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Why use Fourier expansion in Quantum Field Theory? I have just begun studying quantum field theory and am following the book by Peskin and Schroeder for that.
So while quantising the Klein Gordon field, we Fourier expand the field and then work only in the momentum space. What is the need for this expansion?
| First of all, this is just a change of basis, which is up to us to make. Furthermore we should always choose a basis that makes our calculations easier, and hopefully makes things more intuitive. For a simpler example - just try finding the volume of a sphere in cartesian coordinates, its just a bad choice.
Second of all, you don't have to use a Fourier basis, to my knowledge everything -loops renormalization etc can be done in a position basis.
Now as to why the Fourier basis is a convenient choice:
(1) It simplifies derivative terms in the Lagrangian - as usual the Fourier basis turns derivative expressions into algebraic ones, which are much easier to manipulate.
(2) It it more intuitive - written in terms of a Fourier basis the Feynman rules are in terms of momentum. So for example at the vertices momentum is conserved - its just a nice tidy way to think about whats happening at the vertex.
(3) Even if you start in position space, one method for doing the integrals you will encounter when writing for your loop expressions will be going to momentum space - so you sort of cut this step out from the outset.
(4) (following up on Vibert's comment) Plane waves are the basis we do the experiment in. That is, we send in wave packets highly localized in p space, i.e. this is the exact solution we perturb around.
| {
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What happens if an atom absorbs a photon of energy higher than first excited state but lower than second excited state? Since the energy levels of atoms are quantized, I was wondering what happens if an electron is hit by a photon whose energy is higher than electron's first excited state but lower than second excited state. Does it excite to the first excited state? If yes, what happens to the remaining energy?
| Let the electron ground state have energy $E_g$, let the first excited state have energy $E_1$, and let the second excited state have energy $E_2$.
Let the energy of the photon be given by $E_p = hf$.
Now it isn't the energy of the exited states that is important in transitions, but the energy differences between states.
So instead I'll assume you mean the following condition: $(E_1 - E_g) < hf < (E_2 - E_g)$.
Now as the atom absorbs $hf$, it causes the electron to transition to the first excited state. There is therefore excess energy of $hf - (E_1 - E_g)$. This excess energy is either converted to kinetic energy in the atom, or is reflected as a new photon with lower frequency and energy $E' = hf'$.
| {
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3D: Get linear velocity from position and angular velocity I want to find out the linear velocity of a point in 3D space, (Euclidean), given:
*
*Its position
*Its angular velocity
*The point it's rotating around (fulcrum)
(This is a problem I need to solve for 3D graphics programming with a physics engine).
The position of the point and position of the pivot point will be 3-value vectors, $x$, $y$ and $z$.
The angular velocity will also be a 3-value vector, representing Euler angles.
What operation(s) would I need to perform to calculate the linear velocity of the point?
The 3d/physics engine has various high level mathematical operations including matrix, vector and quaternion operations, so hopefully what I need is among those.
| Let $\vec r_0(t)$ denote the point around which the object is rotating and $\vec r(t)$ the position of the object. Then the fact that the particle is rotating around the point $\vec r_0(t)$ can be formalized by the mathematical statement that
$$
\vec r(t) - \vec r_0(t) = R(t) \vec c
$$
for some constant vector $\vec c$ and time-dependent rotation $R(t)$. It follows that
$$
\dot{\vec r}(t) - \dot{\vec r}_0(t) = \dot R(t) \vec c = \dot R(t)R(t)^{-1}(\vec r(t) - \vec r_0(t)) = \vec \omega(t)\times (\vec r(t) - \vec r_0(t))
$$
(the last equality is a standard result about rotations) so we have the final result
$$
\dot{\vec r}(t) =\dot{\vec r}_0(t)+\vec \omega(t)\times (\vec r(t) - \vec r_0(t))
$$
which is what you were looking for I believe.
Cheers!
| {
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Why does Venus transit so slowly? I have calculated that because Venus is $d = 12,103.6~\mathrm{km}$ in diameter and moves at $v = 35.02~\mathrm{km}/\mathrm{s}$, it would take
$$ t=\frac{d}{v} = \frac{12,103.6~\mathrm{km}}{35.02~\mathrm{km}/\mathrm{s}} = 345.62~\mathrm{s} = 5~\mathrm{min}~46~\mathrm{s} $$
for Venus to appear totally in front of the Sun. This time would be from the edge of Venus being against the edge of the Sun to when the opposite edge of Venus is "in touch" with the same edge of the Sun.
But now I think this in reality takes more than just $6$ minutes (about $20$ minutes). If this is true, then why does this measurement not agree with theory?
| Your calculation looks like it would be correct for an observer who was stationary with respect to the Sun. However, the Earth is also moving, and I think this accounts for at least most of the discrepancy.
Earth's orbital period is about 365 days while Venus' is 225, so the angular velocity of Earth about the Sun is 0.625 times that of Venus. Now, draw an imaginary line from an observer on Earth to the edge of the Sun. Imagine that Venus is a disk rather than a sphere. Assuming Earth and Venus are moving parallel to one another (which isn't quite true but is probably close enough), we can say that the intersection between this line and this disk moves at about $35.02 \times (1-0.625)$ km/s across the surface of the disk. Thus, the time in which Venus only partially eclipses the Sun should be
$$
\frac{12,103.6~\mathrm{km}}{35.02~\mathrm{km}/\mathrm{s}\times (1-0.625)} = 921.65~\mathrm{s} = 15~\mathrm{min}~22~\mathrm{s}.
$$
That's not quite 20 minutes, but it's a lot closer than 6 minutes.
As Chris White comments, most of the rest of the discrepancy can probably be accounted for by noting that the disk of Venus doesn't travel straight through the centre of the disk of the Sun, and consequently it intersects the edge of the sun at an angle rather than perpendicularly, as shown in the image below. This means that the disk of Venus has to travel more than its diameter between the first and second contacts between the edges of the two disks.
| {
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What's the exact gravitational force between spherically symmetric masses? Consider spherical symmetric$^1$ masses of radii $R_1$ and $R_2$, with spherical symmetric density distributions $\rho_1(r_1)$ and $\rho_2(r_2)$, and with a distance between the centers of the spheres $d$. What is the exact force between them? I know point masses are a good approximation, but I'm looking for an exact formula. This would be useful for a gravity-simulation toy software.
--
$^1$ Assume for simplicity the idealization where tidal or centrifugal forces do not deform the spherical symmetric, i.e., the various mass parts are held in place by infinitely strong and rigid bonds.
| If you are looking only for the Newtonian gravitational force, the other answers to this question are correct. The spherically symmetric mass distributions can be replace by the total mass at the center of mass and then the Newtonian gravitational force can be computed for these two masses.
However, if you want EXACT calculations, you must use general relativity. In general relativity it is also true that a spherically symmetric mass distribution can be replaced with the Schwarzchild black hole of the same mass at the same center. However this is for a single spherical distribution. It is probably not true for two spherical distributions or for two Schwarzchild black holes. There is no know exact general relativity solution to the two black hole problem. All we have is very good numerical simulations of general relativity for two black holes.
| {
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Superconducting gap, temperature dependence: how to calculate this integral? Tinkham (page 63) states that the temperature dependence of the gap energy of a superconductor $\Delta(T)$ can be calculated using the following integral:
How can this actually be carried out? I am not sure how to approach this problem or re-arrange the equation for finding $\Delta(T)$ numerically.
A URL containing a little more inforamtion about this Eqn: http://katzgraber.org/teaching/ss07/files/burgener.pdf (slide 35)
| I have not tried it on this specific equation, but in principle you can solve problems like this by a combination of numerical integration and a root finding algorithm. For a given variable $x$ you wish to determine for a fixed value $v$ of the integral, the root finding algorithm will find a solution to the equation
$v-integral(x)=0.$
The root finding algorithm will try to match the variable in such a way that the equation is satisfied, while the integral is evaluated numerically. In your example, $\Delta$ corresponds to the variable $x$ while $1/N(0)V$ takes on the role of v.
| {
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Difference between torque and moment What is the difference between torque and moment? I would like to see mathematical definitions for both quantities.
I also do not prefer definitions like "It is the tendancy..../It is a measure of ...."
To make my question clearer:
Let $D\subseteq\mathbb{R}^3$ be the volume occupied by a certain rigid body. If there are forces $F_1,F_2,....,F_n$ acting at position vectors $r_1,r_2,...,r_n$. Can you use these to define torque and moment ?
| Moment is the more general term which means quantity evaluated when something is multiplied by its moment arm (perpendicular distance).
Some examples of moments:
*
*Moment of force (torque): $\vec{r} \times \vec{F}$
*Moment of rotation (velocity): $\vec{r} \times \vec{\omega}$
*Moment of impulse: $\vec{r} \times \vec{J}$
*Moment of momentum (angular momentum): $\vec{r} \times \vec{p}$
So is torque equivalent to moment of force? In my opinion no, because the above moments require a generating vector (force, rotation, impulse, and momentum) to be present. But you can have torque without a force, but with a force couple. I prefer to use the term pure torque instead of force couple because in this case a torque vector $\vec{\tau}$ can stand on its own, without needing to define the details of the force couple (force, separation, and direction).
So torque can have one of two meanings depending on the context
$$ \text{(torque)} = \begin{cases}
\vec{r}\times \vec{F} & \text{(moment of force)} \\
\vec{\tau} & \text{(pure torque)} \end{cases} $$
For example, a shaft carries a pure torque, but a lever transfers a force moment from one end to the other.
| {
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Is it possible to calculate atmospheric pressure if given temperature (F) and elevation? I am working on a report at work and need to determine the atmospheric pressure for small intervals over a 24 hour period. Searching Google, I've found charts which give a base pressure of 14.65 psia at sea level. This is at 68F. That changes to 13.17 psia at 3000ft above sea level.
What I am looking to do is create a spreadsheet where I enter the elevation as a constant, then provide the temperature (F), then have it calculate the atmospheric pressure. Knowing it is 13/17 psia at 68F is useful only if the temperature is 68F for the entire 24 hour period, but it isn't. Currently it ranges from 30F to 75F but could move either direction substantially depending on time of year.
Is this possible to determine?
| There is a standard model of the atmosphere
It's also built into matlab, I don't know of any spreadsheets but some googling should find one.
| {
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Please explain this statement about Lorentz transformations I'm reading Sternberg's Group Theory and Physics. I have a question about chapter 1.2 Homeomorphisms.
Background:
A Lorentz Metric is defined as $||{\bf x}||^2=x_0^2-x_1^2-x_2^2-x_3^2$
And a Lorentz Transformation $B$ as one which satisfies $||B{\bf x}||^2=||{\bf x}||^2$
Identify ${\bf x}$ with $x=x_0I+x_1\sigma_1+x_2\sigma_2+x_3\sigma_3$ where $\sigma_i$ are the Pauli matrices.
We have $x=x^*$ ($x$ is self-adjoint) and $\det x=||{\bf x}||^2$
Let $A$ be any 2x2 matrix and define the action of $A$ on $x$ as $x \to AxA^*$ and denote the corresponding action on the vector ${\bf x}$ as $\phi(A){\bf x}$
$AxA^*$ is also self-adjoint and $\det AxA^*=(\det A)^2\det x$
Therefore if $A \in SL(2,\mathbb{C})$ then $\phi(A)$ is a Lorentz transformation and also a homeomorphism.
Ok so far, here begins my question.
The next bit of the text says: suppose $A \in SU(2)$, then $AIA^*=I$
so if ${\bf e}_0 = \left( \begin{array}{c}
1 \\
0 \\
0 \\
0 \end{array} \right)$ then $\phi(A){\bf e}_0 = {\bf e}_0$
I see why this is so. Now here's the bit I don't understand.
If a Lorentz Transformation $C$ satisfies $C{\bf e}_0 = {\bf e}_0$ then $C$ also carries the space of vectors $\left( \begin{array}{c}
0 \\
x_1 \\
x_2 \\
x_3 \end{array} \right)$ onto itself.
I don't see why this is so. Can someone please explain the logical steps which I'm missing? Thanks.
| Hints:
*
*Note that the time coordinate $x^0=\frac{1}{2} {\rm tr}(x)$ is given by the trace.
*Note that the trace ${\rm tr}(x)$ is invariant under the $SU(2)$ action.
*The subspace with $x^0=0$ is invariant under $SU(2)$.
*Note that the stabilizer/isotropy subgroup of $e_0$ is $SU(2)$.
| {
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Does gravitational force attract bodies with mass or with energy? On my textbook is written that gravitational force is the force that attracts bodies with mass. But I've seen on a book that It actually attracts bodies with energy. I'm having a class tomorrow and I would like to know some argumments to use with - against my professor.
| In relativity gravity is not a force anymore, instead it is geometry of spacetime. Therefore gravity (produced by energy which is equivalent to mass) would also attract massless particles without any force acting.
| {
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Why can we use Gauss' law to compute electric field? For simplicity I'm considering only the sphere case.
In the Gauss' Law formulation we have some field $E$ introduced by charges $Q$ inside some sphere, then we compute flux and integrate, and we get result $Q/e_0$. Right.
But this formulation doesn't take into account any possible outer charges, because $E$ used in this law comes only from $Q$. I suppose that any outer field effects will cancel out, because flux coming from it must go through surfaces with opposite normals. But we didn't make any proof of that, as far as I have seen in Gauss' Law proofs.
So, when we measure electric field inside the sphere with uniformly distributed $Q$, we make some assumptions on uniformity of $E$ inside this sphere and we consider hollow Gauss' sphere with radius smaller than radius of the outer sphere. And then we use $E$ to compute total flux going through inner sphere and conclude that it must be zero. Right.
But why can we use this $E$? We don't know that there aren't any effects from outer charges fields. So why can we use Gauss' Law?
| First of all we use Gauss's law because it is a law of nature!
Second: Gauss's law (together with $\nabla\times\vec{E}=0$, which is only strictly valid in electrostatics) is equivalent to Coulomb's law for the electric field of a distribution of charges. You can prove one from the other.
Third: You a right about external fields cancelling. Gauss's law essentially says that electric field lines only end on charges. So if you have a volume that doesn't contain a charge, any field line which enters will also leave and the total flux through the closed surface is zero.
Fourth: In the sphere problem is the sphere a conductor? If so it screens anything happening outside the sphere. If the sphere is not a conductor then the field inside will depend on what is happening outside, but the total flux through the sphere will only depend on what's inside.
| {
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Equivalence between QFT and many-particle QM My understanding from my QFT class (and books such as Brown), is that many-particle QM is equivalent to field quantization. If this is true, why is it not an extremely surprising coincidence? The interpretation of particles being quanta of a field is -- at least superficially -- completely different from the quantum mechanical description of N point particles.
| As remarked in many references, QFT and QM (many-particle or not) are disjoint theories. The textbooks by Mandl & Shaw, Landau & Lifshitz, Sakurai... report some of the fundamental differences between QFT and QM.
The "interpretation of particles being quanta of a field" is misleading. There is a partial formal analogy [*] for some simple cases --e.g. a collection of independent particles in an external harmonic potential $V(x)$-- but this analogy is lost in the general case, generating well-known difficulties for QFT [Quantum Mechanics: Myths and Facts]:
Whereas the formal mathematical analogy between first quantization and QFT (which implies the irrelevance of the number operator $\hat{N}$) is clear, there is one crucial physical difference: Whereas in first quantization there is really no reason to attribute a special meaning to the operator $\hat{N}$, there is an experimental evidence that this is not so for QFT. The existence of particles is an experimental fact! Thus, if one wants to describe the experimentally observed objects, one must either reject QFT (which, indeed, is what many elementary-particle physicists were doing in the early days of elementary-particle physics and some of them are doing it even today [51]), or try to artificially adapt QFT such that it remains a theory of particles even with general interactions (such as those in (98)).
