units - What is the meaning of speed of light $c$ in $E=mc^2$?

$E=mc^2$ is the famous mass-energy equation of Albert Einstein. I know that it tells that mass can be converted to energy and vice versa. I know that $E$ is energy, $m$ is mass of a matter and $c$ is speed of light in vacuum.

What I didn't understood is how we will introduce speed of light?

Atom bomb is made using this principle which converts mass into energy; in that the mass is provided by uranium but where did speed of light comes into play? How can speed of light can be introduced in atom bomb?

Answer

c is a priori not the speed of light. It is the speed of massless particles. The way it comes about is as follows: You construct the Lorentz-transformations as the symmetry transformations of Minkowski space. The group has one parameter, that's c. You have to fix it by physical means. You can look at the dynamics of massive particles and massless particles and find that massive particles will approach c asymptotically only at infinite energy, and massless particles always move with c.

Since to our best knowledge photons are massless, c is also the speed of light. Also, that was historically Einstein's motivation, which is why it's usually motivated in textbooks this way. However, should it turn out one day that photons do have very tiny masses, then c will still be there, it will just no longer be called the speed of light.

electrostatics - Electric field falls off faster than $frac{1}{r^2}$ for large distances

An excerpt from a book;

The electric field due to a charge configuration with total charge zero, is not zero; but for distances large compared to the size of the configuration, its field falls off faster than $\frac{1}{r^2}$ , typical of field due to a single charge. An electric dipole is the simplest example of this fact.

1. 
   

   Why is the field not zero even if the net charge is zero? Wouldn't the field cancel out?

   
   
   


2. 
   

   What does it mean by - the field falls off faster than $\frac{1}{r^2}$ for large distances.

   
   

Does it mean that the field intensity decreases at a faster rate at large distances? If yes, then why does that happen?

3. 
   

   Why is this typical of a field due to a single charge?

   
   



4. 
   

   How is the dipole an example of this fact?

   
   

I would appreciate if the answer is aimed for a highschool student with only basic knowledge of electrostatics and doesn't involve complicated equations.

thermodynamics - Is $delta Q – delta W$ a state function?

I know that internal energy, $Q+W$ is a state function. But $$dU=\delta Q-\delta W,$$
is the change in internal energy, where $dU$ is change in internal energy, $\delta Q$ is the heat supplied to the system and $\delta W$ is the work done by the system. Is this a state function too?

Answer

The quantity $dU=\delta Q-\delta W$ is not a state function simply because it is not a function, it is a differential of a function. In that case a differential of the state function $U$.

The fact that there is a function $U$ such that the differential $\delta Q-\delta W$ is the differential of $U$ means that the differential is exact or $U$ is a state function, i.e., it has a definite value for every point in the space of configuration. In another words, the change in energy depends only on the initial and final states.

quantum mechanics - If wavefunction is just a probability function, how does an electron interfere with itself

I have read lots of quantum mechanics books.

The chapters that are talking about De Broglie, lots of them name the chapter as "Wave-particle duality" and says: "Electrons are both waves and particles". So I start to think that (for example) an electron sometimes becomes wave sometimes becomes particle.

But when I start to read the chapters about Schrödinger and wavefunction thing. I see that that wave nature thing does not belong to electron itself. It is about its location. So there is no wave particle duality. Electron is always a particle. But the location of electron is represented as wave, because of uncertainty principle. So electron can't interfere with itself because it is always a particle. The interference is about its location function.

1-) Are these all true?

2-) If true, if an electron is always a particle, in double slit experiment how does an electron interfere with itself without observer, but there is no interference pattern with observer?

particle physics - Weak interaction, parity violation, and the observer-dependence of helicity

It is said that the Weak Interaction only couples to left-handed particles which a negative spin (left-handed). However some sources say that spin or helicity is dependent on the observer's position and velocity relative to the particle.

What I don't understand following this principles, is that an observer (1) might see a left-handed particle couple the W,Z bosons. However, if the spin of that particle is relative to the motion of an observer, an additional observer 2 seeing this particle but this time with a positive spin (right-handed) will not see the particle couple to the W,Z bosons because the weak interaction couples to left-handed particles only, I cannot understand this because the particle does couple or does not couple with the W,Z boson and this event should not depend on the observer's relative motion.

Answer

The vocabulary "left-handed" and "right-handed" is used for two distinct concepts: one is chirality, and one is helicity.

• 
  

  Chirality distinguishes between two inequivalent matrix representations of the even part of the algebra of Dirac matrices. The concept of chirality applies to the field operators that are used to construct the model, rather than to particles. When people say that the $SU(2)$ weak interaction couples only to left-handed fermions, they're talking about chirality, not helicity.

  
  


• 
  

  Helicity distinguishes between the two different possible orientations of the angular momentum of a massless particle. Because of electroweak symmetry breaking (EWSB), most elementary particles have a significant mass. Neutrinos remain (nearly) massless. Helicity is Lorentz-invariant only for a massless particle.

  
  

The relationship between field operators and particles is not as straightforward as perturbative calculations sometimes portray. Field operators are used to construct the model's observables and states. Particles are phenomena that the model predicts.

Here are a few related posts, from newest to oldest:

So is it not $CP$ instead of $C$ that is responsible for changing a particle to its antiparticle?

What is polarisation, spin, helicity, chirality and parity?

How can we measure chirality in experiments?

What's the difference between helicity and chirality?

electromagnetism - Why geologist do not consider molten iron in the magma to be the source of Earth's magnetic field?

This was a question that came in a exam. How come is this possible?

Actually, a question on the source of Earth's magnetic field was asked here on physics SE, which matches my opinion.

The source of Earth's magnetic field is the molten iron and metals in the earth's core. The spinning of these liquid metals produces the same effect as electric current in a coil which produces a magnetic field.

Yes, that is true. Circulating ions of iron and nickel in a highly conducting liquid of Earth's core forms current loops and produces a magnetic field.

There is something to do with geologist though I'm not getting it. How come geologist doesn't seem to agree with this?

Answer

The magma has temperature between 700 and 1300 Celsius degrees. The Curie temperature of iron is at 770 degrees Celsius. Above that temperature, iron loses magnetism. Note that right above 770 °C, iron is still solid because the melting point is around 1500 °C.

So magma almost never can be magnetic because it's just too hot for that. Incidentally, if it gets melted and then refreezes, it typically doesn't regain magnetism. The situation is even clearer for Earth's core whose temperature is 6000 °C or so. There can't be any magnetism for iron and nickel that is this hot.

Geomagnetism is driven by the dynamo effect which is a loop of magnetic loops creating currents and vice versa that is partly being sustained by radioactivity.

quantum mechanics - Delayed choice experiment clarification (concrete setup)

I know that clarifications about delayed choice experiment was asked million times, and I understand the idea, but I was not able to find the discussion of this particular situation anywhere, though I tried hard. This setup is described in Brian Greene's Fabric of the Cosmos book (note my question is about modification of this setup, not about it exactly as described in book). Here is a picture of setup from the book

So photon goes from source to the beam splitter, then there are two down-converters that split photon into two, each of which goes in different directions: one follows the previous path (that one is named "signal") and the other goes into different direction to the detector ("idler" photon). Two signal photons then go to detector. Without down-converters we see interference pattern in detector - that seems clear. With down-converters it is claimed that we don't see interference pattern because detecting idler photons reveals information about which path photon has "chosen": if photon detected in top detector - photon has chosen top path, if not - then bottom one. So interference pattern disappears (so is claimed in the book).

Then book follows to description of modification of this setup to erase which path information - this part is clear for me and it's not what I'm interested about. What is not clear is:

What if we remove idle photon detectors from the setup, but leave down-converters in place? Then it seems "which path" information is not revealed and interference pattern should appear again? But that does not make much sense, because we can decide to put or not a detector at much later time (after corresponding singal photon hit the screen). So depending on if we see pattern or not we should know what happened in the future, which of course cannot happen. Is that assumption about removing idler detectors leads to interference pattern incorrect? If so, why? What will happen if we switch idler detectors on and off by our choice? I suspect this has something to do with the fact we now have two photos instead of one (two photons from down-converters). Maybe those photons have different phases and combined they do not form any pattern?

What is happening to rotational kinetic energy when moment of inertia is changed?

I know this question is asked here a lot, but I just had to ask this to finalise the concept.

When a system lets say a rod of length $L$ and mass $M$ is rotating with angular speed $omega_1$ its initial angular momentum is $L1 = (1/12)ML^2\omega_1$ and its initial kinetic energy is $KE = (1/24)ML^2{\omega_1}^2$.

Now after some time the rod is folded in half its angular momentum kept conserved i.e. without applying any external force or torque, its new angular velocity becomes $\omega_2 = 4\omega_1$ and its new kinetic energy becomes $KE_2 = (1/6)ML^2{\omega_1}^2$.

This is 4 times the original kinetic energy when no external force works, since the rod is folded, you can even say melted and formed into a smaller and denser rod, it has not undergone compression/expansion of any sort but still there is change in kinetic energy.

The most sense I could make out of this was that all the particles while rotating felt a centripetal force and the particles of half of the rod under this force went in its direction and did some work which appears as the change in kinetic energy. I have written the same and a proof as an answer here.

Now if I am write in my concept where did this energy come from, tension was providing the centripetal force but no work was done against tension as the rod was folded in half not compressed. If I am wrong then where did the energy come from?

