Saturday, October 31, 2015

reference frames - Can you tell your absolute speed in space?


Normally in relativity your speed can only be known relative to another object,


given that as one approaches light speed more energy is required to accelerate faster, based on the energy consumption profile you would be able to calculate at what % of the speed of light you were going?


apart from being a useful 'speedometer' i thought it interesting that one of the effects of the light speed barrier is you can determine your absolute speed relative to yourself, your proper speed i guess it would be called if this speed barrier did not exist then you would have no way to know your speed


now since you know your own speed, you also know the absolute speed of any other object you encounter from the relative difference


following on from this, if you knew your speed you could slow your ship down by the exact speed you knew yourself to be going and be truly at rest, with respect to the reference frame of the universe


my question is am i right? it seems counter intuitive to derive seeming absolutes in in a relativistic universe





newtonian gravity - Escape Velocity question


I would like some help in understanding the derivation of the formula for escape velocity. Why does $t$ approach infinity? And why is $v(t)$ approaching zero? I know that $r(t)$ is infinite because the gravitational field is infinite. But I don't understand why $t$ (time) and velocity would be zero.


I don't think I understand what $v(t)$ is. Is it velocity at a given point in time? And when time increases, you are farther away from the body producing the gravitational field and thus velocity needed for escape is lower? ....


Here is a photo of the derivation:




From the wikipedia page: http://en.wikipedia.org/wiki/Escape_velocity



Answer



You are right in saying that v(t) is the velocity at a given time.


1) The first equation is the equation for conservation of energy for an object in a gravitational field. The left side is the change in kinetic energy and the right side is the change in gravitational potential energy.


2) For escape velocity, we consider what happens a looooong time after the rocket is launched, so we let t go to infinity. For a rocket launched exactly at escape velocity, it has just enough energy to reach a distance infinitely far away from the planet, so r goes to infinity. If the rocket has "just enough energy" to go infinitely far, that means that once it is at r = infinity, its velocity must be 0, since there would be no more leftover energy. Hence v(t) also approaches 0.


3) The article substitutes these values in to the energy equation and solves for the initial velocity, which is the escape velocity.


renormalization - How do people go about looking for asymptotic safety in quantum gravity?


Do we have (proposed?) methods to look for fixed points in the renormlaization group flow of the Einstein-Hilbert action? My understanding of the RG is still somewhat sketchy at this point and I am having trouble understanding how one would go about searching for a fixed point in a theory that's non-renormlaizable.



Answer



Here are two papers on the subject:


http://arxiv.org/abs/0805.2909 Investigating the Ultraviolet Properties of Gravity with a Wilsonian Renormalization Group Equation


From the abstract: We review and extend in several directions recent results on the asymptotic safety approach to quantum gravity. The central issue in this approach is the search of a Fixed Point having suitable properties, and the tool that is used is a type of Wilsonian renormalization group equation. We begin by discussing various cutoff schemes, i.e. ways of implementing the Wilsonian cutoff procedure. We compare the beta functions of the gravitational couplings obtained with different schemes, studying first the contribution of matter fields and then the so-called Einstein-Hilbert truncation, where only the cosmological constant and Newton's constant are retained. In this context we make connection with old results, in particular we reproduce the results of the epsilon expansion and the perturbative one loop divergences.


http://arxiv.org/abs/1601.01800 The Gravitational Two-Loop Counterterm is Asymptotically Safe


Abstract: Weinberg's asymptotic safety scenario provides an elegant mechanism to construct a quantum theory of gravity within the framework of quantum field theory based on a non-Gau{\ss}ian fixed point of the renormalization group flow. In this work we report novel evidence for the validity of this scenario, using functional renormalization group techniques to determine the renormalization group flow of the Einstein-Hilbert action supplemented by the two-loop counterterm found by Goroff and Sagnotti. The resulting system of beta functions comprises three scale-dependent coupling constants and exhibits a non-Gau{\ss}ian fixed point which constitutes the natural extension of the one found at the level of the Einstein-Hilbert action. The fixed point exhibits two ultraviolet attractive and one repulsive direction supporting a low-dimensional UV-critical hypersurface. Our result vanquishes the longstanding criticism that asymptotic safety will not survive once a "proper perturbative counterterm" is included in the projection space.


electricity - How can current pass through (resistance less) connecting wires?


This is not a duplicate of : Will current pass without any resistance?. I read it but my question isn't answered there.


I'm a physics tutor for high school students and this is my understanding of how current flows:



Across any resistor if there is a potential difference, there will be electric field across that element from the point of higher potential to the point of lower potential. Now since resistor (conductor) contains free electrons they flow (drift) in the direction opposite to electric field and thus we have current.




It implies that, if there is no potential difference between two points, there can't be electric field between them, so no drifting of electrons, hence no current. I have been using this logic to explain why we remove certain resistors in circuits (eg. in a balanced wheatstone bridge).


Question: Consider the following circuit:


enter image description here


if I apply Ohm's law between point a and point b then


$V_a - V_b = \Delta V = IR_{ab} = 0$ since $R_{ab} = 0$


which implies $V_a = V_b$. So from above mentioned logic, there shouldn't be any current flow between them. Then how is the current flowing?


What exactly is wrong with my thinking? How can current pass through connecting (resistance less) wires?




symmetry - Derivation of correction to canonical stress energy tensor due to addition of total divergence to Lagrangian


It is mentioned in almost every text book that equations of motions are not modified if we add a total divergence of some vector $$\partial_\mu \ X^{\mu}$$ to Lagrangian but canonical stress energy tensor:


$$T^{\mu}_{\ \ \ \ \nu} = \frac{\delta {\cal L}}{\delta (\partial_\mu \phi_s)} \partial_\nu \phi_s - \eta^{\mu}_{\ \ \ \nu} \ {\cal L}$$


is modified. However, I was unable to find any text book where a derivation of such correction to canonical stress energy tensor is thoroughly performed. Can anyone, please, either provide a reference where such derivation is made or illustrate it.


The following thread: Symmetrizing the Canonical Energy-Momentum Tensor but no derivation was provided.



One of the comments provided a reference to Bandyopadhyay PhD thesis (http://research.physics.illinois.edu/Publications/theses/copies/Bandyopadhyay/Chapter_3.pdf, end of page 18, paragraph after Eq. 3.9). For simplicity I repeat the statement from there. If we perform a gauge transformation of a Lagrangian


$${\cal{L}}(x) \rightarrow {\cal{L}}(x) + \partial_{\mu} Z^{\mu}(x)$$


where $Z^{\mu}$ is zero at the boundary then treating $Z^{\mu}$ as independent vector field in the Lagrangian results in the following transformation of the canonical stress energy tensor: $${\Theta^{\mu}}_{\nu} \rightarrow {\Theta^{\mu}}_{\nu}+\left( \partial_{\nu}Z^{\mu}-{\delta^{\mu}}_{\nu}\partial_{\alpha}Z^{\alpha}\right)$$


Fine. As an exercise (but with not much physics), let's take a standard EM Lagrangian:


$${\cal L}=-\frac{1}{4}F^{\mu \nu}F_{\mu \nu}$$


and add to it a total divergence constructed from vector potential $A_{\mu}$. We can construct at least 3 such terms with the proper "dimension": $${\cal L}_1=\partial_{\mu} \left(A^{\mu} \partial^{\nu}A_{\nu} \right)$$ $${\cal L}_2=\partial_{\mu} \left(A^{\nu} \partial^{\mu}A_{\nu} \right)$$ $${\cal L}_3=\partial_{\mu} \left(A^{\nu} \partial_{\nu}A^{\mu} \right)$$


and add (any of) them to Lagrangian with some constant multipliers $c_1,c_2,c_3$: $${\cal L} \rightarrow {\cal L} + c_1 {\cal L}_1 + c_2 {\cal L}_2 + c_3 {\cal L}_3$$


As we added total divergence we can integrate it out to infinity using Gauss theorem and provided that the field is zero at the infinity (which is always assumed?) the equation of motions should not change? But we can "forget" about Gauss theorem and just apply Euler Lagrange equations to new Lagrangian. As all of the ${\cal L}_i$ contains the second derivatives of $A_{\mu}$ and unless we use some appropriate values of $c_i$ so that to have the second derivatives cancelled the equation of motions and canonical stress energy tensor will be updated to reflect second derivatives of the field (for example, http://relativity.livingreviews.org/Articles/lrr-2009-4/, paragraph 2.1.1, pp. 11, 12). The additional terms are clearly field dependent and thus above-mentioned gauge transformation approach to find a transformation of stress energy tensor cannot be applied. Or am I missing something?


Back to the original question. The procedure to correct canonical EM stress energy tensor is well-known. Does that mean that we could add a divergence of some field to standard EM Lagrangian so that to obtain corrected (symmetric canonical EM stress energy tensor)? If yes, then what shall we add?




homework and exercises - Rocket Altitude Calculation




Basic data


I was launching a rocket model and I tried to calculate the reached altitude.



  • The engine (C6-0) impulse is 10 Ns


  • Total weight is 65,7 g (includes the engine)


I calculated speed = 152 m/s


$$\vec F \cdot t = \Delta m \vec v$$


Then I calculated the altitude 1007 m which seems too much to me. I guess something about 200 m (you may see the video)


$$y_{\max} = \frac{v_0^2 \ }{2 g}$$


Drag


I guess, I have to consider drag



  • Diameter of rocket 2,5 cm


  • Drag coefficient 0,05 (I guess)


$$F_D\, =\, \tfrac12\, \rho\, v^2\, C_D\, A$$


But what about now, what is really achieved altitude?




(source: estesrockets.com)



(source: estesrockets.com)


Rocket Altitude Calculation




Answer



I made a simple Excel spreadsheet to calculate this. Some simplifying assumptions:


Mass = 66 gram (during thrust), 33 gram (after burn) Cd = 0.5 (like for sphere) rho = 1.22 (air) Simple numerical (Newton) integration of equation of motion (0.1 second time step)


Resulting curve:


enter image description here


Height of about 300 m, total flight time just under 14 seconds. Based on the video (which didn't show the descent) I think that time to peak was about 6 seconds - close to that predicted by this.


Excel file is at http://www.floris.us/physicsSE/rocket.xlsx


It was just a "rough" calculation... I know much better calculators exist "out there".