[*] Partial because only some aspects of the particle approach are covered by the analogy whereas others are ignored.
| {
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Why is electric potential scalar? I can't conceptually visualize why it would be so. Say you have two point charges of equal charge and a point right in the middle of them. The potential of that charge, mathematically, is proportional to the sum of their charges over distance from the point ($q/r$). But intuitively, my thought process keeps going back to the concept of direction and how the electric field at that point would be zero. So why would the electric fields cancel while the electric potentials just add up algebraically?
| An explanation based on the definition of scalar quantities in physics.
To see why electric potential energy is a scalar quantity you need to understand the following:
A physical quantity is a scalar property of a system, when its value and its effects do not depend on the orientation of the system.
Let us assume we have a system of three electrically charged particles carrying electric charge $+Q_A$, $-Q_B$ and $+Q_C$. Let us also assume the three particles are at positions $A$, $B$ and $C$ with position vectors $\bf {r}_A$, $ \bf {r}_B $ and $\bf {r}_C$ with respect to some arbitrary origin O. We can write the total potential energy of the system of three charged particles as
$E=\frac{1}{4\pi\epsilon_0}(\frac{ Q_AQ_C}{| \bf{r}_A-\bf {r}_C |}-\frac{ Q_AQ_B}{|\bf {r}_A-\bf {r}_B |}- \frac{Q_BQ_C}{| \bf {r}_C-\bf {r}_B|})$.
We can observe that as long as $|\bf{r}_{\mu}-\bf{r}_{\nu}|$ remain fixed, with $\nu\ne\mu$, the three particles can be placed in an infinitely large number of positions in various orientations, and yet $E$ will have the same value. I.e.
The orientation of the system does not bear any measurable effects on the value $E$.
This is why electrical potential/energy is a scalar quantity.
| {
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Explanation on the resulting forces of two positive point charges Why will the resulting force lines of two positive point charges be like this:
I would expect this:
| First a comment about the following statements made by Kitchi and Wouter:
Lines of force always have to be smooth, there can't be a sharp bend in them like in your second diagram
and
The comment by @Kitchi basically says it all: lines of force should be continuous and differentiable and they should never intersect.
These statements are incorrect without qualification, and here are two reasons why:
*
*The electric field lines of a point charge consist of rays all of whom intersect at the location of the point charge. Moreover, the field is not smooth at the location of the point charge (it is not even continuous there since there is a singularity).
*There is a discontinuity in the electric field in passing through a surface charge with charge density $\sigma$, in fact the discontinuity is proportional to the surface charge density;
$$
(\mathbf{E}_2 - \mathbf E_1)\cdot \mathbf n = \frac{\sigma}{\epsilon_0}
$$
If you want to make some statement about smoothness of electric field lines, (which you should try to avoid calling "lines of force"), then you really need to make some other qualifications that exclude such cases.
Second, these smoothness justifications are, to some extent, missing the forest for the trees so to speak. The crux of the issue is really addressed by richard. The way one obtains the field due to a charge distribution is by invoking the principle of superposition. This immediately rules out the second diagram you drew because if you simply add the fields of the two point charges vectorially, then you will see that the field lines look like you draw them in the first diagram.
| {
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Two ways to form $SU(2)$ singlets? I am trying to reconcile the two ways of forming $SU(2)$ singlets out of a pair of doublets.
Method (1):
If $v=\begin{pmatrix}v^1\\ v^2\end{pmatrix}$ and $w=\begin{pmatrix}w^1\\ w^2\end{pmatrix}$ are two $SU(2)$ doublets, then I can form a singlet by taking the antisymmetric combination:
$$(v\otimes w)_\text{singlet}=\epsilon_{ij}v^iw^j=v^1w^2-v^2w^1$$
Method (2): Using the same objects from method 1, I can form a singlet by simply taking the conjugate-transpose of one of them, say $v^*=\begin{pmatrix}v_1^*& v_2^*\end{pmatrix}$, and contracting them straight-up:
$$(v\otimes w)_\text{singlet}=(v^*)_i w^i=v_1^*w^1+v_2^*w^2.$$
So are these two different ways to form singlets? Is there a way to understand this from a more general point of view, say $SU(3)$? Also, in elementary quantum mechanics, why is it that when we form spin-singlets out of wavefunctions using method (1), and not using method (2)?
--
As a note to myself and others, $(v^*)_i$ and $\epsilon_{ij}v^j$ are two different objects. But, both transform as in the same way; namely as $\bar{\bf{2}}$.
| OP's two methods are isomorphic. In general one is only interested in classifying representations modulo isomorphism. The point is that for the Lie group $SU(2)$, the spinor representation ${\bf 2}$ and the complex conjugate spinor representation $\bar{\bf 2}$ are equivalent representations ${\bf 2}\cong \bar{\bf 2}$. The equivalence is precisely given by multiplying with the epsilon symbol.
In contrast, the representation ${\bf 3}$ and complex conjugate representation $\bar{\bf 3}$ of the Lie group $SU(3)$ are not equivalent.
| {
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Einstein Field Equations and Electromagnetic Stress-Energy Tensor My question is: if we write Einstein field equations in this form:
$$R_{\mu\nu} - \dfrac{1}{2}g_{\mu\nu}R=8\pi \dfrac{G}{c^4}T_{\mu\nu}$$
Then the left hand side is one statement about the geometry of space-time and the right hand side is one statment about the distribution of momentun and energy right? My point is: what if we use the electromagnetic stress-energy tensor as the energy-momentum tensor?
My thought was: if I understood correctly, does this says that electromagnetic fields can also change the geometry of space-time making it bend as does the presence of matter?
Sorry if it makes no sense, or if it's completely nonsense, it's just a thought that came out, I'm just starting to study those things.
| Yes, all energy-momentum distributions have an effect on the geometry of spacetime, including the the energy-momentum due to the electromagnetic field. For example, the Reissner-Nordstrom solution (which describes a charged, non-rotating spherical mass) has extra terms in it which are dependent on charge. This is because the EM field generated by the charge also has an effect on the geometry of spacetime.
Also note that when you're working with the EM field, the Einstein Field Equations simplify a bit. The stress-energy of the EM field is traceless, which implies $R=0$. So the EFE's simplify to just:
$$R_{\mu \nu} = \frac{8 \pi G}{c^4} ~T_{\mu \nu}$$
| {
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Why can't we store light in the form of light? We can store cold (ice), heat (i.e. hot water bag) and electrical charge (batteries). We can even "store" a magnetic field in a magnet. We can convert light into energy and then, if we want, back to light. But we can't store light in form of light in significant amounts. What is the explanation of that in physics terms?
| It's hard to store light as light because the most common way light interacts with matter is through absorption and emission, which is how mirrors work. However light rays can be bent by gravity, so it would be possible to arrange several massive stars in a way such that a light ray would move in a loop around the stars without energy loss.
| {
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What causes insects to cast large shadows from where their feet are? I recently stumbled upon this interesting image of a wasp, floating on water:
Assuming this isn't photoshopped, I have a couple of questions:
*
*Why do you see its image like that (what's the physical explanation; I'm sure there is an interesting one)?
*Why are the spots surrounding the wasp's legs circle shaped? Would they be square shaped if the 'feet' of the wasp were square shaped?
| That is a truly amazing picture! I am by no means an expert, but I have an idea.
When the wasp stands on the water, it is curved down slightly. The light that hits these parts will then be bent more outwards than if it just hit regular water. This happens at every side of the circle, so the light is always bent out, and doesn't reach the bottom at those points.
Therefore, you get this effect.
Also, to answer your second question, the feet are so small compared to the shadow that they play no big role in the shape.
| {
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"timestamp": "2023-03-29T00:00:00",
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Gamma Ray LASER Theory and Technology I am aware that a similar question has been asked by someone else in the past, but in a very general form. Due to the physics interest and technology, in this question I put emphasis on the detail of the physics part and the question becomes very specific.
We know the power of visible-light-laser and the effects it can have on matter: Industry, Medicine, Military, Entertainment and many other applications.
Given the large amount of energy of $\gamma$-photons, one can extrapolate and see the new applications of LASER designed to amplify $\gamma$-light, emitted by nuclear isomers, which might be called “GLASER” (gamma light amplification by stimulated emission of radiation).
The trick is to excite nuclei to a metastable state, to achieve “population inversion” and then stimulate them to decay simultaneously. The excitation can be achieved either by soft neutron bombardment or by synchrotron irradiation, in order to excite the nuclei into an angular momentum state that does not match the one of the ground state. These nuclei can remain at that state for sufficiently long time, so population inversion can be achieved.
QUESTION:
How can the nuclei be stimulated to decay and emit their $\gamma$-photon so that to achieve GLASER? This could depend on how the nuclei are excited in their isomeric states to begin with.
| There are a large number of problems, and a large number of proposed solutions for Graser operation. I published a paper on laser photochemical separation of nuclear isomers to get a population inversion in the 1970s with George Baldwin.
He moved onto LASL where he continued Graser work for some time. His reviews are quite informative.
While technically a nuclear isomer the Th-229m does not produce a very high energy photon. It is more like an X-ray and not as interesting or problematic as a high energy state.
| {
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Derivative with respect to the nuclear coordinates in the Born–Oppenheimer approximation Reading few sources on the Born–Oppenheimer approximation I don't understand one particular thing.
If you look for example here (PDF, 70 KB) and focus attention on equations 14 and 15 than it is clear that
$$
\nabla_{A}^2 \left( \psi_k(\mathbf{r}; \mathbf{R}) \chi_k(\mathbf{R}) \right) =
\psi_k(\mathbf{r}; \mathbf{R}) \nabla_{A}^2 \chi_k(\mathbf{r}; \mathbf{R})
+ 2 \nabla_{A} \psi_k(\mathbf{r}; \mathbf{R}) \nabla_{A} \chi_k(\mathbf{r}; \mathbf{R})
+ \chi_k(\mathbf{R}) \nabla_{A}^2 \psi_k(\mathbf{r}; \mathbf{R})
$$
where
$$
\nabla_A^2 = \frac{\partial^2}{\partial X_A^2} + \frac{\partial^2}{\partial Y_A^2} + \frac{\partial^2}{\partial Z_A^2}
$$
and
$$
\mathbf{R} = \{ \mathbf{R_i} \}_{i=1}^N = \{ (X_i, Y_i,Z_i) \}_{i=1}^N
$$
is a set of all nuclear coordinates.
But honestly I don't undestand why is that so. The fact is that $\psi_k$ depends implicitly only on $\mathbf{r}$ and parametrically on $\mathbf{R}$ (that's why I think they are delimited by $;$ and not just $,$). As far as I know this parametric dependence means that for each set of nuclear coordinates $\mathbf{R}$ there is a complete set of electronic wave functions $\{ \psi_k(\mathbf{r}) \}_{k}$ which are functions of electronic coordinates only. And then of course when you differentiate $\psi_k(\mathbf{r}) \chi_k(\mathbf{R})$ twice with respect to $\mathbf{R_A}$ you get just $\psi_k(\mathbf{r}) \nabla_{A}^2 \chi_k(\mathbf{R})$ because $\psi_k(\mathbf{r})$ is constant with respect to $\mathbf{R}$.
And one more thing - the linked resource (and many others) claimed that the chain rule is used going from 14 to 15. I don't see any usage of chain rule but I see a usage of product rule.
Seems like I don't understand what's going on here but this is critical step because non-adiabatic coupling terms come from this expansion.
| For each $\mathbf{R}$ there is a complete set of electronic functions $\{\psi_k(\mathbf{r};\mathbf{R})\}_k$, and what these functions are depends on the value of $\mathbf{R}$. As $\mathbf{R}$ is changed continuously, each element of $\{\psi_k(\mathbf{r};\mathbf{R})\}_k$ varies continuously; thus, it is meaningful to speak of the derivative of $\psi_k(\mathbf{r};\mathbf{R})$ with respect to the components of $\mathbf{R}$, and this derivative is generically not zero.
To make this more concrete, consider an infinite chain of atoms in one dimension, with nearest-neighbor distance $a$. We have $\mathbf{R}_n = R_n = na$ for $n$ integer, and one possible set of electronic basis wavefunctions is $\{\sin kx,\,\cos kx\mid k = 2\pi/ma,\,m\text{ integer}\}$. Now imagine varying the interatomic distance by increasing the value of $a$: clearly, each electronic basis function will change (their wavelengths $\lambda = \frac{2\pi}{k} = ma$ will all increase).
| {
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Influence of charged particle's own electric field on itself I read this in my textbook: A charged particle or object is not affected by its own electric field.
Since I find this completely unintuitive and my mind is yelling "wrong! wrong! how could a particle even distinguish between its own field and the external fields?" I would really like to hear an explanation about why this is true, if it is.
(A corollary question that comes to my mind after thinking about this is... What about gravity and mass considering the analogous situation? And the other two forces?)
| This is the problem with classical electrodynamics. The self-energy of a charged point particle is infinite due to its interaction with its on potential.
Since,
$$W=\frac{q^2}{8\pi\epsilon_o}\int_0^{\infty}\frac{1 }{r^2}dr $$
diverges.
| {
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How to determine the direction of a wave propagation? In the textbook, it said a wave in the form $y(x, t) = A\cos(\omega t + \beta x + \varphi)$ propagates along negative $x$ direction and $y(x, t) = A\cos(\omega t - \beta x + \varphi)$ propagates along positive $x$ direction. This statement looks really confusing because when it says the wave is propagating along $\pm$ x direction, to my understanding, we can drop the time term and ignore the initial phase $\varphi$ while analyzing the direction, i.e. $y(x, 0) = A\cos(\pm\beta x)$, however, because of the symmetry of the cosine function, $\cos(\beta x)\equiv \cos(-\beta x)$, so how can we determine the direction of propagation from that?
I know my reasoning must be incorrect but I don't know how to determine the direction. So if we don't go over the math, how to figure out the direction of propagation from the physical point of view? Why $-\beta x$ corresponding to the propagation on positive x direction but not the opposite?
| For a particular section of the wave which is moving in any direction, the phase must be constant. So, if the equation says $y(x,t) = A\cos(\omega t + \beta x + \phi)$, the term inside the cosine must be constant. Hence, if time increases, $x$ must decrease to make that happen. That makes the location of the section of wave in consideration and the wave move in negative direction.
Opposite of above happens when the equation says $y(x,t) = A\cos(\omega t - \beta x + \phi)$. If t increase, $x$ must increase to make up for it. That makes a wave moving in positive direction.
The basic idea:For a moving wave, you consider a particular part of it, it moves. This means that the same $y$ would be found at other $x$ for other $t$, and if you change $t$, you need to change $x$ accordingly.
Hope that helps!
| {
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nuclear fission and half life Why is the alpha, beta or gamma decay of an unstable nucleus unaffected by the chemical situation of an atom, such as the nature of the molecule or solid in which it is bound? The chemical situation of the atom can, however, have an effect on the half life in electron capture. Why is this so?
| A qualitative argument:
The $\alpha$, $\beta$ or $\gamma$ decay are phenomena that are associated with the weak nuclear, and strong nuclear forces which rule in the nucleus. The electromagnetic force, although it plays an important part in the stability of the nucleus, it does not completely determine what and how will happen inside the nucleus. The $\beta$ decay is determined by the weak nuclear force which goes something like this:
$n\rightarrow p+e^-+\bar\nu_e$
$p\rightarrow n+e^++\nu_e$
The $\gamma $ decay results from a rearrangement of the nucleons form higher to lower energy shell configuration, with the emission of the excess energy as $\gamma$ light. The energies involved here are from several eV (Th-229) to several MeV (Co-60).