Extra: I also tried the analogy of this question in translatory motion, suppose there is body of mass $m$ moving with velocity $v$ suddenly, its mass becomes $m/2$ then its velocity becomes $2v$ and its KE becomes $4KE_1$ there is no need to explain the energy conservation here since mass suddenly does not disappear into thin air, however moment of inertia can be changed and hence the question.

Addendum : Since folding the rod seems to bring about unnecessary questions about ways of folding, you can imagine that if rod was melted and formed into a longer rod all the while the system was rotating and angular momentum conserved, then new length becomes $2L$ new angular velocity becomes ${\omega_1}/4$ and new KE becomes $(1/96)ML^2{\omega_1}^2$. This time the energy becomes ${1/4}^{th}$ of the initial, where did this energy go to ? Certainly movement against centripetal force takes place, but since there is no extension in existing rod, energy can not be stored as spring energy in it, or so I think.

homework and exercises - Calculating torque in 3D?

Say you have a sphere, and you have several torque vectors acting on it, all at different points. Say you have the vector (6i + 3j + 5k) originating from point A, and the vector (3i + 1j + 9k) originating at point B, and (7i + 2j + 9k) acting on point C.

Summing the vectors gives you (16i + 6j + 23k) which is the resultant moment/torque vector. But at what point does the moment act on - A,B, or C?

The point it acts on has to matter right? I mean if you think of the moment vector as an axis the sphere revolves around, placing it in the center of the sphere and rotating the sphere around that is clearly different from placing it at the far left of the sphere and rotating it around that.

Answer

So you know about how to get the effective moment of all the forces

$$ \vec{M} = \sum_{i} \vec{r}_i \times \vec{F}_i $$

and the total forces

$$ \vec{F} = \sum_i \vec{F}_i $$

To get the location where the moments balance out (the line of action of the combined force) you do the following

$$ \vec{r} = -\frac{\vec{M} \times \vec{F}}{\vec{F} \cdot \vec{F}} $$

for example a force $\vec{F}=(1,0,0)$ located at $\vec{r}=(0,y,z)$ creates a torque of $\vec{M}=(0,z,-y)$. To recover the location of the force do $$r = -\frac{ (0,z,-y) \times (1,0,0) }{(1,0,0)\cdot (1,0,0)} =- \frac{(0,-y,-z)}{1} = (0,y,z) $$

forces - Black hole gravity vs parent star gravity

In the cases of black holes that form from supernova and collapse of a massive star, I understand that in most of these cases, the star loses significant amounts of mass from the explosion. Presumably, after this point as the remaining mass becomes more dense as it further collapses unto itself, it eventually becomes a black hole with gravitational force greater than that of its parent star. But, if gravity is based on mass, how can the black hole have greater gravitational force than the star from which it is formed?

Answer

For a given mass the gravitational attraction remains the same -- but only if you are far away.

For example, the surface gravity of Sol, our sun, is $274$ $ m/s^2$, about 28 times the surface gravity of Terra, which is $9.8$ $ m/s^2$.

But as the material is compacted, the surface gravity increases: this is because the effective mass can be treated as concentrated at a point in Newtonian gravitation: $F=GMm/R^2$, and here $M$ is the constant mass of the (remanant) of the star, while $m$ is the observer, and $R$ is the distance from the surface to the center of the star.

As the star becomes smaller, the distance between the surface and the center shrinks. For Sol the effective radius would shrink from $5\times 10^5$ $km$ to about $3$ $km$, the Schwarzchild radius. This puts you $100,000$ times closer, so the gravitational force would be $10^{10}$ times greater, and would vary measurable from your feet to your head.

So it all depends upon your distance. The earth would receive the same pull as always, minus the supernova and the missing mass, of course.

gravity - Gravitational waves detectors; are they all similar?

Are the gravitational waves detectors all working on the same principle/effect?

Answer

The fundamental effect of gravitational waves exploited by all detectors is the same: one tries to detect minute oscillatory changes in distances between different parts of the device. These changes are of the order of 1 part in 10²⁰ or smaller, so detecting them is a real challenge.

Different types of detectors do differ significantly when it comes to the techniques they employ to measure these minute changes. Interferometers look out for relative changes in phase of two light beams travelling in perpendicular directions. Weber bars exploit resonance to magnify vibrations caused by gravitational waves of a given frequency. Pulsar timing arrays compare timing of pulsar signals coming from different directions looking out for small differences caused by tiny expansion and contraction of space through which the signals propagate. High frequency detectors examine microwaves circulating in a loop looking out for polarization changes caused by spacetime contraction and expansion in different directions. See this wikipedia article for details.

Depending on how good given detector's isolation from unwanted influences is, it may require substantial effort to extract gravitational wave signature from data collected by these detectors. For example, LIGO phase loops lose lock every time a freight train passes by. The Einstein@Home project allows volunteers to donate the computing power of their home computers to this effort.

optics - What causes blurriness in an optical system?

The way I understand the purpose of a typical optical system is that it creates a one to one mapping between each possible incident ray and a point on a sensor plane. This is like a mathematical function. If there was no mapping, and each ray was free to strike any point on a sensor there would be no image formed on it and it would be just blurry average light. This would be like having a camera sensor without a lens.

Now there's a very simple concept that creates this one to one mapping, pin-hole camera. In a pin-hole camera there's no blurriness possible, as long as the hole is small enough, each point opposing the hole is mapped onto one specific ray. This means this type of camera can never have a blurry image, no matter where it's focal point is. This can be proven, geometrically.

In an optical system that uses lenses to create this mapping however things are not always ideal, because blurriness does happen. Which indicates that mapping is not one to one, and that some sensor points share rays with each other creating local averages, i.'e blurriness. It is often claimed that it happens because the focal point is not at the "right place". If you consider the pin-hole model as the ideal you will understand that this is not true. Changing focal point alone will only make image seem smaller or larger. From geometrical optics alone I don't see what could possibly cause the sharing of rays. It seems to me there's more to it and that i'm not the only one confused.

So what does REALLY create the blurriness? Is there some sort of imperfection in lenses that causes them to send multiple rays to the same point on a sensor and that somehow becomes more visible at certain focal distances? This is the only explanation i have.

Answer

To add some details to Eoin's answer.

Your description of imaging as a mapping is a good one to begin with and it will get yoi a long way. However, even in ray optics (see my answer here for more info), i.e. when the propagation of light can be approximated by the Eikonal equation (see here for more info), the mapping of points one-to-one between the object and image plane as you describe can only happen in very special conditions. In general, a bundle of rays diverging from one point will not converge to a point after passing through an imaging system made of refracting lenses and mirrors. One has to design the system so that the convergence is well approximated by convergence to a point. As Eoin said, this non-convergence is the Ray theory description of aberration: spherical, comatic, astigmatic, trefoil, tetrafoil, pentafoil and so forth are words that are used to describe aberration with particular symmetries (spherical aberration is rotationally symmetric about the chief ray, coma flips sign on a $180^o$ rotation about the chief ray, trefoil flips sign on a $120^o$ rotation and so forth). There is also chromatic aberration, where the image point position depends on wavelength so that point sources with a spectral spread have blurred images. Lastly, the imaging surface, comprising the points of "least confusion" (i.e. those best approximating where the rays converge to a point) is always curved to some degree - it is often well approximated by an ellipsoid - and so even if convergence to points is perfect, the focal surface will not line up with a flat CCD array. This is know as lack of flatness of field: microscope objectives with particularly flat imaging surfaces bear the word "Plan" (so you have "Plan Achromat", "Plan Apochromat" and so forth).

Only very special systems allow for convergence of all ray bundles diverging from points in the object surface to precise points in the image surface. Two famous examples are the Maxwell Fisheye Lens and the Aplanatic Sphere: both of these are described in the section called "Perfect Imaging Systems" in Born and Wolf, "Principles of Optics". They are also only perfect at one wavelength.

An equivalent condition for convergence to a point is that the total optical path - the optical Lgagrangian is the same for all rays passing between the points.

Generally, lens systems are designed so that perfect imaging as you describe happens on the optical axis. The ray convergence at all other points is only approximate, although it can be made good enough that diffraction effects outweigh the non-convergence.

And of course, finally, if everything else is perfect, there is the diffraction limitation described by Eoin. The diffraction limit simply arises because converging plane waves with wavenumber $k$ cannot encode any object variation that varies at a spatial frequency greater than $k$ radians per second. This, if you like, is the greatest spatial frequency Fourier component that one has to build an image out of. Images more wiggly than this Fourier component of maximum wiggliness cannot form. A uniform amplitude, aberration-free spherical wave converges to an Airy disk, which is often taken as defining the "minimum resolvable diffraction limited distance". However, this minimum distance is a bit more complicated than that. It is ultimately defined by the signal to noise as well, so an extremely clean optical signal can see features a little bit smaller than the so-called diffraction limit, but most systems, even if their optics is diffraction limited, are further limited by noise to somewhat less than the "diffraction limit".

What does temperature look like at the subatomic level?

I am trying to get a better understanding of the definition of temperature at the subatomic level. I have a background in molecular biology with some college physics, but no deep quantum mechanics background.

Everything I've found on the web (Wikipedia, Google Scholar) seems to use 'temperature' very loosely as just "agitation of particles": more movement/agitation of particles equals higher temperature. But what exactly does this mean?

