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Anisotropic vs Isotropic Harmonic Oscillator IS there any quantum mechanical process which can take over an anisotropic commensurate harmonic oscillation to an isotropic one?
Mathematically, this kind of transformation is available.
http://dx.doi.org/10.1063/1.1666379
But, is there any physical situation where this kind of transformation can take place?
Any response towards this questions will be highly appreciated.
| Ultracold atoms, and in particular Bose-Einstein condensates, are frequently trapped in magnetoptical potentials which have a quadratic form. If you had a two-dimensional system and changed the trap frequencies in the two directions you would be able to control the anisotropy of the potential very precisely (and of course when the frequencies are equal you would have an isotropic quadratic potential). You can see some examples of how trap potentials can be realized here https://doi.org/10.1016/bs.aamop.2017.03.002
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On the Euclidean action for QCD The Euclidean action for QCD reads, (see e.g., Eq. (45) in "ABC of instantons" by Novikov, Shifman, Vainshtein, and Zakharov)
$$S_E=\int d^4 x\left[\frac{1}{4}G^a_{\mu\nu}G^a_{\mu\nu}+\psi^\dagger(-i\gamma_\mu D_\mu-im)\psi\right].\tag{45}$$
Here $\gamma_\mu$ are gamma matrices in Euclidean space, i.e., $\{\gamma_\mu,\gamma_\nu\}=2\delta_{\mu\nu}$. Sometimes, people also use the notation $\bar\psi$ to replace the above $\psi^\dagger$.
Now, we know that the operator $-i\gamma_\mu D_\mu$ is hermitian but $-im$ is not (actually it is anti-hermitian). Then the fluctuation operator as a whole, $-i\gamma_\mu D_\mu -i m$ is not hermitian. Won't this be a problem? For instance, the eigenvalues for this operator are in general complex.
How do people treat this in,for instance, lattice QCD?
| I think you might be missing something due to the notation used in that article. In Lorentzian signature the quantity $\psi^\dagger \gamma_0 \psi$ is a scalar. Going to Euclidean signature with $\gamma_0 = \gamma_4$, we have that $\psi^\dagger \gamma_4 \psi$ is again a scalar. However, note that the paper defines a new quantity see Equation (44) to distinguish the Euclidean space quantities from Lorentzian quantities:
$$ \hat{\bar\psi} = i \bar \psi = i \psi^\dagger \gamma_4 \qquad \hat \psi = \psi $$
Hence,
$$ \bar \psi \psi = -i \hat{\bar \psi} \hat \psi \implies (-i) \times \hat{\bar \psi} \hat \psi = \bar \psi \psi \quad \text{ is hermitian}$$
However, the easiest answer the question is that in Equation (45) they establish the relationship between the Euclidean and the Lorentzian Lagrangian. Since the Lorentzian Lagrangian is hermitian so must be the Euclidean one since by definition $\mathcal{L}_E$ is constructed from $\mathcal{L}_L$ by sprinkling $i$'s so that schematically
$$ \mathcal{L}_E[\phi_E] = \mathcal{L}_L[\phi_L]$$
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Will state of water change in certain condition? Imagine I have an iron tank with a $20~\mathrm{pm}$ hole on it. Then I completely fill it with water and use a pump to get the water out of that hole. What will come out, water or gas?
| Pressure alone does not break chemical bonds. Expose liquid water suddenly to a vacuum and it will vaporize as a molecule; it will not dissociate into atoms. This is true regardless of the initial pressure of the water.
The hole that you have made is too small for water molecules. It is even too small for a hydrogen atom, which is the smallest atom possible (with a radius of about 50-60 pm).
The hole size that you propose is more akin to the sizes of interstitial sites in solid lattices. For examples of the lattice structures of iron, see the links below.
https://en.wikipedia.org/wiki/Allotropes_of_iron
https://www.tf.uni-kiel.de/matwis/amat/iss/kap_4/backbone/r4_2_2.html
At this point, you should recognize that we cannot make a contiguous "hole" that is 20pm in radius or diameter through an iron wall. Indeed, this because of the way that atoms pack in solids, we cannot make such a hole through any solid wall that is thicker than two atoms stacked on in layers. The "holes" through the one layer of atoms are blocked by the atoms that cover in the next layer.
In essence, you have a solid tank. Nothing will transport through the wall.
| {
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Are there any satellite orbit earth perpendicular to the sun and what is the name of this kind of orbit? I think we could put satellite to orbit earth in such a way that it always see the sun. Which is orbiting along the path of earth orbiting the sun, like a wheel perpendicular to the sun
I don't know the specific name of this type of orbit so I can't find detail about it, are there any satellite orbit earth this way?
| This is called a “sun synchronous” orbit.
If the Earth were a perfect sphere, orbits around it would have a fixed orientation in space. As the Earth goes around the Sun, an orbit would have the right orientation only twice a year.
But the Earth isn’t a perfect sphere: it bulges in the middle. The gravity of that bulge pulls inclined orbits toward the equator. That puts a torque on the angular momentum of the orbit. Much like an inclined top, this torque caused the orbit to precess, rotating in space.
With just the right inclination, this precession will happen once per year so that the satellite’s orbit stays properly aligned with the Sun.
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Why are topological insulators interesting? Why are topological insulators interesting? Meaning, why should an undergraduate or graduate student start working on this? What are the technological applications? I am not sure how to answer these questions and wikipedia does not help since it does not explain why so many people work on this. I am especially interested in applications through photonics but any answer would be appreciated. Thanks in advance.
| This is to complement yu-v answer.
As this is not an "easy" topic (I mean, the effects are not quite seductive at simple sight compared to the sexy name and invariants are always presented in research papers in, to my taste, an obscure manner), a good reference is an appropriate starting point.
One suggestion to start is Tkachov. It is very basic and you only need the courses of Quantum Mechanics; perhaps at graduated level to have a better grasp of the contents.
Asbóth is also a good one, but it appears that contains some typos in the derivations.
The fun could start with Prodan's book, but it's aimed for a mathematical-mature audience.
I would even dare to recommend Kitakawa review. It is aimed for topological effects in quantum walks, but the rudiments of this latter are quite straightforward.
Perhaps it is wise to survey first at the literature and then come back and ask a more precise question on the topic.
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A doubt in trigonometric approximation used in the derivation of mirror formula The following text is from Concepts of Physics by Dr. H.C.Verma, from chapter "Geometrical Optics", page 387, topic "Relation between $u$,$v$ and $R$ for Spherical Mirrors":
If the point $A$ is close to $P$, the angles $\alpha$,$\beta$ and $\gamma$ are small and we can write
$$\alpha\approx\frac{AP}{PO},\ \beta=\frac{AP}{PC}\ \ \ \text{and} \ \ \gamma \approx\frac{AP}{PI}$$
As $C$ is the centre of curvature, the equation for $\beta$ is exact whereas the remaining two are approximate.
The terms on the R.H.S. of the equations for the angles $\alpha$,$\beta$ and $\gamma$, are the tangents of the respective angles. We know that, when the angle $\theta$ is small, then $\tan\theta\approx\theta$. In the above case, this can be imagined as, when the angle becomes smaller, $AP$ becomes more and more perpendicular to the principal axis. And thus the formula for the tangent could be used.
But, how can this approximation result in a better accuracy for $\beta$ when compared to $\alpha$ and $\gamma$? I don't understand the reasoning behind the statement: "As $C$ is the centre of curvature, the equation for $\beta$ is exact whereas the remaining two are approximate." I can see the author has used "$=$" instead of "$\approx$" for $\beta$ and he supports this with that statement. But why is this so? Shouldn't the expression for $\beta$ be also an approximation over equality? Is the equation and the following statement really correct?
| For the angle $\beta$ the author is apparantly thinking of AP as being the arc length along the circle rather than the length of the straight line joining A to P. Then
$$
{\rm arclength}({\rm AP}) = \beta R
$$ exactly.
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What determines if a photon is displayed as light or not? A photon is the particle of light. But a photon is also a force carrier and plays a role in many other situations. What I don’t understand is what determines if it shows up as light or not? Is it just a high concentration or pattern? Why don’t photons show up in all situations?
| Photons are present in all electromagnetic interactions. Sometimes they are virtual (e.g., in the case of electrostatic interactions), and sometimes real (e.g., in the case of electromagnetic radiation). For a photon to be perceived as light, it must a) be real; b) have the right range of wavelength (or frequency, or energy); and c) must be detected by your eye (or another light detector).
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Why does light spread out? So we know the light that's emitted from a torch (flashlight) must be moving in straight lines, so why does it spread out when moving? Why does it cover larger area?
|
So we know the light that's emitted from a torch (flashlight) must be moving in straight lines, so why does it spread out when moving? Why does it cover larger area?
As the comments says, straight lines are not parallel lines. They can have different directions, and they do:
The flash light, (or any light source from heated filaments) can be considered as an accumulation of point sources.
Point sources of light can be modeled by rays that are opening as $1/r^2$
The light is an overlap of point sources, and depending on the geometry of the flashlight and the filament, it is evident that a multitude of directions will appear.
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What are $a$ and $a^*$ called in the context of a classical harmonic oscillator? Consider a harmonic oscillator defined by the coupled differential equations
\begin{align}
\begin{split}
\dot{X} &= \omega Y \\
\dot{Y} &= - \omega X \, .
\end{split} \tag{1}
\end{align}
Defining new variables $a = X + i Y$ and $a^* = X - i Y$, produces a new uncoupled system of equations
\begin{align}
\begin{split}
\dot{a} &= - i \omega \, a \\
\dot{a}^* &= i \omega \, a^* \, .
\end{split} \tag{2}
\end{align}
In classical physics [1] (or just in the mathematical context of this transformation used to solve a pair of coupled differential equations) what are the variables a and $a^*$ called?
[1]: In the context of quantum mechanics, the variables $a$ an $a^*$ would in fact be operators and would be called the "raising" and "lowering" operators.
| I would call $a$ and $a^*$ the complex amplitude of the oscillator. Or I guess $a$ is the complex amplitude itself and $a^*$ is the complex conjugate of the amplitude but the distinction is unimportant as they carry the same information (just like in the quantum case).
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Internal force disintegrating a solid body? Let $M$ be a block on a frictionless surface. Now let us mentally divide (not physically) the block into 2:1 ration (i.e $1/3$ of the left be called $M_1$ and $2/3$ right be called $M_2$). So $M_1$ applies force $F_1$ on $M_2$ and $M_2$ applies force $F_2$ on $M_1$ and by 3rd law they are equal. Hence acceleration of $M_1$ would be $2a$ and that of $M_2$ would be $a$. Shouldn't this deform the block?
| Actually you haven't truly addressed as to what kind of material we are dealing with over here, so let me dissect it into two types:
*
*Totally rigid
*Jelly like
Totally Rigid Bodies
For objects made up of materials behaving as such as soon as an internal force tries to deform the object an opposite restoring force is generated which balances it. So here you were missing a restoring force in your calculation.
Jelly Like Bodies
In these when an internal force acts on the particle then a deformation is caused which leads to greater or lesser volume than originally it would have.
Also the forces aren't unidirectional and hence the deformation occurs in all direction.
$$\underline {\text {Reality}}$$
No real body is a perfect example of both the given case and hence there always is some kind of deformation.
A significant effect of this can be seen in stars (main internal force being gravity).
[Note: It must be noted that since internal forces cancel each other therefore the center of mass doesn't accelerate even a bit(unless an external unbalanced force acts). ]
Suggested Materials:
*
*Does a particle exert force on itself?
*Neutron Stars
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How can atomic configurations represent excited states of atoms? My lecture notes on condensed matter physics talk about pseudopotentials of atoms where the core electrons are replaced by an effective potential. This is in the context of DFT. In the lecture notes, my lecturer talks about transferability, i.e. the ability of the pseudopotential to work in various atomic configurations. My lecturer proposes a test of the transferability of a pseudopotential in the following way: Devise a series of atomic configurations representing (approximations to) excited states of the atom; then compute the energy difference between them for both the all-electron case and the pseudopotential approximation.
What is it meant by that atomic configurations represent excited states of the atoms? I thought only the electrons could be excited, and there should be different excited states for all configurations. Can you explain what is meant by this?
| It is true that in the context of pseudopotential theory (which has a much broader scope than DFT) people use the term "atomic configuration" in a slightly ambiguous way.
Strictly speaking what they are speaking about is the electronic configuration ot the atom. For instance, the ground state of a neutral sodium atom is $1s^22s^22p^63s^1$ or, ore briefly $[{\mathrm{Ne}}]3s^1$. However the concept of transferability of pseudopotentials has to do with the possibility of an accurate description of the interaction between atom and valence-electrons even for electronic configurations different from the reference atomic ground state. This is a key requirement if one has to describe properly the electronic states in condensed phases where the atomic configuration, i.e. the local environment around an atom induces electronic configurations different from the isolated atom electronic ground state. For example, a $[{\mathrm{Ne}}]3p^1$ excited state, or, less physical, but often used in the context of pseudopotentials, a fractional occupation like $[{\mathrm{Ne}}]3s^{0.8}3p^{0.2}$.
In this sense, in this context atomic configuration becomes synonymous of electronic configuration.
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How do gluons bind the quarks together within the hadrons and mesons? I was trying to know about the strong nuclear force within the nucleus and the books and websites told me that it is the gluons which carry the force, and now I am curious how the gluons carry this force
| As in your profile you say you are a ninth class student in India, it means that your physics background is still at understanding the classical physics level.
The strong nuclear force and the strong force of particle physics belong to the quantum mechanical studies, which I doubt it is taught in your year.
In classical physics force can be defined as the change in momentum in time, $F=dp/dt$. At the level of nuclei and particles, this is the only reasonable definition of a force, because any interaction between quantum mechanical entities happens by the exchange of an elementary particle which transfers energy and momentum and quantum numbers ( which you will eventually learn about) between the quantum mechanical entities, be they nuclei or particles.
Think of two boats in a quiet sea. If one boat throws a ball to the other, they interact with the $dp/dt$ the ball carries. That is how forces are to be defined in the micro level of nuclear and particle physics. In this answer here , there is an illustration of the boats with a boomerang thrown, to show an analogy with attractive forces. ( the feynman diagrams are for later in your education, if you still like physics and pursue it)
The elementary particles carrying the force to first order in most interactions are:
the photon, for electromagnetic interactions
the Z and W for weak interactions
the gluon for strong interactions
For nuclei, which are bound states of protons and neutrons, the strong nuclear force, is a spill over force from the strong force which has the gluon as a force carrier . Interaction of nuclei can be modeled by having the composite hadron particle pion as a force carrier.
Protons (and neutrons and other hadrons) are an entity with three valence quarks held together by an innumerable number of gluons as force carriers, because gluons are very attractive to quarks and to each other. ( the word comes from glue)
Here is an illustration of a proton:
The three valence quarks are lost in the plethora of gluons and quark antiquark pairs created by the mechanism of strong interactions.
The strong force is so strong that new quantum mechanical models have been developed to explain the hadrons ( bound states of quarks by gluons), called QCD on the lattice.
| {
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How was the mass of the moon first calculated? How was the mass of the moon first calculated? How do we measure it now?
| Although this might be more suitable for the History of Science and Mathematics page, I thought I would add a few brief details on the physical details behind these measurements.
Newton
The first to calculate (or, estimate) the mass of the moon was probably Newton himself. As mentioned by Baso, Newton used measurements of tides to find the ratio of the lunar and solar masses. This is realised through the tidal force equation:
$$ f_{S} = \frac{2G M_S M_E}{R_E^3},$$
$$f_{M} = \frac{2G M_M M_E}{R_E^3}. $$
Where we take the absolute value. Then Newton considered the case of when the moon was aligned with the sun, and when it was not, to calculate the relative effects. The details can be found in Morin (chapter 10.3). This method, however, estimated a lunar mass significantly larger than the real value ($\approx \frac{M_E}{41}$), even resulting in the Earth-moon centre of mass lying outside of Earth. As Kevin Brown mentions in his website (which I definitely recommend):
The problem is that the height of tides near various land masses is a complicated function of many different factors and resonance effects, so it can’t be used to give a simple estimate of the tidal forces.
Pre-Moon Landing
Prior to the Apollo missions, measurements had to be taken indirectly, and one of the method was to use the fact that the Earth-moon system orbits their common centre of mass, which causes a periodic "wobble" in Earth's solar orbit.
It was observed that every month there is a 6.3'' parallax angle of the sun's position, which repeats periodically. So it became obvious that this could be used to determine the lunar mass. This indeed turned out to be a good method, and was the method until the moon landings.
