I am a sub-50 cuber right now. I am currently using the Fridrich method, I take about 2-5 secs to solve the cross and I am learning the 21 PLL algorithms. I am learning one or two PLL every day. I am using the 2 look OLL that uses 10 algorithms in two steps.
With all these have only a record of 32 secs and a average of 46 secs. Do you have any tips on becoming a sub 30 cuber or even sub 20 or sub 15?
I know the projection along a diameter is an SHM but is circular motion itself an SHM? If we consider the mean position to be the center of the circle then the centripetal acceleration is proportional to the distance and in opposite direction of the position of the particle. So shouldn't it be an SHM?
Setup
Introducing this spinorbital notation:
\begin{align} \Psi_1=\chi_{(r1)}\alpha_{(\omega1)} = 1 \\ \Psi_1=\chi_{(r1)}\beta_{(\omega1)} = \bar{1} \end{align}
and the Slater's determinant, for a system of two electrons, is:
\begin{align} |\Psi\rangle = \frac{1}{\sqrt{2}} \Big| \begin{array}{cc} 1 & \bar{1} \\ 2 & \bar{2} \\ \end{array} \Big| = \frac{|1\bar{2}\rangle-|\bar{1}2\rangle}{\sqrt2} \end{align}
Question
Based on the example, what would be the Slater's determinant representation for an excited state like $\Psi_a^r$ (i.e. with parallel spins like: $|\bar{1}\bar{2}\rangle$) and $\Psi_{ab}^{rs}$ (feel free to use any number) ?
Answer
The slater determinant is only a "trick" to get a total antisymmetric wave function. This is required by the Pauli principle.
For understand this you need to think in indistinguishability of particles. So a any allowed state of a particle need to be assigned equally at each indistinguishable particle in your system. So if you have to state allowed: $|1\rangle$ and $|2\rangle$, then your state of the system is:
$$ |\psi\rangle=\frac{|1\rangle|2\rangle - |2\rangle|1\rangle}{\sqrt2} $$
For a $N$ allowed states is something like:
$$ |\psi \rangle= \frac{1}{\sqrt{N}}\sum_{\sigma \in S_n} p(\sigma) \bigotimes_{i=1}^N |\sigma_i\rangle $$
where $\sigma$ is a permutation of $(1,2,...,N)$, $p(\sigma)$ is the parity of the permutation $\sigma$, and $\sigma_i=\sigma(i)$ is the i-th number of the permutation $\sigma$. The big "O" with an "x" inside is:
$$ \bigotimes_{i=1}^n |\sigma _i\rangle= |\sigma _1\rangle \otimes |\sigma _2\rangle \otimes ...\otimes |\sigma _n\rangle = |\sigma _1,\sigma _2,...,\sigma _n \rangle $$ where $\otimes$ is a tensor product.
Note that the tensor product is not commutative. Then, if you fixing some basis as the position ones, you get:
$$ \psi (x_1,x_2,...x_n)= \frac{1}{\sqrt{N}}\sum_{\sigma \in S_n} p(\sigma) \prod_{i=1}^n \psi_i(x_{\sigma_i}) $$
The non-commutativity goes to $i$ labels, that maintains the order of the permutation $\sigma$. This is precise the formula of the determinant as you can see.
If you want to deal with extra state, you need to do the enumeration $(n,l,m_l,s)\rightarrow i$. e.g. the states allowed is $n=p_1,p_2$ and $s=\pm \frac{1}{2}$. We can enumerate: $(p_1,+\frac{1}{2})\rightarrow 1$, $(p_2,+\frac{1}{2})\rightarrow 2$, $(p_1,-\frac{1}{2})\rightarrow 3$ and $(p_2,-\frac{1}{2})\rightarrow 4$. Some properties of determinant can help you to understanding the structure of this antisymmetric state.
I was reading the altitude records for hot air balloons on Wikipedia, and noted that the max hot air balloon altitude was about 60,000ft. It didn't really say if there was a reason why. I know that closed envelope balloons frequently climb much higher.
I was wondering... If you have a pressurized cabin (for keeping the operator conscious), and you mix oxygen into the fuel to heat up the air, is there really any reason that a balloon cannot ascend higher? Does the thinning of the air prevent it from holding heat as well? Do other atmospheric events come to play in not allowing a hot air balloon to ascend higher? I don't understand hot air balloons that well and could be missing a major component =)
The first law and second laws of motion are obviously connected. But it seems to me that the third law is not related to the first two, at least logically.
(In Kleppner's Mechanics the author states that the third law is a necessity to make sense of the second law. It didn't make sense to me, though. I'll post the excerpt if anyone would like to see it.)
EDIT: Excerpt from Introduction to Mechanics by Kleppner & Kolenkow (1973), p. 60:
Suppose that an isolated body starts to accelerate in defiance of Newton's second law. What prevents us from explaining away the difficulty by attributing the acceleration to carelessness in isolating the system? If this option is open to us, Newton's second law becomes meaningless. We need an independent way of telling whether or not there is a physical interaction on a system. Newton's third law provides such a test. If the acceleration of a body is the result of an outside force, then somewhere in the universe there must be an equal and opposite force acting on another body. If we find such a force, the dilemma is resolved; the body was not completely isolated. [...]
Thus Newton's third law is not only a vitally important dynamical tool, but it is also an important logical element in making sense of the first two laws.
I have been going in circles for a few days now with this word-search puzzle and I still cannot find a clever solution to the problem. Is there a systematic approach that doesn't involve guessing? The problem is as follows:
At a kindergarten's playroom in Taichung a teacher assembled the following configuration using alphabet cubes forming a stack (see the figure as a reference) where it can be read the word
DOS BANDOS(the Spanish word for two sides). Calculate the number of different ways joining neighboring letters can be read the phraseDOS BANDOS.
The possible solutions in my book are:
$\begin{array}{ll} 1.&1536\\ 2.&1280\\ 3.&256\\ 4.&768\\ 5.&1024\\ \end{array}$
In my initial attempt what I tried to do is to draw a circle over each time I could identify the word being asked but in the end, I got very confused and I felt that I counted a possibility twice, hence I couldn't even understand if my attempt was right. During this process I could identify immediately that the word could be read from the top to bottom, there I counted four cubes, going from left to right, hence $4$ ways and from right to the left $2$ ways which together would account for $6$. This is summarized in the drawing from below where the circles are painted with blue color.
In the end that how far I went. As the more I looked at the stack I started to get confused on which zig zag lines are allowed and which do are already counted.
Therefore can somebody help me with this riddle?. To be honest I have little experience with these kinds of problems so I'd like somebody could be as much as detailed possible and include some drawing (perhaps using mine as a reference) and justify a method for solution.
It is very important for me to get a visual aid, because I really don't feel that solely a paragraph alone would be enough to understand, even as hard as I could. So really please if you can help me with this, include some sort of drawing or schematic so I can understand how to calculate the number of ways.
Again, I am looking for an answer which can solve this problem and that it can be extended to similar problems. Does it exist a way? I've been told a hint which mentions that I should consider turns to the left or right. But I don't know how to use this information.
There is also, in the bottom of the truncated pyramid it can be read the word DOS. Would this count if going from bottom to the top?.
Overall I hope somebody can take time and answer these questions. Because I really need help with this one and I'm confused.
I've a question regarding the Hodge star operator. I'm completely new to the notion of exterior derivatives and wedge products. I had to teach it to myself over the past couple of days, so I hope my question isn't trivial.