Friday, October 30, 2015

homework and exercises - Calculate water flow rate through orifice


I'm not very good with fluid physics, and need some help. Imagine the following setup with water contained in-front of a wall with an opening on the bottom:


Fluid image


How do I calculate the water flow $Q$?. I have made some re-search and found I need to (partially) calculate the pressure across the opening (orifice). But I don't know the pressure on the back side of the orifice. Can this be solved in any way?


Note: I'm not saying "please give me the solution, I'm lazy". I want to figure it out myself. But since, in this case, I only found formulas involving calculating pressure drop, I canno't use them to solve the problem. Therefore I'm turning my face to you, to see if there's another way to solve this problem.


Update: The "tank" holding the water is actually a big lake, and the opening is how much the water gate have opened. I need to very precisely calculate how much water flows through the opening.




electromagnetism - Why are so many forces explainable using inverse squares when space is three dimensional?


It seems paradoxical that the strength of so many phenomena (Newtonian gravity, Coulomb force) are calculable by the inverse square of distance.


However, since volume is determined by three dimensions and presumably these phenomena have to travel through all three, how is it possible that their strengths are governed by the inverse of the distance squared?


The gravitational force and intensity of light is merely 4 times weaker at 2 times the distance, but the volume of a sphere between the two is 8 times larger.


Since presumably these phenomena would affect all objects in a spherical shell surrounding the source with equal intensity, they travel in all three dimensions. How come these laws do not obey an inverse-cube relationship while traveling through space?



Answer



This is not paradoxical and it is not necessary for any physical phenomenon to a priori have to obey any particular law. Some phenomena do have to obey inverse-square laws (such as, particularly, the light intensity from a point source) but they are relatively limited (more on them below).


Even worse, gravity and electricity don't even follow this in general! For the latter, it is only point charges in the electrostatic regime that obey an inverse-square law. For more complicated systems you will have magnetic interactions as well as corrections that depend on the shape of the charge distributions. If the systems are (globally) neutral, there will still be electrostatic interactions which will fall off as the inverse cube or faster! The van der Waals forces between molecules, for instance, are electrostatic in origin but go down as $1/r^6$.


It is for systems with a conserved flux that the inverse-square law must hold, at least at large distances. If a point light source emits a fixed amount of energy per unit time, then this energy must go through every imaginary spherical surface we think up. Since their area goes up as $r^2$, the power per unit area (a.k.a. the irradiance) must go down as $1/r^2$. In a simplified picture, this is also true for the electrostatic force, where it is the flow of virtual photons that must be conserved.



blackbody - Why is the black body radiation so important?


In the derivation of the black-body radiation formula the assumption is made that the system is an electromagnetic cavity, so that it can be considered in thermal equilibrium.


Leaving aside the fact that I don't see why one would make that assumption (what are its mathematical consequences in the derivation?),


most everyday sources are not in thermal equilibrium, so how can we adjust the black-body radiation formula for them?


Or is it still applicable?


Also: in a black-body, the spectral energy density is not a constant, which means that some frequencies contribute more to the energy than others: shouldn't we naively expect that each frequency would carry an equal weight? What is the physical meaning of there being "preferred" frequencies?



Answer



Blackbody radiation is characteristic of every object in thermodynamic equilibrium and black bodies at constant uniform temperature.



At any temperature objects emit thermal radiation. EM radiation is emitted because inside the object, due to thermal motion of particles charged particles/dipoles start to oscillate, electromagnetic radiation is emitted because of these vibrations. If the object is a black body at constant uniform temperature, the radiation is called blackbody radiation. The energy emitted by any object is always finite with certain distribution over the frequencies with peak at some frequency. We cannot naively expect the energy emitted with all the frequencies carrying equal weight. This is a phenomenon which happens and is observed. This is explained quantum mechanically, infact this led to the development of quantum mechanics.


So a cavity with a small hole with EM radiation inside it is appropriate to study mathematically and is a near perfect blackbody because the hole allows negligible radiation to enter the cavity so that it affects negligibly the thermal equilibrium condition and we can have a very near thermal equilibrium and observe blackbody radiation from it. Rayleigh and jeans couldn't explain blackbody spectrum at higher frequencies, their law predicted infinte spectral radiance at infinite frequencies. Planck gave the solution to the ultraviolet catastrophe(infinite spectral radiance at infinite frequencies) and explained the spectrum of blackbody radiation by assuming the energy of the oscillators inside the cavity to be series of discrete values but not continuous which eventually results in spectral radiance going to zero at higher and infinite frequencies with peak at some frequency.


Radiations emitted by ordinary objects can be approximated as blackbody radiation, they are nearly in thermal equilibrium.


One of the importance is that to know the temperature of a star, the relation between the temperature and wavelength of the peak, called wien's displacement law, evaluated from planck's radiation formula, is used approximating the radiation to be blackbody radiation.


quantum mechanics - Spin and its connection to magnetic field


How the intrinsic spin of an electron results in a magnetic field? How does spin couples to the magnetic field? I mean to ask how does the phenomenon of Spin orbit coupling happens physically? Is the electron really spinning or not?



Answer



The electron is not spinning. Elementary particles are considered to be point-like particles, meaning that they do not have an internal structure. The spin is, as you say, an intrinsic property of particles. It is a pure quantum mechanical property that particles just have. The spin induces a spin magnetic moment: $$\vec{\mu_{s}}=g\frac{q}{2m}\vec{S}$$ So if an external magnetic field is applied, it will exert a torque on the particle's magnetic moment depending on its orientation with respect to the field. $$\vec{\tau}=\vec{\mu}\times\vec{B}$$


Thursday, October 29, 2015

Doppler redshift in special relativity


I came across this exercise in Elementary General Relativity by Alan MacDonald:


A source of light pulses moves with speed v directly away from an observer at rest in an inertial frame. Let $ \Delta t_e $ be the time between the emission of pulses, and $ \Delta t_o $ be the time between their reception at the observer. Show that $ \Delta t_o = \Delta t_e + v\Delta t_e $.


Based on my understanding of special relativity, the space-time interval between two events as measured from two inertial frames of reference should be the same. Therefore, $$ \Delta t_e^2 = \Delta t_o^2 - \Delta x^2 $$ $$ \implies \Delta t_e^2 = \Delta t_o^2 - v^2\Delta t_o^2 $$ $$ \implies \Delta t_o = (1 - v^2)^{-1/2}\Delta t_e $$


which is not the same relation. What is wrong with my reasoning?



Answer



Your answer is right assuming $\Delta t_e$ is the interval between emission as measured by the emitting source itself. The given answer is right assuming $\Delta t_e$ is the time between emission as measured by the observer. It seems as though this problem is aiming at a lower level than your current understanding of relativity; you put too much thought into it.


Renormalization of field strength


I'm revisiting the elementary algorithms of renormalization that are taught in a classroom setting and find that the procedure taught to students is as follows:



  1. Write down the bare Lagrangian: $\mathcal{L}_0$

  2. Renormalize the field strengths: $\phi_0\rightarrow Z^{1/2}\phi_r$

  3. Renormalize coupling constants: e.g. $Z^2\lambda_0\rightarrow Z_\lambda \mu^{2\epsilon}\lambda_R$

  4. For perturbation theory rewrite $Z=1+\delta$, where $\delta$'s are counterterms.


I have thought long and hard about steps 2 and 3 (especially in the context of operator renormalization), and have come to the following conclusions which I would like verified.



In step 3, the renormalization of coupling constants is actually the renormalization of the operator product multiplying the coupling constant. And, in step 2, when they renormalize the field strength, it is not the renormalization of a single operator; they are actually renormalizing another composite operator: the kinetic term $\partial_\mu\phi\,\partial^\mu\phi$. The renormalization of a single operator $\phi$ actually corresponds to tadpoles, the one-point function.


Am I interpreting this correctly?



Answer



In renormalization, one considers a family of Lagrangian densities, with arbitrary factors for each renormalizable monomial in fields and derivatives; in case of symmetries only of the symmetric ones. Thus it is determined by the field and symmetry content only. For perturbation theory, these factors are then written as the renormalized finite term plus the diverging counterterm. The particular way of generating these factors (by scaling fields or coupling constants) is immaterial.


As the regularization scale is sent to infinity, the counterterms are left to diverge in such a way that the renormalization conditions give finite results. The family of renormalized theories is now parameterizied by the parameters used in the renormalization prescription.


homework and exercises - How feasible is it to power your home with a bicycle?



This article claims "60 Minutes On This Bicycle Can Power Your Home For 24 Hours!" How feasible is this claim? Sounds great - but the skeptic in me says "yeah, right."


Hope this is an acceptable forum to post this question - if not, can someone please migrate it?




homework and exercises - Free Body Diagram of Fluid Statics Problem



I need help determining the forces, and the direction of them in this fluid statics problem.



A hollow cylinder with closed ends is $300\,\rm mm$ in diameter and $450\,\rm mm$ tall. It has a mass of $27\,\rm kg$ and a tiny hole at the bottom. It is lowered slowly into water and then released. Calculate:


a. The pressure of the air inside the cylinder.
b. The height to which the water will rise.

c. The depth to which the cylinder will sink.



I was already given the answers to these problems for clarification, but I think I'm going to start with the free body diagram (FBD) to part c.


Now I know there is a weight from the cylinder going down and a buoyant force going up because of the water and the submerged volume of the cylinder. Now, does the air trapped in the cylinder also create a buoyant force, or some kind of force also pushing up? And how would I calculate this force if there is one? After that, I'm sure I can get the rest of the calculations.



Answer



Question a


The maximum theoretical buoyancy of the cylinder is the maximum cavity volume:


$$V = \pi R^2L$$


where $R$ is the cylinders radius and $L$ is it length. I will ignore the "tiny hole at the bottom" as it is poorly defined. Assuming the air inside the cylinder is incompressible what buoyancy force does the cylinder provide, that is what can it 'float'?


$$V = \pi R^2L = 0.031809\,\mathrm{m^2} = 31.809\,\mathrm{L}$$



From this the mass ($M$) it can float in the ideal case specified above is


$$m = \rho V \approx 997.0 \times 0.031809 = 31.7\,\mathrm{kg}$$


So, no we have cleared that up, you want to do the following force balance to get the equilibrium pressure inside the cylinder for the 'real' case of compressible air. If you consider the system to be the can plus the air inside, the only forces are the weight of the cylinder and the pressure between the air and the liquid. That is,


$$m \mathtt{g} - \pi R^2P = 0$$


where $P$ is the pressure inside the cylinder at equilibrium, $\mathtt{g}$ is the gravitation constant and $m$ is the weight of the cylinder/weight you want the massless cylinder to float (i.e. $27\,\rm kg$).