With the parallax angle known, the distance of the Earth to the CoM, $D_E$ is:
$$ D_E = (1.5 \times 10^{11}) \times \tan(6.3'') $$
Let $D$ be the Earth-moon distance, then:
$$ \frac{M_E}{M_M} = \frac{D}{D_E} -1 = 82.2... $$
Knowing the mass of the Earth from Kepler's third law gives us the mass of the moon.
References:
*
*David Morin - Introduction to Classical Mechanics
With Problems and Solutions (ch 10.3)
*https://www.mathpages.com/home/kmath469/kmath469.htm
*https://en.wikipedia.org/wiki/Parallax
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What is the cross-section size of a photon?
How "wide" is a photon, if any, of its electromagnetic fields? Is there any physical length measurement of these two orthogonal fields, $E$ and $M$, from the axis of travel? When a photon hits a surface, and is absorbed by an electron orbital, this width comes into play, as there could have been more than one electron that could have absorbed the photon?
This is not my personal query, I found it while I was surfing the web, and found it interesting, so I posted it here.
| For a back of the envelope calculation, it is often useful to associate with a particle it's Compton Wavelength. This is generally the most accurately you can know an object's position due to the uncertainty principle.
Other related length scales are usually more useful, factoring in features of the particular interaction in question. This can be expressed as a Mean Free Path, basically the flight time between interactions or a Scattering Cross-Section, measure of the rate of interactions.
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Is Griffiths simply wrong here? (Electrostatic Boundary Conditions)
In the above illustration, shouldn't $E_{above}$ and $E_{below}$ be in opposite directions? If not, how did Griffiths end up the following equation? From the above directions, shouldn't the flux add up?
| To support that Griffiths is correct, I would start with quoting from Bob Jacobsen's answer:
He’s picked a sign convention where the field upward is positive everywhere. That means that $E_{below}$ is defined such that a positive value means "the E vector points up" and a negative value means "the E vector points down".
Adding to this, I would also consider the convention that the direction of the area vector is positive along the upward direction and vice-versa.
Now, let's start from Gauss' law which says that $\boxed{Q_{enc}=\Phi\cdot\epsilon_0}$, where $Q_{enc}$ is the charge enclosed by the Gaussian pillbox and $\Phi$ is the electric flux.
Applying this to the Gaussian pillbox in the figure, we can write
\begin{align*}
Q_{enc} &=\epsilon_0\left[(E_{above}^\perp)\cdot(A)+(E_{below}^\perp)\cdot(-A)\right]\\ \implies\sigma A&=\epsilon_0A\left[E_{above}^\perp-E_{below}^\perp\right]\\ \implies\frac{\sigma}{\epsilon_0}&=E_{above}^\perp-E_{below}^\perp
\end{align*}
Note that the area vector along $E_{above}^\perp$ is positive and the area vector along $E_{below}^\perp$ is negative due to the chosen convention.
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What is baking and what are the effects? In their experiment, Davisson and Germer had to bake the nickel mass because it was oxidized.
What is baking and what does it do to the lattice of the metal?
| The primary effect of the baking was basically to boil off the oxygen that had built up at the nickel surface. When the metal is exposed to air, the oxygen reacts with the surface atoms, and this eventually produces a layer of metal oxide that is typically tens or hundreds of atoms thick. However, the oxygen remains relatively volatile, so by heating the metal, Davisson and Germer got rid of much of the oxygen, producing something closer to a pure nickel crystal surface.
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I dont understand the work equation I don't understand how work = force * displacement as if a force of say 1 Newton was to be applied to two objects of different mass until the object reached a displacement of say 1 meter, surely the object of less mass would displace 1 meter in less time (due to faster acceleration) meaning the force would be applied for less time resulting in less work. I know there is something fundamentally wrong with my understanding of this but I'm not sure exactly what. any help would be greatly appreciated.
| Let’s take a little look at the definition of work.
The work done by a force in moving some system through a certain displacement is defined to be the force times displacelment.
Let’s take gravity as an example.
If you are holding a rock at a certain height and you let it fall, the gravitational force will be acting during the motion of the rock (it is by the existence of such force that the rock falls!)
You can imagine doing this experience in two different ways, one, in vacuum, and two, in the presence of an atmosphere. Of course there will be a different in the time the rock takes to reach the ground, because in the presence of an atmosphere you will have to account the air drag.
Although, gravity does the same amount of work in both cases, provided you let it fall at the same height and you are doing both experiences in the same place (so the acceleration due to gravity will be the same).
The work done by the force will be the same! Same force, same displacement.
For time, we can consider the impulse of the force which is defined as the force times the time interval (this time interval is the amount of time the force is acting on the system).
By that you can take the conclusions you stated in your question, except that the amount of work done is the same!
I hope that this help you! :)
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How are topological qubits encoded in the Majorana-fermion-based platform for quantum computing? Where is “the two level system” of a topological qubit encoded in the Majorana-fermion-based platform of quantum computing?
If the Hamiltonian in a topological quantum field theory is absent (H=0), then what dictates the evolution of a topological system in the circuit model of quantum computation?
| There are many proposals for using Majorana zero modes to encode a qubit; one of the most popular is called the hexon, discussed in this paper: https://arxiv.org/pdf/1610.05289.pdf.
The main idea here is that a one-dimensional topological superconductor hosts Majorana zero modes at its ends. Two Majorana modes combine into a Dirac fermion, so the ground state is doubly degenerate; either the fermion built from the two Majoranas at the ends is occupied or unoccupied. Having $2N$ Majorana zero modes gives a Hilbert space of dimension $2^N$. The architectures discussed in many of these papers are variants on this concept designed to protect the system from unwanted effects like quasiparticle poisoning.
So the qubits in the Majorana models consist of occupying or deoccupying the fermions created by these Majorana zero modes. However, the Hamiltonian is not $0$ -- we're not dealing with a TQFT, we're dealing with a condensed matter system. To implement a quantum circuit we perform an adiabatic evolution of the Hamiltonian, taking it along some closed path in parameter space. While the Hamiltonian returns to its original form, the states evolve into one another. If the zero modes are kept far apart during this evolution, this evolution is topologically protected; it's a quantum gate.
If we did want to use TQFT language, you would consider your surface punctured with several holes; these holes are defects which bind the zero modes. Protected operations then generate a representation of the braid group.
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Is the car braking time formula $ T = v / (\mu_s \, g) $ valid only for uniformly accelerated motion? I'm wondering if the car braking time formula is valid only for uniformly accelerated motion.
$$ T = \frac{v} {\mu_s \, g} $$
with $ v $ average speed, $ \mu_s $ static friction coefficient between the wheel and the ground, $ g $ gravitational acceleration on the earth.
I derived it in this way ($ F_{s, max} = \mu_s \, N = \mu_s \, m \, g $ maximum static friction force; $ N $ normal force, $ m $ car mass):
$$ F_{s, max} = m \, a $$
$$ \mu_s \, m \, g = m \, \frac{v} {T} $$
$$ T = \frac{v} {\mu_s \, g} $$
where $ a $ is the average acceleration of the car.
Thank you in advance.
| While the derivation you've used assumes uniform acceleration, it is also possible to show that the $T$ you have found is a lower bound on the stopping time of the car, even without assuming uniform acceleration. Roughly speaking, even if the acceleration varies with time, its magnitude can be no greater than $\mu_s g$, which implies that the stopping time can be no less than the $T$ you have found.
More formally: assume the frictional force and the acceleration vary with time. The magnitude of the frictional force $F_\text{fr}(t)$ is no greater that $\mu_s$ (the coefficient of static friction) times the normal force $N$:
$$
|F_\text{fr}(t)| \leq \mu_s N = \mu_s m g
$$
assuming the car is on level ground. This means that the acceleration of the car is bounded by
$$
|a(t)| = |F_\text{fr}(t)/m| \leq \mu_s g.
$$
If the car has a positive velocity $v$ initially, then as the car brakes we have $a(t) < 0$, and so $a(t) > - \mu_s g$. Using calculus, we then have
\begin{align*}
\Delta v &= \int_0^T a(t) \, dt \geq \int_0^T (- \mu_s g) \, dt \\
0 - v &\geq - \mu_s g T \\
\mu_s g T &\geq v \\
T &\geq \frac{v}{\mu_s g}.
\end{align*}
Thus, no matter what the car does, it will not be able to stop more quickly (i.e., in less time) than the $T$ you have calculated assuming uniform acceleration.
| {
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Why does an open ended wire cause the load to the oscillator to increase? Using an ammeter between the wall and signal generator, I noticed that when the lead from a single channel is attached to the signal generator, the power into the signal generator increases by about 100 milliwatts.
When the wire leads are removed from the signal generator, the input power into the signal generator decreases by 100 milliwatts.
Why would an open ended wire cause a load? Is there a way for that wire not to add to the input load?
The lead wire from the signal generator is not attached to anything on the open end.
| It appears that someone who is familiar with RF answered my question elsewhere. His answer was that the attached BnC cable to the signal generator is not an open circuit as I has assumed but because of the Mhz frequency, it actually appears to the signal generator as a mismatched circuit (impedance mismatch) drawing a load.
The signal generator is rated at a 50 ohm impedance and the BnC connector is also 50 ohms. So perhaps it simply sees it as a matched impedance load?
| {
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Why does a weather vane arrow point in the direction of the wind? It seems that a weather vane will rotate in order to minimize energy and thus orient itself parallel to the wind.
What I do not understand is why it is implied that the weather vane arrow should point in the direction of the wind.
I do not understand why the arrow pointing in the opposite direction of the wind is also not a minimum energy solution.
| When the wind blows perfectly parallel to the wind vane's long axis, there is no rotational force on the wind vane. When the wind direction is not parallel to the long axis of the wind vane it will exert a turning force on the vane until the wind vane is parallel to the wind direction.
The force of the wind on any part of the wind vane depends on its (surface area) X (its distance from the fulcrum/rotational axis). Any area of the wind vane that is exactly on the fulcrum will result in no rotation.
In designing your wind vane therefore, the part that you desire to be pointing away from the direction from which the wind is blowing should have the greatest product of (surface area) X (distance from the fulcrum).
In the image posted with your question, the rooster and the arrow head at their positions and sizes are merely decorative. A wind vane with only a tail of similar shape to that in your image, attached to the fulcrum would work just fine. Users would need to understand that it is pointing in the direction in which the wind is going, not where it is coming from. It also follows that if the arrowhead were large enough in area and far enough from the fulcrum, then it would point away from the direction from which the wind is coming.
| {
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Properly reporting instrument readings As a first approximation, the uncertainty ($\delta X$) associated to a mensurand can be expressed as $\delta X= \Delta X / 2$ with $\Delta x$ being the resolution of the instrument.
There is also a recommendation indicating that the numerical expression of the result and its uncertainty saying that "Results should be rounded to be consistent with the uncertainty given" (EURACHEM/CITAC Guide CG 4).
So, if we have an instrument like a ruler with a resolution of 1 arbitrary unit ($\Delta X =1$ ) and when measuring an object, the end of said object is very close to the 3 unit mark, why should the reported value be 3 $\pm$ 0.5
and not 3.0 $\pm$ 0.5 or even 3.1 $\pm$ 0.5?
| This is because the first part of the value represent the significant digits that are certain. The answer must contain extra significant digits on top of the ones that are certain. Having taken the reading, you are 100% sure that the ones digit is $3$ and not $4$ or $2$. Writing $3.0$ would mean that you are sure of the length to be $3.0$ which you are clearly not. Same goes with writing $3.1$
| {
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Is Snell’s Law valid in this case? When light travels in a perpendicular path from one medium to another medium of different optical density, is Snell’s law valid?
$\sin i$ and $\sin r$ are both 0, right?
So it isn’t valid.
Is this correct?
| In this case where the incident angle is $0^\circ$ to the interface and thus from $$ n_{1}\sin{\theta_1}=n_2\sin{\theta_2}$$ we get that $\theta_1 = \theta_2$ I'd simply instead reason and say that Snell's law is not applicable in this case as it'd result in the forbidden $\frac{0}{0}$ formulation. Instead as Snell's law, if we consider these cases, it might be more valid to refer to it as Snell's model.
| {
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A formula which relate spring constant to its physical properties like length and area of cross section I know spring is a mechanical analogue of capacitor. And length of a spring is equivalent to plate seperation of capacitor. So what is the mechanical equivalent of area of cross section of capacitor in spring? And is there a similar formula like $C=\epsilon_0 {A \over d}$ for spring constant $k$.
| A simple example is that the stiffness of an elastic rod is
$$
k = \frac{EA}{L_0},
$$
where $A$ is the cross-sectional area of the rod, $L_0$ is its unstretched length, and $E$ is the Young's modulus of the material of the rod.
Note, however, that this is only for a rod that stretches uniformly along its length; think of a rubber band rather than a coiled spring. When a spring stretches, the coils of wire bend slightly relative to each other, rather than the wire simply stretching.
| {
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Problem involving gravitational potential energy
Two uniform solid spheres of equal radii $R$, but mass $M$ and $4M$ have a centre separation of $6R$. The two spheres are held fixed. A projectile of mass $m$ is projected from the surface of the sphere $M$ and towards the second sphere along the line joining the centres of the two spheres. Obtain an expression for the minimum speed $v$ of the projectile so it reaches the surface of the second sphere.
I was looking at the answer of this problem and noticed that the neutral point (i.e. the point where the forces between the two spheres exactly cancel out) had been calculated and conservation of energy had been applied at the neutral point $N$ and at the surface ($E_s$ being the the mechanical energy at the surface).
$$E_s= \frac 12 mv^2-\frac {GMm}R-\frac{4GMm}{5R} $$
$$E_N= -\frac{GMm}{2R}-\frac{4GMm}{4R}$$
Equating $E_s$ and $E_N$ gives $v=\sqrt\frac{3GM}{5R}$.
What I did not understand was that while writing the mechanical energy at the neutral point $N$ they assumed the kinetic energy of the projectile to be zero.
If the kinetic energy of the particle at $N$ is zero, implying that the particle is stationary, then how would it reach the surface of the second sphere since there is no force pulling it towards $4M$?
| The null point...
... it is unstable i.e. the force on that point is indeed zero but what if we displace the object by $\mathbf{\delta x}$?
What I mean to say is that the system is in unstable equilibrium.
The force on null point $N$ may be zero but the potential energy at the point is maximum suggesting unstable equilibrium.
What is unstable equilibrium?
Assume a pendulum in two positions
(i) normal equilibrium where hinge is above free end
(ii) equilibrium where hinge is below free end and it is perfectly inverted
The first case is stable equilibrium and the second state is unstable equilibrium.
In unstable equilibrium if we displace the object by very small displacement it will not return to its original state or position.
Similarly
In your question we just need the object to reach the unstable point after that even a small negligible displacement toward other sphere (4M) will result in the gravitational force of 4M to take over.
| {
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Is there any book that treat time-dependent perturbation theory with rigorous mathematics? I am searching for rigorous mathematics books or notes for time dependent perturbation theory.
For introductory quantum mechanics there is the excellent book spectral theory and quantum mechanics by Valter Moretti and for time-independent perturbation theory there is the book Perturbation theory for linear operators from Kato which is a good book.
Is there any book that treat time dependent perturbation theory
with rigorous mathematics?
| During my bachelor, I found quite useful the book Quantum Mechanics Vol.2 by Cohen-Tannoudji, which treates the basics of time dependent perturbation theory really well in my opinion, and with some examples. However, there are other books more focused on a time dependent analysis of Quantum Mechanics like Introduction to Quantum Mechanics: A time dependent perspective by David J. Tannor, which is also quite good.
I hope you find these references useful!
| {
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Is it possible to land on sun? The Sun is, of course, quite hot. A space probe built by normal materials probably would melt or burn even it’s still quite far from the Sun.
However, from a theory point of view, is it possible to build a space probe that could land on the Sun?
| If, by "land on the sun", you mean "land of the surface of the sun", then you might consider the following from https://solarsystem.nasa.gov/solar-system/sun/in-depth/:
Surface
The surface of the Sun, the photosphere, is a 300-mile-thick
(500-kilometer-thick) region, from which most of the Sun's radiation
escapes outward. This is not a solid surface like the surfaces of
planets. Instead, this is the outer layer of the gassy star.