The events that take place in the outside the nucleus part of the atom/molecule is simply rearrangements of the electrons in the various energy levels. Very low energy involved. If you can understand that the electromagnetic interaction of the protons inside the nucleus cannot have an effect in a $\beta$ decay, then you can understand why the electrons outside the nucleus do not have any effect on what is going on inside the nucleus.
| {
"language": "en",
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Finding the force of friction of a moving object and its change when it accelerates to a constant speed If an object is moving at a constant speed the force of friction must equal the applied (horizontal) force, and for it to be accelerating or decelerating, the force of friction and the applied force must be unequal. Also, I know that $f = \mu N$.
This is what I do not understand: if the applied force is greater than the friction, wouldn't that mean the object would continue to accelerate infinitely? Shouldn't the friction force change to equal the applied force to prevent this? If so, how do I work out how the friction force $f$ changes?
Here is a sample situation: say I have a box with mass $10$ kg and I apply a horizontal force $50$ N, and the coefficient of kinetic friction is $0.5$. How long does it take for the box to finish accelerating and reach a constant velocity? Then, if I increase the force to $60$ N, how long does it take to reach constant velocity again?
| There are a number of wrong assertions in your statement. You say, "surely ... continue accelerating to infinity." Since $ a = \frac {F}{m} $, as long as you apply net force F, you only get constant acceleration a. The acceleration value not only does not go to infinity, but actually does not change, unless the net force changes. When there is no net force, the acceleration goes to zero!
What changes is the speed of the object ( $V = V0 + at$), as long as the acceleration is present.
Using the values you provide, the corresponding accelerations would be $$a1 = \frac {50- (.5)(10)(9.8)}{10} = 0.10 \text{ } m/{s^2} $$ $$a2 = \frac {60 - (.5)(10)(9.8)}{10} = 1.10 \text{ }m/{s^2}$$.
Starting from V0 = 0 and applying the 50N force for 60 seconds, the object would attain a speed of $$V1 = 0 + (.1)(60) = 6\text{ } m/s$$ If we now change the force to 60N, and apply it for 30 seconds, the new speed would be $$V2 = 6 + (1.1)(30) = 39 \text{ } m/s$$
Since the coefficient for static friction (not given) is normally higher than kinetic friction, the object most likely will not move with the 50N force. If this is the case, then the final velocity would be 33 m/s
As you may have noticed, I had to arbitrarily specify the length of time of the applied forces, otherwise an answer could not be obtained.
Note: acceleration due to gravity = 9.8 m/${s^2}$.
| {
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Bat hitting a ball When a bat hits a ball, consider two cases:
1) The batsman goes for a defense, and stonewalls it, to reduce its speed.
2) the batsman goes for a shot, e.g. a home-run, etc.
in which case will the bat have the highest chance of breaking due to the impact?
And am I right in assuming that the force on the bat due to the reaction force of the ball is the only thing that causes it to break?
| Every one might have studied about material science . When a body (bat)undergoes repeated stress due transfer of kinetic energy by the ball.this in turn develop micro structural crack in the bat,which keeps on developing when a ball travels with a kinetic energy which is required to develop and break the bat is strikes by the bat.It eventually breaks
| {
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Are two waves being in phase the same as saying that the two waves are coherent? If two waves are coherent, is it the same as them being in phase? Please correct if I'm wrong.
| While I agree with BebopButUnsteady, there is a simpler way to summarize what he said. Coherence and relative phase difference are not the same. Take two signals,
$$s_1 = \sin(\omega_1t + \theta_1)\\
s_2 = \sin(\omega_2t + \theta_2)$$
Two waves can be coherent and out of phase. But they cannot have a constant relative phase difference and be incoherent. The crux of the issue is the phase delay, theta, and frequencies, omega, of the two waves. If the two waves have the same frequency $\omega_1 = \omega_2$ and are phase aligned, $\theta_1 = \theta_2$, they are coherent and in phase. If the two have different phase offsets $\theta_1 \neq \theta_2$, but are of the same frequency, $\omega_1 = \omega_2$, then they are coherent and out of phase. If they have different frequencies, $\omega_1 \neq \omega_2$, they are incoherent regardless of the phase shift.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What fraction of peak horsepower do typical 4 door passenger vehicles use? I was surprised when I looked at the power rating of the engine used on a Humvee. It's only ~190 horsepower, which is exceeded by many sedan engines.
So an obvious question is why doesn't my Camry SE burn more gas than a Humvee and I think it's just because during regular use, it doesn't use anywhere near 250 HP.
Does anyone know about how much mechanical power a typical sedan engine puts out during highway driving?
Also, can someone speculate what the torque vs RPM graphs of a sedan engine and a Humvee engine look like and what the typical operating points are?
| You might be interested in my answer to If an electric car were to drive without having to stop, would the range be greatly affected by the speed at which the vehicle is moving?. From just air resistance I calculated that at 70 mph the engine in my Ford Focus is generating about 17 bhp (12.5 kW).
Obviously you need to add in losses in the transmission. Apart from anything else the above result predicts a top speed of about 150 mph and I can assure you my Focus isn't that fast! This graph suggests that the rolling resistance is about 0.8 of the aerodynamic resistance at 70 mph, and that would make the total power about 30 bhp - reassuringly close to Martin's estimate.
| {
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What Pauli matrices should I use for this two spin system? Consider the Hamiltonian $H = -J_\text{F}S^{(1)}_zS^{(2)}_z + J_{AF}S^{(1)}_zS^{(2)}_z$, describing the graph
Here, F means ferromagnetic and AF means antiferromagnetic interactions. I am having problem with the value of $S^{(1)}_zS^{(2)}_z$.Someone suggested to me that
$$S^{(1)}_z=\frac{1}{2}\begin{pmatrix}
-1 & 0 &0 &0 \\
0&-1 &0 &0 \\
0 &0 &1 &0 \\
0 &0 &0 &1
\end{pmatrix},\quad S^{(2)}_z=\frac{1}{2}\begin{pmatrix}
-1 & 0 &0 &0 \\
0&1 &0 &0 \\
0 &0 &-1 &0 \\
0 &0 &0 &1
\end{pmatrix},$$ and therefore $$S^{(1)}_z\cdot S^{(1)}_z=\frac{1}{4}\begin{pmatrix}
1 & 0 &0 &0 \\
0&-1 &0 &0 \\
0 &0 &-1 &0 \\
0 &0 &0 &1
\end{pmatrix}.$$
On the other hand, from page 7 of these notes on Pauli spin matrices, I know that for two spin systems $$\Sigma_z = \begin{pmatrix}
2 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & -2 \\
\end{pmatrix}.$$ I asked the person but never got a reply. I don't see what is the difference between $S^{(1)}_z\cdot S^{(1)}_z$ and $\Sigma_z$; when I should use what?
| The operator $\Sigma_z$ is the $z$ component of the total spin, and therefore is the sum of the quantities $S^{(1)}_z$ and $S^{(2)}_z$, whereas the operator $S^{(1)}_z\cdot S^{(2)}_z$ is their product. You can easily check that for the matrices in your question the identity
$$\Sigma_z=-2(S^{(1)}_z+ S^{(2)}_z)$$
holds; the factor of $-2$ is due to differing conventions in sign and normalization between the ones in your post and the ones in the link.
For your particular system you want the product, as that will have the states $|\uparrow\rangle|\uparrow\rangle$, $|\downarrow\rangle|\downarrow\rangle$, $|\uparrow\rangle|\downarrow\rangle$ and $|\downarrow\rangle|\uparrow\rangle$ as eigenstates, with eigenvalues $+1,+1,-1$ and $-1$ resp., which is how you want to model (anti)ferromagnetic interactions in the first place.
The sum is useful in situations where you want the total number of excitations. This is the case, for example, in the Jaynes-Cummings model, where you use spin matrices to model two-level atoms even though the states are not necessarily spin states (and more probably electronic eigenstates). Thus taking $|\uparrow\rangle=|\text{excited}\rangle$ and $|\downarrow\rangle=|\text{ground}\rangle$, the operator $S^{(1)}_z+ S^{(2)}_z$ counts the total number of excited atoms, which is of course the atomic part of the Jaynes-Cummings hamiltonian.
You should also notice that your hamiltonian simplifies to $H=(J_\text{AF}-J_\text{F})S^{(1)}_z\cdot S^{(2)}_z$: only a ferromagnetic or an antiferromagnetic interaction is possible between any two spins. They either like being parallel or not. The two edges in your graph combine to a single one, of either F or AF character. The graph formalism is useful when you have multiple spins with varying interactions.
| {
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Potential Inside Conducting Cube A cubical box with sides of length L consists of six metal plates. Five sides of the box { the plates at $x=0, x=L, y=0, y=L, z=0$ - are grounded. The top of the box (at z = L) is made
of a separate sheet of metal, insulated from the others, and held at a constant potential $V_0$. Find the potential inside the box.
I am attempting to set up this problem with Legendre polynomials. Since the top portion is grounded, this affects the "potential" region I can take, and I am unclear on how about to go setting it up.
I have set up the Laplace as follows:
$1\over x$${d^2x\over dx^2} +$ $1\over y$${d^2y\over dy^2} +$$1\over z$${d^2z\over dz^2} = 0$
In which at this point, must each component equal a constant?
which gives
$k_x^2+k_y^2+k_z^2 = 0$
$1\over x$${d^2x\over dx^2} = k_x^2$
$1\over y$${d^2y\over dy^2} = k_y^2$
$1\over z$${d^2z\over dz^2} = k_z^2$
where z is given by:
$Z(z) = sinh(k_zz)$
but at this point, how do I solve the coefficients?
Or should I be using the general form and find coefficients:
$x = Acos(k\theta) + Bsin(k\theta)$
$y = Ccos(m\phi) + Dsin(m\phi)$
$z = Ee^{(k^2 +m^2)^.5z}$ + $Fe^{-(k^2 +m^2)^.5z}$
I'm stuck on finding the coefficients using this method.
| The basic steps:
*
*Write the potential $V(x,y,z)$ as a series in products of appropriate orthogonal functions. Often one uses orthogonal functions that arise from the solution to Laplace's equation by separation of variables.
*Impose the boundary conditions (there are six of them) to solve for the coefficients in the series expansion. For example, the $z=L$ boundary condition would be
$$
V(x,y,L) = V_0
$$
All "grounded" sides are at zero potential; this is just the mathematical definition of the term "grounded." So, for example, at $z=0$ you would have
$$
V(x,y,0) = 0
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why planets are orbiting only in one plane? Since gravity is three dimensional why planets are orbiting only in one plane around sun.
| It is believed that the planets formed when a cloud of gas underwent gravitational collapse. Any small angular momentum that the cloud started out with has to be conserved, and since the cloud's moment of inertia becomes smaller, it spins faster. Also, the centrifugal force "throws out" the edges of the cloud and makes it more disc-like.
As vartec said, pluto is an exception:
which has lead to speculation that it was gravitationally captured by our solar system rather than formed with the rest of the planets.
| {
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How to describe heat transfer between two solid materials? A general equation for dealing with heat transfer between one material and a region of insulating material. I've seen basic heat transfer equations for one material, but I'd love to see an explanation of how to do two.
| The one-dimensional heat equation for a solid can be written as:
$$
\rho C_p\frac{\partial T}{\partial t}= -\frac{\partial}{\partial x} \left( k\frac{\partial T}{\partial x} \right) +\sigma
$$
where $\sigma$ is the source term and $ \dot q =-k\frac{\partial T}{\partial x}$ is the diffusive heat flux. At the boundary the temperature and the flux must be continuous (if we consider contact resistance negligible, otherwise a gap in temperature could be possible) that is:
$$T_1=T_2$$ $$\dot q_1=\dot q_2 $$
| {
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Will we feel right up inside the airplane if the airplane accelerates toward earth at 20m/s^2? Suppose you are in the airplane and the airplane falls toward the earth with acceleration of $20m/s^2$(double of gravitational acceleration $g$). This double acceleration by airplane will cancel the gravitational acceleration and create gravitational acceleration upward from the earth. If this happens, you will be pushed to the ceiling of the airplane if there is such a airplane that falls vertically from its rest position. While falling, the environment in the airplane will be the same as on the earth because in both case you will be pushed in the direction of the acceleration. My question is if this happens, will we feel right up while while being pushed to the ceiling of airplane?
Will we feel just standing on earth?
| if this happens, you will be stuck to the ceiling of the airplane but it does not mean that you are pushed up. Actually, the airplane and the person in that airplane, both are falling towards earth simultaneously but the airplane has acceleration=2g, and the person has acceleration =g. Due to this, airplane falls faster than person and ceiling of airplane will strike the person.So, person in airplane will not feel right up and he will not feel like one feels while standing on earth.
| {
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Are electrons moving? If so relative to whom? and why is there no magnetic field? electrons on earth moving with us, due to rotation of earth, revolution of earth, sun and our galaxy right?
Then, why is there no magnetic field around a piece of copper wire?
| To good approximation Earth is an Inertial System. Charges at rest with us in this inertial system do not produce a magnetic field, as Maxwell's equations hold in any Inertial system. Viewed from another Inertial system magnetic fields of electrons and protons cancel, as explained in answer (1). Viewed in the Earth's Intertial system protons are needed to compensate the Coulomb field produced the electrons at rest with respect to us. You see here that what is a magnetic field to the one inertial system is a Coulomb field as viewed from the other system.
| {
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Place each foot on a scale: can you add the two to find your weight? I frequent a blog from a British psychologist, and every Friday he likes to pose an interesting puzzle or riddle. The Monday after that he posts the answer. They're good fun, and IANAP but this week's answer made my it-might-not-be-quite-as-simple-as-that detector go off.
My question boils down to this: let's say I have two identical scales, and I stand on the scales with one foot on each scale. The scales read W1 and W2. Does my weight equal W1 + W2?
| Yes, this is not the hardest problem ever, but here's the mechanics calculation that leads to the yes answer.
Draw a free body diagram of your body as you are standing still with one foot on each scale. You experience three forces (I will label their magnitudes): (1) The force due to gravity pulling you down, $W$ (aka your weight), (2) the normal force $N_1$ of scale 1 pushing up on one foot, and (3) the normal force $N_2$ of scale 2 pushing up on your other foot. Since your body is not accelerating, these forces balance by Newton's Second Law;
$$
W = N_1+N_2
$$
Now the question is, how are these forces related to what the scales read? Well, each scale reads the force of the corresponding leg that pushes down on it. Let's call the magnitude of these forces $W_1$ and $W_2$. As it turns out, Newton's Third Law tells us that the magnitude of the force that each scale exerts on each foot (the normal force) equals the corresponding magnitude of the force that each foot exerts on the scale;
$$
W_1 = N_1, \qquad W_2 = N_2
$$
It follows that
$$
W = W_1 + W_2
$$
as desired.
| {
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Where are we : On level ground or on a ramp - moving in a train? Let's say we are traveling in a train. The path has two parts: one at ground-level and the other moving up on the ramp. The ramp has an inclination of $\arctan\frac{a}{g}$ with the horizontal, where $a$ is the acceleration of the train on level ground and $g$ is the acceleration due to gravity.
The train does not accelerate on the ramp, but moves with a constant velocity.
