Given that the second law of thermodynamics states that entropy always increases and the universe is currently not in a state of maximal entropy does that not imply that it must have started at a point in the past? And that it could not have existed forever?
Thermodynamics was developed in the mid 1800's while the big bang was first proposed in the 1920's if the above argument is correct did anyone else argue it before general relativity?
We can imagine many changes to the laws of physics - you could scrap all of electromagnetism, gravity could be an inverse cubed law, even the first law of thermodynamics could hypothetically be broken - we've all imagined perpetual motion machines at one time or another.
However, the second law of thermodynamics seems somehow more 'emergent'. It just springs out of the nature of our universe - the effectively random movement of physical objects over time. Provided you have a Universe whose state is changing over time according to some set of laws, it seems like the second law must be upheld, things must gradually settle down into the state of greatest disorder.
What I'm particularly wondering is if you can prove in any sense (perhaps using methods from statistical mechanics)? Or is it possible to construct a set of laws (preferably similar to our own) which would give us a universe which could break the second law.
Answer
The short answer is that such a universe cannot be envisaged, not with relevance to our known physics.
Entropy as defined in statistical thermodynamics is proportional to the logarithm of the number of microstates of the closed system, the universe in your question. You would have to devise a universe where the number of microstates diminishes with time.
The great multiplier of microstates in our universe is the photon, which is emitted at every chance it gets, and thus increases the number of microstates. Photons are emitted by electromagnetic interactions and by all bodies consisting of atoms and molecules due to the black body radiation effect. Each emitted (or absorbed, because the state of the atom that absorbed it has changed) photon defines a new microstate to be added to the number of microstates, whose logarithm defines entropy. A universe without electromagnetism would not have atoms.
It is worth noting that all biological systems decrease entropy, as does the crystallization of materials, but this is possible because the systems are open and the energy exchanges create a large number of microstates thus obeying in the closed system the entropy constraint.
What books are recommended for an advanced undergraduate course in electrodynamics?
The vector potential $A$ is perpendicular to $B = \nabla \times A$, by definition, and hence, in a plane wave, it is either in the direction of $E$ or the direction of propagation. I suspect it is in the direction of propagation.
What is its direction?
Answer
While for vectors $\vec{B}$ and $\vec{C}$, the cross product $\vec{B}\times\vec{C}$ is indeed perpendicular to both of the vectors, it is simply not the case that the curl of a vector field is orthogonal to the vector field. Do not read too much into the cross product notation.
In particular, you can add any constant vector field to $\vec{A}$ without changing the fields. So we can make it be nonorthogonal by adding a constant of our choice. When someone tells you a vector potential points in a particular direction they are simply making a gauge choice, and a different choice of gauge can result in the vector potential pointing in a different direction.
This means your question simply isn't well defined. We can find the direction of the electric field by seeing the force per unit charge of stationary charges, and we can find the magnetic field by finding the force on moving charges that move in three linearly independent directions. But there is no classical experiment to find the direction the vector potential points, so it isn't a scientific question.
Seeing a video of astronaut Chris Hadfield playing a guitar on the International Space Station made me wonder if a guitar or other stringed instrument played in zero-G would sound any different than on earth.
It seems that on earth, the downward pull of gravity could cause an asymmetrical oscillation of the guitar string, with a larger amplitude as the string moves downward due to the pull of gravity. (of course, it wouldn't take zero-G to test this since a guitar could be held vertically)
With the amount of tension that a guitar string is under, perhaps this effect is so minuscule so as to be unnoticeable?
Answer
The effect of gravity is miniscule, and here's why:
The speed of sound in a string is basically $$ v = \sqrt{\frac{T}{\lambda}}, $$ where $T$ is the tension and $\lambda = M/L$ is the mass per unit length. The frequency of a plucked string will then be this sound divided by the length of the oscillator: $f = v/L$. Combining and rearranging tells us the tension keeping the string taught will be $$ T = MLf^2. $$
Gravity can alter this tension only by something of the order of $$ T_\mathrm{grav} = Mg, $$ give or take some factor of order unity depending on orientation. Thus the intrinsic tension overwhelms the effect of gravity by a factor of something like $$ x \equiv \frac{T}{T_\mathrm{grav}} = \frac{Lf^2}{g} \approx \frac{(1\ \mathrm{m})(500\ \mathrm{Hz})^2}{10\ \mathrm{m/s^2}} = 2.5\times10^4. $$
Since frequency goes as the square root of tension, moving the string from free fall to standing still in the Earth's gravitational field will change the frequency by something like one part in $2x = 5\times10^4$.
In summers when we switch off the air conditioner, the room seems to instantly get hot again. But in winter, when we switch off the heater the room seems to remain hot for some time. Why this difference?
First question: Is the question I asked valid in the first place? Is it really true? Second question: If so, is it because the exchanged heat between the bodies between which the temperature gradient exists is also infinitesimal and hence negligible? The explanation I provided is not convincing for me as, in physics, "infinitesimal" is not always synonymous with "negligible".
We all know that gravitation force between two small (not heavenly) bodies is negligible. We give a reason that their mass is VERY small. But according to inverse square law, as $r\to 0$, then $F\to \infty$. But in real life we observe that even if we bring two objects very close, no such force is seen.
Why is this so?
Answer
The inverse-square law holds for spherically symmetric objects, but in that case the main problem is that $r$ is the distance between their centers. So "very close" spheres are still quite a bit apart--$r$ would be at least the sum of their radii.
For two spheres of equal density and size just touching each other, the magnitude of the gravitational force between them is $$F = G\frac{M^2}{(2r)^2} = \frac{4}{9}G\pi^2\rho^2r^4\text{,}$$ which definitely does not go to infinity as $r\to 0$ unless the density $\rho$ is increased, but ordinary matter has densities of only up to $\rho \sim 20\,\mathrm{g/cm^3}$ or so.
Tests of Newton's law for small spheres began with the Cavendish experiment, and this paper has a collection of references to more modern $1/r^2$ tests.
Consider a heat reservoir which gains heat $Q$ irreversibly at temperature $T$ from the surroundings which is at temperature $T_0$. The entropy change of reservoir is then given by $\frac{Q}{T}$, while that of the surroundings is $-\frac{Q}{T_0}$.
My question is, how is this possible? According to the Clausius inequality, the entropy change of a irreversible process is greater than that due to heat transfer. Please help, thank you!
My question is even though photons have no (rest) mass, do they emit a external force due to EM radiation causing electrons to be excited and jump to higher energy shells which electrons have mass thus photons can emit a kinetic force? I am new so I would like to get the record straight on this issue.
I am looking for a computer program which simulates the Schrödinger equation (say for a single particle) in two dimensions and for potentials and initial states specified by the user. Typical situations I have in mind are the double slit experiment, harmonic oscillator, Aharonov–Bohm, tunneling effect etc. Unfortunately I was only able to find the one dimensional case on the web.
So are there some 2d programs available?
PS: I don't think my question is a duplicate of this one. I only ask about a single particle, not about thousands. On the other hand I want freedom in my choice of potential. Also I want a numeric solution of a PDE, while the other question is about (as far as I understand, which is not ver far) a Monte Carlo simulation. Anyway these are completely different requirements.
In John Terning, Modern Supersymmetry, $R$-symmetry is introduced as below: "The SUSY algebra is invariant under a multiplication of the supercharges by a phase, so in general there is a linear combination of U(1) charges called $R$-charge, that doesn't commute with the supercharges."
He the proceeds to write down the following commutator relations: $$[Q_{\alpha}, R] = Q_{\alpha}, $$ $$[Q_{\alpha}^{\dagger}, R] = - Q_{\alpha}^{\dagger}$$
How are the above relations derived?
Paper is an extremely flexible material, at least when it is in sheet form. It will deform significantly according to the pressure applied and it is easy to fold.
Therefore, it's extremely counterintuitive that a sheet of paper could cut through human skin and probably through stiffer/harder materials, since when the skin applies a pressure on the paper, one would expect it to fold or bend. Yet it is easy to have a severe cut from paper, through both the epidermis and the dermis. How is that possible? Certainly the width of the sheet of paper plays a big role: the smaller it is, the sharper it is, but also the more flexible it becomes and the less it should sustain an applied pressure without folding up!
I can think of other materials such as thin plastic films and aluminium foils. My intuition tells me the plastic foil would not cut through skin but the aluminium foil would, although I am not sure since I did not try the experiment. If this hold true, what determines whether a material would be able to cut through skin? A hair for example, which is flexible and thinner than a paper sheet, is unable to cut through the skin. What makes paper stand out? What is so different that makes it a good cutter?
Maybe it has to do with its microscopic properties and that it contains many fibers, but I highly doubt it because the aluminium foil does not contain these and yet would probably also cut.
Answer
Paper, especially when freshly cut, might appear to have smooth edges, but in reality, its edges are serrated (i.e. having a jagged edge), making it more like a saw than a smooth blade. This enables the paper to tear through the skin fairly easily. The jagged edges greatly reduce contact area, and causes the pressure applied to be rather high. Thus, the skin can be easily punctured, and as the paper moves in a transverse direction, the jagged edge will tear the skin open.
Paper may bend easily, but it's very resistant to lateral compression (along its surface). Try squeezing a few sheets of paper in a direction parallel to its surface (preferably by placing them flat on a table and attempting to "compress" it laterally), and you will see what I mean. This is analogous to cutting skin with a metal saw versus a rubber one. The paper is more like a metal one in this case. Paper is rather stiff in short lengths, such as a single piece of paper jutting out from a stack (which is what causes cuts a lot of the time). Most of the time, holding a single large piece of paper and pressing it against your skin won't do much more than bend the paper, but holding it such that only a small length is exposed will make it much harder to bend. The normal force from your skin and the downward force form what is known as a torque couple. There is a certain threshold torque before the paper gives way and bends instead. A shorter length of paper will have a shorter lever arm, which greatly increases the tolerance of the misalignment of the two forces. Holding the paper at a longer length away decreases this threshold (i.e. you have to press down much more precisely over the contact point for the paper to not bend). This is also an important factor in determining whether the paper presses into your skin or simply bends.
