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About virtual displacement
Thornton Marion
The varied path represented by $\delta y$ can be thought of physically as a virtual displacement from the actual path consistent with all the forces and constraints (see Figure above).
The varied path $\delta y$, in fact, need not even correspond to a possible path of motion
Doesn't the second quote contradict the first. The first says the virtual path is a possible path, the second says it need not be?
| Tl;DR: A virtual displacement is by definition frozen in time and hence never corresponds to an actual path of motion.
The other hallmark of a virtual displacement is that it obeys the constraints. For more information, see e.g. this, this, this & this related Phys.SE post.
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Proton Electron Merger Can somebody explain what would happen if an electron & a proton, very close to each other are left to "fall" to each other in a straight line?
| If they are simply falling directly towards each other, they can't combine. To combine, they would need to form a neutron, but a neutron has slightly more mass. The extra mass would have to come from another particle, or source of energy - for example smashing them together forcefully enough.
So as they can't combine, they would remain as a proton and electron. They would be attracted to each other because of having an opposite charge, but when they got "too close", the nuclear interactions would become dominant (more powerful) and cause them to repel each other.
Another way of looking at the energy needed to merge, is its the energy needed to overcome that repulsion when they get very close.
So they would end up close but not too close. Attracted electrically, but unable to get very much closer or merge.
So it would remain as a hydrogen atom - a proton with a single bound electron.
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True or False: energy is conserved in all collisions Using introductory physics, how would you answer this question? (I have a disagreement with my instructor and I’m curious to hear your input)
One of us says true because the question doesn’t specify “kinetic energy,” or a “system” and all energy is always conserved. The other says false because “only perfectly elastic collisions conserve energy. Otherwise energy will be lost to sound or light”
What’s your opinion?
| Yes it is conserved, and this does not depend on the course level. The tricky part it can still be transferred to parts of the system that are not the colliding objects.
Energy can escape the system of colliding particles via sound or heat, converted to mechanical work resulting in a permanent deformation of the colliding particles, be converted to internal energy if the bodies are not strictly rigid, etc.
Sooooo… it’s a matter of what is included (or allowed) in your definition of energy, and what you are allowed to ignore as forms of energy. Every possible form of it or limited to one or more specific form of energy, like kinetic energy?
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What is the difference between circular motion and rotation? I've tried so hard to understand the difference, yet no progress. There is a lot of answers here, on Quora, on Youtube,... but everyone give a different answer.
So can you please give a simple yet satisfactory answer?
Someone says that rotation is only about an axis that oass through the center of mass, other say that the axis can be anywhere inside the body but outside no because if it's outside it will be circular motion, but then if you search Wikipedia about Parallel Axis Theorem, they'll say : If the body "ROTATES" about an axis outside of it, you can use the Parallel Axis Theorem to...
So who's right?
And one more question : In circular motion, the kinetic energy formula for a body is $\frac{1}{2} MV^2$ or $\frac{1}{2} Iω^2$ (like in rotation)? I mean can we use the equation $x=\frac{1}{2} at^2 + Vt + X$ or $θ=\frac{1}{2}θ"t^2 + θ't + Θ$ (like in rotation)?
So many questions yet no one gives me a good answer, I hope that someone can here.
And what about this picture here, is it rotation? "The disc (D) can oscillate freely around a horizontal axis (A), perpendicular to its plane and passing through a point O of its periphery."
https://i.stack.imgur.com/iBodB.jpg
https://i.stack.imgur.com/AJMhI.jpg
| This is semantics, not physics. Whether you call it “rotation,” “circular motion,” “circular rotation,” “circulation,” “circumambulation,” “rotary motion,” or any other permutation/synonym, the key is to clearly define what you mean. That way, everyone can be clear about the physics, regardless of their semantic opinions.
I recommend you focus on understanding the physics. You’ll get tripped up if you rely on memorizing definitions which are defined with vague terminology.
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Addition of velocities vs. Addition of forces Imagine two strings tied to a box.
Case 1: Two strings are pulled with the same $u$ velocity. The box will also move with velocity $u$.
Case 2 : Tension along $\text{a}$ string is $T$. Therefore total force acting on the box is $T+T=2T$. (Box is accelerating)
I think my problem is obvious. Both velocity and force are vectors. But why we can not get the velocity of the box in the first case as $u+u=2u$? (This is obviously wrong, but why?)
| You are walking down the street with your friend. Now you hold hands. Are you now moving twice as fast?
If addition of velocity worked like that, you would get total nonsense. Whenever two moving objects would stick to each other, they would move faster.