Can we comment where we are (sitting inside the train of course!) when we have only a pendulum hanging on the roof to observe. (windows are blackened)
| If the train is an ideal isolated system, the answer is no, we can't tell where we are. As zhermes pointed out, the equivalence principle states that the gravitational force experienced by a body at rest or moving with a constant velocity is the same as the pseudo-force experienced by that body is a non-inertial frame of reference. This simply states that the inertial mass is equal to the gravitational mass. To see that this is so, instead of thinking that the train is accelerating, imagine that you place another gravitational field that pulls on the pendulum in the same direction as the inertial force. Also, read about Einstein's thought experiment involving his famous elevator. In some ways its similar to your question.
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Capacitance of a capacitor Why does the capacitance of a capacitor increase if the distance between the two plates of a parallel plate capacitor is decreased? I think, with decreasing distance between the two plates, the force of attraction between the charges on the two plates will increase, and as a result more charge will be stored.
| 1) Capacitance is the ratio of the stored charge Q (+ on either plate, - on the other) to the voltage between the plates.
2) The electric field between the plate is proportional to the plate surface charge density $\sigma = Q/A$, with $A$ the plate area.
3) So if one moves the plates closer together while holding the charge fixed (no external circuit connections), a) the electric field also stays fixed (by 2 above), b) the voltage (= electric field x spacing) decreases because the spacing is decreasing, and c) the capacitance ($C=Q/V$) increases.
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Have negative pressures any physical meaning? Some cubic thermodynamical equations of state predict negative pressures, have negative pressures any physical meaning? Could they be related to negative mass?
| Thermodynamically pressure is related to the work done in changing the volume. Assuming an adiabatic change the change in the internal energy of a system is given by the work done on it:
$$ dU = \delta W = PdV $$
If you take a volume of an ideal gas and let it expand by $dV$ it does work on it's surrounds and it's internal energy decreases. This is the usual (positive) pressure.
But consider the dark energy that we believe pervades the universe. If we take some volume of vacuum and let it expand by $dV$ then the internal energy goes up. This may seem odd, but it goes up because dark energy density is a constant everywhere so a bigger volume has more dark energy. Anyhow, if the internal energy of our system goes up when the volume increases we must have done work on it i.e. the sign of $PdV$ is negative. Since the sign of $dV$ is positive, because the volume increases, we conclude that the sign of $P$ must be negative i.e. dark energy has a negative pressure.
Dark energy is not equivalent to negative mass in the $E = mc^2$ sense. The energy density is positive. The reason it shows a negative pressure is due to it's equation of state and not to it being in some sense negative energy.
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Path of an electron through an electric circuit When a potential difference is applied across a conductor, and if an electron moves from the negative terminal of the battery and reaches the positive terminal, then I want to know if the electron will remain at the positive terminal or will it again move toward the negative terminal through the battery?
| In an ideal electrochemical cell case, it should remain at the positive terminal.
The salt bridge should only allow for the ions to flow.
However, practical batteries have other physical phenomenon that restrict the flow of current, and hence introduce this 'internal resistance'. There could be a wide variety of reasons depending on the type of cell used. I believe one of the most common reason is polarisation. But that too, shouldn't cause the electrons to flow through the cell.
To quote a good explanation from University of Illinois Physics Dept Q&A page:
You might wonder why the electrons don't just flow back through the battery, until the charge changes enough to make the voltage zero. The reason is that an electron can't move from one side to the other inside the battery without a chemical reaction occurring. In other words, inside the battery plain electrons can’t travel around because it takes too much energy to put a plain electron in solution. Electrons can only travel inside the battery via charged chemicals, ions, which can dissolve off the electrodes. The chemical reaction is what pushes the electrons inside toward the negative end, because the electrodes at the two ends are made of different materials, which have different chemical stabilities. So overall, electrons flow AROUND the circuit, toward the negative end inside the battery, pushed by the chemical reaction, and toward the positive end in the outside circuit, pushed by the electrical voltage.
Electrical current can flow in the other way in the battery too, if the battery is hooked up to something with a bigger voltage difference (a battery charger, for example).
EDIT: As to why there is current flow inside the battery: Electrons are not necessary for current to flow. The flow of ions does happen inside the battery.
| {
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Algorithm for identifying planes in a Bravais Lattice I have a lattice with Lattice Vectors $(\vec{t}_1,\vec{t}_2,\vec{t}_3)$ which are NOT orthogonal in general.
How can I identify the atoms/unit cells that belong to a plane - that is normal to a given direction.
I do recognise that the lattice might not be periodic in ANY direction - only specific ones.
I worked out a way to calculate the periodicity of the lattice planes:
1. Given the direction $\vec{t}$, construct the corresponding reciprocal lattice vector G.
2. Project $\vec{G}$ in the direction of $\vec{t}$ and take the inverse of the length of the projected vector.
i.e. Distance between lattice planes normal to the direction $d = \vec{G}\cdot \frac{\vec{t}}{\vert\vec{t}\vert}$
My question, once again, is to find an algorithm that identifies the atoms in the crystal planes thus formed.
| For "simple" planes that "fit" into one or a few unit cells of the lattice, the task is relatively simple, in that you just identify all atoms that belong to the plane in one such block and then use the periodicity of the crystal.
For the most general case, I'm not 100% sure what the best way would be. Here's an idea.
Let $T$ be the matrix whose columns are your lattice vectors. Let $P$ be the matrix whose columns are three vectors $\vec{u}, \vec{v}, \vec{d}$ where $\vec{u}$ and $\vec{v}$ lie in the plane you're interested in and $\vec{d}$ is perpendicular to that plane.
Then any point can be expressed as
$$\begin{pmatrix}x\\ y\\ z\end{pmatrix} = P \cdot \begin{pmatrix}\alpha \\ \beta \\ \gamma \end{pmatrix} = T \cdot \begin{pmatrix}h\\ k\\ l\end{pmatrix}$$
Let's make the coordinate origin such that the plane goes through it. Then, a point that lies on the plane is characterized by $\gamma = 0$ in the above representation. On the other hand, lattice vectors are characterized by integer values of $h$, $k$, and $l$. Hence, we solve the above equation for $(h,k,l)$ and get
$$\begin{pmatrix}h\\ k\\ l\end{pmatrix} = T^{-1} P \begin{pmatrix}\alpha\\ \beta \\ 0\end{pmatrix}$$
The lattice points that lie in the plane are then those points for which we can find values $\alpha$ and $\beta$ such that the resulting $h, k$ and $l$ are integers. Haven't thought much about how we could do that, though. I guess it depends on the particular plane, and some "inspection".
| {
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Soft Mass and Physical Mass in Softly-broken SUSY In softly broken SUSY, the bare mass parameters may be specified at e.g. the GUT scale, and then we can run these down to another scale using RGEs, similar in form to the RGEs for gauge couplings, with a 1-loop and 2-loop differential beta function. Once these parameters have been run to the desired scale, tree-level physical masses are computed via diagonalization of the mass matrices depending on how the mixings were determined. Next, additional accuracy is gained by working out the loop corrections to the physical masses with self-energy graphs.
I am confused by why the physical masses that you determine from diagonalizing the soft parameters are considered tree-level masses if they are determined from 2-loop RGEs. Could you have determined the physical masses at the high scale and computed the self-energy loop corrections there, and then renormalized the physical mass down to the lower scale?
| The physical masses should be independent of the renormalization scale. We, however, only calculate a finite number of loop corrections, resulting in a scale dependence in the physical mass. This scale dependence can be used to estimate the error in the mass calculation from the missing higher orders.
In principle, one could calculate the sparticle mass spectrum at any scale. In practice, though, one wants to minimize the effect of the missing orders. This is typically achieved, in the MSSM, by minimizing the scale dependence of the one-loop EWSB condition
$$
\mu^2 + 1/2 m_Z^2 = \frac{1}{\tan^2\beta-1}(m_{H_1}^2 - \tan^2\beta m_{H_2}^2)
$$
which is achieved by choosing the scale $\mu=\sqrt{m_{\tilde{t}_1}m_{\tilde{t}_2}}$ - the sparticle mass spectrum is typically calculated at this scale. This scale, however, must be found iteratively, because, writing the dependence explicitly, $\mu=\sqrt{m_{\tilde{t}_1}(\mu)m_{\tilde{t}_2}(\mu)}$.
| {
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2 following gates, permutation matrix I have a circuit that has 4 wires and 2 following each other Toffoli gates.
I have permutation matrix for each Toffoli gate (A and B).
Do I have to multiply that 2 matrices to get the entire permutation matrix of that 2 Toffoli gates?
And if I do that is $A\times B$ or $B\times A$?
| Yes, of course !
This nevertheless suppose that no time is lost between the two gates, but it is a usual assumption in basic lectures.
If the state vector $\left| \Psi_0 \right>$ enters at the left of the circuit, then reach the gate A, represented by matrix $A$, then the B one represented by matrix $B$, and end up at the right of the logic circuit as $\left| \Psi_1 \right>$, then the final state is $\left| \Psi_1 \right> = \left( B \cdot A \right) \left| \Psi_0 \right>$, since A applies first, then B.
In your case, I suppose $\left| \Psi_0 \right> = \alpha_{0000} \left| 0 0 0 0 \right> + \alpha_{0010} \left| 0 0 1 0 \right> + \alpha_{0100} \left| 0 1 0 0\right> + ... $ has 16 entries, since you have 4 wires. Each wire usually represents the time evolution of a single q-bit. A Toffoli gate is usually a 3 q-bits gate, so you have to correctly generalise the Toffoli gate given on the Wikipedia page. Well, it means that you must put extra diagonal 1's in front of the untouched q-bit. You also have to change block-wise for the second gate I presume. (Please tell me if this last point is unclear for you, it is simpler to understand on a black-board than on internet actually :-) !)
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"as measured in a local Lorentz frame"? I've seen the phrase "as measured in a local Lorentz frame" tagged on the end of so many sentences.
What does it mean precisely?
To give an explanation with an example, consider the context of measuring the speed of an object, if this applies.
| I think it's safe to say that this phrase means the same thing as "local inertial frame." Assuming that this is actually is a safe assumption, I'll describe, physically, what a local inertial frame is. I can give a lot of mathematical detail in an addendum if you wish.
In special relativity, we assume that spacetime is flat, and in this case, any observer moving with a constant three-velocity below $c$ is called an inertial observer. The laws of physics look the same to all such observers.
In general relativity, the spacetime is not necessarily flat, but in small regions of spacetime, the spacetime looks approximately flat, and freely falling massive observers, which move along timelike geodesics, when making measurements in small regions of spacetime around them, will see physics that is approximately the same as that seen by an inertial observer in flat spacetime as in special relativity.
One then says that such observers are making observations from reference frames that are locally inertial.
| {
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What is the maximum power available from a magnetic field? I just want to validate something I inferred from studying Griffiths (1999).
The instantaneous magnetic field $\vec{b}(t)$ at a distance $r$ from a long infinite conductor carrying a current $i(t)$ is
$$\vec{b}(t) = \frac{\mu}{2\pi r} \hat{\psi} \cdot i(t)$$
The energy density per unit volume is defined in terms of $\vec{b}(t)$ as
$$w_V(t) = \frac{b^2(t)}{2\mu_0}$$
Question 1: Does it immediately follow that the power density available per unit volume is $\frac{dw_V(t)}{dt}$?
Question 2: If I place an inductor in that time-varying magnetic field and connect it to some electrical load, I'll be able to capture that power. Assuming the surface normals $\vec{A}$ of the turns of the inductor coils are perfectly collinear with $\hat{\psi}$, what might be some energy loss mechanisms that prevent me from getting all that power, aside from electrical series resistance in the inductor?
This is a self-study question, not a homework.
| First of all, your formula for the energy density is missing a term for the electric field:
$$
w(t)=\frac{\varepsilon_0 e^2(t)}{2}+\frac{b^2(t)}{2\mu_0 }
$$
You will always have an electric field if the magnetic field is time-varying.
However, this is not that important for the more conceptional question you seem to be struggling with. I think a good analogy to help you here is to think of this energy as a resorvoir of water with an inflow and an outflow. The change in the amount of water in the reservoir is the inflow minus the outflow. But the flow through the reservoir can be much larger than the change in water in the reservoir. For instance, if the inflow is equal to the outflow, there is no change in water in the reservoir even though there is a flow through the reservoir.
Similarly, the electomagnetic energy flowing through a unit volume can be much larger than $dw(t)/dt$. What it is in your example and how that energy flow could be "captured" is a question I'll leave to someone else.
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microcanonical distribution We know that in an isolated system, the density matrix is the microcanonical distribution matrix. That this the possibility for all the states with energy in a certain interval is a constant?
But how can I deduce this from the postulate of equal probability?
| The assertion that the density matrix for an isolated system is that of the microcanonical ensemble implies the postulate of equal a priori probabilities since, as you indicate, it assigns equal probabilities to each of the energy eigenstates of the system.
I would then ask you the following question: On what grounds do you assert that the density matrix of an isolated system is microcanonical?
One answer you could give is that the microcanonical density matrix is precisely the one that maximizes the (von-Neumann) entropy of the system.
So then the question becomes: Why is the density matrix that which maximizes the entropy?
The best answers to this come from understanding the von-Neumann entropy in the context of information theory. More on that here. You might also find the following related Physics.SE question interesting:
Why is (von Neumann) entropy maximized for an ensemble in thermal equilibrium?
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Eye sensitivity & Danger signal Why are danger signal in red, when the eye is most sensitive to yellow-green?
You can check luminosity function for more details...
| I believe you should have googled first:
google hits
Especially the second link very clearly explains the main reason:
The primary reason why the color red is used for danger signals is that red light is scattered the least by air molecules. The effect of scattering is inversely related to the fourth power of the wavelength of a color. Therefore blue which has the least wavelength of all the visible radiations is scattered the most and red which has the highest wavelength of all the colors we can see is scattered the least. So red light is able to travel the longest distance through fog, rain, and the alike.
Also, red is a color we inherently perceive as one that is associated with danger. I guess it has been genetically programmed as hot objects are red in color and also red is the color of blood. These reasons make danger signals the most effective when they are painted in red.
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Deriving torque from Euler-Lagrange equation How could you derive an equation for the torque on a rotating (but not translating) rigid body from the Euler-Lagrange equation? As far as I know from my first class in Classical Mechanics, there is no potential defined for a rotating body, so the only term I see in the Lagrangian is the $\frac{1}{2}I\omega^2$. Since there is no $\theta$, I'm just getting that torque always equals $0$.
| In my view, you don't derive torque from the Euler-Lagrange equations. It arises naturally from them, as Lagrangian is a formalism of classical mechanics that can explain the behavior of physical systems both in their translations and rotations. For instance, consider a simple pendulum. It's only degree of freedom is $\phi$, the angle denoting its deviation from the vertical.
So you would write the Euler-Lagrange equations as
$$ \frac{\partial \mathcal{L}}{\partial \phi} - \frac{d}{dt}\left( \frac{\partial \mathcal{L}}{\partial \dot \phi}\right) = 0.$$
Let's look at dimensions. $\frac{\partial \mathcal{L}}{\partial \phi}$ has units of energy per radian; that is, $\rm N \cdot m$. The first is canonically the generalized force; in this case, it is a torque.
Then, $\frac{\partial \mathcal{L}}{\partial \dot \phi}$ has units of energy per rotational speed; and thus in this case, the generalized momentum is an angular momentum. So rotational dynamics can arise naturally from minimizing the action of a system. In the case of a simple pendulum, the generalized torque is $-mgl\sin\phi$, as you would expect from the potential $-mgl\cos\phi$.
For a rotating particle with potential $V$, it has Lagrangian
$$ \mathcal{L} = T - V = \frac{1}{2}mR^2\dot\phi^2 - V.$$
Then the generalized force (i.e., torque) is
$$ \tau = \frac{\partial\mathcal{L}}{\partial \phi} = \frac{\partial V}{\partial\phi},$$
which is what we expect from rigid-body dynamics.
| {
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A charged sphere with pulsing radius
Radius increases and decreases periodically (as a pulse).And so does the charges on the surface of sphere.