Paper is made of cellulose short fibers/pulp, which are attached to each other through hydrogen bonding and possibly a finishing layer. When paper is bent or folded, fibers at the folding line separate and detach, making the paper much weaker. Even if we unfold the folded paper, those detached fibers do not re-attach to each other as before, so the folding line remains as a mechanically weak region and decreasing its stiffness. This is why freshly made, unfolded paper is also more likely to cause cuts.
Lastly, whether a piece of paper cuts skin easily, of course depends on its stiffness. This is why office paper is much more likely to cut you than toilet paper. The paper's density (mass per unit area), also known as grammage, has a direct influence on its stiffness.
I've been trying to get to grips with SpaceTime.
As I understand it, we move at a set rate through spacetime. Any increase in our rate of travel through space results in a decrease in our rate of travel through time (via standard vector maths, pythagoras)
Where I get confused is the implication for our perception of time. Do we perceive time as the magnitude of our movement through spacetime, and the speed of someone stationary as the 'time' component of our vector through spacetime? Wouldn't this mean that, as a diagonal line is longer than a straight one of the same height, that things around us should slow down as we move more quickly?
Say this triangle is our vector in Spacetime. If $a$ is the 'time' component of spacetime, and $b$ is the 'space' component, does that mean that the $c$ component is of fixed magnitude? Do we perceive things around us moving at a rate of $a/c$? (ie, slower than they occur now)
I know other questions about spacetime have been asked, but I can't find one that explains this aspect.
Answer
You've hit on one of the fundamental "weirdnesses" of the Minkowski "metric". Its name "metric" is a bit misleading if you're a mathematician: it does not fulfil two out of the of the three axioms of the distance function defining a metric (in the topologist's words) space. There are null vectors, i.e. nonzero vectors $X$ for which $\langle X,\,X\rangle=0$ and it is NOT subadditive i.e. it does not fulfill the triangle identity. So the intuition you have from Euclidean (and general metric) space that the sum of the lengths of two of a triangle's sides must be longer than the third alone is WRONG!
The names "metric", "inner product" and "norm" in connexion with Minkowski spacetime reflects (1) the "structural" likeness that these operations as algebraic operations have to genuine metrics, inner products and norms (written on a page, they look a great deal like the genuine ones) and (2) the fact that many of the main theorems of Riemannian geometry also hold when we replace genuine inner products with nondegenerate ones (i.e. the matrix representing the two form is nonsingular) such as the Minkowski "inner product".
I recall many years ago reading Bert Mendelson in his wonderful introductory undergrad textbook "Topology" making the memorable and intuitive statement:
".... the [triangle] inequality $\mathrm{d}(x,\,z)\leq \mathrm{d}(x,\,y) + \mathrm{d}(y,\,z)$ may be thought of as asserting the transitivity of [the] closeness [relationship]; that is, if $x$ is close to $y$ and $y$ is close to $z$, then $z$ is close to $x$." (Chapter 2, Section 2, p 31 in my edition)
This wonderful statement - this transitivity - is precisely what is shattered in the twin paradox. The spacefaring twin at the halfway point ($y$) in his journey can be close to his beginning point ($x$) (he ages very little - the path has only short proper time), and at the end of his journey at ($z$) he can be near to his halfway point (again, he ages very little coming back), but ($z$) in spacetime is far, far away from ($x$): the Earthbound twin is now a decrepit, bent up old man.
PS if you study topology and they try to get you to read Munkres, I highly recommend Bert Mendelson's book to be read together with the heavier Munkres.
I understand that the fundamental representation of $U(1)$ amounts to a multiplication by a phase factor, e.g. EM. I thought that when it is extended to higher dimensional representations, it would just become a phase factor times the identity matrix.
Can someone explain where the hypercharge comes into the $U(1)$ generator matrices in $SU(2)\times U(1)$ model, e.g. $Y = -(1/2) I$ in $(2, -1/2)$ representation?
I don't quite understand where the "$-1/2$" comes from. Where do all these hypercharges come from?
What is the logic behind choosing a particular value like $-1/2$?
Can one use the Lagrangian formalism for all classical systems, i.e. systems with a set of trajectories $\vec{x}_i(t)$ describing paths?
On the wikipedia page of Lagrangian mechanics, there is an advertisement, which says that it also works for systems for which energy and momentum is not conserved. It's unreferenced but it sounds nice, but I wonder if there are other problems one might encounter. Does this statement already mean that all systems can be described by a Lagrangian?
I have found that, at least in some dissipative systems, you have to introduce non-standard Lagrangians, which are not of the form $L=T-V$ and so there is no clear kinetic and potential term. However, from a Newtonian point of view, there is still the $T:=\sum \frac{m}{2}\vec{x}_i'(t)^2$ term. Does that mean that you have a kinetic term, but it just isn't part of the Lagrangian?
Also, if it's possible to write down the Lagrangian (even if there is not such thing as a conserved energy), what is the meaning of the Legendre transform of this Lagrangian? Usually, it would be the Hamiltonian, but now, is it just any random function without any use?
I don't mind some differential geometry speak btw.
Lastly, are there extension of this, i.e. variational principles, which are far away from the Lagrangian ideas?
Answer
Roughly speaking, physicists thought a lot about this right before the revolutions of relativity and quantum theory. Heinrich Hertz reduced all of classical mechanics to a kind of Lagrangian and Hamiltonian framework and a new principle of least curvature. See Hertz, The Principles of Mechanics, out of copyright, http://www.archive.org/details/principlesofmech00hertuoft and Whittaker, Analytical Dynamics, pp. 254ff. Their thoughts turned out to be very helpful for general relativity, Wave Mechanics, and Quantum Field Theory.
Hertz's ideas of least curvature are very close to Lagrange's ideas.
All classical mechanics can be put into the Lagrangian framework: if energy is not conserved (for example, if the system is an open system, if friction is present, etc.) then merely one adjusts to allowing a time-varying Lagrangian.
But the practical utility of this formulation is sometimes low: questions about Statistical Physics require a different way of looking at the phase space and the system: its laws of motion are almost irrelevant and the kind of information about the trajectories of the parts of the system which the Lagrangian equations gives you are almost useless, one instead wants to know things such as their auto-correlation functions, which are almost independent of the particular trajectory or initial condition chosen.
I was reading a couple of Earth-Moon related questions (Mars just collided with Earth! A question of eccentricity, Could the earth have another moon?) and they got me thinking about planet-moon systems in general.
Binary star systems are pretty common. The types of the two stars in the binary can vary pretty widely (main sequence, puslar/neutron star, black hole, white dwarf, giant phases, etc, etc), but some are formed of a pair of roughly equal mass (within a factor of <10).
I can't say I've ever heard of a binary planet system, though. Of course a planet with a moon is sort of a binary, but I've never heard of an equal mass binary planet. I think the closest thing in the solar system would be the Pluto-Charon system, with a mass ratio of about 10:1.
Is there any reason a binary planet would be unstable? Obviously this is a three-body system, which has some inherent instability, but Earth-Moon-Sun seems pretty stable. Would increasing the mass of the Moon to match that of Earth make the system unstable?
How about gas-giants? I think a Jupiter-Jupiter binary close to a star would be short lived because of three-body interactions, but what about further out? Would, for instance, a double-Jupiter or double-Saturn be stable in our solar system? Or is there some tidal effect that would cause the orbit to decay and the binary planet to merge?
As an aside, it seems that binary asteroids aren't terribly difficult to find... perhaps we just haven't seen any binary planets yet because they're only stable relatively far away from their star, making them difficult to detect outside our own solar system?
Answer
They are sometimes also called double planets and they're more widespread in fiction than in observations. I don't think that there is any new instability that would appear for the system of double planets orbiting a star and that wouldn't be present for other, more asymmetric pairs of planets. Obviously, the tidal forces would be really large if the planets were close enough to each other. But because the tidal forces go like $1/r^3$, it's enough to choose the distance that is 5 times larger than the Earth-Moon distance and the tidal forces from the other Earth would be weakened 125 times and would already be as weak as they are actually from the Moon now (with a lower frequency).
One must realize that the systems with two or several planets are rather rare and the condition that the mass of the leading two planets is comparable is even more constraining.
Imagine that each of the two planets has a mass that is uniformly distributed between 1/20 of Earth's mass (like Mercury) and 300 Earth masses (like Jupiter) on the log scale. The interval goes from the minimum to the maximum that is 6,000 times heavier. That's more than 12 doublings, $2^{12}=4,096$. So if you pick the first planet to be at a random place on that interval of masses (uniformly at the log scale), the probability that the second planet's mass (which is independent) differs by less than the factor of $\sqrt{2}$ from the first mass is about $1/12$.
Only $1/12$ of systems that look like a pair of planets will be this symmetric. And the number of pairs of planets - even asymmetric ones – is rather low, indeed. The reason is really that during the violent eras when Solar-like systems were created, rocks had large enough velocities and they flew in pretty random directions so they were unbound at the end. It's just unlikely to find two large rocks in the same small volume of space: compare this statement with some high-temperature, high-entropy configurations of molecules in statistical physics.
It's also rather unlikely that a collision with another object creates two objects that will orbit one another. After all, the two-body orbits are periodic so if the two parts were in contact during the collision, they will collide again after one period (or earlier). Equivalently, the eccentricity of the orbit is likely to be too extreme which will lead to a fast reunification of the two new planets. Moreover, even if something would place the two newly created planets from a "divorce" on a near-circular orbit, perhaps a collision with a second external object (good luck), it's very unlikely that such an orbit will have the right radius, like the 1 million km I was suggesting in the case of the hypothetical double Earth above. If the two objects are too close, the tidal forces will be huge and (at least for some signs of the internal angular momentum) they will gradually make the planets collapse into one object again. And if the energy with which the planets are ejected from one another is too high, no bound state will be created at all. So the initial kinetic energy of the newborn 2 planets would have to be almost exactly tuned to their gravitational potential energy (without the minus sign) and that's generally unlikely, too.
