Tuesday, June 21, 2016

everyday life - How can we see the moon while it's between the Earth and the Sun?



I know this sounds like (and probably is) a stupid question, but I can't figure it out.



As far as I know, the crescent shape of the moon is when the moon is on the sunny side of the Earth, but that wouldn't explain how it can be seen from the night.


Horrible, not-to-scale and probably completely wrong diagram of what I think here:


diagram


How is it that we can see clear to the other side of the Earth?



Answer



Here's a slightly more accurate diagram:


Diagram of crescent moon visible from the night side of Earth


It's still not quite to scale — the Moon is actually a lot further away from the Earth than shown here — but it should suffice to demonstrate that the moon can indeed be seen from the night side of the Earth even when it's nearly between the Earth and the Sun. Note how, in the orientation shown in the diagram, the side of the Moon visible to Earth is almost completely dark, with only a thin crescent lit by the Sun.


The diagram also illustrates some notable properties of the Sun-Earth-Moon geometry:





  • The crescent Moon is only visible (at night) shortly after sunset or before sunrise. Indeed, for the Moon to be visible at midnight, it must be at least (approximately) half full. There's some fuzziness here due to the fact that the Earth's axis of rotation doesn't perfectly match the axis of the Moon's orbit, but even so, you can never see a thin crescent Moon at midnight — unless you're close enough to the Earth's poles that you can see the Moon over the pole.




  • The lit side of the Moon's disk at night always faces (more or less) downwards, towards the Sun. At the equator, it faces (almost, again due to axial tilt) straight down; the further north or south you go, the more the Moon's crescent is tilted, but even so, it never faces upwards while the Sun is below the horizon.




Of course, another reason why we can see the crescent Moon is that the Moon is actually quite visible even during the day:


Daytime Moon and Clouds by Alana Sise
Daytime Moon and Clouds by Alana Sise @ Flickr, used under the CC-By 2.0 license.



newtonian mechanics - Is 'restoring force' a particular type of force?


I have a question about the restoring force in elastic band or rope which confusing me for a long time.


As I was told in high school physics, for an elastic band (or spring), if Hooke's law holds, we have $F = k\Delta x$. What's confusing me is: should F be the total force acting on the object or the restoring force only? Or I ask this way: is there anything called "restoring force" existing independently, just like gravity, friction or tension?



To my understanding, restoring force should be the total force which is pointing to the equilibrium point. For example, if we consider a bungee cord, we should always count the tension of the cord as well as the gravity so the restoring force at any time should be the total force of tension and gravity; hence, when we apply Hooke's law, we should always have $F$ being the total force, not just the tension. Is that correct?


This is pretty confusing to me because there use many terms in the book. Sometimes they said it is the tension in Hooke's law, sometimes they say the restoring force and sometimes the total force....



Answer



"Restoring" forces refer primarily to forces that try to return a system to equilibrium. So a spring has a restoring force of $F = -k\Delta x$. This means that if you choose the origin as being $x = 0$, then compressing the spring would correspond to a negative $x$ (displacing the spring to the left), and stretching the spring would correspond to a positive $x$ (stretching the spring to the right). In that sense, by extending the spring, a positive $\Delta x$ creates a negative force ($-1 \times \Delta x$) that acts to restore the spring to equilibrium (pulling back on the spring extension) and by compressing the spring, you would have a negative $\Delta x$ ($-1 \times \Delta x$), which creates a positive force that restores the spring equilibrium.


So Hooke's Law is actually $F=-k \Delta x$


Hope that helps.


quantum mechanics - Path integral vs. measure on infinite dimensional space


Coming from a mathematical background, I'm trying to get a handle on the path integral formulation of quantum mechanics.


According to Feynman, if you want to figure out the probability amplitude for a particle moving from one point to another, you 1) figure out the contribution from every possible path it could take, then 2) "sum up" all the contributions.