I can't get what is gonna happen.the EM waves are produced perpendicularly to motion of the charges,but here where is the perpendicular ? Will the EM waves get produced into the space of sphere?
Also, can two EM waves intersect each other or not? [as Electric fields lines can't intersect.]
| Results with an accurate simulation, near field radiating only :
Simulation of a sphere with a pulsating charge - text in french page 2
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What is the speed of acceleration of the inflation of the universe? Is the inflation speed of the universe accelerating or is it a constant speed of expansion proportional to distance between objects.
| As the universe stands today its not said to be "inflating" but it is "expanding". I think one way to make the distinction is to say that during inflation the radius of the causal horizon ($\frac{1}{aH}$ in FRW universe) falls (linearly) whereas during expansion it increases (being equal to half the conformal time for EdS universe).
I would love to know if some other answers can make it more precise.
Also about how much and for how long did inflation happen - I think Wikipedia gives a number in the first paragraph of the entry on cosmological inflation.
I would love to know if there is any theoretical derivation of these mind-boggling numbers.
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Direction of the Area Vector (with regards to magnetic dipole) I'm learning about torque on a conductive coil in a magnetic field. I have been taught that $\vec\tau = \vec\mu \times \vec{B}$, where $\vec\mu$ is the magnetic dipole moment. Also, $\mu = I\vec{A}$, where $\vec A$ is the area vector of the loop.
To find the direction of the area vector, I am told to use the right hand rule with regards to the current in the loop (curl your fingers in the direction of current, and your thumb points in the direction of the area vector).
My question is: Why does this give the correct direction for the area vector? Is the area vector just defined to be this way to avoid nasty usage of minus signs, or is there some other reason for this?
My guess is that whoever formalized this law/equation (not sure what correct term is for this instance) started with the direction of torque, and worked backwards defining the direction of $\vec\mu$ and $\vec{A}$ to reduce or eliminate stray minus signs in the equations. However, this is, of course, just a guess; I want to know what the true reason is.
| The area vector is typically (in the treatments I have encountered) simply defined this way and then all other facts are written in such a way that they are consistent with this convention.
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How Do Electrons "Resonate Against the Restoring Force Of Positive Nuclei?" While reading "Surface Plasmon Resonance," I came across the following:
"The resonance condition is established when the frequency of light photons matches the natural frequency of surface electrons oscillating against the restoring force of positive nuclei."
1.) What is the "restoring force of positive nuclei"?
2.) How do electrons resonate against the restoring force of the positive nuclei?
I realize that electrons and protons attract each other and that they both behave like waves, but the concept of a "restoring force" eludes me.
| This is too long for a comment, so here it goes:
What causes a valence electron of a large atom to feel an effective attractive potential that is similar to a harmonic oscillator?
I hope people following this forum at least understand that in order to have electrons around nuclei display a stable solution and form an atom, a potential is necessary.
This potential is very complicated since it is the result of a many body problem.
Fortunately we have mathematical tools to help us estimate the potential of a valence electron by the following logic:
The unknown real potential must be well behaved since the atom is stable and therefore it can be expanded in a Taylor series expansion in the variable of interest, call it r, the radial variable for the potential.
The first few terms are dominant in the series for r<1. The potential must be symmetric because no large asymmetries have been observed in atoms, thus the coefficient of the r term must be zero. The first power that is the strongest in the expansion ( by the form of a Taylor expansion) is the r^2 which is the function of a harmonic oscillator potential, well studied with the Schrodinger equation.
In a first approximation in quantum mechanics, all attractive symmetric potentials have as a dominant term the harmonic oscillator functional form thus one can use the language of a "restoring force" , because a spring is a harmonic oscillator and the terminology carries over.
To address the "resonate" part: a harmonic oscillator restores i.e. pulls back the particle and it does so even in classical physics with a certain frequency ( think of a spring again). In quantum mechanics there still is a frequency associated with the position of the electron's orbital. When the disturbance is of the same frequency as the frequency of the potential solution for the electron one can have a resonance phenomenon.
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How old is SUN ☉? How do we know/calculate the exact age of sun ☉ ? ie. 4.57 billion years. What is the way to calculate it?
|
The Sun was formed about 4.57 billion years ago from the collapse of part of a giant molecular cloud that consisted mostly of hydrogen and helium and which probably gave birth to many other stars. This age is estimated using computer models of stellar evolution and through nucleocosmochronology. The result is consistent with the radiometric date of the oldest Solar System material, at 4.567 billion years ago. Studies of ancient meteorites reveal traces of stable daughter nuclei of short-lived isotopes, such as iron-60, that form only in exploding, short-lived stars. This indicates that one or more supernovae must have occurred near the location where the Sun formed. A shock wave from a nearby supernova would have triggered the formation of the Sun by compressing the gases within the molecular cloud, and causing certain regions to collapse under their own gravity. As one fragment of the cloud collapsed it also began to rotate due to conservation of angular momentum and heat up with the increasing pressure. Much of the mass became concentrated in the center, while the rest flattened out into a disk which would become the planets and other solar system bodies. Gravity and pressure within the core of the cloud generated a lot of heat as it accreted more gas from the surrounding disk, eventually triggering nuclear fusion. Thus, the Sun was born.
Wikipedia
Also see The age of the Sun and the relativistic corrections in the EOS.
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Types of photon qubit encoding How many types of qubit encoding on photons exist nowadays? I know only two:
*
*Encoding on polarization:
$$ \lvert \Psi \rangle = \alpha \lvert H \rangle + \beta \lvert V \rangle $$
$$ \lvert H \rangle = \int_{-\infty}^{\infty} d\mathbf{k}\ f(\mathbf{k}) e^{-iw_k t} \hat{a}^\dagger_{H}(\mathbf{k}) \lvert 0 \rangle_\text{Vacuum} $$
$$ \lvert V \rangle = \int_{-\infty}^{\infty} d\mathbf{k}\ f(\mathbf{k}) e^{-iw_k t} \hat{a}^\dagger_{V}(\mathbf{k}) \lvert 0 \rangle_\text{Vacuum} $$
*Time-bin:
$$ \lvert \Psi \rangle = \alpha \lvert 0 \rangle + \beta \lvert 1 \rangle $$
$$ \lvert 0 \rangle = \int_{-\infty}^{\infty} dz\ f\left(\frac{t -z/c}{\delta t_{ph}}\right) e^{-i w_0 (t-z/c)} \hat{a}^\dagger(z) \lvert 0 \rangle_\text{Vacuum} $$
$$ \lvert 1 \rangle = \int_{-\infty}^{\infty} dz\ f\left(\frac{t -z/c+\tau}{\delta t_{ph}}\right) e^{-i w_0 (t-z/c+\tau)} \hat{a}^\dagger(z) \lvert 0 \rangle_\text{Vacuum} $$
Is there anything else?
| Yeah there is a couple of 'em - off the top of my head I can think of:
\begin{align}
|\uparrow\;\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|\downarrow\;\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|g\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|e\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|L\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|R\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|H\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|V\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|0\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|1\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|+\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|-\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}
\end{align}
But these are only the naming conventions in the northern hemisphere.
In the southern hemisphere they are labeled;
\begin{align}
|\downarrow\;\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|\uparrow\;\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|e\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|g\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|R\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|L\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|V\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|H\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|1\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|0\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|-\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|+\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}
\end{align}
So as you can see you have to be careful about who you are talking to, otherwise you could get your calculations all backwards.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
How do Doppler Effect and Time Dilation differ?
Above, I have drawn a diagram showing Doppler Effect (here we are using space-time but in a non-relativistic sense. Time and distance are the same for A and B).
Edit: I am adding a relativistic space-time diagram below this with
lines of simultaneity drawn. I am also editing the description to be
more relavent to the updated diagram.
The diagram shows the frame of a stationary observer B. A travels with velocity c/2. A emits flashes of light every second (according to the time of observer B - flashes of light are shown as dashed lines and are emitted where the lines of simultaneity meet the worldline of A and hence they are emitted every second in the time of B)
We see that in the frame of B, we begin to see the light one second after it is emitted. light is continuous and after the lag shown, light from a time delta t' (according to the time in frame B) is observed across time delta t (which is 2 x delta t').
This seems to indicate that what we OBSERVE in frame B seems to depend on only the slope of A. Is this indeed what we will see in B?
| Is important to realize time dilation and the Doppler effect are 2 frame-dependent parts of a single relativistic phenomenon: the constancy of the magnitude of four-velocity (which is always ||c||). Requiring that all inertial observers see a four-velocity with magnitude ||c|| leads to frame dependent observations of time dilation and Doppler effect.
For a Doppler radar, the relativistic Doppler effect can be derived by forcing the scattered radar's (read: photon's) four-acceleration to be orthogonal (in Minkowski space) to the targets 4-velocity, that is:
[(E, pc) - (E', p'c)]*(c, v) = 0.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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The rule breaker, emissivity + reflectivity = 1 If emissivity and reflectivity are inversely proportionate, why does glass have a high emissivity of around 0.95-0.97 as well as being very reflective for IR Radiation?
normally it works but not with glass!
Can anyone explain this?
| Based on the experiments you describe in the comments, it seems like you might very well have a reflectivity of 20-30% in your window, for the spectral region where your camera measure. The question is where you got the high emissivity numebers from. It seems likely that the problem is that you're assuming the emissivity and reflectivity is the same throughout the infrared region. The high emissivity might be for another part of the spectrum than where your camera measure.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Divergent sum in lightcone quantization of bosonic string theory I had the following question regarding lightcone quantization of bosonic strings - The normal ordering requirement of quantization gives us this infinite sum $\sum_{n=1}^\infty n$. This is regularized in several ways, for example by writing
$$
\sum_{n=1}^\infty e^{- n \epsilon } n = \frac{1}{\epsilon^2} - \frac{1}{12} + {\cal O}(\epsilon^2)
$$
Most texts now simply state that the divergent part can be removed by counterterms. David Tongs notes (chap. 2 page 29) specifically state that this divergence is removed by the counterterm that restore Weyl invariance in the quantized theory (in dimensional regularization).
I would like to see this explicitly. Is there any note regarding this? Or if you have any other idea how one would systematically remove the divergence above, it would be great!
| Note that $n$ is really the momentum in the $\sigma$ direction so it has the units of the world sheet mass. The exponent $-n\epsilon$ in the regulator has to be dimensionless so $\epsilon$ has the units of the world sheet distance.
Consequently, the removed term $1/\epsilon^2$ has the units of the squared world sheet mass. This are the same units as the energy density in 1+1 dimensions. If you just redefine the stress energy tensor on the world sheet as
$$T_{ab} \to T_{ab} + \frac{C}{\epsilon^2} g_{ab}$$
where $C$ is a particular number of order one you may calculate (that depends on conventions only), it will redefine your Hamiltonian so that the ground state energy is shifted in such a way that the $1/\epsilon^2$ term is removed.
This "cosmological constant" contribution to the stress-energy tensor may be derived from the cosmological constant term in the world sheet action, essentially $C\int d^2\sigma\sqrt{-h}$. Classically, this term violates the Weyl symmetry. However, quantum mechanically, there are also other loop effects that violate this symmetry – your regulated calculation of the ground state energy is a proof – and this added classical counterterm is needed to restore the Weyl (scaling) symmetry.
It's important that this counterterm and all the considerations above are unable to change the value of the finite leftover, $-1/12$, which is the true physical finite part of the sum of positive integers. This is the conclusion we may obtain in numerous other regularization techniques. The result is unique because it really follows from the symmetries we demand – the world sheet Weyl symmetry or modular invariance.
| {
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"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
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} |
Is there a symbol for "unitless"? I'm making a table where columns are labelled with the property and the units it's measured in:
Length (m) |||| Force (N) |||| Safety Factor (unitless) ||| etc...
I'd like not to write "unitless" on several columns...and I'm quite surprised I can't seem to find a symbol for it. Any suggestions?
| Units I usually keep in brackets []. Like 70 [kg], 60 [GPa], 5.2 [ms^-1] and for no unit I would say 1.5 [-].
| {
"language": "en",
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What is wrong with these ways of determining the mean occupation number? Could anyone point out what went wrong in this argument?
Setup:
We have a system with 2 energy levels say with energies $0,e$ respectively.
We consider the grand canonical ensemble for the system occupied by identical spinless fermions. Then there are clearly 4 feasible microstates --
*
*no particles, $E=0$;
*1 particle in $E=0$ level, $E=0$;
*1 particle in $E=e$ level, $E=e$;
*2 particles one in each level, $E=e$
Clearly the corresponding grand canonical function is $$\Xi = 1+f+f\exp(-\beta e)+f^2\exp(-\beta e) $$ where $f\equiv \exp(\mu\beta)$.
Now I wish to find the mean occupation number of a state with energy $e$.
Here is where my arguments go all wrong, please help me:
Method 1:
$$n(e)=1\times \text{probability of state (3)}+2\times \text{probability of state (4)}\\
= {f\exp(-\beta e)+2f^2\exp(-\beta e)\over \Xi}\\
={1+2f\over 1+f}\cdot{1\over f^{-1}\exp(\beta e)+1}$$
BUT...
Method 2:
If I use the F-D distribution directly, shouldn't I get $$n(e) = {\text{degeneracy of energy level $E=e$}\over f^{-1}\exp(\beta e)+1}\\= 2\cdot{1 \over f^{-1}\exp(\beta e)+1}$$??
What went wrong?
Thanks a lot in advance!!
Christie
| In your Method 1, you forgot that the state (2) also has the occupation number $n=1$ (it has a vanishing energy but you're computing the mean occupation number, not mean energy), so it must be added to the numerator. This changes $1+2f$ in the numerator on the following line to $2+2f$. The simple fraction simplifies to $2$ and you get an agreement with Method 2.
| {
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"source": "stackexchange",
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Is lattice enthalpy positive or negative? I've learnt that the lattice enthalpy (defined as the energy change from a solid ionic lattice to separate gaseous ions) is always positive, obviously. However, I've seen it explained as the opposite other places, so it's negative.
What is correct?
| When it is explained as opposite, it is usually the heat RELEASED when a crystal is formed, and since the word "released" already makes up for the sign of heat(i.e. if heat is -10 that means the heat released or evolved is +10) , it is taken as positive. Nonetheless, some books also present a negative value to it but it doesn't matter unless you are careful about it in calculations.
| {
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"timestamp": "2023-03-29T00:00:00",
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How does the freezing temperature of water vary with respect to pressure? I know when the pressure is reduced, the boiling temperature of water is reduced as well. But how does the pressure affect the freezing point of water?
In a low-pressure environment, is water's freezing temperature higher or lower than $0\sideset{^{\circ}}{}{\mathrm{C}} \, ?$
| When high pressure is applying on a gas, it gets converted into liquid form. Similarly, if more pressure is applied to the liquid, force of attraction increases so that the liquid is converts into solid state. As the pressure increases the rate of crystallization also increases. i.e., the freezing point also increases.
| {
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"source": "stackexchange",
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Difference between slanted indices on a tensor In my class, there is no distinction made between,
$$
C_{ab}{}^{b}
$$
and
$$
C^{b}{}_{ab}.
$$
All I know, and read about so far, is the distinction of covariant and contravariant, form/vector, etc. etc. But what is this slanted business all about?
| Each of the indices in a tensor have a particular left-right ordering. This ordering cannot be changed unless the tensor has some particular symmetry that permits it (or rather, that equates different components on interchange).