Another, more mathematical way of looking at it. The energy of an object of mass $m$ moving at velocity $v$ is $$E = \frac12 mv^2.$$ Now mentally imagine dividing this object in two objects of masses $m_1$ and $m_2$ (like person = body + head). We obviously have $$m = m_1+m_2.$$ Now imagine these two parts are moving with velocities $v_1$ and $v_2$ respectively. But since mentally dividing an object cannot change its energy, we must have $$\frac12 mv^2 = \frac12\left(m_1v_1^2 + m_2v_2^2\right).$$ Solving this will lead you to $$v=v_1=v_2$$
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Doubt concerning biot savarts law Why do we calculate $dB$ for an infinitesimal part of a wire instead of a point?What is the reason behind that?Why can't we determine the magnetic field of a point with respect to a point of the wire instead of an infinitesimal part?
Also why is $dB$ proportional to $dl$ where $dl$ is the infinitesimal length of the wire?An infinitesimal quantity doesn't have any definite value,so how can we increase or decrease $dl$ for $dB$ to be proportional to it?Please give an intuitive explanation since it's really bothering me.
| As already mentioned by @Lili FN in the comments, "points" of a wire do not give rise to a magnetic field. But a small length of wire does.
This makes sense intuitively since it's actually the current (density) $\vec{j}(\vec{r})$ that gives rise to a magnetic field $\vec{B}$ (Ampere's Law).
And in the $1d$ case (infinitely thin wire), integrating over the direction of the current, or integrating over small bits of the wire $\mathrm{d}\vec{l}$ is equivalent.
Looking at the whole equation makes it easier to see why "$\mathrm{d}{B}$ is proportional to $\mathrm{d}{l}$". In the Biot-Savart law you're really integrating over the current at a given position $I(\vec{r})$. The $\frac{\mathrm{d}\vec{l} \times \vec{r}}{|r|^3}$ bit of the integral is just giving you a direction for the magnetic field $d\vec{B}$, while it's amplitude is dictated by $I(\vec{r})$. Just as Ampere's law hints at.
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Why is the gravitational potential energy lost not subtracted from the required work done in the given problem?
An elastic string of natural length $l \;\text{m}$ is suspended from a fixed point $O$. When a mass of $M \;\text{kg}$ is attached to the other end of the string, its extension is $\frac {l}{10} \;\text{m}$. Some work is done to produce an additional extension of $\frac{l}{10} \;\text{m}$. Show that the work done in producing this additional extension is $\frac{3Mgl}{20} \;\text{J}$.
My Attempt. I tried to apply the work-energy principle which says that the change in total energy of an object equals the work done on it. So, the required work done should be the elastic potential energy (EPE) gained minus the gravitational potential energy (GPE) lost, which gives unmatched $\frac{Mgl}{20} \;\text{J}$. Later, I found out that if I simply ignore the GPE I will get the desired answer. But why the GPE can be ignored? Isn't the additional GPE loss got stored in the EPE?
Comment. It is a high school mechanics problem, so please do not over-complicate things. Thank you in advance.
| You are 100 percent correct when you say that additional GPE loss got stored in EPE and actually the question requires you to only calculate this change i.e. from initial $l/10$ extension to additional $l/10$ extension.
Just consider the EPE initially for an extension of $l/10$. This will give you a value $Mgl/20 \text{J}$.
Now consider further extension of $l/10$ (so total $l/10 + l/10 =l/5$).
With total extension of $l/5$, now the final EPE will come out to be $Mgl/5 \text{J}$.
Finally the change between the final EPE and the initial EPE ($Mgl/5-Mgl/20$)$\text{J}$ which will give you $3Mgl/20 \text{J}$ as the desired answer. We are not ignoring the GPE. It is already taken into account when we calculate EPEs for both the final as well as the initial string extensions.
Hope it clarifies.
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Why doesn't charge escape from capacitor? A charged spherical capacitor kept in air do not loose charge because air is a bad conductor and increase in charge results in Corona Discharge. Is it because the nucleus of air molecule repels the charge in sphere and after a limit the repulsion is less than the attraction by sphere leading to Corona Discharge. But if the same charged spherical capacitor is kept in vacuum, what would happen. Shouldn't charge repel each other and escape the spherical surface because now there is no nucleus of air molecule to repel those charges. Will the capacitance decrease or not.
| It does escape - via a quantum tunnelling effect.
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Does gravity get stronger the higher up you are on a mountain? So I saw this article stating that gravity is stronger on the top on the mountain due to there being more mass under you however I have read some questions other people have asked and most of the responses state that the mass is concentrated at the middle of the earth meaning gravity doesn't get stronger the higher up you go. I would like to know which one of these it is as the article is a pretty reliable source. Here is the link to the article https://nasaviz.gsfc.nasa.gov/11234
| If you are a satellite 6870 km above the center of the Earth, and directly below you there is flat land, you will experience some gravity. If you move on to another point, also 6870 km above the center, but this time there is a huge mountain below you, then this time you will feel a gravity which is slightly larger.
