particle physics - status of +4/3 scalar as explanation of $tbar t$ asymmetry

One of the early proposals for the Tevatron asymmetry on $t \bar t$ was a "fundamental diquark" with a charge (and hypercharge) +4/3, either in a triplet or a sextet colour. I am interested on the current status of this proposal:

Generically, has some particular model of been favoured in ulterior research?

For starters, you can see the short report in JHEP 1109:097,2011, at the end of section 2, points 5 and 6. I only become aware of the survival of this hypothesis after seeing http://arxiv.org/abs/1111.0477 last week.

thermodynamics - How can my water cool down more quickly?

I have a cup and I can only pour hot water inside, I wanna know whether the heat will dissipate more quickly with more water or less water? How about the occasion when my cup is well covered?

special relativity - Discreteness of Spacetime and Violation of Lorentz symmetry

It is usually said that existence of discrete spacetime violates Lorentz symmetry. What quantity is used to quantify such violation? I mean could someone points a reference for a derivation that shows such analysis.

My other question is: if Lorentz symmetry is violated, does that imply space-time is discrete? or not necessarily?)

Is there still no known origin of the law of inertia?

To quote Feynman at about the 21 minute mark of the first Messenger Lecture on The Character of Physical Law,

...that the motion to keep it going in a straight line has no known reason. The reason why things coast forever has never been found out. The law of inertia has no known origin.

This lecture was given in year 1964. I'm curious if there has been any progress since then to understanding the origin of the law of inertia. If yes, if a layman explanation can be provided.

Edit 1, adding the definition of scientific law for discussion in comments. From Kosso (2011, pp 8):

One more term should be clarified, ‘‘law’’. Theories differ in terms of their generality. The big bang theory, for example, is about a singular, unique event. It is not general at all, despite being about the entire universe. The theory of gravity, either the Newtonian or relativistic version, is very general. It is about all objects with mass and their resulting attraction. The most general theories, including the theory of gravity, are laws. In other words, laws are theories of a particular kind, the ones that identify whole categories of things and describe their relations in the most general terms. Laws start with the word ‘‘all’’, as in, All this are that, All massive objects are attracted to each other.

Being a law has nothing to do with being well-tested or generally accepted by the community of scientists. A theory is a law because of what it describes, not because of any circumstances of confirmation. And a theory is or is not a law from the beginning, even when it is first proposed, when it is a hypothesis. The status of law is not earned, nor does it rub off; it is inherent in the content of the claim. So neither ‘‘theoretical’’ nor ‘‘law’’ is about being true or false, or about being well-tested or speculative. ‘‘Hypothetical’’ is about that kind of thing.

See Kosso (2011) for the definitions of the terms Theory, Fact, and Hypothesis, if needed.

Edit 2, I acknowledge I do not know what definition Feynman held when using the term "law" in the Messenger Lecture (as I had quoted above). It seems he also referred to it as the principle of inertia (The Feynman Lectures on Physics, Volume I, Chp 7, Sec 3 - Development of dynamics):

Galileo discovered a very remarkable fact about motion, which was essential for understanding these laws. That is the principle of inertia—if something is moving, with nothing touching it and completely undisturbed, it will go on forever, coasting at a uniform speed in a straight line. (Why does it keep on coasting? We do not know, but that is the way it is.)

An interesting side note, according to user Geremia (link):

Galileo, Newton, or even the medieval physicist Jean Buridan (1295-1358), who developed the notion of impetus, were not the first to discover the law of inertia.

The first was John Philoponus ("The Grammarian"), who lived in the late 5th and 2nd ½ of 6th century A.D.

Edit 3, I agree that no "Laws" of physics have a "known" reason. But that is not the point of my question. My question is whether or not any progress has been made on understanding the origin (i.e. the mechanisms underlying) the law of inertia. For example, Darcy's Law can be derived from the Navier–Stokes equations. The Navier-Stokes equations arise from applying Isaac Newton's second law to fluid motion. I suppose this regression to more fundamental mechanisms or reasons can go ad infinitum (as explained here by Feynman. He also addresses the "why" question, Aaron Stevens).

Edit 4, I am not making Feynman into a Pope nor am I appealing to his authority. He has simply made a statement about the current understanding of the law of inertia. Of course, I attributed his statement to him. I then asked a question about his statement. I made no assumption as to whether his statement was correct or not. If anyone cared to make an answer pointing out his statement is incorrect I would be grateful to hear it.

visible light - Is white a single color?

We have seen the Newton disc, when it is rotated it produces white color. So I want to know whether white exists as a single color or is that a combination of multiple colors?

homework and exercises - How many atoms are there in our solar system?

Including all objects gravitationally bound to the Sun, how many atoms are there in our solar system?

Answer

Giving that most of the solar system's mass is concentrated in the sun, you may say that the order of magnitude of the number of atoms in the sun and in the solar system is the same. Thus, we may find this number by using the sun's mass and dividing it by the hydrogen's mass, because the sun is composed of it almost entirely: $$\frac{M_s}{M_h}=\frac{2\cdot10^{30}}{1.67\cdot10^{-27}}=1.2\cdot10^{57}$$

So the order of magnitude of the number of atoms in the solar system is $10^{57}$.

gravity - Negative Energy and Wormholes

Apparently to create wormholes you need negative energy/matter. Say you had negative matter/energy, how would it be applied towards making a wormhole?

Answer

There are two related but distinct questions:

1. 
   

   how do you keep a wormhole stable?

   
   


2. 
   

   how do you make the wormhole in the first place?

   
   
   

Courtesy of Matt Visser we can give one answer to the first question. Matt's example is to make the wormhole cube shaped, and in that case all you need to do is construct a cube from string i.e. the twelve edges of the cube are made from string. However the string would have to have a negative tension, and indeed it would have to have the ridiculously high negative tension of $−1.52 \times 10^{43}$ Joules/metre. This is where your exotic matter comes in since the tension in any string made from normal matter would always be positive.

The second question is harder. Matt's analysis applies to a time independant wormhole, i.e. one that has existed for an infinite time. Constructing your cube of exotic string would warp spacetime in the manner required for a wormhole, but calculating what happens as you tie the strings into a cube is probably impossible at present.

Response to comment:

This is going to be a bit hard to explain, but the space inside the cube doesn't exist. It isn't part of the manifold on which the universe exists. If you travelled towards the cube you wouldn't hit anything - you'd just keep going without feeling anything as you past where it's wall is, but now you'd be travelling in the other region of spacetime on the other side of the wormhole.

Re your comment it doesn't sound like there's a lot of control, the wormhole Matt describes is not the same as the sort of wormhole Sci-Fi writers use to allow interstellar travel. As far as I know there is no theoretical support for the interstellar travel type wormhole. The wormhole Matt describes connects two regions of spacetime but makes no statement about the global topology, so the region of spacetime the other side of the wormhole need not be, and almost certainly isn't, some distant region of the universe around us. The wormhole does not allow FTL travel to e.g. Alpha Centauri. It just allows travel (at up to the speed of light) to the new region of spacetime on the other side of the wormhole.

mathematics - 5 weights on modified balance scale

This question is follow up version of What's the fewest weights you need to balance any weight from 0,5 to 40 pounds, on a modified balance scale with one arm twice as long as the other.

     
^{(Picture courtesy of Jamal Senjaya)}

In the original question, the answer is 6 weights and lots of solutions are possible, it is somehow shown that it is not possible to get 5 weights from 0 to 40.

In this question, you have 5 weights only (any positive value for each weight is possible) to weigh from 0 to X pounds exactly with the 0.5 pound increment as the original question.

What is the maximum value of X? and you need to provide all 5 weights' weight with it.

Answer

This answer originally reported a hand-made X = 22 pounds solution but, while I was finding better solutions and preparing an analysis, Nopalaa’s program found a likely-maximal solution that can balance up to X = 38 pounds. So this answer now applies my analysis to Nopalaa’s solution.

Here are Nopalaa’s weights and the only two weighings that absolutely require the mystery weight to sit on the long side of the scale, where everything weighs double what it would on the short side.

              2   5.5   8.5   9   9.5

                 Short side        .        Long side (multiply by 2)
                                   .
              Reference weights    .    Mystery weight   Reference weight
                                   .
                 8.5 + 9.5         .          3.5      +      5.5
                                   .
             2 + 5.5 + 9 + 9.5     .          13
                                   .

                                   .  (Every other weighing can be made with
                                   .   the mystery weight on the short side.)  

That mystery weights 3.5 and 13 require placing them on the long side, while all other weighings do not, is reflected in a tree diagram of weight combinations, a portion of which is previewed here.