I've found the following formulas on the internet, which seem to match the definitions of the two books (Carroll and Baez & Muniain) that I own. For a general $p$-form on a $n$-dimensional manifold:
\begin{equation} v=\frac{1}{p!} v_{i_1 \ldots i_p} \mathrm{d}x^{i_1} \wedge \cdots \wedge \mathrm{d}x^{i_p} \end{equation}
the Hodge operator is defined to act on the basis of the $p$-form as follows:
\begin{equation} *\left( \mathrm{d}x^{i_1} \wedge \cdots \wedge \mathrm{d}x^{i_p} \right) = \frac{1}{q!} \tilde{ \varepsilon}_{j_1,\ldots,j_q}^{i_1,\ldots,i_p} \mathrm{d} x^{j_1} \wedge \cdots \wedge \mathrm{d} x^{j_q} \end{equation}
where $q=n-p$ and $\tilde{ \varepsilon}$ is the Levi-Civita tensor. Up until here everything is fine, I managed to do some exercises and get the right answers. However, actually trying to calculate the curvature does cause some problems with me.
To give a bit of background. I'm working with a curvature in a Yang-Mills theory in spherical coordinates $(r, \theta, \varphi)$. Using gauge transformation, I've gotten rid of time-dependence, $r$ dependence and $\theta$ dependence. Therefore, the curvature is given by:
\begin{equation} F = \partial_\theta A_{ \varphi} \; \mathrm{d}\theta \wedge \mathrm{d} \varphi \end{equation}
Applying the Hodge operator according to the formula above gives:
\begin{equation} * \left(\mathrm{d} \theta \wedge \mathrm{d} \varphi\right) = \frac{1}{(3-2)!} \tilde \varepsilon^{\theta \varphi}_r \mathrm{d}r=\mathrm{d}r \end{equation}
such that:
\begin{equation} *F = (\partial_\theta A_{ \varphi}) \mathrm{d} r \end{equation}
However, three different sources give a different formula. Specifically they give:
\begin{equation} *F = (\partial_\theta A_{ \varphi}) \frac{1}{r^2 \sin \theta} \mathrm{d} r \end{equation}
It is not clear to me where they get this from. Something is being mentioned about the fact that the natural volume form is $\sqrt{g} \; \mathrm{d} r \wedge \mathrm{d} \varphi \wedge \mathrm{d} \theta$ with $\sqrt{g}=r^2 \sin \theta$, which I agree with. However, I do not understand why that term is incorporated in the Hodge operator.
Boaz and Muniain define the Hodge operator as:
\begin{equation} \omega \wedge * \mu = \langle \omega , \mu \rangle \mathrm{vol} \end{equation}
But I don't see how that formula is applicable to calculating the Hodge operator on the curvature. Could anybody tell me where I am going wrong or provide me a source where they explain this?
Answer
It seems the resolution to OP's question lies in the difference between
the Levi-Civita symbol, which is not a tensor and whose values are only $0$ and $\pm 1$; and
the Levi-Civita tensor, whose definition differs from the Levi-Civita symbol by a factor of $\sqrt{|\det(g_{\mu\nu})|}$.
I was reading a text in which Electric polarization in terms of Berry phase was defined as
$P=\frac{e}{2\pi}\sum_{n}\int A_n (k) dk$
under gauge transformation $P\rightarrow P+ne$ (which means polarization can't be defined because it is not gauge invariant quantity) but in the text they defined polarization as $P mod(e\times integer)$ I am having trouble understanding this $mod$ argument ,
How can polarization is defined as $Pmod(e\times integer)$?
I recently learned that the strength of a Magnetic field around a conductor is proportional to the current flowing in it. So if we have a Mercury wire at absolute zero and pass a current through it (Resistance = 0) and then toss some iron filings at it, will the filings cylindrically float around the conductor (along the Magnetic field lines) due to the extreme strength of the field or just behave the way they do for an ordinary conductor (like Copper), in which case we must rest the filings on cardboard and then give it a jerk to align it along the lines.
Answer
First thing: even with huge fields, iron will not "float" like that in circular motion. Iron will always be attracted to the region where the magnetic field is the most intense: which is in contact with the conductor/superconductor.
In addition, there is a limit to the magnetic field you can achieve with any superconductor: it is called the critical field. There are 2 types of critical fields, Hc_1 and Hc_2, which correspond to 2 things:
The "floating" of magnets above superconductors, are due to these vortices, which cannot move in space as freely as usual magnetic flux lines in vaccuum can. This phenomenon "traps" the field exactly as it is, and the magnet with it, and only a force above a certain threshold (usually higher than the weight of the magnet used for the demonstration) can move the magnetic flux lines inside the superconductor.
This "floating" is the only difference in principle between superconductors and usual conductors.
One last thing: you can have objects, even living creatures like frogs, floating inside a copper coil, but this happens only with very high fields (more than 10 Teslas for water, see http://en.wikipedia.org/wiki/Magnetic_levitation#Diamagnetism)
Hope I answered your question :-)
This is a very simple question, but I can't find an answer anywhere.
Earth's gravity at sea level is 1 (approximately 9.8m/s^2)
What happens being under sea level in a cavern/chasm/ravine, how much does gravity increase normally here on Earth?
Inspired by Polyomino Z pentomino and rectangle packing into rectangle
Also in this series: Tiling rectangles with N pentomino plus rectangles
Tiling rectangles with T pentomino plus rectangles
Tiling rectangles with U pentomino plus rectangles
Tiling rectangles with V pentomino plus rectangles
Tiling rectangles with W pentomino plus rectangles
Tiling rectangles with X pentomino plus rectangles
The goal is to tile rectangles as small as possible with the F pentomino. Of course this is impossible, so we allow the addition of copies of a rectangle. For each rectangle $a\times b$, find the smallest area larger rectangle that copies of $a\times b$ plus at least one F-pentomino will tile. Example shown, with the $1\times 1$, you can tile a $3\times 3$ as follows.
Now we don't need to consider $1\times 1$ any longer as we have found the smallest rectangle tilable with copies of F plus copies of $1\times 1$.
There are at least 17 more solutions. More expected. I tagged it 'computer-puzzle' but you can certainly work some of these out by hand. The larger ones might be a bit challenging.
Answer
Here are a couple of $1 \times n$ solutions, which I believe to be optimal. They seem to follow some pattern(s), some of which are generalizable (spoiler ahead):
the rectangles are at the edges of the board; often one rectangle per edge, sometimes more, and sometimes (the $n$ = 12, 13 and 15 case) in the middle. The cases $n$ = 4, 6, 14 and 16 are extra 'nice' and can be generalized to $n=10k+4$ and $n=10k+6$ (though the solutions won't necessarily be optimal). Here's a nice visualization of the inductive step:
Note that these solutions also work for all $n$ not divisible by 5; e.g. two 1x7 rectangles fit in a single 1x14 rectangle, so we can just reuse that solution. There is a smaller one, though (see below).
The solutions for $1 \times 10$, $1 \times 18$ and $1 \times 20$ (all below) also seem to form some kind of generalizable family.
As @JaapScherphuis notes in the comment, there's a (probably non-optimal) generalizable solution for $1 \times n$, $n$ is even, rectangles, just like for the W-pentomino. By halving the rectangles, we can also obtain solutions for odd $n$, and the parts with just rectangles and no F-pentominos can be shortened.
Even $n$ fit in a $2n+1 \times 3n$ rectangle (left)
Odd $n$ fit in a $2n+1 \times 4n$ rectangle (right)
$1 \times 3$:
$8 \times 6 = 48$
$1 \times 4$:
$6 \times 6 = 36$
$1 \times 5$:
$11 \times 10 = 110$
$1 \times 6$:
$8 \times 8 = 64$
$1 \times 7$:
$20 \times 9 = 180$
$1 \times 8$:
$22 \times 9 = 198$
$1 \times 9$:
$13 \times 12 = 156$
$1 \times 10$:
$20 \times 11 = 220$
$1 \times 11$:
$13 \times 28 = 364$
$1 \times 12$:
$17 \times 17 = 289$
$1 \times 13$:
$15 \times 25 = 375$
$1 \times 14$:
$16 \times 16 = 256$
$1 \times 15$:
$17 \times 25 = 425$
$1 \times 16$:
$18 \times 18 = 324$
$1 \times 18$:
$19 \times 28 = 532$
$1 \times 20$:
$21 \times 30 = 630$
Some minimal solutions for $2 \times 2k+1$; first $2 \times 3$:
$10 \times 11 = 110$
$2 \times 5$:
$18 \times 20 = 360$
$2 \times 9$:
$30 \times 30 = 900$
$2 \times 11$ (a nice one, but IMHO equally nice as $2 \times 9$):
$36 \times 36 = 1296$
The last two are part of a generalizable solution, see the bottom of the post.