If you are a person standing on the surface of the Earth, 6370 km above its center, you will feel some gravity. If from there, you climb a 4 km high mountain, after that you will be 6374 km above Earth's center. Because your distance from most of Earth increases, the gravity you feel is slightly smaller.
(In both examples, the latitude should be the same before and after, otherwise the oblateness of Earth's shape (and in the second example also the related centrifugal force due to Earth's rotation) will influence the result.)
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In 2D space, how to calculate the direction to hit a moving projectile from a position? Imagine a 2D space. A is where our missile is, and B is where our target is currently moving with a velocity of $v_{2}$. B will come close to A in a certain time and then move away like a comet to earth. Our missile can travel at a speed of $s_{1}$, where the target speed is $|v_{2}|$. The magnitude and direction of $v_{2}$ won't change. Similarly, for our missile, whose speed is constant, we need to calculate $v_{1}$ such that once the missile is launched, it will hit the moving target.
So we have $A$, $B$, $v_{2}$ and $s_{1}$. How can one find $v_{1}$ to hit the target?
The missile and target move with the equation
$$A = A + v_{1}\cdot\mathrm dt$$
$$B = B + v_{2}\cdot\mathrm dt$$
How do I calculate $v_{1}$?
| If I understood the question correctly, there is a point $\vec A_0$, from which a mass point is launched at time $t_0$ at some constant velocity $\vec v_1$ which should hit another mass point launched from $\vec B_0$ at the same time with velocity $\vec v_2$ and $v_1 := |\vec v_1|$ is known, but not the direction.
The trajectories can be parameterised as
$$
\vec A(t) = \vec A_0 + v_1t \begin{pmatrix} \cos(\theta) \\ \sin(\theta)\end{pmatrix}~, \qquad \vec B(t) = \vec B_0 + \vec v_2t~,
$$
then the problem is reduced to finding $t$ and $\theta$, such that $\vec A(t) = \vec B(t)$. Plugging in the parameterisation yields
$$
\vec A_0 + v_1t \begin{pmatrix} \cos(\theta) \\ \sin(\theta)\end{pmatrix} = \vec B_0 + \vec v_2t \qquad \Leftrightarrow \qquad \vec A_0 - \vec B_0 = \vec v_2 t - v_1 t \begin{pmatrix} \cos(\theta) \\ \sin(\theta)\end{pmatrix}
$$
$$
\Leftrightarrow A_{0x} - B_{0x} = v_{2x} t - v_1 t \cos(\theta)~,~ A_{0y} - B_{0y} = v_{2y}t - v_1t\sin(\theta)~.
$$
$$
\Rightarrow \frac{A_{0x} - B_{0x}}{v_{2x} - v_1\cos(\theta)} = t~.
$$
$$
\Rightarrow A_{0y} - B_{0y} = v_{2y} \frac{A_{0x} - B_{0x}}{v_{2x} - v_1\cos(\theta)} - v_1 \frac{A_{0x} - B_{0x}}{v_{2x} - v_1\cos(\theta)} \sin(\theta)~.
$$
$$
\Rightarrow (A_{0y} - B_{0y}) (v_{2x} - v_1 \cos(\theta)) = (A_{0x} - B_{0x}) (v_{2y} - v_1 \sin(\theta))~.
$$
This equation is a little nasty, and to my knowledge not solvable directly for $\theta$. However, $\theta$ can only be in $[0,2\pi)$, so it should be fairly easy to solve this numerically for given $\vec A_0,\vec B_0,v_1,\vec v_2$. As soon as you have a solution for $\theta$,
$$
\vec v_1 = v_1 t\begin{pmatrix} \cos(\theta) \\ \sin(\theta)\end{pmatrix}
$$
should give you what you called $v_1$.
Remark: Sorry for not using the symbols you suggested, but I needed to distinguish between $\vec v_1$ and $v_1 := |\vec v_1|$ as well as between $\vec A(t),\vec B(t)$ and $\vec A_0,\vec B_0$ for my calculation.
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Why does water contract on melting whereas gold, lead, etc. expand on melting? My book mentions that water contracts on melting, but the book doesn't give any reason why it does so. It is mentioned that:
$1\,\mathrm g$ of ice of volume $1.091\,\mathrm{cm}^3$ at $0^\circ\mathrm C$ contracts on melting to become $1\,\mathrm g$ of water of volume $1\,\mathrm{cm}^3$ at $0^\circ\mathrm C$.