  M is on                       X                                     X
short side       'v'v'v'v'v'v'v'''v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'''v'v'v'v'v'v'v'v'v'
                  0   1   2    3.5    5   6   7   8   9  10  11  12  13  14  15  16  17
  M is           ^^^.^^^^^^^^^^^^^.^^^^^^^^..^^^^.^^^^^^^^^^....^^^.^^^.^..^^......^...^
   on   R = 2    /\/_\/\\/|\\|/||/_|||\///\__\|||_///|||\///____|||_///_|__\/______|___/

  long    = 5.5  _|__|____/__/_\\\_|||________///_\/_|||________///_____|__________/
  side    = 8.5  _|__\_____________/||_______________//|________________/
          = 9    _|_________________/|_________________/
          = 9.5  _|__________________/
                  0                                                || 1 space = 1/4 pound

The story behind the diagram begins with a table of six ways that each reference weight R can contribute to balancing mystery weight M.

                   Amount of mystery weight M balanced by reference weight R


                            R is on the       R is       R is on the
                             short side     not used     long side

   M is on the short side       - R            0           + 2R

   M is on the long side       + R/2           0            - R

The choices shown in this table are diagrammed with a pair of decision trees that meet at a number line. The top tree grows downward as it represents the possibility that M is on the short side; the bottom tree grows upward as it represents the possibility that M is on the long side.

 MAGNIFIED:
                             R is on same      R is      R is on other
            Branches when    (short) side    not used    (long) side
            M is on the               __________|_________________________
            short side               /          |                         \
                                    /           |                          \

            Number line  .  .  .  -R  .  .  .  +0  .  +R/2  .  .  .  .  .  +2R  .  .  .  .

                                    \           |      /

            Branches                 \__________|_____/
            when M is on                        |
            the long side    R is on same      R is      R is on other
                              (long) side    not used    (short) side


 UNMAGNIFIED:
                    M is on        _____________|___________________________
                  short side      /             |                           \
                                  v'''''''''''''v'''''''''''''''''''''''''''v

 (e.g, R = 7)                    -7             0    +3.5                  +14
                                  ^.............^......^
                    M is on       \_____________|______/
                   long side                    |                  || 1 space = 1/4 pound

Here is the tree diagram for 2 reference weights — .5 and 1 pound — that can balance up to X = 2 pounds.   Note that M = 1, 1.5 or 2 pounds may balanced only with M on the short side while M = .5 requires M to sit on the long side.

                    M is on                      .5  1.5          (Only the leaves of the
                  short side              v'v'v'v'''v'v'v'''v      top tree are displayed

                                                0 : 1   2          because that tree is
                    M is on               ^.^^^.^^^^               a double-sized mirror
                   long side     R = .5   \_|/\_\/|/               copy of the lower tree.)
                                 R =  1     \___|_/
                                                0                  || 1 space = 1/4 pound


 SUPER-MAGNIFIED,
  WITH FULL TOP:                         0
                            _____________|_____________________________

          R = 1            /    -1       |               +2            \
                          /         _____|_____________                 \
  M is             ______/_________/____ |      +1     \           ______\______________
   on     R = .5  / -.5  |   +1   /     \|              \         / -.5  |      +1      \
  short          /       |       /       \               \       /       |               \
  side           v   '   v   '   v   '   v   '   '   '   v   '   v   '   v   '   '   '   v

                                         0      .5       1      1.5      2              (3)

  M is           ^   .   ^   ^   ^   .   ^   ^   ^   ^

   on            \  -.5  |   /   \       \   /-.5|   /
  long    R = .5  \______|__/     \ -.5  |\_/____|__/
  side                   \         \_____|_/     /
                          \     -1       | +.5  /
          R = 1            \_____________|_____/                   || 1 space = 1/16 pound
                                         |
                                         0

Having these diagrams in mind made it easy to manipulate the resultant number line alone, without rendering full trees, and easy enough to find a solution with X = 29 by hand.   With the above layouts as reference, though, here instead are Nopalaa’s weights balancing up to X = 38 pounds.   Much detail is obscured by overlapped branches but it is clear that many irregular mystery-weight gaps exist before reaching decision tree leaves.

              (The top line displayed has the leaves of the top tree (M on short side),
               identical to the bottom tree's leaves, but in reverse order and stretched
               twice as wide.  The bottom tree's streak of 1/4 weights from -19 to -6.75
               corresponds to the top longest streak of 1/2 weights, from 13.5 to 38.)

 M is on                                                                                                                                                               X                                     X
short side     v'''''''v'''''''''''''v'v'''''v'''v'v'v'''v'v'v'v'''''''v'v'v'v'v'v'v'v'v'v'''v'v'v'v'''''v'v'v'v'v'v'v'v'''v'v'v'v'v'v'v'v'v'v'v'v'v'''v'v'v'v'v'v'v'v'''v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'''v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'v'''v'v'v'v'v'''v'''''v'v'''v'''v'v'v'''v'v'''''v'''v'''v'v'''''v'''''''''''''''v'''''''''''''''''''v'''''''v'''''''''''''''v
                                                                           -19                                            -6.75                          0   1   2    3.5    5   6   7   8   9  10  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  30  31  32  33  34  35  36  37  38
  M is         ^.......^...^.........^.......^..^^.^.^..^^.^^^.^.^^..^.^^^^^.^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.^^^^^^^^^^^^^^^^^^.^^^^^^^^.^^^^^^^^^^^^^.^^^^^^^^..^^^^.^^^^^^^^^^....^^^.^^^.^..^^......^...^
   on   R = 2  \_______|___/         \_______|__\/_\_\__||_|//_/ /\__\_\\\|\_|/|||\/\/\/\\|\||/||//\//\/\\/|\\|/||/|/|\/\/\/\\|_|\/||||\///\\/\|||_/|/|//\/_\/\\/|\\|/||/_|||\///\__\|||_///|||\///____|||_///_|__\/______|___/

  long    = 5.5        \_____________________|__________/\_\_\____________\__\_|\|_|______/_/\/\|\_|__|____/__\\\/\|\|_|______/_\_/\|||_|||____///_/\\|__|__|____/__/_\\\_|||________///_\/_|||________///_____|__________/
  side    = 8.5                              \_________________________________|_\_\____________/__\__\____________|_|_\____________/|/_|\\___________/__|__\_____________/||_______________//|________________/
          = 9                                                                  \___________________________________|_\_______________/__\________________|_________________/|_________________/
          = 9.5                                                                                                    \_____________________________________|__________________/
                                                                                                                                                         0

(scroll sideways for most of diagram)                              || 1 space = 1/4 pound

quantum field theory - Expected consequence of the term $theta F^{munu}tilde{F}_{munu}$ in the Standard Model

There is a coupling in the standard model of the form $\theta F^{\mu\nu}\tilde{F}_{\mu\nu}$ where $F^{\mu\nu}$ is the QCD field strength. I've read that $\theta<10^{-8}$, which is speculated to be related to Peccei-Quinn symmetry.

1. 
   

   What would be the consequences of this term if $\theta$ were not small? In particular, I know that this term is CP-violating. How will this CP-violation be manifested in experiments?

   
   
   


2. 
   

   How do we know $\theta<10^{-8}$?

   
   

Answer

To answer point $2$, $\theta$ is derived from the neutron elecric dipole moment that is measured to be like $<10^{-18}\,\text{e}\cdot\text{m}$. You can prove that the dipole moment is proportional to $\theta$ and from that you can get your upper limit for $\theta$. See Donoghue, Golowich, Holstein - "Dynamics of the Standard Model", pages 44-45 for more details.

As far as your first question is concerned, a value of $\theta$ different from $0$ would make $CP$ not a symmetry for QCD, but I don't really know what effect this could have in detail.

Why is the center of buoyancy located at the center of mass of the displaced fluid volume?

Consider Archimedes' principle

Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object.

I do not understand why the center of buoyancy is the center of mass of the volume of the displaced fluid.

In principle it could be another point, for istance the center of mass of the object (not the displaced fluid) or maybe a point of the displaced fluid but not the center of mass.

Is there an simple way to understand why the center of buoyancy located in the center of mass of the displaced volume of fluid?

Answer

Let us suppose we remove the object and fill the space left (in the fluid) with the same fluid. Assume this portion of fluid become solid without changing its volume or density. It will be in equilibrium with the fluid. Now suppose the buoyancy force on this solidified portion is off the center of mass. This would imply non vanishing resultant force or torque and the solidified portion would not be in equilibrium. A kind of inverse argument can also be given. Suppose the center of buoyancy coincides with the center of mass of an object immersed in a fluid. Then you would never observe resultant torque (by means of rotations) on any object. And that is empirically not true.

quantum field theory - What's wrong with this QFT thought experiment?

In quantum field theory, the propagator $D(x-y)$ doesn't vanish for space-like separation. In Zee's book, he claims that this means a particle can leak out of the light-cone. Feynman also gives this interpretation.