Usually when you want to "sum up" an infinite number of things, you do so by putting a measure on the space of things, from which a notion of integration arises. However the function space of paths is not just infinite, it's extremely infinite.


If the path-space has a notion of dimension, it would be infinite-dimensional (eg, viewed as a submanifold of $C([0,t] , {\mathbb R}^n))$. For any reasonable notion of distance, every ball will fail to be compact. It's hard to see how one could reasonably define a measure over a space like this - Lebesgue-like measures are certainly out.


The books I've seen basically forgo defining what anything is, and instead present a method to do calculations involving "zig-zag" paths and renormalization. Apparently this gives the right answer experimentally, but it seem extremely contrived (what if you approximate the paths a different way, how do you know you will get the same answer?). Is there a more rigorous way to define Feynman path integrals in terms of function spaces and measures?




Answer



Path integral is indeed very problematic on its own. But there are ways to almost capturing it rigorously.


Wiener process


One way is to start with Abstract Wiener space that can be built out of the Hamiltonian and carries a canonical Wiener measure. This is the usual measure describing properties of the random walk. Now to arrive at path integral one has to accept the existence of "infinite-dimensional Wick rotation" and analytically continue Wiener measure to the complex plane (and every time this is done a probabilist dies somewhere).


This is the usual connection between statistical physics (which is a nice, well-defined real theory) at inverse temperature $\beta$ in (N+1,0) space-time dimensions and evolution of the quantum system in (N, 1) dimensions for time $t = -i \hbar \beta$ that is used all over the physics but almost never justified. Although in some cases it was actually possible to prove that Wightman QFT theory is indeed a Wick rotation of some Euclidean QFT (note that quantum mechanics is also a special case of QFT in (0, 1) space-time dimensions).




Intermezzo


This is a good place to point out that while path integral is problematic in QM, whole lot of different issues enter with more space dimensions. One has to deal with operator valued distributions and there is no good way to multiply them (which is what physicist absolutely need to do). There are various axiomatic approaches to get a handle on this and they in fact do look very nice. Except that it's very hard to actually find a theory that satisfies these axioms. In particular, none of our present day theories of Standard model have been rigorously defined.




Anyway, to make the Wick rotation a bit clearer, recall that Schrödinger equation is a kind of diffusion equation but for an introduction of complex numbers. And then just come back to the beginning and note that diffusion equation is macroscopic equation that captures the mean behavior of the random walk. (But this is not to say that path integral in any way depends on the Schrödingerian, non-relativistic physics)



Others


There were other approaches to define the path-integral rigorously. They propose some set of axioms that path-integral has to obey and continue from there. To my knowledge (but I'd like to be wrong), all of these approaches are too constraining (they don't describe most of physically interesting situations). But if you'd like I can dig up some references.


diffusion - Do particle velocities in liquid follow the Maxwell-Boltzmann velocity distribution?


The Maxwell-Boltzmann velocity distribution arises from non-reactive elastic collisions of particles and is usually discussed in the context of the kinetic theory (for gases). There are various offhand remarks, for example here (slide 5), that state without reference that particles observe a similar velocity distribution in liquid. Is that true? References?



The main reason I'm curious is that it seems as though the mean free path would be extremely short in liquid vs. gas. I'm actually most curious about the nature of the collisions in liquid vs. gas, i.e., are collisions in liquid still (on average) elastic?


EDIT: The linked post on particle velocity in liquids is definitely interesting and weighs in on this question, and I appreciate the distinction made between local position fluctuation vs. long range movement. Still, let me frame this question in a few different ways.




  1. Gillespie proposed a stochastic framework for simulating chemical reactions which is formulated on the idea that non-reactive elastic collisions serve to 'uniformize' particle position so that the assumption of well-mixedness is always satisfied (see page 409 in the linked version). This is formulated from kinetic theory. A corollary to this is that a non-reactive collision between two molecules that are able to react does not induce local correlation, i.e., two particles able to react with each other that just collided, but didn't react are no more likely to react with each other in the next dt than any other particle pair in the volume. Gillespie's algorithm is commonly used in biology where biochemical species are modeled in the aqueous environment of cells. Is this valid, and if so why?