Question b


You can have a go at. For b you will need to think about the pressure and volume of the air inside the cylinder initially $P_0$ and $V_0$ and how they relate to the air inside the container during its compressed state $P_1$ and $V_1$, where you have worked out the pressure $P_1$ above.


Hint: assume one of the ideal gas relationships to get the unknown volume $V_1$. Use this to get the depth.


Question c


You will need to think about the volume of water required to float the cylinders mass of $27\,\rm kg$ plus the water inside it. From this volume you will be able to extract the length of cylinder that needs to be submerged in order to provide sufficient buoyancy to float.



Wednesday, October 28, 2015

How does Newtonian mechanics explain why orbiting objects do not fall to the object they are orbiting?



The force of gravity is constantly being applied to an orbiting object. And therefore the object is constantly accelerating. Why doesn't gravity eventually "win" over the object's momentum, like a force such as friction eventually slows down a car that runs out of gas? I understand (I think) how relativity explains it, but how does Newtonian mechanics explain it?



Answer



Newtonian mechanics explains that they do fall toward the object they're orbiting, they just keep missing.




Quick and dirty derivation for a circular orbit.


Let the primary have mass $M$ and the satellite mass $m$ such that $m \ll M$ (it can also be done for other cases, but this saves on mathiness).


Assume we start with an initial circular orbit on radius $r$, velocity $v = \sqrt{G\frac{M}{r}}$. The acceleration of the satellite due to gravity is $a = G\frac{M}{r^2}$ which means we can also write $v = \sqrt{\frac{a}{r}}$. The period of the orbit is $T = \frac{2\pi r}{v} = 2\pi \sqrt{\frac{r}{a}}$.


Chose a coordinate system in which the initial position is $r\hat{i} + 0\hat{j}$ and the initial velocity points in the $+\hat{j}$ direction. Chose a short time $t \ll T$ and lets see how far from the primary the satellite ends up after that time.


If we have chosen $t$ short enough, we can approximate gravity as having uniform strength through the time period (and we shall show later that that is justified).


The new position is $\left(r - \frac{1}{2}at^2\right)\hat{i} + vt\hat{j}$ which lies at a distance $$ r_2 = \sqrt{r^2 - r a t^2 + \frac{1}{4}a^2 t^4 + v^2 t^2} $$ pulling our at factor of $r$ we get $$ r_2 = r \sqrt{1 - \frac{a}{r} t^2 + \frac{1}{4}\frac{a^2}{r^2} t^4 + \frac{v^2}{r^2} t^2} $$ and converting all the $\frac{a}{r}$ and $\frac{v}{r}$ terms into expressions of the period we get $$ r_2 = r \sqrt{1 - \left(2\pi\frac{t}{T}\right)^2 + \frac{1}{4}\left(2\pi\frac{t}{T}\right)^4 + \left(2\pi\frac{t}{T}\right)^2}$$ Finally, we drop the $(t/T)^4$ term as negligible and note that the $(t/T)^2$ terms cancel so the result is $$r_2 = r$$ or the radius never changed (which justified the constant magnitude for acceleration, and a small enough $t$ justifies both the constant direction and the dropping of the fourth degree term).



Tuesday, October 27, 2015

newtonian gravity - Inverse Square Law and extra space dimensions


Newton's famous Inverse Square Law says that in $n=3$ dimension of space, force is inversely proportional to the square of the distance between a source and a target.


I understand that for higher dimensions, this can be generalized as thus:


$$F\propto1/r^{n-1}$$


Where $n$ is the dimension of the space.


Why is this so? Is there a rigorous derivation of this from a deep fundamental theory? Or is there a heuristic argument why this is so?



Answer



You can get this more "intuitively" (idiosyncratically): the flux of this force in closed surface is equal to the quantity of source inside (is a Gauss's Law). This source could be a mass or a charge. The physical picture is: the pressure applied in a closed surface by the field-force is proportional to the quantity of source inside.



You can get the force-field produced by a point source with suitable choices of surface (a sphere concentric with the source). Then for any dimension you can see that your field obey the $\frac{1}{r^{d-1}}$ because the area of this surface ($d$-sphere, $S_2$) grow with $r^{d-1}$ (for $d>2$).



Yes, exist a more "rigorous" (Standard) derivation. Actually we need to check first that this law imply a potential that obey the Laplace's equation: $\nabla^2 V(x)=0$. Any point source of this force will produce a potential that is a Green's function of $\nabla^2$ for suitable boundary condition ($V=0$ at $\infty$).


For three dimensions, the Green's function is $\frac{1}{r}$, this imply $\frac{1}{r^2}$ for the force. For $d>2$, the Green's function is $\frac{1}{r^{d-2}}$ and imply a force that is $\frac{1}{r^{d-1}}$. For $d=2$ is a logarithm and for $d=1$ is linear with $r$.


quantum field theory - Is the empty space really empty?



I've come across another article in "list verse" which says that the empty space is not actually empty at least for a while. I've tried to find about this, so I googled it .It also quotes a word "Quantum foam" .But all the explanations are way too hard to understand .Any one who knows about this could help . Thank you in advance....enter image description here



Answer



First let us address "emty space".


Empty space is a theoretical concept, a space where there is no matter and no energy.


In our universe, no matter how far away one goes in space, it is not empty. It contains the cosmic microwave background radiation, cool photons, which is at a temperature of 2.7 K .


Within quantum mechanics and elementary paricles, the "empty space" , ignoring the CMB existence and assuming empty space exists, can be mathematically populated. What does this mean? A mathematical model exists where every elementary particle in this table populates "empty space" with the ground state of the solution for the field of that particle. This means that the existence of that particle, an electron on the way to the sun, for example, appears as an excited state on this ground state solution for electrons, and a probability distribution of its trajectory in space accompanies it, as the Heisenberg uncertainty principle , HUP


hup



defines a region around the trajectory where the probability of finding the electron at x, constrains the probability of its momentum being p.


The quantum foam is an extension of this format, taking the HUP and examining a region of space for a tiny time interval delta(t) and saying then there must be a delta(E) for that energy.It then goes further, saying that the positron field also exists at that (x,y,z,t) and a "virtual" electron positron loop could be drawn existing for very small time with the energy allowed by the hup.


This is a misuse of the Feynman diagrams where such loops exist, misuse because in empty space no interactions are happening, and virtual particles are just lines in Feynman diagrams representing mathematical forms to be integrated over. A closed loop cannot be integrated over if the limits of the integral are (0,0). A real particle interaction is needed with real outgoing particles to be able to define intermediate virtual "particles", which have the name but not the mass of its name.


virtual


So the "quantum foam" picture is a handwaving hypothesis, based on the HUP, but really not calculable in any way so as to give anything measurable larger than zero in empty space. It is only within interactions that the effect of virtual loops is observed as measurable, as in the Casimir effect and the Lamb shift.


What increases the resistance at collector in photoelectric effect?


The photoelectric experiment uses a setup like this:


https://dc.edu.au/wp-content/uploads/1.png


The graph of photoelectric current $I$ against potential applied at the anode/collecting plate $V$ looks something like this:


https://i.stack.imgur.com/R17EJ.png


Once we reach the saturation current, further increasing the collector potential $V$ does not further increase the photoelectric current $I$. Rather, resistance simply increases ($R = \frac{V}{I}$). But what is the underlying physical change in the electrodes that increases the resistance as $V$ increases? Are there more negative charges on the cathode, inhibiting the current? Is that the reason why an increasing potential $V$ causes $R$ to increase?




Monday, October 26, 2015

quantum field theory - Why is there a minus sign in this wave equation derivation?


My book on quantum mechanics suggests a derivation of the wave equation


$$\left(\Delta - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \right) \psi(\bar{r},t) = 0$$


from the photon energy-impulse relation


$$E^2 = c^2p^2$$


using the substitutions


$$\bar{p} \to -i\hbar \nabla$$ and $$E \to i \hbar \frac{\partial}{\partial t}$$


I suppose the $\hbar$s are there for the non-photon generalisation, but is there a reason for the minus sign in the $p$ substitution? Removing it would yield the same result.




Answer



The relative sign is not just a convention. Once you decide that $E$ is represented by $i\hbar \partial/\partial t$, there must be a minus sign in the formula for $p$, namely $p=-i\hbar \partial / \partial x$. Or vice versa.


First of all, there has to be $i$ or $-i$ in all the formulae because $\partial/\partial x$ is an anti-Hermitian operator (because of the minus sign in the integration by parts) and we need Hermitian operators (which has real, measured eigenvalues) for the energy, momentum, and others. What about the signs?


The only sign convention that was chosen in the early days of quantum mechanics was one for the energy; indeed, one could have replaced $i$ by $-i$ in that equation because $i$ and $-i$ play the same algebraic role: exchanging $i$ and $-i$ is an "outer automorphism" of complex numbers. But once this sign is fixed, all other signs are fixed, too. That includes minus signs in momentum, angular momentum, gauge transformations, Schrödinger's picture, Heisenberg's picture, Feynman's path integral, and any other formula of quantum mechanics. There is only one way to define quantum mechanics given a classical limit (with its sign conventions) we need to get.


The relative minus sign in $p,E$ may become invisible if you only act with second derivatives - squared momentum, squared energy - but it is visible if you act with first powers of the operators.


De Broglie's wave - or a wave associated with a particle - is proportional to $$ \exp \left[ \frac{i}{\hbar} (\vec p\cdot \vec x-Et)\right] $$ Note that if you differentiate with respect to $x$ and $t$, and multiply the result by $i\hbar$, you get $-p$ and $E$, respectively. (Omit the vector signs if you want just one-dimensional space with one $x$ and one $p$.)