We see radiation from the photosphere as sunlight when it reaches
Earth about eight minutes after it leaves the Sun. The temperature of
the photosphere is about 10,000 degrees Fahrenheit (5,500 degrees
Celsius).
| {
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Why does $\phi=\phi^*$ imposed on complex scalar field Lagrangian miss out $1/2$ factors? If we require the reality condition $\phi=\phi^*$ on the Lagrangian for a complex scalar field is $$\mathcal{L}=(\partial^\mu\phi^*)(\partial_\mu\phi)-m^2(\phi^*\phi),$$ two degrees of freedom $\phi$ and $\phi^*$ is reduced to one. For consistency, I would expect it should give $$\mathcal{L}=\frac{1}{2}(\partial^\mu\phi)(\partial_\mu\phi)-\frac{1}{2}m^2\phi^2.$$ But this procedure misses the $1/2$ factors. Why? Did I mess up some normalization?
| The standard convention is to divide each term in the Lagrangian with its symmetry factor. Therefore the kinetic term for a real (complex) scalar field is
with (without) a symmetry factor $\frac{1}{2}$, respectively. A complex scalar field $\phi= \frac{\phi^1+i\phi^2}{\sqrt{2}}$ can be viewed as 2 real scalar fields.
| {
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60 kg on earth is 60 kg on the moon I'm writing a trivia quiz and intend this question, which dates from a high school physics test I took in 1972.
An astronaut tips the scale at 60 kg while on earth, what will she be if she steps on the scale on the moon? Answer 60 kg.
Kg measures mass, which is constant.
The question is not in pounds, or ask about weight. Yes, it is a trick question, but I think it is an entertaining one. The point is, I've looked around the internet and the discussion all focus on in common usage, weight is mass and nobody knows what a Newton is.
I'm certain of my answer, but I'd like to be ready for some blowback.
| Unfortunately, scales do not measure mass. They measure the force applied by the mass, under the influence of the local gravity. The fact that they are labelled in "mass" is irrelevant. All that happens is that the readout is "adjusted", so that 588 N is shown as 60 kg.
Hence, on the moon your scales will show about 10 kg.
EDIT
I disagree with the comments: even if you are using balance scales, you are measuring force, not mass. Consider the following 2 graphics
The first shows a normal balance, the second an unequal-arm balance. In the second image, it makes no sense to say that 30 kg balances 60 kg. The balance is achieved by having equal torque (moment of rotation) on both sides. Torque is given by $\tau=r\times F$ where $F$ is force, not mass. Hence balance is achieved by using forces of 588 N and 294 N.
| {
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Derivation of the Euler-Lagrange equation for fields In the derivation of the Euler-Lagrange equation in Peskin and Schroeder, p.15, we have:
$$\delta S = \int d⁴x \left[ \frac{\partial L}{\partial \phi}\delta \phi -\partial_\mu \left(\frac{\partial L}{\partial (\partial_\mu) \phi}\right)\delta \phi + \partial_\mu \left(\frac{\partial L}{\partial (\partial_\mu \phi)} \delta \phi \right)\right]. \tag{2.2}
$$
They argue that the last term can be turned into a surface integral, and since the initial and final field configurations are given, $\delta \phi$ is zero at the temporal beginning and end of this region. If we restrict our considerations to deformations $\delta \phi$ that vanish on the spatial boundary of our region as well, then the surface term is zero.
So I guess they used the divergence theorem on the last term, but why doesn't this argument apply to the second term as well?
| The last term is a total divergence. The second one is not.
| {
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From photon flux to number of photons inside a cavity Consider a light beam with photon flux $\Phi$ entering a cavity. The latter consists of two mirrors, from which photons can enter/exit at rates $\kappa_1$, $\kappa_2$, respectively. How do I obtain the number of photons inside the cavity in the stationary state? What happens if the light is detuned by $\Delta$ from cavity resonance?
| In the interaction picture, the equations of motion for the cavity (assumed single mode by the OP) read:
$$ \dot{a}(t) = -(\kappa_1 + \kappa_2)a(t) + \eta(t). $$
The constants are defined as in the question and $\eta$ is the drive strength, which is related to $\Phi$ (if I understand ''photon flux'' correctly it is $\eta \propto \sqrt{\Phi}$. Otherwise more information about the quantum state is required). Note the absence of a detuning $\Delta_{\mathrm{cav}}$, since we are in an interaction picture that measures frequencies relative to the cavity mode frequency. We note that this approach even works for pulsed drives.
Since the equations are linear, they can be solved by Fourier transforms, to give the solution in the frequency domain:
$$a(\Delta) = \frac{\eta(\Delta)}{i\Delta + \kappa_1 + \kappa_2}.$$
The steady state photon population in the cavity for a steady state drive at detuning $\Delta$ is then given by $\langle a^\dagger(\Delta) a(\Delta)\rangle$.
| {
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Mode expansion of string On Polchinski, the mode expiation of an open string is given as
$$X^i(\tau,\sigma)=x^i+\frac{p^i}{p^+}+i \sqrt{2\alpha'}\sum_{n\neq 0} \frac{1}{n}\alpha_n^ie^{-\frac{\pi i n c \tau}{\ell}}\cos\frac{\pi n \sigma}{\ell}\tag{1.3.22}$$
Giving the B.C. $\partial_a X^i=0$ at $\sigma = 0, \ell$.
However, In David Tong, he writes the general solution to the wave function(EOM) and says $$X^\mu = X_L^\mu(\sigma^+)+ X^\mu_R(\sigma^-)$$
and
$$X_{L/R}^\mu=\frac{1}{2}x^\mu + \alpha' p^\mu \sigma^\pm + i \sqrt{\frac{\alpha'}{2}}\sum_{n\neq 0}\frac{1}{n}\alpha^\mu_n e^{-in\sigma^\pm}\tag{1.36}$$
then he impose the B.C. after mode expansion for to arrive Neumann,$\alpha_n^a=\bar \alpha_n^a$, and Dirichlet, $x^I=c^I, p^I=0,\alpha_n^a=-\bar \alpha_n^a$.
Is the mode expansion of David Tong is the most general one and once we impose the B.C. and apply lightcone gauge then we get the one on Polchinski? If that is true, there are 2 questions I have:
*
*I could not get the $\cos\frac{\pi n \sigma}{\ell}$ term when I try this.
*The mode expansion seems reasonable but where does the $\alpha'$ comes? I feel it comes from EOM but I am not sure.
| 1) $\sigma^\pm = \tau \pm \sigma$ so you get something of the form
$$
\sum_n e^{-in\tau}\Big(e^{-in\sigma} + e^{in \sigma} \Big)
$$
which gives you the cosine.
2) The $\alpha'$ (and $l$) is just a convenient normalisation.
| {
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Is the equation for dispersive power applicable for all dispersive elements? The following text is from Concepts of Physics by Dr. H.C.Verma, from the chapter "Dispersion and Spectra", page 434, topic "Dispersive Power":
The dispersive power of a material is defined as the ratio of angular dispersion to the average deviation when a light beam is transmitted through a thin prism placed in a position so that the mean ray (ray having the mean wavelength) passes symmetrically through it.
After the definition of dispersive power, the author has derived the following expression based on the assumptions - the angle of prism is small and so the angle of deviation is small:
$$\omega=\frac{\mu_v-\mu_r}{\mu_y-1}\tag{20.1}$$
This equation itself may be taken as the definition of dispersive power.
In the above equation, $\omega$ is the dispersive power, $\mu_v,\mu_r$, and $\mu_y$ are the refractive indices of violet, red and yellow components of light respectively.
The equation for dispersive power was derived for a specific case - a thin prism (a prism with a small refracting angle). Then how could it be "taken as the definition of dispersive power"? This statement from the book seems to imply that it must also be applicable for dispersing elements other than "thin" prisms. Is the equation really valid for other dispersing elements like a prism of large refracting angle, or a glass sphere, or a grating?
Note: I haven't included the complete derivation of the formula as I thought it will increase the size of the post tremendously. However, I hope I have explained the main point clearly. If not, kindly ask your queries in the comments.
The Wikipedia article on Dispersion doesn't offer any explanation regarding dispersive power.
My search results on dispersing power didn't fetch any thing from reliable sources. So I decided to ask it here.
| Dispersive power is used to study dispersion of visible light in optical instruments. The lack of sources indicate that this is a quantity that isn’t used nowadays.
That being said, the above derived definition is a good one because it purely depends on the material and not its geometry. And you’d want to find the maximum deviation between the highest and lowest wavelengths that you are interested in normalised by the mean deviation (this makes it more reliable to compare between materials).
| {
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Photon vs electromagnetic waves Suppose an electron makes a single transition from higher to lower energy level releasing energy. Would that energy be released in exactly one photon equal to $h\nu$? Also, is saying "one photon is released" equivalent to saying "one electromagnetic wave of $\nu$ frequency is released"?
| We know that the ocean is made of water molecules. We also see that ocean has waves. Now does that mean that each water molecule is a water wave?
Physically realisable EM waves are analogous to the water waves where many photons collectively behave in a manner where EM waves emerge. But what is astonishingly peculiar about photons is that the energy of the individual photons somehow manifests itself as the frequency of the collective that is the EM wave!
And the reason why this happens can be read here.
| {
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What is the reason to believe that the laws of physics are same in all frames of reference? The first postulate of Special Relativity is that the laws of physics must be the same in all frames of reference i.e. invariant of coordinate transformations. I know this might be moot to ask but after reading a critique's paper on Special Relativity, I thought the question needs to be answered.
Is there any supportive evidence which suggests so other than the evidence of common sense and intuition?
As we know common sense and intuition are easily defied in most of the physics theories like Quantum Mechanics we should not rely upon such assertions to formulate an entire theory of Universe.
| If special relativity were not true, the everyday world we experience would be very different, in very obvious ways. I will provide just one simple example:
Special relativity is embedded in Maxwell's equations and tells us (among other things) that if we insert a magnet into a coil of wire, so as to induce a current to flow in it, there will be no way to tell from looking at the meter measuring that current whether it was the coil that moved or the magnet that moved.
If this was not the case, then the output of an electric generator would depend on whether the armature inside it were rotating and the case was fixed in position, or the armature was fixed and the case were rotating. No such effect exists.
In addition, the output of an electric generator would depend on its state of motion relative to some fixed reference frame attached to the universe as a whole. As the generator, which is fixed to the surface of the (rotating) earth, moved relative to that frame, its output would vary according to its angle relative to that frame. This effect does not exist either.
| {
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Gauge ghosts & unphysical states in gauge theory I have a general question about a statement from Wikipedia about ghost states as occuring in gauge theory:
"In the terminology of quantum field theory, a ghost, ghost field, or gauge ghost is an unphysical state in a gauge theory."
I learned gauge theory up to now with mainly mathematical beckground. My main reference is this https://arxiv.org/abs/1607.03089 paper by A. Marsh.
Question: What is concretely an "unphysical" state or field from viewpoint of gauge theory?
| A third view on this. Ghosts are fields that are introduced when you remove the gauge redundancy in the path integral.
In a path integral of a gauge theory you integrate over equivalent fields due to the gauge redundancy. You can fix the gauge and isolate the infinite volume of the gauge group but you are left with a determinant that you can then rewrite as the exponential of an action with Grassmann fields. These Grassmann fields are the ghosts. They appear in your theory and interact with the original matter fields of your theory, but resulting type of interaction ensures that they can never appear as external fields. They are not physical fields, i.e. fields that correspond to an observable degree of freedom.
I am aware that the use of the two words path integral may scare off mathematicians, but this is imo the most physical explanation of what ghosts are.
| {
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Entropy change in the free expansion of a gas Consider the adiabatic free expansion of a gas since there is no external Pressure hence Work done on the system is 0 and since the walls are insulated (hence adiabatic) the heat absorbed is 0. However since this is a irreversible process then entropy change > 0 hence dQ > 0 . However there is no heat absorption. What am I missing ?
| The reason is explained by others. Mathematically, for free expansion:
$ΔS_{system}=nR\ln\left(\frac{V_{2}}{V_{1}}\right)\ ;\ ΔS_{surroundings}=0$
Hence overall entropy of universe increases.
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Conservation of Energy stored in electric field? Let's say due to some particle process, an electron is created at time $t >0$. And from this moment on, the electric field will start to propagate to infinity at the speed of $c$. But we know that the energy stored in the electric field is proportional to the volume integral of $E^2$. Then wouldn't this mean more energy is being stored in the field as time passes? How is the total energy conserved and what offsets the continual increase in stored field energy?
| Lets say ... an electron is created - there's your problem; you just violated charge conservation.
You can't create a single charge out of nothing.
You might create a pair of charges (e.g., by decay of a neutral boson), one positive and one negative, but then you'd have two opposing fields whose energy content would exactly balance.
| {
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Does friction do work or dissipate heat? I know there are a bunch of similar questions but I read through them all and they don't answer my question.
Let's say I give a box on a floor an initial "kick" of force such that it has kinetic energy $KE$. Due to friction between the box and the floor, the box will slide to a halt. This means the friction must supply work equal and opposite to the objects energy: $W = -KE$.
However, we know that friction is an irreversible process. This means there is an entropy increase $S > 0$. But according to the classical definition of entropy, $S = \frac{Q}{T}$. Since work does not appear in this equation, this would imply there had to be a heat transfer at some point, but where? Is the frictive force also causing heat?
| 1) Work is done by the friction forces until the box stops.
2) Box kinetic energy is transformed to increased temperature (internal energy) of the sliding surfaces.
3) The cooling to the neighbourhood is an irreversible process, increasing entropy.
| {
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Why COM and COG are different things? In the formula for Newton's law of gravitation i.e.
$$F = G(m_1)(m_2)/r^2 $$
here $r$ is the distance between COM of the two objects , so we consider that the object is a single point located at it's COM and at this point gravitational force of attraction acts . Then why is sometimes COM and COG different for an object ? Please give some practical example .
| The COM for a rigid body is a fixed position that does not change. The center of gravity can depend on orientation, and can be in a different place on the rigid body, when in a non uniform gravitational field. In such a situation, the COG will always be closer to the main attractive body than the COM. For example, a satellite in orbit around Earth would have it's COG slightly lower than it's COM, because gravitational attraction decreases with distance, so the lower part of the satellite has slightly higher gravitational attraction to Earth than the upper part. see; https://en.wikipedia.org/wiki/Center_of_mass#Center_of_gravity for more information.
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Is there some other measurement that describe the ability to absorb some spicified range of frequencies of sound? There are a variety of materials could be used to absorb sound in order to soundproof.
Absorption coefficient is used to describe the absorb sound ability of the material.
Is the following guess true:
some materials are good at absorbing some spicified range of frequencies of sound (such as little girl screaming), while some other materials are good at absorbing some other spicified range of frequencies of sound, such as car's engine.
If yes, is there some other measurement that describe the ability to absorb some spicified range of frequencies of sound, like little girl screaming?
| The absorption coefficient is usually specified by the manufacturer over some frequency range, in dB of loss. High frequencies are commonly absorbed by light, fluffy materials with a lot of air in them. Low frequencies are commonly absorbed by heavy materials with little or no elasticity.
| {
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Unique characterization of Ideal gas In Thermodynamic state - Wikipedia, it defines a thermodynamic state as:
A thermodynamic state of a system is its condition at a specific time, that is fully identified by values of a suitable set of parameters known as state variables.
In the part explaining state functions, it says
In the most commonly cited simple example, an ideal gas, the thermodynamic variables would be any three variables out of the following four: mole number, pressure, temperature, and volume. Thus the thermodynamic state would range over a three-dimensional state space.
I would think this is not a result in thermodynamics. Is this just an assumption made about ideal gas, or can it be derived from considering the model of the ideal gas statistically?
| Your doubt is well founded. A three-dimensional space of thermodynamic states is justified only in the case of specific classes of systems. Generalizing a result valid for the ideal gas to every thermodynamic system is an unduly step. It is possible that somewhere in the text there was an explicit indication that all the statements were intended for the so-called simple systems. I.e. systems which, by definition, exchange work with the outside only through changes of volume. In practice, fluids without electric or magnetic effects.
Even a two-component perfect gas will require one additional variable (related to composition) to specify uniquely the thermodynamic state (notice that the equation of stat for pressure is not enough to discover this point).
In general, the proper choice of the state variables is strictly connected with the physics of the thermodynamic system. One has to discover experimentally what is a set of variables large enough to provide a unique description of the thermodynamic state.
A systematic way to find the independent variables is to analyze the independent processes which may change the energy of the system, i.e. the independent contributions to the differential of the internal energy. If we have neglected some process (and its related state variables) we will find that the incomplete set of thermodynamic variables will not specify uniquely the internal energy of the system, with possible apparent violations of the first principle of thermodynamics.
| {
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Constraining the 2-point correlation function Consider the two-point function
$$
\langle\mathcal{O}_1(x_1)\mathcal{O}_2(x_2)\rangle=f(x_1,x_2)
$$
If the operators are in a CFT, we can constrain this function using the symmetries of the theory. Using translational symmetries and the symmetries of the Lorentz group we have
$$
f(x_1,x_2) = f(X_{12})
$$
where $X_{12} := (x_1-x_2) $
When we impose dilatation symmetry we get
$$
\langle\mathcal{O}_1(x_1)\mathcal{O}_2(x_2)\rangle=\frac{C_{12}}{|x_1-x_2|^{\Delta_1+\Delta_2}}
$$
where $\Delta_1,\Delta_2$ the dilatation weights of the operators.