I searched on the internet but I failed to find any useful insight. Could someone please explain why water contracts on melting?
| The reason is attributed to the hydrogen bonds that determine the structure of ice. The molecules of water in ice are arranged in a cage-like fashion, with rather hollow spaces in between them, thanks to the hydrogen bonds shown as dotted lines:
When ice melts, the energy supplied helps in breaking these hydrogen bonds, and the molecules of water come closer, hence increasing the density of water, or 'contracting it'. This process goes on till the water reaches about 4 degrees Celsius, at which water has the highest density. After this, water expands like any other liquid.
This property has an interesting, but equally important application in nature. As ice is less dense than water, it floats up to the surface of the water bodies and acts as a thermal insulator to the underwater world. This is a boon to aquatic animals and fish during harsh winters.
It also explains other properties of water like the unusually high latent heat of fusion and specific heat.
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Why does twisting a cork make it easier to remove from a bottle? When we want to remove a cork from a bottle first we turn the cork. Turning in one direction makes it easier to remove in the axial direction.
Does anyone know something more about this?
| Let me take a shot:
The cork has made chemical bonds with the glass. These are the same for all dS of the cork surface: the difference is that in rotating the cork because of the circular motion, a small d(theta) brings the surface unstuck, the resistive forces will not add( different directions). For the axial direction the surface is continuous and the forces needed are additive. Once it becomes unstuck then axial force is effective, because it takes time for the bonds to form, between cork and bottle.
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How strong is the force of ice expanding when freezing? Why does water contract on melting whereas gold, lead, etc. expand on melting? reminded me about something I've been wondering myself for some time.
We know that water expands as it freezes. The force is quite formidable - it can cause solid steel pipes to rupture. But nothing is limitless. If we created a huge ball of steel and placed a small amount of water inside it (in a small, closed cavity) and then froze it - I don't think the big ball would rupture.
But what would we get? Compressed ice? Can this even be done? Can you compress ice? Or would the water simply never freeze? Or freeze only partially? What if we kept cooling it, down to absolute zero (or as close as we can get)?
What happens when water should expand, but there is no room for it to do so, and the container is too strong to be deformed?
| The other answers are excellent in explaining the theory.
There is a good Youtube video "Can You Stop Water From Expanding When It Freezes Into Ice" showing a practical experiment trying to contain freezing ice in a steel pipe, and then explaining the pressures required to contain freezing ice.
I don't have the reputation to comment, which is why I wrote this as an answer. Apologies if this was wrong.
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What's the difference between a perfect fluid and an ideal gas? This is how I understand it at the moment:
*
*A perfect fluid is a collection of non-interacting particles, which are as a whole characterised by energy and pressure.
*An ideal gas is also a collection of non-interacting particles, but here the ideal gas law holds. There, we have pressure, volume and temperature (let's assume a fixed number of particles for both cases), s.t. by applying the ideal gas law, again two parameters remain.
Furthermore, the stress-energy tensor of a perfect fluid can be seen e.g. here on Wikipedia, but I haven't found the stress-energy tensor of an ideal gas.
| Contrarily to the existing answers, and to the citation in one of them, the terminology is not uniform. According to the answer to a previous similar question and to Landau&Lifshitz textbook on Mechanics of Fluids, a perfect fluid is a fluid described by the Euler equation, continuity equation, without viscosity and thermal conduction. It may be compressible. Basically, according to this point of view, the perfect character of a fluid pertains to its dynamics.
The ideal gas was introduced as a prototype system in thermodynamics, and statistical mechanics of equilibrium. Therefore rheological properties are not directly related to this model, although in some cases (for instance within the kinetic theory of gases) some model for the transport coefficients is added to the ideal gas approximation.
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Why can't the speed of gravitational waves be greater than the speed of light if the universe can expand faster than the speed of light? Since the expansion speed of the universe can be greater than the speed of light, why can't gravitational waves, which also uses space as the medium, travel faster than the speed of light?
| Gravitational waves are solutions to the linearized field equations
$\Box h_{\mu \nu} = 0,$
where $\Box$ is the d'Alembertian differential operator.
They are of the form
\begin{align}
h_{\mu \nu}= \text{Re}[H_{\mu \nu} e^{i k_{\rho} x^{\rho}}] ,
\end{align}
where $H_{\mu \nu}$ is just a complex symmetric matrix setting the polarization of the wave and $k_{\rho}$ is its wave-vector.
For the above ansatz to satisfy the equations, we plug it into the equations to get the condition $k_{\mu} k^{\mu} =0$. This is just the statement that the wave-vector must be null, meaning the wave propagates at the speed of light.
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Wigner classification of particles vs generic Hamiltonian spectrum Wigner tells us we should associate infinite dimensional unitary irreps of the Poincaré group with particle states. His classification using eigenvectors of the spacetime generators $P^\mu$ and the method of little groups tells us particles are asssociated with two numbers $m^2$ and spin $S$.
However if we look at a generic spectrum of the Hamiltonian for an interacting theory we get something along the lines of
This includes single particle states, bound states, and a continuum of multi-particle and quasi-particle states.
