is a duplicate of the other questions, since they don't have answers that fit this.

Answer

There is another nice way of seeing this mathematically. It is not too hard to show that in the body frame, there are two conserved quantities: the square of the angular momentum vector $$ L^2 = L_1^2 + L_2^2 + L_3^2 $$ and the rotational kinetic energy, which works out to be $$ T = \frac{1}{2}\left( \frac{L_1^2}{I_1} + \frac{L_2^2}{I_2} + \frac{L_3^2}{I_3} \right). $$ (Note that the angular momentum $\vec{L}$ itself is not conserved in the body frame; but its square does happen to be a constant.)

We can then ask the question: For given values of $L^2$ and $T$, what are the allowed values of $\vec{L}$? It is easy to see that $L^2$ constraint means that $\vec{L}$ must lie on the surface of a sphere; and it is almost as easy to see that the $T$ constraint means that $\vec{L}$ must also lie on the surface of a given ellipsoid, with principal axes $\sqrt{2TI_1} > \sqrt{2T I_2} > \sqrt{2T I_3}$. Thus, the allowed values of $\vec{L}$ must lie on the intersection of a sphere and an ellipsoid. If we hold $L^2$ fixed and generate a bunch of these curves for various values of $T$, they look like this:

Note that for a given value of $L^2$, an object will have its highest possible kinetic energy when rotating around the axis with the lowest moment of inertia, and vice versa.

Suppose, then, that an object is rotating around the axis of its highest moment of inertia. If we perturb this object so that we change its energy slightly (assuming for the sake of argument that $L^2$ remains constant), we see that the vector $\vec{L}$ will now lie on a relatively small curve near its original location. Similarly, if the object is rotating around its axis of lowest inertia, $\vec{L}$ will stay relatively close to its original value when perturbed.

However, the situation is markedly different when the object is rotating about the intermediate axis initially (the third red point in the diagram above, on the "front side" of the sphere. The contours of slightly perturbed $T$ near this point do not stay near the intermediate axis; they wander all over the sphere. There is therefore nothing keeping $\vec{L}$ from wandering all over this sphere if we perturb the object slightly away from rotating about this axis; which implies that an object rotating about its intermediate axis is unstable.

homework and exercises - Period $T$ of oscillation with cubic force function

How would I find the period of an oscillator with the following force equation?

$$F(x)=-cx^3$$

I've already found the potential energy equation by integrating over distance:

$$U(x)={cx^4 \over 4}.$$

Now I have to find a function for the period (in terms of $A$, the amplitude, $m$, and $c$), but I'm stuck on how to approach the problem. I can set up a differential equation:

$$m{d^2x(t) \over dt^2}=-cx^3,$$

$$d^2x(t)=-{cx^3 \over m}dt^2.$$

But I am not sure how to solve this. Wolfram Alpha gives a particularly nasty solution involving the hypergeometric function, so I don't think the solution involves differential equations. But I don't have any other leads.

How would I find the period $T$ of this oscillator?

Answer

Since $$\frac1 2mv^2+U(x)=U(A)$$ We have $$dt=\frac{dx}v=\frac{dx}{\sqrt{2(U(A)-U(x))/m}}=\frac{dx}{\sqrt{c(A^4-x^4)/(2m)}}$$ Then $$\frac T4=\int_0^{\frac T4}dt=\int_0^A\frac{dx}{\sqrt{\frac{c}{2m}(A^4-x^4)}}$$ Thus $$T=4\int_0^A\frac{dx}{\sqrt{\frac{c}{2m}(A^4-x^4)}}$$

mathematical physics - How a physicist perceives gauge fields

Facts

• The mathmatical framework of a gauge theory is that of principal $G$-bundles, where $G$ is a Lie group (representing some physical symmetry); let ${\pi}\colon{P}\to{M}$ be such a bundle.


• A connection 1-form on $P$ corresponds to a gauge field, and the associated to it curvature 2-form corresponds to the field strength tensor.



• A (local) section of $\pi$ serves as a (local) reference frame (alias, gauge) for an observer on $M$. In particular, if $s\colon{U}\to{P}$ is such a section, where $U$ is an open subset of $M$, then an observer on $U$ describes elements of $\Omega^{k}\left({\pi}^{-1}(U),\mathfrak{g}\right)$ by pulling them back by $s$.