The relative sign between $\vec p\cdot \vec x$ and $Et$ in de Broglie's formula above is physically necessary because only $Et - \vec p \cdot \vec x$ is the correct Lorentzian inner product of the vectors $(E,\vec p)$ and $(t,\vec x)$: the relative minus sign comes from the opposite signs of space and time in the signature of spacetime. Note that the doubly relative sign in $(E,\vec p)$ and $(t,\vec x)$ can't be flipped because in relativity, $\vec pc^2/E$ is the velocity $\vec v$.


The argument above is for a relativistic interpretation of $E$. However, the non-relativistic kinetic energy is defined just as a shifted relativistic energy, $$ E_{\rm nonrel} = E_{\rm rel} - mc^2 $$ so the coefficient (and sign) in front of $E$ remains unchanged and $\vec p$ is totally unchanged. One may also design related arguments analyzing where the de Broglie wave is moving. For it to be moving in the right direction, there has to be a minus sign in $Et-px$. It's related to the fact that the shape of objects moving by velocity $v$ only depends on $x-vt$ because $x=vt$ (without a minus sign, which becomes a minus sign if you move both terms to the same side) is the equation for their center of mass.


The hypersurfaces of constant phase of the de Broglie wave are orthogonal to the world lines of the particles which dictates the sign. After all, the waves' maxima and minima should be moving in the same direction as the particles themselves.


kinematics - Why does work equal force times distance?



My book says:



Energy is the capacity to do work and work is the product of net force and the 1-dimensional distance it made a body travel while constantly affecting it.



This seems quite unmotivated to me.


Why does work equal $F \cdot d$ ?


Where does the distance part come from?



I always thought of time as the one thing we can only measure (not affect) so it justifies why we may measure other things in relation to time. But we have a much greater control over distance (since it's just a term for a physical dimension we can more or less influence as opposed to time).


Edit: That^ doesn't make much sense, but it's been graciously addressed in some of the answers.


Level: High-school Physics.




terminology - What is the difference between Quantum Physics, Quantum Theory, Quantum Mechanics, and Quantum Field Theory?


What is the difference between Quantum Physics, Quantum Theory, Quantum Mechanics, and Quantum Field Theory? Are they the same subject? I believe that they are not the same subject! Maybe there is not big difference between those subjects but I need to know what is main difference between those subjects and what is main intersection? Also I need to know which one is big subject relative to another?



Answer




Quantum mechanics (QM – also known as quantum physics, or quantum theory) is a branch of physics which deals with physical phenomena at microscopic scales, where the action is on the order of the Planck constant. Quantum mechanics departs from classical mechanics primarily at the quantum realm of atomic and subatomic length scales. Quantum mechanics provides a mathematical description of much of the dual particle-like and wave-like behavior and interactions of energy and matter. Quantum mechanics is the non-relativistic limit of Quantum Field Theory (QFT), a theory that was developed later that combined Quantum Mechanics with Relativity.


Quantum field theory (QFT) is a theoretical framework for constructing quantum mechanical models of subatomic particles in particle physics and quasiparticles in condensed matter physics, by treating a particle as an excited state of an underlying physical field.



Some of the relativistic quantum field theories would be QED, QCD, and the Standard Model.



References:


http://en.wikipedia.org/wiki/Quantum_mechanics


http://en.wikipedia.org/wiki/Quantum_Field_Theory


quantum mechanics - Is there an equivalent of probability current for the Wigner distribution?


I know that for a wavefunction, I can derive a probability current $\mathbf{J}$ that satisfies the continuity equation: $$\nabla \cdot \mathbf{J}=-\frac{\partial}{\partial t} \big|\psi\big|^2$$


Can a similar quantity be derived for the Wigner distribution? If so, what is it?



Answer



It's intuitively clear that this current must exist because the integral of the Wigner function is conserved by unitary evolution. This current is known as the Wigner flow, and it exists but it's not particularly pretty. For an example of the Wigner flow in use, see arXiv:1208.2970; in short, it is the current $$ J=\begin{pmatrix}J_x\\J_p\end{pmatrix} =\begin{pmatrix} \tfrac pm W(x,p,t) \\ -\sum_{l=0}^\infty\frac{(i\hbar/2)^{2l}}{(2l+1)!}\frac{\partial^{2l}W(x,p,t)}{\partial p^{2l}}\frac{\partial^{2l+1}V(x)}{\partial x^{2l+1}}\end{pmatrix} ,$$ where $V(x)$ is the system's potential energy, and it obeys the continuity equation $$ \frac{\partial W}{\partial t}+\frac{\partial J_x}{\partial x}+\frac{\partial J_p}{\partial p}=0. $$



Sunday, October 25, 2015

Final theory in Physics: a mathematical existence proof?


Some time ago, I read something like this about the issue of "a final theory" in Physics:


"Concerning the physical laws, we have several positions as scientists





  1. There are no fundamental physical laws. At the most elementary level, the Universe/Multiverse is essentially chaotic and anarchical. There are no such laws.




  2. There are a continuous sequence of more and more precise theories, but there is no a final theory. Physics will be always evolving from one approximate theory to another bigger and more accurate. In the end, we will be also able to find a better theory and additional levels of complexity or reality.




  3. There is a final theory explaining everything, and we will find if and only if:


    i) We are clever enough to find such a theory. ii) We make good and sophisticated enough mathematics. iii) We guess the right axioms/principles/ideas. iv) We interpretate data correctly and test the putative final theory with suitable instruments/experiments. "





Supposing 3) is the right approach...


Question: How could we prove the mere mathematical existence of such a theory? Wouldn't it evade the Gödel's incompleteness theorem somehow since, as a Theory of Everything, it would be explain "all" and though it should be mathematically self-consistent? How could a Theory of Everything be a counterexample of Gödel's theorem if it is so, or not?


Note: The alledged unification of couplings in supersymmetric theories is a hint of "unification" of forces, but I am not sure if it counts as sufficient condition to the existence of a final theory.


Complementary: Is it true that Hawking has changed his view about this question?


SUMMARY:


1') Does a final theory of physics exist? The issue of existence should be tied to some of its remarkable properties (likely).


2') How could we prove its existence or disproof it and hence prove that the only path in Physics is an infinite sequence of more and more precise theories or that the Polyverse is random and/or chaotic at the most fundamental level?


3') How 1') and 2') affect to Gödel's theorems?


I have always believed, since Physmatics=Physics+Mathematics (E.Zaslow, Clay Institute) is larger than the mere sum that the challenge of the final theory should likely offer so hint about how to "evade" some of the Gödel's theorems. Of course, this last idea is highly controversial and speculative at this point.




electromagnetism - Force between two finite parallel current carrying wires


Remark: This is not a homework question...It is pure out of theoretical interest. I asked this the mathematics-community a couple days ago and got no answer, so I figured I'd try here.


Most standard physics textbooks compute the force two infinite wires exert on each other, but they remain silent about the case where the wires are finite. Let's say we have two parallel wires carrying a current of equal magnitude in the same direction, both of which have a length $d$ and also seperated by a distance $d$. I now want to find out the force one wire exerts on another, using the Biot-Savart Law.


Let the left wire be positioned at the origin of the $xy$-plane, going along the $y$-axis, and let the other wire be a distance $d$ to the right. We assume the currents are flowing in the positive $y$-direction. Then we first choose a source element (on the left wire) of infinitesimal length $dy$ described by the position vector $\mathbf{r_0} = y_0 \hat{j}$. This constitutes a current source of $I d\vec{l} = (Idy) \hat{j}$.


We then pick an arbitrary field point $P$ on the other wire with position vector $\mathbf{r_p} = x\hat{i} + y \hat{j}$. Then the position vector $\mathbf{r}$ pointing from the source point to the field point is given as \begin{align*} \mathbf{r} = \mathbf{r_p} - \mathbf{r_0} = x\hat{i} + (y - y_0) \hat{j}, \end{align*} with $\sqrt{x^2 + (y-y_0)^2}$ being the length of this vector. If we now calculate the crossproduct $d\vec{l} \times \mathbf{r}$, we can write it as \begin{align*} dy \hat{j} \times (x\hat{i} + (y - y_0)\hat{j}) = -dy x \hat{k} \end{align*} Now comes the tricky part. I think I need to setup a double integral, because we are working with infinitesimal force elements $d\mathbf{F}$, each which is given as $d\mathbf{F} = I d\vec{l} \times \mathbf{B}$. But we also have that \begin{align*} d\mathbf{B} = \frac{\mu_0 I}{4 \pi} \frac{d \vec{l} \times \hat{r}}{r^2} = \frac{\mu_0 I}{4 \pi} \frac{d \vec{l} \times \mathbf{r}}{r^3} = -\frac{\mu_0 I}{4 \pi} \frac{dyx}{\sqrt{x^2 + (y-y_0)^2}} \hat{k} \end{align*} Hence I need to somehow integrate over $d\mathbf{F}$ and $d\mathbf{B}$. Does anyone have an idea how to do this?