These extra states are parameterised by other discrete and continuous labels. My question is do these extra states form irreps of the Poincaré group? If not, why not? To me, it seems as though his treatment only looks at free theory Hamiltonians, i.e. only looking at the single particle states.
| Wigner's classification looks at the quantum numbers that are associated to the Poincaré group. It says nothing about other quantum numbers.
Consider a free theory with a two particle state $|\boldsymbol p_1,\boldsymbol p_2\rangle$. Wigner will tell you that the quantum number associated to the generator of translations, $P^\mu$, is $p^\mu_1+p_2^\mu$, the center-of-mass energy. Wigner will tell you nothing about the other quantum number, $p_1^\mu-p_2^\mu$, the relative momentum. The total momentum is a quantum number associated to a generator of Poincaré; the relative momentum is not. Wigner tells you about the former, not the latter. The total momentum is associated to a spacetime symmetry; the relative momentum is not.
Wigner's classification is nothing but a statement about representation theory: the states of the theory can be organized into representations of a complete set of commuting observables; the Poincaré group gives you a universal set of such operators, but there is no claim that it is complete. In order to fully label the states of the theory, you also need to diagonalize other operators, which in general do not come from spacetime symmetries (for example, you also often find quantum numbers associated to conserved charges for internal symmetries, such as electric charge; these are again not in the Poincaré group).
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Path independence and Spherically Symmetric Force
This problem is from John Taylor's Classical Mechanics.
I can't figure out how to prove that a series of paths consisting of paths moving radially or in the angular direction. I understand intuitively that it works, but I'm not sure if I know to prove that the series of paths converges to an arbitrary path. My first thought was to segment an arbitrary path $R=R(t)$ from $A=R(0)$ to $B=R(1)$ into $N$ pieces and construct a "spherical" sequence of paths $P=\{AP_2=P_1P_2,\dots,P_{M-1}P_M=P_{M-1}B\}$ such that each path is either stricly radial, or strictly angular (in the $\varphi$ or $\theta$ directions) and each point $R(\frac{i}{N})$ is intersected by some path $P_jP_{(j+1)}$ in $P$. Then using the definition of spherical symmetry,
$$
\sum_{i=1}^{M-1}\int_{i/M}^{(i+1)/M}f(r)\cdot d\mathbf{r}=\sum_{i=1}^{M-1}W_{P_iP_{(i+1)}}\\
=W_{AP_2}+\cdots+W_{P_{M-1}B}
$$
where $W_{P_iP_{i+1}}=0$ if the path $P_iP_{i+1}$ is a movement in an angular direction.
How can I show that this is a reasonable approximation?
| If the path went from A to B in a series of small steps like this
If we focus on one of the steps, the real path in blue can be approximated as a straight line
the work done, against $F$, going from P to R, is $PR\times Fcos\theta$ and this is the same as $F\times PQ$. Since it's true for one of the small steps it'll be true for the whole path
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Why do little chips break off so easily from strong neodymium magnets? I have some strong toy neodymium magnets. Typically after a while little chips start breaking off, unlike from most other small metal objects, like in this image.
It could of course be that neodymium is more brittle than metals used for other objects, or that they often hit each other much harder than in a fall due to their magnetism, or that they are just low-quality, but I was wondering if it could have to do with internal tensions that are not present in non-magnetic objects, maybe due to adjacent domains of different magnetization?
Does anyone know what could cause this?
| Magnetizing a chunk of magnet material induces stresses in the material which are relieved when the material cracks apart. When you combine this with the extreme brittleness of the magnet material, it is natural for the magnet to shed flakes and chips of itself.
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Why ground water is cold during summer and hot during winter? I've heard somewhere that during summer the temperature of ground water is lower than that on the Earth while during winter the temperature of ground water is higher.
How is this possible ?
| Soil is a bad conductor of heat. So, the layers of the top surface of earth act as insulators.
During summer, they insulate the underground water and soil from the heat radiation from the sun, and from the general hotter atmosphere, hence underground water is colder.
In winter, they again act as insulating layer and prevent the heat from underground to escape into the cold outside air. Thus, underground water can be warmer than outside.
One noteworthy point is that the underground water temperature remains fairly constant throughout the year, (for a given geographical region), because the water table is quite below the ground, and hence it is insulated from the outside atmosphere.
It feels colder during summer, because outside water is hotter and it feels warmer during winter because outside water is colder .
| {
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Two point measurement statistics in Quantum systems I am reading a paper related to fluctuations in Quantum thermodynamics. I am unable to understand the math behind equation no. 10 where the probability density function for work distribution is calculated. I can't understand how delta function is arising in the equation.
The same is also done in another article. Can someone help me in understanding how is PDF constructed?