Now if we impose special conformal symmetries we have$$
\left(-2x_{1\mu}\Delta_1-2x_{2\mu}\Delta_2+k_{1\mu}+k_{2\mu}\right)\frac{C_{12}}{|x_1-x_2|^{\Delta_1+\Delta_2}}=0
$$
If we make use of $$
(k_{1\mu}+k_{2\mu})|x_1-x_2|=-(x_{1\mu}+x_{2\mu})|x_1-x_2|
$$
we should be able to derive
$$
\langle\mathcal{O}_1(x_1)\mathcal{O}_2(x_2)\rangle=\frac{C_{12}}{|x_1-x_2|^{2\Delta}}
$$
for $\Delta_1=\Delta_2=\Delta$ and $0$ otherwise.
I can't derive the final equation from the given identity of $k$'s, is there something else that I'm missing?
Edit:$$
k_\mu=x^2\partial_\mu-2x_\mu x^\nu\partial_\nu
$$
The operator associated to special conformal transformations.
| You did all the work but just missed the last step:
As you wrote, from the definition of $k_\mu$ you have
$$
(k_{1\mu}+k_{2\mu})|x_1-x_2|=-(x_{1\mu}+x_{2\mu})|x_1-x_2|,
$$
and so
$$
(k_{1\mu}+k_{2\mu})\frac{C_{12}}{|x_1-x_2|^{\Delta_1 + \Delta_2}}
= (\Delta_1 + \Delta_2) (x_{1\mu}+x_{2\mu})
\frac{C_{12}}{|x_1-x_2|^{\Delta_1 + \Delta_2}}
$$
This means that the constraint from special conformal symmetry becomes
$$
\left(-2x_{1\mu}\Delta_1-2x_{2\mu}\Delta_2+k_{1\mu}+k_{2\mu}\right)\frac{C_{12}}{|x_1-x_2|^{\Delta_1+\Delta_2}}
= (\Delta_2 - \Delta_1) (x_{1\mu} - x_{2\mu})\frac{C_{12}}{|x_1-x_2|^{\Delta_1+\Delta_2}}
=0.
$$
The only way the last equality can be valid for all points $x_1$ and $x_2$ is that $\Delta_1 = \Delta_2$. Now you call this $\Delta$ and you have your result...
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Diffusion equations and classical mechanics In Wikipedia it is stated that the diffusion equation can be derived from the continuity equation. It is not clear to me how the classical mechanics affect the diffusion equation. For example, if the total force on an object would be equal to the second time derivative of momentum instead of first derivative, then how that would change the diffusion equation?
| The diffusion equation can be derived from the continuity equation, the equation that reflects the conservation of mass in continuum mechanics. Changes in the momentum would only have an impact on Newton's second law, which in continuum mechanics is given by the momentum equation, and not on the continuity equation. As a consequence the diffusion equation would stay the same.
If you look at this brief derivation of the diffusion equation you will see the momentum equation is in fact never required:
Derivation of Fick's first law
In a one dimensional steady-state diffusion process, where particles at one point diffuse equally into both directions, the number of particles (amount of constituent) moving in the positive $x$ direction for a discrete system is given by
$$- \frac{1}{2} \left( n(x+\Delta x, t) - n(x, t) \right)$$
and the corresponding diffusive flux $J$ per area element $A$ and time step $\Delta t$ is given by
$$J = - \frac{1}{2 \, A \, \Delta t} \left( n(x+\Delta x, t) - n(x, t) \right) $$
This can be rewritten to
$$J = - \frac{\Delta x^2}{2 \, \Delta t} \left( \frac{n(x+\Delta x, t) - n(x, t)}{A \, \Delta x^2} \right)$$
and introducing the molar concentration $c := \frac{n}{V} = \frac{n}{A \, \Delta x}$ as well as the diffusion constant $D$
$$D = \frac{\Delta x^2}{2 \Delta t}$$
this yields
$$J = - D \left( \frac{c(x+\Delta x, t) - c(x, t)}{\Delta x} \right)$$
which for the case of $\Delta x \to 0$ leads to
$$J = - D \frac{\partial c_i}{\partial x}.$$
For a three dimensional system the gradient replaces the partial derivative
$$\vec J = - D \vec \nabla c.$$
Derivation of Fick's second law
Assuming again a one-dimensional system where now concentration changes over time as well as due to diffusion but the fluid is still at rest
$$\frac{\partial c}{\partial t} + \frac{\partial J}{\partial x} = 0$$
one yields with Fick's first law
$$\frac{\partial c}{\partial t} - \frac{\partial}{\partial x} \left( D \frac{\partial c}{\partial x} \right) = 0$$
which assuming a constant diffusion coefficient $D$ finally yields what is generally known as Fick's second law
$$\frac{\partial c}{\partial t} = D \vec \nabla^2 c.$$
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How to calculate the resistance impulse of a wheel on an axle when it receives torque during acceleration? I am programming a vehicle system for a company, but I cannot find the right formulas to be able to have access to the resistance pulse of a wheel attached to an axle during acceleration / deceleration. All I know is that this value decreases with acceleration and increases with deceleration, but I can't get my finger on it. I am able to find the angular momentum, the rolling coefficient, the friction coefficient and the rolling resistance, but I cannot find a formula allowing me to use these in order to be able to find this value.
The vehicle travels at a speed of $2{km/h}$ with wheels of a radius of $0.59{m}$
The torque sent by the engine with a revolution of 900 rpm during an acceleration ($200{N m}$) sent to the transmission: $ T = 200{N m}$
The transmission multiplies the torque received from the engine by a transmission ratio (8.03) : $ T = (200{N\cdot m}) (8.03) $
The differential again multiplies the torque received from the transmission (3.36):
$ T = (200{N\cdot m}) ((8.03)(3.36)) $
I do not take into account in this version the loss due to friction.
In this case, the Fz is the force (in newton) of the suspension which is connected to the vehicle of $14014{N}$.
$ Fz = 14014{N} $
$ Vx{(m/s)} = (2{km/h} (3.6)) $
$ r = 0.59{m} $
$ rpm = 8.9917 $
$\Omega{(rad/s)} = r * \omega = {0.59m * (rpm({8.9917})/9.5492)}$
I'm trying to find the Fx with the formulas mentioned above as well as the context.
$ Fx = ?$
| To answer this problem, first do a free body diagram that includes all forces and motions acting.
Now assume that the axle bears some fraction of the total mass of the vehicle, $m_{\rm axle}$.
Also assume no slipping, which means $v - \Omega R =0$ or it terms of accelerations
$$ \dot{\Omega} = \frac{\dot{v}}{R} $$
where $R$ is the radius of the tyre, and $\dot{\square}$ designates the time derivative.
Now form the three equations of motion with care taken to consider what is considered a positive or a negative sense.
$$ \begin{aligned}
F_x & = m_{\rm axle}\, \dot{v} \\
F_y - m_{\rm axle}\, g & = 0 \\
- \tau + R F_x & = I (-\dot{\Omega})
\end{aligned}$$
where $I$ is the mass moment of inertia of the tyre and wheel.
The solution of the above equations is
$$ \begin{aligned}
\dot{v} & = \frac{R\, \tau}{I+m_{\rm axle} R^2} \\
F_x & = m_{\rm axle}\, \dot{v} \\
F_y & = m_{\rm axle}\, g \\
\end{aligned} $$
| {
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Normalization of the action in Special Relativity The action for a massive point particle in Special Relativity is given as
$$A =-mc^2\int d\tau,$$
Where $\tau$ represents the proper time, and $m$ represents the (rest) mass. From what I could understand, the Action must not change with respect to the reference frame, and hence it can be written as $$something\int d\tau$$ but why must the something need to be proportional to mass?
| Since action has the units of angular momentum, the proportionality constant needs the units of energy by dimensional analysis. It must also be Lorentz-invariant, so is $mc^2$ times some real number. This number's modulus can be varied without changing the resulting equations of motion, but @Cryo's answer shows a modulus of $1$ recovers the usual $\int(T-V)dt$ outcome of Newton's physics. The sign, on the other hand, is fixed by the requirement that the action be minimized.
| {
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Can a Black Hole become a normal mass again? It is well known how Black hole forms. But once it is formed are there any circumstances which can lead a black hole to becoming a normal mass again?
| Apart from the possibility of BH relics as outlined above, if it were possible to remove or ‘destroy’ a BH event horizon, then that might qualify. Of course, to avoid the cosmic censorship hypothesis (no naked singularities please) you need:
*
*a BH that doesn’t have a singularity, i.e. something like a Bardeen black hole, also known as a ‘regular black hole’. BBH’s are thought to have a de Sitter core. Some people think astrophysical black holes are ‘regular’ i.e. predicted singularities just show where general relativity fails.
The event horizon of the no-singularity BH is still specified by mass $M$, spin $J$ and charge $Q$ (as per Kerr-Newman):
$M^2≥Q^2+J^2/M^2 (1)$
If you violate Eqn (1) then – no more event horizon. So theoretically, if the BBH can absorb enough angular momentum and charge, and re-stabilise in a steady state, then the event horizon disappears and – maybe – you have a ‘normal mass’ again.
There is a 2013 paper that thinks the accretion process from a BH disk might be able to achieve this, and suggest a test of the hypothesis via study of the energy of BH mergers. Not sure whether this has been falsified or not yet.
| {
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Are reflectance $R$ and transmittance $T$ independent of the direction in which the light travels? If we have two different optical mediums $1$ and $2$, and we know the reflectance and transmittance when the light gets to the interface from the first one to the second, $R_{1 \rightarrow 2}$ and $T_{1 \rightarrow 2}$, would they have the same value when the light gets to the interface on the contrary sense, from $2$ to $1$?
$$R_{1 \rightarrow 2}\stackrel{?}{=} R_{2 \rightarrow 1}$$
$$T_{1 \rightarrow 2}\stackrel{?}{=} T_{2 \rightarrow 1}$$
For example, for the particular case of glass prism: if a light beam enters and exits it, will T and R take the same values when the light enters (air to glass) and when it leaves (glass to air)?
| This question can be answered by looking at the Fresnel equations for reflectance and transmission, which are shown below in the case of normal incidence and ignoring polarization.
$$r = \frac{n_1-n_2}{n_1+n_2} \\ t = \frac{2n_1}{n_1+n_2}$$
and the total reflectance and transmission are $R=\vert r \vert^2$ and $T=\frac{n_2}{n_1}\vert t \vert^2$.
We can see that the complex reflectance coefficient $r$ changes its sign when you swap $n_1$ and $n_2$, but the overall reflectance $R$ is unchanged. This tells you that get the phase shifts reflecting from $1\rightarrow2$ and $2\rightarrow1$ have opposite signs. Thus $r_{12}=-r_{21}$ but $R_{12}=R_{21}$. This makes sense since waves reflecting from a "denser" medium get a phase shift, but when going to a "thinner" one you don't.
For transmission, again we see that swapping $n_1$ and $n_2$ does change the complex transmission coefficient, but not the total transmission. So $t_{12} \neq t_{21}$ but $T_{12} = T_{21}$
This is because:
$$T=\frac{n_2}{n_1}\vert \frac{2n_1}{n_1+n_2} \vert^2=\frac{4n_1 n_2}{(n_1+n_2)^2}$$
So swapping $1\rightarrow2$ does nothing to the overall $T$.
Edit: Per request I will discuss the case of non-normal incidence, now $1\rightarrow2$ and $2\rightarrow1$ are inequivalent in general. In the case of total internal reflection we have the extreme case of light being able to enter the higher-index material, but not able to leave.
Let's ignore polarization again, an only consider reflection (we can always get $T=1-R$ anyways).
Now the expression for $r$ is:
$$r=\frac{n_1 \mathrm{cos}(\theta_i)-n_2 \mathrm{cos}(\theta_t)}{n_1 \mathrm{cos}(\theta_i)+n_2 \mathrm{cos}(\theta_t)}$$
Where $\theta_i$ is the incident angle (normal incidence is $\theta_i=0$), and $\theta_t$ is the transmitted angle given by Snell's law $n_1 \mathrm{cos}(\theta_i) = n_2 \mathrm{sin}(\theta_t) $.
Now we can rewrite this expression for $r$ purely in terms of the incident angle $\theta_i$.
$$r=\frac{n_1 \mathrm{cos}(\theta_i)-n_2 \sqrt{1-\left(\frac{n_1}{n_2} \mathrm{sin}(\theta_i)\right)}}{n_1 \mathrm{cos}(\theta_i)+n_2 \sqrt{1-\left(\frac{n_1}{n_2} \mathrm{sin}(\theta_i)\right)}}$$
Now we can see that swapping $1\rightarrow2$ does change things so that $r_{12} \neq r_{21}$ and $R_{12} \neq R_{21}$.
As a concrete example, take $\theta_i=\frac{\pi}{4}$, or 45 degrees, and $n_1=1$ and $n_2=2$. Then $r_{12}\approx -0.45$ or $R_{12}\approx 0.20, meaning we get incomplete reflection going from the low-index to higher index.
Now compare this to $n_1=2$ and $n_2=1$, we get $r_{12} = e^{-1.23i}$, or $R_{21}=1$, so total internal reflection from the higher index to lower index!
| {
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If magnetic field lines don't exist, what are these iron filings doing around a magnet? Obviously the iron filings can be seen aligning themselves along the virtual magnetic field lines produced by the permanent magnet, the virtual magnetic field line is made of electromagnetic field due to the alignment of electrons in the magnet but why the patterns, why lines? Do these lines have thickness? Are they due to interference pattern?
| What is happening when iron filings form a "field line" pattern, is actually an energy minimisation process, somewhat akin to the reason that solar systems form out of rotating clouds of dust. Each individual iron particle becomes magnetised by the applied field, much more so than the surrounding medium. There are then forces acting on the adjacent iron particles which can be treated as lots of small magnets. The minimum energy configuration has the long axis of the particle lined up with the applied field, and particles grouped together into filaments which have roughly the same shape as the imaginary magnetic field lines.
The energy may not be fully minimized -- there is friction with the supporting paper to consider, and also the likelyhood that the pattern gets stuck in a local energy minimum rather than the global one.
You get some truly weird patterns if you use a ferromagnetic liquid and a magnet. (Actually, lots of really tiny more or less spherical ferromagnets suspended in oil).
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Boyle's Law and hot air balloon A bit dumb question because it is really difficult to imagine it. :-
Pressure is force per area. Talking about gases, the pressure is said to be the force molecules exert on walls of let's say a balloon.
Usually in examples of Boyle's law, our teachers mention hot air balloon, that the size of balloon increases as pressure decreases.
But at as height increases pressure decreases because there are less molecules above us.
How will the pressure of molecules outside the balloon effect the pressure of molecule inside the balloon.
| There are three forces at play here caused by the internal pressure, external pressure and the elastic balloon. The internal pressure balances out the other two to maintain equilibrium. For ease of calculations, I am assuming that no air is able to diffuse through the rubber and escape. Thereby the internal pressure stays constant throughout.
Boyle's law is applicable for both internal and external pressure. Hence when external pressure decreases, the volume increases. The rubber then applies greater force to try to compensate for this loss of external pressure.
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Why isn’t the center of the Earth cold? If the pressure of the Earth is keeping the inner core solid, keeping it rigid to take up the least space, and temperature is dependent on how much the atoms are moving, why isn’t the inner core cold? If the pressure is so high that it’s forcing the inner core to be solid then the atoms can’t move around and thus they can’t have temperature.
| Your argument would require that all solids must be cold, because all solids have constituent atoms that are constrained to remain in their solid lattice positions. But clearly not all solids are cold, so there is something wrong with your argument. That thing is that atoms or molecules constrained within a solid structure can still vibrate and oscillate around their equilibrium positions. So they do have an internal energy and this is where the heat is stored.
| {
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Are there any black hole neutron star binary? Has a black hole-neutron star binary aver been observed?
I mean observed in any way: gravitationally, through eclipse, or any other means.
EDIT Thanks to the comment to this question, we know that some system are known from GW observation.
Are these observations the only ones?
| A quote from the website of the BlackHoleCam European Research Council Project (as of 17/2/2020):
Although pulsar-BH systems can provide unique benchmarks of theories of gravity, they are expected to be very rare and to date not a single pulsar-BH system has yet been found.
In principle, a pulsar-BH system could be used to confirm the black hole nature of the companion and measure the BH mass and spin, by analysing the precise timing of the (radio) pulses from the pulsar.
If the neutron star is not a pulsar, then there is little to give these binaries away, except when they merge.