What is wrong, then, with this thought experiment:

Bob and Alice synchronize their watches and space-like separate themselves by some distance. Bob tells Alice that at exactly 3 o'clock, he intends to test his new particle oven, which claims to make a gazillion particles. Since Bob's oven makes so many particles, there is a pretty good chance that Alice will detect a particle right at 3 o'clock. If she detects such a particle, Bob will have transmitted a piece of information (whether or not the oven works) instantaneously - way faster than the speed of light.

special relativity - If I am travelling on a car at around 60 km/h, and I shine a light, does that mean that the light is travelling faster than the speed of light?

The title says it all.

If I was on a bus at 60 km/h, and I started walking on the bus at a steady pace of 5 km/h, then I'd technically be moving at 65 km/h, right?

So my son posed me an interesting question today: since light travels as fast as anything can go, what if I shined light when moving in a car?

How should I answer his question?

Answer

If I was on a bus at 60 km/h, and I started walking on the bus at a steady pace of 5 km/h, then I'd technically be moving at 65 km/h, right?

Not exactly right. You would be correct if the Galilean transformation correctly described the relationship between moving frames of reference but, it doesn't.

Instead, the empirical evidence is that the Lorentz transformation must be used and, by that transformation, your speed with respect to the ground would be slightly less than 65 km/h. According to the Lorentz velocity addition formula, your speed with respect to the ground is given by:

$$\dfrac{60 + 5}{1 + \dfrac{60 \cdot 5}{c^2}} = \dfrac{65}{1 + 3.333 \cdot 10^{-15}} \text{km}/\text h \approx 64.9999999999998\ \text{km}/\text h$$

Sure, that's only very slightly less than 65 km/h but this is important to your main question because, when we calculate the speed of the light relative to the ground we get:

$$\dfrac{60 + c}{1 + \dfrac{60 \cdot c}{c^2}} = c$$

The speed of light, relative to the ground remains c!

reflection - Can we keep a photon bouncing of mirrors for a long time (eg days)?

If a photon is sent into a box with perfect mirrors aligned in a way that it won’t ever be reflected out, is it possible to hold the photon for extended lengths of time with its quantum properties intact, and measure it when we want by opening a path into the detector?

quantum mechanics - Why do we think Spin is angular momentum as opposed to some other quantity?

What leads us to the conclusion that spin is angular momentum? Could it not be some other quantity? Sorry if this is a rookie level question.

forces - Work done against gravity

The work done against gravity is $mgh$, well at least that's what my textbook says. I have a question: I can apply a force say 50N, so total work done = $mgh + mah$. Where $ma$ = Force. But the truth is irrespective of the force applied, the work done against gravity is always $mgh$. Why?

For example, when I move an object with a force, the work done is more, so work depends on the Force. But in case of gravity it always depends upon the weight

Answer

If I take a mass $m$ and apply a force $F$ (greater than $mg$) to it for a distance $h$ upwards then I will do work of:

$$W = Fh \tag{1}$$

The force $F$ has to be greater than the force due to gravity, $mg$, or the object won't move upwards, so let's write the force I apply to the mass as:

$$F = mg + F'$$

then equation (1) becomes:

$$\begin{align} W &= (mg + F')h \\ &= mgh + F'h \end{align}$$

and the first term $mgh$ is work done against gravity while the second term is the work done to increase the velocity of the mass i.e. after the distance $h$ the velocity of the object will be given by:

$$\tfrac{1}{2}mv^2 = F'h$$

or:

$$v = \sqrt{\frac{2F'h}{m}}$$

So if you apply any force $F$ over a distance $h$ then subtract off the increase in the kinetic energy you'll be left with an amount of energy equal to $mgh$. That's why the work done against gravity is always $mgh$.

group theory - Invariant tensors in a general representation and their physical meaning

I'm trying to use tensor methods to find invariant elements of representations. Specifically I'm looking at representations of $SU(5)$.

I can show that the invariant element in $5\otimes\bar{5}$ (or equivalently the $1$ in the $5\otimes\bar{5} =$1$\oplus 24$ representation?) is $\delta_i^j$: this is straightforward because $X \in SU(5)$ acts by $[X\delta]_j^i = X_\lambda^i\delta_j^\lambda - X^\lambda_j\delta^i_\lambda = 0$.

1. 
   

   I'm wondering how we find the $1$ more generally. E.g. how do we find the invariant tensor in a decomposition $5\otimes10\otimes10$ etc. is there a general method for this?

   
   


2. 
   

   Secondly I'm wondering what is the physical content of a $1$ representation generally?

   
   


3. 
   

   Thirdly I'm trying to find the branching of such tensors under various subgroups of $SU(5)$.

   
   

Answer

Short answers

1. Apply the Young calculus (per ACuriousMind's suggestion in the comments). For finding the multiplicity of the trivial representation in a tensor product of representations of $SU(n)$, note that each irreducible representation $D$ of $SU(n)$ has a unique conjugate irreducible representation $\bar D$ such that the Young calculus allows $D\otimes \bar D$ to include a rectangular Young diagram of full height $n$ (which is invariant under $SU(n)$). As suggested by Wooster in the comments, in order for $D\otimes\bar D$ to accommodate such a Young diagram, the joint symmetry types of $D$ and $\bar D$ must be compatible with (i.e. have nonzero overlap with) some tensor power/outer product of the fully antisymmetric tensor $\epsilon_{i_1,\dots,i_n}$. In $SU(n)$, Young diagrams of this type correspond to the invariant or trivial representation.


2. If you view the $n$ dimensional representation of $SU(n)$ as a sort of single-particle Hilbert space, then the invariants formed from tensor products of this representation can be thought of as '$SU(n)$-neutral' many-particle states. More abstractly you could interpret representations $SU(n)$ very differently as a sort of gauge theory where 'particle number' is gauged.


3. The branching problem has been solved in a number of special cases. For example, there is an explicit formula for the branching of representations for $SU(n)\rightarrow SU(n-1)$. For low-rank representations, the Young calculus is a powerful general-purpose tool for determining branching. One strategy is to decompose the fundamental representation of $SU(5)$ into representations of $H\subset SU(5)$, and then iteratively compare how tensor products decompose. As an example, consider the problem of decomposing rank 2 representations of $SU(5)$ into representations of $SU(2)\subset SU(5)$. The fundamental (vector) representation of $SU(5)$ breaks up as $5_5 \rightarrow 2_2\oplus (1_2 \oplus 1_2 \oplus 1_2)=2_2\oplus 3\times 1_2$. Next, we have $5_5\otimes 5_5=10_5^A\oplus 15_5^S \rightarrow (2_2\oplus 3\times 1_2)\otimes (2_2\oplus 3\times 1_2)=(2_2\otimes 2_2)\oplus 3 \times (2_2\otimes 1_2) \oplus 3\times (1_2\otimes 2_2)\oplus 9\times 1_2=3_2^S\oplus 1_2^A\oplus 3\times 2_2^S\oplus 3\times 2_2^A\oplus 3\times 1_2^A\oplus 6\times 1_2^S$. Grouping terms according to symmetry, we see that $10_5^A\rightarrow 4\times 1_2^A\oplus 3\times 2_2^A$, and $15_5^S\rightarrow 3_2^S\oplus 3\times 2_2^S\oplus 6\times 1_2^S$.

Background on the Young calculus

In physics, irreducible representations are often labeled by their dimension. This notation is compact, but obscures the underlying algebraic structure. Young diagrams provide a more transparent notation based on a deep result, Schur-Weyl duality, that relates irreducible representations of $GL(n)$ to those of the permutation group $S_r$ on $r$ symbols (here $r$ is the rank of a tensor representation). Ultimately, Schur-Weyl duality comes from the fact that finite dimensional representations of $GL(n)$ can all be constructed out of tensor products of a single fundamental representation (this is the analogue of the $\frac{1}{2}$ representation of $SU(2)$ from elementary quantum mechanics). For now, all you need to know is that there is a 1-1 correspondence between representations of $GL(n)$ and the set of all Young diagrams with maximum height $n$. Young diagrams greatly simplify the task of decomposing tensor products of representations of $GL(n)$, as well as many subgroups of $GL(n)$ with 'similar' structure (e.g. $U(n)$, $SL(n)$, $SU(n)$, etc.). They also make it easier to notice certain partial solutions to the branching problem, such as determining how representations of $GL(n)$ decompose into representations of $GL(n-1)$.

Let $r$ be a positive integer. Young diagrams are associated with partitions of $r$: sequences of integers $\lambda_1\geq\lambda_2\geq\cdots\geq\lambda_k\geq 0$ such that $\sum_j \lambda_j = r$. Given a partition $(\lambda_1,\dots,\lambda_k)$, draw a Young diagram as follows: (i) draw a horizontal row of $\lambda_1$ boxes, (ii) draw a horizontal row of $\lambda_{j+1}$ boxes starting from the left below the $j$th row, $1\leq j

As mentioned above, each diagram with at most $n$ rows corresponds to an irreducible representation of $GL(n)$. Again, this fact is useful because $GL(n)$ is closely related to many other groups of interest in physics. A Young diagram can be thought of as an efficient way to keep track of the symmetrization of tensor indices: after placing tensor indices $i_1$ through $i_r$ in the squares of a Young diagram, the corresponding irreducible tensors are symmetric (even) under permutations that preserve rows, and antisymmetric (odd) under permutations that preserve columns. There is a general formula for the dimension of a $GL(n)$ representation labelled by a Young diagram, but in practice the dimension can be computed more efficiently for low rank using the decomposition rules for tensor products, to be explained now.