This recent news article reports that DARPA is doing work with "quantized inertia", despite their claim that it's not widely accepted by physicists:
The Defense Advanced Research Projects Agency (DARPA) recently awarded a $1.3 million contract to an international team of researchers to study quantized inertia, a controversial theory that some physicists dismiss as pseudoscience.
Quantized inertia (QI) is an alternative theory of inertia, a property of matter that describes an object’s resistance to acceleration. QI was first proposed by University of Plymouth physicist Mike McCulloch in 2007, but it is still considered a fringe theory by many, if not most, physicists today. McCulloch has used the theory to explain galactic rotation speeds without the need for dark matter, but he believes it may one day provide the foundation for launching space vehicles without fuel.
–"DARPA Is Researching Quantized Inertia, a Theory Many Think Is Pseudoscience", Motherboard (2018-10-02)
Questions:
Is it true that most physicists do not accept quantized inertia theory?
Would quantized inertia imply that the conservation of momentum is only approximate?
I also found a paper on arXiv:
It is shown here that if we assume that what is conserved in nature is not simply mass-energy, but rather mass-energy plus the energy uncertainty of the uncertainty principle, and if we also assume that position uncertainty is reduced by the formation of relativistic horizons, then the resulting increase of energy uncertainty is close to that needed for a new model for inertial mass (MiHsC, quantised inertia) which has been shown to predict galaxy rotation without dark matter and cosmic acceleration without dark energy. The same principle can also be used to model the inverse square law of gravity, and predicts the mass of the electron.
–"Quantised inertia from relativity and the uncertainty principle", M.E. McCulloch (2016-10-13)
If I push or hit an object in space (vacuum and no gravitation) in direction what is not going trough its centroid, will it rotate or move along in straight line?
I expect that on earth it will depend on what is less difficult for the object (rotation or linear movement). So the object will do some kind of combination of both movements (rotating and also moving along the direction of impulse or force).
But how could an object "decide" what to do in space, where is not resistance?
Answer
Any linear force not going through the centre of mass will create torque, which I hope you know, is related to how far from the centre of mass the line of force is.
So, if you manage to hit the object exactly at its centre of mass, i.e. the line of force is directly passing through the centre of mass, then it will show NO ROTATION. It will go straight ahead in a line.
But, if you fail to do so, i.e. the line of force misses the centre of mass, it will show BOTH kind of motions, Rotational and Linear. It will go straight ahead in a line as in previous case, but will also rotate. How much is the speed of rotation depends on how badly you missed the centre of mass.
But in both cases, the total momentum will be (has to be) same.
I've been trying to solve this by myself without success. I am trying to find the velocity of a falling object as its height varies (change in velocity as it falls). Although it is often assumed that the acceleration is constant, I know that it isn't due to Newton's Law of Gravitation, where a higher height will cause acceleration to be lower. So I started off with this.
$$a=G\frac{M}{\left(R+h\right)^2}$$
Then, I got a relation with velocity:
$$a=\frac{\text{d}v}{\text{d}t}=\frac{\text{d}v}{\text{d}h}\times\frac{\text{d}h}{\text{d}t}=v\times\frac{\text{d}v}{\text{d}h}$$
So to find velocity in terms of h, I integrated both sides w.r.t. h and came to $$\int_{}^{}a(h)dh=\frac{1}{2}v(h)^2+C$$
But I didn't know what to do with that constant and after doing some research I realised that I had to make it in a definite integral form, with set values. So I had final position $h_f=0$ since that would be when the object comes to a stop (hits the ground) with $h$ as a changing variable and initial velocity $v_0=0$ since at release, the velocity is 0. So I tried: $$\int_{h}^{0}a(h)dh=\frac 12\left(v_0^2-v(h)^2\right)=-\frac{1}{2}v(h)^2$$
I integrated $a$ from above: $$\int_{h}^{0}\frac{GM}{(R+h)^2}dh=\left[-\frac{GM}{R+h}\right]_h^0=-\frac{GM}{R}+\frac{GM}{R+h}$$
Putting this with the equation of velocity above gave:
$$v(h)=±\sqrt{2\left(\frac{GM}{R}-\frac{GM}{R+h}\right)}$$
I assumed that the velocity would be negative since the object is falling. But when I tried to graph this, I realised that as $h$ gets smaller, or as the object falls, the velocity also decreases. This doesn't really make sense to me because wouldn't the velocity increase as an object keeps falling, meaning that as $h$ decreases, the velocity should increase? Does this velocity represent something else? Or did I simply do something wrong?
Also, how would I get to obtaining a height or velocity function against time?
EDIT: Okay, so I've tried solving this with an initial height set in mind and adding a negative in front of the acceleration. So, $h_0=50\times10^3$m and I put the final height as an unknown (negative values would be ignored). Doing this allowed me to $$\int_{50\times10^3}^{h}-\frac{GM}{(R+h)^2}dh=\left[\frac{GM}{R+h}\right]_{50\times10^3}^h=-\frac{GM}{R+h}+\frac{GM}{R+50\times10^3} =\left[\frac{1}{2}v(h)^2\right]_{50\times10^3}^h=\frac{1}{2}v(h)^2$$
This gave me a new equation of, $$v(h)=\sqrt{2\left(\frac{GM}{R+h}-\frac{GM}{R+50\times10^3}\right)}$$
This new equation starts off at 0 velocity when $h=50\times10^3$ and increases as $h$ decreases. This seems more of a reasonable result to me. Does this mean that my equation needs to have another variable, $h_0$, that also affects the velocity?
Answer
From energy conservation
$$\dfrac{mv^2}{2}=\frac{GMm}{R+h}-\frac{GMm}{R+h_o}$$
We immediately obtain your solution
$$v=\sqrt{2GM\left(\frac{1}{R+h}-\frac{1}{R+h_o}\right)}\tag{A}$$
Or
$$-\frac{dh}{dt}=\sqrt{\frac{2GM}{R+h_o}}\cdot\sqrt{\dfrac{h_0-h}{R+h}}$$
By integrating
$$-\int_{h_o}^h\sqrt{\frac{R+h}{h_0-h}}\cdot dh=\sqrt{\frac{2GM}{R+h_o}}\space \int_{t_o}^t dt$$
We obtain the exact (but not explicit) solution for $h(t)$
$$(R+h_0)\arctan\left(\sqrt{\dfrac{h_0-h}{R+h}}\right)+\sqrt{(h_0-h)(R+h)}=\sqrt{\frac{2GM}{R+h_o}}\space(t-t_o)\tag{B}$$
For a verification, simplify for $\space h\ll R$
Remember $\space\arctan(x)\approx x\space$ for $\space x\ll1$
$$\left(R+h_0\right)\sqrt{\dfrac{h_0-h}{R+h}}+\sqrt{(h_0-h)(R+h)}=\sqrt{\frac{2GM}{R+h_o}}\left(t-t_o\right)$$
And also $\space R+h\approx R\space$ for $\space h\ll R\space$ and $\space R+h_o\approx R\space$ for $\space h_o\ll R$
$$R\space\sqrt{\dfrac{h_0-h}{R}}+\sqrt{(h_0-h)R}=\sqrt{\frac{2GM}{R}}\left(t-t_o\right)$$
We obtain
$$\sqrt{(h_0-h)}=\sqrt{\frac{GM}{2R^2}}(t-t_o)$$
Or
$$h_0-h=\frac{g\space(t-t_o)^2}{2}$$
This is a well known formula where $g$ is the gravitational acceleration
$$\space g=\dfrac{GM}{R^2}$$
The solution for $v(t)$ consists of $(A)$ and $(B)$. It is not explicit, but is exact and can be solved numerically to any degree of precision.
EDIT: Integration steps
First, lets derive a specific reduction rule for integration. Note that
$$\int\dfrac{dx}{x^2+1}=\arctan{(x)}+C_1=-arctan{\frac{1}{x}}+C_2\tag{1}$$
Take a derivative
$$\dfrac{d}{dx}\left(\frac{x}{x^2+1}\right)=\frac{1}{x^2+1}-\frac{2x^2}{(x^2+1)^2}$$
And integrate it
$$\dfrac{x}{x^2+1}=\int\frac{dx}{x^2+1}-\int\frac{2x^2dx}{(x^2+1)^2}+C^{'}$$
By rearranging and using $(1)$, we obtain the reduction rule
$$-\int\dfrac{2x^2dx}{(x^2+1)^2}=\arctan\dfrac{1}{x}+\frac{x}{x^2+1}+C\tag{2}$$
Now define
$$x=\sqrt{\dfrac{R+h}{h_o-h}}\tag{3}$$
Solve for h
$$h_o-h=\dfrac{R+h_o}{x^2+1}\tag{4}$$
Differentiate (3)
$$\dfrac{dx}{dh}=\frac{1}{2x}\left(\frac{R+h}{(h_o-h)^2}+\frac{1}{h_o-h}\right)$$
Combine with (3) and (4)
$$\dfrac{dx}{dh}=\frac{1}{2x}\cdot\frac{x^2+1}{R+h_o}(x^2+1)$$
Rearrange
$$dh=(R+h_o)\dfrac{2xdx}{(x^2+1)^2}\tag{5}$$
From (3) and (5)
$$\int\sqrt{\dfrac{R+h}{h_o-h}}dh=\int xdh=(R+h_o)\int\dfrac{2x^2dx}{(x^2+1)^2}$$
By using (2) we obtain
$$-\int\sqrt{\dfrac{R+h}{h_o-h}}dh=(R+h_o)\left(\arctan\dfrac{1}{x}+\frac{x}{x^2+1}+C\right)$$
Finally by using (3) we obtain the solution (B). The lower integration limit $h=h_o$ defines $C=0$.
There seems to be an obvious contradiction between the predictions of the physics of black holes and the Big Rip, a predicted event about 16.7 Gyr in the future where local groups, galaxies, solar systems, planets, and even atoms will be torn apart. More loosely bound items will be torn apart before more tightly bound items, but would this include black holes, which are thought to casually prohibit escape across the event horizon?