• Specifically, let $A\in\Omega_{con}^{1}\left(P,\mathfrak{g}\right)$ be a connection 1-form and ${F^A}\in\Omega_{hor}^{2}\left(P,\mathfrak{g}\right)^{\operatorname{Ad}}$ the associated curvature 2-form. Then, our observer on $U$ interprets ${A_s}:={s^*}A$ and ${F_s^A}:={s^*}{F^A}$ as the gauge field and the field strength tensor, respectively, in that specific gauge, viz. $s$.

Conclusion: In a certain gauge, any connection 1-form and any curvature 2-form define, respectively, a unique element of $\Omega^{1}\left(M,\mathfrak{g}\right)$ and $\Omega^{2}\left(M,\mathfrak{g}\right)$.

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Now, a physicist's approach on gauge theory is to define the space of gauge fields as $\Omega^{1}\left(M,\mathfrak{g}\right)$, when a particular gauge has (implicitly) been chosen. So my question is the following:

"does the whole $\Omega^{1}\left(M,\mathfrak{g}\right)$ qualify as the set of gauge fields, in the sence that there is one-to-one correspondence between $\Omega_{con}^{1}\left(P,\mathfrak{g}\right)$ and $\Omega^{1}\left(M,\mathfrak{g}\right)$, when a gauge is chosen? Or are there elements in $\Omega^{1}\left(M,\mathfrak{g}\right)$ that do not qualify as gauge fields, in the sence that they are not pullbacks of connection 1-forms by the chosen gauge?"

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Probably, the question is of mathematical nature and can be answered by rigorous calculations. If one's mathematical background is such that they can provide a rigorous proof, then they are kindly requested to do so (or at least provide a sketch of that proof). However, if there is an intuitive way to approach the answer, I would be glad to know about that.

Answer

It is not true in general that a connection form $A$ on the bundle descends to a well-defined form on all of $M$. In fact, it does so if and only if the bundle is trivial - since the existence of a global section of a principal bundle by which we can pull back $A$ necessarily implies that $P$ is trivial.

Instead each connection form $A$ descends to local $A_i\in\Omega^1(U_i,\mathfrak{g})$ where the $U_i$ are a cover of $M$ that trivializes $P$, i.e. there are sections $s_i : U_i \to \pi^{-1}(U_i)\cong U_i\times G$ with $s_i(x) = (x,1)$ and we define $A_i := s_i^\ast A$. With regards to the transition functions $t_{ij} : U_i \cap U_j \to G$, these fulfill $$ A_i = \mathrm{ad}_{t_{ij}}(A_j - t_{ij}^\ast \theta),$$ where $\theta$ is the Cartan-Maurer form, or, in a more familiar notation, $$ A_i = t^{-1}_{ij}A_j t_{ij} - t_{ij}^{-1}\mathrm{d}t_{ij}.\tag{1}$$ Conversely, every system of $A_i$ that fulfills these relations defines a connection form on $P$. When the bundle is trivial, this relation is vacuous since we only need on $U_i = M$, and we recover the claimed bijection between connection forms on $P$ and $A\in\Omega^1(M,\mathfrak{g})$ for this special case.

Now, a gauge-transformation is a fiber-preserving automorphism $g : P\to P$ that descends to local functions $g_i : U_i\to G$ with $g_i = g_j t_{ij}$, or, in the trivial case, a function $g : M\to G$, which act on the local connection forms much as the transition functions do (which is the reason the physics literature has trouble distinguishing these two conpcets at times). Such a transformation amounts to choosing another point in every fiber $\pi^{-1}(x) \cong G$ as the identity, which we may do, since the fiber carries a group action but is not naturally a group itself, and so has no distinguished identity element.

If you now note that the sections $s_i$ include such a choice of identity in their definition, then you can see that we may view such a gauge transformation as changing the sections by which we defined the connection form. So the physicist, who usually does not care about the connection form on $P$ but about its local description on the $U_i$, declares those $A_i$ which are related by such transformations to be equivalent, and quotienting them out of the space of $\Omega^1(U_i,\mathfrak{g})$ with the correct transformation property (or equivalently quotienting the space of connection forms on $P$ by gauge transformation $P\to P$) yields the space of gauge equivalence classes.