Answer



Let's call the circuit in the origin circuit one and it's line element $\mathrm{d}l_1=(0,\mathrm{d}y_1,0)$ and the one to it's right $r_2=(d,y_2,0)$ then the force between them is $\mathrm{d}F_{12}=i \mathrm{d}l_2 \times B_1$ where



$$B_1=\frac{\mu_0i}{4\pi}\int_{l_1} \frac{\mathrm{d}l_1\times \Delta r}{(\Delta r)^3}$$


and $\Delta r=(d,(y_2-y_1),0)$ so we have that


$$\mathrm{d}l_1\times \Delta r=(0,0,-\mathrm{d}y_1 d)$$


so we get as you wrote:


$$B_1=-\frac{\mu_0 i d}{4 \pi}\int_{0}^{d} \frac{\mathrm{d}y_1}{(d^2+(y_2-y_1)^2)^{\frac{3}{2}}}$$


Ok now let's call $y_2-y_1=t$ so $\mathrm{d}t=-\mathrm{d}y_1$ then we can write


$$B_1=\frac{\mu_0 i d}{4\pi}\int_{y_2}^{y_2-d}\frac{\mathrm{d}t}{(d^2+t^2)^{\frac{3}{2}}}$$


we now make the substitution


$$t=d\cdot \sinh(u)$$


and we obtain



$$\mathrm{d}t=d\cdot \cosh(u)\mathrm{d}u$$


and then


$$B_1=\frac{\mu_0 i d}{4\pi}\int \mathrm{d}u \frac{d \cosh(u)}{d^3 \cosh(u)^3}$$


in which we used


$$\cosh(u)^2-\sinh(u)^2=1$$ $$B_1=\frac{\mu_0 i }{4\pi d}\int \frac{\mathrm{d}u}{\cosh(u)^2}$$


now $\frac{1}{\cosh^2(u)}$ is the derivative of $\tanh(u)$ so


$$\int \frac{\mathrm{d}u}{\cosh(u)^2}=\tanh(u)$$


we get then


$$B_1=\frac{\mu_0 i }{4\pi d}\tanh\left(a\sinh\left(\frac{y_2-y_1}{d}\right)\right)+\text{const}$$


where we have substituted back all parameters



$$u=a\sinh\left(\frac{t}{d}\right) \\ t=y_2-y_1$$


so knowing that (where $a\sinh(x)$ is the inverse function of $\sinh(x)$):


$$\tanh(a\sinh(x))=\frac{x}{\sqrt{x^2+1}}$$


finally


$$B_1=\frac{\mu_0 i }{4\pi d} \frac{\frac{y_2-y_1}{d}}{\sqrt{(\frac{y_2-y_1}{d})^2+1}}+\text{const}=\frac{\mu_0 i }{4\pi} \frac{1}{\sqrt{(y_2-y_1)^2+d^2}}+\text{const}$$


now we calculate it between $y_1=0$ and $y_1=d$ which yields


$$B_1=\frac{\mu_0 i }{4\pi} \left[ \frac{1}{\sqrt{(y_2-d)^2+d^2}}-\frac{1}{\sqrt{y_2^2+d^2}}\right]$$


to calculate the force we take $B_1=(0,0,B_1 \hat{z})$ and we operate the following:


$$\mathrm{d}F_{12}=i\mathrm{d}l_2 \times B_1=i(B_1\mathrm{d}y_2,0,0)$$


now we have to integrate on the circuit two:



$$F_{12}=\frac{\mu_0 i^2 }{4\pi} \int_{0}^{d}\mathrm{d}y_2\left[ \frac{1}{\sqrt{(y_2-d)^2+d^2}}-\frac{1}{\sqrt{y_2^2+d^2}}\right]=\frac{\mu_0 i^2 }{4\pi} \left[I(y_2-d)-I(y_2)\right]$$


and know we do the same trick as before


$$t=y_2-d \ \text{or}\ t=y_2\ \text{for the second piece}$$ $$t=d\cdot \sinh(u)$$ $$\mathrm{d}y_2=\mathrm{d}t=d\cdot \cosh(u)\mathrm{d}u$$


then:


$$I=\int \mathrm{d}u\cdot d \cdot \cosh(u) \frac{1}{\sqrt{d^2\cosh^2(u)}}=u=a\sinh\left(\frac{t}{d}\right)$$


we finally get


$$F_{12}=\frac{\mu_0 i^2 }{4\pi}\left[a\sinh\left(\frac{y_2-d}{d}\right)-a\sinh\left(\frac{y_2}{d}\right)\right]_0^d$$


which curiously enough is zero for this choice of parameters! I hope that helped!


Saturday, October 24, 2015

solid state physics - Is crystal momentum really momentum?


Almost every solid state physics textbook says crystal momentum is not really physical momentum. For example, phonons always carry crystal momentum but they do not cause a translation of the sample at all.


However, I learned that in indirect-band-gap semiconductors, we need phonons to provide the crystal momentum transfer to make happen electron transitions between the top of the valance band and the bottom of the conduction band. Along with absorbing or emitting photons, of course.


Photons do carry physical momentum. For the purpose of momentum conservation, it seems that phonons do carry physical momentum as well.


How can we explain this?


========================================


To put it more specifically, I drew a graph to tell the story:



K (capital) is crystal momentum.



For such transition, photon provides most of the energy transfer (and a little momentum transfer hk, k in lower case), phonon provides most of the momentum transfer (and a little energy).


Similar graphs can be found in most solid state physics textbooks. The picture tells me, either the photon participating in the transition carries crystal momentum, which value is equal to physical moemntum hk, or the crystal momentum itself is a kind of physical momentum.


However, one can prove that a phonon does not carry physical momentum (here I quote Kittel's "Introduction to Solid State Physics"):



So, how do we explain the momentum transfer in the electron transition aforementioned?




particle physics - What are the details around the origin of the string theory?


It is well-known even among the lay public (thanks to popular books) that string theory first arose in the field of strong interactions where certain scattering amplitudes had properties that could be explained by assuming there were strings lurking around. Unfortunatelly, that's about as far as my knowledge reaches.



Can you explain in detail what kind of objects that show those peculiar stringy properties are and what precisely those properties are?


How did the old "string theory" explain those properties.


Are those explanations still relevant or have they been superseded by modern perspective gained from QCD (which hadn't been yet around at the Veneziano's time)?




Answer



in the late 1960s, the strongly interacting particles were a jungle. Protons, neutrons, pions, kaons, lambda hyperons, other hyperons, additional resonances, and so on. It seemed like dozens of elementary particles that strongly interacted. There was no order. People thought that quantum field theory had to die.


However, they noticed regularities such as Regge trajectories. The minimal mass of a particle of spin $J$ went like $$ M^2 = aJ + b $$ i.e. the squared mass is a linear function of the spin. This relationship was confirmed phenomenologically for a couple of the particles. In the $M^2$-$J$ plane, you had these straight lines, the Regge trajectories.



Building on this and related insights, Veneziano "guessed" a nice formula for the scattering amplitudes of the $\pi+\pi \to \pi+\rho$ process, or something like that. It had four mesons and one of them was different. His first amplitude was the Euler beta function $$ M = \frac{\Gamma(u)\Gamma(v)}{\Gamma(u+v)}$$ where $\Gamma$ is the generalized factorial and $u,v$ are linear functions of the Mandelstam variables $s,t$ with fixed coefficients again. This amplitude agrees with the Regge trajectories because $\Gamma(x)$ has poles for all non-positive integers. These poles in the amplitude correspond to the exchange of particles in the $s,t$ channels. One may show that if we expand the amplitude to the residues, the exchanged particles' maximum spin is indeed a linear function of the squared mass, just like in the Regge trajectory.


So why are there infinitely many particles that may be exchanged? Susskind, Nielsen, Yoneya, and maybe others realized that there has to be "one particle" of a sort that may have any internal excitations - like the Hydrogen atom. Except that the simple spacing of the levels looked much easier than the Hydrogen atom - it was like harmonic oscillators. Infinitely many of them were still needed. They ultimately realized that if we postulate that the mesons are (open) strings, you reproduce the whole Veneziano formula because of an integral that may be used to define it.


One of the immediate properties that the "string concept" demystified was the "duality" in the language of the 1960s - currently called the "world sheet duality". The amplitude $M$ above us $u,v$-symmetric. But it can be expanded in terms of poles for various values of $u$; or various values of $v$. So it may be calculated as a sum of exchanges purely in the $s$-channel; or purely in the $t$-channel. You don't need to sum up diagrams with the $s$-channel or with the $t$-channel: one of them is enough!


This simple principle, one that Veneziano actually correctly guessed to be a guiding principle for his search of the meson amplitude, is easily explained by string theory. The diagram in which 2 open strings merge into 1 open string and then split may be interpreted as a thickened $s$-channel graph; or a thick $t$-channel graph. There's no qualitative difference between them, so they correspond to a single stringy integral for the amplitude. This is more general - one stringy diagram usually reduces to the sum of many field-theoretical Feynman diagrams in various limits. String theory automatically resums them.


Around 1970, many things worked for the strong interactions in the stringy language. Others didn't. String theory turned out to be too good - in particular, it was "too soft" at high energies (the amplitudes decrease exponentially with energies). QCD and quarks emerged. Around mid 1970s, 't Hooft wrote his famous paper on large $N$ gauge theory - in which some strings emerge, too. Only in 1997, these hints were made explicit by Maldacena who showed that string theory was the right description of a gauge theory (or many of them) at the QCD scale, after all: the relevant target space must however be higher-dimensional and be an anti de Sitter space. In AdS/CFT, much of the original strategies - e.g. the assumption that mesons are open strings of a sort - get revived and become quantitatively accurate. It just works.


Of course, meanwhile, around mid 1970s, it was also realized that string theory was primarily a quantum theory of gravity because the spin 2 massless modes inevitably exist and inevitably interact via general relativity at long distances. In the early and mid 1980s, it was realized that string theory included the right excitations and interactions to describe all particle species and all forces we know in Nature and nothing could have been undone about this insight later.


Today, we know that the original motivation of string theory wasn't really wrong: it was just trying to use non-minimal compactifications of string theory. Simpler vacua of string theory explain gravity in a quantum language long before they explain the strong interactions.


everyday life - If water is not a good conductor, why are we advised to avoid water near electricity (no wet hands near circuits etc.)?


How can water be a medium to conduct current while its ionisation is so negligible that, in principle, no current should flow?




Friday, October 23, 2015

classical mechanics - Are point transformations necessarily canonical?


A point transformation $ Q(q,t)$ only modifies the location coordinates, while a canonical transformation, in general, also modifies the momentum coordinates $Q(q,p,t)$.


In the context of the Lagrangian formalism, it is usually argued that we have absolute freedom to perform point transformations. I was wondering if this freedom carries over to the Hamiltonian formalism.


Specifically, do arbitrary point transformations leave the form of Hamilton's equations unchanged? In other words, are transformations in phase space which only involve the location coordinates $q$ necessarily canonical?




The defining condition for canonical coordinates is $$ \{Q,P\} = 1 $$ Therefore, we can check for a point transformation $$ Q = Q(q)$$ $$ P = p$$ if this condition is fulfilled \begin{align} \{Q,P\} & = \frac{\partial Q}{\partial q} \frac{\partial P}{\partial p} - \frac{\partial Q}{\partial p} \frac{\partial P}{\partial q} \\ &= \frac{\partial Q(q)}{\partial q} \frac{\partial p}{\partial p} - \frac{\partial Q(q)}{\partial p} \frac{\partial p}{\partial q} \\ &= \frac{\partial Q(q)}{\partial q} \, . \end{align} In general, I think, this is not equal to one or any other constant and therefore point transformations aren't canonical.