The possibilities as I can conceive of them:
That last one is a big stretch, but it was my main take-away from reading this one really weird (and now defunct) alternative physics website. I found it insightful, regardless of whether it has physical merit or not. This question has been asked in other places before, but didn't seem to have a very informed set of answers.
Answer
The generic thing that happens is that the black hole horizons merge with the cosmological horizon.
To see why this is so, you can consider the case of deSitter Schwarzschild, described in Anon's answer:
$$ dS^2 = f(r) dt^2 - {dr^2 \over f(r)} - r^2 d\Omega^2 $$
$$ f(r) = 1- {2a\over r} - br^2 $$
This has a Nariai limit, described on the Wikipedia page, where the black hole is as large as possible, and this limit is instructive because it has the following properties:
The last point is the most relevant, because for two black holes, you know what happens when they meet--- they merge. For a typical observer, which hasn't fallen into a black hole, a deSitter black hole will be attracted to other deSitter black holes and merge. When they become big enough, they are a cosmological horizon, and then the falling of matter into the black hole turns into inflation of the universe into the cosmological horizon smoothly.
If you have a little black hole, and you aren't in it, unless you accelerate, this black hole will merge with the cosmological horizon in your patch. The merger process is an irreversible joining of horizons, analogous to any other such merger. This is the generic situation. To reach the Nariai limit, you need to keep two black holes on opposite sides of the Einstein static universe, and dump all the dust in the universe into them exactly symmetriclally. This is a very unstable process, a small deformation will lead the two black holes to merge into one cosmological horizon, that then eats all the dust.
What is each mathematical step (in detail) that one would take to get from:
$E^2 - p^2c^2 = m^2c^4$
to
$E = \gamma mc^2$,
where $\gamma$ is the relativistic dilation factor.
This is for an object in motion.
NOTE: in the answer, I would like full explanation. E.g. when explaining how to derive $x$ from $\frac{x+2}{2}=4$, rather than giving an answer of "$\frac{x+2}{2}=4$, $x+2 = 8$, $x = 6$" give one where you describe each step, like "times 2 both sides, -2 both sides" but of course still with the numbers on display. (You'd be surprised at how people would assume not to describe in this detail).
Answer
Starting with your given equation, we add $p^2 c^2$ to both sides to get $$ E^2=m^2 c^4 + p^2 c^2$$ now using the definition of relativistic momentum $p=\gamma m v$ we substitute that in above to get $$E^2 = m^2 c^4 +(\gamma m v)^2 c^2=m^2 c^4 +\gamma^2 m^2 v^2 c^2$$ Now, factoring out a common $m^2 c^4$ from both terms on the RHS in anticipation of the answer we get $$E^2=m^2 c^4 (1+\frac{v^2}{c^2}\gamma^2)$$ Now using the definition of $\gamma$ as $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ and substituting this in for $\gamma$ we get $$E^2=m^2 c^4 \left(1+\frac{\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}\right)$$ and making a common denominator for the item in parenthesis we get $$E^2=m^2 c^4 \left( \frac{1}{1-\frac{v^2}{c^2}} \right)=m^2 c^4 \gamma^2$$ Taking the square root of both sides gives $$E=\pm \gamma mc^2$$ Hope this helps.
My text book reads
If a particle is acted upon by the forces which are conservative; that is, if the forces are derivable from a scalar potential energy function in manner $ F=-\nabla V $.
I was just wondering what may be the criteria for force to be expressed as negative gradient of scalar potential energy and HOW DO WE PROVE IT?
Answer
Your Question all but includes the right search term for an Answer from Wikipedia, "Conservative Forces", which gets you to http://en.wikipedia.org/wiki/Conservative_Forces. There's even what you ask for, a proof. There's also another link to http://en.wikipedia.org/wiki/Conservative_vector_field, which gives some quite good visualizations that will probably help. Loosely, there mustn't be any vortices in the force field for there to be a scalar potential energy that generates the force vector field as $\nabla\!\cdot\!\phi(x)$.
I've heard a couple of scary stories from experienced accellerator physiscists about something called neutron clouds. Apparently, if you have an experiment like a fixed-target experiment that produces a lot of neutrons with the correct energy, they don't just dissipate or get caught in surrounding matter. Instead, they hang around due to their large half-life (~15 minutes). The rumor goes that they actually form clouds, that wander around the facility, and that in the early days of some CERN experiment, people didn't think about the effect, and got a nasty (although not accute) dose when they entered the collision hall just after shutting down the beam.
The description of the behavior of these clouds varies in different accounts. Sometimes they just pass through everything, but sometimes they're supposed to behave like a real gas, being held back by walls (but creeping through small openings).
Answer
Thermal neutrons capture on hydrogen and carbon with reasonable (i.e. not large, but significant) cross-sections (this is the delayed event detection methods of most organic liquid scintillator anti-neutrino detectors--i.e the one that don't dope their scintillator with Gadolinium).
So though a "cloud"--meaning a localized diffuse gas--of neutrons can develop in the neighborhood of a strong source (size of the cloud is driven by how far they go as they thermalise), their dissipation is driven by their mean capture time, not their half-life.
Confession: Here I am presuming that the mean capture time is significantly shorter than the half-life, but I haven't measured it in a "near the laboratory" setting. In organic liquid scintillator the capture time is on order of $200\text{ }\mu\text{s}$, but air has a lot less hydrogen and carbon in it. Note that the neutrons also go into the ground, the building, nearby vehicles and passers-by (if any) where they may find things to interact with.
At my grad-school we had a 2 Curie (i.e. huge) AmBe source. The source vault would register unusually high back-grounds on a survey meter for a few minutes after it was returned from the moderator tank to the shielded vessel, so that may be a rough measure of the time scale. It also says something about the strength of the radiation field: a few times the in-the-basement background level.
Shielding methodology for strong neutron sources generally incorporates a great deal of boron in various layers to help suck up the thermal neutron flux; not incidentally this means that most of the capture gammas are generated inside the shielding. Borated plastics are common as are borated concretes. These days Gadolinium is cheep enough that I imagine we'll start seeing it used in shielding design. The source vault in grad school was built of borated cinder block---two layers with a meter air-gap between.
Another not-very-quantitative story that might shed some light on this.
I was friends with one of the Radiation Safety guys at JLAB. Part of his job was monitoring the radiation level at the fence around the secure area with the accelerators, experimental halls, etc. Mostly they just put out general purposes detectors and compared the results with background reading from nearby, but early on they built a more sophisticated detector out there to understand the various contributions to the dose (probably trying to tune their Monte Carlos, those guys are really big into modelling). He told me two interesting things
The fence was about 40 meters from the beam dumps.
I'm trying to learn the renormalization group, but I am confused about renormalization schemes. The general idea of RG is that physical predictions are independent of "the scale at which a theory is defined", and that every renormalization scheme will throw in such an arbitrary scale, which we can then vary.
My problem is that my textbooks will say this, and then, a hundred pages later, suddenly declare that some random quantity is "the scale that defines the theory". I still have no idea what that phrase means! Usually, this quantity is the parameter $\mu$ that appears in dimensional regularization, but it's unclear how to interpret that quantity.
To make my question more specific, I have a restricted version of the question for each of the three renormalization schemes I know about.
Answer
Question 1
Nontrivial RG flow is the result of explicit breaking of classical theory scale invariance in corresponding quantum field theory. If there is no dimensionful parameters in classical lagrangian of corresponding theory (the generalization on the presence of masses is straightforward, but is not relevant here), we naively expect that after scaling transformation, $$ \Phi (x) \to e^{\sigma \epsilon}\Phi (e^{\epsilon}x), \quad x \to e^{\epsilon}x, $$ with $\sigma$ being the canonical dimension of $\Phi$ field and $\epsilon$ being the continuous parameter of transformation, the action will be unchanged. Corresponding symmetry defines dilatation current conservation law: $$ \partial_{\mu}D^{\mu} =0 $$ This naive law is completely broken by infinities of QFT (such breaking is called the trace anomaly), because of which regularization enters the game. I.e, we introduce the dimensionful parameter by hand, and initial free from dimensionful parameters lagrangian begins to contain the one, called $\mu$. Since it is dimensionful, it is called the scale of theory. However, it is unphysical, and we can't say that it is the scale on which theory is defined: for example, we can choose it so that it coincides with square of transfer momentum, but it is only formal correspondence which depends on our wish.
In general, because of the presence of scale the dilatation current conservation law is modified by quantum corrections. For example, for massless QED $$ \partial_{\mu}D^{\mu} \sim \beta (\alpha)F_{\mu \nu}F^{\mu \nu}, \quad \alpha \equiv \frac{e^{2}}{4 \pi} $$ This leads to nontrivial behaviour of lagrangian parameters (like couplings) with changing of $\mu$.
What's about your question, $\mu$ as the scale on which theory is defined? The answer is dimensional transmutation phenomena, which occurs because of described above breaking of scaling invarince. Precisely, by solving RG equation (here $\alpha$ is the running coupling) $$ \mu\frac{d\alpha}{d\mu} = \beta (\alpha (\mu)) $$ we obtain that $$ \alpha (\mu) = f(\mu , \mu_{0}, \alpha (\mu_{0})) $$ We can invert this relation and use dimensionful parameter $\mu_{0}$ instead of $\alpha $ in perturbation theory (and this is what people call the dimensional transmutation). Such parameter is really physical scale: it defines set of theory parameters. For example, for QCD it defines the strong coupling scale, which is closely related to the confinement and chiral symmetry breaking scale $\Lambda_{QCD}$. The latter determines the scale at which effective theory which describes hadrons interactions works.
Question 2
1. General remarks
The scheme of renormalization precisely defines renormalization constants, including their finite part. In general, the renormalization constants are given as (for example, for dimensional regularization) $$ \tag 0 Z_{i} = a_{i} + \sum_{j = 1}\frac{c^{(i)}_{j}}{\epsilon^{j}}, $$ We have the freedom for choosing $a_{i}$, while $c^{(i)}_{j}$ are completely fixed by the structure of infinities in theory. The renormalization group states that the scheme dependence of the physical observables is absent.