The reason I'm asking is because the use of "particles" in relation to temperature seems to just mean atoms. The increase in agitation of atoms is equal to an increase in temperature. But I am asking because I don't know if this is true.

So atoms are made out of protons/neutrons/electrons. Protons and neutrons are composite particles, each made up of 3 elementary particles: quarks. Also, each of these examples I've mentioned are matter particles, but other particles like photons are massless. So how do they fit into temperature?

Basically, how do the different subatomic particles (both composite and elementary) relate to temperature?

Answer

The thermodynamic definition of temperature has been found to be emergent from the underlying particulate nature of matter. It is connected with an average over the kinetic energy of individual particles.

Here v is velocity of a molecule, m its mass, k_B the Bolzman constant and T the temperature

The kinetic energy requires to have a degree of freedom, which is fine in gases. In solids the degrees of freedom are the rotations and vibrations of the molecules, as the molecules themselves are bound and thus do not have degrees of freedom in space. The same for the internal constituents of molecules, atoms , etc. They exist in a bound state and a temperature cannot be defined for them. Their only contribution comes into contributing to the mass of the molecules.

One can stretch the definition by using the kinetic energy of a particle in the formula, and derive a temperature. All one is saying is that "this would be the temperature of an ensemble of particles that have this kinetic energy on average"

Another stretch of definitions is found here.

Thus at the subatomic level there does not exist a temperature for the bound quarks and gluons as no kinetic degree of freedom exists.

In the comment the quark matter subject has been broached. This is a hypothetical state of matter where the energies are such that the QCD asymptotic freedom behavior emerges. This can happen in two ways :

1) during the Big Bang ,

The earliest phases of the Big Bang are subject to much speculation. In the most common models the universe was filled homogeneously and isotropically with an incredibly high energy density and huge temperatures and pressures and was very rapidly expanding and cooling. Approximately 10−37 seconds into the expansion, a phase transition caused a cosmic inflation, during which the universe grew exponentially.[18] After inflation stopped, the universe consisted of a quark–gluon plasma, as well as all other elementary particles.

The temperatures here are defined by the kinetic energies of the hypothesized particles and it is supplied by the energy of the universe as it evolves after the Big Bang

2) and is searched for in ion ion collisions at LHC.

In these heavy-ion collisions the hundreds of protons and neutrons in two such nuclei smash into one another at energies of upwards of a few trillion electronvolts each. This forms a miniscule fireball in which everything “melts” into a quark-gluon plasma.

The temperature here is defined by the kinetic energy of quarks and gluons in the plasma that have degrees of freedom as for a while they are asymptotically free. The energy is supplied by the accelerator.

Are there aspects of General Relativity that have yet to be tested?

Good evening everyone, I am new in the field of General Relativity and I have been reading and learning about the subject in recent months.

For example, I read several articles about experiments designed to test the Principle of Equivalence and other aspects of this theory.

So I asked myself:

Are there aspects of General Relativity that have yet to be tested? Are there any experiments or projects under development? How can I contribute as an engineering student?

My questions are not innocent. This year I have to work on a project of initiation to research in my engineering school and I wondered in what type of project I could start, and above all if it is wise to get into such a field at my level.

I am very curious and wish to deepen my knowledge in the field. Last year I studied the Three-body problem from a newtonian point of view and in particular the calculation of Lagrange points and the study of their stability in the Sun-Earth orbital plane. Would there be one aspect of this problem that needs to be explored from a relativistic point of view as much in mathematics, physics and computer science?

I hope my English is sufficiently well written and understandable.

Answer

A good general resource on this kind of thing is the review article by Will, The Confrontation between General Relativity and Experiment, https://arxiv.org/abs/1403.7377 . He updates it every few years. The most recent version predates the direct detection of gravitational waves (although they were already verified observationally based on systems like the Hulse-Taylor pulsar [PSR B1913+16]).

If objects like Sag A* are correctly modeled as Kerr (i.e., spinning) black holes, then we have not yet observed whether they have event horizons, as predicted by GR. As safesphere noted in a comment, this may happen in the fairly near future.

GR incorporates the equivalence principle, and therefore it predicts that nothing special happens at an event horizon in terms of the local properties of space -- no firewalls or anything else crazy. This has not been tested.

There is quite a bit of work by theorists like Joshi suggesting that astrophysical collapse might not actually lead to a black hole but rather to some other object such as a timelike singularity. That is, the cosmic censorship conjecture is looking weaker and weaker. This has not been tested.

If it turns out that people like Joshi are right in their suspicions, then we could potentially observe singularities (because they wouldn't be hidden behind event horizons). This would allow us to test a prediction of GR, which is that singularities should be a generic thing that happens for most initial conditions. If we don't ever get such an opportunity, then the only singularity we'll ever have a chance to observe is the big bang.

solid state physics - Interpretation Born-Von Karman boundary conditions

The cyclic Born-Von Karman boundary condition says that if we consider a one dimensional lattice with length $L$, and if $\psi(x,t)$ is the wavefunction of an electron in this lattice, then we can say that $\psi(x+L,t) = \psi(x,t)$ for every $x$. Applying this boundary condition leads to correct solutions for $\psi$. This boundary condition can also be generalized to three dimensional lattices and can be applied when working with phonons instead of electrons.

I wonder if there is a reasoning why these boundary conditions can be applied. They appear to me as elegant boundary conditions, but I don't see any reason why these conditions could be applied. Can we derive the Von Karman boundary conditions or are they just an experimental result?

Answer

In a real-life misshapen blob of metal, strictly speaking the cyclic boundary conditions cannot be applied, since the blob only has a trivial group of spatial symmetries. However, the blob is approximately invariant under lattice translations (with the only mismatch occurring with the extremely small number of atoms at the surface) so it is tempting to associate the much larger symmetry group of lattice translation operators with the system.

Doing so, one obtains a system which is considerably more mathematically convenient than the otherwise rigorously-correct picture of a misshapen blob of atoms. As such, the Born-von Karman boundary conditions can be considered an approximate model to the behavior of real materials. That said, it's a pretty good approximation for macroscopically-sized materials.

That said, I don't have an explicit proof of why this approximation doesn't cause the math to explode and give wrong answers, so someone better versed than me could probably elaborate on this.

electromagnetism - What is the theoretical justification for a fluid flow's being irrotational?

I am not a fluid dynamicist, and I really just began thinking about this problem as my curiousity drew me into building an answer for the question What really allows airplanes to fly?.

It is very clear from the answers to the following questions:

When is a flow vortex free?

Does a wing in a potential flow have lift?

that viscous flows cannot be everywhere irrotational. Moreover, some handwaving justification is given in answer to the first question that low viscosity fluid can be irrotational.

Now, the spin angular momentum per unit volume of a fluid is the vector $\rho\,\nabla \wedge \vec{v}$. So the assumption of irrotational flow is the assumption of lack of spin angular momentum.

I can accept that it is reasonable in some cases to accept this to be true. But is there a deeper theoretical justification as to why and when a fluid's spin angular momentum should be nought?

In electromagnetism, we have specific equations - the Ampère and Faraday laws i.e. $\nabla\wedge\vec{H} = \vec{J}+\partial_t\vec{D}$ and $\nabla\wedge\vec{E} = -\partial_t\vec{B}$ - for the "source" of such vorticity in the fields. At an elementary level, we can see that in electrostatics conservation of energy behests that $\oint\vec{E}\cdot \mathrm{d} \vec{r} = 0$, so we immediately grasp irrotationalhood for the electrostatic field.

Is there any such analogue in fluid dynamics?

The Navier-Stokes equation doesn't seem "split" the velocity field up into curl and divergence terms like the Maxwell equations do, or to put it more pithily, the Maxwell equations say that the source distribution must be an exact form $\mathrm{d} \vec{F} = \mu_0 \vec{J}$ and we can thus "invert" $\mathrm{d}$, to within a constant (which we can set to nought the grounds that it would have infinite energy).

Are there any other ways to "split" the Navier-Stokes equation like Maxwell's equations are to shed intuitive light on the nature of an irrotational flow?

Answer

it's just a simplifying assumption. It's in fact rarely true. The main problem is that the Navier-Stokes equations are just tremendously more complex than the Maxwell equations--they have chaotic solutions, they couple nonlinearly, they behave wildly differently in the viscous and inviscid limits, and most strikingly, they have no proven existence and uniqueness theorem.

So, when learning them, it's best to work out simplified, specialized cases, like the steady flow, irrrotational case (which can then be solved using methods more analogous to E&M), and then to work from there.

Importance of Kronecker product in quantum computation

To get product state of two states $|\phi \rangle$ and $|\psi \rangle$, we use Kronecker product $|\phi \rangle \otimes |\psi \rangle$. Instead of Kronecker product $\otimes$, can we use Cartesian product, or any other products available in literature? But we do not do so. Here Kronecker product is more efficient than any other products. My question is why Kronecker product? Any physical reasoning or any problem in mathematical formulation for which Kronecker product is so important? The founders of quantum physics did not form it as they wished. Definitely they got some ideas that convinced them the efficiency of Kronecker product. What were them?

I asked the question at my very first class of quantum information theory. Till now I did not get any satisfactory answer but the course is going to be finished. Thank you for your help.

Answer

ACuriousMind's Answer pretty much summed up the reasons, which are essentially mathematical.