Another one for $3 \times 4$:
$14 \times 17 = 238$
and for $3 \times 5$:
$22 \times 30 = 660$
This night, I found a new one for $2 \times 7$ by combining the $2 \times 9$/$2 \times 11$ padding with the inner pattern for the $1 \times 16$. This is a hand-made solution so I'm not sure it's minimal:
38x38 = 1444
In general, the generalizable solutions for $1 \times (10k+4)$ resp. $1 \times (10k+6)$ give rise to solutions for $2 \times (10k+9)$ resp. $2 \times (10k+11)$; the $2 \times 7$ solution mentioned above is basically a subdivision of the $2 \times 21$ one minus some extraoneous padding.
The length of the red line is $10k+4$ or $10k+6$; this makes the purple line $10k+9$ or $10k+11$ and the total solution $(3n+3) \times (3n+3)$.
Which is the correct algorithm to determine the torque needed to rotate a spacecraft to a given quaternion ? I have a set of quaternions of a spacecraft and the time difference (delta T) between the quaternions.
I've tried estimating the angular velocity and angular acceleration as: 1. Find quaternion q so q = q1/q0
len=sqrt(q.xq.x+q.yq.y+q.z*q.z); angle=2*atan2(len, q.w); axis = q.xyz()/len;
Using this algorithm, the computed angular acceleration (a) is not correct.
What I'm doing wrong? Is this the correct way to obtain the angular acceleration and torque from a sequence of two quaternions?
There are a bunch of experiments that claim to show that the electron is highly spherical by measuring the electron electric dipole moment. See e.g.:
However, according to the Cartesian expansion of a charge distribution (see http://en.wikipedia.org/wiki/Multipole_expansion), if we assume the electron is made up of two small balls of negative charge $q$ separated by a small distance $2\mathrm{d}z$, and calculate the dipole moment in the centre of the two balls, the dipole moment is zero.
$$(-q)(-\mathrm{d}z)+(-q)(\mathrm{d}z)=0$$
which means even a highly non-spherical shape produces a dipole moment of zero.
So if the dipole moment is measured to be tiny, why does that imply the electron is highly spherical?
Some similar questions on this site:
In the double-slit experiment, we emit a photon that is in a state of superposition (wave form) which travels through both slits to interfere with itself. When we measure which slit it went through, it "collapses" to a particle at that point removing the interference from the other wave.
From that point to the detector wall, does the photon now go back into a wave form to travel from the collapsed point to the detector wall? Is a photon always in a state of superposition while traveling through space?
Answer
It's tempting to think of the light as a little ball (the photon), and since little balls have a definite position the little ball has to be in a superposition of a state where it goes through one slit and a state where it goes through the other. However this is not a good description of what actually happens.
The light is not a photon, and it's not a wave - it is an excitation in a quantum field. As a general rule it is a good approximation to model the light as a wave when studying it's propagation and a particle when studying its interactions, but these are only working approximations and should not be taken as literally true.
The light propagating in our double slit experiment does not have a position in the sense that macroscopic objects have a position. To ask the position of the photon is a meaningless question because there is no position. The light travels though both slits because the excitation in the quantum field spans both slits. When the light interacts with something, e.g. it hits the photographic plate or CCD, then the interaction happens at a point (though the position of the interaction is subject to Heisenberg's uncertainty principle). The interaction looks like a localised exchange of energy, so it looks like a photon.
It is often claimed in quantum gravity circles (I'm thinking of LQG in particular) that it is "not possible" to discretize spacetime. The issue, as I understand it, is that should one attempt to assign each point on a spatial lattice a value of e.g. the spatial metric $\gamma_{ab}$, the lattice spacings (both across time and space) will not be coordinate-invariants. This is pretty obviously true.
I work in numerical relativity, however, and we do precisely this all the time, maintaining parameters during the simulation such that the lattice spacings are indeed functions of spacetime. Should we seek to express quantities in a new "frame" we just transform those functions.
There appears to be no particular difficulty with this approach. One might, I suppose, run into trouble if a particular transformation were sought that brought previously irrelevant ultraviolet modes into the simulation domain, but this can be dealt with in a problem-dependent way by just choosing a high enough resolution at the outset. I am presumably, therefore, misunderstanding something about the difficulty the LQG people are getting at. What is it?
My infix won't tell you a thing,
But you can find me from the rest:
My suffix sounds exactly like me,
And my prefix is already half known.
In wood or wire, what am I?
Answer
You are a:
Knot
My infix won't tell you a thing,
"no"
My suffix sounds exactly like me,
"not" is pronounced just like "knot"
And my prefix is already half known.
"kno" is half of "known"
In wood or wire, what am I?
Knots in wood, and knots in wire are both very common.
I may also be found in your body.
You can also have a knot in a muscle.
I have naive question about Einstein action for field-free case: $$ S = -\frac{1}{16 \pi G}\int \sqrt{-g} d^{4}x g^{\mu \nu}R_{\mu \nu}. $$ It contains the second derivatives of metric. When we want to get the Einstein equation (which doesn't contain the third derivatives), we must use variational principle. The variation of "problematic" factor $R_{\mu \nu}$ (which contains the second derivatives) is equal to $$ \delta R_{\mu \nu} = D_{\gamma}(\delta \Gamma^{\gamma}_{\mu \nu}) - D_{\nu}(\delta \Gamma^{\lambda}_{\mu \lambda}). $$ So the corresponding variation of action may be rewritten in a form $$ \delta_{R_{\mu \nu}} S = -\frac{1}{16 \pi G}\int d^{4}x \sqrt{-g}\partial_{\lambda}(g^{\mu \nu}\delta \Gamma^{\lambda}_{\mu \nu} - g^{\mu \lambda}\delta \Gamma^{\sigma}_{\mu \sigma}). \qquad (1) $$ Then one likes to say that it is equal to zero. But why it must be equal to zero? It isn't obvious to me. After using the divergence theorem $(1)$ becomes $$ \delta_{R_{\mu \nu}} S = -\frac{1}{16 \pi G}\int dS_{\lambda} \sqrt{-g}(g^{\mu \nu}\delta \Gamma^{\lambda}_{\mu \nu} - g^{\mu \lambda}\delta \Gamma^{\sigma}_{\mu \sigma}). $$ Why it must be equal to zero? It is metric, not physical field, even if Christoffel symbols refer to the gravitational field, so I don't understand why it must be equal to zero at infinity.
Answer
It seems that OP is pondering the following.
What happens in a field theory [in OP's case: GR] if spacetime $M$ has a non-empty boundary $\partial M\neq \emptyset$, and we don't impose pertinent (e.g. Dirichlet) boundary conditions (BC) on the fields $\phi^{\alpha}(x)$ [in OP's case: the metric tensor $g_{\mu\nu}(x)$]?
I) Firstly, it should stressed that when people say that the infinitesimal variation $\delta S_0$ of the action $S_0[\phi]$ [in OP's case: the Einstein-Hilbert action $S_{EH}$] vanishes on-shell, i.e. when the Euler-Lagrange equations [in OP'case: Einstein's field equations] are satisfied, it is implicitly assumed that the infinitesimal variations $\delta\phi^{\alpha}(x)$ of the fields $\phi^{\alpha}(x)$ only take place in the interior/bulk of spacetime $M$ away from the boundary $\partial M$. In such cases, the infinitesimal variation $\delta S_0$ clearly vanishes on-shell, as part of the stationary action principle, aka. Hamilton's principle.