The up-down positions of indices tells us about whether the index is associated with using a basis vector (up) or a basis covector (down) for that index to help extract the component. Let $v$ be a one-index tensor. $v^a$ are the components associated with a set of basis vectors $e_{(a)}$, and $v_a$ are the components associated with a set of basis covectors $e^{(a)}$. In general, $v_a \neq v^a$ for an arbitrary coordinate system. Each index can be associated with basis vectors or basis covectors, and we need not use all the same kinds of basis elements for all the indices of a tensor.
In a space with a metric, we can convert back and forth between using basis vectors and basis covectors to extract components of tensors (we can raise or lower indices more or less at will), so we tend to associate all such combinations of up-down indices with the same inherent object. Still, whether a particular index is up or down for a given situation depends on what's convenient or necessary to use.
Edit: So strictly speaking, writing an indexed object with two indices lined up on top of each other doesn't really have meaning. Still, physicists often do this anyway for, as an example, Christoffel symbols--it's relatively rare that they're used in any other way than as $\Gamma^a_{bc}$. Still, when it comes to the Riemann tensor or other such objects, it's best to think of each index has occupying a whole column--nothing else should come above or below that index, to give room for it to move freely up or down when contracted with the metric.
| {
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$\beta^+$ decay We've been discussing radioactive decay at school, and I grasped everything except for $\beta +$ decay. When I googled radioactive decay, I immediately found out they dumbed down radioactive decay for us, which is probably why they didn't care to explain what they did, they just showed some calculations. We have never discussed neutrino' s and antineutrino's, they leave that out of the equation, which is no problem since they have negligible mass and no charge.
So we've been taught that a proton and electron form a neutron, which I have also discovered is not true (I'm discovering a lot of new things :P). I learned that this is caused by the spontaneous (?) change from up to down quarks and vice versa.
However, at school I must keep to the 'rules' and by those rules I don't really understand $\beta +$ decay. I see $\beta -$ decay as follows:
An electron flees a neutron and leaves a proton. That's why you get an atom with a higher atomic number.
However, how would this work with beta + decay? Can it even be dumbed down to this kind of high school thinking?
| $\beta^+$-decay can be explained as a result of nuclear protons collision:
$p^+ p^+\to p^+ p^+ W^- W^+ \to (p^+ W^-)\: p^+ W^+\to n\: p^+ e^+ \nu_e$
Compare to $\beta^-$-decay:
$n \to p^+ W^- \to p^+ e^- \bar{\nu}_e$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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How to express a Hamiltonian operator as a matrix Suppose we have Hamiltonian on $\mathbb{C}^2$
$$H=\hbar(W+\sqrt2(A^{\dagger}+A))$$
We also know $AA^{\dagger}=A^{\dagger}A-1$ and $A^2=0$, letting $W=A^{\dagger}A$
How can we express $H$ as $H=\hbar \Big(\begin{matrix}
0 & \sqrt2 \\
\sqrt2 & 1
\end{matrix} \Big)$
So far I've shown that if we consider the eigenvalues of $W$,
$$W|\psi \rangle=w|\psi \rangle$$
It implies that $A|\psi \rangle$ and $A^{\dagger}|\psi \rangle$ are also eigenvectors of $W$ with eigenvalue $1-w$. Using $A^2=0$ we find that $w=0$ or $1$
I'm not entirely sure how you go about expressing operators as matrices, as the majority of my course has been using wave function notation, I'd really appreciate if someone could explain the next steps here just so I can have a more rigorous understanding of it.
| The matrix element $O_{ij}$ of an operator are defined by
$$O_{ij} = \langle i | \hat{O} | j \rangle,$$
and it is traditional that the $i$ index labels the row and $j$ labels the column. This way matrix multiplication works as you would expect:
$$ (O P)_{ij} = \sum_k O_{ik} P_{kj}, $$
which you can show by inserting a complete set of states.
| {
"language": "en",
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"source": "stackexchange",
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Kinematics with non constant acceleration A particle experiences an acceleration described by
$$
a=kx^{-2}
$$
where x is the displacement from the origin and k is an arbitrary constant.
To what value does the velocity v of the particle converge to as x approaches infinity if the particle starts at some point x0?
If I approach this problem with energy, then
$$
W = \int F \mathrm{d}x
$$
$$
= \int_{x_0}^\infty mkx^{-2} \mathrm{d}x
$$
$$
= mk(-\infty^{-1}+{x_0}^{-1})
$$
$$
W = K = mk{x_0}^{-1}
$$
$$
\frac{1}{2}mv^2 = mk{x_0}^{-1}
$$
$$
v = \sqrt{2k{x_0}^{-1}}
$$
How would I solve this problem with pure kinematics? (there appears to be some sort of cyclical dependency where acceleration affects velocity, velocity affects displacement, and displacement affects acceleration)
Likewise, two particles experience accelerations described by
$$
a_1=k_1x^{-2}
$$
$$and $$
$$
a_2=-k_2x^{-2}
$$
where x is the distance between the two particles
What two velocities do the particles reach as x approaches infinity if the two particles are initially separated by some x0?
| The solve the first part with just kinematics, use the chain rule:
$$ a(x) = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = \frac{dv}{dx}v,$$ and then $a(x)dx = vdv$.
Integrating both sides ($x_0$ to infinity on the left and $v_0$ to $v_f$ on the right), we get
$$\frac{k}{x_0} = \frac{v_f^2 - v_0^2}{2},$$
or
$$v_f = \sqrt{\frac{2k}{x_0} + v_0^2}.$$
Solving the two particle scenario is no more complicated than the single particle version as long as you pay attention to signs for particle 2.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is it possible to "add cold" or to "add heat" to systems?
Amanda just poured herself a cup of hot coffee to get her day started.
She took her first sip and nearly burned her tongue. Since she didn't
have much time to sit and wait for it to cool down, she put an ice
cube in her coffee and stirred it with a metal spoon. After a moment,
she felt the spoon warm up, but when she took another sip, the coffee
was cooler. She was pretty sure, the ice added cold to her coffee,
and the coffee added heat to her spoon.
Would you agree?
| The most important thing about feeling is that its relative and depends on the temperature of sensor point on skin. That's the reason we need thermometer to measure temperature with which everyone could agree.
Energy transfer is involved here, but it has very little to do with Amanda's feeling. Amanda's tongue and mouth was at high temperature due to initial sip which caused her to feel next sip cooler which was really cool due to addition of ice.
In case of spoon, spoon can't get hotter because heat of coffee (and some of heat of spoon) was consumed in melting ice. But, Amanda felt the spoon warmer because her fingers were relatively cold due to ice touch.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Observer effect, do this mean literally someone or just any interaction with other matter? I am a layman and was wondering, the quantum observer effect. The regular notion to laymen seems to be literally "if you look at it", but as I am coming to understand the world I live in better I feel it means just coming in contact with something.
Is this an ongoing question? Whether a particle that is not interacting with anything undergoes changes in state? Though now that confuses me too, If it is in two states at once. What denote say spin to the left from spin to the right, what denotes the origins of the calculations we make?
| This "observation" is actually just a more simple way to state the uncertainty principle. For a one dimensional case involving momentum and position it would be as follows.
$$\Delta x \Delta p \ge h$$
This means that if we know the position exactly then we don't know the momentum.
For energy and time it would be
$$\Delta E \Delta t \ge \frac{\hbar}{2}$$
Basically what these mean is that there is a certain amount of maximum precision that we can make. If we knew the energy exactly ($\Delta E = 0$) then we know that $\Delta t$ is AT LEAST $\frac{\hbar}{2}$. This is the problem with quantum mechanics (and is misinterpreted by many people). Observing is actually an act of measuring. There is a wave particle duality that goes on in qm. Think of a particle existing along a wave. It is not too difficult to find the momentum of a wave, but by measuring this you know that it is a wave, and lose information about where it is. The converse is true too, that if you measure it's position you have an inherent uncertainty in the momentum. I suggest looking up the double slit experiment if you would like a visual example of this. Many videos of this exist.
| {
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"timestamp": "2023-03-29T00:00:00",
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Towing of asteroid I recently studied that NASA has planned to tow and place it in the orbit of the moon.
My doubt is when asteroid is placed in the orbit near moon.since the gravitational field of earth is very high.what will it revolve around the moon or the earth.
Can anyone clarify my doubt ??
| There are stable orbits around the Moon, also Nasa, Esa, and many more have satellits orbiting the Moon, and they don't "decay" towards the Earth.
*
*Wiki List of lunar probes Look in the link for orbiters
To look for the physics and no only examples search for Hill sphere:
*
*One google entry
*Hill sphere Wiki
*Sphere of influence Wiki
Acording to the last link, the Moon's sphere of influence is 38 Moon radii or $66.1\cdot 10^{3}km$ plenty of space to host an asteroid.
| {
"language": "en",
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Classical results proved using quantum mechanics Are there any results in classical mechanics that are easier to show by deriving a corresponding result in quantum mechanics and then taking the limit as $\hbar\rightarrow0$?
(Are there classical results that were first discovered through taking the classical limit of quantum mechanics even if they are now easier to demonstrate, in hindsight, using classical mechanics?)
Update: One semi-example is the derivation of classical adiabatic invariants from the quantum adiabatic theorem in some textbooks. But the quantum adiabatic theorem isn't really any easier.
| I just came across an example that I'll post as an answer to my own question.
In Brillouin scattering sound passes through a medium causing periodic variations in refractive index and hence forms a diffraction grating that can scatter light. Because the sound waves are moving, the scattered light is doppler shifted. All of these effects can be described classically and so there is a classical argument to determine the frequency of the scattered light. Nonetheless, most discussions of Brillouin scattering I've seen treat the effect as an interaction between phonons and photons. For example, if an incoming phonon has frequency $\omega_s$, and the incoming and outgoing photons have frequencies $\omega_i$ and $\omega_o$ then conservation of energy gives
$\hbar\omega_o = \hbar\omega_i+\hbar\omega_s$.
We don't even need to take the limit as $\hbar\rightarrow 0$.
The classical argument gives the same result but the quantum argument is much simpler (once you already know how to quantise the relevant systems). Higher order scattering from the grating can be viewed as $n$-phonon interactions.
| {
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Does spatial coupling prohibit resonances due to an external source field? The harmonic oscillator coupled to a sinodial external source
$$\frac{\partial^2 x(t)}{\partial t^2}+\omega_0^2 x(t)=F_0\sin(\omega_\text{ext}\ t),$$
has the solution
$$x(t)=x(0)\cos(\omega_0 t)+C \sin(\omega_0 t) +\frac{F_0}{\omega_0^2-\omega_\text{ext}^2}\sin(\omega_\text{ext} t),$$
with a constant of integration $C$. So here we have a resonance for $\omega_\text{ext}=\omega_0$.
If we consider the Klein-Gordon equation with non-vanishing coupling and with a source function $F$
$$\frac{\partial^2 \phi(t,x)}{\partial t^2}-c^2\nabla_x^2\ \phi(t,x) +\omega_0^2\ \phi(t,x)=F(s,x,t),$$
(one dimensional, three dimension, whatever you can deal with)
is there also an family of functions $F(s,t,x)$, (family w.r.t. the parameter s), such that we have a singularity at $s=\omega_0$?
Maybe you go about similar as with the Helmholtz equation, except than you get a shifted coefficient $\omega_0+k^2$.
I think that could be the case as a damping term $\delta\cdot x'(t)$ also destroys the resonance, at least for a sinusoidal source.
| Let us denote the field amplitude $\varphi$ (because $x(x)$ does not make sense).
If you drop the term $\omega_0^2\varphi(x,t)$, then your KG equation will look like a field sourced with an external source (force). The corresponding (space) Fourier harmonic $\varphi_{k(\omega_0)}(t)$ will have a growing solution if the force is periodic and starts acting at $t=0$.
EDIT: The mass term will shift the spectrum and it may make low-frequency force non resonant.
Edit2: The external source $F(s,x,t)$ can be represented as due to a point-like particle, i.e., $F(s,x,t)=\delta(x-x_0)F(s,t)$, then, after Fourier transformation, you will obtain driven oscillator equations for $\chi(k,t)$.
| {
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If photons can be absorbed by electrons, wouldn't that mean light has a charge? I am a biochemistry and molecular biology major. If photons can be absorbed by electrons, wouldn't that mean light has a charge? Electrons only attract positive charges. Isn't it?
| Light does not have a charge. "Charge" is just what we call the tendency of something to affect and be affected by electromagnetic fields, which are what light is made of.
The reason positive and negative charges are attracted to each other is that they both give off these fields. An electron creates an electric field that attracts positively charged things, and a proton creates an electric field that attracts negatively charged things. Moving charges create magnetic fields. Also, a changing electric field creates a magnetic field and vice versa.
If a charge is accelerated, it creates a combination of changing electric and magnetic fields that sustain each other, this is called an electromagnetic wave. Light, radio waves, and X-rays are all just different frequencies of these waves.
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Could some design of a propeller be used in both air and water? Propellers in water are smaller in diameter. They also move more slowly. On the other hand, aircraft propellers are larger in diameter, have narrower blades and operate at very high speeds. An aircraft propeller would break apart in water, while a water propeller would produce little to no thrust in air.
Rotational speed can be easily adjusted while moving between environments. However, is there such a thing as a propeller shape that is halfway between the two designs? Or would it simply be extremely inefficient in both environments?
| There's two important differences between air and water: Air is compressible, and the densities are about a factor of ~1000 apart - 1 kg/m³ vs 1 t/m³!
For most concerns where you use propellers, compression plays no role because the pressure diferences are very low. The densities, however play a large role.
The thrust can be described as $F = \dot m * \Delta v$, with $\dot m$ beeing the mass flow - kg&/s or such - and $\Delta v$ the difference in velocity a volumeelement of fluid is accelerated.
So, to achieve a similiar thrust, the same propeller would have to move 1000 times more air than water by volume. Hence the often larger and faster spinning propellers for planes.
On the other hand, in a heavier medium each wing of the propeller is subject to stronger torque (all else beeing equal):
$$Q = \rho V_{a}^2 D^3 f_q(\frac{ND}{V_a})$$
(Source)
$\rho$ is density, $V_a$ rate of advance (how much the propeller moves forward per revolution), D is diameter and N number of revolutions.
Without going into the math it can probably be shown that in a heavier medium the propeller will experience somewhat more torque for the same thrust - I'm to lazy to try now. The propeller will be built with more robust (and possibly heavier) material than would be the case id it's an only air propeller.
That said, I believe that a propeller for both media is entirely possible, though challenging.
However, a propeller for a both media will need a drivetrain that can accomodate speed roughly a factor 1000 apart (that's not trivial).
One other reason why don't see a propeller for both media is that there is no vehicle that could make use of one.
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Physical Interpretation of the Integrand of the Feynman Path Integral In quantum mechanics, we think of the Feynman Path Integral
$\int{D[x] e^{\frac{i}{\hbar}S}}$ (where $S$ is the classical action)
as a probability amplitude (propagator) for getting from $x_1$ to $x_2$ in some time $T$. We interpret the expression $\int{D[x] e^{\frac{i}{\hbar}S}}$ as a sum over histories, weighted by $e^{\frac{i}{\hbar}S}$.
Is there a physical interpretation for the weight $e^{\frac{i}{\hbar}S}$? It's certainly not a probability amplitude of any sort because it's modulus squared is one. My motivation for asking this question is that I'm trying to physically interpret the expression
$\langle T \{ \phi(x_1)...\phi(x_n) \} \rangle = \frac{\int{D[x] e^{\frac{i}{\hbar}S}\phi(x_1)...\phi(x_n)}}{\int{D[x] e^{\frac{i}{\hbar}S}}}$.
| One interesting interpretation comes from the Wick rotation, where you interpret $it/\hbar=\beta=1/(k_BT)$ - an imaginary time turns many quantum equations into similar equations of thermodynamics / statistical mechanics.