So, in summary, we answer the question like this: A gauge field is a collection of local gauge potentials $U_i \to \mathfrak{g}$ with the compatibility condition (1), which are in bijection to principal connections on $P$, but physically those that are related by a gauge transformation are declared equivalent and essentially indistinguishable.

quantum mechanics - Is the momentum operator well-defined in the basis of standing waves?

Suppose I want to describe an arbitrary state of a quantum particle in a box of side $L$. The relevant eigenmodes are those of standing waves, namely $$ \left=\sqrt{\frac{2}{L}}\cdot \sin \left(\frac{n\pi x}{L}\right)$$ In this basis, the operate $\hat{p}^2$ is diagonal by construction, so every state has a definite energy.

But suppose I want to construct this operator from the matrix elements of $\hat{p}$? These are ($\hbar=1$) $$\left=-\frac{2imn((-1)^{m+n}-1)}{L\cdot(m^2-n^2)}$$ which vanishes for $m$ and $n$ with the same evenness, making this an off-diagonal "checkered" matrix.

I tried taking a finite portion of this matrix and squaring it in order to obtain $\hat{p}^2$, but from what I could tell, this yields an on-diagonal checkered matrix, contrary to the construction that it would be strictly diagonal.

I assume this has something to do with the fact that the wavefunctions obtained by applying $\hat{p}$ do not satisfy the same boundary conditions as the basis functions (do not vanish). Would this problem be solved I had taken the the infinite matrix in its entirety before squaring? Or is this entire problem (decomposing $\hat{p}$ in this basis) ill-defined?

Answer

The Hilbert space for a particle in a box has no associated momentum operator (as a momentum operator implies that the state space is invariant under space translations). So your attempt do define one leads to artifacts.

[Edit] In general, the Hamiltonian must be a densely defined operator on the physical Hilbert space. But a square well potential is too singular as an operator on the space of all square integrable functions on $R$ (i.e., no densely defined subspace is mapped by it into the Hilbert space), hence its Hilbert space cannot be the standard Hilbert space.

[Edit2] Note that the usual formula for the momentum is not a canonical momentum on the restricted space of wave functions defined only in the box, as it fails to satisfy either the commutation relation or does not preserve the boundary conditions.

The appropriate replacement for the momentum operator is the mode counting operator $\hat n$ with $\hat n|n\rangle=n|n\rangle$; see also the derivation of Qmechanic.

special relativity - What is the correct relativistic distribution function?

I am trying to figure out the proper way to model a velocity/momentum distribution function that is correct in the relativistic limit. I would like to determine/know two things:

1. Is there an analytical form for an anisotropic relativistic momentum distribution (i.e., the relativistic analog of the bi-Maxwellian distribution)?


2. What does temperature (i.e., kinetic temperature) mean in the relativistic limit?
   
   
   
   • Temperature cannot be a Lorentz invariant, can it?
   
   
   • It certainly cannot be invariant if the average particle thermal energies correspond to relativistic thermal velocities, correct?
     
     
     
     • So how can just a simple scalar temperature be a proper normalization factor as a Lagrange multiplier in the Maxwell-Jüttner distribution, for instance?
     
     
     
     
   
   
   
   
   


3. Extra: Is there an appropriate relativistic version of the $\kappa$ distribution (see this arXiv PDF for reference, or e-print number 1003.3532 if you do not trust links)?

I am aware of the Maxwell-Jüttner distribution of particle species $\alpha$, given by: $$ f_{\alpha} \left( p \right) = \Lambda \ exp \left[ - \Theta_{0} \ m c^{2} \ \gamma \left( p \right) - \sum_{i=1}^{3} \Theta_{i} \ c \ p^{i} \right] $$ where $\Lambda$, $\Theta_{0}$, and $\Theta_{i}$ are Lagrange multipliers, $p$ is the relativistic momentum, and $\gamma\left( p \right)$ is the Lorentz factor. The $\Theta_{\nu}$ terms are 4-vector components with units of inverse energy.