If you want to grasp the "physical significance", then I suggest you should work through an example: think of two quantum systems, each with three base states: $\left.\left|1\right.\right>$, $\left.\left|2\right.\right>$ and $\left.\left|3\right.\right>$. The set of linear superpositions in one of these quantum spaces is the set of unit magnitude vectors of the form $\alpha_1\,\left.\left|1\right.\right>+\alpha_2\,\left.\left|2\right.\right>+\alpha_3\,\left.\left|3\right.\right>$, where $\alpha_1^2+\alpha_2^2+\alpha_3^2=1$. Your states are going to be $3$-component vectors and they live in three dimensional spaces.

Now when we combine these two systems, the base states don't combine in a Cartesian product to give a six dimensional space. No, individually, each quantum system stays in its own space spanned by $\{\left.\left|1\right.\right>, \,\left.\left|2\right.\right>,\, \left.\left|3\right.\right>\}$ whilst the other one can be in any state in its own space spanned by its own versions of $\{\left.\left|1\right.\right>, \,\left.\left|2\right.\right>,\, \left.\left|3\right.\right>\}$.

So, with system 1 in state $\left.\left|1\right.\right>$, system 2 can be in any state of the form $\alpha_1\,\left.\left|1\right.\right>+\alpha_2\,\left.\left|2\right.\right>+\alpha_3\,\left.\left|3\right.\right>$. So the set of combined quantum states where system 1 is in state $\left.\left|1\right.\right>$ is a three dimensional vector space. A different 3-dimensional vector space of combined states arises if system 1 is in state $\left.\left|2\right.\right>$ with system 2 in an arbitrary $\alpha_1\,\left.\left|1\right.\right>+\alpha_2\,\left.\left|2\right.\right>+\alpha_3\,\left.\left|3\right.\right>$ state. Likewise for the set of combined states with system 1 in state $\left.\left|3\right.\right>$.

So our combined system has nine base states: it is a vector space of 9 dimensions, not 6. Lets write our base states for the moment as $\left.\left|i,\,j\right.\right>$, meaning system 1 in base state $i$, system 2 in base state $j$. Now, write a superposition of these states as a nine dimensional column vector stacked up as three lots of three: the first 3 elements are the superposition weights of the $\left.\left|1,\,j\right.\right>$, the next 3 the weights of $\left.\left|2,\,j\right.\right>$ and the last three the weights of the $\left.\left|3,\,j\right.\right>$. This is what a matrix representation of a general combined state will be.

Now, suppose we have a linear operator $T_1$ that acts on the first system alone, and a linear operator $T_2$ that acts on the second alone. These operators on the individual states have $3\times 3$ matrices. Then an operator on the combined system has a $9\times 9$ matrix. If you form the matrix Kronecker product $T_1\otimes T_2$, then this is the matrix of the operator that imparts the same $T_1$ to the three $\left.\left|i,\,1\right.\right>$ components, the three $\left.\left|i,\,2\right.\right>$ components and the three $\left.\left|i,\,3\right.\right>$ components and likewise imparts the same $T_2$ to the three $\left.\left|1,\,j\right.\right>$ components, the three $\left.\left|2,\,j\right.\right>$ components and the three $\left.\left|3,\,j\right.\right>$ components. This is what ACuriousMind means when he says:

we want every action of an operator (which are linear maps) on the individual states to define an action on the combined state - and the tensor product is exactly that, since, for every pair of linear maps $ T_i : \mathcal{H}_i \to \mathcal{H}$ (which is a bilinear map $(T_1,T_2) : \mathcal{H}_1 \times \mathcal{H}_2 \to \mathcal{H}$) there is a unique linear map $T_1 \otimes T_2 : \mathcal{H}_1 \otimes \mathcal{H}_2 \to \mathcal{H}$.

I work through a further detailed example for two coupled oscillators in my answer here.

classical mechanics - Is there a rotational equivalent to newtons laws?

Newtons three laws of motion appears to apply only for linear motion:

1. 
   

   An object remains at rest or moves in a straight line at uniform velocity unless a force is applied.

   
   


2. 
   
   

   Force is mass times acceleration.

   
   


3. 
   

   Every action causes an equal and opposite reaction.

   
   

Is there a rotational equivalence? For example:

1'. Every body rotates around a fixed axis at uniform angular velocity unless a torque is applied

2'. Torque is Moment of Inertia times angular acceleration

3'. When one body exerts a torque on another; there is an equal and opposite torque applied on the first body by the second.

First are these actually correct; if not, what are the correct equivalence; and who formulated them?

Answer

There is a rotational equivalence, but it is not what you stated. The problem, as pointed out by @curiousOne, is that conservation of angular momentum does NOT imply rotation about the same (fixed) axis. But I think a simple restatement like this could work:

• if no torque acts on a body, its angular momentum will remain unchanged


• rate of change of angular momentum is proportional to applied net torque


• when two bodies interact, the torque that A applies to B is equal and opposite to the torque that B applies to A, so that the angular momentum of the combined system (A+B) is preserved.

I believe that addresses the objections raised to your earlier version. Note that "axis of rotation is unchanged" is fundamentally different from "angular momentum is unchanged".

newtonian mechanics - Does $y$-motion really have nothing to do with $x$-motion?

I am watching a Physics 1 for physical science majors on coursera.com, and in one of the concept tests there is a question that goes like this; "A bullet is fired horizontally from a rifle on the Moon (where there is no air). The initial speed of the bullet when it leaves the gun barrel is $V_{0}$. Assume that the ground is perfectly level (and endless)."

And then there's 3 statements, on of which is; III) The time it takes for the bullet to hit the ground increases as $V_0$ is increased.

Apparently this statement is false.

The lecturer shows a simulator here where a cannon shoots a ball in a air free environment. And when the lecturer checks the time that it took for the ball to hit the ground after several shots with different initial velocities, the time is the same. Not really understanding this, I tried it for myself because I thought; "Surely, the time it takes for the ball to hit the ground is increased as the initial velocity is increased, at least if I were to increase the initial velocity by a large amount." So I did. And sure enough the time for the ball to hit the ground did increase.

How is that, and isn't it inconsistent with what the lecturer says about $x$-motion having nothing to do with $y$-motion?'

The lecturer has also showed the equation; $$x = x_0+ v_{0x}t+\frac{1}{2}a_xt^2$$ $$ a = 0 \space \vec{} \space x=x_0+v_0\cos(\theta) t$$ $$a = -g \space \vec{} \space y = y_0+v_0\sin\theta-\frac{1}{2}gt^2$$ Showing that you can treat $x$- and $y$-motion independently.

Answer

The time it takes is independent of the initial velocity if and only if the barrel is horizontal to the ground. If the barrel is horizontal, then the initial velocity will be solely in the $x$-direction and the only variable affecting the $y$-direction will be the acceleration due to gravity.

However, if the barrel is at an angle, then the initial velocity will have a vertical component. If $\theta$ is the angle from the ground, we can express the initial velocity in the $y$-direction as:

$v_{0, y} = v_0\sin(\theta)$

We can also express the height $y$ above the ground using the free-fall equation:

$y=v_{0,y}t - \frac{1}{2}gt^2=v_0\sin(\theta)t - \frac{1}{2}gt^2$

Notice that an increase in $v_0$ results in an increase in the amount of time it takes for the object to reach the ground. However, if $\theta=0$, then $v_0\sin(\theta)t$ will always equal $0$ because $\sin(0) = 0$. If $v_0\sin(\theta)t=0$, we can cancel it out of the equation and get $y = -\frac{1}{2}gt^2$, which is independent of the initial velocity.

energy conservation - Why doesn't this perpetual motion machine using the buoyant force work?

I realize this isn't possible, but I can't see why not, especially if you change the model a little bit so that the balls simply travel through a tube of water on the way up, rather than exactly this model.

Please be clear and detailed. I've heard explanations like "the balls wouldn't move" but that doesn't do it for me - I really do not see why the balls on the right would not be pulled/pushed up, and the rest of the chain wouldn't continue on.

Answer

The balls are entering the water well below the surface. The pressure there is much higher than at the surface. The work needed to push the balls into the water at this depth cancels the work gained when they float back up.

We can ignore the gravitational force on the balls since gravity pulls down as much as up as you traverse the loop.

Mathematically, if the balls enter the water at depth $d$, the pressure is $g \rho d$ with $g$ gravitational acceleration and $\rho$ the density of water.

The work done to submerge the balls is then the pressure times their volume, or $W_{ball} = g \rho V d$.

The force upwards on the ball is the weight of the water they displace which is $g \rho V$, and the work the water does on the balls is this force times the distance up that they travel, or $W_{water} = g \rho V d$.

The work the ball does on the water is the same as the work the water does on the ball. No free energy.