II) Secondly, if the sole purpose of the infinitesimal variation $\delta\phi^{\alpha}(x)$ is just to locally (re)derive the equations of motion (=the Euler-Lagrange equations) in an interior/bulk point $x_0$ of spacetime $M$, it is enough to choose localized variations $\delta\phi^{\alpha}(x)$ with support in sufficiently small compact neighborhoods around this point $x_0$. In particular, one may assume that $\delta\phi^{\alpha}(x)$ and all its (higher) derivatives vanish at the boundary $\partial M$ for such variations, and still derive the equations of motion.
III) Thirdly, if we do not impose adequate BC, then the global notion of a functional derivative
$$ \tag{1} \frac{\delta S_0}{\delta\phi^{\alpha}(x)}$$
may not exists, i.e. there may not exists a globally defined function $^1$ $$\tag{2} f_{\alpha}(x) ~=~f_{\alpha}(\phi(x), \partial \phi(x), \partial^2 \phi(x),\ldots ;x)$$
such that
$$ \tag{3} \delta S_0 ~=~\int_M \! d^{n}x ~f_{\alpha}(x)~\delta\phi^{\alpha}(x)$$
for all allowed infinitesimal variations $\delta\phi^{\alpha}(x)$. In plain English, the problem is that we cannot use the usual integration-by-parts argument when deriving the Euler-Lagrange expression since we have not imposed sufficient BC to ensure that boundary terms vanish. Then one must typically amend the bulk action $S_0$ with a boundary action $S_1$ [in OP's case: The Gibbons-Hawking-York boundary action $^2$ $S_{GHY}$], which only lives on the boundary $\partial M$. The total action then reads $S_0+S_1$.
--
$^1$ If the function (2) happens to exist, it is unique, and we will call it the functional derivative of $S_0$, and denote it by the symbol (1).
$^2$ See also these Phys.SE posts for more on the Gibbons-Hawking-York boundary term.
Does a similar easy to read introduction like David Mc Mohan's Demystified books exist for LQG? I'd like something easier to start with than a paper but deeper than a popular book.
Assuming that dark matter is not made of WIMPs (weakly interacting massive particles), but interacts only gravitationally, what would be the possible mechanism giving mass to dark matter particles? If they don't interact weakly, they couldn't get mass from interacting with the Higgs field. The energy of gravitational interactions alone does not seem to be sufficient to account for a large particle mass. Would this imply that dark matter consists of a very large number of particles with a very small mass, perhaps much smaller than of neutrinos? Or do we need quantum gravity to explain the origin of mass of dark matter?
Answer
I think this question contains a misconception unfortunately caused by popular science descriptions of the Standard Model.
The question seems to assume there needs to be some concrete source that particles "get" mass from, as if mass is a resource like money and the Higgs field is giving it out. But that's not right. In a generic field theory there is no issue adding a new field $\psi$ whose particles have mass. The only thing you have to do is make sure the Lagrangian has a term proportional to $\psi^2$.
You might protest that this violates the conservation of energy because the mass has to "come from" somewhere, but that's not right. Mass is the energy price for creating a particle. I don't create money by changing the pricetag of an item in a store.
The reason science popularizers say that mass must come from the Higgs mechanism is because of a peculiarity of the Standard Model (SM). The symmetries of the SM forbid a term such as $\psi^2$ for any field $\psi$ in the SM, so we need a trick to get a mass term. In brief, the Higgs field $\phi$ allows us to write terms like $\phi \psi^2$ which do respect the symmetry. This is an interaction term, but we can set up the Lagrangian so the Higgs field $\phi$ acquires a constant part, yielding the $\psi^2$ mass term we wanted.
However, once you start speculating about dark matter models, especially dark matter that does not interact with the electroweak force at all, these constraints don't apply and generically there is nothing forbidding a $\psi^2$ term. There's no need for any special mechanism for "giving" mass. You just treat mass exactly like you did in high school, intro mechanics and quantum mechanics: write it down, call it $m$ and call it a day.
Following these questions: Faulty computers, Knights and jokers I wonder, what if the conditions are the following:
- We have T Knights (truth-tellers), L Knaves (liars) and R Jokers (random-tellers). $T > 0$. We know all 3 numbers, but we do not know who is who.
- We can chose any two of them and ask the first about the second one of the 3 possible questions "Is he a Knight/Knave/Joker?". We can repeat this questioning procedure any number of times.
- We need to find one Knight.
How N, M and K must be related to make the task possible to complete?
It is clear for me, that $T+L>R$. Otherwise $T$ jokers can simulate Knights, $L$ Knaves and we would never be able to distinguish between them and find a real Knight.
Also we know already that if $L = 0, T > R$ the task is possible. Also it is obviously possible if $T>R+L$, because we can treat Knaves as Jokers.
So the question is what happens when $L>R-T$ and $L \ge T - R$. For example, when $L = 1, T = R = 10$.
Answer
Let's assume the jokers are intelligent, coordinated and aware of our strategy. If they fail to meet these conditions they only make it easy for us.
Edited for clarity This works for any $R
Step one is to identify the knave(s) by the fact that they will say 'yes' to 2 of the three questions for any given person ("Is bob a knave?", "Is bob a knight?", "Is bob a joker?"). If you conclusively identify a knave you have won - you just take the reverse of their statements to be true.
Ask each person if their neighbour is a knight, a knave, a joker. Knights will say 'yes' once. Knaves will say yes twice. An identified knave is effectively a truth teller - you can use them to identify everyone else.
This means that the jokers must use L jokers to impersonate the knaves. You now have a set of at least 2L alleged knaves who you know are not knights. This means the rest of the people - everyone who has not acted as a knave - have a majority of knights. You then ask all of those people about someone outside this group. The majority will be telling the truth. Continue this until you identify a liar, who you can use to identify everyone else.
Edit again: The jokers must impersonate the entire group of knaves. If you know there are 2 knaves and you see three people acting as knaves, ask each one about the identity of the other 'knaves'. The two genuine knaves will identify the joker (one you account for their backwards answers) and by extension themselves.
Yet another edit If you don't know the values of T,L,R you can still do it so long as $R Next, have the apparent knights identify all other apparent knights. They will, again, form groups. Have the knight groups identify the knave groups. They will form what I'll call "coalitions" - groups of knights and knaves who mutually identify each other and identify anyone in another coalition as a joker. The largest coalition must be the one with knights and knaves, given $R
Is there anyway to see by inspection that a form like $$a(x^2 )^{-3} (g _{μσ} x_{\rho} x_{ ν} + g_{μρ} x_{σ} x_{ ν} +g_{νσ} x_{ρ} x_{ μ} + g_{ νρ} x_{ σ} x_{ μ} ) $$ may be equivalent to (i.e reduced down to or reexpressed) $$ b(g _{μν} x_{ ρ} x_{ σ} + g_{ ρσ} x_{ μ} x_{ ν} )(x^2 )^{ −3} ?$$ where $g_{\mu \nu}$ is the metric tensor (diagonal).
I have tried to put in various permutations of $\mu \nu \rho \sigma$ and from $1111$ and $2222$ for example, I obtained the constraint that $a/b = 2$ but I am not really sure what this means. If I try the combination $1221$ e.g then it implies $b=0$, which seems to contradict my first result.
Does this mean that the two forms are not equivalent?
Answer
Indeed, if no values of $a$ and $b$ work for across different sets of indices, then the forms are not equivalent.
In fact, these two forms are not equivalent even under the restriction of the metric being diagonal (and thus are not equivalent under a general metric). The diagonal case is easy to analyze, and you gave a good set of indices to do it: $\mu\nu\rho\sigma = 1221$. Then the large parenthesized part of the first expression becomes $g_{11} x_2 x_2 + g_{22} x_1 x_1$. The middle terms drop out because they involve off-diagonal parts of the metric. However, both terms in the second expression also involve off-diagonal metric coefficients, so the second expression is identically $0$.