There are a number of candidate systems that have been suggested as the source of detected gravitational wave signals from merging compact objects, but no optical counterpart has yet been reported that might offer some confirming evidence.
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Electron absorption by materials Is it possible to define whether or not a material has absorbed electrons after it has collided with an accelerated electron beam? If so is it possible to calculate where they gathered after the collision?
|
Is it possible to define whether or not a material has absorbed electrons after it has collided with an accelerated electron beam?
It will depend on a large number of conditions: energy of beam, the type of material, in general the boundary conditions of the experiment.
If so is it possible to calculate where they gathered after the collision?
In general if a neutral material absorbs electrons, it will become negatively charged and there are instruments developed that can measure charge. It will depend on the type of material and the topology of the situation/experiment. Metals have charge on the surface, for example.
It is not advisable to expose random material, including humans, to accelerator beams.
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If sound passes through material, vibration is produced. So are electromagnetic waves produced too? Sound means vibration of molecules and vibration produces electromagnetic waves. So, this means that sound produces electromagnetic waves directly.
Is this possible?
| I am deeply surprised by the neglect of the energy balance. In the previous answers. All sound is dispersed within a body. The energy difference between the incoming sound and the outgoing sound turns into heat.
The increase in temperature of the body is accompanied by an increase of electromagnetic radiation. So your guess is right, the body is out of its thermal equilibrium and responsible for this this the sound.
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Is tension a conservative force? Are forces such as tension (from an in extensible string), normal reaction, and applied force from us, non conservative forces? If so why?
I have read in few books that these forces are labeled as nonconservative, but most of the time they are internal and do zero work and thus mechanical energy of the system remains constant, shouldn't they be conservative?
| First, you need to know how a conservative force is defined. A force is conservative if the work done by it is path independent.
Now coming to your doubt, according to the work-energy theorem,
$$
\text{work (done by all the forces} = \Delta\left(\text{kinetic energy}\right).
$$
If you know, potential energy is defined only for conservative forces so,
$$
\text{work (done by a conservative force)} = -\Delta U,
$$
the negative of the change in potential energy.
So, take an example where tension and gravity (conservative) is acting on a pendulum. Applying the work-energy theorem,
$$
\text{work (done by tension)} + \text{work (done by gravity)} = \mathrm{K.E.} \text{ (final)} - \mathrm{K.E.} \text{ (initial)}
\tag{1}
$$
which implies
\begin{align}
\text{work (gravity)} & = - \left( U\text{ (final)} - U \text{ (initial)}\right)
\\
\text{work (tension)} & = 0.
\end{align}
Putting these in $(1)$ we get
\begin{align}
\mathrm{K.E.} \text{ (final)} + U \text{ (final)} = \mathrm{K.E.} \text{ (initial)} + U \text{ (initial)},
\end{align}
i.e., mechanical energy is conserved. Now, if the tension had done work this equation would be invalid and the mechanical energy would not have been conserved.
Observe that even though gravity (conservative force) is doing work the total mechanical energy is conserved but if the tension had done any work the mechanical energy of the system wouldn't have been conserved. This equation of conservation of mechanical energy is only valid if the work done by tension is zero, and is invalid if tension starts doing any work, hence a non conservative force.
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Dynamicity inside a stationary water drop I was doing some experiments with water drops on lampblack when I saw this.
You can see the full video here.
Inside a water drop which is perfectly still from the outside, you can see some moving things, which I suppose are some lampblack flakes from the surface. I am not bothered about how those flakes got in there, but I am interested in their motion.
Why are they even moving?
Is it some kind of Brownian motion?
I am amazed by the dynamicity inside such a seemingly inactive water droplet.
If I had not seen this, I would have considered the water drop as a simple sphere for the rest of my life.
This observation, at least for me, raises an important question.
Is anything in the world really NOT dynamic?
| Based on the similar movement of inhomogeneities on soap bubbles, I'd say the flakes' movement is caused by air currents around the drop and perhaps thermal convection.
Source: https://youtu.be/LM3p3X92mWI
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Why does the same proportion of a radioactive substance decay per time period? (half life) Just wondering, if decay is random, why does the activity half every half life, as in, why does it have to reduce by the same proportion in the same time period?
| A couple of answers above hit it well. Here is a slightly different perspective.
From an visual standpoint, consider a pointillist painting. If you look at any single dot up close, the painting makes no sense. Stand back, and order falls into place.
The term “random” does not mean without order. It means that nothing we know up to this point with this particular perspective enables us to predict its function going forward in time.
| {
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How long will last the CBMR? I am confused about the duration of the cosmic background microwave radiation.Will it eventually be replaced by same radiation from discrete directions?
| The current model of the expansion of the universe is dominated by $\Omega_{\Lambda}$. In the distance future
1-$\Omega_{\Lambda}$ << 1.
At this era the expansion of the universe will become exponential, and at some point the temperature of the CMB will become less than the temperature of Hawking radiation of black holes. When that happens, near a black hole the radiation will become mostly the Hawking radiation.
I will later calculate the time when this happens.
ADDED
Step 1. In the future, the scale factor (a) gets larger, and in the Friedmann equation with four $\Omega$s (with subscripts: R, M, k, and $\Lambda$) have the following approximate values.
$\Omega_{\Lambda}$ ~ 1, and the other 3 $\Omega$s << 1.
Assuming a=10 (z=-0.9) should be adequate for this purpose. The corresponding time (since the big bang) is
(Eq 1) $t_1 = 51 Gyr$.
This is calculated using
http://www.astro.ucla.edu/~wright/CosmoCalc.html
with $H_0$=70 (km/s)Mpc (1/$H_0$~14 Gyr) and $\Omega_m$=0.315. Note: the corresponding value of $1/H(t_1)$ is
$$(Eq 2) \frac {1} {H(t_1)} = 0.8 Gyr.$$
So the Friedmann equation for $t > t_1$ becomes (approximately)
$$(Eq 3) H(t) = \frac {da/dt} {a} = \frac {1} {0.8 Gyr}$$
Step 2: (TO BE COMPLETED - NEEDS INPUT FROM @Janko Bradvica) Select a value for the mass of a black hole to use for the remainder of the discussion.
I want to develop the answer to the question considering only a single value for the mass of a black hole. Three are three plausible choices. I would like Janko to choose one.
(a) The black hole at the center of the Milky Way. This is the simplest, but also the most likely to be unrealistic.
(b) A black hole with the total mass of the Milky Way (including dark matter). This is moderately realistic and only moderately complicated.
(c) A black hole with the total mass of the Local galaxy cluster (which contains the Milky Way including dark matter). This is the most realistic, but also the most complicated.
Step 3. The remainder of the work will involve solving (Eq 3) and calculating a value for $t_2$ (time when the black hole temperature is greater than the CMB temperature) and $a(t_2)$, and then corresponding values for the temperatures $T_{CMB}(t_2)$ and $T_{Hawking}(t_2)$.
| {
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How do I decompose my overall rotation curve for M31 into a rotation curve for the disk and halo? How do I decompose my grand rotation curve into rotation curves for the disk and halo, like the image below?
The picture is from this link pages.pomona.edu/~tmoore/RotCurve
The data points are for NCG 2403, my rotation curve is for NCG 224. I want to plot the total (which I drew from my data), halo, and disk rotation curves.
| You cannot without additional information - at a minimum, how much lumnosity comes from with radius $r$ in the galaxy studied.
The rotation curve arises from the sum of all gravitating matter that is present in the system.
In principle you can estimate the contribution of the disk to the rotation curve by measuring how much mass is contained within a radius $r$ and then working out what rotation speed at $r$ that would produce. Note, that in your simulation, "disk" refers to the baryonic matter in the disk, whose mass can be estimated by observing how much luminosity it produces and then using an assumed "mass-to-light" ratio. "Halo" refers to the dark matter halo. The mass of this (and its distribution with radius) can only be inferred by adding it to the baryonic matter in order to bridge the gap between the rotation curve predicted by just the baryonic matter and the observed rotation curve.
Having said that - the disk line in the plot is entirely smooth, so it is obviously not showing measurements. It is just a smooth functional fit to some observational data (that isn't shown) that might be modelled in conjunction with the rotation curve.
The purpose of the simulation appears to be to demonstrate how a rotation curve can be modelled by combining the mass distribution present in the luminous matter, with that assumed to be present in a smooth, dark halo.
| {
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Help using the definition of Hermitian operator in $\int\psi^*(\hat F-\left)^2\psi dr$ In my lecture the professor said that the mean value of a physical quantity- since it must be real- must satisfy the following condition:
$$\begin{align}
\left<F\right>=\left<F\right>^*
\end{align}
$$
$$\begin{align}
\int\psi^*(\hat F\psi)\,\mathrm dr=\int\psi(\hat F\psi)^*\,\mathrm dr.....(\alpha)&
\end{align}
$$
With this, he's definiting a hermitian operator, as the operator $\hat F$ that makes the previous condition true.
Then he used it in the following passage:
$$\begin{align}
\int\psi^*(\hat F-\left<F\right>)^2\psi\,\mathrm dr=\int\psi^*(\hat F-\left<F\right>)(\hat F-\left<F\right>)\psi\,\mathrm dr=\int[(\hat F-\left<F\right>)\psi]^*[(\hat F-\left<F\right>)\psi]\,\mathrm dr&
\end{align}
$$
I really don't see why the last step is true. The professor said he's only applying the definition of hermitian conjugate given in $(\alpha)$ to the part of the operator(that is to $(\hat F-\left<F\right>)$) but I still don't see it.
This is the only definition of hermitian operator we have seen, so please try to explain it without using matrices or brackets or any other notation, or without seeing the operator as a matrix, I know some about this stuff, but I would like to understand it the way he's doing it.
Any help would be really appreciated.
| It is usually assumed that $\psi$ can be expressed as a linear sum $\sum_i a_i\psi_i$ of eigen function $\psi_i$ with eigen values $\lambda_i$. So $F \psi = \sum_i a_i\lambda_i\psi_i$. This converts the problem into complex number algebra.
In effect, this is a rephrasing of - turn the operator into a matrix - but I hope it unpacks it so the idea can be seen in terms of explicit functions and simple differential operators.
| {
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Why can terms of a spacetime metric only have 2 differential factors? In my general relativity textbook, the following equation is given:
$$ds^2 = g_{\alpha \beta}(x)dx^{\alpha}dx^{\beta}$$
Which describes the line element $ds^2$ for a metric. $g_{\alpha \beta}$ is a matrix that is the metric itself. This equation implies that a given term of the equation for a line element can only have two differential factors. Either two different linear variables (For example: $dtdr$) or one variable squared (For example: $d\theta^2$).
Intuitively, this makes some sense. Our line element is $ds^2$, so that exponent of 2 could explain why, when you multiply things out, you can only get two differential factors per term, just like squaring any polynomial.
But this doesn't really make sense to me from a more formal perspective. I suppose the issue is that I don't really understand what the exponent of 2 is actually doing. So what is it doing? And what is the exact reason that we can only have two differential factors per term?
(Reading back this question, I'm not sure if everything is clear, so let me know if I can clarify anything or otherwise improve this question).
| The expression for the line element is a generalization of the Pythagorean Theorem of 2D Euclidean space,
$$c^2=a^2+b^2,$$
to 4D curved spacetime, on an infinitesimal scale. So the squares shouldn’t be surprising.
In 3D Euclidean space, you have probably seen that infinitesimal arc length along a curve is just
$$ds^2=dx^2+dy^2+dz^2.$$
In the flat (Minkowskian) spacetime of Special Relativity it is just
$$ds^2=-dt^2+dx^2+dy^2+dz^2.$$
For a general 4D curved spacetime, your expression, written out in $txyz$ coordinates, is
$$\begin{align}ds^2&=g_{tt}(t,x,y,z)\,dt^2+g_{xx}(t,x,y,z)\,dx^2+g_{yy}(t,x,y,z)\,dy^2+g_{zz}(t,x,y,z)\,dz^2 \\
&+2g_{tx}(t,x,y,z)\,dt\,dx+2g_{ty}(t,x,y,z)\,dt\,dy+2g_{tz}(t,x,y,z)\,dt\,dz\\
&+2g_{xy}(t,x,y,z)\,dx\,dy+2g_{xz}(t,x,y,z)\,dx\,dz+2g_{yz}(t,x,y,z)\,dy\,dz.
\end{align}$$
The essential idea is that curved space(time) should be, over sufficiently small regions, very similar to flat space(time), in the same way a small patch of the Earth’s spherical surface seems like a plane. This means that you can't have something weird with different powers like
$$ds^2=-A dt^3+B dx^4+C dy^5+D dz^6.$$
| {
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Due to incompressibility of water same volume of water should flow through pipe of different cross section . is that really true This seems true when I slightly compress the opening of pipe the speed of water flowing through pipe increases and same amount of water flows out but as I further decrease the opening of pipe. Instead of increasing speed of water to maintain same volume flow rate.the amount of water coming out of it decreases and same volume rate stops. Why is it ?
| Yes, due to incompressibility, the same amount of water flows into and out of any given section of a tube - that is correct. But due to viscosity and a no-slip condition on the tube wall, the pressure of the water drops downstream (if the tube width and flowrate is constant).
If you have a constant pressure potential available (high pressure at the tube inlet, atmospheric at the outlet), the flowrate will adjust with the geometry of the tube. The longer and thinner the tube, the more effect from viscosity (more pressure drop for a given flowrate) and a lesser flow at your constant pressure drop.
| {
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Why do springs have a linear relationship? Why does:
F = k*(change in position)
Why can't the relationship be quadratic or higher ordered?
| It's an approximation motivated by calculus. Since by definition force is $0$ in the equilibrium position we label as $x=0$, the lowest-order approximation of $F$ as a Taylor series of $x$ is of the form $F=-kx$, unless $k=0$. (The sign indicates a force opposing the perturbation from a stable equilibrium with $k>0$.) Equivalently, the energy stored relative to equilibrium is, to lowest order, $\tfrac12kx^2$. (Since energy needs to be minimal in the stable equilibrium $x=0$, a quadratic force wouldn't work; maybe a cubic one would, but that requires two coincidentally zero coefficients, so forget it.)
Of course, these approximations work best for small $|x|$. It's a bit like a pendulum's small-angle approximation. In real life, the relationship is far from being linear at large $x$; we call this behavior anharmonic. The above logic motivates adding a cubic (quartic) term to the force (potential). Eventually, the spring undergoes permanent damage at some finite $x$ and energy, and this changes their future relationship. This doesn't even require you to snap it in two. So what really happens depends on the history of the spring. This is an example of hysteresis.
| {
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Is there an approximation for the Lorentz factor for very large velocities? I am aware of the approximation generally used for low speeds to calculate the Lorentz factor, that being,
$$\gamma \approx 1 + \frac{1}{2} \left(\frac{v}{c} \right)^2$$
But I need the exact opposite thing -- is there any suitable approximation for when v is extremely close to c?
| For ultra-relativistic particles, $c-v$ basically stops being an experimentally accessible observable. Unless you are extremely careful about timing, you assume that the beam is traveling at $c$ and measure the Lorenz factor by comparing the kinetic energy per particle to the particle mass, $\gamma = E/mc^2$. (If you care about the difference between total energy $E=\gamma m c^2$ and the kinetic energy $K=(\gamma-1)mc^2$, you're not ultra-relativistic yet.)
I first understood this when I got to Jefferson Lab, which has two antiparallel 1 GeV electron linacs connected like a racetrack. The electrons are injected at 50 MeV. After a lap they're at 2 GeV. After five laps, they're at 10 GeV. The accelerator feeds beam to four different halls at once, each of which can request a different energy by accepting the beam after a different number of passes around the track. So at any given instant while the beam is on, the linac might have beam bunches with five or six different energies interleaved. And --- here's what's interesting to you --- the fast ones don't have different timing than the slow ones. Even the electrons from the 50 MeV injector already have $\gamma = 100$, and there are a bunch of 1% issues that make it hard to distinguish between their speed and exactly $c$.
Search term: "XY problem."
| {
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How is a 25-year-old can of soda now empty without having been opened or poked? I just discovered in my parents' basement a Sprite can from 1995* and also a Coca-Cola can probably from the same year.
Both cans are unopened and have no visible damage or holes. The Coca-Cola can feels "normal", but the Sprite can is empty!
You can hear in my video that there is no liquid sloshing around in it.
I can't find any way that the soda could have escaped.
We also don't see any mess near where the cans were, but I don't know for certain that the cans stayed in the same place for 25 years, so maybe there could have been a mess of liquid leaked out somewhere else if the cans had been stored somewhere else earlier. But nothing feels sticky or looks like there has been a leak of any kind.
What are possible explanations for how a carbonated drink could disappear from a sealed aluminum can?