Eqn 10 looks like this(taken from article):
$$
P(Q, t)=\sum_{n, m} \delta\left[Q-\left(E_{m}-E_{n}\right)\right] P_{n \rightarrow m}^{\tau} P_{n}^{0}
$$
where we measure the system energy at the beginning of the process to find it in some state $|n\rangle$ with associated probability $P_{n}^{0}$ and again after some time $\tau=\gamma t$, which is rescaled with respect to the damping constant $\gamma$, to find it in a state $|m\rangle .$ The probability to move from the initial state $|n\rangle$ to the final state $|m\rangle$ is given by $P_{n \rightarrow m}^{\tau} .$
| The Delta function is constraining energy to be conserved.
The sum is over all transitions $n\rightarrow m$. However, the term $\delta[Q - (E_m - E_n)]$ is picking out only the terms where the heat generated ($Q$) is equal to the energy lost in the transition ($E_m - E_n$).
| {
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How to deduce existence of photons by applying quantum mechanics to Maxwell's equations Watching this video lecture of Peter Higgs https://youtu.be/QtudlGHoBQ8?t=372, he says (roughly) at one point that Paul Dirac applied quantum mechanics to Maxwell's field equations and deduced the existence of photons (previously predicted by Einstein in the photoelectric effect?).
Question. Mathematically, how does one obtain the existence of photons (or photon-like particles) by applying quantum mechanics to Maxwell's equations of electromagnetism?
Disclaimer. I'm a mathematician by training, with pedestrian knowledge in Physics.
| Historically, quantum mechanics was first postulated by Planck as a way to solve the ultraviolet catastrophe that arose when trying to apply thermodynamics to electromagnetism. So, it's not so much that we have quantum mechanics and applied it to E&M, we had E&M and statistical mechanics/thermodynamic and the mismatch between them gave the first hint at QM.
That said, I'll outline a plausible path from Maxwell's equations to being able to deduce in quantum field theory that photons exist as particles.
First, derive the Hamiltonian from the electromagnetic stress-energy tensor,
$$H = \int \mathrm{d}^3x \left[\frac{\epsilon_0}{2}\mathbf{E}^2 + \frac{1}{2\mu_0} \mathbf{B}^2\right].$$
Second, write the Hamiltonian in terms of the vector potential and electric potential using $\mathbf{E} = -\nabla \Phi - \partial_t \mathbf{A}$ and $\mathbf{B} = \nabla \times \mathbf{A}$ to get
$$H = \int \mathrm{d}^3x \left[\frac{\epsilon_0}{2}(\nabla \Phi + \partial_t \mathbf{A})^2 + \frac{1}{2\mu_0} (\nabla \times \mathbf{A})^2\right].$$
Third, split the Hamiltonian's terms into contributions from the electric field's solenoidal and divergencful parts using Helmholtz decomposition
$$H = \int \mathrm{d}^3x \left[\frac{\epsilon_0}{2}(\nabla \Phi + \partial_t \mathbf{A}_D)^2 + \frac{\epsilon_0}{2}(\partial_t \mathbf{A}_S)^2 + \frac{1}{2\mu_0} (\nabla \times \mathbf{A}_S)^2\right].$$
Identify the momentum canonically conjugate to $\mathbf{A}_S$ is $\mathbf{\Pi}_S = \partial_t\mathbf{A}_S$, and $\mathbf{E}_D = -\nabla \Phi - \partial_t \mathbf{A}_D$ to get
$$H = \int \mathrm{d}^3x \left[\frac{\epsilon_0}{2}\mathbf{E}_D^2 + \frac{\epsilon_0}{2}\mathbf{\Pi}_S^2 + \frac{1}{2\mu_0} (\nabla \times \mathbf{A}_S)^2\right].$$
Finally, Fourier transform into mode-space (i.e. Fourier transform the spatial dimensions, not time)
\begin{align}
H &= \int \mathrm{d}^3k \left[\frac{\epsilon_0}{2}\mathbf{E}_D^* \cdot \mathbf{E}_D + \frac{\epsilon_0}{2}\mathbf{\Pi}_S^* \cdot \mathbf{\Pi}_S + \frac{1}{2\mu_0} |\mathbf{k} \times \mathbf{A}_S|^2\right] \\
&=\int \mathrm{d}^3k \left[\frac{\epsilon_0}{2}\mathbf{E}_D^*\cdot \mathbf{E}_D + \frac{\epsilon_0}{2}\mathbf{\Pi}_S^*\cdot \mathbf{\Pi}_S + \frac{k^2}{2\mu_0} \mathbf{A}_S^*\cdot \mathbf{A}_S\right].