In the isotropic limit, one can set all $\Theta_{i}$ $\rightarrow$ 0. This leads to the canonical form of the isotropic relativistic momentum distribution function is given by: $$ f_{\alpha} \left[ \gamma \left( p \right) \right] = \Lambda e^{ - \Theta_{0} \ m c^{2} \ \gamma \left( p \right) } $$ where $\Theta_{0}$ was shown¹ to be the inverse of a temperature.

The definition of $\Lambda$, however, has led to multiple results, as stated by Treumann et al. [2011]:

the correct (non-angular-dependent part of the) relativistic thermal-equilibrium distribution should become the modified-Jüttner distribution. (The ordinary Maxwell-Jüttner distribution function was derived by F. Jüttner, 1911, who obtained it imposing translational invariance in momentum space only.)

An attempt⁴ was made to derive $\Lambda$ by imposing Lorentz invariance on only momentum space, ignoring the spatial coordinates of the volume integral. However, Treumann et al. [2011] note that:

This is either not justified at all or it is argued that the particles are all confined to a fixed box which is unaffected by the Lorentz transformation and invariance. However, the momentum and configuration space volume elements the product of which forms the phase-space volume element, are not independent, as we have demonstrated above. Even in this case of a fixed outer box, the particle's proper spaces experience linear Lorentz contractions when seen from the stationary frame of the observer, i.e., from the box-frame perspective. The consequence is that the extra proper Lorentz factor $\gamma\left( p \right)$ in the phase-space volume element cancels thereby guaranteeing and restoring Lorentz invariance...

They go on to show that the correct Lagrange multipliers are: $$ \Theta_{0} = \frac{ 1 }{ T } \\ \Lambda = \frac{ N^{0} }{ 4 \pi \ m^{2} T^{2} } \left[ 3 K_{2}\left( \frac{ m c^{2} }{ T } \right) + \frac{ m c^{2} }{ T } K_{1}\left( \frac{ m c^{2} }{ T } \right) \right]^{-1} $$ where $N^{0}$ is the scalar part of the particle current density 4-vector (i.e., number density), $K_{i}(x)$ is the second order modified Bessel function, and $T$ is a scalar temperature. Notice there is an additional term (i.e., $K_{1}(x)$) in the normalization factor $\Lambda$, which is why they called this the modified Maxwell-Jüttner distribution. This accounts for Lorentz invariance in the phase-space element, not just momentum-space.

Regardless of its accuracy, the distribution function in Treumann et al. [2011] still only assumes an isotropic distribution and I am still a bit confused how the temperature is just a scalar. In plasma physics, it is more appropriate to think of it as kind of a pseudotensor derived from the pressure tensor or 2nd moment of the distribution function. So am I supposed to interpret relativistic temperatures through the energy-momentum tensor or something else? See more details about velocity moments here: https://physics.stackexchange.com/a/218643/59023.

In many situations, plasmas can be described as either a bi-Maxwellian or bi-kappa [e.g., Livadiotis, 2015] velocity distribution functions. The bi-Maxwellian is given by: $$ f\left( v_{\parallel}, v_{\perp} \right) = \frac{ 1 }{ \pi^{3/2} \ V_{T \parallel} \ V_{T \perp}^{2} } \ exp\left[ - \left( \frac{ v_{\parallel} - v_{o, \parallel} }{ V_{T \parallel} } \right)^{2} - \left( \frac{ v_{\perp} - v_{o, \perp} }{ V_{T \perp} } \right)^{2} \right] $$ where $\parallel$($\perp$) refer to directions parallel(perpendicular) with respect to a quasi-static magnetic field, $\mathbf{B}_{o}$, $V_{T_{j}}$ is the $j^{th}$ thermal speed (actually the most probable speed), and $v_{o, j}$ is the $j^{th}$ component of the bulk drift velocity of the distribution (i.e., from the 1st velocity moment).