  2. On a microscopic scale, suppose we are interested in two 'A' particles diffusing in one dimension in an aqueous environment. The two A particles collide but don't react. What is their behavior immediately after the collision? Is it a 'reflection' which conserves velocity and might correspond to a Neumann BC? In a gas that approach seems natural, but in liquid the short mean-free path makes me think that diffusive forces would rapidly dissipate any momentum from the A-A collision, which might imply the A particles collide and 'stop'. How should I be thinking about this?





Just to bring it back to the original question, I think both (1) and (2) depend on the statistical velocity behavior of particles in liquid.



Answer



I think Maxwell-Boltzmann distribution should be valid for molecules in liquid too, at least according to classical statistical physics, because the factor $e ^{−\beta p^2/2m}$ in the Gibbs-Boltzmann probability density does not depend on potential energy and is the same whether the molecule is in gas, or a liquid. I do not know if there is a measurement supporting this theoretical result.



are collisions in liquid still (on average) elastic?



Elastic collision means that appreciable change in the kinetic and potential energy of two bodies happens to them only during short time interval and the energy long after that is the same as the energy long before that - the interaction of the two molecules is thought of as a scattering process. In liquids the interaction of the molecules may not be idealizable in this way, as the molecules are believed to be in incessant complicated motion constantly influencing each other (Brownian motion...) This does not seem to be a reason to abandon the Boltzmann statistics, however.


Monday, June 20, 2016

quantum field theory - Are the path integral formalism and the operator formalism inequivalent?



Abstract


The definition of the propagator $\Delta(x)$ in the path integral formalism (PI) is different from the definition in the operator formalism (OF). In general the definitions agree, but it is easy to write down theories where they do not. In those cases, are the PI and OF actually inequivalent, or is it reasonable to expect that the $S$ matrices of both theories agree?


For definiteness I'll consider a real scalar field $\phi$, with action $$ S_0=\int\mathrm dx\ \frac{1}{2}(\partial \phi)^2-\frac{1}{2}m^2\phi^2 \tag{1} $$


Path integral formalism


In PI we insert a source term into the action, $$ S_J=S_0+\int\mathrm dx\ \phi(x)\ J(x) \tag{2} $$


The mixed term $\phi J$ can be simplified with the usual trick: by a suitable change of variables, the action can be written as two independent terms


$$ S_J=S_0+\int \mathrm dx\;\mathrm dy\ J(x)\Delta_\mathrm{PI}(x-y)J(y) \tag{3} $$ and this relation defines $\Delta_\mathrm{PI}(x)$: to get the action into this form we have to solve $(\partial^2+m^2)\Delta_{\mathrm{PI}}=\delta(x)$, i.e., in PI the propagator is defined as the Green function of the Euler-Lagrange equations of $S_0$. This definition is motivated by the the fact that when $S_J$ is written as $(3)$ the partition function can be factored as $$ Z[J]=Z[0]\exp\left[i\int \mathrm dx\;\mathrm dy\ J(x)\Delta_\mathrm{PI}(x-y)J(y) \right] \tag{4} $$ which makes functional derivatives trivial to compute. For example, if we differente $Z[J]$ two times we get $$ \langle0|\mathrm T\ \phi(x)\phi(y)|0\rangle=\Delta_\mathrm{PI}(x-y) \tag{5} $$


In this formalism, the propagator is always the Green function of the differential operator of the theory.