\end{align}
By inspection you can now see that the electromagnetic field consists of two components that are continuum harmonic oscillators (the solenoidal terms) that obey $\omega = k / \sqrt{\mu_0\epsilon_0}$, and one component that is not (the divergenceful term). The harmonic oscillator type terms are what give rise to the states of definite particle number via the ladder operator formalism, and hence, photons.
The divergenceful (non-photon supporting) term deserves some commentary. In the Weyl gauge it's a kinetic term (because it's the square of a time derivative of a coordinate), in the Coulomb gauge it's a potential term (the square of a space derivative of a coordinate), and it's mixed in other gauges. Regardless, parameterizing it in terms of the electric field satisfies gauge invariance, and is not a problem quantum mechanically since it has no canonically conjugate counterpart in the Hamiltonian. That means the states of definite $\mathbf{E}_D$ are also eigenstates of the Hamilton.
Of course, this gets more complicated when you start including sources, $\rho$ and $\mathbf{J}$, but only a little so. For more details, I would recommend Chapter 8 of Weinberg's The Quantum Theory of Fields (Volume 1).
| {
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Missing complex conjugate in (1/2,1/2) representation of Lorentz group Ticciati QFT I've been working through some computations involving representations of the Lorentz group (now using the fantastic Ticciati QFT textbook).
After some work, Ticciati gives the following formula
$$D^{j_{1},j_{2}}(X_{i})=D^{j_{1}}(T_{i})\otimes \mathbf{I}_{2j_{2}+1}+\mathbf{I}_{2j_{1}+1}\otimes D^{j_{2}}(T_{i}),$$
where $X_{i}$ are the generators of the Lorentz group written in the mathematicians convention without the extra factor of i, and the $T_{i}$ are the $\mathfrak{su}(2)$ matrices.
I've computed $D^{0,1/2}(X_{k})$ and $D^{1/2,0}(X_{k})$, obtaining $-\frac{i}{2}\sigma_{k}$ in each case. This result agrees with what Ticciati gets in eq (6.7.9).
The issue arrives when I compute $D^{1/2,1/2}$. I know that I will have to perform a change of basis using the Clebsch-Gordon coefficients, however, I only end up with the correct matrix if I add a complex conjugate to the formula Ticciati gives: $$D^{j_{1},j_{2}}(X_{i})=D^{j_{1}}(T_{i})\otimes \mathbf{I}_{2j_{2}+1}+\mathbf{I}_{2j_{1}+1}\otimes (D^{j_{2}}(T_{i}))^{*}.$$
I found another post: Proof that $(1/2,1/2)$ Lorentz group representation is a 4-vector Which does the same, however, the author of this post has not explained why this complex conjugate appears.
I tried to derive the formula using the complexified Lorentz algebra $$A_{k}=\frac{1}{2}(X_{k}+iB_{k}), \qquad C_{k}=\frac{1}{2}(X_{k}-iB_{k}),$$ and then embedding reps into the product space $\mathbf{C}^{(2j_{1}+1)(2j_{2}+1)}$ by writing $$D^{j_{1},j_{2}}(A_{k})=D^{j_{1}}(T_{k})\otimes \mathbf{I}_{2j_{2}+1}$$ and $$D^{j_{1},j_{2}}(C_{k})=\mathbf{I}_{2j_{1}+1}\otimes D^{j_{2}}(T_{k}).$$ Unfortunately I still get the same issue! There must be something that I'm not understanding! Any help would be appreciated.
*Edit: I give here an explicit calculation for $X_{1}$ using the complex conjugate.
$D^{1/2,1/2}(X_{1})=\frac{-i}{2}\sigma_{1}\otimes \mathbf{I}_{2}+\mathbf{I}_{2}\otimes (\frac{-i}{2}\sigma_{1})^{*}=\begin{pmatrix}
0 & 0 & \frac{-i}{2} & 0 \\
0 & 0 & 0 & \frac{-i}{2} \\
\frac{-i}{2} & 0 & 0 & 0 \\
0 & \frac{-i}{2} & 0 & 0 \\
\end{pmatrix}+\begin{pmatrix} 0 & \frac{i}{2} & 0 & 0 \\
\frac{i}{2} & 0 & 0 & 0 \\
0 & 0 & 0 & \frac{i}{2} \\
0 & 0 & \frac{i}{2} & 0 \\
\end{pmatrix}$
Then the author in the post I linked above uses the following matrix to change basis (if someone could explain where this matrix is obtained that would be very helpful!):
$U=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 0 & 0 & 1 \\
0 & 1 & 1 & 0 \\
0 & i & -i & 0 \\
1 & 0 & 0 & -1 \\ \end{pmatrix}$
Then I get
$U^{-1}D^{1/2,1/2}(X_{1})U=\begin{pmatrix} 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 \\
0 & 0 & 1 & 0 \\ \end{pmatrix}=-iJ_{1}$
| Check the definition of $S$ in Eq. 6.7.1, you can observe first that it is defined using the complex generators for the Lorentz group, $T_r, \bar{T}_r$, which are given before Homework 6.3.9. (where the $X_i$ and $B_i$ are defined), and second that the complex elements $\bar{T}_r$ are sent to $\tau_r$ not $\bar{\tau}$ so the conjugation is included therein.