The bi-kappa distribution function is given by: $$ f\left( v_{\parallel}, v_{\perp} \right) = A \left[ 1 + \left( \frac{ v_{\parallel} - v_{o, \parallel} }{ \sqrt{ \kappa - 3/2 } \ \theta_{\parallel} } \right)^{2} + \left( \frac{ v_{\perp} - v_{o, \perp} }{ \sqrt{ \kappa - 3/2 } \ \theta_{\perp} } \right)^{2} \right]^{- \left( \kappa + 1 \right) } $$ where the amplitude is given by: $$ A = \left( \frac{ \Gamma\left( \kappa + 1 \right) }{ \left( \pi \left( \kappa - 3/2 \right) \right)^{3/2} \ \theta_{\parallel} \ \theta_{\perp}^{2} \ \Gamma\left( \kappa - 1/2 \right) } \right) $$ and where $\theta_{j}$ is the $j^{th}$ thermal speed (also the most probable speed), $\Gamma(x)$ is the complete gamma function and we can show that the average temperature is just given by: $$ T = \frac{ 1 }{ 3 } \left( T_{\parallel} + 2 \ T_{\perp} \right) $$ if we assume a gyrotropic distribution (i.e., shows symmetry about $\mathbf{B}_{o}$ so that the two perpendicular components of a diagonalized pressure tensor are equal).

In summary, I would prefer a relativistically consistent bi-kappa distribution but would be very happy with the bi-Maxwellian version as well.

After several conversations with the R. Treumann, he and his colleague decided to look into an anisotropic Maxwell-Jüttner distribution. I also referred him to this page and he decided to try and remain consistent with the original Maxwell-Jüttner distribution normalization to avoid further confusion.

His new results can be found in the arXiv paper with e-print number 1512.04015.

Summary of Results
One of the interesting things noted by Treumann and Baumjohann is that one cannot simply take the expression for energy and split the momentum into parallel and perpendicular terms as has occasionally been done in the past. Part of the issue is that the normalizations factors, i.e., temperature-like quantities, are not relativistically invariant. The temperature in this case is more akin to a pseudotensor than a scalar (Note: I use pseudotensor very lightly/carelessly here).

They use the Dirac tensor, from the Klein-Gordon approach, to define the energies. They treat the pressure as a proper tensor, with an assumed inverse, to define what they call the temperature tensor.

Unfortunately, the equation cannot be reduced analytically, but it is useful none-the-less given the alternative is to assume the unrealistic case of an isotropic velocity distribution in a relativistic plasma.

1. Israel, W. "Relativistic kinetic theory of a simple gas," J. Math. Phys. 4, 1163-1181, doi:10.1063/1.1704047, 1963.


2. Treumann, R.A., R. Nakamura, and W. Baumjohann "Relativistic transformation of phase-space distributions," Ann. Geophys. 29, 1259-1265, doi:10.5194/angeo-29-1259-2011, 2011.


3. Jüttner, F. "Das Maxwellsche Gesetz der Geschwindigkeitsverteilung in der Relativtheorie," Ann. Phys. 339, 856-882, doi:10.1002/andp.19113390503, 1911.


4. Dunkel, J., P. Talkner, and P. Hänggi "Relative entropy, Haar measures and relativistic canonical velocity distributions," New J. Phys. 9, 144-157, doi:10.1088/1367-2630/9/5/144, 2007.


5. Livadiotis, G. "Introduction to special section on Origins and Properties of Kappa Distributions: Statistical Background and Properties of Kappa Distributions in Space Plasmas," J. Geophys. Res. Space Physics 120, 1607-1619, doi:10.1002/2014JA020825, 2015.

Answer

The following answer is largely taken from the arXiv paper with e-print number 1512.04015 by Treumann and Baumjohann (from here on I abbreviate references to this paper as TB15).

It is well known that the Maxwell-Jüttner distribution works well for a momentum/energy distribution with an isotropic, scalar temperature, $T$. This distribution is generally applied to hot, relativistic plasmas. Herein lies the problem and the source of the question. Hot plasmas, even in non-relativistic cases, are rarely isotropic.

As an aside and to clarify definitions, TB15 state:

The meaning of classical here is not that quantum effects are excluded which, in a hot plasma is anyway the case, because of the tinniness of the thermal quantum length $\lambda_{q} = \sqrt{2 \pi \hbar^{2} / m_{e} T}$. It implies that particle numbers are conserved. This inhibits pair creation and annihilation and thus restricts to temperatures below, roughly, $T_{e} < 2 m_{e} c^{2} \approx 1 \ MeV$.