Maximizing entropy in QM (Sakurai's Modern Quantum Mechanics)

There is section in Sakurai's "Modern Quantum Mechanics 2nd edition" page 188 that is quite confusing as to what he is doing. In the section on "Quantum Statistical Mechanics" he defines a quantity $\sigma = -\text{tr}(\rho \text{ln}\rho)$, where the entropy is given as $S = k \sigma$. Then in order to maximize $\sigma$ he states the following "Let us maximize $\sigma$ by requiring that $$\delta \sigma = 0.~~~~~~~~~~~~~(1)$$

However, we must take into account the constraint that the ensemble average of $H$ has a certain prescribed value. In the language of statistical mechanics, $[H]$ is identified with the internal energy per constituent, denoted by $U$: $$[H] = \text{tr}(\rho H) = U.$$ In addition, we should not forget the normalization constraint $\sum_{k} \rho_{kk} = 1$ (where $\rho$ is the density operator). So our basic task is to require (1) subject to the constraints $$\delta(H) = \sum_{k} \delta \rho_{kk} E_k = 0$$ and $$\delta(\text{tr} \rho) = \sum_{k} \delta \rho_{kk} = 0.$$

We can most readily accomplish this by using Lagrange multipliers. We obtain $$\sum_{k} \delta \rho_{kk}[(\text{ln}\rho_{kk} +1) + \beta E_{k} + \gamma] = 0$$ which for an arbitrary variation is possible only if $$\rho_{kk} = \text{exp}(- \beta E_{k} - \gamma - 1).$$"

Question: What is the reasoning for the requirement (1) in order to maximize $\sigma$, and how are the constraints obtained?

general relativity - Is time going backwards beyond the event horizon of a black hole?

For an outside observer the time seems to stop at the event horizon. My intuition suggests, that if it stops there, then it must go backwards inside. Is this the case?

This question is a followup for the comment I made for this question: Are we inside a black hole?

Food for thought: if time stops at the event horizon (for an outside observer), for inside, my intuition suggests, time should go backwards. So for matter, that's already inside when the black hole forms, it won't fall towards a singularity but would fall outwards towards the event horizon due to this time reversal. So inside there would be an outward gravitational force. It would be fascinating if it turns out that all this cosmological redshift, and expansion we observe, is just the effect of an enormous event horizon outside pulling the stuff outwards.

So from outside: we see nothing fall in, and see nothing come out.

And from inside: we see nothing fall out, and see nothing come in.

Hopefully the answers make this clear, and I learn a bit more about the GR. :)

Answer

It's easy to forget that, in the context of relativity, there is no universal time. You write:

For an outside observer the time seems to stop at the event horizon. My intuition suggests, that if it stops there, then it must go backwards inside. Is this the case?

But your intuition doesn't seem to take into account that, for an observer falling into the hole, time doesn't stop at the event horizon.

The point is that one must be much more careful in their thinking about time within the framework of general relativity where time is a coordinate and coordinates are arbitrary.

In fact, within the event horizon, the radial coordinate becomes time-like and the time coordinate becomes space-like. This simply means that, to "back up" inside the event horizon is as impossible as moving backwards in time outside the event horizon.

In other words, the essential reason it is impossible to avoid the singularity once within the horizon is precisely that one must move forward through time which, due to the extreme curvature within the horizon, means moving towards the central singularity.

Why does the Lorentz transformation in special relativity have to be like this?

Basically I think Albert Einstein (A.E.) was trying to find a transformation that:

1. Always transform a constant-velocity movement into a constant-velocity movement.


2. Always transform a light-speed movement into a light-speed movement.


3. If an object with a speed $v$ in frame $A$ is rest in frame $B$, then any rest object in $A$ has a speed $-v$ in $B$.

A.E. gave:

$$x'=\frac{x-ut}{\sqrt{1-u^2/c^2}},$$

$$t'=\frac{t-(u/c^2)x}{\sqrt{1-u^2/c^2}}.$$

But there are more than one transformation that can do this.

Multiply a factor, we can get another one:

$$x'=\frac{x-ut}{\sqrt{1-u^2/c^2}}(1+u^2),$$

$$t'=\frac{t-(u/c^2)x}{\sqrt{1-u^2/c^2}}(1+u^2).$$

It satisfies all three postulates given. So why can't the latter be the Lorentz transformation?

Answer

You have discovered the well-kept secret that in 2 dimensions, the transformations which keep light-rays fixed include conformal transformations, not just Lorentz transformations. You need to pick a subset of the conformal transformations which form a group, and which are compatible with reflections.

The ones you used are not the good ones, because if you use your transformation with speed u, and invert it, it is not the transformations with speed -u. The factor you get is not multiplicative, so if you compose two transformations with u and u', you don't get something in the group. If you keep transforming, your coordinates just get a bigger and bigger scale factor.

But there is another subgroup of the 1+1 d conformal group which is a group which obeys all of Einstein's speed-of-light postulates:

$$ x' = e^{k\alpha} ( \cosh(\alpha) x - \sinh(\alpha) t) = ( \sqrt{1+v\over 1-v})^k{x-vt\over \sqrt{1-v^2}}$$ $$ t' = e^{k\alpha} ( \sinh(\alpha) t - \cosh(\alpha) x )= (\sqrt{1-v\over 1+v})^k{t-vx\over\sqrt{1-v^2}}$$

This transformation scales by the rapidity (relativistic analog of 2d rotation angle) with the only 1d scale factor that forms a group. These transformations are the alternate Lorentz transformations you want. Their orbits are relativistic analogs of geometric spirals, not circles, and they form a one-dimensional group, and they reduce to Galilean transformations at low velocities ($c=1$ in the formulas above).

In his derivations of the Lorentz transformations, Einstein implicitly used reflection symmetry, by assuming that the transformation for -v will be the same as the transformation to speed v with just the sign on x reversed. This assumption allows you to kill this possibility, because it is asymmetric, the scale for positive velocity transformations is inverse to the scale with negative velocity transformations.

particle physics - Are There Strings that aren't Chew-ish?

String theory is made from Chew-ish strings, strings which follow Geoffrey Chew's S-matrix principle. These strings have the property that all their scattering is via string exchange, so that the amplitudes are entirely given by summing world sheets and by topology. This leads to a successful theory of gravity, we know, uniqueness, we know etc.

I would like to know exactly how much of the Chew-Mandelstam S-matrix philosophy, that proto-holographic principle which was so important for the development of string theory historically, is logically necessary for deriving the string interactions. This is obscured in more modern treatments, because these take for granted certain things that were historically difficult and came directly from the S-matrix.

One of those things which are taken for granted is that strings interact only by exchanging other strings, and that this interaction respects world-sheet duality. In a recent exchange (here: What are the details around the origin of the string theory?), Lubos Motl suggested that this may be derived only from the assumption that there is some sort of string excitation, and that this string interacts consistently. I would like to poke at this claim with examples, to determined exactly how many extra assumptions are hidden behind this claim.

The assumption that strings exchange strings with Doleh-Horn-Schmidt duality is equivalent to the assumption that the only interaction between strings is the topological exchange diagram from string theory. This assumption not only fixes the string interaction given the spectrum, it fixes the spectrum and even the dimension of space time, it fixes everything. So it is very weird physics. I believe that this assumption is Chew/Mandelstam physics, and that you can't derive strings without using it.

Stupid rubber band model

There are N scalar fields of mass m, and N U(1) gauge fields. Scalar k has a charge +1 under gauge field number k, and a charge -1 under gauge field k+1 (with periodic boundaries, so scalar N and 1 are interacting).

Each U(1)'s is broken from a separate SU(2) at some high energy by a technicolor mechanism which is irrelevant, but which embeds the U(1)'s in an asymptotically free theory at high energies, so that the theory is completely consistent.

These N scalars can form a matter-string, composed of a type 1 particle, bound to a type 2 particle, bound to a type 3 particle, etc around a loop closing back on particle 1. I will call this the rubber-band.

Further, the N different scalars are each separately in the fundamental representation of an SU(N) gauge theory, so that individual quanta of these scalars have infinite mass by confinement. The coupling of the SU(N) will be infinitesimal at the rubber-band scale, so that it will not affect the dynamics.

The rubber band is one of the baryons of the theory. If you just look at rubber-band dynamics, if you collide two rubber bands, you can exchange U(1) quanta between corresponding points on the rubber band, and get long range Van-der-Waals, splitting and rejoining interactions. But there is no duality.

Properties

Given a "baryon" in the model, it contains N scalars. If the scalars are not numbers 1...N, they don't bind together, and if the scalars are 1...N, then they bind together, releasing some U(1) photons and making a closed bound rubber-band at a lower energy by N times the binding energy of each pair (approximately). The isolated particles are infinitely massive, so the lightest baryonic state in the theory is the rubber band.

The rubber band has oscillation excitations. For large N, you get a spectrum which is by the frequency of a 1-dimensional fixed length string. There are modes which correspond to the N particles sitting on top of each other, rotating with 1 unit of angular momentum, moving apart and shrinking, etc.

The rubber band effective theory holds for energies above the bound-state baryon mass and below the mass of a cut rubber band, and it holds well above this too. But the interactions of the rubber band have nothing to do with string theory.

Questions

1. Did I miss a failed conservation law which makes the rubber band decay?


2. Would you consider the rubber band to be a relativistic string?


3. How close to the real string spectrum can you get by mocking up rubber bands?


4. Why do you say that string theory is only based on the assumption of "relativistic strings", when the relativistic rubber band is a "relativistic string" and does not obey duality?


5. Is it obvious, given this model, that Chew bootstrap is necessary for deriving string theory, and that duality does not "obviously" follow from the existence of a string picture?

nuclear physics - What is the half-life threshold for an isotope to be considered stable?

What minimum half-life an isotope should have to be considered stable?

electromagnetism - What is $c$ in the Lorentz force expression?