Operator formalism


In OF the propagator is defined as the contraction of two fields: $$ \Delta_\mathrm{OF}(x-y)\equiv \overline{\phi(x)\phi}(y)\equiv \begin{cases}[\phi^+(x),\phi^-(y)]&x^0>y^0\\ [\phi^+(y),\phi^-(x)]&x^0

In general, $\Delta_\mathrm{OF}$ is an operator, but if it commutes with everything (or, more precisely, if the propagator is in the centre of the operator algebra) we can prove Wick's theorem, which in turns means that $$ \langle 0|\mathrm T\ \phi(x)\phi(y)|0\rangle=\Delta_\mathrm{OF}(x-y) \tag{7} $$ i.e., the propagator coincides with the two-point function. This makes it very easy to see, for example, that $$ \Delta_\mathrm{OF}=\Delta_\mathrm{PI} \tag{8} $$


In this theory, the fact that the propagator is a Green function is a corollary and not a definition. The theorem may fail if the assumptions are not satisfied.


The discrepancy


The positive/negative frequency parts of $\phi$ are the creation and annihilation operators, which in OF usually satisfy $$ [\phi^+(x),\phi^-(y)]\propto\delta(x-y)\cdot1_\mathcal H \tag{9} $$ and therefore $\overline{\phi(x)\phi}(y)$ is a c-number. This means that the assumptions of Wick's theorem are satisfied and $(8)$ holds.


The relation $(9)$ can be derived from one of the basic assumptions of OF: the canonical commutation relations: $$[\phi(x),\pi(y)]=\delta(x-y)\cdot1_\mathcal H\tag{10} $$


But if we use any non-trivial operator in the r.h.s. of $(10)$ instead of a $1_\mathcal H$, Wick's theorem is violated and in general $\Delta_\mathrm{PI}\neq \Delta_\mathrm{OF}$. One could argue that the r.h.s. of $(10)$ is fixed by the Dirac prescription $\{\cdot,\cdot\}_\mathrm{D}\to\frac{1}{i\hbar}[\cdot,\cdot]$, where $\{\cdot,\cdot\}_\mathrm{D}$ is the Dirac bracket. In the Standard Model, it's easy to prove that $\{\cdot,\cdot\}_\mathrm{D}$ is always proportional to the identity, but in a more general theory we may have complex constraints which would make the Dirac bracket non-trivial - read, not proportional to the identity - and therefore $\Delta_\mathrm{OF}\neq\Delta_\mathrm{PI}$.


Scalar, spinor and vector QFT's always satisfy a relation similar to $(10)$, and therefore OF and PI formalisms agree. But in principle it is possible to study more general QFT's where we use a commutation relation more complex than $(10)$. I don't know of any practical use of this, but to me it seems to me that we can have a perfectly consistent theory where PI and OF formalisms predict different results. Is this correct? I hope someone can shed any light on this.




EDIT


I think it can be useful to add some details to what I said about Dirac brackets, defined as $$ \{a,b\}_\mathrm{DB}=\{a,b\}_\mathrm{PB}-\{a,\ell^i\}_\mathrm{PB} M_{ij}\{\ell^j,b\}_\mathrm{PB}\tag{11} $$ where $\ell^i$ are the constraints and $M^{ij}=\{\ell^i,\ell^j\}_\mathrm{PB}$. As $\{q,p\}_\mathrm{PB}=\delta(x-y)$, the only way to get non-trivial Dirac brackets is through the second term. This may happen if we have non-linear constrains such that the second term is a function of $p,q$. If in some case we have non-linear constrains the matrix $M$ will depend on $p,q$ and $\{p,q\}_\mathrm{DB}$ will be a function of $p,q$. If we translate this into operators, we shall find $$ [\pi,\phi]=\delta(x-y)\cdot1_\mathcal H+f(\pi,\phi)\tag{12} $$ and so $[\pi,\phi]$ won't commute with neither $\pi$ nor $\phi$ as required by Wick's theorem. (This term $f$ is perhaps related to the $\hbar^2$ terms in QMechanic's answer, and the higher order terms in, e.g., the Moyal bracket).




Answer



General comments to the question (v1):




  1. Any textbook derivation of the correspondence between $$\tag{1} \text{Operator formalism}\qquad \longleftrightarrow \qquad \text{Path integral formalism}$$ is just a formal derivation, which discards contributions in the process, cf. e.g. this Phys.SE post.