Since $S = \int L\,dt = \int (T-V)\,dt$ has the dimension of energy times time, this means the exponent can be interpreted as energy times $\beta$, i.e.
$$e^{\frac i\hbar S} \approx e^{-\beta E}$$
which is the summand of the partition function, from which you can then derive other thermodynamic quantities like entropy, heat capacity or the Helmholtz free energy.
Note I've been very hand-wavy about the transition from $L$ to -$E$, you'd have to insert the actual Legendre transformation $L=pq-H$ and really evaluate the path integration. I think we've done that more exactly in lectures at one point, but I can't remember for sure...
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What is Transverse Energy? What is transverse energy? Why we use transverse total energy instead of energy and transverse momentum in place of Total momentum in the particle detectors?
| Transverse momentum, $\vec{p}_T$, is the momentum of an object transverse to the beam. Transverse energy is defined as $E_T = \sqrt{m^2+p_T^2}$ for an object with mass $m$ and transverse momentum $p_T$.
The initial longitudinal momentum in a parton collision is unknown, because the partons that make up a proton share the momentum. We do know, however, that the initial transverse momentum was zero. So we look for missing transverse momentum, defined $E_T^\textrm{miss} = -\sum_i \left|\vec{p}_T(i)\right| \equiv -\sum_{i}p_{T}\left(i\right)$ for visible particles $i$. Finding missing transverse momentum would indicate that new, unaccounted for particle(s) had escaped the detector.
Confusingly, $E_T^\textrm{miss} = -\sum_i p_T(i)$ is commonly called missing transverse energy or MET. Missing transverse energy is equivalent to missing transverse momentum only if the missing particle(s) were massless.
Also, events in which the products have large transverse momentum are more likely to be genuine, interesting events.
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Invariance, covariance and symmetry Though often heard, often read, often felt being overused, I wonder what are the precise definitions of invariance and covariance. Could you please give me an example from field theory?
| It helps to remember that invariant quantities are seen as scalars to the transformation (they have no indices in the target space). In the other hand, covariant quantities are objects that transform in a certain way.
Example: Vectors in $R^{2}$, under rotation $R_{ij}$, transform covariantly since $v'_{i}=R_{ij}v_{j}$, but it's length is invariant since $|v'|^{2}=v'_{a}v'_{a}=R_{am}v_{m}R_{an}v_{n}=v_{m}R^{t}_{ma}R_{an}v_{n}=v_{m} \delta_{mn} v_{n}=v_{n}v_{n}=|v|^{2}$. This means that Newton's second Law transforms covariantly under rotations and the magnitude of the force is invariant.
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What would happen to the Moon if Earth is turned into a black hole? Assume that all of sudden the Earth is turned into a black hole. And the moon revolves around the Earth (before turning into a black hole). What would happen to the Moon after earth changes to black hole will it be sucked to the black hole or continue to revolve the black hole?
| I think that force of attraction between earth and moon is depend on their masses and the distance between them . Since any of these factors are not going to change if earth becomes a black hole. Hence it will continue to revolve in orbit
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Bremsstrahlung vs energy conservation From Wikipedia:
Bremsstrahlung is electromagnetic radiation produced by the
deceleration of a charged particle when deflected by another charged
particle, typically an electron by an atomic nucleus. The moving
particle loses kinetic energy, which is converted into a photon
because energy is conserved.
Isn't energy conserved for the moving particle in an electrostatic potential, $E_{kinetic} + E_{potential} = \frac{mv^2}{2}+\frac{kqQ}{r}$? If so, where does the extra energy for photons come from?
Why don't electrons in atoms radiate away their energy?
| The short answer to your question is "because of quantum mechanics". If you view the atom and the electron orbit around the nucleus as a classical system then indeed the system would not be stable and the electron would lose orbital energy. This is why Niels Bohr proposed a model with quantized energy levels.
This was one of the first serious motivations for quantum mechanics.
If you want an intuitive model to understand this, instead of thinking of the electron as some point orbiting, think of a 3D standing wave. A vibrating water drop or other shell-like membrane around the nucleus vibrating without any friction. This model isn't any more correct than the single orbiting electron but at least then you can see it not losing energy.
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Why must the angular part of the Schrodinger Equation be an eigenfunction of L^2? I was reading about the solution to the Schrodinger Equation in spherical coordinates with a radially symmetric potential, $V(r)$, and the book split the wavefunction into two parts: an angular part and a radial part. When dealing with the angular part of $\psi$, the book claims that the angular part "must" be an eigenfunction of $L^2$ (the square of the angular momentum operator) and since the eigenfunctions of $L^2$ are the spherical harmonics $Y_{lm}$, then the angular part of $\psi$ is just the spherical harmonic equations.
I am confused as to why the angular part of $\psi$ must be an eigenfunction of $L^2$.
| The operator $L^2$ is a symmetry of the Hamiltonian, which means that $[H,L^2] = 0$. This means that we can find simultaneous eigenfunctions to $H$ and $L^2$ since $HL^2 | \psi \rangle = L^2 H | \psi \rangle = \ell (\ell + 1) E | \psi \rangle$. The operator $L^2$ does not depend on the radial coordinate, thus the angular part of the wavefunction for hydrogen can be chosen to be an eigenfunciton of $L^2$. This is useful as it allows to label states with quantum numbers corresponding to the energy and angular momentum of the state.
In general, whenever you have a spherically symmetric potential (as in the case of the hydrogen atom) the spherical harmonics will show up as eigenfunctions of the angular part.
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What is Anderson localization? Could someone give an example worked out in detail? What is Anderson localization, for someone with no previous knowledge on the subject?
I tried to read Anderson's original paper, but it was too terse for me. I have seen a couple of intuitive explanations, e.g. 50 years of Anderson localization on Physics Today. I also read "Localization of waves" by van Tiggelen, but it is more of a review with a lot of formulas and no deductions.
What I need is an introduction to the subject through one example of Anderson localization, worked out in detail.
You can skim over the tedious math. If you point out what needs to be done, I'll work it out (and if I get stuck I'll ask!).
(Related Phys.SE question: Introduction to Anderson localization)
| The toy example is localization on the Bethe lattice (AKA the regular Cayley tree). There is a paper by Abou-Chacra, Thouless and Anderson that discusses this. Or you can just google around.
R. Abou-Chacra et al 1973 J. Phys. C: Solid State Phys. 6 1734
or
R. Abou-Chacra and D.J. Thouless, J. Phys. C 7 (1974), 65.
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How long does a supernova last? Is a supernova over instantaneously? Or, does the (for want of a better word) explosion continue for a while? What is/are the order of timescales involved? What is the duration for which the supernova continues to release copious amounts of energy?
| If you're asking about the duration of the explosion of supernovae, then it's done within a few seconds (the number can still be less) - similar to a nuclear fission. But, the cloud surrounding the explosion (i.e) the matter remnant that is thrown off can still remain expanding for years and it can emit EM radiation in the form of $\gamma$-rays and X-rays which can be detected. This result has a similarity with that of the big bang. The historical light from the big-bang (thought of as) is still detected in the microwave region due to redshifts. For example, SN-1987A has been observed by the Hubble for over 15 years and there's a time lapse sequence for the explosion in Wiki...
(In case of atom bomb, a mushroom-kinda cloud is formed which expands slowly and so, the matter can cover high altitudes, which can be easily tracked by GPS satellites)
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Some questions about anyons? (1) As we know, we have theories of second quantization for both bosons and fermions. That is, let $W_N$ be the $N$ identical particle Hilbert space of bosons or fermions, then the "many particle" Hilbert space $V$ would be $V=W_0\oplus W_1\oplus W_2\oplus W_3\oplus...$ , and further we can define creation and annihilation operators which satisfy commutation(anticommutation) relations for bonsons(fermions).
So my first question is, do we also have a "second quantization theory" for anyons like bosons and fermions?
(2) Generally speaking, anyons can only happen in 2D. Is this conclusion based on the assumption that the particles are point-like?
In Kitaev's toric code model, the quasiparticles are point-like due to the local operators in the Hamiltonian. My question is, in 3D case, whether there exists a simple model whose Hamiltonian contains local operators and spatially extended operators, so that it has both poit-like quasiparticles(say, $\mathbf{e}$) and knot-like quasiparticles(say, $\mathbf{m}$), then the $\mathbf{e}$ and $\mathbf{m}$ particles have nontrivial mutual statistics in 3D?
| No, “this conclusion” is based on the topological properties of rotation groups. Namely, for any $n > 2$ $\mathrm{Spin}(n)$ is the universal cover of $\mathrm{SO}(n)$, whereas for $n = 2$ it is not. That’s why in $n > 2$ any thing has to be controlled by a representation of the Spin group, whereas in $n = 2$ it has not.
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What is the fastest a spacecraft can get using gravity-assist? Assuming normal spacecraft and space objects (no neutron stars, black holes, etc). To what speed can a spacecraft accelerate using gravity-assist?
For example, if a spacecraft is moving at relativistic speeds, it probably won't get seriously sped up by normal-density objects.
| This limitation would be imposed by the relative speed difference between the starting point and the object being used for a gravity assist.
to steal Wikipedia's example: Imagine you have a train coming at you at 50mph. You throw a ball at the front of the train at 30mph.
From your perspective, the ball was moving 30mph on the way in and 130mph on the way out (having bounced off the train). From the perspective of the train, the ball approached at 80mph and left at 80mph.
So the maximum speed, as viewed from the thrower, is a combination of the initial speed and the relative speed of the object in question.
With infinite object in perfect alignment: there is no limit other than relativity. Every exchange takes intertia from the planet and gives it to the spacecraft. The limiting factor is how much can be puller per interaction, and how many useful interactions can occur.
This is why the voyager probes were launced when they were: Jupiter, Saturn, and Uranus were aligned in such a way as to make a "grand tour" of using all three for gravity assists possible.
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Ratio of Size of Atom to Size of Nucleus I have the following problem:
In nuclei, nucleons exists in nuclear energy levels and in atoms, electrons exist in atomic energy levels. The order of magnitude of nuclear energy is 1MeV whereas the energy of atomic energy levels is of the order 1eV. Use this info and the particle in the box model to make an order of magnitude estimate of the ratio $$\frac{\text{size of atom}}{\text{size of nucleus}}$$
So the way I approached this was to consider $$\frac{E_a}{E_n}=10^{-6}\Rightarrow \frac{\frac{h^2}{8m_aL_a^2}}{\frac{h^2}{8m_nL_n^2}}=\frac{m_nL_n^2}{m_aL_a^2}=10^{-6}\Rightarrow \frac{L_a}{L_n}=\sqrt{10^6\cdot m_n/m_a }.$$The mass of the nucleus we assume to be 1u but what about the mass of the atom? In the answers at the end of the book it says that we take $m_n = m_e$. But why is this so? Shouldn't the mass of a typical atom also include the mass of the nucleus, along with the mass of the electron?
Please keep your answers simple since I am only doing A-Level physics.
Thanks in advance.
| In this case, "size of atom" really means "size of the box that is holding the electron in its place". The "box" is provided by the confining electromagnetic force exerted by the nucleus on the electron. Similarly, the box representing the nucleus is provided by the strong nuclear force between protons and neutrons that hold them all in place.
In the formula you provide, the mass $m_a$ that appears really refers just to the mass of the particle in the box, which for the atom is the electron. The mass of the box itself does not affect the energy of the particle inside it: the only thing that matters is the length of the box. Including the mass of the nucleus would be a bit like including the mass of the box in your formula.
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Is it physically meaningful to talk about the 'total temperature' of an object? If I had a semi infinite, 1-D object and a finite 1-D object, both heated at the same constant rate at one end each for the same time period and both begin at the same initial temperature, is it physically meaningful for me to integrate along the length of the object and consider this integral as a function of time and a measure of the 'amount' of heat on the object?
| Temperature is the measurement of kinetic energy per unit particle mass. Since you've added the same amount of heat energy to each object, the finite object will have a higher temperature because its heat energy is distributed across a smaller collection of mass. Taking something's temperature is indeed a meaningful measurement ;)
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When does Thevenin's theorem not apply (modelling a power source with a ohmic internal resistance) Most physics text books say that a power source can be modelled as an EMF with a internal resistance. This is also know as Thevenin's theorem or Norton's theorem. However I have read in some sources that this is not always the case.
When does this model not apply/break down and for what reasons?
| Since you mention batteries in a comment:
The voltage source + resistance model works well in many circumstances. At the extremes:
*
*high frequency / fast rise time waveforms: There will always be some inductance, from wiring if nothing else, which appears as an inductance, an easy add to the model.
*low frequency / long durations: As the battery discharges (or charges), its terminal voltage falls (rises). The simplest model substitutes a (large) capacitor for the fixed voltage source. More accurate models add a variety of distributed resistance and capacitive elements to simulate the electrochemistry inside the cell. More accurate yet is to make these elements non-linear (with capacitance a function of the voltage). People measure and study these characteristics intensively these days.
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What is the most optimal earth's axial tilt in terms of variation of seasons? What is the most optimal earth's axial tilt in terms of variation of seasons?
What would be optimal axial tilt for earth that life would exist and change of seasons would be at minimal level?
| If Earth had 0 inclination to the ecliptic, then life could still exist and there would be no change in seasons. In fact, at latitudes greater than $-23.5^\circ$ and less than $23.5^\circ$, the seasons are already irregular compared with the rest of the planet; they have two points where the Sun shines directly on top of them, effectively making them have two summers. At the equator, there is practically no/minimal variation in seasons already.
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Inserting metal into parallel plate capacitor
The plates of an isolated parallel plate capacitor with a
capacitance C carry a charge Q. The plate separation is d.
Initially, the space between the plates contains only air.
Then, an isolated metal sheet of thickness 0.5d is inserted
between, but not touching, the plates. How does the
potential difference between the plates change as a result
of inserting the metal sheet?
The answer is: The potential difference will decrease. How do I get this result, it seems right but how do I reason it out?
$$V = Ed$$
There was mention about "E is the same outside the conductor. But
the field inside is 0.". But in the formula, E is the field outside conductor right? If so it remains the same? Then $d = 0.5d$?
| Yes d=0.5d, Basically start thinking in terms of work per unit charge. So work you do on $1$ coulomb is $E$x$0.5d$. as field in conductor is $0$ in electrostatic condition , you don't have to do any work at all in that region . so $\Delta V$ is 0 in that region . Hence , $\Delta V_{net}= E_{outside}$x$0.5d$. = V for the capacitor . And calculate the new field in the two new air regions divided by the metal plate , you will find that by principle of superposition , it remains same . So it is because of this calculation that Net field also comes out to be same as before, that you're able to use E , basically think in terms of work and everything will be resolved . And then you will know , you will have to evaluate E everywhere again because now the metal plate has been polarised .
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Alpha particle and helium nucleus The symbol for the alpha particle is α or $α^{2+}$, it can be written as $He^{2+}$.
What I want to know is that, are they same? I mean alpha particle and helium nucleus are same or any subtle difference exists?
| They are the same particles. When $\alpha$-radiation eventually gets stopped by an object (a sheet of paper or simply a meter or so of air will do the trick) it attracts two electrons and becomes elementary Helium.
Most of the world's Helium actually originates from reserves underground where Uranium and other $\alpha$ radiators have been creating He for millenia.