The usual Lorentz force expression I am familiar with is this:

$$\vec F=q(\vec E+\vec v \times \vec B)$$

I have seen some other versions lately that include an extra factor $1/c$:

$$\vec F=q\left(\vec E+ \frac{1}{c} \vec v \times \vec B\right)$$

What is this $c$ and how is it included? I guess other parameters in the expression are also different from the top expression for this to fit?

Example of a text snippet where I have run across this extra parameter:

Answer

In the second formula, $c$ is the speed of light.

Both formulas use different system of units. The first one uses the SI: $q$ in coulombs, $\vec{E}$ in volts per meter and $\vec{B}$ in teslas. The second one uses gaussian units: $q$ in statcoulombs, $\vec{E}$ in statvolts per centimeter and $\vec{B}$ in gauss (being statvolts per centimeter and gauss dimensionally equivalent).

When dealing with electromagnetism, it is common to encounter various systems of units (SI, gaussian, Heaviside...), in wich equations differ in factors of $c$, $4\pi \mu_0$, etc. It's always convenient to make clear what system of units are you using.

elasticity - Compressible material with Poisson ratio of 0.5

According to comments and an answer to this question, it is claimed that:

1. Incompressible materials require the Poisson ratio to be 0.5.



2. A poisson ratio of 0.5 does not imply that the material is incompressible.

It is thus implied that the statement "A material has a Poisson ratio of 0.5 if and only if the material is incompressible" is false, and that causality cannot be reversed. However, the maths seem reversible, and I have not been able to find any compressible material with a Poisson ratio of 0.5, whether that be in practice or theory.

Note: arguably only theory matters; the hypothesis is academic: no real material exists with a Poisson ratio of exactly 0.5.

Can above second statement be proven with theoretical evidence (e.g. by presenting a theoretical material with a Poisson ratio of exactly 0.5 that does change volume for small stress/strain)? Or, alternatively, can the second statement be disproven mathematically (e.g. by proving reversibility of causality)?

Answer

Idealized liquids have a Poisson's ratio of exactly 1/2 and are not incompressible (because no material is incompressible). No solid can have a Poisson's ratio of exactly 1/2.

To explain: If you apply a deviatoric (i.e., not hydrostatic; the 1-D version is shear) load on a liquid, it will deform without resistance and without a change in volume, implying that $\nu=1/2$. One way to see this is to use one of the elasticity relations (which assume a homogeneous isotropic linear elastic material): $G=\frac{3K(1-2\nu)}{2(1+\nu)}$. All stable materials have a positive bulk modulus $K$ (i.e., all stable materials compress to some degree under hydrostatic stress); thus, setting $\nu=1/2$ implies a shear modulus $G$ of zero, which corresponds to a fluid.

Let's look at a couple of the elasticity relations from the Poisson's ratio side: $\nu=\frac{3K-E}{6K}$, where $E$ is the Young's elastic modulus, and $\nu=\frac{3K-2G}{2(3K+G)}$. Thus, elastomers have a Poisson's ratio of nearly 1/2 because their shear and Young's elastic moduli are much smaller than their bulk moduli. When you shear or pull on rubber, for example, it's relatively easy to unkink and uncoil its long polymer chains to obtain shearing or uniaxial deformation. When you apply pressure from all sides, however, you're essentially trying to push C atoms closer to C atoms, which is not easy.

Note that this review (Greaves et al., "Poisson’s ratio and modern materials", Nat Materials 10 2011, DOI: 10.1038/NMAT3134) defines $\nu=\frac{3K-2G}{2(3K+G)}=0$ for gases for unexplained reasons, possibly because they are idealizing gases as exhibiting $K=0$.

What topics do I need to study electromagnetism on the quantum scale?

1. 
   

   What topics do I need to study (in order) so that I can study electromagnetics on the quantum scale?

   
   


2. 
   

   What is the name of the discipline studying electromagnetism on the quantum scale?

   
   


3. 
   

   Do I need to know nuclear physics?

   
   

general relativity - What is the geometric interpretation of the Einstein tensor $R_{mu nu} - frac{1}{2} g_{mu nu} R$

The Riemann curvature tensor $R_{\mu \nu \rho \sigma}$ has the geometric interpretation of giving how much parallel transport fails to close around tiny loops. The Ricci tensor $R_{\mu \nu}$ the Riemann curvature averaged over all directions, as in, if there is negative curvature in some direction there must be positive curvature in another if $R_{\mu \nu} = 0$.

What is the geometric interpretation of the Einstein tensor $$ G_{\mu \nu} = R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R? $$ Is there a way to understand $$ \nabla^\mu G_{\mu \nu} = 0 $$ Intuitively?

quantum mechanics - Why can't $psi(x) = delta(x)$ in the case of Harmonic oscillator?

In the analysis of Harmonic Oscillator, it is claimed that $\langle\hat H\rangle$ cannot be zero, why is it so?

I mean $\hat H = \frac{ \hat p^2 }{2m } + \frac12 k \hat x^2$, and $$\left = \int dx (x\psi(x))^\dagger (x\psi(x)) = 0$$ would imply that $x\psi(x) = 0 \quad \forall x.$ In particular, this is true when $x = 0$, so we have two options; either $\psi(x) = 0$ or $\psi(x) = \delta(x)$.

So, why can't $\psi(x) = \delta(x)$ in the case of Harmonic oscillator ?

Note: $\hat H = \frac{ \hat p^2 }{2m } + \frac12 k \hat x^2 $

Answer

The state $\psi(x) = \delta(x)$ is a perfectly valid state for the harmonic oscillator to occupy. (With caveats, though: it is not normalizable, so it's not a physically-accessible state. Still, it's a perfectly reasonable thing for the mathematical formalism to handle.) As you note, it has a position uncertainty equal to zero, as well as a vanishing expectation value $⟨x^2⟩=0$.

However, it does not have a vanishing momentum uncertainty, and in fact if you expand it as a superposition of plane waves, $$ \delta(x) = \frac{1}{2\pi\hbar} \int_{-\infty}^\infty e^{ipx/\hbar}\mathrm dp = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^\infty A(p) e^{ipx/\hbar}\mathrm dp, $$ you require an even weight $A(p) \equiv 1/\sqrt{2\pi\hbar}$ for all momenta, which means that the momentum-squared expectation value $$ ⟨p^2⟩ = \int_{-\infty}^\infty |A(p)|^2 p^2\mathrm dp = \frac{1}{2\pi\hbar}\int_{-\infty}^\infty p^2\mathrm dp = \infty $$ diverges to infinity. (This result is required by the uncertainty principle, but the derivation here does not rely on it - it's an independent proof of that fact. Still, you can see the consistency in that $\Delta x=0$ and $\Delta p \geq \hbar/2\Delta x$ can only be satisfied by having $\Delta p = \infty$.)

This then implies that the expectation value of the hamiltonian is also infinity: $$ ⟨H⟩ = \frac{1}{2m}⟨p^2⟩ + \frac12 k ⟨x^2⟩ = \infty. $$

---

As for this,

In the analysis of Harmonic Oscillator, it is claimed that $\langle\hat H\rangle$ cannot be zero, why is it so?

this is the zero-point energy of the oscillator, which has been explored multiple times on this site. If you want to ask why this is, you should ask separately, with a good showing of the previous questions here and what it is about them you do not understand.

homework and exercises - Ball flying towards me or me flying towards ball

Suppose a ball is flying towards me at a speed of 10m/s and that, on impact, I feel "x" amount of pain.

If, instead, it was me flying towards the ball at the same speed, with all other conditions being the same, would I feel the same amount of pain?

Answer

Look at it this way:

Suppose you are in a train travelling at 10 m/s. Somebody inside the train throws a ball at you in the opposite direction at 10 m/s. You feel the pain belonging to your first experiment. However, somebody looking at this experiment from outside the train would say that the ball is standing still and you are travelling towards the ball at 10 m/s (= your second experiment). Since it is the same experiment, you will feel exactly the same pain.

homework and exercises - Show that $partial_nu T^{munu} = - j_nu F^{munu}$

In a theoretical physics homework problem, I have to show the following: $$\partial_\nu T^{\mu\nu} = - j_\nu F^{\mu\nu}$$

Where $T$ is the Energy-Momentum-Tensor, $j$ the generalized current and $F$ the Field-Tensor. We use the $g$ for the metric tensor, I think in English the $\eta$ is more common.

I know the following relationships:

• 
  

  Current and magnetic potential with Lorenz gauge condition: $$\mathop\Box A^\mu = \mu_0 j^\mu$$

  
  


• 
  

  Energy-Momentum-Tensor: $$T^{\mu\nu} = \frac1{\mu_0} g^{\mu\alpha} F_{\alpha\beta} F^{\beta\nu} + \frac1{4\mu_0} g^{\mu\nu} F_{\kappa\lambda} F^{\kappa\lambda}$$

  
  


• 
  

  Field-Tensor: $$F^{\mu\nu} = 2 \partial^{[\mu} A^{\nu]} = \partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu}$$

  
  


• 
  
  

  d'Alembert operator: $$\mathop\Box = \partial_\mu \partial^\mu$$

  
  


• 
  

  Bianchi identity: $$\partial^{[\mu} F^{\nu\alpha]} = 0$$

  
  

So far I have set all the definitions into the formula I have to show, but I only end up a lot of terms from antisymmetrisation and product rule. I also drew all what I have in Penrose graphical notation, but I still cannot see how to tackle this problem.

Could somebody please give me a hint into the right direction?