  2. Rather than claiming complete understanding and existence of the correspondence (1), it is probably more fair to say that we have a long list of theories (such as e.g. Yang-mills, Cherns-Simons, etc.), where both sides of the correspondence (1) have been worked out.





  3. The correspondence (1) is mired with subtleties. Example: Consider a non-relativistic point particle on a curved target manifold $(M,g)$ with classical Hamiltonian $$\tag{2} H_{\rm cl} ~=~\frac{1}{2} p_i p_jg^{ij}(x), $$ which we use in the Hamiltonian action of the phase space path integral. Then one may show that the corresponding Hamiltonian operator is $$\tag{3}\hat{H}~=~ \frac{1}{2\sqrt[4]{g}} \hat{p}_i\sqrt{g}~ g^{ij} ~\hat{p}_j\frac{1}{\sqrt[4]{g}}+ \frac{\hbar^2R}{8} +{\cal O}(\hbar^3),$$ cf. Refs. 1 & 2. The first term in eq. (3) is the naive guess, cf. my Phys.SE answer here. The two-loop correction proportional to the scalar curvature $R$ is a surprise, which foretells that a full understanding of the correspondence (1) is going to be complicated.




References:




  1. F. Bastianelli and P. van Nieuwenhuizen, Path Integrals and Anomalies in Curved Space, 2006.




  2. B. DeWitt, Supermanifolds, Cambridge Univ. Press, 1992.





electromagnetic radiation - light point and photosensor


When a laser pointer is pointed at a screen or at a target that spot can be detected because the laser beam falls on the target and bounces back and then is detected by a photosensor.


But, what if it was possible for an electronic eye / photosensor to detect a spot or point of light in just air, without the need of a surface for that spot/point to bounce of. What would be the practical applications of such an invention?



Answer




In order to see something, that "something" must either emit or reflect photons into your eye (or detector). If photons leave a light source in some direction, but are seen/detected "in the air", then the air must have somehow directed those photons toward your eye/detector.


Like this: light


I can only see two ways for this to happen:



  1. Some particles in the air scatter the photons. This happens in many cases.

  2. Magic make the photons change direction. To my knowledge, this has not happened.


Practical implications of the first scenario include





  • guiding the eye toward a star with a laser pointer (photons scatter on particles in the air and thus create a visible ray),




  • studying supernovae looking at the light echo from the light scattering off of surrounding dust particles in their remnants,




  • measuring the size of the broad-line region around supermassive black holes, using the technique known as reverberation mapping (see also this answer),




  • breaking into banks using flour to see laser traps, and, most importantly,





  • summoning Batman.




Practical implications of the second scenario are limited only by your imagination.


Sunday, June 19, 2016

quantum mechanics - Observable: possible outcome of measurement vs (linear) transformation


One of the postulates of quantum mechanics is that every physical observable corresponds to a Hermitian operator $H$, that the possible outcomes of the measurements are eigenvalues of the operator, and that after a measurement the state collapses into the eigenfunction corresponding to the observed eigenvalue.


Some authors state that this is the reason quantum mechanics is "built on" linear algebra. However from a linear algebra standpoint this postulate seems strange. In linear algebra operators are linear transformations that take vectors to vectors; hence I would expect an operator also be a linear transformation, physically realised by a black box that takes in a particle in a state $\psi$ and emits the particle in the transformed state $H \psi$.


In fact, if I look at the creation operator (from say an electron in a harmonic oscillator potential), this clearly can't correspond to an observable because it's not hermitian. I can imagine the black box to be implemented by a machine that fires a photon into the electron. I can also imagine operators that for example rotate the angular momentum of an electron without measuring it. Can I create a physical apparatus that "implements" the position operator (in position space, it takes a state $\psi(x)$ and returns $x \psi(x)$)? Why does QM use operators in these two different senses and how do I distinguish them?





classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?

I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...