For non-relativistic plasmas, dealing with a temperature anisotropy is easy because the kinetic energy is additive, as shown in the example bi-Maxwellian and bi-kappa equations in the above question. In a relativistic plasma, the energy has an additional momentum-dependent factor, $\gamma\left( \mathbf{p} \right)$, which is the Lorentz factor, defined as: $$ \gamma\left( \mathbf{p} \right) = \sqrt{ 1 + \left( \frac{ \mathbf{p} }{ m \ c } \right)^{2} } \tag{1} $$ where $\mathbf{p}$ is the momentum, $m$ is the particle rest mass, and $c$ is the speed of light in vacuum. Then the particle's total energy can be defined as: $$ \epsilon\left( \mathbf{p} \right) = \gamma\left( \mathbf{p} \right) \ m \ c^{2} \tag{2} $$ The problem is that $\epsilon\left( \mathbf{p} \right)$ is no longer additive. It is often the case that anisotropic temperatures are introduced in the following form: $$ \epsilon\left( \mathbf{p} \right) = m \ c^{2} \sqrt{ 1 + \left( \alpha_{\perp} \bar{p}_{\perp} \right)^{2} + \left( \alpha_{\parallel} \bar{p}_{\parallel} \right)^{2} } \tag{3} $$ where $\bar{p}_{\perp (\parallel)} = c \ p_{\perp (\parallel)}/T_{\perp (\parallel)}$ and $\alpha_{\perp (\parallel)} = T_{\perp (\parallel)}/mc^{2}$. There are several issues produced by this format, as TB15 state:

This inhibits a simple definition of a distribution expressed solely in $\gamma$ enforcing the use of the momentum distribution. Moreover, in the exponent of the relativistic distribution the normalisation to temperature becomes arbitrary... A further aspect concerns the question, which $\gamma$ is meant. If the observer is dealing from his inertial system with a moving extended gaseous or plasma body with relativistic bulk momentum $\mathbf{P}$ the problem is different from the one where the observer is embedded at relative rest into the volume with all particles moving relative to him at relativistic momenta $\mathbf{p}$. In the latter case the relativistic effects are intrinsic to the system which is in contrast to the above simpler extrinsic bulk effect...

They use the Klein-Gordon approach because particle spins are ignored. Thus, squaring Equation 2 results in: $$ \begin{align} \epsilon^{2} & = \left[ c \ \mathbf{p} + i \ m \ c^{2} \ \mathbf{e} \right] \cdot \left[ c \ \mathbf{p} - i \ m \ c^{2} \ \mathbf{e} \right] \tag{4a} \\ & = \left( c \ p_{\nu} + i \ m \ c^{2} \ e_{\nu} \right) \delta_{\mu}^{\nu} \left( c \ p^{\mu} - i \ m \ c^{2} \ e^{\mu} \right) \tag{4b} \end{align} $$ where $\mathbf{p} = \left( p_{\perp} \cos{\phi}, p_{\perp} \sin{\phi}, p_{\parallel} \right)$, $\delta_{\mu}^{\nu}$ is the Dirac tensor, and $\mathbf{e}$ is a unit vector that can be arbitrarily chosen. TB15 make several points about the form of Equation 4b:

Of course, the energy maintains its scalar property which is reflected in the scalar dot-product of the two bracketed expressions. The appearance of the imaginary element has no effect on the energy. In particular, it does not imply any damping of particle motion! However, by the above factorisation, the Hamiltonian can be split into two Hamiltonians which are linear in the vector $\mathbf{p}$. This is important to realise. It permits the consistent introduction of a pressure anisotropy...