Given tensor components $T_{\mu\nu\rho\sigma}$ and $S_{\mu\nu\rho\sigma}$, we have $T_{1221} = g_{11} x_2 x_2 + g_{22} x_1 x_1$ while $S_{1221} = 0$, so clearly $T \not\propto S$.
I know very little of physics after Einstein.
I am aware of that Einstein's gravity theory says that the existence of matters creates curvature of a space-time, so that our Earth orbits our Sun. I can grasp this idea.
But I do not see how to use the language of general relativity theory to justify why an apple can fall?
Answer
I don't have time to write this up properly, but I was asked to make this an answer. (I also feel that John Rennie is about to post something much better)
The key point is that it's not space that's curved, it's spacetime. The apple falls because the accelerating downward path through spacetime is the geodesic through curved space, rather than the "straight" hovering path through spacetime.
This question is the second one in a series of Poké-related questions I hope that I can ask.
Who's this Pokémon that:
Answer
I'm pretty sure it's
Clue 1
Generation 2 had 251 total Pokémon (100 new Pokémon combined with Generation 1's 151), almost 255
Clue 2
Rapunzel's parents (and the rest of the kingdom) release lanterns on her birthday each year in Tangled
Clue 3
Arsenal is mocked for usually terminating in 4th position. In your list, the 4th element from the end is an island that was visited in 2010 – the same year Tangled was released (though interestingly not in Greece, where it wasn't released until January of 2011). It could not be indexed from the start, because that destination was visited in 1995, and you were too young to be in university.
Clue 4
Lanturn is weak to Ground type moves (which many Rock type Pokémon have access to), but Brock's Rock types are weak to Lanturn's Water moves
If I move a magnet towards the coil then the magnet will experience a restive force because of Lenz's law. This way mechanical energy will convert into electrical energy.
But what if the magnet and the coil are a sufficiently large distance apart? The change in the magnetic field travels at the speed of light so the magnet will not feel any restive force instantly. Doesn't this violate the energy conservation law? Because we fix the magnet to a particular point before the restive force arrives at the magnet. And we will able to produce arbitrarily large energy in the coil.
If the magnet starts to move at time t=t1 and lets assume that t0 is the time taken for the flux linking the coil to change due to motion of the magnet.
t0 = (distance between the magnet and the coil/velocity of light)
the magnet will begin to experience resistance at time (t1+2t0). so the delay will be 2t0. In this time period (2t0), magnet is free without experiencing any resisting force. During this time period we can draw current arbitrarily large value by making arbitrarily large numbers of turns of the coil. If energy is conserved, then what mechanism stops us to draw any large amount of power from the coil ?
The Lorentz matrix defines the transformation of a four-vector between different frames of reference, such that $$ p^{'\mu} = \Lambda^{\mu}_{\ \ \nu}p^{\nu} $$ where in this example $p^{\mu}$ is the four-momentum.
1) Are Lorentz transformations of this form only valid for constant (not changing in magnitude) velocities?
I guess so, since $\gamma$ is a function of $v^2$. How can we transform between accelerating frames?
2) Is Lorentz invariance a law of nature?
Which physical quantities should we expect to be invariant (forces? charge?)?
3) What are the eigenvectors and the eigenvalues of the general Lorentz matrix?
I mean what is their physical significance? They do not change under Lorentz transformations?
(I know the ones for the boost in the z direction are something like the Doppler shifted frequencies, but what does this mean? They are the same in all frames? What about the eigenvalues for the boost in a random directiom matrix?)
Answer
A Lorentz transformation lets you compute an object's properties in one inertial frame, given its properties in another inertial frame. Inertial frames, by definition, do not accelerate. An accelerating object is always instantaneously at rest in some inertial frame.
Whether such-and-such is a law of nature is an experimental question. We have no evidence that Lorentz invariance is broken, but people are looking. You might look at the participants in this conference to get an idea of the field.
The most general Lorentz matrix is a product of three rotations and three boosts. For pure rotations we may always choose our coordinate system so that the Lorentz matrix has the form $$\left(\array{ 1\\ &1\\ &&\cos\theta & -\sin\theta \\ &&\sin\theta & \cos\theta \\ }\right).$$ A timelike vector, or a vector along the rotation axis, has eigenvalue 1, since they are not affected by rotations. In the plane of rotation the eigenvectors are $(1,\pm i)$; all real vectors in the plane of rotation get rotated. The corresponding eigenvalues are $e^{±i\theta}$.
Similarly, we may always choose our axes so that a boost is written $$ \left(\array{ \gamma & -\gamma\beta \\ -\gamma\beta & \gamma \\&&1\\&&&1}\right) .$$ Some algebra shows that the non-unity eigenvalues of this matrix are $$ \gamma (1\pm\beta) = \sqrt\frac{1\pm\beta}{1\mp\beta} $$ which is, as you say, the relativistic Doppler shift between an observer at rest and an emitter in the boosted frame. You can verify by hand that the corresponding eigenvectors are $(1,\pm1,0,0)$. These are the light-like worldlines on a Minkowski diagram: the paths taken by photons which would be found later to have the associated Doppler shifts.
A boost in a random direction would have the same four eigenvalues: $\gamma(1\pm\beta)$ for light-like vectors parallel and antiparallel to the boost, and unity for vectors in the spacelike plane perpendicular to the boost.
I'm having trouble getting my head around the time dilation paradox.
Observer A and B are at the same "depth" in a gravity well. Observer B then descends into the well. A will observe B's time as going slower than their own. B will observe A's time as going faster than their own.
What happens if B were to ascend the well back to A's depth, would B's local time speed back up to the same rate as A's, but B would be younger (relative to A)?
What about the paradox caused by relative motion (ignoring gravity)? If A is moving relative to B, A and B will both observe the other's time as going slower. If A and B were together initially, then B moves away and returns, do their clocks agree? they can't both be younger than each other :s (i get thats the paradox, but what explanation resolves it?)
Thanks
Answer
Calculations show that younger will be the observer who suffered accelerations/decelerations.
In which case does the Hamiltonian $H$ commutes with the momentum $P$?
Can anybody help me? With an example? (No particular or strange Hamiltonians and no particular momenta are involved).
How can I prove that $[H, P] = 0$?
According to Noether's theorem, every continuous symmetry (of the action) yields a conservation law.
In fluid, there is a local particle number conservation law, which is $$\partial{\rho}/\partial{t}+\nabla \cdot \vec{j} ~=~0,$$ where $\rho$ and $\vec{j}$ is the density and current respectively. I just wonder is there any symmetry associated with this conservation law?
Consider any arbitrary shape with arbitrary rotation axes, global and or local to the shape that are able to describe any orientation of the shape. The shape is also defined with an arbitrary density function. Using
$$dI=r^2dm$$
to obtain the moment of inertia in each of those axes, would then
$$\sum \frac{1}{2}I_i\omega_i^2$$
equal the total rotational kinetic energy of the shape?
Hold a pen (or pencil, ruler etc.) using your two index fingers with your fingers at the ends of the object. Now move your two fingers toward each other. Assuming the frictional force provided by both fingers is identical, they should meet at the center of mass of the object, with your fingers alternating between being stationary and moving. When going backwards from the centre, one finger stays stationary the whole time. Why is this the case?
I have a general 'hand-wavey' explanation (the closer to the centre of mass a finger is, the greater the reaction force and thus the frictional force exerted, the finger with less friction moves until it is holding more weight and the frictional force becomes too large...) but I am looking for a much better detailed (mathematical if necessary) explanation including all the forces at play, torque, moments etc.
Sorry if this is considered a duplicate, but I have further questions that are based on an old post. I was told that air pressure is strong enough to hold up water. Is this true? The vapour pressure of water at room temperature is 0.0313 atm which is far smaller than air pressure. But air pressure is 101325 N/m$^2$ and a tall column of liquid water with a surface area of 0.001m$^2$ and height of 1000m is 997 kg. This translates into a net gravitational force of 9771 N or 9771000 N/m$^2$ which is greater than that of air pressure in this contrived example.