Your question is following: suppose that we have specifit renormalization scheme for which the scale parameter $\mu$ doesn't affect parameters of theory - particularly, the mass parameter, which is fixed by the pole of propagator - why do we introduce the other scheme, for which the mass becomes to run and the RG equations enter the game?
The specific renormalization scheme is called on-shell scheme, while the convenient scheme with the precence of the scale in the expression for the mass is called minimal subtraction. So what's the point?
2. On-shell renormalization scheme: limitations
Let's assume that you use on-shell renormalization scheme. For this scheme $a_{i}$ is not zero, and it is uniquely fixed by specific conditions.
Lets assume the simplest case - scalar theory with self-interaction, and lets concentrate on the mass renormalization. After computing the self-energy by keeping this scheme you have that the propagator is $$ D^{-1}(p^{2}) = p^{2} - m_{\text{pole}}^{2} - \Sigma (p^{2}), $$ where $m_{\text{pole}}$ is the physical mass, for which $$ \tag 1 D^{-1}(m_{\text{pole}}^{2}) = 0 $$ (since it is observable than it doesn't depend on the $\mu$ scale), and $\Sigma (p^{2})$ is self-energy. Eq. $(1)$ expicitly reads $$ \tag 2 \Sigma (p^{2} = m_{\text{pole}}^{2}) = 0 $$ Also, the requirement that the propagator has the unity residue leads to the statement that $$ \tag 3 \left(\frac{d\Sigma (p^{2})}{dp^{2}}\right)_{p^{2} = m_{\text{pole}}^{2}} = 0 $$ This condition, let me remind, in fact is nothing but the requirement that the propagator corresponds to the one-particle state.
Note two things about $\Sigma (p^{2})$ in on-shell scheme. The first one is that it doesn't depend on the scale $\mu$ since the mass $m_{\text{pole}}$ is indeed scale independent, and this result, of course, is regularization scheme independent. The second is that the condition $(3)$ can't be satisfied in the limit of $m_{\text{pole}}^{2} = 0$, since in massless limit Callen-Lemman representation of the propagator (which just binds the pole of the Green function with one-particle state) doesn't contain isolated pole: the one-particle state with zero energy isn't different from many-particle states.
We can't deal with this problem without introducing regularizing scale $m_{\text{reg}}^{2}$. It is indeed unphysical, and in general this is the price for obtaining $\mu$-scale independent quantities in theories with massless states. Note that the most of realistic theories are ones with massless states. For example, QED in on-shell prescription suffers from IR divergences in self-energy because of exactly zero photon mass.
To avoid such singularities, we need to change the renormalization scheme.
3. Extra: the minimal subtraction scheme
For this scheme, all $a_{i}$s in Eq. $(0)$ are zero. So that, Eq. $(1)$ now is $$ D^{-1}(p^{2}) = p^{2} - m^{2} - \Sigma (p^{2}, \alpha , \mu), $$ where $\alpha$ is the set of other couplings which are present in theory (here they are couplings for cubic, quartic terms).
For $p^{2} = m_{\text{pole}}^{2}$ $$ D^{-1}(m_{\text{pole}}^{2}) = m_{\text{pole}}^{2} - m^{2} - \Sigma (m_{\text{pole}}^{2}(m^{2}, \alpha , \mu), \alpha, \mu) = 0, $$ or at the lowest order of perturbation theory $$ m_{\text{pole}}^{2} = m^{2} + \Sigma (m^{2}, \alpha , \mu) $$ In this scheme $\Sigma$ depends on $\mu$ explicitly. But the $m_{\text{pole}}$ doesn't depend on it, so that we come to the statement that $m^{2}$ and $\alpha$ depend on $\mu$.
I think it's clear enough that if you turn your bicycle's steering wheel left, while moving, and you don't lean left, the bike will fall over (to the right) as you turn. I figure this is because the bike's momentum keeps it moving in the direction you were going, and since your wheels have friction against the ground, the top of the bike moves forward relative to the bottom of the wheels. The top of the bike going north while the bottom of the wheels go northwest will understandable cause you to topple.
So to counteract this and keep you from falling over, leaning into the turn is necessary.
But is there also a causal relationship -- that leaning will cause the bike to start to turn? If I start leaning left, I will turn left... but maybe that's because I know that if I don't turn the steering wheel left, the bike will fall over (to the left). I experimented with unruly turns of the steering wheel when I was a kid, and got my scrapes and bruises. Now that I'm a cautious and sedate adult I'm not anxious to experiment that way. :-)
(I also want to ask why airplanes bank into a turn... they don't have the same issues as a bike, i.e. the bottom part has no special friction against the ground. But that would probably make the question too broad.)
Answer
The simple answer is that the angle between the front fork and the vertical causes the force from the ground to create a moment about the axis of rotation that turns the wheel in that direction. This has nothing to do with actually riding the bike, and it will happen even if the bike is stationary.
Basically, if you project the axis of (steering) rotation all the way through the wheel, top to bottom, it will not be coincident with the point of contact with the ground. When the bike leans over, the upward (normal) force from the ground is not in the same plane as the axis of rotation, which causes a moment about that axis.
When the bike begins to turn, the frictional component of the contact force will cause the force to go back into the same plane as the axis of rotation, which causes the wheel to hold its position steady.
And why does this contraction make the orbital more stable?
Answer
I'm assuming you are talking about the average radius of the orbital decreasing as the kinetic energy increases, and not the Lorentz contraction effect.
Lower orbitals have more kinetic energy $T$ by the virial theorem. Specifically for a $1/r$ potential V:
$$2\langle T\rangle=-\langle V\rangle.$$ (See the proof on the Wikipedia page I linked).
So if an electron is closer to the nucleus is has a more negative potential $V$ and thus a higher kinetic energy. At the same time it is more stable because $$\langle E\rangle=\langle T\rangle+\langle V\rangle=-\langle T\rangle,$$ so it is in a lower overall energy state the more kinetic energy it has.
Does the mass of matter falling into a black hole affect the size of an event horizon the moment it passes through it, or when it has been incorporated into the singularity?
why current will not flow from battery if I connect positive terminal of battery to ground even when there is potential difference
Answer
I connect positive terminal of battery to ground...
[The] negative terminal is not connected to anything.
If the negative terminal of the battery is not connected to anything, then no current can flow through the battery. Current can only flow around a complete circuit (i.e., it can only flow around a closed loop).
There is the problem that is bothering me with the black hole evaporation because of Hawking radiation.
According to Hawking theory the black hole will evaporate in finite time because of quantum effects randomly producing particle-antiparticle pairs on the opposite sites of horizon.
However, the gravity causes the time dilatation, which goes to infinity on the horizon itself, so the arbitrarily small amount of time from horizon's perspective is going to be infinitely long from external observer's perspective?
So, from which perspective is the time the black hole will evaporate calculated? If it is calculated from horizon's perspective, it would mean, the black hole will never evaporate from the external observer's perspective?
Answer
If we refer to the evaporating black hole's Penrose diagram
we would see some important differences between the ideal and the evaporating black holes. For the evaporating black hole, there is a moment in spacetime, where the horizon (and the singularity) disappears, mass left at the center of coordinates is zero, and no time dilatation is left. Hense, this event can be seen for every external observer with no problem.
As it turns out, descriptions of this are already all over the internet:
I don't think describing the procedure is necessary, just click on any of the links.
I tried it both with my fingers and with cards a lot and it seems to always work. I only tried artificial light sources, not sun light, for safety reasons.
However, isn't the light of incandescent light bulbs (I – among other kinds of light sources – tried it with several different ones of those) black body radiation which isn't coherent whatsoever? Why does it still work?
I'm not using any optical instruments (glasses, contact lenses, prisms, etc.), btw. Just a light source, my eyes, my fingers, and cards.
If we take the Minkowski metric, $\eta_{\mu\nu}=(1,-1,-1,-1)$, instead of the usual $(-1,1,1,1)$, does this change the form of the Lorentz Transform? I think the standard Lorentz Transform looks like: $$ \left( \begin{matrix} \gamma & -\gamma\beta & 0 & 0\\ -\gamma\beta & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 &0 & 1 \end{matrix} \right) $$
Answer
An interesting question indeed :-) Yes, you can flip the overall sign of the Minkowski metric, and in fact a lot of physicists do this! The sign choice $\operatorname{diag}(-1, 1, 1, 1)$ is conventional in fundamental quantum field theory and in quantum gravity, if I remember correctly, whereas $\operatorname{diag}(1, -1, -1, -1)$ is conventional in particle physics.
This doesn't affect the Lorentz transform, though. If you apply the Lorentz transform to a metric tensor, it computes as $g'_{\alpha\beta} = \Lambda_\alpha^\mu \Lambda_\beta^\nu g_{\mu\nu}$, and so you will automatically come out with the same sign convention that you put in.
I'm studying rigid body motion on Landau but I'm having troubles to understand this proof of the fact that the angular velocity $\vec{\Omega}$ is constant for a rigid body.
My doubt is about the two last equations in the last two lines of the text. If I use the two I get
$\vec{V'}=\vec{V}+(\vec{\Omega}-\vec{\Omega'})\times \vec{r'} +\vec{\Omega} \times\vec{a}$
How are (31.3) derived from this?
Answer
You have
$$ \boldsymbol{v} = \boldsymbol{V'} + \boldsymbol{\Omega'} \times \boldsymbol{r'} $$
and
$$ \boldsymbol{v} = \boldsymbol{V} + \boldsymbol{\Omega} \times (\boldsymbol{r'}+\boldsymbol{a}) $$
You collect the $\boldsymbol{r'}$ terms
$$ \boldsymbol{v} = \boldsymbol{V'} + \boldsymbol{\Omega'} \times \boldsymbol{r'} = \left( \boldsymbol{V}+ \boldsymbol{\Omega} \times \boldsymbol{a} \right) + \left( \boldsymbol{\Omega} \times \boldsymbol{r'} \right) $$
which is solved uniquely when
$$ \begin{align} \boldsymbol{\Omega'} & = \boldsymbol{\Omega} \\ \boldsymbol{V'} &= \boldsymbol{V}+ \boldsymbol{\Omega} \times \boldsymbol{a} \end{align} $$
In my class it was told that ensemble decompositions of a density operator $\rho$ are not unique, but that the ones that exist are related by a unitary operator. I'm trying to prove this, but I get stuck somewhere along the way.