The tensor-product decomposition rules for $GL(n)$ follow from a special sort of 'inverse branching' problem for the permutation group $S_r$. In the end, one obtains the following rules:

Let $L=(\lambda_1,\dots,\lambda_k)$ and $M=(\mu_1,\dots,\mu_\ell)$ be two irreducible representations of $GL(n)$, given by their Young diagrams.

1. Draw the diagrams corresponding to $L$ and $M$. In the diagram for $M$, choose a distinct symbol for each row (e.g. $a$ for the first row, $b$ for the second, $c$ for the third, etc.), and write the symbol in each box of that symbol's row.


2. Find all ways in which $\mu_1$ $a$'s can be added to the Young diagram of $L$ so that no two $a$'s appear in the same column, and the resulting graph is another Young diagram (i.e. the length of rows is nonincreasing).


3. For each larger Young diagram obtained above, find all the ways in which $\mu_2$ $b$'s can be placed with no two in the same column, together with an additional constraint: when reading the added symbols from right to left, top to bottom, the number of $a$'s that have been read must match or exceed the number of $b$'s that appear at any step.


4. Repeat for the $\mu_3$ $c$'s, then $\mu_4$ $d$'s, etc., except that now when imposing the last constraint mentioned in step 3, the recorded number of $c$'s cannot exceed the number of $b$'s (etc.).



5. The tensor product $L\otimes M$ decomposes into a direct sum of all Young diagrams obtained this way.

As an example, consider the following tensor product:

To decompose this, first we label the second diagram with $A$'s and $B$'s:

Next we find all the ways of adding $A$ blocks, and then $B$ blocks, to the Young diagram of $L$ according to the above rules:

$\rightarrow$

Note that diagrams like the ones below are not permitted:

The first two diagrams contain two $A$'s in the same column, while the last one isn't allowed because when reading the added symbols right-left top-bottom, we obtain $ABBA$, which has more $B$'s than $A$'s after the third letter (this is from the rule stated in step 3).

Now it turns out that all irreducible representations of $GL(n)$ remain irreducible when restricted to $SU(n)$. However, some representations of $GL(n)$ that were previously distinct become isomorphic. This comes from the fact that it is possible for two irreducible representations of $GL(n)$ to differ from each other only by powers of the determinant homomorphism: $\det(gh)=\det(g)\det(h)$. Once the determinant is set to unity in $SU(n)$ (or $SL(n)$ for that matter) this distinction vanishes, and representations that only differed in their power of $\det(g)$ are isomorphic. Fortunately there is a simple way to account for this redundancy: under $SU(n)$, the representations $(\lambda_1,\dots,\lambda_n)$ and $(\lambda_1+s,\dots,\lambda_n+s)$ are equivalent. To account for the redundancy, we simply choose $s=-\lambda_n$ and label representations of $SU(n)$ with only $n-1$ non-increasing integers instead of $n$. A consequence of this is that if $\lambda_1=\lambda_2=\dots=\lambda_n$, then $[\lambda_1,\dots,\lambda_n]\cong[0,0,\dots,0]$: tensors corresponding to rectangular Young diagrams are invariant under $SU(n)$. For finding the multiplicity of the trivial representation in tensor products, you can check from the decomposition rules that each irreducible representation $V$ of $SU(n)$ has a unique conjugate $\bar V$ such that $V\otimes\bar V$ includes the trivial representation.

References for further reading:

Group Theory and its Application to Physical Problems (by Morton Hamermesh): chapters 7 and 10.

Theory of Group Representations and Applications (A. Barut & R. Raczka): chapters 7 and 8.

electromagnetism - Physical meaning of wave vector?

Recently I have been studying the Rayleigh-Jeans derivation, and for this problem we need the three dimensional wave equation for electromagnetic radiation:

$$\frac{\partial^2{E}}{\partial{x}^2} + \frac{\partial^2{E}}{\partial{y}^2} + \frac{\partial^2{E}}{\partial{z}^2} = \frac{1}{c^2}\frac{\partial^2{E}}{\partial{t}^2}$$

For our blackbody radiation problem, we need to define a field, E, that is zero at the boundaries of the cavity it is contained within. From a simple perspective, I think that we can use a sinusoidal functional form with no phase shift (correct me if I'm wrong here):

$$\vec{E}(x,y,z,t)=\vec{E_m}\sin(\vec{k}\dot{}\vec{r}-\omega{}t)$$ $$\vec{E} = \vec{E_m}\sin(k_x x +k_y y + k_z z - \omega{}t)$$

My notation here is that k is the wave vector and Em is the magnitude and direction of the wave. Plugging this into the wave equation gives us the following:

$$k_x^2 +k_y^2 + k_z^2 = \frac{w^2}{c^2}$$

Which implies that the magnitude of the wave vector is equal to the wave number, k = w/c. In addition, we should have corresponding variables for the modes of the standing wave:

$$n_x, n_y, n_z$$ and "wavelengths": $$\lambda_x, \lambda_y, \lambda_z$$

We know that there is physically only one wavelength, lambda, so what is the meaning of the x, y and z components? I did some math on this earlier today and if the following is true:

$$n^2 = n_x^2+n_y^2+n_z^2$$ $$k^2 = k_x^2+k_y^2+k_z^2$$

I found that:

$$\lambda^2 \ne \lambda_x^2 + \lambda_y^2 + \lambda_z^2$$

I feel more comfortable with the notion of nx, ny and nz, because they are the number of "half" wavelengths (x, y, or z components) that can fit into the respect x, y and z directions.

In addition, what do the various components of the wave vector, k, mean?

Answer

As the name indicates, $\vec k$ is a vector representing the positional sinusoidal dependence of a plane wave propagating in its direction in space. As such, $\vec k$ can be decomposed into its components $k_x$, $k_y$ and $k_z$ into the x, y and z directions. Therefore the plane wave can be written in exponential form $$\vec{E}(x,y,z,t)=\vec{E_m}\exp(i\vec{k}\dot{}\vec{r}-i\omega{}t)$$ and thus as as a product of "waves" $$\vec{E}(x,y,z,t)=\vec{E_m}\exp(ik_x·x)\exp(ik_y·y)\exp(ik_z·z)\exp(-i\omega{}t)$$ Thus, i.e, when you have a $\vec k$ in z-direction and turn it a little so that it has a small $k_x$ component, you can have an arbitrary long "wavelength" in x-direction $\lambda_x=2\pi/k_x$. The relation $|\vec{k}|=(2\pi/\lambda)=\omega/c$ holds only for the wavelength $\lambda$ in the propagation direction and because of $$k^2=|k|^2 =k_x^2+k_y^2+k_z^2=(2\pi/\lambda_x)^2+(2\pi/\lambda_y)^2+(2\pi/\lambda_z)^2$$ in general $$\lambda^2 \ne \lambda_x^2 + \lambda_y^2 + \lambda_z^2$$ will hold.

cosmology - Is the de Sitter Universe equivalent to the static Einstein Universe?

The de Sitter universe is a flat exponentially expanding universe with a cosmological constant $\Lambda$ and no matter.

Einstein's static universe also has a cosmological constant $\Lambda$ but it also has matter and it is static. Can we make some coordinate transformation to the de Sitter universe to show equivalence of the universes?

Answer

The de Sitter universe is a [...] universe with [...] no matter.

Einstein's static universe [...] has matter

Can we make some coordinate transformation to the de Sitter universe to show equivalence of the universes?

Coordinate transformations cannot change the zero energy-momentum tensor of a de Sitter universe into the non-zero energy-momentum tensor of an Einstein uiverse.

Fluid Dynamics pressure and velocity

I am trying to understand the fundamental concepts of fluid dynamics. Lets say I have

1.The larger end has a diameter of 8cm and Area 50.26cm2.

2.The small side has a diameter of 3cm and Area 7.0685cm2.

3.The water jet exerts a force of 87N at the smaller end.

4. Water Entering the pipe is art 20degrees C

I know that :

1 - There's a change in cross-sectional area: A1 > A2

2 - Thanks to conservation of mass, (1) implies V2 > V1

3 - Thanks to Bernoulli, (2) implies p2 < p1

I am assuming a steady flow and no frictions and that the liquid is water, how can I find the velocity at the smaller end of the nozzle ?

I have calculate the Pressure using P = F/A to get a P2 of 12.3N/cm2.

However I am getting stuck in calculating P1 and V2

I have tried using Bernoulli's Equation however, I do not know P1 which is leaving me with more than one unknown.