*I had kept it as a "collector's item" from when the Houston Rockets won their second championship.
| As others have said, more than likely it was never full in the first place. My grandfather worked for a can manufacturing company and had a large collection of empty sealed beer cans of all different brands. Your can could have been a reject or more likely just a sample.
| {
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Splitting a single particle wave function The wikipedia article on the double slit experiment contains the following animation:
https://upload.wikimedia.org/wikipedia/commons/transcoded/a/a0/Double_slit_experiment.webm/Double_slit_experiment.webm.180p.vp9.webm
Here we can see that part of the wavefunction is reflected back at the electron source. Does this only happen when there are multiple particles? If the experiment is set up so that at any one time, there is at most 1 electron between the electron source and the screen, can this sort of reflection still happen?
More generally: what happens when a wave function of a single particle splits in two parts, with each part propagating in a different direction? Is this even possible?
|
Does this only happen when there are multiple particles?
No. Wave function reflection can happen with single-particle wave functions.
what happens when a wave function of a single particle splits in two parts, with each part propagating in a different direction? Is this even possible?
Yes, this is possible. It just means that there is some probability of finding the particle at various locations in space along those directions.
| {
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How to find the magnetic field of a current using the differential form of Maxwell's equations? To find the magnetic field produced by a long straight wire, one would ise either Biot-Savart law or Ampere's Law in integral form. How do you find this simple result starting from $\nabla \cdot \vec{B} = 0$ and $\nabla \times \vec{B} = \mu_0 \vec{J}$? Let's imagine a current flowing in the $\hat{y}$ direction, then the previous equations are:
$$ \partial_x B_x+ \partial_y B_y + \partial_z B_z = 0$$
$$ \partial_y B_z - \partial_z B_y = 0$$
$$ \partial_x B_y - \partial_y B_x = 0$$
$$ \partial_z B_x - \partial_x B_z = \mu_0 J_y$$
And then? what next?
| Edited in the light of first comment below answer
Your problem arises because Maxwell's equations are local equations, applying at a point. Thus the four equations in your question, judging by the last one, apply to a point at which $\mathbf J$ is non-zero (that is a point in your wire). But you wish to find $\mathbf B$ at a point P some distance away from this point. Your local equations, applying to points where $\mathbf J$ is non-zero clearly aren't going to do the job by themselves.
The solution is to apply the local equations to all points between the wire and the field point, P, even though $\mathbf J$, and hence $\nabla \times \mathbf B$, is zero at these intermediate points. You might surround your wire by a net of infinitesimal two-dimensional cells covering the a plane through which the wire passes normally. Imposing the condition that ∇×⃗=0
for all cells except the central, current-carrying one will then give you a line integral around a loop surrounding the wire equal to $\mu_0 I$. I leave the details to you!
The method just described is in fact Stokes's theorem in disguise. I don't think you can do without it. So without apology, here is the slick treatment.
Integrate $$\nabla \times \mathbf B =\mu_0 \mathbf J$$
over a surface S bounded by a closed loop (in your case a circle of radius r centred on the -current-carrying wire), so $$\int_S (\nabla \times \vec B) \cdot d \mathbf S=\mu_0\int_S \mathbf J \cdot d\mathbf S$$
Now apply Stokes's theorem to the left hand side:
$$\int_s \mathbf B \cdot d \mathbf s=\mu_0\int_S \mathbf J \cdot d\mathbf S$$
in which the left hand side is a line integral around the bounding loop.
But we have here Ampère's law, since the right hand side is simply the current, I, through the loop. This is perfectly general, but in your special case, having chosen a loop centred on the current-carrying wire, we know by symmetry that $\mathbf B$ is of equal magnitude, B, all round the loop, so
$$\int_s \mathbf B \cdot d \mathbf s=2 \pi r B$$
So we have
$$B=\frac{\mu_0 I}{2 \pi r}$$
| {
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Why doesn't Rayleigh scattering happen at low altitude in air? An answer I found online was: At high altitude where molecules are far apart, scattered photons can travel without interfering with each other, thus they fill the sky with blue light. But at low altitude, molecules are so close to each other that scattered photons destructively interfere with each other to cancel each other out, that's why we see air as transparent.
However, if they perfectly cancel out, then wouldn't the energy be not conserved?
Another explanation was simply that the effect is only apparent in a very large column of air since the scattering effect is not that noticeable.
So why doesn't Rayleigh scattering happen at low altitude in air?
| The assumption is wrong. Rayleigh-scattering does happen independent of altitude. That it happens at sea level, too, becomes clear when you look at the other bank of a large river mouth or a nearby coast over a bay: they loose much contrast and everything is blue-ish; the same effect happens in the mountains looking from one peak to the other.
In the sky there is no other light source, so that only scattered light arrives from there which makes it so nicely blue.
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Clarification of the concept "less resistance means less heating" in a wire So my textbook says that the reason cables that are suppose to carry high currents, are thicker that those that are meant to carry lesser current, is that "less resistance (of the wire) means less heating..."? Is this even true?
Isn't CURRENT the reason wires heat up? If we decrease resistance, more current flows, and that should produce more heating!
|
Isn't CURRENT the reason wires heat up? If we decrease resistance, more current flows, and that should produce more heating!
The key thing you're missing here is that current is (almost) constant and is (almost) not dictated by the wire. The "ideal" wire has no resistance and the current through it is solely set by the load you have connected at the other end of the wire.
In the real world, that's not quite true. Sometimes we really do need to consider the effect of the wire - for example, if you have 100km of transmission line carrying power from a power station to a nearby city, losses in the wire will certainly become significant. Also if you put AC down a wire, the AC behaviour as frequency increases is certainly something you need to think about.
In practise though, the point of choosing wires that are thick enough is that the resistance in the cables is very small compared to whatever load is connected. We can therefore usually ignore losses in the cables when we think about the circuit.
The only time where your statement is true is if the wire shorts out the power supply - for example, if you happen to connect a wire directly from live to neutral or earth in a mains cable. In that case there is no other load, only the resistance of the wire. A lot of current flows, and the wire heats up very very fast! In your house, the mains supplies go through circuit breakers which trip if too much current is drawn; and depending on where you are in the world, your mains plug may also contain a fuse. The purpose of these is to cut power rapidly in the event of an electrical short, so that the wires don't heat up and set your house on fire.
| {
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Post-measurement $\psi$ in quantum mechanics I have a question regarding the wave function after a measurement. Everything I found online says that this is the following formula:
$\psi = \frac{M_m\psi}{\sqrt{P(m)}}$
Where $P(m)$ is the probability of observing m, the $\psi$ on the left is the wavefunction AFTER the measurement and the $\psi$ on the right is the original wavefunction. However, I cannot find a good definition on how I would go about calculating $M_m$? The Berkley lecture notes say that this is the measurement operator, but how would I go about finding that for my specific problem?
Also the probability function is $P(m) = |<\psi|\omega>|^2$, how would I find $\omega$ in this case? Is it just the eigenstate at that observable?
| It is instructive if you work with some simple setups. Consider the example of a spin-half system where you’re measuring the spin along the z-axis. We know that the possible outcomes are either up or down. So in general your Say your pre-measurement state is a normalised linear combination:
$$|\psi\rangle=\alpha|u\rangle+\beta|d\rangle$$
What this says is that probability of observing the state to be up when measured is given by:
$$P(u)=\big|\langle u|\psi\rangle\big|^2=|\alpha|^2$$
And the probability to observe a down state under spin measurement is similarly given by:
$$P(d)=\big|\langle d|\psi\rangle\big|^2=|\beta|^2$$
Remember that after measurement of any observable the state will always be in exactly one of the eigenstates. Like in this example, the state after measurement is either up or down. And not a linear combination of the two.
| {
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$\mathcal{N} \ge 2$ Supersymmetry massive supermultiplets In Bertolinis SUSY notes [https://people.sissa.it/~bertmat/susycourse.pdf] we have defined:
$$
\{Q^I_\alpha,\bar{Q}_\dot{\beta}^J\}=2m\delta_{\alpha\dot{\beta}}\delta^{IJ}\tag{3.24}
$$
$$
\{Q^I_\alpha,Q^J_\beta\}=\epsilon_{\alpha\beta}Z^{IJ}
$$
$$
\{\bar{Q}_{I\dot{\alpha}},\bar{Q}_{J\dot{\beta}}\}=\epsilon_{\dot{\alpha}\dot{\beta}}\bar{Z}_{IJ}
$$
Also we define:
$$
a_\alpha=\frac{1}{\sqrt{2m}}Q_\alpha\;\;\;\;\;a_\dot{\alpha}^\dagger=\frac{1}{\sqrt{2m}}\bar{Q}_\dot{\alpha}
$$
Lastly, the central charges $Z^{IJ}$ can be written in the form
$$Z^{IJ}=
\left(\matrix{0 & Z_1 \\-Z_1 & 0\\&&0&Z_2\\&&-Z_2&0\\&&&&\ddots\\&&&&&0&Z_{\mathcal{N/2}}\\&&&&& -Z_{\mathcal{N}/2}&0\\}
\right)\tag{3.28}
$$
(Where the charges are non-zero only for even $\mathcal{N}$)
From these we define the following:$$
a^r_\alpha=\frac{1}{\sqrt{2}}\left(Q_\alpha^{2r-1}+\epsilon_{\alpha\beta}(Q_\beta^{2r})^\dagger\right)
$$
$$
b^r_\alpha=\frac{1}{\sqrt{2}}\left(Q_\alpha^{2r-1}-\epsilon_{\alpha\beta}(Q_\beta^{2r})^\dagger\right)
$$
where $r= 1,\dots,\mathcal{N}/2$
These equations satisfy the oscillator algebra:
$$
\{a^r_\alpha,(a^s_\beta)^\dagger\}=(2m+Z_r)\delta_{rs}\delta_{\alpha\beta}
$$
$$
\{b^r_\alpha,(b^s_\beta)^\dagger\}=(2m-Z_r)\delta_{rs}\delta_{\alpha\beta}
$$
How does one "see" that we need to define those equations for $a^r_\alpha$,$b^r_\alpha$?
| The point is to devise an algorithm that constructs supermultiplets given only the commutation relations.
Since you have a nice basis of harmonic oscillator esque creation/annihilation operators, you can define a supermultiplet by postulating a vacuum state $|s\rangle$ annihilated by $a,b$, so that every state in the supermultiplet is given by hitting $|s\rangle$ with creation operators $a^\dagger,b^\dagger$. Since the $a^\dagger$ and $b^\dagger$ are all fermionic the procedure will terminate at a finite number of steps, leading to a finite-dimensional supermultiplet.
You get different supermultiplets by postulating different $|s\rangle$s. You can label them by Poincare casimirs like helicity, mass squared etc. and central charge
| {
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Gravity, matter vs antimatter I have a simple question regarding matter-antimatter gravity interaction.
Consider the following though experiment:
If we imagine a mass $m$ and an antimass $m^-$, revolving around a large mass $M$
the potential energy of mass $m$ should be:
$$ U_1=-\frac{GmM}{R} $$
and the potential energy of mass $m^-$ should be:
$$ U_2=-\frac{GmM}{R} $$
or:
$$ U_2=\frac{GmM}{R} $$
depending on the sign of the gravity interaction between matter and antimatter.
If the two particles annihilate to energy, then the gravitational field of $M$ will interact with the emitted photons and will change their frequency.
But, as the interaction between gravity and the photons has nothing to do with the question of the gravity between matter and antimatter, can't we simply use the interaction between gravity and photons, and the energy conservation to establish the nature of the gravity interaction between matter and antimatter?
| In addition to John's answer:
There is a subtlety in antimatter. In the standard model it is axiomatic that matter and antimatter have the same sign mass. But as long as gravity is not quantized in a theory of all four forces, it is possible that antiparticles,even having a positive mass, instead of being attracted gravitationally by particles, are repulsed.
An experiment is running at CERN to check the assumption that antiparticles fall under the force of gravity.
informed in a comment that two more experiments are running at CERN to determine the behavior of antimatter to gravity: aegis and gbar.
| {
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Velocity after applying a force in the vacuum I’m sorry for so simple question, but I just need to be sure.
I understand, that the changing of the speed occurs only when the force is applied, I understand that if one punch a ball in the free space it will infinitely move with a constant velocity
Some point-like body with mass $m$ is situated in vacuum, and has initial velocity $v_1=0 \space m/s$.
Some force is now acting on a body for a infinitely short period of the time.
The acceleration that gives the application of this force to body equals $a=5 \space m/s^2$.
The velocity after will be $v_2=0+5 =5\space m/s$?
Also, if the force is acting for a non-infinitely short period of time how to calculate then?
I found this from https://physics.stackexchange.com/a/231120/255554
$$x=( x + \frac{|F| }{2m} t^{2} ) $$
Seems it can be applied for both of my cases, but I don’t know why there is 2 times mass
And, can you, please confirm, if 1 Newton is the force that during 1 second changes the 1 kg body velocity on 1 m/s, then 2 Newtons is the force that changes:
*
*if mass is same: during 1 second velocity on 2 m/s
*if mass is 2 kg: during 1 second velocity on 1 m/s
Am I understanding correctly?
| In order to determine the velocity as a result of the application of the force, you need to know the duration (time) of the application of the force on $m$, or the displacement $x$ of the mass $m$ during the application of the force, and the force as a function of time if not constant.
Your value of $v_2$ is based on a constant acceleration of 5 m/s$^2$ due to a constant force applied for a duration of 1 second, and comes from the equation:
$$v_{f}=v_{i}+at$$
where $v_{i}$ is the initial velocity and $v_f$ is the final velocity
You equation for displacement $x$ is based on the equation
$$x_{f}=x_{i}+\frac{at^2}{2}$$
Where $m_f$ and $m_i$ are the final and initial displacements
Substituting $$a=\frac{F}{m}$$
Gives
$$x_{f}=x_{i}+\frac{F}{2m}t^2$$
Hope this helps.
| {
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The Enigma of Universal Gravitation Forces This is taken from a book called "Physical Paradoxes and Sophisms" by V. N. Lange.
1.22. The Enigma of Universal Gravitation Forces
The law of gravitation can be written $F=\gamma\frac{m_1m_2}{R^2}$.
By analyzing this relationship we can easily arrive at some interesting conclusions: as the distance between the bodies tends to zero the force of their mutual attraction must rise without limit to infinity.Why then can we lift up, without much effort, one body from the surface of another body (e.g., a stone from the Earth) or stand up after sitting on a chair?
| Answers which merely refer to the distance between the centre of mass of one thing and the centre of mass of another are missing the point being made by Lange.
When we sit on a chair, the distance between part of your body and part of the chair does tend to zero, so what happens to the force? If our atoms had point-like particles in them with finite mass, then, even though the masses are small, when those particles met there would be infinite forces. It is quantum theory that prevents these infinities, by showing how the mass is not in the end located in point-like regions of space. In classical physics one can get to the same conclusion by postulating that the density has to be finite everywhere.
(This answer amounts to a further comment on/exposition of the one by Qmechanic.)
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Geodesics through the wormhole To define a traversable wormhole, there should be some conditions on the metric components, such as:
I) No event horizon,
II) Minimum value for the shape function (considering a spherical symmetric solution, we should have $b(r_t)=r_t$ ),
III) The flare--out condition ($b'(r_t)<1)$ ans ...
If this condition satisfies besides the null energy condition, we have a traversable wormhole solution with nonexotic matter.
My question is that, could we describe the spacetime through the throat with the wormhole metric or it is another patch? It seems that on the above conditions, it is not possible to use the mentioned wormhole solution in the throat. But, then, why in the literature this metric is used for throat passage description?
| There's a variety of metrics used for the Morris-Thorne metric. The two most common are the Schwarzschild coordinates and the proper radial coordinateS.
The Schwarzschild coordinates are just done using the classic spherical symmetric coordinates,
$$ds^2 = -e^{2\phi_{\pm}(r)}dt^2 + \frac{dr^2}{1 - b_\pm(r) / r} + r^2 d\Omega^2$$
This is done on two different patches glued together at $r_0$, with the functions $+$ and $-$ depending on which part you are on. But you can otherwise switch to proper radial coordinates, via the coordinate trnsform
$$l = \pm \int_{r_0}^r \frac{dr'}{\sqrt{1 - b_\pm(r') / r'}}$$
In which case you get the new metric
$$ds^2 = -e^{2\psi(l)} dt^2 + dl^2 + r^2(l) d\Omega^2$$
which is defined on the entire manifold.
| {
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What does this notation for spin mean? $\mathbf{\frac 1 2}\otimes\mathbf{\frac 1 2}=\mathbf{1}\oplus\mathbf 0$ In my quantum mechanics courses I have come across this notation many times:
$$\mathbf{\frac 1 2}\otimes\mathbf{\frac 1 2}=\mathbf{1}\oplus\mathbf 0$$
but I feel like I've never fully understood what this notation actually means. I know that it represents the fact that you can combine two spin 1/2 as either a spin 1 (triplet) or a spin 0 (singlet). This way they are eigenvectors of the total spin operator $(\vec S_1+\vec S_2)^2.$ I also know what the tensor product (Kronecker product) and direct sum do numerically, but what does this notation actually represent?