Lets begin by assuming two different decompositions of density operator $\rho$: $\rho = \sum_{j=1}^n{p_j|\psi_j\rangle\langle\psi_j|} = \sum_{k=1}^m{q_k|\phi_k\rangle\langle\phi_k|}$
Now, these two decompositions live in a Hilbert space $\mathcal A$. We can then define a purification of both, using a system described by a Hilbert space $\mathcal B$ of dimension $k=\max (n,m)$, so that we get $|\Psi_1\rangle_{\mathcal A\mathcal B} = \sum_{j=1}^n{\sqrt{p_j}|\psi_j\rangle \otimes |b_j\rangle}$ and $|\Psi_2\rangle_{\mathcal A\mathcal B} = \sum_{k=1}^m{\sqrt{q_k}|\phi_k\rangle \otimes |b_k\rangle}$.
Now, here we can use that as these pure states are purifications of the same density operator, there must be a unitary $U$ connecting them: $(1_A \otimes U_B)|\Psi_1\rangle_{\mathcal A\mathcal B} = |\Psi_2\rangle_{\mathcal A\mathcal B}$.
Here is where I get stuck. I should be able to use this to prove the unitary relation between the $\psi$ and the $\phi$, but it is not obvious to me how I should do this.
Update: after reviewing the comments to the first question, I should have written that the $\psi$ and $\phi$ states do NOT have to be orthonormal, per se.
Answer
We will prove the following Theorem:
Let $\rho=\sum_{i=1}^Np_i\lvert\phi_i\rangle\langle\phi_i\rvert$ be a eigenvalue decomposition and $\sigma=\sum_{i=1}^Mq_i\lvert\psi_i\rangle\langle\psi_i\rvert$ ($M\ge N$). Then, $\rho=\sigma$ if and only if \begin{equation} \sqrt{q_j}\lvert\psi_j\rangle=\sum_i v_{ij}\sqrt{p_i}\lvert\phi_i\rangle\ , \end{equation} with $\sum_j v_{ij}v_{i'j}^*=\sum_k \delta_{ii'}$, i.e., $V\equiv(v_{ij})$ is an isometry.
Proof:
The "if" direction is straightforward: \begin{align} \rho&=\sum_{j}q_j\lvert\psi_j\rangle\langle\psi_j\rvert\\ &=\sum_{i,i',j} v_{ij} v_{i'j}^* \sqrt{p_i}\sqrt{p_{i'}} \lvert\phi_i\rangle \langle\phi_{i'}\rvert \\ &=\sum_i p_i\lvert\phi_i\rangle \langle\phi_{i}\rvert = \sigma\ , \end{align} where in the last step we have used that $\sum_j v_{ij}v_{i'j}^*=\delta_{ii'}$.
To prove the converse, let $$ v_{ij} := \langle\phi_i\lvert\psi_j\rangle\,\sqrt{q_j/p_i}\ .$$ Then, $$ \sum_i v_{ij} \sqrt{p_i}\lvert\phi_i\rangle = \sum_i \lvert\phi_i\rangle\langle \phi_i\lvert\psi_j\rangle\sqrt{q_j} = \sqrt{q_j}\lvert\psi_j\rangle\ ,$$ i.e., $v_{ij}$ is the desired basis transformation. Further, $$ \sum_{ii'}\underbrace{\delta_{ii'}\sqrt{p_ip_{i'}}}_{=:a_{ii'}}\lvert\phi_i\rangle\langle\phi_i\rvert = \sum_ip_i\lvert\phi_i\rangle\langle\phi_i\rvert = \sum_jq_j\lvert\psi_j\rangle\langle\psi_j\rvert = \sum_{ii'}\underbrace{\sum_jv_{ij}v^*_{i'j}\sqrt{p_ip_i'}}_{=:b_{ii'}}\lvert\phi_i\rangle\langle\phi_{i'}\rvert\ . $$ Now, since the $\lvert\phi_i\rangle$ are orthogonal (as they form an eigenbasis), the $\lvert\phi_i\rangle\langle\phi_{i'}\rvert$ are linearly independent, and thus, $a_{ii'}=b_{ii'}$, which implies $\sum_jv_{ij}v^*_{i'j}=\delta_{ii'}$.
If the $\lvert\phi_i\rangle$ do not form an orthonormal basis, we can generalize the theorem by going through an orthonormal basis $\rho=\sum r_k\lvert\chi_k\rangle\langle\chi_k\rvert$: Then, \begin{align} \sqrt{p_i}\lvert\phi_i\rangle&=\sum_k u_{ki}\sqrt{r_k}\lvert\chi_k\rangle\\ \sqrt{q_j}\lvert\psi_j\rangle&=\sum_k w_{kj}\sqrt{r_k}\lvert\chi_k\rangle \end{align} with $\sum_i u_{ki}u_{k'i}^*=\delta_{kk'}$ and $\sum_j w_{kj}w_{k'j}=\delta_{kk'}$. The second equation then yields $\sqrt{r_k}\lvert\chi_k\rangle= \sum_j w_{kj}^*\sqrt{q_j}\lvert\psi_j\rangle$. After inserting this in the first equation, we obtain $$ \sqrt{q_j}\lvert\psi_j\rangle=\sum_i v_{ij}\sqrt{p_i}\lvert\phi_i\rangle $$ with $v_{ij} = \sum_k u_{ki}w_{kj}^*$, i.e., $V=UW^\dagger$ is a partial isometry.
This is a follow-up to a previous question How can you weigh your own head in an accurate way?.
My purpose is not to restart the flurry of more or less humorous jokes (which are not such a bad thing when infrequent) but to try to draw some conclusions, as I found the event instructive, and hopefully to get a real answer.
Beyond the jokes, I think the question was also interesting from a more social point of view: how fallacies operate, how people vote, what is the effect how the fast answer inducement by site rules.
It is clear now, at least to me, that the real physical problem was to determine the mass density distribution of a (partly ?) solid object without using destructive means. Of course the concept of destructiveness may depend on the concerned body, and that was one source of jokes.
Still, it is a relevant topic to review the means, if any, for achieving such a purpose, as it seems that there are not that many. This is actually the question I am asking.
But I would like also, at the same time, to review some of the answers given to the previous qiestion.
One that attracted my attention was the micro-satellite solution. Does it really work ? So we could precisely ask: given a precise knowledge of the gravity field created by a solid object, can we deduce from it the mass density distribution in the solid. The answer seems to be no. Take the trivial case of a sphere with a uniform radial density distribution. The gravity outside the sphere depends only on the total mass, so that it tells us nothing about the internal density distribution. Is that an exceptional degenerate case, or is it a strong hint that the analysis of the gravity field is not enough ?
The slicing solution I proposed was an interesting mistake, certainly a silly one. But it is ever so tempting to believe the solution is near when you have a set of equations, and easy ones to boot. A cute trap. It becomes more obvious when you try to do it in continuous rather than discrete form.
It should have been obvious, as we know that the torque produced by weighing a massive object depend only on total mass and distance of the center of mass to the axis.
What was less obvious, at least to me (but I have not done much physics since college), was the use of the moment of inertia. One could also think of measuring moments of inertia after virtual slicing of the object as previously with torque. Unfortunately the moment of inertia depends only on 3 values, the previous two and a reference moment of inertia (for a given direction of the rotation axis). It is a useful remark for computing moments of inertia, but shows that there is no hope there for solving our problem.
This was remarked by @Ben Crowell in another question (see below).
More generally, it seems there is no hope from any discretized measurement of a quantity that depends polynomially on the distance. One cannot get more unknowns than the degree of the polynomial.
I will not comment on Compton scattering, and other uses of indirect physical phenomena, if only for lack of competence. There is also the fact that I would like to know of a solution involving only mechanics.
One technique I heard of is measuring wave propagation across the object. I hinted to that in a comment on the Compton scattering solution. I have no expertise on that, but I think that is how the structure of the planet is analyzed by geophysicists (earthquake waves). I also heard of underground analyses by similar means, using explosives. But I am not sure about the type of information that can be obtained in this way. Does one get density distribution. Does it always work ?
The other technique I thought of is ultrasound based medical imagery. Could it measure densities.
Now since we are considering waves, could we get something out of the mesurement of gravity wave propagation, assuming it is possible to do the necessary measurements. But I have no idea what this actually entails or means.
Interestingly, a simpler version of the problem has already been discussed to some extent two months ago in How can I determine whether the mass of an object is evenly distributed? Some ideas were repeated for the recent question, but new ideas emerged too (gravity field analysis) even if they have weaknesses, and are somewhat hard to use in most situations.
Then, what techniques have been used to know the density distribution of Earth, the Sun or possibly other bodies. Is there anything systematic that we overlooked.
So, are there means of solving the problem purely with mechanics and gravity?
A related question is whether it is easier, or as hard, to determine whether the mass density distribution is uniform.
More generally, I am wondering whether there is a way of characterizing the properties of phenomena that can help determine the density distribution inside a solid object.
Alternatively, could one prove it is not achievable by purely mechanical means.
We know that the sum of the masses of the quarks in a proton is approximately $9.4^{+1.9}_{-1.3}~\text{MeV}/c^2$, whereas the mass of a proton is $\approx931~\text{MeV}/c^2$. This extra mass is attributed to the kinetic energy of the confined quarks and the confining field of the strong force. Now, when we talk about energetically favourably bound systems, they have a total mass-energy less than the sum of the mass-energies of the constituent entities. How does a proton, a bound system of quarks with its mass-energy so much more than its constituent entities, remain stable? The strong force and other energetic interactions supposedly contribute this mass-energy by the mass-energy equivalence principle, but how exactly does this occur?