Any Help would be appreciated.

R

Answer

You can use the continuity equation in the form $A_1 V_1=A_2 V_2$. However, there's not enough information given in your problem. You need at least one more quantity (a flow rate, velocity, or a pressure somewhere). The temperature of the water is spurious information.

I will note that the formulation of this problem looks very strange, not to use a harsher term: There is no "jet" in this problem, and it is entirely unclear what the phrase "the water jet exerts a force of 87N at the smaller end" could possibly mean.

Update after OP was updated with the final image:

Ahh, this is an entirely different problem now. You had left out crucial information. For this problem you need to use conservation of momentum in finite volumes to determine your flow rate. Choose a control volume that has inflow in the horizontal direction from the jet emanating from the pipe. You can use conservation of momentum in the horizontal direction to link the force you were given to the flow rate. This provides the missing information above.

newtonian mechanics - Understanding the different kinds of mass in gravity

On this site, the Phys.SE question Is there a fundamental reason why gravitational mass is the same as inertial mass? has been asked. See also this Phys.SE question. The 'answer' provided on this forum has been that the curvature of spacetime explains both. The answer is still cryptic for me as I am more a concrete thinker.

Newton said $F=ma$. I can use this formula to measure inertial mass. Experimentally I can measure the motion of an object while applying a constant force to it. Newton also said $F=\frac{GMm}{r^2}$. In this case, what simple experiment will allow the measurement of gravitational mass?

cipher - The Writing's on the Wall

In the course of your travels, you stumble onto a cryptic message:

You lament that the three bottom lines are badly degraded and nearly unreadable. You're certain that the information they contain is necessary to decode the message.

The smaller squares scattered throughout the message intrigue you. You're positive they're not punctuation. They remind you of something else from your college days.

It also occurs to you that the symbols' square shape is somehow significant.

After contemplating the problem for a while, inspiration strikes and you're able to decode the message into a short phrase. In light of your circumstances, the phrase deeply depresses you. You carry on with your journey anyway, your heart heavy.

What is the decoded message?

Answer

Taking the lines with squares on them, we get this message:

$(K(AK^T)^{-1}+E^TH^T(HKO((AO)^{-1}C^T+O^{-1}K^{-1}L^T)+T^T))^T+S$

Superscript T has a specific meaning: it's the transpose of a matrix. Assuming letters represent matrices, the bottom three lines can be guessed:

• $K$ is [symme]tric



• $H$ an[d] $A$ a[r]e [orthog]onal


• $I$ i[s the] identi[ty]

This means that $K=K^T$, $A^{-1}=A^T$, and $H^{-1}=H^T$. Also, $I$ can be multiplied anywhere without changing the result. Now we can evaluate the expression using matrix mathematics:

$(K(AK^T)^{-1}+E^TH^T(HKO((AO)^{-1}C^T+O^{-1}K^{-1}L^T)+T^T))^T+S\\ =(K(AK)^{-1}+E^TH^T(HKO(O^{-1}A^{-1}C^T+O^{-1}K^{-1}L^T)+T^T))^T+S\\ =(KK^{-1}A^{-1}+E^TH^T(HK(OO^{-1}A^{-1}C^T+OO^{-1}K^{-1}L^T)+T^T))^T+S\\ =(A^{-1}+E^TH^THK(A^{-1}C^T+K^{-1}L^T)+E^TH^TT^T)^T+S\\ =(A^T+E^TKA^{-1}C^T+E^TKK^{-1}L^T+E^TH^TT^T)^T+S\\ =(A^T+E^TK^TA^TC^T+E^TL^T+E^TH^TT^T)^T+S\\ =(A^T)^T+(E^TK^TA^TC^T)^T+(E^TL^T)^T+(E^TH^TT^T)^T+S\\ =A+CAKE+LE+THE+S\\ =THE+CAKE+IS+A+LIE$

So the message is "The cake is a lie."

atoms - Can a standing wave exist on a spherical surface?

I've often seen the DeBroglie wave illustrated by a two dimensional surface as a standing wave, but then the 'electron cloud' surrounding an atom is hardly two dimensional and furthermore held to the uncertainty principle. So I'm having trouble visualizing how electron 'shells' or 'clouds' might appear from a DeBroglie perspective. Do DeBroglie waves actually propagate around the nucleus of an atom as a spherical standing wave?

More generally I'm having trouble visualizing or conceiving standing waves with dimensionality any higher than two for any system. Do standing waves exist in any physical system along a spherical, closed surface?

As a hypothetical example I can imagine a perfectly spherical, gravitating body (a planet) covered in an ocean with no land mass, and no atmosphere. On such a planet imagine a comet striking the ocean. Waves would radiate outward in circular rings and eventually interfere with themselves, but I can't see that they would ever be able to reinforce as standing waves by any means. Doesn't the geometry of a sphere forbid standing waves?

I'm not even sure how I would approach this from a mathematical analysis.

logical deduction - 100 Cards with numbers in a room

The scenario is like this:

There's a room, inside the room there are 100 cards with numbers [1,100] on them (No duplications), the cards are ordered randomly.

Two persons stand outside the room, They can decide on a tactic before the following:

• One person enters the room, he can see the order, he can make one switch between places of two numbers. (e.g: (3,1,2) ---switch 3 and 1---> (1,3,2)).


• After the switch (Or not switching) the person flips the cards and exits the room. (Other than the switch, the person does not change the order)


• The second person enters the room, A number between [1,100] is called out of the void, he has 50 guesses to get that number.

A few notes:

• The cards are in a perfect line, you can't order them vertically or something like that.


• The cards are identical, There are no particular features for specific cards or something like that.


• You can't leave anything in the room (items / human parts / ...)


• The two persons have no idea about the number that will be called.


• A guess = Flips a particular card.


• This is not the same as the prisoners, Since the first person doesn't know about the number the second person is supposed to find.

Answer

This is the same as 100 Prisoners' Names in Boxes

The algorithm is... Mentally number the cards 1 to 100 from left to right for example. turn the card with the (mental) number that you are supposed to find. If you found it, you are done.
If not, turn the cart with the (mental) number same as on the card you just turned.
The first person that goes into the room needs to make sure that there are no cycles more than 50 cards.
And this can be done easily since it can be max 1 cycle that is 51 or more. You can just break that cycle by changing 2 cards.
If you make sure there are no 51+ cards cycle you are always sure that you start in the right cycle and get to the card you need in 50 guesses or fewer.

material science - Why doesn't the Young's modulus change when an alloy is strengthened?

There is one thing about Young's modulus that I find unexpected and confusing.

When certain solid materials, pure metal, steel or an alloy of a certain composition, gets strengthened by cold working or by heat treating, the Young's modulus stays exactly the same as before even though the yield strength of that material gets doubled, and the elongation gets reduced by an order of magnitude.

Take maraging steel 350 for example. Annealed yield strength = 830 MPa ... Annealed elongation = 18% ... Annealed Young's modulus = 190 GPa

Aged yield strength = 2300 MPa ... Aged elongation = 4% ... Aged Young modulus = 190 GPa

This seems crazy to me. Strength triples and elongation is reduced to less than a quarter, yet the Young's modulus doesn't change one bit? I don't understand.

If the definition of Young's modulus is the ratio between stress and strain, when steel after aging gets 300% stronger, and that strength is achieved at 20% elongation, how could the Young's modulus possibly not get massively changed too?

Answer

Here’s a diagram relating pre and post aging performance:

pre vs post aging performance

For that material, you can see how yield point elongation decreases while the slope of the linear part (modulus) doesn’t change much.

mathematics - Nine gangsters and a gold bar

One night nine gangsters stole a gold bar. When the time came for dividing the bar, they faced a problem: two of the criminals put guns to each other's faces. Now it's up to fate whether one of them lives, they both live or both die.

While these two are dealing with each other, the others decide to continue dividing the gold bar. What is the minimal amount of pieces they should divide the bar into, so that no matter how things pan out, everyone can be given an equal share?

Scenario 1: Both gangsters blow each other's brains out. The gold must be divided evenly among the seven remaining gangsters.

Scenario 2: One gangster is quicker on the draw, and manages to take out his opponent. The gold must be divided evenly among the eight remaining gangsters.

Scenario 3: The duelling gangsters discuss their differences, come to a mutually beneficial agreement, and put away their guns. The gold must be divided evenly among all nine gangsters.

Answer

18 pieces:

3,7,9,14,16,18,23,24,25,30,31,32,38,40,42,47,49,56

In 9 parts:

{3,23,30} {7,49} {9,47} {14,42} {16,40} {18,38} {24,32} {25,31} {56}

In 8 parts:

{3,18,42} {7,56} {9,24,30} {14,49} {16,47} {23,40} {25,38} {31,32}

In 7 parts:

{3,7,24,38} {9,14,18,31} {16,56} {23,49} {25,47} {30,42} {32,40}

---

We need to be able to split it into 9 parts of 56, so it can't hurt to make 9 pieces of 56 and then split those further. Since we need to do better than 19 pieces, we can have at most 18 pieces. This means that most of our 56s are split into exactly two parts (we can have an extra piece for every 56 we don't split).