They then construct a pressure tensor assuming a fixed scalar number density, $N$, and write it in terms of the temperature as: $$ \begin{align} \mathbb{P} & = P_{\perp} \delta_{\mu}^{\nu} + \left( P_{\parallel} - P_{\perp} \right) \delta_{3}^{3} \tag{5a} \\ & = N \left[ T_{\perp} \delta_{\mu}^{\nu} + \left( T_{\parallel} - T_{\perp} \right) \delta_{3}^{3} \right] \tag{5b} \end{align} $$ where direction $3$ is assumed to be along the quasi-static magnetic field direction, $\mathbf{b} = \mathbf{B}/B$ and Equation 5b applies for an ideal gas. Here is where things become difficult because the temperature is not merely a scalar quantity anymore, thus we cannot simply divide by it as was done in Equation 3 above. If the pressure tensor is invertible, which it should be, then we have: $$ \begin{align} \mathbb{P}^{-1} & = N^{-1} \left[ T_{\perp}^{-1} \delta_{\nu}^{\mu} + \left( T_{\parallel}^{-1} - T_{\perp}^{-1} \right) \delta_{3}^{3} \right] \tag{6a} \\ & = N^{-1} \Theta \tag{6b} \end{align} $$ where $\Theta$ is the "inverse temperature tensor." Now we can rewrite Equation 4b as: $$ \begin{align} \epsilon^{2} & = \left[ c \ p_{\nu} + i \ m \ c^{2} \ e_{\nu} \right] \Theta_{\mu}^{\nu} \Theta_{\lambda}^{\mu} \left[ c \ p^{\lambda} - i \ m \ c^{2} \ e^{\lambda} \right] \tag{7a} \\ & = \frac{ m^{2} c^{4} }{ T_{\perp}^{2} } \left( \frac{ p^{2} }{ m^{2} c^{2} } + 1 \right) \left( \sin^{2}{\theta} + A^{2} \cos^{2}{\theta} \right) \tag{7b} \end{align} $$ where $A = T_{\perp}/T_{\parallel}$ and $\mathbf{p} = \left( p \sin{\theta}, p \cos{\theta} \right)$ (i.e., just reduce the momentum to parallel and perpendicular). If we define $\beta_{\perp} = m \ c^{2}/T_{\perp}$ and $\psi = A^{2} - 1$ (i.e., isotropy implies $\psi = 0$), then we can write the anisotropic, relativistic momentum distribution for an ideal gas as: $$ \begin{align} f\left( \mathbf{p} \right) & = C_{o} exp\left( - \beta_{\perp} \sqrt{ \left( 1 + \frac{ p^{2} }{ m^{2} c^{2} } \right) \left( 1 + \psi \cos^{2}{\theta} \right) } \right) \tag{8a} \\ & = C_{o} exp\left( - \beta_{\perp} \ \gamma\left( p \right) \sqrt{ \left( 1 + \psi \cos^{2}{\theta} \right) } \right) \tag{8b} \end{align} $$

The normalization is with respect to the scalar density, $N$, and the approach is given in TB15. They start by defining $\beta = \beta_{\perp} \sqrt{ 1 + \psi \cos^{2}{\theta} }$ and then show: $$ N = 4 \ \pi \ C_{o} \ \left( m \ c \right)^{3} \int_{0}^{1} \ dx \ \frac{ K_{2}\left( \beta\left( x \right) \right) }{ \beta\left( x \right) } \tag{9} $$ where $K_{2}\left( z \right)$ is the second order modified Bessel function of argument $z$ and the integral does not have an analytical solution. One can rewrite Equation 9 in the following form for $C_{o}$ given by: $$ C_{o} = \frac{ N \ \beta_{\perp} }{ 2 \ \pi \ \left( m \ c \right)^{3} } \left[ \int_{1}^{1 + \psi} \ dz \ \frac{ K_{2}\left( \beta_{\perp} \sqrt{z} \right) }{ \sqrt{z \left( z - 1 \right)} } \right]^{-1} \tag{10} $$ which also cannot be analytically solved.

In the limit of small anisotropy (i.e., $\psi < 1$), one can approximate the integral in Equation 9 as: $$ \int_{0}^{1} \ dx \ \frac{ K_{2}\left( \beta\left( x \right) \right) }{ \beta\left( x \right) } \approx \frac{ 1 }{ 2 \ \beta_{\perp} \sqrt{ \psi } } \left\{ \left[ K_{1}\left( \frac{ \beta_{\perp} }{ 2 } \right) \right]^{2} - \left[ K_{1}\left( \frac{ 1 }{ 2 } \beta_{\perp} \sqrt{ 1 + \psi } \right) \right]^{2} \right\} \tag{11} $$

All of the above assume a perpendicular anisotropy, but analogous forms can be found for parallel anisotropies. TB15 also comments on the generalization to include an external potential fields, but I leave that to the article.