Answer
You say:
Now, when we talk about energetically favourably bound systems, they have a total mass-energy less than the sum of the mass-energies of the constituent entities.
and this is perfectly true. For example if we consider a hydrogen atom then its mass is 13.6ev less than the mass of a proton and electron separated to infinity - 13.6eV is the binding energy. It is generally true that if we take a bound system and separate its constituents then the total mass will increase. This applies to atoms, nuclei and even gravitationally bound systems. It applies to quarks in a baryon as well, but with a wrinkle.
For atoms, nuclei and gravitationally bound systems the potential goes to zero as the constituents are separated so the behaviour at infinity is well defined. If the constitiuents of these systems are separated to be at rest an infinite distance apart then the total mass is just the sum of the individual rest masses. So the bound state must have a mass less then the sum of the individual rest masses.
As Hritik explains in his answer, for the quarks bound into a baryon by the strong force the potential does not go to zero at infinity - in fact it goes to infinity at infinity. If we could (we can't!) separate the quarks in a proton to infinity the resulting system would have an infinite mass.
So the bound state does have a total mass less than the separated state. It's just that the mass of the separated state does not have a mass equal to the masses of the individual particles.
You can look at this a different way. To separate the electron and proton in a hydrogen atom we need to add energy to the system so if the added energy is $E$ the mass goes up by $E/c^2$. As the separation goes to infinity the energy $E$ goes to 13.6eV. If we try to separate the quarks in a proton by a small distance we have to put energy in and the mass also goes up by $E/c^2$ just as in any bound system. But with the strong force the energy keeps going up as we increase the separation and doesn't tend to any finite limit.
Yesterday we have studied the Lorentz transformation in school. So we have two frames of reference, $S$ and $S'$ . $S$ is stationary and $S'$. $S'$ has a constant velocity $v$, relative to the $S$ frame. $v$ is directed along the Ox axis. Ox is parallel to Ox' and Oy is parallel to Oy'.
If we apply the Galilran Transformations we get:
$x = x' + ut' $ $y = y'$ $z = z'$ $t = t'$
$ x' = x - ut $ $y'=y$ $z'=z$ $t' = t$
Now, our physics teacher, assumed that:
$ x=k(x'+ut')$ $ x'=k(x-ut)$ with k being a constant.
Why did he do that? I didn't understand. I undrstood that the length of an object depends o the frame of reference and that the speed of light is the same in the two frames.
Assuming the above facts, we can derive the $k$ constant:$$\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}$$
But why did we make that first assumptikn? I didn't get the logic. Could somebody explain, please?
What is the meaning, mathematical or physical, of the anti-commutator term? $$\langle ( \Delta A )^{2} \rangle \langle ( \Delta B )^{2} \rangle \geq \dfrac{1}{4} \vert \langle [ A,B ] \rangle \vert^{2} + \dfrac{1}{4} \vert \langle \{ \Delta A, \Delta B \} \rangle \vert^{2},$$ where $\Delta A, \Delta B, A$ and $ B$ are operators.
The inequality is still true, and the anti-commutator term "strengthens" the inequality, but why does it appear?
Answer
Dear Rodrigo, it's an interesting stronger version of the uncertainty principle for general operators $A,B$ that I've never seen before but I just verified it holds. Just to be sure, the anticommutator is simply $$\{A,B\}\equiv AB+BA.$$ I like when the braces are only used for pairs of Grassmannian objects but people use it as a bookkeeping device to simplify $AB+BA$ in all situations. Nothing difficult about the notation. Note that the commutator and anticommutator appear totally symmetrically in the inequality, a fact we will derive.
To see why the stronger inequality holds, open Wikipedia here
http://en.wikipedia.org/wiki/Uncertainty_principle#Mathematical_derivations
where only the simpler version of the inequality (without the squared anticommutator) is proved by combining two inequalities. The first one, $$ ||A\psi||^2 ||B\psi||^2 \geq |\langle A\psi|B\psi\rangle|^2 $$ remains unchanged. However, the second inequality from the Wikipedia article may be strengthened to a full-fledged equality $$ |\langle A\psi|B\psi\rangle|^2 = \left| \frac{1}{2i} \langle \psi | AB-BA | \psi \rangle \right|^2 + \left| \frac{1}{2} \langle \psi | AB+BA | \psi \rangle \right|^2 $$ This identity simply says that the squared absolute value of a complex number is the sum of the squared real part and the squared imaginary part (which was omitted on Wikipedia). Combining the previous two inequalities, one gets your "stronger" uncertainty principle.
(Of course, the equation derived above is uselessly weak unless the expectation values of $A,B$ vanish themselves. It can be strengthened into yours by repeating the same prodedure for $\Delta A = A-\langle A\rangle$ and similarly $\Delta B = B-\langle B\rangle$ instead of $A,B$.)
I wrote what the anticommutator means mathematically and why the inequality is true. Now, what does the anticommutator term mean physically? I don't know what this question mean. It's a term in an equation that I can read and explain for you again. The precise answers in physics are given by mathematics. So I guess that the answer you want to hear is that it means nothing physically, it's just pure mathematics. This fact doesn't mean that it can't be useful.
Well, in normal cases, the stronger version is not "terribly" useful because the anticommutator term is only nonzero if there is a "correlation" in the distributions of $A,B$ - i.e. if the distribution is "tilted" in the $A,B$ plane rather than similar to a vertical-horizontal ellipse which is usually the case in simple wave packets etc. Maybe this is what you wanted to hear as the physical explanation of the anticommutator term - because $AB+BA$ is just twice the Hermitean part of $AB$, it measures the correlation of $A,B$ in the distribution given by the wave function - although the precise meaning of these words has to be determined by the formula.
I am just a high school student trying to self study, so please excuse me if you find the question silly.
I don't really need to explain my question.I am just asking if we can make a source of light such that it only emits one ray and if our eye is not in the middle of this ray we can't see the source of light.
Maybe it seems so easy, but it is not!
How can we obtain the normalization constant $N$ for a set of eigenfunctions with different domains?
For example, we have
$\psi_{1}=N(f_{1}e^{-\kappa x}+g_{1}e^{\kappa x}),\hspace{1cm}x\in[0,1],\hspace{.2cm} \nonumber \\ \psi_{2}=N(f_{2}e^{-\kappa x}+g_{2}e^{\kappa x}),\hspace{1cm}x\in[0,1], \hspace{.2cm} \nonumber \\ \psi_{3}=N(f_{3}e^{-\kappa x}+g_{3}e^{\kappa x}),\hspace{1cm}x\in[-1,0], \\ \psi_{4}=N(f_{4}e^{-\kappa x}+g_{4}e^{\kappa x}),\hspace{1cm}x\in[-1,0].\nonumber$.
we can normalize each wavefunction by the integral $\int^{x2}_{x1}\psi^{*}\psi dx=1$, but that way, the other eigenfunctions are not normalized to one!
This is related to the scaling solution of the hydrodynamic equations. I get a relation for the scaling parameter $b$:
$\ddot{b} = -\omega^2(t)*b + \omega_0^2/b^3$
When the trap for the condensate in suddenly switched off $\omega(t)$ goes to zero, so you get the equation
$\ddot{b} = \omega_0^2/b^3$ with initial conditions $b(0)=1$ and $\dot{b}(0) = 0$.
What would be the solution for $b(t)$?
Answer
The equation is $$\frac{d^2b}{dt^2} = \frac{\omega_0^2}{\, b^3 \, }$$ Multiply both sides of the equation by $\frac{db}{dt}$ and obtain $$\frac{d^2b}{dt^2} \, \frac{db}{dt} = \frac{\omega_0^2}{\, b^3 \, } \, \frac{db}{dt}$$ which can be interpreted as $$\frac{db}{dt} \, \frac{d}{dt}\left(\frac{db}{dt}\right) = \Big(\omega_0^2\, b^{-3}\Big) \, \frac{db}{dt}$$ which by going backwards with chain rule is the same as $$\frac{1}{2}\, \frac{d}{dt}\left(\frac{db}{dt}\right)^2 = \frac{d}{dt} \Big(\omega_0^2\, \frac{b^{-2}}{-2}\Big)$$ $$\frac{1}{2}\, \frac{d}{dt}\left(\frac{db}{dt}\right)^2 = -\, \frac{1}{2}\frac{d}{dt} \Big(\omega_0^2\, b^{-2}\Big)$$ and after cancelling the one half $$ \frac{d}{dt}\left(\frac{db}{dt}\right)^2 = -\,\frac{d}{dt} \Big(\omega_0^2\, b^{-2}\Big)$$ Integrate both sides with respect to $t$ $$ \left(\frac{db}{dt}\right)^2 = E_0 - \omega_0^2\, b^{-2}$$ $$ \left(\frac{db}{dt}\right)^2 = \frac{E_0 \, b^2 - \omega_0^2}{b^2}$$ $$ b^2 \, \left(\frac{db}{dt}\right)^2 = {E_0 \, b^2 - \omega_0^2}$$ $$ \left(b \, \frac{db}{dt}\right)^2 = {E_0 \, b^2 - \omega_0^2}$$ $$ \left(\frac{1}{2} \, \frac{d(b^2)}{dt}\right)^2 = {E_0 \, b^2 - \omega_0^2}$$ $$ \left(\frac{d(b^2)}{dt}\right)^2 = {4 \, E_0 \, b^2 - 4 \, \omega_0^2}$$ When $\frac{db}{dt}(0) = 0$ and $b(0) = 1$ we arrive at $E_0 = \omega_0^2$. Change the dependent variable by setting $u = b^2$ and the equation becomes $$ \left(\frac{du}{dt}\right)^2 = {4 \, E_0 \, u - 4 \, \omega_0^2}$$ or after taking square root on both sides $$ \frac{du}{dt} = \pm \, \sqrt{\, 4 \, E_0 \, u - 4 \, \omega_0^2 \, }$$ This is a separable equation $$ \frac{du}{ 2\, \sqrt{\, E_0 \, u - \omega_0^2 \, }} = \pm \, dt$$ $$ \frac{d\big(E_0 \, u - \omega_0^2 \big)}{ 2\, \sqrt{\, E_0 \, u - \omega_0^2 \, }} = \pm \, E_0 \, dt$$ $$ d \Big( \sqrt{\, E_0 \, u - \omega_0^2 \, } \Big) = \pm \, E_0 \, dt$$ Integrate both sides $$\sqrt{\, E_0 \, u - \omega_0^2 \, } = C_0 \pm E_0 \, t $$ Square both sides $${\, E_0 \, u - \omega_0^2 \, } = \big(\, C_0 \pm E_0 \, t \,\big)^2 $$ and solve for $u$ $$ u = \frac{1}{E_0} \, \big(\, C_0 \pm E_0 \, t \, \big)^2 + \frac{\omega_0^2}{E_0}$$ Return back to $u = b^2$ $$b^2 = \frac{1}{E_0} \, \big(\, C_0 \pm E_0 \, t \, \big)^2 + \frac{\omega_0^2}{E_0}$$ so finally $$b(t) = \pm \, \sqrt{ \, \frac{1}{E_0} \, \big(\, C_0 \pm E_0 \, t \, \big)^2 + \frac{\omega_0^2}{E_0} \, }$$ However, we know that $E_0 = \omega_0^2$ so $$b(t) = \pm \, \sqrt{ \, \frac{1}{\omega_0^2} \, \big(\, C_0 \pm \omega_0^2 \, t \, \big)^2 + 1 \, }$$ Thus, $b(0) = 1$ is possible when $C_0 = 0$ and finally $$b(t) = \sqrt{ \, \omega_0^2 \, t^2 + 1 \, }$$
In my class it was told that ensemble decompositions of a density operator $\rho$ are not unique, but that the ones that exist are related by a unitary operator. I'm trying to prove this, but I get stuck somewhere along the way.