Does the 1/2 refer to the states? Or to the subspaces? Subspaces of what exactly (I've also heard subspaces many times but likewise do not fully understand it). Is the equal sign exact or is it up to some transformation?
And finally is there some (iterative) way to write a product of many of these spin 1/2's as a direct sum?
$$\mathbf{\frac 1 2}\otimes\mathbf{\frac 1 2}\otimes\mathbf{\frac 1 2}\otimes\dots=\left(\mathbf{1}\oplus\mathbf 0\right)\otimes\mathbf{\frac 1 2}\dots=\dots$$
| This is actually the decomposition for the tensor product of irreducible representation of SU(2). We can set your $1/2$ as $j$, which means the (2j+1) dimimension irreducible representation of SU(2). Generally, Clebsch–Gordan series gives:
$$D^{\left(j_{1}\right)} \otimes D^{\left(j_{2}\right)}=\bigoplus_{J=\left|j_{1}-j_{2}\right|}^{j_{1}+j_{2}} D^{(J)}$$
thus, it can explains the reason behind $\frac{1}{2} \otimes \frac{1}{2}=0 \oplus 1$. Physically, the $1/2,0,1$ here just means the $S=1/2,0,1$. And the dimension of representation, i.e. $j$, means the number of states in this space, e.g. $s_z=-1, 0, 1$ for $S=1$.
| {
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Intuitive explanation why rate of energy transfer depends on difference in energy between two materials? The temperature of an object will decrease faster if the difference in temperature between the object and it's surroundings is greater.
What is the intuitive explanation for this?
| I think it’ll be helpful to think in terms of the kinetics of the constituent particles. When an object is at a higher temperature, the kinetic energy of its constituents is higher. They are more in motion when compared to the ones with a lower temperature.
Now if there’s a high temperature object in contact with a lower temperature one, there will be transfer of kinetic energy at the interface. Energy will be transferred in both directions except that the net transfer will be from higher to lower. This is because there are more ways for energy to be transferred from the higher to the lower. But as the temperature difference decreases, the rate of transfer from high to low and low to high are closing in. Until they reach equilibrium where the transfer of energy from either sides are now equal. Thus there will be no more net heat transfer on the average.
| {
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Quantization of complex scalar field I'm learning Peskin's qft now and I'm a little confused about problem 2.2 .
Suppose I write the field $\phi(x)$ as:
$\phi(x) =\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{p}}} (a_{p}e^{-ipx}+b_{p}e^{ipx})$
I know that $b_p$ should be written as $b_p^\dagger$ because it annihilate antiparticle, otherwise $b_p$ creates particle with negative energy.
However, when I calculate the Hamiltonian, all I got is:
$\int \frac{d^3p}{(2\pi)^3}E_{p}(a^\dagger_{p}a_{p}+b^\dagger_{p}b_{p})$
In my result, the $b$ particles create positive energy as $a$ particle did. I'm not sure if I did something wrong in calculation or there are some other explanation in the result.
| I am afraid what is wrong is in your initial writing.
A complex scalar field should be written as
$\phi(x) = \int \frac{d^3p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}} (a_p e^{-ipx} + b_p^\dagger e^{ipx})$
And, by complex conjugation,
$\phi^*(x) = \int \frac{d^3p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}} (a_p^\dagger e^{ipx} + b_p e^{-ipx})$
Clearly $a_p^\dagger \neq b_p^\dagger$ as these operators create particles of opposite charge. However in both cases $\omega_p = \sqrt{\vec p^2 +m^2} \gt 0$.
To get the normal ordering of the $b's$ operators, at the end of the demonstration you have to apply the equal-time commutation relations
$[b_p, b_{p'}^\dagger] = (2 \pi)^3 \delta^3 (\vec p - \vec p')$
Note: Your notation of the $b$ operator as creator is confusing. The notation used in literature for a complex scalar field and reported here is a generalization of the notation applied to a real scalar field.
| {
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Must a rainbow's observer also be able to observe the sun? I think I may recall sometimes a rainbow going away when a cloud comes between myself and the sun. I know that the appearance of the rainbow is location dependent. But do we need the sun, rainbow, observer triangle for the effect to manifest?
| While observing a rainbow your back is on sunside, but you dont face the sun (into your eyes). Because light undergoes total internal reflection within each raindrop. What you see is the refracted ray undergoing dispersion.
| {
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Where does energy go in joining capacitors of different capacitance charged by different potential, hypothetically assuming no resistance in circuits? I don't understand why there is any change in initial and final energy since we have already assumed a perfectly conductive circuit. I mean, theoretically at least, there should be no change in energy.
Now, considering there is a change in energy at all, is it because electrons accelerate in moving from one capacitor to another, so energy gets dissipated in the form of electromagnetic radiation?
Image source: NCERT Physics Textbook for Class XII Part I, page 82
| There is a short period of time through which the electrons move from the charged capacitor to the uncharged capacitor. Through this time, assuming the wire has zero resistance, the energy is lost through the magnetic field generated by the accelerating electrons.(We know the electrons accelerate because the potential difference changes, implying the existence of an electric field, which exerts an opposite electric force). If that electric potential is changing, then the force itself is also changing.
But.
You can also think about this in an electrostatic context. Consider two positive charges, when the charges repel and move away from each they lose the potential energy they had. The same is happening to the electrons that were stored in the initially charged capacitor. It is that energy.
| {
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Why do positive charges attract negative charges? From school and/or university we know that there is Coulomb's law that allows us to calculate the amount of force between two electrically charged particles.
However, I have never found any information on the origin of this force.
Is there any explanation as to why this force exists? Is it really fundamental (i. e. it is not the result of any other forces like Archimedes force or the friction force for example)? Is there any evidence here? Does it propagate at light speed or another speed and why?
| The electromagnetic interaction is one of the four fundamental forces of the universe. It just so happens that there are two types of charges that interact under this interaction. If you put two particles with the same type of charge together, they repel each other. Two with different types of charge, they attract each other. Nothing in particular pushed us to call them positive and negative, just that the math worked best and simplest that way. Electric charges aren't really "opposite", just different, we named them that way because of the mathematical relationships we found at first.
Down the line, one discovers that the electromagnetic force is mediated at the speed of light by a particle acting as a carrier, called the photon, or the quantum of light. It is the particle responsible of transmitting and mediating all electromagnetic phenomena, such as electrostatic forces, like you asked about, all the way to electromagnetic waves, such as radio waves and light.
All of these phenomena take place at the speed of light, though in classical physics we disregard the lag that takes place to simplify the scenarios we study for simple systems.
| {
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Has a 3D chart of nuclides ever been published or proposed $(N, Z, S)$? What information might it show? Phys.org's 'Strange' glimpse into neutron stars and symmetry violation leads to the new Nature Physics Letter Measurement of the mass difference and the binding energy of the hypertriton and antihypertriton and that led me to Wikipeida's Hypertriton which says:
Normal nuclei are formed only of protons and neutrons. To study them, scientists arrange the various nuclides into a two-dimensional table of nuclides. On one axis is the number of neutrons N, and on the other is the number of protons Z. Because the antihyperon introduces a third component (strangeness), the table becomes three-dimensional.
However the article only shows the more familliar two-dimensional Z vs N Chart of Nuclides something like that shown below.
which leads me to ask:
Question: What would a 3D chart of nuclides actually look like (neutron and proton number and strangeness; $N$, $Z$, $S$)? Has one been made? If so, what information is entered for each entry?
Example of a more conventional 2D $Z$ vs $N$ chart:
click for larger, Source
| (This anecdote kind of straddles the line between a comment and an answer.)
I saw the beginning of such a table in a conference presentation a decade ago. The format was the same as the usual (Z,N) chart of nuclides, but the data were measured lifetimes for hypernuclei where one baryon was a $\Lambda$. The heaviest nuclei on this chart had mass number $A\lesssim5$ --- it was just the low-mass corner of the table of isotopes.
I suppose you could construct such a table for nuclei where one baryon is a $\Sigma$ with some charge quantum number. Nuclei with more than one strange baryon are unlikely to be experimentally accessible. So the presentation wouldn't really by three-dimensional; it'd be a series of similar-looking two-dimensional diagrams. Nuclides with one $\Lambda$ hyperon, nuclides with one $\Sigma^+$ hyperon, etc.
I only vaguely remember this conference presentation, but I think the presenter was describing work done in Jefferson Lab's Hall B. That might be enough information to point a motivated sleuth towards an actual publication.
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Can momentum never be zero in quantum mechanics? I have seen Zetilli's QM book deals with $E>V$ and $E< V$ (tunnelling) in case of the potential wells deliberately avoiding the E=V case, so I thought maybe something is intriguing about this and made this up.
Suppose the total energy of the particle is equal to its potential energy.Then its kinetic energy should be zero, (speaking non-relativistically). But Kinetic energy operator is $\hat{T}=\hat{p}^2/2m$ (where $\hat{p}=-i\hbar\frac{\partial}{\partial x}$), So clearly since Kinetic energy is 0 here, momentum eigenvalue will also vanish.
Now, Putting $E=V$ in time-independent Schrodinger equation (1D) we get,
$$\frac{\partial^2\psi}{\partial x^2}=\frac{2m(E-V)}{\hbar^2}\psi\implies\frac{d^2\psi}{d x^2}=0\implies\psi=Ax+B$$ where $A$ and $B$ are arbitrary constants. Since, the wave function must vanish at $\pm\infty$, $A=0$,hence the wave function equals a constant=$B$ and is not normalizable.
So, a particle with no momentum(or kinetic energy), gives a physically unrealizable wave function!
Does this imply $E=V$ is a restricted critical case or momentum cant be zero in quantum mechanics or did i just go wrong somewhere?
| You are not wrong, but it is worth noting that the same thing is true of any momentum eigenstate (or closely related unbound eigenstate of a Hamiltonian with a potential well in it). Explicitly
$$
-i\hbar\frac{\partial}{\partial x} \psi(x) = p\,\psi(x)
$$
then
$$
\psi = A e^{i \frac{p}{\hbar}x}
$$
which is not normalisable either. This means that we can never truly realise a momentum eigenstate, but we can still use them as a basis for physically realisable states usng the rigged Hilbert space formalism.
So yes we cannot realise a state with exactly zero momentum, but this is not a special property of the zero momentum state.
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How do we know not all photons are absorbed? Only those of specific energies? When a photon hits an electron in an atom, its energy has to be equal to the difference in energy between the current shell and a shell with a higher energy level, otherwise it is not absorbed at all.
How do we know not all photons are absorbed? Wouldn't at least some energy of the photon be absorbed since it is an oscillation in the EM field?
| Here is the spectrum of light coming from the sun.
Solar spectrum with Fraunhofer lines as it appears visually.
The spectrum seen has all the wavelengths ( frequency = c/wavelength), and the dark lines are the absorption lines. Absorption and emission lines are one of the reasons quantum mechanics had to be invented.
Wouldn't at least some energy of the photon be absorbed since it is an oscillation in the EM field
The photon is a point elementary particle of energy=to $hν$, it is not a classcical electromagnetic oscillation. It is described by a wavefunction which is not measurable. The complex conjugate square of the wavenctions give the probability of finding the photon at (x, y,z,t) with energy E , and the dark lines show an overwhelming probability for the specific frequency/energy photon to be absorbed.
Light is composed by zillions of photons superimposed, but a photon is not light. (a buildingis made of bricks, but a brick is not a building is a classical analogue)
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BEC in two dimensions with interactions For noninteracting 2D system of bosons, many textbooks have a statement that no BEC exists as the system is capable to accommodate infinite number of bosons when chemical potential $\mu \rightarrow 0 $. But what happens if interaction is taken into account?
| Bose-Einstein condensation per se is a non-interacting effect, solely driven by particle statistics. In $d=2$ dimensions, free space, the energy density is such that particles can still be accommodated in the excited states for $T \neq 0$ thereby not triggering the macroscopic occupation of the ground state.
This does not mean that you cannot superfluidity (SF) in 2D. Superfluidity is not the same thing as BEC. SF just means that you have a critical velocity $v_{\mathrm{c}}$ below which the fluid experiences no dissipation. But $v_{\mathrm{c}}$ depends on the interaction strength, so if you want in 3D the BEC is a "boring SF" with $v_{\mathrm{c}} = 0$.
You can still have a mechanism that allows SF in a 2D fluid. It's called the BKT transition, and is driven by the energy favourability of creating free/bound vortices. The vortex size is determined by the interaction length (healing length of the wavefunction), so it only plays a rôle for an interacting gas. You don't consider BKT in a 3D gas because they are unstable anyway (line vortex instability).
Interesting literature:
*
*this paper, where they connected the BEC and BKT phase transitions by changing the interactions strength of a 2D/3D trapped atomic gas.
*this paper, where they look at a BEC on the surface of a sphere, where, depending on the radius of curvature, you can have BEC or BKT.
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Outside temp of vessel of is 40 Deg C and Water inside it is 86 Deg C.. Why? I was boiling water in a steel utensil.
Used Fluke's IR Gun (59 Max) to measure the temperature of Water inside AND the surface temp of vessel outside.
The boiling water read 86 Deg C whereas the outside surface temp of vessel read just 40 Deg C. Why is that? If I touched the outside of the vessel it would 'feel' extremely hot but measurement shows just 40 Deg C. So what's going on here?
| When the surface is reflecting (like the metal) you are actually measuering then temperature of what can be seen in the reflection, not the reflecting body.
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"timestamp": "2023-03-29T00:00:00",
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Why is the speed of light in vacuum a universal constant? While getting familiar with relativity, the second postulate has me stuck. "The speed of light is constant for all observers". why can't light slow down for an observer travelling the same direction as the light?
| The constancy of speed of light was first predicted by Maxwell. He had discovered for equations, which we now call Maxwell equations.
Maxwell equations
$$\nabla\cdot{E}=\frac{\rho }{\epsilon_{o}}$$
$$\nabla\times{{E}}=-\frac{\partial B}{\partial t}$$
$$\nabla\cdot B=0$$
$$\nabla\times B=\mu_o j+\mu_o\epsilon_o \frac{\partial E}{\partial t}$$
These four equations represent the Maxwell equations.
First let’s see what are the conditions/constraints of the vacuum equations. In a vacuum there is no
charge, so ρ just becomes zero. Also the is no change in current in vacuum, which means $\frac{\partial j}{\partial t}$ will be zero
Electromagnetic Waves and Maxwell Equations
Integrating the second Maxwell equation with respect to time we get
$$ B=-\int{\nabla\times{ E}}{dt}$$
Now let's put this expression for B into the fourth equation. Note we are allowed to interchange the positions of the integral and curl in this case.
$$-\int\nabla\times{\nabla\times E}dt=\mu_o\epsilon_o\frac{\partial E}{\partial t}$$
Now this equation simplifies as
$$-\frac{\rho}{\epsilon_o}+\nabla^2 E=\mu_o\epsilon_o\frac{\partial^2 E}{\partial t^2}$$
Now as we have written the equations for vaccum, $\rho =0$ hence our equation just becomes
$$\nabla^2 E=\mu_o\epsilon_o\frac{\partial^2 E}{\partial t^2}$$
Now this equation is very similar to the standard wave equation which is
$$v^2\nabla^2 \phi = \frac{\partial^2 \phi}{\partial t^2}$$
Thus by comparison, we get
$$c=\frac{1}{\sqrt{\mu_o\epsilon_o}}$$
Alright but what does this have to do with the constancy of speed of light in vaccum?
The fact is that $\mu_o$ and $\epsilon_o$ are independent of reference frame, and hence, the speed of light in vaccum is a fundamental constant irrespective of the reference frame in question. This was the start of Theory of Relativity as proposed by Einstein.