Now, what will our pieces be mod 9? We must be able to group them into 7 or 8 groups that sum to 0, or make 9 pairs (mostly) that sum to 2. If we have a 2, we can group it with a 7 to make 0. Then we pair the 7 with a 4, and repeat with a 5 and 6. If we do this 3 times, we can use the 6s to make another 0.

To make 63s and 72s, we should repeat at an interval of 9, so that we can swap the pairs to add 9 (e.g. {7,56} {16,47} {25,38} -> {16,56} {25,47}). The other two numbers that are swapped out need an extra 27.

Suppose we start with a 56 (freeing up one cut) and see what numbers we need.

    56-> 7->49->14->42
 9->47->16->40->23->33
18->38->25->31->32->24

So far, this makes 6 63s with 9 18 24 33 42 left over. We also need to make a 27 and a 72 out of 24 33 42. If we split 33 into 3 and 30, both of these are resolved. This results in the final answer with 18 pieces.

electromagnetism - Experiments looking for monopoles

Background: (skip it if you know it)

In the easiest formulation of classical electromagnetism magnetic monopoles do not exist. In fact, the Maxwell's equation $\nabla \cdot \vec{B}=0$ implies (using Gauss' Theorem) that the surface integral of the flux of $\vec{B}$ over the bounday of any finite surface is zero. Therefore no isolated magnetic charges (i.e monopoles do exist).

However Dirac discovered that even if a single monopole existed in the universe, we could then explain in a rather easy way why the electric charge is quantized. Note that since the charge is not an observable its quantization is completely different than the quantization of energy or momentum in QM, for example.

Furthermore, in the recent developings of QFT, the theoretical model we usually assume implies that every time a gauge symmetry is broken monopoles (and other kind of topological defects such as solitons) arise. Since in the hot big bang model it is usually belived that many gauge symmetries were broken in a primordial of the universe, monopoles of various kinds (not just magnetic monopoles but also Yang-Mills ones) could (at least this is what teorists say) have been produced.

Up to date, not a single monopole has been found by experiments.

With this in mind I ask the following

Question:

Which experiments are currently being carried over to look for magnetic, and other types of, monopoles?

Answer

I work on an experiment that involves trying to create magnetic monopoles in a type of material called a spin ice (Dy2Ti2O7) so called because its spins obey the same rules as ice water. The interesting part of the structure is a tetrahedra of rare earth atoms where in the ground state you have two spins pointing in and two spins pointing out across its four corners. If you excite the system you can flip one of the spins in the system so you have for example 3 in 1 out spins. This can give a net charge at one tetrahedra. To balance this an opposite spin flip will occur elsewhere in the lattice so you effectively have a dipole. The interesting thing about this is that you will have a chain of connected spin flips (a Dirac string) between the two tetrahedra with net charge but it doesn't cost any energy to flip the spins between these two sites. This means that each part of the dipole can propagate entirely independently of the other meaning that it is a site of charge that is not energetically linked to an opposite site of charge making this a monopole. You can look for these features using neutron diffraction (that being the experiment, to answer your question). If you want to read some more on the topic Castelnovo et al. is a good place to start as is Morris et al.

The most famous experiment looking for monopoles in condensed matter is the Stanford monopole experiment.

mazes - Haisu: Pink Rose

HAISU is a portmanteau of three Japanese words - 'hairu', to enter, 'su', number, and 'hausu', an English borrow word meaning house, of course.
Together, we get a meaning of 'enter number house', which I have roughly translated to English as 'Room Count'.

The rules are simple - draw a path from the O to the X, passing through every cell in the grid exactly once. The grid is divided into several rooms. When your path passes over a cell with the big number N, it must be the Nth time you have entered the room. If a room has a small number m in the top left corner, you must enter that room a total of m times. An example Haisu puzzle and its unique solution are shown below.

Hopefully this example puzzle clarifies the rules. Your actual challenge is this!

Answer

Here is my answer

Only good way to enter 5 times I could find. Also must hit the 1 on first entrance.

Logical prolongation of lines that shouldn't touch each others.

The path coming from the entrance will be forced to take this path eventually.

Only way to hit the 3 on the 3rd entrance and respect the current flow of the lines.

Further deductions based on straight forward paths.

Some of my logical deductions were not so logical after all. Here is the result after some minor revisions.

The Diamond Game

Barkeep:

Well hello there, come to play the famous Diamond Game?

The rules are notoriously simple. Here's an ace of diamonds card (it's larger than normal), and here are a pile of pennies. We each take turns to place pennies on the card. Whoever is first not to be able to fit a penny onto the card loses.

Tell you what, you go first this time. Put a tenner on it. If you win, you'll get twenty!

How can the player guarantee that they win every time, regardless of how the barkeep plays?

---

For the purposes of the game, assume that both players are able to exactly follow any rule or process, regardless of its complexity.

Answer

Actually, I heard about this puzzle and resolved it once, so it's easy for me, sorry.

To ensure, that you'll win,

you must start at the very center of the card. The other player will put a coin at a random position. You just have to put yours symmetrically to the other's, with the center of your first. Since, there will be for every opponent coin a symmetrical place, you will win with this method.

newtonian mechanics - Direction of friction on particle placed on a rotating turntable

If a particle is placed on a rotating turntable then the particle has a tendency to slip tangentially with respect to the underneath surface... So the friction should act tangentially to the particle... Why then does it act towards the center... Please explain with respect to inertial frame

Friction opposes relative slipping between surfaces... When the particle is initially placed on the rotating turntable... the surface under it has a tangential velocity. So, in order to oppose this slipping friction must act tangentially... How then does it act radially inward?

What is the difference between thermodynamic free energies and the Landau free energy?

How and why is the Landau free energy any different from thermodynamic free energies?

It is written on page 140 of Nigel Goldenfeld's book Lectures on Phase Transitions and The Renormalization Group that

The Landau free energy has dimensions of energy, and is related to, but, as we will see, is not identical with the Gibbs free energy of the system.

The explanation in section 5.6 is quite elaborate and too complicated. Please help me with a simple understanding of why the Landau free energy is not the Helmholtz free energy or the Gibbs free energy, and how it is related to the thermodynamic free energies.

Answer

The Landau free energy, also called the Landau-Ginzburg Hamiltonian, is treated in an adhoc and rather confusing manner in a lot of textbooks. But in the modern view, it has a simple interpretation as an effective Hamiltonian attained by integrating out degrees of freedom.

Suppose we have a spin system, such as an Ising magnet. We can describe the state of the system by a magnetization field $\phi(x)$, noting that this field doesn't make sense if we examine length scales smaller than the lattice spacing $a$. We can write a sum over all spin states by an integral over field configurations, as long as the integral is cut off at the distance scale $a$.

If the Hamiltonian is $H[\phi]$, then the thermodynamic free energy $F$ obeys $$Z = e^{-\beta F} = \int_{\Delta x\, >\, a} \mathcal{D}\phi \, e^{-\beta H[\phi]}$$ which is just a rewording of the standard identity $F = - k_B T \log Z$. In the Wilsonian view, the thermodynamic free energy is acquired by integrating out all microscopic degrees of freedom. The result only depends on macroscopic quantities like temperature, pressure, and the external field. This is useful because the entire point of thermodynamics is to ignore the microscopic details and focus on macroscopic quantities that are easy to measure. For example, using just the function $F$, we can determine the equilibrium magnetization by minimizing it.

Now, the Landau free energy $H_L$ satisfies $$Z = \int_{\Delta x \, > \, b} \mathcal{D}\phi \, e^{-\beta H_L[\phi]}$$ where $b$ is a mesoscopic distance scale, larger than $a$ but still much smaller than a macroscopic length. In the Wilsonian view, the Landau free energy is the effective Hamiltonian acquired by integrating out degrees of freedom on lengthscales $a < x < b$. The point of the Landau free energy is that it represents a compromise between the completely microscopic $H$, which has too much detail to be useful, and the completely macroscopic $F$, which tells us nothing about, e.g. position dependence. Like $H$, $H_L$ is a functional, but it's a functional of "fewer variables".

The above explains why $H_L$ can be called a Hamiltonian, but why is it also called a free energy? Usually, the starting point for applying Landau theory is the saddle point approximation, which states that typical equilibrium field configurations minimize $H_L$. Since we're minimizing $H_L$, we're treating it like we would a free energy, which is why it's sometimes called the Landau free energy.