Lets begin by assuming two different decompositions of density operator $\rho$: $\rho = \sum_{j=1}^n{p_j|\psi_j\rangle\langle\psi_j|} = \sum_{k=1}^m{q_k|\phi_k\rangle\langle\phi_k|}$
Now, these two decompositions live in a Hilbert space $\mathcal A$. We can then define a purification of both, using a system described by a Hilbert space $\mathcal B$ of dimension $k=\max (n,m)$, so that we get $|\Psi_1\rangle_{\mathcal A\mathcal B} = \sum_{j=1}^n{\sqrt{p_j}|\psi_j\rangle \otimes |b_j\rangle}$ and $|\Psi_2\rangle_{\mathcal A\mathcal B} = \sum_{k=1}^m{\sqrt{q_k}|\phi_k\rangle \otimes |b_k\rangle}$.
Now, here we can use that as these pure states are purifications of the same density operator, there must be a unitary $U$ connecting them: $(1_A \otimes U_B)|\Psi_1\rangle_{\mathcal A\mathcal B} = |\Psi_2\rangle_{\mathcal A\mathcal B}$.
Here is where I get stuck. I should be able to use this to prove the unitary relation between the $\psi$ and the $\phi$, but it is not obvious to me how I should do this.
Update: after reviewing the comments to the first question, I should have written that the $\psi$ and $\phi$ states do NOT have to be orthonormal, per se.
Answer
We will prove the following Theorem:
Let $\rho=\sum_{i=1}^Np_i\lvert\phi_i\rangle\langle\phi_i\rvert$ be a eigenvalue decomposition and $\sigma=\sum_{i=1}^Mq_i\lvert\psi_i\rangle\langle\psi_i\rvert$ ($M\ge N$). Then, $\rho=\sigma$ if and only if \begin{equation} \sqrt{q_j}\lvert\psi_j\rangle=\sum_i v_{ij}\sqrt{p_i}\lvert\phi_i\rangle\ , \end{equation} with $\sum_j v_{ij}v_{i'j}^*=\sum_k \delta_{ii'}$, i.e., $V\equiv(v_{ij})$ is an isometry.
Proof:
The "if" direction is straightforward: \begin{align} \rho&=\sum_{j}q_j\lvert\psi_j\rangle\langle\psi_j\rvert\\ &=\sum_{i,i',j} v_{ij} v_{i'j}^* \sqrt{p_i}\sqrt{p_{i'}} \lvert\phi_i\rangle \langle\phi_{i'}\rvert \\ &=\sum_i p_i\lvert\phi_i\rangle \langle\phi_{i}\rvert = \sigma\ , \end{align} where in the last step we have used that $\sum_j v_{ij}v_{i'j}^*=\delta_{ii'}$.
To prove the converse, let $$ v_{ij} := \langle\phi_i\lvert\psi_j\rangle\,\sqrt{q_j/p_i}\ .$$ Then, $$ \sum_i v_{ij} \sqrt{p_i}\lvert\phi_i\rangle = \sum_i \lvert\phi_i\rangle\langle \phi_i\lvert\psi_j\rangle\sqrt{q_j} = \sqrt{q_j}\lvert\psi_j\rangle\ ,$$ i.e., $v_{ij}$ is the desired basis transformation. Further, $$ \sum_{ii'}\underbrace{\delta_{ii'}\sqrt{p_ip_{i'}}}_{=:a_{ii'}}\lvert\phi_i\rangle\langle\phi_i\rvert = \sum_ip_i\lvert\phi_i\rangle\langle\phi_i\rvert = \sum_jq_j\lvert\psi_j\rangle\langle\psi_j\rvert = \sum_{ii'}\underbrace{\sum_jv_{ij}v^*_{i'j}\sqrt{p_ip_i'}}_{=:b_{ii'}}\lvert\phi_i\rangle\langle\phi_{i'}\rvert\ . $$ Now, since the $\lvert\phi_i\rangle$ are orthogonal (as they form an eigenbasis), the $\lvert\phi_i\rangle\langle\phi_{i'}\rvert$ are linearly independent, and thus, $a_{ii'}=b_{ii'}$, which implies $\sum_jv_{ij}v^*_{i'j}=\delta_{ii'}$.
If the $\lvert\phi_i\rangle$ do not form an orthonormal basis, we can generalize the theorem by going through an orthonormal basis $\rho=\sum r_k\lvert\chi_k\rangle\langle\chi_k\rvert$: Then, \begin{align} \sqrt{p_i}\lvert\phi_i\rangle&=\sum_k u_{ki}\sqrt{r_k}\lvert\chi_k\rangle\\ \sqrt{q_j}\lvert\psi_j\rangle&=\sum_k w_{kj}\sqrt{r_k}\lvert\chi_k\rangle \end{align} with $\sum_i u_{ki}u_{k'i}^*=\delta_{kk'}$ and $\sum_j w_{kj}w_{k'j}=\delta_{kk'}$. The second equation then yields $\sqrt{r_k}\lvert\chi_k\rangle= \sum_j w_{kj}^*\sqrt{q_j}\lvert\psi_j\rangle$. After inserting this in the first equation, we obtain $$ \sqrt{q_j}\lvert\psi_j\rangle=\sum_i v_{ij}\sqrt{p_i}\lvert\phi_i\rangle $$ with $v_{ij} = \sum_k u_{ki}w_{kj}^*$, i.e., $V=UW^\dagger$ is a partial isometry.
Let's say we have two planets at a stand still within reasonable distance of each other. They will accelerate towards each other and subsequently collide.
If instead we give them a sufficient (but finite) initial velocity in opposite directions orthogonal to the path between them, they will instead enter into a orbit around each other. In this orbit they will experience continuous acceleration. Thus, for a finite initial velocity I get in return a continuous (and thus infinite, given infinite time) acceleration in return.
Acceleration is work and work takes energy. The energy is kindly supplied by gravity. Is it correctly understood that energy is continuously put into the system, in order to maintain the orbit? And that gravity is thus an infinite source of energy? And that a system of planets + gravity, if given the right initial condition, constitutes a perpetual motion machine? Even if so, it could still be principially impossible to extract surplus energy from the system, eg. for practical uses.
Suppose I am provided with a positive is charged....and someone tells me to find which type of charge is present in it? Then how can we detect that the body is positively or negatively charged? Is there a device which can the charge?
I'm well aware this is a very active area of research so the best answer one can give to this question may be incomplete.
Topological states in condensed matter are well-known, even if not always recognized as such. The most famous example is likely the quantum Hall effect. In this case, time-reversal symmetry is broken by an external $\vec{B}$ field.
In the past decade, it was realized that spin-orbit coupling can be used to break time-reversal symmetry as well. This leads to topologically preserved states in so-called topological insulators.
However, I've overheard that some condensed matter theorists believe spin-orbit coupling may not be necessary for breaking time-reversal symmetry in topological insulators. Apparently, there are some other mechanisms proposed in which this breaking is not (or at least not primarily) due to spin-orbit coupling. I've heard from a fairly well-respected condensed matter physicist that he believed spin-orbit coupling was important in all realistic topological insulators, but probably not essential for the theory.
Being a relative novice in the area, I don't know of any other mechanism by which time-reversal symmetry could be broken. Besides spin-orbit coupling effects, is there any other way that topologically protected states could exist with 0 $\vec{B}$ field? If so, how realistic are these? If not, what is meant when people claim spin-orbit coupling is not fundamental to topological insulators, and what would be a more fundamental way to look at it? Any references are certainly appreciated.
Answer
The short answer: graphene is a counterexample.
The longer version: 1) You do not need to break the time reversal symmetry. 2) spin-orbit coupling does not break the time-reversal symmetry. 3) In graphene, there are two valleys and time inversion operator acting on the state from one valley transforms it into the sate in another valley. If you want to stay in one valley, you may think that there is no time-reversal symmetry there.
A bit more: It seems that time-reversal symmetry is not a good term here. Kramers theorem (which is based on time-reversal symmetry) says that state with spin up has the same energy as a state spin down with a reverse wavevector. It seems that in your question you use time-reversal symmetry for $E_{↑}({\bf k})=E_{↓}({\bf k})$ which is misleading and incorrect in absence of space-reversal symmetry.
Do you still need a citations or these directions will be enough?
UPD I looked through the papers I know. I would recommend a nice review Rev. Mod. Phys. 82, 3045 (2010). My answer is explained in details in Sec. II.B.II, Sec. II.C (note Eq. (8)) Sec. III.A, IV.A. The over papers are not that transparent. Sorry for the late update.