Experimental Evidence
After the predictions, Michaelson and Morely performed the famous experiment which proved that the speed of light was to be independent of reference frame or else the Earth wouldn't be moving
| {
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Why dont we consider the cohesive forces between the surface and inner molecules while formulating excess pressure inside a droplet? As the surface of a liquid will have imbalanced cohesive forces leading to the phenomenon of surface tension,we just consider the forces among the surface molecules which act tangentially and dont consider the component of cohesive force towards the center of the droplet while formulating excess pressure inside a droplet due to surface tension.Is there any valid reason for this or is it just a convention for the sake of simplicity?
| The surface tension is due precisely to the cohesive forces that you say are ignored. Surface tension is defined to be the surface free-energy $F=E-TS$ per unit area of the surface, and so takes into account the unsatisfied bonds that are present because the inside and outside environments of a surface molecule are different- i.e because the bulk is pulling the surface molecules inwards. The tangential bonds differ very little from the ones in the bulk. As long as there is an energy that depends on the surface area it acts like a tangential force because increasing the area costs energy -- just as stretching a rubber band costs energy. This is very clear when you use the virtual work method of computing the excess pressure by saying that
$$
\sigma \delta {\rm Area}= P_{\rm excess} \delta {\rm Volume}
$$
| {
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Two versions of Diffusion coefficient I found two versions of the Diffusion coefficient, first:
$$D=\frac{\pi \lambda }{8}\overline{c}$$
Where $ \overline{c}$ ist the particles mean thermal velocity and $\lambda$ the particles mean free path.
(Found in W. C. Hinds, Aerosol Technology. Wiley Interscience (1999). S.156)
and second the version from my lecture (also found on wikipedia):
$$D=\frac{ \lambda}{3} \langle v\rangle$$
Where $\langle v\rangle$ is the particles mean thermal verlocity.
Hinds gives two more equations (S.154):
$$c_{\text{rms}}=\sqrt{\frac{3kT}{m}}$$
$$\overline{c}=\sqrt{\frac{8kT}{\pi m}}$$
With $c_{\text{rms}}$ being the root mean square velocity of a particle, $k$ being the Boltzmann-constant, $T$ being the temperature and $m$ the particles mass.
My first thought was that maybe there is a typo in the book or the lecture notes so I calculated
$$\frac{c_{\text{rms}}}{\overline{c}}=\sqrt{\frac{3\pi}{8}}$$
which led to
$$\overline{c}=\sqrt{\frac{8}{3\pi}}c_{\text{rms}}$$
so I replaced $\overline{c}$ in
$$D=\frac{\pi \lambda }{8}\overline{c}$$
just to come to
$$D=\frac{\pi \lambda }{8}\sqrt{\frac{8}{3\pi}}c_{\text{rms}}$$
Which (set $\langle v\rangle=c_{\text{rms}}$, due to the potential typo) because of the square root isn't
$$D=\frac{ \lambda}{3} \langle v\rangle$$
Where did I make a mistake?
| So to start out, you're looking at the RMS and mean of the Maxwell-Boltzmann distribution; you can look them up on Wikipedia, where they match Hinds' expressions.
In the two expressions you have, you see they'd cancel out were you to get rid of the square root. So, I believe $\lambda$ to be defined differently in your source material: In the one in Hinds, $\lambda = \bar{c}\tau$, whereas on the page on Wikipedia you found, I take it $\lambda = c_\text{rms}\tau$.
| {
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Conservation of energy of 2 identical Rolling Disks with and without friction My physics book claims that if two identical disks moving at the same velocity travel up nearly identical hills, with the second hill not having friction, then the disk rolling up the first hill will travel to a greater height. Given that the disks started with the same kinetic energy while rolling at the base of the hill, shouldn't they reach the same height (i.e. same potential energy) as a result?
| Cars can drive up hills. Their tires' rotation combines with friction to move forward.
When the friction's removed, e.g. due to the road being icy, the tires' rotation doesn't have the same effect.
That said, the textbook's a bit off: friction means that the disc's rotation matters, but this doesn't necessarily mean that it goes further as it could instead go shorter if it's rotating like a car's tire when driving in reverse.
I'd speculate that the textbook meant to compare a disc moving with perfect slippage vs. perfect traction, because if there's perfect traction, that'd imply that the disc is rotating like a car's tire driving in that direction. However, merely saying that there's friction isn't the same thing.
| {
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Decay of electron? Have we detected any decays of electrons to an electron neutrino and $W$-boson in Fermilab or in CERN? Are neutrinos the only possible stable leptons inside an electroweak field?
| A very interesting question. If we ignore charge conservation law, then electron could decay into neutrino and photon :
$$ e^− \to ν_e + \gamma $$
Current estimates gives that life time of electron $\gt 10^{26} \,\text{years}$. Feynman diagram of such electron decay :
It is calculated that energy needed to break electron into neutrino + photon is on the order $\approx 10^{22} \,\text{GeV} $. So it's by $1'000$ times greater than Plank energy !
As far as I know CERN achieved biggest record of it's collision energies $\approx 10^6 \text{GeV}$ at 25 Nov 2015.
No need to say that this is only very very tiny amount of energy required for actually breaking the electron. Probably for this event to occur we need a particle accelerator with diameter comparable to our Sun planetary system size or even greater. Which would cost some gazillions of USA yearly budget. So as for now technologically we will not be able to test if charge conservation law holds or not for electron.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why complex numbers are used in electronics? The impedance of a capacitor or an inductor is imaginary. How do we know these quantities are imaginary?
| A capacitor (with capacitance $C$) is fully described by
the differential equation between current $I(t)$ and voltage $V(t)$:
$$I(t)=C\frac{dV(t)}{dt} \tag{1}$$
Suppose you have an AC voltage with frequency $\omega$
connected to the capacitor. By using the complex calculus this is
$$V(t)=V_0 e^{j\omega t} \tag{2}$$
Then, by plugging voltage (2) into differential equation (1),
you get the current through the capacitor
$$I(t)=C V_0 j\omega e^{j\omega t} \tag{3}$$
The impedance is defined to be
$$Z=\frac{V(t)}{I(t)}.$$
From (2) and (3) you get the impedance of the capacitor
$$Z=\frac{1}{j\omega C}.$$
From the $j$ you see, this is a purely imaginary value.
The impedance of a inductor (with inductance $L$) can be derived
in a very similar way, except that here you begin with the
differential equation
$$V(t)=L\frac{dI(t)}{dt}.$$
From that you finally get the impedance of the inductor as
$$Z=j\omega L.$$
| {
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Why can vector components not be resolved by Laws of Vector Addition? A vector at any angle can be thought of as resultant of two vector components (namely sin and cos).
But a vector can also be thought of resultant or sum of two vectors following Triangle Law of Addition or Parallelogram Law of Addition, as a vector in reality could be the sum of two vectors which are NOT 90°.The only difference here will be that it is not necessary that components will be at right angle.
In other words why do we take components as perpendicular to each other and not any other angle (using Triangle Law and Parallelogram Law).
| The vectors can be arbitrary and there are good reasons explained in the other answers. But I would like to approach this form another angle.
Often you as the person doing the problem setup can choose whatever coordinate system you like. Then, unless there is a apparent benefit of not choosing a orthogonal basis, you would probably choose a orthogonal basis for expediency and ease of decomposing of vectors. In the same way as you would like to prefer choosing the length of your all your basis vector to be 1. Again you could choose different values for different axes, but it makes reasoning about your results easier if you chose easily understandable bases. Again unless you have a reason to do otherwise.
Changing between bases is easy enough.
| {
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Why is the electric potential at infinity zero? As per net results, the potential at infinity is considered to be zero. Apart from considering this as a physics law, is there any proper reason why we consider potential at infinity to be zero?
| By definition the potential energy is chosen to be zero at infinity. It can also be defined to be zero at the ground. Generally speaking the work $W$ done for moving a body against a force $\vec F(\vec r)$ from point $A$ to $B$ is given by the difference of potential energies
$$
W = U(\vec r_A) - U(\vec r_B).
$$
That is to say the potential energy at point $B$ is
$$
U(\vec r_B) = U(\vec r_A) - \int_{\vec r_A}^{\vec r_B}\vec F(\vec r)\cdot d\vec r\,.
$$
As we can see, we can choose our reference point $\vec r_A$ and potential energy there $U(\vec r_A)$ arbitrarily since it cancels out with the integral.
| {
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How can you (computationally) calculate the halfchain entanglement entropy of a spinchain? I am simulating a (small) spinchain with exact diagonalization and dynamics. I would like to track the entanglement entropy of half the chain with the other part of the chain.
I have the vectors of my state $|\Psi(t)>$ in the basis $|s_1, s_2, .., s_i, .. , s_n>$ which are the spins at each sites at a certain time. I know you can get the density matrix of this by taking the outer product $\rho = |\Psi(t)>< \Psi(t)|$ which is a $n \times n$ matrix, but this is a pure state so entanglement entropy will be zero, so now I want the density matrix of the subsystem which includes the first (or last) half of the spin chain. This probably includes tracing out certain parts of the density matrix $\rho$ to $\rho_A$, but how do I do this specifically, what parts?
This is specifically a computational question because I am simulating in Python, but it's not the coding part that's troublesome, I really have no idea how to get the "first half of the chain" out of this full density matrix.
| It all boils down to a two-partite system,
$$
\vert\psi\rangle = \sum c_{ij} \vert i,j\rangle\ ,
$$
where $i$ and $j$ are all indices in the left and right part, respectively. Then, the reduced density matrix is given by
$$
\rho = \sum_{ii'} \left(\sum_j c_{ij}c^*_{i'j}\right)\vert i\rangle\langle i'\vert\ .
$$
That is, in the standard basis $\rho = CC^\dagger$, where $C=(c_{ij})$ is the coefficients of the pure state $\vert\psi\rangle$, arranged in a matrix.
So computationally, what you want to do is to take your coefficient vector $c_{ij}$, reshape it into a matrix $C$ (this comes at zero computational cost as it just changes the way the object is indexed), and then compute its singular values. The square of those singular values are then the eigenvalues of $CC^\dagger$, and from those, you can easily compute the entropy.
| {
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Why can a partial derivative be added to a hamiltonian in canonical transformations? In canonical transformations, how come we allow hamiltonian to change by a partial derivative of time?
$$H'(P, Q, t) = H(p, q, t) + \frac{\partial F}{\partial t}.$$
Here $F$ is the generating function.
I mean geometrically that is not how a function should be transformed when there is a change of variables. Geometrically it should be $$H'(P, Q, t) = H(p, q, t).$$
In Lagrangian mechanics it is indeed so
$$L'(Q, \dot{Q}, T) = L(q, \dot{q}, t).$$
| In a canonical transformation, the new hamiltonian could have nothing to do with the initial hamiltonian, it just have to preserve Hamilton's equations. So in the new variables $(Q,P,t)$ you have to have that
$$\dot{Q} = \frac{\partial K}{\partial P} \qquad \dot{P} = -\frac{\partial K}{\partial Q}$$
where $K$ is the new Hamiltonian. Whenever this happens
$$K(Q,P,t) = H(q(Q,P),p(Q,P),t)$$
we call the transformation completely canonical (with the added bonus that the transformation is time independent), and it's a particular type of canonical transformation.
| {
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Why do antiferromagnets occur at lower temperature than ferromagnets? The minimal model for describing magnets is the Heisenberg Hamiltonian
$$H = -\frac{1}{2}J\sum_{i,j} \mathbf{S}_i \cdot \mathbf{S}_j$$
Where $i,j$ are nearest neighbors and the factor of $1/2$ is for double counting.
If $J$ is positive, spins will want to align to save energy (ferromagnets), and if it is negative they will anti-align (antiferromagnets). Ultimately $J$ comes about from Pauli exclusion and electrons not wanting to sit in the same orbital (Coulomb repulsion).
But if I look at a table of ferromagnets here, I see transition temperatures up to 1400 K. On the other hand, the highest transition temperature for antiferromagnets is a measly 525 K, with most being below room temperature.
Why do antiferromagnets generally occur at significantly lower temperatures than ferromagnets?
One can argue that maybe $\vert J\vert$ is larger in ferromagnets than antiferromagnets (as one of the current answers does), but this just begs the question. Why should that be the case (assuming it is true)? I don't see an experimentally-verified theoretical basis for asserting $\vert J_{\mathrm{AFM}}\vert < \vert J_{\mathrm{FM}}\vert$.
This question came up in a class I am teaching to talented senior undergraduates.
| It is the value of the exchange parameter that is smaller for antiferromagnets than for ferromagnets.
For example, Iron has $J$ of roughly 0.3 eV and La$_2$CuO$_4$ of 0.13 eV. Iron has a 1000K transition temperature and La$_2$CuO$_4$ of about 325K.
| {
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Chromatic aberration in lenses
We all know that concave lenses act as diverging lenses, but while searching for remedies for chromatic aberration, I observed a concave lens (in the picture see flint glass) acting as a converging lens, could someone please explain why?
| Actually, a lens does not necessarily behave as diverging or converging under all conditions.
In the picture you have posted, if you notice carefully, the incident ray is converging.
Usually, we consider diverging incident light rays as "real" objects (because the rays will emerge from a point in front of the lens) and converging light rays as virtual objects.
Also, note that diverging refracted (or reflected rays) are virtual and converging refracted (or reflected rays) are real.
Now, coming to your question, assume the flint glass was absent. The image would be obtained a bit closer. But when you insert the flint glass the image shifts further away, obviously because the flint glass diverged the rays, but not enough to make it a virtual image.
This can also be seen by the quantity "Power of a lens". The converging power of the crown glass is more than the diverging power of the flint glass.
You can apply similar methods and make a plane mirror produce a real image, and so on.
And it does not just stop there. The mere shape of a lens does not say whether it is converging or diverging. Even a convex lens placed in a fluid of proper refractive index will behave like a concave lens!
| {
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Collision Thought Experiment I am thinking of a system of isolated, moving, and colliding particles in a frictionless box. The collisions were inelastic. My question is:
Supposing that the initial total momentum of the system is not zero. Since the collisions are inelastic, kinetic energy will not be conserved (converted to heat, sound, etc.). But since the isolated system experiences no net external force, the momentum could not change (Newton’s Second Law). Hence, total momentum of the system will remain constant. If I let the particles in the system collide for awhile, eventually, they will build up heat, but then again the momentum could not change. The particles would not stop until the box overheats.
What’s the problem with the scenario and my line of thinking?
| In many-particle problems one has to distinguish the constants characterizing the movement of the system as a whole (it energy, the three components of the center-of-mass momentum, and the three components of the angular momentum), and the relative movement of the parts in the system.
So, indeed, the relative movement of the particles will eventually cease and its temperature will increase, but the seven integrals of motion (constants) mentioned above will conserve (if there are no external forces).
Remark
The concept of the particles that convert their kinetic energy into heat is suspect, since heat is the kinetic energy of the particle movement. So I suppose what is meant as particles is really macroscopic objects.
| {
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Interplay of energy between photon, electronic energy levels and kinetic energy I can understand that atoms have quantized energy levels for its electrons, but an atom's translational kinetic energy is continuous. As such, why is the absoprtion spectrum not continuous? That is to say, why can't the excess energy from the photon beyond what is needed to promote electrons be simply converted into kinetic energy?
On a related topic, can an excited atom simply move its electron to a lower energy level and convert that into translational kinetic energy of the atom (the same atom itself)? If so, why can't it convert only a fraction of that energy and emit the remaining fraction as a photon (thereby also producing a continuous emission spectrum)? After all, an excited atom can collide another atom and pass that energy as kinetic energy to that other atom.
| Because both momentum and energy have to be conserved when the atom absorbs the photon.
Suppose we have an atom with a mass $m$ and the energy difference between the initial and final levels is $E$. The photon energy is $hf$, so conservation of energy gives us:
$$ hf = E + \tfrac12 m v^2 \tag{1} $$
where $v$ is the speed of the atom after absorbing the photon. However a photon also has a momentum $hf/c$ so conservation of momentum gives us:
$$ \frac{hf}{c} = mv \tag{2} $$
and combining equations (1) and (2) we find:
$$ E + \tfrac12 m v^2 = mvc $$
which gives us a quadratic for the final velocity of the atom:
$$ v^2 - 2vc + \frac{2E}{m} = 0 $$
or:
$$ v = c\left(1 \pm \sqrt{1 - 2E/mc^2} \right) $$
If we assume $E \ll m$ we can expand the square root using a binomial approximation to get:
$$ v = c\left(1 \pm \left(1 - \frac{E}{mc^2} \right) \right) $$
and we can ignore the solution greater than the speed of light since it's unphysical, so we end up with:
$$ v = c \frac{E}{mc^2} \tag{3} $$
(I've written it this way because $mc^2$ is the rest energy of the atom so the equation makes clear that the key factor is the ratio of the excitation energy to the rest energy.)
So there is one, and only one, possible velocity the atom can have after absorbing the photon. That's why the absorption line is sharp and not a continuum.
The velocity after absorbing the photon is generally negligibly small. For example consider a hydrogen atom absorbing the $10.2$ eV photon required for the $1s \to 2p$ transition. Using equation (3) gives us $v \approx 3$ m/s, and the kinetic energy associated with this velocity is only about $10^{-8}$ eV.
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