Also, in the experiment of inverting a cup with water where there is a flat sheet cover on the opening of the cup, does the flat layer simply suppress amplitude oscillations of the water at the interface between the water and layer and thus stop Rayleigh-Taylor instabilities from growing? Does the sheet have to be perfectly flat? What is stopping the sheet from falling down? Are there important water-surface interactions that keep the surface from falling? Is this valid for any liquid (e.g. oil) and any surface (e.g. denser than air) that is not porous? Is it still consider the Rayleigh-Taylor instability once the evolution become nonlinear?
Answer
Yes, it is true, depending on the size of the vessel.
If you fill a vessel with water and place a massless, impermeable membrane across its mouth, then whether or not the water will fall out when you invert the vessel depends on the balance of fluid pressure pushing out and atmospheric pressure pushing in. The water will stay in the cup if $$P_{atm} \geq \rho_{water}gD$$
where $D$ is the depth of the cup. Roughly speaking, the trick works if $ D\leq 10 \text{ m}$, which is a pretty reasonable requirement.
Note that the pressure balance does not depend on the presence of the membrane. However, the presence of the membrane prevents the onset of the Rayleigh-Taylor instability by not allowing the air to "bubble up" through the water.
Because the membrane resists deformation, there is no instability; in the absence of such protection, the arrangement of water-above-air becomes unstable to perturbations in the air/water interface and the experimenter gets all soggy.
Can anyone explain where the following expression for the electric field vector comes from? $$ \mathbf E(\mathbf r,t) = -\nabla \phi(\mathbf r,t) - \frac{\partial}{\partial t}\mathbf A(\mathbf r,t) $$
where $\phi$ and $\mathbf A$ are the scalar and vector potentials, respectively.
Presumably it can it be derived from Maxwell's Equations?
Most people own me
Although I am yours, you don't often use me
Others use it more frequently than you yourself
What am I?
Answer
is it
name
because
our name : we don't use it , others use it often(to call us)
[..quoting from Page 11 of Polchinski Vol2..]
Given $1+1$ conformal bosonic fields $H(z)$ one has their OPE as, $H(z)H(0) \sim -ln(z)$
Then from here how do the following identities come?
$e^{iH(z)}e^{-iH(z)} \sim \frac{1}{z}$
$e^{iH(z)}e^{iH(0)} = O(z) $
$e^{-iH(z)}e^{-iH(0)} = O(z) $
$\langle \prod_{i} e^{i\epsilon_i H(z_i)} \rangle_{S_2} = \prod_{i
I have pretty much no clue how these are derived!
(..I have done many OPE calculation earlier but this one beats me totally!..)
It would be great if someone can either show the derivation or give a reference where this is explained!
Also I am curious if there is a generalization of this to arbitrary even dimensional complex analytic complex manifolds...
Answer
I think there is a typo in the first formula. Let me propose this (partial) answer for the $3$ first formulae:
Because $H(z)H(0) \sim -ln(z)$, we may write the OPE for any pair of operators $F(H), G(H)$ functions of $H$ (in analogy with formula $2.2.10$ p.$39$ vol $1$)
$$:F::G: = e^{- \large \int dz_1 dz_2 ln z_{12} \frac{\partial}{\partial H(z_1)}\frac{\partial}{\partial H(z_2)}} :FG:\tag{1}$$
This gives, for $F = e^{i \epsilon_1 H(z_1)}, G = e^{i \epsilon_2 H(z_2)}$
$$:e^{i \epsilon_1 H(z_1)}::e^{i \epsilon_2 H(z_2)}: = (z_{12})^{\epsilon_1 \epsilon_2} :e^{i \epsilon_1 H(z_1)}e^{i \epsilon_2 H(z_2)}:\tag{2}$$
So, we have :
$$:e^{iH(z)}::e^{-iH(0)}:~ = \frac{1}{z}~:e^{i H(z)}e^{-i H(0)}: ~\sim \frac{1}{z}:e^{i H(0)}e^{-i H(0)}: \sim \frac{1}{z}\tag{3}$$
$$:e^{iH(z)}::e^{iH(0)}:~ = z~:e^{i H(z)}e^{i H(0)}: ~\sim z~:e^{2i H(0)}: \sim O(z)\tag{4}$$
$$:e^{-iH(z)}::e^{-iH(0)}:~ = z~:e^{-i H(z)}e^{-i H(0)}: ~\sim z~:e^{-2i H(0)}: \sim O(z)\tag{5}$$
[EDIT]
For the last equation, I think it is the same reasoning that the one done in Vol $1$, page $173,174$, formulae $6.2.24$ until $6.2.31$
[EDIT 2]
The formula $1$, and the formula $(2.2.10)$ are not formulae ad hoc. These are the consequence of a definition of the normal ordering, and the definitition of the contractions. These are the consequence of the general formulae $2.2.5$ to $2.2.9$, , for instance :
$$F = :F:+ ~contractions \tag{2.2.8}$$ $$:F::G: = :FG:+ ~cross-contractions \tag{2.2.9}$$
Now, we may specialize to holomorphic fields $Y(z)$, so that $Y(z)Y(0) \sim f(z)$, and write :
$$:F::G: = e^{ \large \int dz_1 dz_2 f(z_{12}) \frac{\partial}{\partial Y(z_1)}\frac{\partial}{\partial Y(z_2)}} :FG:\tag{6}$$ where $F$ and $G$ are functions of $Y$
The specialization to an holomorphic field does not change the logic and the calculus done in $2.2.5$ to $2.2.9$
Which part of the earth would be the safest in case of the major solar flare?
Will it be near equator or something like that?
Answer
Just to confirm that CuriousOne is correct as regards direct radiation damage to your body.
From Solar Flares
Solar flares are gigantic explosions associated with sunspots, caused by the sudden release of energy from “twists” in the sun’s magnetic field. They are intense bursts of radiation that can last for anywhere from minutes to hours. Solar flares and CMEs pose no direct threat to humans—Earth’s atmosphere protects us from the radiation of space weather. (If an astronaut out in space is bombarded with the high-energy particles from a CME, he or she could be seriously injured or killed. But most of us won’t have to worry about that situation.) We could, however, feel the effects of CMEs indirectly, through the disruptions to our technology—some of which could have devastating and lingering effects on civilization.
As an example of this, you could read Quebec Blackout and as we become more reliant on satellites, we spend a lot of effort hardening their electronic components against radiation. We also try to predict Solar weather, so we can turn things off before the flare affects us.
However, taking into account what I have learned from @RobJefferies comments below, your best chance of minimising indirect effects are, as you say in your post, near the equator.
Past the Tolman–Oppenheimer–Volkoff limit, gravity overpowers neutron degeneracy pressure and neutron stars collapse, possibly to black holes. This essay by Graeme Heald suggests that a quark star could form under the event horizon of a black hole, with quark degeneracy pressure preventing the collapse to a singularity. (The Penrose singularity theorem article once claimed it doesn't apply to fermions, "It does not hold for matter described by a super-field, i.e., the Dirac field.")
Is such a quark star possible? (Or any other degeneracy-pressure-supported object under an event horizon?) If so, what's the minimum degeneracy pressure required to resist collapse for a given mass / Schwarzschild radius?
I have a few (possibly very stupid) questions relating to position vectors; more specifically my confusion about them.
Following Halliday and Resnick's text, we define vectors by their magnitude and direction, but not by their 'location' in space. They give as an example the displacement vector, and draw three of the same vector in different locations to emphasize that shifting the vector does not change it.
Then in the next chapter we are introduced to position vectors relative to a given origin, which is a vector extending from the origin to the position of the particle.
How are we to think about position vectors? It seems like location is also important, even though we only defined vectors as having magnitude and direction.