But, for example, in Vol. 1 of Landau-Lifshitz they write:



"Since Lagrange's equations are unchanged by the [point transformation $Q_i = Q_i(q,t)$], Hamilton's equations are also unchanged."



This seems like a reasonable argument. However, the calculation above shows that does not seem to be the case. Is Landau-Lifshitz wrong or did I make a mistake in my reasoning?


I think it would be really strange if point transformations weren't canonical because in almost all Classical Mechanics textbook some like the following is mentioned:



[S]ince there are twice as many canonical variables (q,p) as generalized coordinate, the set of possible transformations is considerably larger. This is one of the advantages of the canonical formalism. (Calkin, Lagrangian and Hamiltonian Mechanics)



E.g. to quote once more Landau-Lifshitz




"The enlargement of the class of possible transformations is one of the important advantages of the Hamiltonian treatment."




Answer





  1. Yes, a point transformation
    $$q^i \longrightarrow Q^i~=~f^i(q,t)\tag{1}$$ is a canonical transformation (CT) with type 2 generating function $$ F_2(q,P,t)~=~\sum_{i=1}^nf^i(q,t)P_i, \tag{2}$$ cf. Ref. 1.





  2. Alternatively, it is a straightforward exercise to show that a point transformation (1) when extended to the phase space/cotangent bundle $$ (q^i,p_j) \longrightarrow (Q^i,P_j)~=~\left(Q^i(q,t), \sum_{k=1}^n\frac{\partial q^k}{\partial Q^j}p_k\right) \tag{3}$$ is a symplectomorphism. i.e. it preserves the fundamental/canonical Poisson bracket relations. Here it is important to realize that the canonical momenta $p_j$ transform as components of a covector/1-form, cf. e.g. this & this Phys.SE posts.




References:



  1. H. Goldstein, Classical Mechanics; eq. (9.26).


quantum field theory - Gordon Identity confusion



For the Gordon identity


$$2m \bar{u}_{s'}(\textbf{p}')\gamma^{\mu}u_{s}(\textbf{p}) = \bar{u}_{s'}(\textbf{p}')[(p'+p)^{\mu} -2iS^{\mu\nu} (p'-p)_{\nu}]u_{s}(\textbf{p}) $$


If I plug in $\mu$=5, what exactly does the corresponding $(p'+p)^{5}$ represent? 4 vectors can only have 4 components so is this just an exponential?


Thanks




faq - Book recommendations




Every once in a while, we get a question asking for a book or other educational reference on a particular topic at a particular level. This is a meta-question that collects all those links together. If you're looking for book recommendations, this is probably the place to start.


All the questions linked below, as well as others which deal with more specialized books, can be found under the tag (formerly ).


If you have a question to add, please edit it in. However, make sure of a few things first:



  1. The question should be tagged

  2. It should be of the form "What are good books to learn/study [subject] at [level]?"


  3. It shouldn't duplicate a topic and level that's already on the list


Related Meta: Do we need/want an overarching books question?




Why do two beams of light pass through one another without interacting?


Why is it that if I have two torches, each shining rays in perpendicular directions at perpendicular screens, such that the beams cross, then the images on the respective screens are independent of whether the alternate torch is switched on? In other words, why does light seem to pass through itself without interacting?


Apologies if this question is basic, and apologies if there is a similar question somewhere on the site but I haven't managed to find an answer.


Thanks, A.




Thursday, October 22, 2015

collision - Why are ions and electrons at different temperatures in a plasma?


In plasmas, the collision rate among ions or electrons is much larger than the collision rate between ions and electrons. Why is that so?




classical mechanics - Motivating the Legendre Transform Mathematically


If I begin with a functional of the form



$$J[y] = \int_a^b f(x,y,y')dx$$


and find its Euler-Lagrange equations


$$\frac{\partial f}{\partial y} - \frac{d}{dx}\frac{\partial f}{\partial y'} = 0 = \frac{d}{dx}\frac{\partial f}{\partial y'} - \frac{\partial f}{\partial y}.$$


I end up with a second order ODE


$$\frac{d}{dx}\frac{\partial f}{\partial y'} - \frac{\partial f}{\partial y} = \left( \frac{\partial ^2 f}{\partial y' \partial y'} \right) \frac{d^2 y}{dx^2} + \left(\frac{\partial ^2 f}{\partial y \partial y'}\right) \frac{dy}{dx} + \frac{\partial ^2 f}{\partial x \partial y'} - \frac{\partial f}{\partial y} = 0$$


Now every higher order ODE can be broken up into a system of first order ODEs in $y$ and the derivative $M = \frac{dy}{dx}$, giving


$$ \frac{d}{dx}\frac{\partial f}{\partial y'} - \frac{\partial f}{\partial y} = \left( \frac{\partial ^2 f}{\partial y' \partial y'} \right) \frac{d M}{dx} + \left(\frac{\partial ^2 f}{\partial y \partial y'}\right)M + \frac{\partial ^2 f}{\partial x \partial y'} - \frac{\partial f}{\partial y} = 0. $$


From this perspective, Hamilton's equations


$$\left\{ \begin{array} & \frac{dy}{dx} = \phantom{-}\frac{\partial \mathcal{H}}{\partial p}\\ \frac{d p}{dx} = - \frac{\partial \mathcal{H}}{\partial y} \end{array}\right.$$


are merely a system of first order equations making my above system of first order ODEs look more symmetric, after a suitable change of variables.



My question is, looking at


$$\frac{d}{dx}\frac{\partial f}{\partial y'} - \frac{\partial f}{\partial y} = \left( \frac{\partial ^2 f}{\partial y' \partial y'} \right) \frac{d^2 y}{dx^2} + \left(\frac{\partial ^2 f}{\partial y \partial y'}\right) \frac{dy}{dx} + \frac{\partial ^2 f}{\partial x \partial y'} - \frac{\partial f}{\partial y} = 0$$


it should be possible to see why the Legendre transformation arises, first because its a transformation using derivatives to change variables but also because it should just make some terms go to zero in this second order ode so that everything looks nicer, but how do you see this explicitly?


It'd be great if you could use my notation, ie. $J[y]$ etc... as you see I snuck in $\mathcal{H}$ in above which shouldn't really be there, I'd love to see how that comes about in my notation - thanks!




special relativity - Can one create mass from energy?


Due to $ E =m c^2 $, one can convert mass to energy. A classic example would be matter/anti-matter annihilation to produce energy (photons, etc.).


Can one do the reverse? So could one do something to photons to create mass?



Answer



The conversion between mass and energy isn't even really a conversion. It's more that mass (or "mass energy") is a name for some amount of an object's energy. But the same energy that you call the mass can actually be a different type of energy, if you look closer. For example, we say that a proton has a specific amount of mass, about $2\times 10^{-27}\text{ kg}$. But if you look into the structure of a proton, about half that mass (or more, depending on conditions) is actually kinetic energy of the gluons.


Of course, that's probably not what you had in mind. To more directly answer your question, it is possible to produce matter from two colliding photons, although the probability is not especially high. You need energetic photons, and lots of them, to create an appreciable number of detectable matter particles. Wikipedia's article on matter creation has more information and links.


homework and exercises - How do I show for an ideal transformer $M^2=L_1L_2$?


I've been stuck on this problem for about an hour.



In an ideal transformer, the same flux passes through all turns of the primary and of the secondary . Show that in this case $M^2=L_1L_2$, where $M$ is the mutual inductance of the coils, and $L_1$, $L_2$ are their individual self-inductances.



My first instinct was the following: If we isolate the response of each coil to the other one, we get the equations $N_1\Phi_1=MI_2$ and $N_2\Phi_2=MI_1$, where $\Phi$ is magnetic flux going through a single turn of the coil and $I$ is current. Likewise, if we isolate the response of each coil to itself, we get the equations $N_1\Phi_1=L_1I_1$ and $N_2\Phi_2=L_2I_2$. I would now naturally want to say "Now just multiply $N_1\Phi_1$ and $N_2\Phi_2$ both ways, and get the desired result!" i.e.,



$$(N_1\Phi_1)(N_2\Phi_2)=(MI_2)(MI_1)=(L_1I_1)(L_2I_2)\implies M^2=L_1L_2$$


But that's incorrect, because the response of the coils to themselves and to the other coil are different quantities (i.e. not necessarily equal).


Realizing this, I write down the more general equations.


$$N_1\Phi_1=MI_2+L_1I_1$$ $$N_2\Phi_2=MI_1+L_2I_2$$


Now even with the hypothesis that $\Phi_1=\Phi_2=\Phi$, I still can't see how the identity $M^2=L_1L_2$ follows. Could you help clear up my confusion?


*Taken from D. Griffiths Introduction to Electrodynamics (4th ed.), Problem 7.58.



Answer



In a perfect transformer, if we run a current through either the primary or secondary coils, we are guaranteed in the quasistatic case that $\Phi_1=\Phi_2$ for each individual turn of the coils. Now suppose we ran a current $I_1$ through the primary coil, and waited long enough so that $I_1$ was steady and $I_2$ (current induced on second coil) was zero. we could multiply both sides of that equation by $N_1/I_1$ ($N_1$ is the number of turns in the primary coil) and see that


$$\frac{N_1}{I_1}\Phi_1=\frac{N_1}{I_1}\Phi_2\longrightarrow L_1=\frac{N_1}{N_2}M$$


Where we have used the identities $N_2\Phi_2=M I_1$ ($M$ is mutual inductance) and $N_1\Phi_1=L_1I_1$ ($L_1$ is the self-inductance of the primary coil). Likewise, if we had reversed the roles of the primary and secondary coil, we would get the analagous result



$$L_2=\frac{N_2}{N_1}M$$


Multiplying those two results together gives the desired identity.


$$L_1L_2=\frac{N_1}{N_2}\frac{N_2}{N_1}M^2=M^2$$


It's interesting to note that, in the general case (i.e. two coils that aren't part of an ideal transformer), if we run a current through the first coil, we get $\Phi_1\geq\Phi_2$. This is because all the field lines pass through the first coil itself, but not all the field lines have to pass through the second coil (they could pass outside of it). Propogating this inequality down the previous steps shows that in the general case,


$$L_1L_2\geq M^2$$


quantum mechanics - Understanding Feynman Diagram of pair-production


The below Feynman Diagram can be found on the Wikipedia page for Pair Production. I am having trouble understanding the significance of the "Z", as well as what appears to be an "X" over the start of the second photon. I also do not understand why there is an "e" labeling the interaction. Any help would be greatly appreciated.


enter image description here



Answer



Energy and momentum conservation do not allow a single photon to break up into two particles. This is clear if you look at the center of mass system of the products : they have an invariant mass given at least by twice the rest mass of the electrons, if they are produced at threshold. The incoming photon does not have a center of mass since its mass is zero. So a single photon cannot decay into a pair of particles.