But why is this valid? You definitely can't get the right answer to any thermodynamic question by minimizing $H$, because it doesn't take into account thermal effects; you instead have to minimize $F$. Minimizing $H_L$ gives the right answer precisely when thermal effects are negligible on distance scales greater than $b$. This is true when $b$ is much greater than the system's correlation length $\xi$, which is why Landau theory does such a good job, and usually not true at a critical point where $\xi$ diverges, which is why Landau theory fails to describe continuous phase transitions.

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

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

I don't understand this explanation.

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

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

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

Answer

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

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

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

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

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

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

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

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

quantum mechanics - Countable Matrix Representation

In my quantum mechanics class, my professor explained that the Hamiltonian along with position and momentum operators can be represented by matrices of countable dimension. This is especially usefull in harmonic oscillator problems. My professor explained that the eigenvalues of the Hamiltonian are (of course) the discrete allowed energies of the system, while the eigenvalues of the position operator are all possible positions, a continuum. How can a countable matrix have an uncountable number of eigenvalues? Why do the Hamiltonian and the position operator have the same dimension but different numbers of eigenvalues?

Answer

The countable and uncountable infinities are "different cardinals" according to set theory but in physics, the bases of that size produce equally large Hilbert spaces: the Hilbert-space is infinite-dimensional and all infinite-dimensional Hilbert spaces are isomorphic to each other (in other words, there is only "one unified kind of infinity" when it comes to the dimension of a Hilbert space). Quantum mechanics offers you infinitely many examples.

Perhaps the simplest example are Fourier expansions. Consider a particle in an infinite well so that the wave function $\psi(x)$ is only nonzero for $0\lt x \lt +\pi$. The operator $x$ has a continuous spectrum i.e. an uncountable number of eigenvalues and eigenstates (the basis of $x$-eigenstates is uncountable).

On the other hand, the operator $p^2 = -\hbar^2 \partial^2 / \partial x^2$ has a discrete spectrum and a countable set of eigenvalues and eigenstates. The eigenstates are standing waves $\sin (nx)$ for positive integer $n$ and the eigenvalues are $n^2$.

Nevertheless, every ("smooth enough" and/or $L^2$-normalizable etc.) function $\psi(x)$ that is nonzero in that interval – every combination of uncountably many wave functions $\psi(x) = \delta(x-x_0)$ – may also be written as a linear combination of the standing waves, $\sin(nx)$. This fact is what makes the Fourier series possible. (Normally, I would talk about periodic functions and complex exponentials but the sines in a well may be more beginner-friendly.)

There is really no contradiction with the different cardinality of the sets because the two sets, the uncountable basis of $x$ eigenstates and the countable basis of $p^2$ eigenstates, are not being identified via a one-to-one map. Instead, the map between one basis and the other is a general linear transformation that mixes them, and the different cardinality places no restrictions on such linear transformations of infinite-dimensional vector spaces.

Quite generally, cardinal numbers (the science about distinguishing many types) as well as most other related results in set theory (I mean especially Gödel's theorems) are completely inconsequential in physics. They're just some "recreational subtleties" in mathematical logic and physics doesn't find any of these operations relevant. So a physicist may do state-of-the-art string theory and interpret it in all corners of physics without even "knowing" that the real numbers are uncountable. The uncountability is unphysical. A physicist is generally agnostic about the existence of real numbers that cannot be constructed, about the validity of the axiom of choice, and other problems that cannot be operationally performed by an experiment. A physicist's typical reaction is that these questions are "philosophy" – they are empirically undecidable (we know that the axiom of choice is undecidable even in major axiomatic systems of set theory) so he doesn't care about the answers.

electromagnetism - Do light waves precisely follow null geodesic paths in General Relativity?

In special relativity one may show that a plane wave solution of Maxwell's equations (in a vacuum), of the form $A^a=C^a\mathrm{e}^{\mathrm{i}\psi}$ has the following properties: The normal $k:=\mathrm{d}\psi$ to the surfaces of constant $\psi$ is a null vector and the integral curves of $k$ are null geodesics. Here $A$ is the electromagnetic vector potential, $C$ is a constant vector and $\psi$ is some function.

This analysis is possible because of the relatively simple form of the Maxwell equations in flat space, $\partial^a\partial_a A^b=0$ (Lorenz gauge assumed). However, in curved spacetime, we have an extra term involving the Ricci tensor that is irrelevant for SR: $$\nabla^a\nabla_aA^b=R^b{}_aA^a,$$ where $\nabla$ is the Levi-Civita connection of our spacetime $(\mathcal{M},g)$ and $R_{ab}$ its Ricci curvature.

The textbook treatment is to now look at solutions of the form $A^a=C^a\mathrm{e}^{\mathrm{i}\psi}$ where the covariant derivatives of $C$ are "small." In order to obtain the condition for $\mathrm{d}\psi$ to be null and autoparallel ($\nabla_a\psi\nabla^a\psi=0$), one must ignore the term $\nabla_b\nabla^b C^a$ as well as the Ricci tensor term. The missing details may be found in [1], sections 4.2 and 4.3. This approximation is called the geometric optics approximation.

Ref. [2] gives the following characteristic lengths for ray optics (section 2.8):

1. 
   

   The wavelength $\lambda$.

   
   


2. 
   

   The typical length $L$ over which the amplitude, polarization and wavelength of the wave vary significantly.

   
   


3. 
   

   A typical "radius of curvature," which can be taken to be $$R:=\lvert\text{typical component of the Riemann tensor in a typical local inertial system}\rvert^{-1/2}$$

   
   

The region of validity for geometric optics is then $\lambda\ll L$ and $\lambda \ll R$.

Question:

Since one must ignore terms in the preceding analysis, do light rays not actually follow null geodesics in GR? Is $\mathrm{d}\psi$ even null? Furthermore, how do wave-like solutions of Maxwell's equations behave over length-scales greater than those given in [2] (i.e., over length scales where the curvature of $\mathcal{M}$ can vary greatly and rapidly). In particular, to what extent do they travel along null geodesics?

Please note: The argument that massless particles travel along null geodesics in flat space so the same must be true (by the Equivalence Principle) in curved space is not an answer to this question. I'm asking about wave-like solutions to the Maxwell equations. Light, classically, is just a wave solution of the vacuum Maxwell equations. Any answer should include (or reference) a rigorous analysis of Maxwell's equations. This is not a question that can be answered well by stating a few equations everyone knows and nitpicking at the words in the OP. (Apparently one member of the site did not have this impression, so I'm making it more clear.)

References:

[1] Wald, R.M. General Relativity. Chicago University Press, 1984.

[2] Straumann, N. General Relativity. Springer, 2013.

Answer

For clarity I think is best to start with Minkowski spacetime.

The equation we are trying to solve to understand the radiation of a point particle is: $$\square A^{b}=j^{b}$$

with the gauge $\nabla_{a}A^{a}=0$ and $j^{b}$ is the current density.

The potential $$\begin{eqnarray} A^{b}(t,x)&=&\int G^{b}_{a}(t,x,t'x')j^{a}(t',x')dx'^{3}dt\\ &=&\int\delta_{a}^{b}\delta(t−t′−|x − x′|)∕|x − x′|j^{a}(t',x')dx'^{3}dt' \end{eqnarray}$$

where $G^{b}_{a}$ is the Green function with support in the past light cone. In fact, the potential $A^{b}(t,x)$ only depends in the single event $(t',x')$ in the past which is the intersection between the null cone from $(t,x)$ and the world line of the particle.

Now in curved spacetime the generalization $$\begin{eqnarray} A^{b}(t,x)&=&\int G^{b}_{a}(t,x;t'x')j^{a}(t,x)dV\\ &=&\int\delta_{a}^{b}\delta(\gamma(t,x,t'x'))∕\Gamma(x − x ′)|j^{a}(t',x')\sqrt(g)dx'^{3}dt' \end{eqnarray}$$

where $\gamma$ is the null geodesic between the two points $(t,x),(t',x')$ and $\Gamma$ is the distance with respect the induced metric of a suitable spacelike surface that contains $x,x'$ does not work.

In general curved spacetimes the retarded Green function would depend on the whole causal past cone and not only the past light cone. This dependence comes from the interaction with the curvature and is related with the extra terms that you point out that vanish for Minkowski.

Therefore the potential is not only defined by the information that travel along the null geodesics but depends on the whole past of the particle. Nevertheless, singularities of the field travel along null geodesics globally. This is the content of the propagation of singularity theorems for linear hyperbolic systems and is related with the geometric optics limit.

As you required rigorous analysis I will point you to some papers with appropriate calculations:

Section 1.4 of http://relativity.livingreviews.org/open?pubNo=lrr-2011-7&page=articlese1.html

http://arxiv.org/abs/1108.1825

http://arxiv.org/abs/gr-qc/0008047

Also notice that my answer is just about electromagnetism in curved spacetime. To talk about General Relativity we would need to solve also for the Einstein's Equations. The point particle will affect the metric as self force corrections to the background metric. These type of corrections are treated in depth in the first reference.

thermodynamics - Active Matter Systems

Active matter is composed of large numbers of active "agents", each of which consumes energy in order to move or to exert mechanical forces. Due to the energy consumption, these systems are intrinsically out of thermal equilibrium.