Answer

Let's look at different terms from differentiating $T^{\mu\nu} $.

The first from differentiating $ g^{\mu\alpha} F_{\alpha\beta} F^{\beta\nu}$ is $$\partial_\nu g^{\mu\alpha} F_{\alpha\beta} F^{\beta\nu}= g^{\mu\alpha} F_{\alpha\beta} (\partial_\nu F^{\beta\nu}) +(\partial^\nu F^{\mu\beta}) F_{\beta\nu}= - \mu_0 F_{\alpha\beta} j^\beta +(\partial^\nu F^{\mu\beta}) F_{\beta\nu}$$

The first term is exactly what you want, the second cancels against the stuff you get from differentiating $g^{\mu\nu} F_{\kappa\lambda} F^{\kappa\lambda}$:

$$\partial^\mu F_{\kappa\lambda} F^{\kappa\lambda}=2 F_{\kappa\lambda} (\partial^\mu F^{\kappa\lambda})=-2 F_{\kappa\lambda} (\partial^\kappa F^{\lambda\mu}+\partial^\lambda F^{\mu\kappa}) =-4 (\partial^\nu F^{\mu\beta}) F_{\beta\nu}$$ where in the second equality sign we have used Bianchi identity and in the last equality we have used $$ F_{\kappa\lambda} \partial^\kappa F^{\lambda\mu} \underset{\text{relabel indecies}}= F_{\nu\beta}\partial^\nu F^{\beta \mu} \underset{\text{antisym. of $F$}}= F_{\beta\nu}\partial^\nu F^{\mu\beta} $$ This exactly cancels the second term in the first equation.

classical mechanics - Liouville's theorem and conservation of phase space volume

It can be proved that the size of an initial volume element in phase space remain constant in time even for time-dependent Hamiltonians. So I was wondering whether it is still true even when the system is dissipative like a damped harmonic oscillator?

Answer

The interplay of Hamiltonian and Lagrangian theory is based on the following general identities, where $L$ is the Lagrangian function of the system, $$\dot{q}^k = \frac{\partial H}{\partial p_k}\:,\qquad(1)$$ $$\frac{\partial L}{\partial q^k} = -\frac{\partial H}{\partial q^k}\:.\qquad(2)$$ Above, the RH sides are functions of $t,q,p$ whereas the LH sides are functions of $t,q,\dot{q}$ and the two types of coordinates are related by means of the bijective smooth (with smooth inverse) relation, $$t=t\:,\quad q^k=q^k\:,\quad p_k = \frac{\partial L(t,q,\dot{q})}{\partial \dot{q}^k}\:.\qquad(3)$$ Finally, the Hamiltonian function is defined as follows $$H(t,q,p) = \sum_{k}\left.\frac{\partial L}{\partial \dot{q}^k}\right|_{(t,q,\dot{q}(t,q,p))}\dot{q}(t,q,p) - L(t,q,\dot{q}(t,q,p))\:.$$ Suppose that the following E-L hold, $$\frac{d}{dq}\left(\frac{\partial L}{\partial \dot{q}^k}\right) - \frac{\partial L}{\partial q^k} = Q_k(t,q,\dot{q})\:, \quad \frac{d q^k}{dt}= \dot{q}^k \quad (4)\:.$$ The functions $Q_k$ take the (e.g. dissipative) forces into account, which cannot be included in the Lagrangian. For a system of $N$ points of matter with positions $\vec{x}_i$, if the degrees of freedom of the system are described by coordinates $q^1,\ldots,q^n$ such that $\vec{x}_i= \vec{x}_i(t,q^1,\ldots,q^n)$, one has: $$Q_k = \sum_{i=1}^N \frac{\partial \vec{x}_i}{\partial q^k} \cdot \vec{f_i}$$ $\vec{f}_i$ being the total force, not described in the lagrangian, acting on the $i$th point.

One easily proves that, in view of the general identities (1) and (2), a curve $t \mapsto (t, q(t), \dot{q}(t))$ satisfies the EL equations (4), if and only if the corresponding curve $t \mapsto (t, q(t), p(t))$ (constructed out of the previous one via (3)), verifies the following equations: $$\frac{dq^k}{dt} = \frac{\partial H}{\partial p_k}\:, \quad \frac{dp_k}{dt} = -\frac{\partial H}{\partial q^k} + Q_k\:.\quad(5)$$ In the absence of the terms $Q_k$, these are the standard Hamilton equation. If $Q_k\equiv 0$, even if $H$ explicitely depend on time, the solutions of Hamilton equations preserve, in time, the canonical volume: $$dq^1 \wedge \cdots \wedge dq^n \wedge dp_1 \wedge \cdots \wedge d p_n\:.$$ In the presence of dissipative forces which cannot be included in the Lagrangian, the term $Q_k$ show up and the volume above generally fails to be preserved. However this is not the whole story. Let us consider the damped harmonic oscillator. In the absence of dissipative force, the Lagrangian reads $$L(x, \dot{x}) = \frac{m}{2} \dot{x}^2 -\frac{k}{2} x^2\:.$$ The dissipative force $-\gamma \dot{x}$ takes place in the EL equations due to the presence of the term: $$Q = - \gamma \dot{x}\:.$$ In this juncture, passing to the Hamiltonian formulation, the canonical volume is not preserved along the solutions of the equation of motion. However, sticking to the damped oscillator, there is a way to include the dissipative force in the Lagrangian function. As a matter of fact, this new Lagrangian produces the correct equation of motion of a damped oscillator $$L(t,q,\dot{q}) = e^{\gamma t/m}\left(\frac{m}{2} \dot{x}^2 -\frac{k}{2} x^2\right)\:.$$ In this case the Hamiltonian function turns out to be $$H(t,q,p) = e^{-\gamma t/m}\frac{p^2}{2m} + e^{\gamma t/m}\frac{k}{2}x^2\:.$$ As the general theory proves, the canonical volume is preserved by the solutions of Hamilton equations referred to that Hamiltonian function, regardless the fact that the system is dissipative.

It is important to notice that, with the second Lagrangian, $p$ ceases to be the standard momentum $mv$ differently from the first case.

fluid dynamics - What cools a drink?

When you stick ice in a drink, AFAICT (the last physics I took was in high school) two things cool the drink:

• The ice, being cooler than the drink, gets heat transferred to it from the drink (Newton's law of cooling). (This continues after it melts, too.)


• The ice melts, and the cool liquid mixes with the drink, which makes the mixture feel cooler.

My first question is, to what extent does each of these cool the drink: which has a greater effect, and how much greater?

Secondly, how much of the cooling by the first method (heat transfer) is without melting the ice? That is, is there any significant amount of heat transfer to each speck of ice before it's melted, and how much does that cumulatively affect the drink's temperature?

I suppose all this will depend on a bunch of variables, like the shape and number of ice cubes and various temperatures and volumes. But any light that can be shed, let's say for "typical" situations, would be appreciated.

Answer

There are three processes to take into account:

1. The warming of ice towards the melting point if it was originally below $0^{\circ} C$.


2. The melting of ice itself


3. The warming of the resulting water

The 1. and 3. part is addressed by heat capacity of ice and water respectively and the amount of heat will be directly proportional to temperature difference and weight of the water/ice. The proportionality constant (actually it also depends on the temperature but not very strongly so let's just ignore that) is called specific heat. For water it is about twice as large as that of ice at temperatures around $0^{\circ} C$.

As for the 2. part, this has to do with latent heat. Simply put, this is an amount of heat you need to change phases without changing temperature. Less simply put, when warming you are just converting the heat into greater wiggling of water molecules around their stable positions in the crystal thereby increasing their temperature. But at the melting point that heat will instead go into breaking chemical bonds between molecules in the ice lattice.

Now, latent heat is really big (you need lots of energy to break those bonds). To get a hang on it: you would need the same amount of heat to warm water from $0^{\circ} C$ to $80^{\circ} C$ as you would need to melt the same amount of ice.

Now, presumably you want your drink cold in the end so that temperature for 3. will be close to $0^{\circ} C$ and also the ice cubes should be pretty warm (no use in producing ice cubes of e.g. $-50^{\circ} C$, right?). This means that these processes won't contribute much cooling. It's fair to say that melting of the ice takes care of everything.

Note: we can also quickly estimate how much ice you need by neglecting the processes 1. and 3. Say you are starting with a warm drink of $25^{\circ} C$ and you want to get it to $5^{\circ} C$. So, reusing the argument about the $80^{\circ} C$ difference being equivalent to a latent heat of the same mass, we see that you need four times less ice than water to get the job done.

radioactivity - Can an ensemble of meta-stable systems be prepared so their survival probability drops approx. linearly right after preparation?

In this answer dealing with details of decay theory (incl. references) it is shown that

[Given] a system initialized at $t = 0$ in the state [...] $| \varphi \rangle$ and left to evolve under a time-independent hamiltonian $H$ [... its] probability of decay is at small times only quadratic, and the survival probability is slightly rounded near $t = 0$ before of going down [exponentially].

Is it correct that therefore it is also possible to prepare (initialize) an entire ensemble of $N \gg 1$ such states $| \varphi \rangle$, such that their survival probability is at small times only quadratic ?

Is it instead possible at all to prepare an ensemble of $N$ states (which would likewise "evolve under the Hamiltonian $H$") such that their survival probability is (at least to a good approximation) not quadratic but rather drops linearly as a function of the duration since completion of the preparation ?