The diagram is the appropriate Feynman diagram which shows how pair production can happen if there exist electric fields, which are displayed by the Z charges of a nucleus , positive for nucleus negative for the surrounding electrons, and the X showing the vertex of the virtual photon from the field just covers that it could be either an interactions with the field of an electron, or an interaction with the field of the nucleus.


Feynman diagrams are iconal representations that have a one to one correspondence with the integrals that can calculate the crossections for the interactions.


The e internal line is a virtual electron, a function integrated over, until it picks up a definite momentum and energy vector hitting the field. The gamma from the X to the virtual electron is also virtual.


A classical example of the lowest order Feynman diagram for a simpler reaction that explains the function of a virtual particle :


virtual


A virtual particle has all the quantum numbers of its name except it is off mass shell, its mass appearing in the propagator under the integral.


In the picture in this question one can see the spectator electron in the creation of the electron positron pair. The nucleus is heavy and does not appear in the bubble chambers as in conserving momentum it does not have a high enough velocity to travel any length before loosing its energy.


Wednesday, October 21, 2015

electromagnetism - Why is divergence and curl related to dot and cross product?


I've been reading Griffith's intro to electrodynamics and I've been a bit confused on his explanation of divergence and curl. I don't understand how divergence is the dot product of a gradient acting on a vector function and curl is the cross product of gradient acting on a vector function. Does it relate to the fact that one uses sine while other uses cosine? Just to clarify, I understand the concept of divergence and curl from a purely conceptual standpoint, it's just this mathematical definition that I can't wrap my head around.





quantum field theory - Time reversal symmetry and T^2 = -1


I'm a mathematician interested in abstract QFT. I'm trying to undersand why, under certain (all?) circumstances, we must have $T^2 = -1$ rather than $T^2 = +1$, where $T$ is the time reversal operator. I understand from the Wikipedia article that requiring that energy stay positive forces $T$ to be represented by an anti-unitary operator. But I don't see how this forces $T^2=-1$. (Or maybe it doesn't force it, it merely allows it?)


Here's another version of my question. There are two distinct double covers of the Lie group $O(n)$ which restrict to the familiar $Spin(n)\to SO(n)$ cover on $SO(n)$; they are called $Pin_+(n)$ and $Pin_-(n)$. If $R\in O(n)$ is a reflection and $\tilde{R}\in Pin_\pm(n)$ covers $R$, then $\tilde{R}^2 = \pm 1$. So saying that $T^2=-1$ means we are in $Pin_-$ rather than $Pin_+$. (I'm assuming Euclidean signature here.) My question (version 2): Under what circumstances are we forced to use $Pin_-$ rather than $Pin_+$ here?


(I posted a similar question on physics.stackexchange.com last week, but there were no replies.)




EDIT: Thanks to the half-integer spin hint in the comments below, I was able to do a more effective web search. If I understand correctly, Kramer's theorem says that for even-dimensional (half integer spin) representations of the Spin group, $T$ must satisfy $T^2=-1$, while for the odd-dimensional representations (integer spin), we have $T^2=1$. I guess at this point it becomes a straightforward question in representation theory: Given an irreducible representation of $Spin(n)$, we can ask whether it is possible to extend it to $Pin_-(n)$ (or $Pin_+(n)$) so that the lifted reflections $\tilde R$ (e.g. $T$) act as an anti-unitary operator.



Answer




The answer for CPT transformations is obvious. A CPT transformation is a 180 degree rotation in the Euclidean theory, so CPT followed by CPT is a 360 degree rotation, which gives you a minus sign on fermionic states, and a plus sign on bosonic states. This is true for all Lorentz invariant theories, or even theories that spontaneously violate Lorentz invariance and keep a CPT unbroken.


The case where you have a T symmetry can be understood from the above. It is not always true that the T symmetry squares to 1 on bosons and -1 on fermions, it is only true for those fermions and bosons whose CP symmetry doesn't square to -1. This is true for normal realizations of CP symmetry, but not for unusual cases, where some parity symmetry can have crazy phases, because it mixes with another discrete symmetry of the theory.


For example, start with electromagnetism with the usual parity, and consider a real pseudoscalar $\phi_3$, which can be made a pseudoscalar by coupling it to a parity invariant fermion using $\gamma_5$, so there is a term in the action.


$$ \phi_3 \bar{\psi}\gamma_5 \psi $$


Consider a second complex scalar $\phi_1$ coupled to $\phi_3$ as follows:


$$ (\phi_1^2 \phi_3 + \bar\phi^2 \phi_3) $$


Now the square of $\phi_1$ needs to get a minus sign under any hypothetical parity transformation. This can be arranged either by multiplying by i, or by swapping the real and imaginary parts of $\phi$. To exclude the latter, you can add a term to the action which is


$$ A(\phi^4 + \bar\phi^{4}) + iB(\phi^4 - \bar{\phi}^4) $$


This is invariant under a parity transformation, but only by sending $\phi\rightarrow \pm i \phi$. The square of parity is -1, so that the T operator on this field $\phi$ squares with a negative sign.


This type of thing is easy to cook up in nonrenormalizable theories. The theory above shows that even renormalizability doesn't exclude these shenanigans.



electromagnetism - Why does electrical impedance have as many parameters as it has?


The impedance of a circuit is written:


$$ Z(R,L,C,f) = R + j ( 2 \pi f L - \frac{1}{2 \pi f C} ). $$


What is the line of thinking that frequency, resistance, capacitance, and inductance are the only parameters? From a physical perspective, why is the definition of impedance parameterized exclusively by these four variables? This model can be verified with experiments, but what goes to show that I couldn't make another circuit element that affects the inductance without changing any of those four properties?




electromagnetism - Why do we use potential for quantizing the electromagnetic field?



For quantizing the electromagnetic field authors go to its potential and then find themselves facing to the problems of degree of freedom from gauge transformation.


Why we can't simply quantize electromagnetic field itself: decompose it to wave planes and promote normal modes to quantum harmonic oscillator?




homework and exercises - Preventing a block from sliding on a frictionless inclined plane


I want to demonstrate what force $F$ you would have to exert on an inclined plane of angle $t$, mass $M$ to prevent a block on top of it with mass $m$ from sliding up or down the ramp. I worked out an answer, but I figure it must be wrong because it doesn't factor the mass $M$ of the inclined plane into the force needed for the block on top to stay still.


Here is my logic:




  • The components normal to the force of gravity on the block on top are $mg \cos(t)$, $mg \sin(t)$

  • In particular, the component down the ramp is $mg \sin(t)$ as can be demonstrated with a visual

  • Thus, we want $F$ to have a component in the direction opposite the vector down ramp with equivalent force so that $F_{\rm ramp (net)} = 0$

  • So we want it to be true that $mg \sin(t) = F \cos(t)$

  • So $F = mg \tan(t)$.


Intuitively, this makes some sense: A steeper slope seems like it would require more force to counteract the component of gravity acting down the ramp.


However, this answer and explanation completely disregards $M$, the mass of the ramp itself.


Can somebody explain where I am going wrong, or if I obtained the right answer, why it is not dependent on $M$?




Answer



I'm not sure I entirely understand the question. This is what I think you're asking; please ignore the rest of this answer if I've misunderstood you.


If the plane is stationary (and I assume there is no friction) then a block on the plane will feel a force down the plane of $mg \space \sin\theta$, so it will accelerate down the plane. If we push the plane so it accelerates at the same rate as the block, then the block won't move relative to the plane. What force on the plane is required to do this?


Assuming I've understood you correctly, the mistake you've made is to assume that the force you apply to the plane, $F$, is the same as the force the block exerts on the plane.


Plane


The force $f$ that the block exerts on the plane is not the same as the force $F$ that you exert on the plane.


When you apply a force F to the inclined plane it starts accelerating at some acceleration $a$ given by $a = F/(M + m)$. If you want the block to stay still relative to the plane the acceleration $a$ along the plane must be the same as the acceleration down the plane due to gravity:


$$ a \space \cos\theta = g \space \sin\theta $$


or:


$$ \frac{F}{M + m} \space \cos\theta = g \space \sin\theta $$



so:


$$ F = (M + m)g \space \tan\theta $$


Tuesday, October 20, 2015

classical mechanics - In a four mass six spring vibration, how is the kinetic energy represented



This is from Hobson, Riley, Bence Mathematical Methods, p 322. A spring system is described as follows (they are floating in air like molecules):


enter image description here


The equilibrium positions of four equal masses M of a square with sides 2L are $R_n=\pm L_i\pm L_j$ and displacements from equilibrium are $q_n=x_ni+y_nj$. According to the text,



"The coordinates for the system are thus x1, y1, x2, . . . , y4 and the kinetic energy of matrix A is given trivially by $MI_8$ where $I_8$ is the 8x8 identity".



What does that mean? The velocity doesn't even appear. How does that relate to energy?



Answer



An engineer would call it the "mass matrix" not the "kinetic energy matrix". The KE is given by $\frac 1 2 \mathbf{v}^T \mathbf{M} \mathbf{v}$ where $\mathbf{v}$ is the vector of the velocity components $\dot x_1, \dots, \dot x_4, \dot y_1, \dots, \dot y_4$ and $\mathbf{M} = M \mathbf{I}_8$ - i.e. an $8\times8$ diagonal matrix with all the diagonal terms equal to $M$.


"Kinetic energy matrix" seems a silly name IMO, because as you said it doesn't fully represent the kinetic energy of the system.