How do we think about the addition of a displacement to a position vector? We define displacement vectors as the difference between two position vectors. Mathematically this is the same as saying that the final position vector is the result of adding the displacement vector to the initial position vector. The vector algebra doesn't care which ones are assigned the label of 'position' or 'displacement'; do we simply agree that when we add a 'displacement' vector to a 'position' vector, we get a 'position' vector?
Thanks
Answer
This is one of those things that (intentionally) gets conflated, though it may be better if we were more consistent about keeping them separate.
So, points don't form a vector space. It makes no sense to ask "what's the location of New York plus the location of DC". However, given two points we can subtract them and get a displacement, and we can add that displacement to points to get new points. The mathematical structure for this is called (among other things) a torsor.
Your text is, however, accurate. If we choose a particular point to be our origin, call it $O$, then we can make a vector $r = P - O$ and call it a position vector for $P$. Now the difference between a position vector and a "regular" vector is that it changes when we change what we consider the origin. When we perform a translation on a system, position vectors change, regular vectors do not.
If I throw a object in empty space, I apply a force to throw that. Then it gains some acceleration and it's speed increases.
So will it's speed keep on increasing, or it will get stable? If yes, what happened to the force I applied. No change in speed, so acceleration = 0, then $F = 0$.
1-In QM every observable is described mathematically by a linear Hermitian operator. Does that mean every Hermitian linear operator can represent an observable?
2-What are the criteria to say whether some quantity can be considered as an observable or not?
3-An observable is represented by an operator via a recipe called quantization if it has an analog in classical mechanics. If not, such as spin since it has no classical analog, do we use data from experiment to guess what this operator could look like? are there any other methods for finding that?
4-Are there other observables, besides spin, which also have no classical analog?
Answer
Questions 1,2:
An observable is an element that is obtained from experiments. You can take this as the definition of an observable. The fact that we make an operator and give it some properties does not change/influence the outcome of an experiment. It just so happens that the theory we have ascribes linear, hermitian operators to explain experiments. With this in mind, it is easy to say that not all linear, hermitian operators we cook up describe observables.
Question 3
Initially, the classical-quantum correspondence was used, but people quickly realized that it was of limited use. The modern view is that nature can be described by Group Theory (especially the Poincare Group) and everything that is observed follows from there. With this in mind, you don't have to guess about the existence of the Spin Operator, it comes up naturally. What is more important though, is the representations of the operator. When you relate theory and experiments, remember that you are dealing with the representations of an operator. An operator cannot be measured and is useless by itself unless you specify the basis.
Question 4
I don't know the answer to this, but I can tell you that we never measure spin by itself, but the interaction of a spin with something else. Why? In my view, that is the definition of a measurement.
In QED, one can relate the two-particle scattering amplitude to a static potential in the non-relativistic limit using the Born approximation. E.g. in Peskin and Schroeder pg. 125, the tree-level scattering amplitude for electron-electron scattering is computed, and in the non-relativistic limit one finds the Coulomb potential. If one allows for 1/c^2 effects in the non-relativistic expansion, one also finds spin-dependent interactions (e.g. spin-orbit, see Berestetskii, Lifshitz, Pitaevskii pg. 337).
Are there any alternative methods for calculating a two-particle non-relativistic potential?
I'm sure this question is a bit gauche for this site, but I'm just a mathematician trying to piece together some physical intuition.
*Question:*Is the statistical interpretation of Quantum Mechanics still, in any sense, viable? Namely, is it completely ridiculous to regard the theory as follows: Every system corresponds to a Hilbert space, to each class of preparations of a system corresponds to a state functional and to every class of measurement procedure there is a self-adjoint operator, and finally, a state functional evaluated at one of these self-adjoint operators yields the expected value of numerical outcomes of measurements from the class of measurement procedures, taken over the preparations represented by the state?
I am aware of Bell's inequalities and the fact that the statistical interpretation can survive in the absence of locality, and I am aware of the recent work (2012) which establishes that the psi-epistemic picture of quantum mechanics is inconsistent with quantum predictions (so the quantum state must describe an actual underlying physical state and not just information about nature). Nevertheless, I would really like a short summary of the state of the art with regard to the statistical interpretation of QM, against the agnostic (Copenhagen interpretation) of QM, at present.
Is the statistical interpretation dead, and if it isn't...where precisely does it stand?
An expert word on this from a physicist would be very, very much appreciated. Thanks, in advance.
EDIT: I have changed the word "mean" to "expected" above, and have linked to the papers that spurred this question. Note, in particular, that the basic thing in question here is whether the statistical properties prescribed by QM can be applied to an individual quantum state, or necessarily to an ensemble of preparations. As an outsider, it seems silly to attach statistical properties to an individual state, as is discussed in my first link. Does the physics community share this opinion?
EDIT: Emilio has further suggested that I replace the word "statistical" by "operational" in this question. Feel free to answer this question with such a substitution assumed (please indicate that you have done this, though).
Answer
The statistical interpretation of quantum mechanics is alive, healthy, and very robust against attacks.
The statistical interpretation is precisely that part of the foundations of quantum mechanics where all physicists agree. In the foundations, everything beyond that is controversial.
In particular, the Copenhagen interpretation implies the statistical interpretation, hence is fully compatible with it.
Whether a state can be assigned to an individual quantum system is still regarded as controversial, although nowadays people work routinely with single quantum systems. The statistical interpretation is silent about properties of single systems, one of the reasons why it can be the common denominator of all interpretations.
[Added May 2016:] Instead of interpreting expectations as a concept meaningful only for frequent repetition under similar conditions, my thermal interpretation of quantum mechanics interprets it for a single system in the following way, consistent with the practice of thermal statistical mechanics, with the Ehrenfest theorem in quantum mechanics, and with the obvious need to ascribe to particles created in the lab an approximate position even though it is not in a position eigenstate (which doesn't exist).
The basic thermal interpretation rule says:
Upon measuring a Hermitian operator $A$, the measured result will be approximately $\bar A=\langle A\rangle$ with an uncertainty at least of the order of $\sigma_A=\sqrt{\langle(A−\bar A)^2\rangle}$. If the measurement can be sufficiently often repeated (on an object with the same or sufficiently similar state) then $\sigma_A$ will be a lower bound on the standard deviation of the measurement results.
Compared to the Born rule (which follows in special cases), this completely changes the ontology: The interpretation applies now to a single system, has a good classical limit for macroscopic observables, and obviates the quantum-classical Heisenberg cut. Thus the main problems in the interpretation of quantum mechanics are neatly resolved without the need to introduce a more fundamental classical description.
I am working on a CMS - related project where the ROOT trees contain both reconstruction-level and generation-level particle variables (like mass). However, I don't know the basic difference between the two or when we prefer using one type of variable over the other?
Answer
Real data goes $$\text{Collisions} \to \text{Detector} \to \text{Trigger} \to \text{Reconstruction}$$
Simulated data goes $$\text{Event generator} \to \text{Simulated detector} \to \text{Trigger} \to \text{Reconstruction}$$
Generator-level is the information from the first step, before the events are passed through the simulated detector. This is also sometimes called "MC Truth" or "Truth level".
Reconstruction-level is the information from the last step, after the reconstruction algorithms have been run over the data from the simulated detector.
You can compare the two in order to quantify things like efficiencies and resolutions.
A photon, a neutrino (if it has zero rest mass) move at $c$ but what about charged particles? If the answer is no, is there a fundamental reason or just because of the radiation it emits?
Yet another Tokyo Metro puzzle based on the map! Puzzles 1, 2, and 3 are avaliable.
Riddle:
From the western of the U
Along the airport line,
Go along until you hit
A shining water line.
Then head upon to the first
And turn around with glee,
Don't forget to go on to
The algae, algae line.
From there you go until you hit
Upon a certain Diet
And then along the Astoria line
Through 3 more stations plus.