In particular, if an ensemble of $2~N$ states $| \varphi \rangle$ had been given and (in the process of an extended preparation procedure) half of those (i.e. $N$ systems) had decayed, do the remaining/surviving $N$ systems together then constitute such an ensemble? What exactly is the survival probability of these given, momentarily remaining/surviving $N$ systems; as a function e.g. of $t_{\text{(extended prep.)}} := t - \tau_{1/2}$, where $\tau_{1/2} = \tau~\text{Ln}[2]$ is the specific overall "half-life" duration?

thermodynamics - Why isn't temperature measured in Joules?

If we set the Boltzmann constant to $1$, then entropy would just be $\ln \Omega$, temperature would be measured in $\text{joules}$ ($\,\text{J}\,$), and average kinetic energy would be an integer times $\frac{T}{2}$. Why do we need separate units for temperature and energy?

Answer

One reason you might think $T$ should be measured in Joules is the idea that temperature is the average energy per degree of freedom in a system. However, this is only an approximation. That definition would correspond to something proportional to $\frac{U}{S}$ (internal energy over entropy) rather than $\frac{\partial U}{\partial S}$, which is the real definition. The approximation holds in cases where the number of degrees of freedom doesn't depend much on the amount of energy in the system, but for quantum systems, particularly at low temperatures, there can be quite a bit of dependence.

If you accept that $T$ is defined as $\frac{\partial U}{\partial S}$ then the question is about whether we should treat entropy as a dimensionless quantity. This is certainly possible, as you say.

But for me there's a very good practical reason not to do that: temperature is not an energy, in the sense that it doesn't, in general, make sense to add the temperature to the internal energy of a system or set them equal. Units are a useful tool for preventing you from accidentally trying to do such a thing.

In special relativity, for example, it makes sense to set $c=1$ because then it does make sense to set a distance equal to a time. By doing that, you're simply saying that the path between two points is light-like.

But $T=\frac{\partial U}{\partial S}$ measures the change in energy with respect to entropy. Entropy and energy are extensive quantities, whereas temperature is an intensive one. This means that it doesn't very often make sense to equate them without also including some non-constant factor relating to the system's size. For this reason, it's very useful to keep Boltzmann's constant around.

My personal favorite way to do it is to measure entropy in bits, so that $k_B = \frac{1}{\ln 2} \,\mathrm{bits}$ and the units of temperature are $\mathrm{J\cdot bits^{-1}}$. Having entropy rather than temperature as the quantity with the fundamental unit tends to make it much clearer what's going on, and bits are a pretty convenient unit in terms of building an intuition about the relationship to probability theory.

quantum field theory - What are some approaches to discrete space-time used in modern physics?

This thought gave rise to some new questions in my mind.

What are the consequences for:

1. How would it affect duality i.e. particle, wave property of photons?


2. How does this statement affect the information theoretical aspect (entropy) of the universe? Update: Given a volume V of space, is the entropy (maximum information that can be store) in this volume changed when this statement is applied?


3. How is a black hole affect by this statement? Update How is entropy changed inside the black hole?


4. Could one consequence be that the universe is hologram, since the construction isn’t continues?


5. Would the smallest quantified space be planck's constant? Is there an equivalent constant for time?

I hope to get some of your feedbacks regarding this statement.

Answer

Let's try and make things more precise, step-by-step.

1. There's no such thing as "particle-wave duality": the name-of-the-game is "Quantum Field Theory". This paradoxical notion of a possible "duality" only happens when you don't use the appropriate framework to describe your Physics. Therefore, it makes no sense to speculate on what would happen if spacetime were quantized/discrete: in this scenario, the question would be: "Would a quantized/discrete spacetime affect Quantum Field Theory?" And the answer to this question is "No." The reason being that different physical theories have different domains of validity, given by the characteristic energy of the phenomena they describe.


2. What is the "information theoretical aspect of the universe"?! This is not even appropriately defined, let alone "well defined".


3. The black hole is the stereotypical object in a quantum gravity theory. So, when you quantize spacetime, you should look at black holes to see what happens. We already know that black holes have Entropy. So, the very first question should be: What does your particular quantization scheme yields for black hole Entropy? The current state-of-the-art, as far as i know, is that all different schemes of quantization of spacetime yield a reasonable answer to this question.


4. This question, again, is not even appropriately defined, let alone "well defined". Holography has a very precise and well defined meaning in Physics, which is not related to the hologram in a credit card, for example. So, holography does play a role in quantum gravity, the more famous statement being that of AdS/CFT. But, as it stands, your question does not have meaning.


5. This has already been stablished a long time ago: if you quantize spacetime, the smallest unit of spacetime is given in terms of Natural units.

Can elementary particles be nicely classified into "force transmitters" and "force emitters"?

This well-received answer begins with

First of all, you can't compare photons with electrons. They are different types of particles (spin 1 vs spin1/2; force transmitter vs force emitter).

I'd never heard of assigning a binary distinction of whether a particle emits a force, or transmits one, so I'd like to ask the following:

1. 
   

   Does this classification work?

   
   


2. 
   

   Is it exclusive and comprehensive?

   
   


3. 
   
   

   Is it used in some way?

   
   


4. 
   

   Are force emitters always also force responders?

   
   

At first I thought that there may be some interaction between photons and gravity that blurs the line, but then again perhaps gravity just affects the space that the photons propagate in rather than deflects the photons themselves.

quantum mechanics - Is the Von Neumann entropy related to heat transfer?

The Von Neumann entropy of a QM system, as far as I understand it, is a measure of the information a particular observer has about that system. Is this definition of entropy directly related to heat transfer in a sense analogous to the classical viewpoint where $\Delta S\geq\frac{Q}{T}$ ?

Note that I'm not asking about the equivalence of the mathematical form. My question is whether the amount of information one has affects the evolution of the system i.e., would the physics between two truly identical states (assuming that's even possible) be different if the level of knowledge of its starting conditions was different between the two?

In the classical sense, information is an accounting tool and it doesn't affect how the state evolves over time. Also in the classical sense, entropy may be discussed among physicists and amateurs like me as akin to the amount of knowledge one has of the state but, as I just mentioned, physics doesn't care about how much information there is. Information is not a physical quantity, but entropy clearly is.

This Nature News article suggests there isn't an academic consensus answering the question I posed. I want to see what this community has to say.

http://www.nature.com/news/battle-between-quantum-and-thermodynamic-laws-heats-up-1.21720

quantum mechanics - Conserved charges and generators

For the Klein Gordon field, the conserved charge for translation in space is given by: $$\vec{P}=\frac{1}{2}\int d^{3}k \, \vec{k}\{a^{\dagger}_{k}a_{k}+a_{k}a^{\dagger}_{k}\}$$

If we were to find the generators for a space translation, we would find that, $$P_{j}=i\partial_{j},$$ where $j=1,2,3$.

If we act both of the above operators on the field $\phi$, the result matches! My question is whether both of these, the generators and the conserved charges of a symmetry are always the same thing? What would be a simple way to see this connection?

Answer

OP is wondering whether the conserved charge associated to a continuous symmetry always generates the symmetry itself. We can say, in full generality, that the answer is

Yes.

Let us see how this works.

Classical mechanics.

We use a notation adapted to classical field theory rather than point-particle mechanics, but the former includes the latter as a special sub-case so we are losing no generality.

Consider a classical system which may or may not include gauge fields and/or Grassmann odd variables. For simplicity, we consider a flat space-time. Assume the system is invariant under the infinitesimal transformation $\phi\to\phi+\delta\phi$. According to Noether's theorem, there is a current $j^\mu$ $$ j^\mu\sim \frac{\partial\mathcal L}{\partial\dot\phi_{,\mu}}\delta\phi $$ which is conserved on-shell, $$ \partial_\mu j^\mu\overset{\mathrm{OS}}=0 $$

This in turns implies that the associated Noether charge $Q$ $$ Q\overset{\mathrm{def}}=\int_{\mathbb R^{d-1}} j^{0}\,\mathrm d\boldsymbol x $$ is conserved, $$ \dot Q\overset{\mathrm{OS}}=0 $$

In Ref.1 it is proved that the charge $Q$ generates the transformation $\delta\phi$,

$$ \delta\phi=(Q,\phi) $$

where $(\cdot,\cdot)$ is the DeWitt-Peierls bracket. This is precisely our claim. The reader will find the proof of the theorem in the quoted reference, as well as a nice discussion about the significance of the result.

Furthermore, a similar statement holds when space-time is curved, but this requires the existence of a suitable Killing field (cf. this PSE post).

Moreover, for standard canonical systems, Ref.1 also proves that $(\cdot,\cdot)$ agrees with the Poisson bracket $\{\cdot,\cdot\}$.

Quantum mechanics.

This is in fact a corollary of the previous case. Ref.1 proves that, up to the usual ordering ambiguities inherent to the quantisation procedure, the DeWitt-Peierls bracket of two fundamental fields agrees with the commutator $[\cdot,\cdot]$ of the corresponding operators.

If we assume that the classical conservation law $\partial_\mu j^\mu\equiv 0$ is not violated by the regulator (i.e., if the symmetry is not anomalous), then we automatically obtain the quantum analogue of our previous result, to wit

$$ \delta\phi=-i[Q,\phi] $$

as required.

References

1. Bryce DeWitt, The Global Approach to Quantum Field Theory.