The "stiffness matrix" (or whatever the mathematicians who wrote your book call it!) can similarly be written as an $8\times8$ matrix $\mathbf{K}$, though it's not so simple as the mass matrix. The potential energy stored in the springs is then given by $\frac 1 2 \mathbf{x}^T \mathbf{K} \mathbf{x}$


Reading a textbook or web page on matrix methods for modelling multi-degree-of-freedom (MDOF) systems, written for engineers or physicists rather than for mathematicians, might help to understand the basic ideas.


gauge theory - How is the Chern-Simons action well-defined?


The Chern-Simons action $$ S = \int_M A \wedge \mathrm{d} A + \frac{2}{3}A \wedge A \wedge A $$ is not obviously gauge invariant. It is usually stated that under a gauge transformation, the action changes by a total derivative, which we neglect, plus some additional term which depends on the winding of the gauge transformation around the manifold. This shift of the action by an integer is unimportant if we choose the normalisation of the action correctly.


However, this seems to ignore the fact that in general $A$ is not globally defined. For instance, in the context of Dirac monopoles, we see that we need to define two different gauge fields on two different patches of the 2-sphere, and then tie them together with a 'gauge transformation' (namely the transition function of the fibre bundle on which $A$ is a connection) along the equator.


My question is: how do we even define the above integral in the first place, if there are several 'local versions' of $A$ on part of the manifold $M$?


In particular, even in the Abelian case for which there is no 'winding term' associated with a gauge transformation, the two different gauge fields on a particular overlap still differ by a total derivative term, and when we integrate this across the overlap, we generate boundary terms that don't seem to vanish in general. This seems to imply that our action depends on which local form of the connection $A$ we choose to integrate on the overlaps.



Answer



The expression of the Chern-Simons functional as an integral over a 3-apace is just a shorthand notation. The integration in the Chern-Simons functional differs from the integration of differential forms.


The integration can formulated by means of the theory of Deligne-Beilinson cohomology. (Please, see the following very clear review by Frank Thuillier, and the second chapter of his thesis.


Although the formulation of this theory by the mathematicians Deligne and Beilinson was for completely different intentions, this theory found major applications in geometric quantization and topological field theory. In fact, the need to deeply understand several phenomena such as the Aharonov-Bohm effect, the $B$-field in string theory, etc. led physicists to (partially) rediscover this theory, such as the works by Wu and Yang and Orlando Alvarez.



In Deligne-Beilinson cohomology the gauge potential $\mathbf{A}$ , the Chern Simons characteristic class $\mathrm{CS}(\mathbf{A})$ etc., are examples of Deligne-Beilinson (DB) cocycles. Among other things, the Deligne-Beilinson theory allows an unambiguous integration of (DB) cocycles over homology cycles of a compact manifold $M$. The integral values will be well defined modulo integers.


For smooth manifolds the Deligne-Beilinson cohomology can be constructed by means of the Cech-de Rham bi-complex, clearly described in Orlando Alvarez's reference above. Its basic objects are the Deligne-Beilinson (DB) cochains consisting of a collection of local data: $$\mathbf{\omega}^{(n)} = (\omega^{(n)}_{\alpha}, \eta^{(n-1)}_{\alpha \beta}, \zeta^{(n-2)}_{\alpha \beta \gamma}, …., c^{(0)}_{\alpha \beta , ….}, n^{(-1)}_{\alpha \beta , ...})$$ (the last terms in the decomposition (with de Rham degree of $-1$) are integral Cech cochains)


For example an Abelian gauge potential has the following components: $$\mathbf{A} = (A_{\alpha}, \psi_{\alpha \beta}, n_{\alpha \beta \gamma} )$$ Where, the individual objects in the collections are Cech cochains of local differential forms with decreasing form degree and increasing Cech degree defined on multiple intersections of a good cover of $M$.


The coboundary operator $D$ in the DB cohomology (satisfying D^2=0) is $$ D = \tilde{d} + \delta$$ Where $$ \tilde{d} = (-1)^c d$$ ($d$ is the de-Rham differential, $c$ is the Cech degree) and , $\delta$ is the Cech differential given for example on a Cech 2-cochain:


$$\delta \psi_{\alpha \beta} = \psi_{\alpha \beta} + \psi_{\beta\gamma} + \psi_{\gamma\alpha} $$ The following should be understood: The action of the operator $\tilde{d}$ on global forms (such as gauge field strengths) results in zero and its action on integral Cech cochains is an injection of their numerical values.


The above decomposition allows a natural definition of a cup product just by adding up the different components of all possible wedge products of two BD cochain components with the same de Rham and Cech degrees.


Example: the coboundary of the Abelian gauge potential is given by (in shorthand): $$D\mathbf{A} = (0, \delta A + \tilde{d}\psi, \delta \psi + \tilde{d}n, \delta n)$$ Thus the cocycle condition for the vector potential: $$D\mathbf{A} = 0$$ produces the gauge transformation in the double intersections $, \delta A + \tilde{d}\psi = 0$ (second term),the third term $\delta \psi + \tilde{d}n$ is a reformulation the consistency condition in a triple intersection, the last term asserts that the Cech cocycles are constants on every triple intersection. The integrality requirement of the Cech cocycles is equivalent to the Dirac's quantization condition.


The above relations can be expressed by means of the following tic-tac-toe diagram \begin{array}{c c|c c c c c c c } \Omega^2 & F & F_{\alpha} \\ & & \uparrow_{d} \\ \Omega^1 & & A_{\alpha} & \xrightarrow{\delta} & \delta A_{\alpha}= d\psi_{\alpha\beta} \\ & & & &\uparrow_{d} \\ \Omega^0 & & & & \psi_{\alpha\beta} & \xrightarrow{\delta} & \delta\psi_{\alpha\beta} = n_{\alpha\beta\gamma}\\ \hline & & & & && n\\ & & U_{\alpha}& &U_{\alpha\beta}& &U_{\alpha\beta\gamma}\\ \end{array} As mentioned in Orlando Alvarez's article, using the above relations, the flux of the connection $\mathbf{A}$ can be expressed as: $$\int F = 2 \pi \sum_{U_{\alpha \beta \gamma}} n_{\alpha \beta \gamma}$$ In the case when $ \mathbf{A}$ is a flat connection (as in the Aharonov-Bohm effect): $$dA = 0$$ We have further relations as depicted in the following tic-tac-toe diagram \begin{array}{c c|c c c c c c c } \Omega^2 & 0 & 0 \\ & & \uparrow_{d} \\ \Omega^1 & A & A_{\alpha} & \xrightarrow{\delta} & \delta A_{\alpha}= d\psi_{\alpha\beta} \\ & & \uparrow_{d} & &\uparrow_{d} \\ \Omega^0 & & \psi_{\alpha} & \xrightarrow{\delta} & \psi_{\alpha\beta} + c_ {\alpha\beta} & \xrightarrow{\delta} & 0\\ \hline & & & & && \\ & & U_{\alpha}& &U_{\alpha\beta}& &U_{\alpha\beta\gamma}\\ \end{array}


Giving rise to the possibility of existence of constant Cech 2-cocycles $c_{\alpha \beta}$. There is no integrality constraints on $c_{\alpha \beta}$, since the $\psi$s are gauge transformations of a vector potentials therefore defined modulo $2 \pi \mathbb{Z}$, then also the $c$s are given modulo $2 \pi \mathbb{Z}$


The holonomy of the flat connection can be expressed solely in terms of these constant Cech cocycles (Please see the exact definition of the integral in the sequel) $$\mathrm{hol}(A) = e^{i \oint A} = e^{i \sum_{U_{\alpha \beta }} c_{\alpha \beta }}$$



This relation is a special case of the general statement that the integral of a BD cocycle over a homology cycle is defined modulo $2 \pi \mathbb{Z}$.


The above two cases give rise to the characterizations of the BD first cohomology groups $H_D^1(M, \mathbb{Z})$ in terms of Cech-de Rham cohomology groups $ H^2(M, \mathbb{Z}) $ and the space of one forms on $M$ modulo integers:


$$0 \rightarrow \frac{\Omega^1_{\mathbb{Z}} (M)}{ \Omega^1(M)}\rightarrow H_D^1(M, \mathbb{Z}) \rightarrow H^2(M, \mathbb{Z}) \rightarrow 0$$ The right hand space gives rise to Aharonov-Bohm type connections while the left hand one is responsible for vector potentials of the magnetic monopole type.


Now, to perform the integration of a BD cocycle, one needs to choose a polyhedral decomposition of the integration cycle as Orlando Alvarez did and integrate each component of the BD cocycle on the corresponding polyhedron. This definition is independent of the good cover, polyhedral decomposition or gauge transformations modulo $2 \pi \mathbb{Z}$ .


The integral of the vector potential viewed as BD one cocycle has the form (generalization Orlando Alvarez equation 2.8) $$\oint_{\Lambda} A = \sum_{\alpha} \int_{\Lambda_{\alpha}} A_{\alpha } + \sum_{\alpha \beta} \int_{\Lambda_{\alpha \beta}} \psi_{\alpha \beta} + \sum_{\alpha \beta \gamma} \int_{\Lambda_{\alpha \beta \gamma}} n_{\alpha \beta \gamma} $$ Where $(\Lambda_{\alpha}, \Lambda_{\alpha \beta}, \Lambda_{\alpha \beta \gamma})$ is a polyhedral decomposition of the trajectory $\Lambda$.


(Actually the last term is not needed since $n_{\alpha \beta \gamma}$ are integers)


Similarly, the integral of the Chern Simons class expressed by means of the BD components of the vector potential is given in Thuillier review equation 45: $$\int_M \mathrm{CS}(\mathbf{A}) = \sum_{\alpha} \int_{M_{\alpha}} A_{\alpha }\wedge dA_{\alpha} + \sum_{\alpha \beta} \int_{M_{\alpha \beta}} \psi_{\alpha \beta} \wedge dA_{\beta} + \sum_{\alpha \beta \gamma} \int_{M_{\alpha \beta \gamma}} n_{\alpha \beta \gamma} A_{\gamma} + \sum_{\alpha \beta \gamma\delta} \int_{M_{\alpha \beta \gamma\delta}} n_{\alpha \beta \gamma} \psi_{\gamma \delta} $$


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 \...