Now which station are you at?
Note: For the purposes of my question, when I refer to free fall assume it takes place in a vacuum.
From my (admittedly weak) understanding of the equivalence principle, falling in a gravitational field is physically indistinguishable from floating in interstellar space. This would make sense to me if gravity simply caused an object to move at a constant velocity. Moving at a constant speed, or floating in space, are just two different ways of describing an inertial frame, and are fundamentally no different. But free falling in a gravitational field means accelerating continuously, and doesn't an accelerating body experience a force? Then isn't free falling fundamentally different from floating in space?
Answer
Before telling you why an observer in free fall does not feel any force acting on him, there are a couple of results that should be introduced to you.
Newton's second law is only valid in inertial frames of reference:
To measure quantities like the position, velocity, and acceleration of an object, you need a coordinate system $(x,y,z,t)$. Now the coordinates that an observer in uniform motion(constant velocity) uses is what we call an inertial frame of reference, and the coordinates of an observer in non-uniform motion is what we call a non-inertial/accelerated frame of reference
Now $\vec{F}=m\vec{a}$ is only valid in inertial frames of reference. This means that only observers in uniform motion are allowed to make valid inferences about an object being acted upon by a force(and hence being accelerated) and so on, while non-inertial frames of reference are not justified in making inferences about an object being accelerated or not.
Fictitious forces
For example, consider the case of two observers, one who is at rest on the ground and the other who is in an accelerated car(say moving in the positive x-axis with constant acceleration) that is passing by the observer resting on the ground. The observer in the car will discover a very peculiar situation in his frame of reference, when he holds his medal by a string, he immediately observes that the medal starts to move backwards in the negative x-direction and the string that is holding the medal makes an angle with the vertical. If he has a ball in his hand and lets it go, he observes that the ball starts to accelerate backwards(negative x-direction) until it hits the back of his car. So it appears as if there is some mystical force in this observer's frame that has no obvious origin, which acts upon all objects and accelerates them backwards. This observer will further note that this mystical force is proportional to mass, or in other words, the acceleration of any object is independent of it's mass, so that if you hold two different masses in your hand and let them go, they will hit the back of the car at the same time.
But the observer who's at rest on the ground will object! he will argue(rightly) that there is no mysterious force that is accelerating the objects in the car. The fact that any object "appears to accelerate" backwards is a simple consequence of these two following facts:
1)The car is accelerating onward in the positive x-direction.
2)The objects, when they are let go, they are moving with constant velocity(they both have the same velocity) in the positive x-direction according to the ground observer, and following Newton's first law, they will continue to do so, but the car is still accelerating onward, so they eventually hit the back of the car at the same time.
So as you can see in the above example, an observer in accelerated frame, when making inferences about an object being accelerated or not, will arrive at the wrong conclusions, since Newton's law are only valid in an inertial frame. If he makes inferences, he concludes the existence of some fictitious force with no obvious origin, that is proportional to the mass, but this is just an artifact of the observer being in an non-inertial frame and using Newton's laws to make inferences about the motion of objects. This fictitious force can simply be explained due to the combined result of the acceleration of the car and the inertia of the bodies inside the car that were just let go.
(There is one complication which I ignored in this example, namely gravity, actually when the masses are let go inside the car, their trajectories are not going to be straight lines, but sections of a parabola, But if you performed the above example in space-free gravity, the example holds exactly true).
The litmus test for an inertial frame of reference
Newton's first law is the litmus test to tell apart whether you're accelerating or not. If you are floating in space, and there is an object in your hand, and you let it go(at rest), it will stay at rest. But if you're accelerating(like the case of the car), and let the mass go, it will start to mysteriously accelerate with a force that is proportional to the mass.
Einstein's big idea
The fact that gravity has no obvious origin, and is proportional to mass, prompted him to suggest that maybe gravity is just another fictitious force, that results from us, observers who are at rest on the ground being in an accelerated frame of reference.
But to ultimately prove this is true, he had to find a frame of reference in which this force of gravity disappears, just as we concluded that the mystical force in the car's frame of reference is fictitious, by switching to the frame of reference of an observer who's standing on the ground.
And Einstein found such a frame! Switch to a freely falling frame of reference and this mystical force of gravity suddenly disappears; you feel weightless. Put a scale down at your feet and it will read zero. Try to hold a ball by a string that is attached to your hand and the tension on the string immediately disappears, and it becomes loose as you start to freely fall, and so on. In such a frame there's no force of gravity, just as there is no mystical force when you switch from the car to the ground frame of reference.
Newton's explanation
Newton will argue that gravity is not fictitious, but real. The fact that you feel no force acting on you when you're in free fall can be explained like this:
According to Newton, an observer in free fall is being acted upon by the force of gravity, so he's accelerating, so his frame of reference is not inertial and any inferences he makes about motion using Newton's laws are incorrect. Since the freely falling observer is accelerating, in his frame, there appears a fictitious force that acts on him upward and is proportional to his mass, but gravity acts on him downward and it's proportional to his mass as well! Therefore, they will cancel each other out, and he feels no force, even though he's accelerating!
Einstein responds
Einstein used the litmus test to tell whether, being in free fall, one is in an inertial frame of reference or not. You hold a certain mass in your hand, and let it go, and it stays at rest with respect to you. This case is totally equivalent to the observer floating in space we described above.
One the other hand, if you're on the ground holding an object and then let it go, it does not stay at rest, but rather it starts to accelerate downward with a force that is proportional to its mass. This case is totally equivalent to the case of an observer in a car letting go of masses that we described above.
He called this the equivalence principle.
So yeah, gravity is indeed fictitious.
Now after you have been introduced to the relevant concepts, the response to your assertion that "But free falling in a gravitational field means accelerating continuously. And doesn't an accelerating body experience a force?" is something like this:
To draw valid inferences about acceleration of any object, you have to be in an inertial frame of reference, otherwise you're led to the wrong conclusions, just as we demonstrated above. Your assertion that a body in a gravitational field is accelerating, and hence should experience a force, is false in the Einsteinian sense. That is because, as we have noted above, you made this assertion being on the ground, and an observer on the ground according to Einstein is in an accelerated/non-inertial frame of reference, so his inferences about a body in a gravitational field being accelerated will be false. Only freely falling observers are justified in making claims about acceleration of objects because they are in an inertial frame of reference.
But even ignoring Einstein and sticking to Newton's worldview, an observer in a free fall experiences no force at all, because gravity(which is real according to Newton) and the fictitious force exactly cancel each other out, even though he is accelerating!
So as you can see, in either cases, whether Newtonian or Einsteinian, an observer in free fall does not feel any force acting on him.
Credits should be given to this video.
Most images you see of the solar system are 2D and all planets orbit in the same plane. In a 3D view, are really all planets orbiting in similar planes? Is there a reason for this? I'd expect that the orbits distributed all around the sun, in 3D.
Has an object made by man (a probe) ever left the Solar System?
Answer
Nic and Approximist's answers hit the main points, but it's worth adding an additional word on the reason the orbits lie roughly in the same plane: Conservation of angular momentum.
The Solar System began as a large cloud of stuff, many times larger than its current size. It had some very slight initial angular momentum -- that is, it was, on average, rotating about a certain axis. (Why? Maybe just randomly! All of the constituents were flying around, and if you add up those random motions, there'll generically be some nonzero angular momentum.) Because angular momentum is conserved, as the cloud collapsed the rotation rate sped up (the usual example being the figure skater who pulls in her arms as she spins, and speeds up accordingly).
Further collapse in the direction perpendicular to the plane of rotation doesn't change the angular momentum, but collapse in the other directions would change it. So the collapse turns the initial cloud, whatever its shape, into a pancake. The planets formed out of that pancake.
By the way, you can see the signs of that initial angular momentum in other things too: not only are all of the planets orbiting in roughly the same plane, but so are most of their moons, and most of the planets' rotations about their axes as well.