This is the Wikipedia's definition of active matter.

I would like to ask the question that, How are these systems different from a canonical ensemble? In canonical ensemble, sytem exchanges energy with a heat bath. And we treat system plus surroundings as a microcanonical ensemble and derive all quantities of interest. I'm not getting the exact motivation behind this formulation. Any help is appreciated. Thank you

quantum mechanics - SPDC and single photon production

In Spontaneous Parametric Down Conversion, incoming photons are split into entangled photon pairs by passing through a non-linear optical medium. Since I get a photon pair as an output then how can I say that I am producing single photons?

Answer

There are indeed some subtle but important differences between an SPDC-based heralded source and a true single photon source. In order to understand these differences, consider what a true single-photon source really means.

A true single-photon source emits a single excitation at a specified frequency when demanded. So mathematically, the output state would be a true Fock state with photon number $n=1$. The probability that the detector would record any other number ($n=0$ or $n=2$ etc.) of photons in an instant is exactly zero. Thus, the probability $p(n)$ that $n$ photons would be detected at a given instant in a detector takes the value unity at $n=1$ and zero for all other values of $n$, implying that the variance $\Delta n$ is zero.

Lets contrast this with a weak coherent source, which is routinely used as an approximation to a single-photon source. A weak coherent source is obtained by attenuating the output from a laser to the point where the mean number of photons per second becomes unity, i.e, $\bar{n}=1$. However, such a source differs in a very important way from a true single-photon source. This is because the light has been attenuated in a way that does not alter the statistical distribution of the photons. The photon statistics of a weak coherent source remain Poissonian, i.e $(\Delta n)^2=\bar{n}$, which is in sharp contrast with the highly sub-Poissonain statistics of a true single photon source with $\Delta n=0$.

The experimental consequences of this difference can be understood by looking at the photon number distribution of a weak coherent source, which is just be the Poisson distribution with $\bar{n}=1$. The salient feature of this distribution is that $p(0)$ takes on a large value, followed by a small value for $p(1)$, and exponentially smaller values for the rest. Clearly, this distribution is sharply different from that of a true Fock state for $n=1$. The large probability $p(0)$ that no photon is detected implies that in an experiment, a large number of detection time windows are wasted without an arrival of a photon. As a result, a large error may be introduced in the measurement due to dark counts.

Now in order to avoid this problem, quite often a single photon source is simulated by means of what is called a heralded single photon source based on SPDC. The idea here is to utilize the strong arrival time correlations of the photon pairs produced from SPDC, which means that one would keep the detector active only conditional to a separate detection of the other photon in the pair. It is then hoped that the ensemble of measurements would approximate the ensemble of measurements from a true single photon source.

However, there are problems with this claim as have been systematically analyzed in Phys.Rev. A 90, 053825 (2014). The essential point becomes clear if you consider the process of SPDC more closely. The Hamiltonian for the process is of the form $H=\epsilon_{0}\chi^{(2)}\hat{a}_{p}\hat{a}^{\dagger}_{s}\hat{a}^{\dagger}_{i}+ c.c$, where the first term signifies the SPDC forward process involving annihilation of the pump photon and the creation of the signal and idler photons, and the second term denotes sum frequency generation, which ensures hermiticity of the Hamiltonian. The time evolved output state $|\psi\rangle$ would then be of the form,

$|\psi\rangle=e^{iHt}|0\rangle_{s}|0\rangle_{i}=c_{0}|0\rangle_{s}|0\rangle_{i}+c_{1}|1\rangle_{s}|1\rangle_{i}+c_{2}|2\rangle_{s}|2\rangle_{i}+\cdots$

, where the labels $s,i$ denote the signal and idler modes and the pump has been treated as a classical field. Clearly, this output state from SPDC is not just a two-photon Fock state. In fact, the efficiency of the SPDC process is very low (typically less than $10^{-8}$) and consequently, the probability $|c_{0}|^2$ that no photons are produced is almost unity. This is followed by a small probability $|c_{1}|^2$ that a pair of photons is produced, which we use to produce a heralded source. What is crucial however, is that the higher order terms $|c_{2}|^2$, etc. while being small, are certainly not zero. It is precisely these higher order terms that prevent a heralded SPDC-based source from accurately simulating true single-photon source.

Finally, I note that the design and implementation of a true single photon source remains an experimental challenge to this day, and several groups are working actively on this.

word - The Riddle - I Could be Seen in Many Places

In whole, you may see me in the competition, or on the street.

Remove my front part, you may see me in the body, or on the lawn.

Remove my rear part, you may see me in the prison, or in the grocery.

Remove my middle part, you may see me in the sea, or on the stage.

Remove my middle part again, you may see me in the baseball game, or in the university.

Who/What am I?

Hint 1:

Chances to see in places(based on my thoughts :P):

competition: LOW to HIGH, positive correlate to competition schedule.
street: LOW, but vary in cities or countries.

body: VERY HIGH.
lawn: MEDIUM.

prison: HIGH.
grocery: LOW, but vary in locations.

sea: MEDIUM.
stage: MEDIUM.

baseball game: LOW to MEDIUM.
university: HIGH.

Hint 2:

The answer could be a noun or an adjective. Also the answer has 2 main meanings which are almost on the contrary.

Answer

We'll see where this goes...

My guess is:

BADASS

In whole, you may see me in the competition, or on the street.

You could be a BADASS competitor or a gangster

Remove my front part, you may see me in the body, or on the lawn.

You have an ASS and a donkey could be on the lawn

Remove my rear part, you may see me in the prison, or in the grocery.

BAD people go to prison and they could rob grocery stores?

Remove my middle part, you may see me in the sea, or on the stage.

BADASS = BASS like a sea bass or a bass guitar

Remove my middle part again, you may see me in the baseball game, or in the university.

BADASS = BS like a bachelor of science or blown save in baseball

Answer to clue #2

Badass in a noun and adjective. A badass is someone who is good or you could have a literal bad ass (flat/lumpy) XD

electromagnetism - Is it possible to kill a human with a powerful magnet?

I'm asking in terms of physics. Can powerful magnetic induction rearrange spins of my body in such way I will die? How?

Or maybe it can rip all iron from me, which would make my blood cells useless? How many teslas should such magnet have? Are there other ways to kill people with magnetic induction only?

Answer

I don't know much about the topic, but here are some research points you can get started with.

For strong magnetic fields, the most notable effect seems to be visual effects (source), called phosphenes (magnetophosphenes in the specific case of magnetic causes) caused by inductance of electric currents in the retina (source).

"Studies" seem to have suggested that 50T fields cause tissue damage, for unspecified reasons (weak source). I could not locate these studies. However, the implication is that immediate death / severe damage is not caused at even 50T fields (for reference, MRIs generally run in the 1.5-3T range).

There are related questions here:

There is an interesting discussion on Reddit:

There is also a field of study called bioelectromagnetics dedicated to biological effects of magnetic fields, which can serve as a good starting point for research:

"Transcranial magnetic stimulation", referenced in both the Reddit and Wikipedia pages, uses small fields in the range 1-10mT to affect the polarization of neurons in the brain.

It seems that the pattern of change of a magnetic field has a more pronounced effect than the strength of the field. Static fields do significantly less (or no) damage, while at high frequencies a weak magnetic field could certainly do significant damage, e.g. a microwave oven.

Primary causes of damage from non-static fields mostly seem to be due to heat, or due to induced electrical current; for example, from the ReviseMRI link above:

A more serious consequence of electric currents flowing through the body is ventricular fibrillation (though these levels are strictly prevented in MRI). ... As a general guide, the faster the imaging or spectroscopy sequence, the greater the rate of change of the gradient fields used, and the resultant current density induced in the tissue is higher.

It would doubtless take an extremely strong magnet, higher than anything we could produce, to pull the iron out of your body (conjecture, no source). Note also that there is only about 3-5 grams of iron (something like 2 cm³) in the human body (source, unreferenced source), mostly bound to hemoglobin.

Count Iblis pointed out, in question comments, that there is a nice discussion of magnetars and strong magnetic fields here, which provides nice overviews and plenty of interesting information (although a bit dated):

From there:

Fields in excess of 10⁹ Gauss, however, would be instantly lethal. Such fields strongly distort atoms, compressing atomic electron clouds into cigar shapes, with the long axis aligned with the field, thus rendering the chemistry of life impossible. A magnetar within 1000 kilometers would thus kill you via pure static magnetism -- if it didn't already get you with X-rays, gamma rays, high energy particles, extreme gravity, bursts and flares...

As for long term effects of more commonly encountered field strengths, there is generally little association between magnetic fields and cancer (source, source).

I hope this helps. Sorry I do not know a direct answer. It certainly depends on more than just the field strength, however.