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How can Ohm's law be correct if superconductors have 0 resistivity? Ohm's law states that the relationship between current ( I ) voltage ( V ) and resistance ( R ) is
$$I = \frac{V}{R}$$
However superconductors cause the resistance of a material to go to zero, and as I understand it, as $R \to 0$, $I \to \infty$. Does this present a problem for Ohm's law?
| Ohm's law is generally NOT correct, it's called a law for historical reasons only!! It's a law in the same sense in which Hooke's law is a law... it holds only for certain systems under certain conditions, but it's widely known because it's simple and linear!
It's not just superconductors, diodes are a neat everyday example of Ohm's law failing to hold. But it fails for every material under sufficiently extreme circumstances.
Check out this I-V graph for a diode.
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Polarization of sound Sound can't be polarized because the vibration of such type can't be polarized i.e, it can't be limited or controlled by any barriers and so polarization is not possible in them.
This is what my teacher answered me when i asked the question. But i didn't understand what did he mean by "the vibration can't be controlled or limited." Does the word cant be limited or controlled make sense here? Moreover can anybody explain in details and more clearly to me?
| It sounds like your teacher's explanation might have been a little misleading. The reason sound can't be polarised is that it is a longitudinal wave, unlike light which is a transverse wave. (Those links have some animated diagrams that should help to make the difference clear.)
"Transverse" means that if a beam of light is coming towards you, the electromagnetic field is vibrating either from side to side or up and down. Unpolarised light is doing a mixture of those two things, but a polarising filter puts it into a more "pure" state, so that it's only going side to side, or only going up and down. (Or diagonally or whatever. There's also a third possibility, called circular polarisation, which is a special combination of the two.) On the other hand, "Longitudinal" means that if a sound wave is coming towards you, the air molecules are vibrating forwards and backwards, not side to side or up and down. Sound waves cannot be polarised because because they don't have any side-to-side or up-and-down motion, only front-to-back.
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Elementary question on pion-proton scattering cross-section
Is $E_k$ the energy of the outgoing or incoming pion?
The first peak is supposed to be a delta baryon.
What does the graph tell us, experimentally? A pion of kinetic energy x comes in, then we look at the graph and find the cross-section, so then we have the probability of what happening? Or is $E_k$ the net loss in kinetic energy of the pion, and this graph tells us the distribution of its energy loss?
How is this graph constructed, what is measured?
| This is a plot where the incoming pion beam is varied and the total interaction cross section is measured , showing higher scattering cross section at the resonances, as the beam kinetic energy is varied. The interactions are measured by looking how many pions are left in the beam direction after the beam has passed has passed the target. The difference is due to their having interacted , from elastic scattering and changing the angle to creating more particles , they are missing in the count.
This is related. explaining how similar plots can be made after the interaction:
In particle physics E_k is the kinetic energy of an outgoing pion from a pi proton scattering.Such plots are usually constructed by scattering a beam of pions with fixed energy and looking at two body interactions with the protons at rest, identifying the outgoing proton and pion by ionization. One would have plotted the invariant mass distribution and then the resonances would have been easily labeled.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is the (non-relativistic) stress tensor linear and symmetric? From Wikipedia:
"[...] the stress vector $T$ across a surface will always be a linear function of the surface's normal vector $n$, the unit-length vector that is perpendicular to it. [...] The linear relation between $T$ and $n$ follows from the fundamental laws of conservation of linear momentum and static equilibrium of forces, and is therefore mathematically exact, for any material and any stress situation. The components of the Cauchy stress tensor at every point in a material satisfy the equilibrium equations (Cauchy’s equations of motion for zero acceleration). Moreover, the principle of conservation of angular momentum implies that the stress tensor is symmetric."
What is the more detailed explanation for this, and/or where can I find it?
| One important thing to remember is that there is the supposition that there is no force that acts on the bulk of the object that's capable of exert net torque (i.e., exchange angular momentum).
This idea changes when you think on something like the interplay between matter and electromagnetic field. Although you can adapt the image above to show that the full stress tensor of the whole system must be symmetric, it's not necessary the stress tensor of it's part should be symmetric too.
| {
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Gas Laws And Adiabatic Process Air at 20 degrees Celsius is compressed adiabatically from 1 bar to 10 bar, what will its temperature be?
With
$$P_1 = 1,$$ $$P_2 = 10$$
$$T_1 = 293K,$$ $$T_2 = unknown$$
using $$\dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}$$
my solution was
$$\dfrac{1}{10}=\dfrac{293}{T_2}$$
giving
$$T_2 = 2930$$
My physics tutor said this is wrong and I should use $P_1V_1^\lambda=P_2V_2^\lambda$ and then use $\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}$ to find the temperature.
The only problem I now have is that $V_1 = V_2 = unknown$ which is making me think that he may have forgot to add that information to the question.
I would like to know if there is anyway to solve this (I do not want the answer just a point in the right direction).
thanks in advance!
| *The only problem I now have is that $V1=V2$=unknown which is making me think that he may have forgot to add that information to the question. *
I belive that's the problem. You can't have $V1=V2$ in an adiabatic process (suposing that the number of particles is constant).
Your tutor already pointed the right direction, combine $P_1V_1^\lambda=P_2V_2^\lambda$ with the ideal gas law, and this should be enough to eliminate the $V$ dependence. ($V_1/V_2=T_1/T_2$ already comes from ideal gas law)
| {
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Are gravitational time dilation and the time dilation in special relativity independent? There are two kinds of time dilation:
*
*One because the other clock moves fast relative to me (special relativity).
*Another one because the other clock is in a stronger gravitational field (general relativity), or accelerating rapidly (equivalence principle).
So are these two effects are totally independent? Is it possible to derive the general relativity case from the special relativity case?
| In the calculation of the relativistic precession of Mercury there is a factor of 3. This includes 1 part kinematic time dilation, and 2 parts gravitational time dilation. They are not mutually exclusive.
| {
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How to determine the direction of medium's displacement vectors of a standing wave? Consider the following problem taken from a problem booklet. My questions are:
*
*What is displacement vector?
*And how to determine the direction of displacement vector at a certain point?
*Where is the position with zero displacement vector?
| Just because a wave is a standing wave doesn't necessarily mean that the particles themselves do not move, in face if the particles themselves didn't move there wouldn't be any wave motion at all. For a longitudinal wave (made of particles that oscillate in the direction of wave propagation like here as sound waves) particles oscillate left and right but have no net displacement.
Take a look at this site on standing waves in compression/longitudinal motion, perhaps it will help you understand what the answer is and why it is the correct one.
http://www.acs.psu.edu/drussell/Demos/StandingWaves/StandingWaves.html
| {
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What counts as "observation" in Schrödinger's Cat, and why are superpositions possible? So if I understood correctly, Schrödinger's Cat is a thought experiment that puts a cat inside a box, and there's a mechanism that kills the cat with 50% probability based on a quantum process. The argument is that the cat now must be in a superposition of dead and alive.
This makes sense at first, but the state of the cat inside the box will affect the outside world in an observable way, right? For example if the cat dies, it might meow loudly which would be audible. If it didn't meow, it would produce a thud on the ground when it dies. And even if the ground was very solid, the redistribution of mass inside the box will affects its gravity field which means the whole universe theoretically immediately observe's the cat's death.
So extending this argument to all superpositions, the different states would cause different effects on the rest of the universe, usually a slight change in the gravity field is the minimum. This gravity perturbation would propagate throughout the universe, and even all the experimenters go to sleep with thick, thick earplugs, somebody or something in the universe is going to inadvertently observe the event and the superposition immediately collapses. Thus superpositions cannot exist beyond an extremely short amount of time.
What's wrong with my reasoning?
| Superpositions can exist without problem until infinity. Basically everything including me only exists if your look (measure it). QM is very weired and that why most who are responsible for QM like Schrödinger and Einstein felt very uncomfortable with its philosophical side effects. Thats where the cat comes in. From the scientist point of view it does matter when its decided if the cat dies or not.
| {
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Evolution principle of the physical laws I wanted to know if there is a physical theory that considers that the laws of physics undergo an evolutionary process. That see the law of physics or the absence of them, as something dynamic, and that with time they slowly converge to something we know today. A kind of simulated annealing of the physical laws.
| Yes, evolution has influenced physics, though it isn't wide-spread. Cosmologoical natural selection draws upon Darwin's natural selection. It postulates that a black hole, upon collapse, spawns a new Universe, with parameters similar to its Universe, though slightly mutated. By this, it is meant that the fundamental parameters in the parent Universe are $p=(h,e,c,G,\ldots)$ and in the child Universe $p\prime=(h+\delta h,e+\delta e,c+\delta c,G+\delta G,\ldots)$.
Consequently, stable Universes which permit black holes survive and proliferate, whilst those that don't perish. The population of Universes ought to be dominated by those that have the best conditions for black holes.
In string theory, there is the so-called string landscape of $\sim10^{500}$ vacua, which preserve or break different symmetries, and result in different laws of nature. I'm not sure an analogy with evolution is appropriate, so I'll leave it to an expert.
Also, I'm sure that there physical systems that exhibit emergent complexity (rather than evolutionary features per se). For example, complicated features of special relativity, like time dilations, contractions, planes of simultaneity etc, emerge from two simple assumptions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Eigenfunctions in a harmonic oscillator This assignment is about the one dimensional harmonic oscillator (HO).
The hamiltonian is just as you know from the HO, same goes for the energies, but I get that the wavefunction of the particle, at $t = 0$, is given by:
$\psi \left( x,\,\,t=0 \right)=A\left( {{\phi }_{0}}\left( x \right)+2i{{\phi }_{2}}\left( x \right) \right)$
Now I have to calculate A, so that $\psi$ is normalized.
My question is, what is ${{\phi }_{0}}\left( x \right)$ and ${{\phi }_{2}}\left( x \right)$ given by ?
Otherwise I guess is just doing the usual:
$\int_{-\infty }^{\infty }{{{\left| \Psi \left( x,\,\,t \right) \right|}^{2}}}dx=1$
To figure out A.
| The explicit eigenfunctions of the Harmonic oscillator hamiltonian are given here, but I would highly discourage you from explicitly doing an integral using these expressions to determine $A$. It is significantly easier to use the fact that the eigenfunctions are orthogonal;
$$
\int_{-\infty}^\infty dx\,\phi_m^*(x)\phi_n(x) = \delta_{mn}
$$
If you use this fact, then the integral on the left hand side of the $t=0$ normalization condition you wrote down will be very easy. Try this out, and if you still have trouble we can give you more guidance since this is a homework question.
| {
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How do objects change their axis of rotation? If I hold a pencil at its end and spin it, throwing it upwards, it will spin about its end, but will soon start spinning around its center. How is this?
I would draw the following torque diagram for while it's in the air:
*
*Object: uniform thin rod with length $\ell$ and moments $I_{center}=\frac 1 {12} m\ell^2$ and $I_{end}=\frac 1 3 m\ell^2$)
*Center of rotation some small distance $d$ from the end
*Torque $m\vec g$ downward, at center of mass, with $\theta = 90°$ and $r = \frac \ell 2 - d$
*Possibly wind resistance $\vec D$ upward, at center of mass, with $\theta = -90°$ and $r = \frac \ell 2 - d$
Thus, $\vec \tau = \sum {\vec r \times \vec F} = \sum {rF~sin\theta} = \left (\frac \ell 2 - d\right)(m\vec g - \vec D)$. I could see how this might cause it to spin, but how does the center of rotation to move?
EDIT: Here's a picture because apparently I wasn't clear. It definitely is spinning around the end (or close to it, anyway) before I release the pencil.
| Not sure if I am interpreting your description of the problem correctly, but if I take the initial conditions (using nice round numbers) as a rod of $r = 1\:\mathrm{m}$ length, pivoting about one end at at $\omega = 1\:\mathrm{rad/s}$ at $t=0$, I can decompose the motion as a COM motion of $v=r\,\omega/2$ with a spin about the COM of $\omega$. If I plot the motion (ignoring gravity), I get:
where $t=0$ is the horizontal line at the bottom and the "pencil" is moving upwards after release. While the initial conditions are set by the spin around the end, the following motion is still mechanics about the COM.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
} |
North and south of magnetic field
The current I is flowing upward in the wire in this figure. The direction of the magnetic filed due to the current can be determined by the right hand rule.
Can we determine the north and the south of the magnetic field produced by the current I by using a hand rule?
| The concept of magnetic poles only makes sense with respect to permanent magnets/dipoles/solenoids, where the field lines point from one end of the object to the other. In the case of the magnetic field produced by a current, the field lines are in circular alignment around the wire. Hence, there are no endpoints.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do you determine the heat transfer from a $p$-$V$ diagram? I doubt this question has been addressed properly before, but if there are similar answers, do direct them to me.
I am currently studying the First Law of Thermodynamics, which includes the p-V diagram and of course, $\Delta U = Q_{to} + W_{on}$.
My question is: how do I determine the heat supplied to the system from a given $p$-$V$ diagram? I know $W_{on} = -p(V_2-V_1)$, but is there a way to find $Q_{to}$ for a given path?
| One way is to find out the internal change energy of the system and infer the heat transfer to the system from that and the work done: $$\delta Q_\text{to}=dU-\delta W_\text{on}.$$
If you have a handle on the system's entropy, on the other hand, then you can use the Gibbs relation, $$\delta Q_\text{to}=TdS,$$to find the heat delivered.
In general, though, any method is going to require more information about the system, because for the same process (and thus the same work performed) systems with different entropies $S=S(U,V)$ (or equivalently with different energy functions, $U=U(S,V)$) will heat up by different amounts.
| {
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"timestamp": "2023-03-29T00:00:00",
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Angular momentum conservation while internal frictional torque is present So this appears in a problem which looks simple enough in its context; It's something like this:
Two discs, A and B, are mounted coaxially on a vertical axle. The discs have moments of inertia $I$ and $2I$ respectively about the common axis. Disc A is imparted an initial angular velocity $2\omega$ using the entire potential energy of a spring compressed by a distance $x_1$. Disc B is imparted an angular velocity $\omega$ by the same spring compressed by a distance $x_2$. Both the discs rotate in the clockwise direction.
When disc B was brought in contact with A, they acquire a common angular velocity in time $t$. The average frictional torque by the other during this period is: ?
The answer ($2I\omega/3t$) was obtained by applying angular momentum conservation and is exactly what confused me. How can we apply angular momentum conservation when friction is present?
| The angular momentum of each disk individually is not conserved, however the total angular momentum of both disks is conserved because there are no external torques acting.
Start by calculating the total angular momentum of both disks (I'm going to replace "w" by "v" since "w" is confusingly close to "$\omega$"):
$$\begin{align}
L_{total} &= I_a \omega_a + I_b \omega_b \\
&= I.2v + 2I.v \\
&= 4Iv
\end{align}$$
Now we bring the disks into contact and they settle down to a constant speed $\omega_{final}$. The total angular momentum is now:
$$\begin{align}
L_{total} &= I_a \omega_{final} + I_b \omega_{final} \\
&= 3I\omega_{final}
\end{align}$$
Because angular momentum is conserved we just equate our two expressions fo $L_{total}$:
$$ 4Iv = 3I\omega_{final} $$
so:
$$ \omega_{final} = \frac{4}{3}v $$
Now, torque $\times$ time is the angular impulse, and we know that the impulse is equal to the change of momentum. So if we calculate the change in momentum of disk A this is equal to the torque times the time i.e. $Tt$, and we know the initial angular momentum of disk A is $2Iv$ so:
$$\begin{align}
Tt &= I_a 2v - I_a \omega_{final} \\
&= 2Iv - \frac{4}{3}Iv \\
&= \frac{2}{3}Iv
\end{align}$$
And dividing both sides by $t$ gives the answer:
$$ T = \frac{2}{3} \frac{Iv}{t} $$
| {
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