The above expressions only apply to bosons. To include fermions, one would need to include spin in the calculations which would greatly increase the complexity of the problem. As stated by TB15:

We note that a similar calculation can also be performed for Fermionic plasmas. Then, however, one must refer to Dirac’s method of splitting the relativistic energy. This leads to the use of the Dirac matrices and substantially more involved expressions including the spin of the particles. Fermionic effects of the anisotropy will be expected only at very low temperatures and in very strong magnetic fields. Such situations might occur in the Quantum-Hall effect and when dealing with the interiors of highly magnetised objects like pulsars and magnetars... An interesting conlusion which can be drawn is that in relativistic media the temperature and its inverse, usually called $\beta = 1/T$ must be understood as vectors. This is the consequence of the pressure tensor.

As a final conclusion, they note:

One might, moreover, think of that the split of the energy would provide two different versions of the distribution. This is, however, not the case. The linearity in the momentum of the two energy-vectors has no classical meaning. Energies are scalars and do not transform like vectors. Multiplication with the inverse relativistic temperature vector does not release this property. The linearity would lead to a linear dependence of the exponent in the distribution which is not Gaussian anymore and thus loses the property of a probability. Moreover, each of the energy vectors becomes complex. This complexity resolves in quantum mechanics in the operator formalism of the Klein-Gordon and Dirac equations but has no meaning in classical physics.

Since the posting of this question and answer there have been two refereed publications on the topic, one by Treumann and Baumjohann [2016] (doi:10.5194/angeo-34-737-2016) and one by Livadiotis [2016] (doi:10.5194/angeo-34-1145-2016) (both papers are Open Access so there should be no paywall). Treumann and Baumjohann [2016] correct their arXiv result and find the same normalization factor as the original paper by Jüttner [1911].

Livadiotis [2016] is more rigorous in its treatment giving the probability distribution: $$ P\left( \mathbf{p}, \Theta \right) = C_{o} \ e^{ - \Theta^{-1} \ \sqrt{ 1 + \sum_{i=1}^{\delta} \tfrac{\Theta}{\Theta_{i}} \left( \tfrac{p_{i}}{m \ c} \right)^{2} } } $$ where $\delta$ is the number of degrees of freedom, $\Theta_{i} = \tfrac{k_{B} \ T_{i}}{m \ c^{2}}$ is the i^th component of the normalized temperature, $k_{B}$ is the Boltzmann constant, and $\Theta = \tfrac{1}{\delta} \ \sum_{i=1}^{\delta} \Theta_{i}$, and $C_{o}$ is given by: $$ C_{o} = \pi^{\left( \tfrac{ 1 - \delta }{ 2 } \right)} \ 2^{-\left( \frac{ \delta + 1 }{ 2 } \right)} \ \left( m \ c \right)^{-\delta} \ \frac{ \sqrt{ \Theta } }{ K_{\tfrac{ \delta + 1 }{ 2 }}\left( \tfrac{ 1 }{ \Theta } \right) } \ \left[ \prod_{i=1}^{\delta} \ \Theta_{i}^{-1/2} \right] $$ where $K_{n}\left( x \right)$ is the Macdonald function (it is also a modified Bessel function), $m$ is the particle mass, and $c$ is the speed of light.

In the specific case of $\delta = 3$ and $\left\{ T_{i} \right\}_{i=1}^{3} = \left( T_{\parallel}, T_{\perp} \right)$, $C_{o}$ will reduce to: $$ C_{o} \rightarrow \frac{ \sqrt{ \Theta \ A } \ \beta_{\perp} }{ 4 \pi \ \left( m \ c \right)^{3} \ K_{2}\left( \tfrac{ 1 }{ \Theta } \right) } $$ where $A = T_{\perp}/T_{\parallel}$ and $\beta_{j} = \tfrac{ m \ c^{2} }{ k_{B} \ T_{j} }$.