Then Eq. 6.7.6 tells you what the representation matrix of each generator is, so you should be able to verify that $D_{1/2,1/2}(X)= X$, for a generic element $X\in\mathfrak{so}(1,3)$. You need to use both rules, which you do not seem to be using.
| {
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What are the degrees of freedom of a dumbbell? Edit 1: May be I should modify my question after getting the answers. I see why $(X_c, Y_c, Z_c, \theta, \phi)$ are legitimate Dof's of the dumb-bell, I never had any problem with that.
Please consider now any rigid body in general. It has $6$ Dof's, two examples of which are mentioned below:
*
*$(X_c, Y_c, Z_c, \alpha, \beta, \gamma)$: $(X_c, Y_c, Z_c)$ are CoM co-ordinates, and $(\alpha, \beta, \gamma)$ are the Eulerian angles (or, the pitch, roll and the yaw) of the rigid body.
*Take a line through any point $(x_1, y_1, z_1)$ inside the rigid body, fix a line through the point using two angles $\theta, \phi$ (or fix a unit vector along the line, using two components). The last $Dof$ is obtained by specifying an angle around this line as axis, $\theta'$ that a point not on this line makes with, say the horizontal.
My question is more about the constraint relations. There are $3N$ co-ordinates to begin with, and each constraint relation eliminates one redundant co-ordinate. If this is true, because a rigid body has $6$ Dof's, after considering all the constraint relations (minus all the redundant constraints), I will have only $6$ (out of the set of $3N$ co-ordinates to start with). So can I have the Dof's to be any $6$ co-ordinates out of the initial $3N$, for example $(x_1, y_2, z_3, x_4, x_5, z_6)$ - is that sufficient to locate the rigid body?
If not, the constraint relations only lowers the number of co-ordinates, doesn't eliminate (remove) though. The reduced set of co-ordinates doesn't have to be any of those before reduction, it could be any, just one less in number.
| For simplification I will take the 2D space .
The constraint equation is:
$$ \left( x_{{2}}-x_{{1}} \right) ^{2}+ \left( y_{{2}}-y_{{1}} \right) ^
{2}-{L}^{2}
=0\tag 1$$
thus you obtain three generalized coordinates
solving equation (1) for $x_2$
$$x_2=x_1\pm\sqrt{L^2-(y_2-y_1)^2}$$
you get two solution for $x_2$ but this doesn't affected the equations of motion . to solve the equations of motion the initial conditions must fulfilled the constraint equation. you get this two initial conditions configurations one with $+x_2~$ and one with $-x_2$ , but notion is wrong with this two solution.
| {
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Amount of force required it to tip over a cone Say I have a cone of height $h$, radius $r$, and mass $m$.
How can I determine the amount of force required to tip it over (to have it fall completely to the other side), say exerted (horizontally) at the top of the cone? And in addition, how does the position at which I exert the force affect the amount of force I will need to tip it over?
Any advice on how I would aproach this problem is appreciated.
| You will need to start by locating the center of gravity (or mass). As the cone tips, that will move up along an arc centered on the tipping point, and the torques will decrease, and reach zero when the center of gravity is over the tipping point. For a minimum driving force, it should be applied at a right angle to the line joining the top and the tipping point. You also need enough friction to keep the cone from sliding along the surface.
| {
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Can nothing other than light have speed independent of source? Is it impossible for there to be some phenomenon that travels at a different speed than light to have speed independent of the source? Because if there were such a phenomenon there would be competing formulas for time dilation, correct?
To be clear, this is not a faster-than-light question; the phenomenon could have constant speed lower than light and still be problematic if I'm not mistaken.
| Fundamentally, the speed of light has less to do with light, and more to do with a universal speed limit, which is given by the value of $c$.
Provided an object is massless, it would move at the speed given by $c$, and would meet the criteria that its speed is independent of the motion of source or observer.
Anything with mass would not move at $c$, and nothing can move faster than $c$, and so we would conclude that $c$ would be the only speed something would move, such that this speed is a constant, independent of the motion of the source and observer.
| {
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Can a metal sheet roof be vibrated at audio frequencies? So I'm renting a cheap place for now due to certain circumstances. Problem is outside noise is excessive because there is no ceiling and the roof is only a metal sheet... Lately I play white noise to get some peace but it's not as effective as I want it to be... So I was wondering if I built my own speaker (come) to attach to the metal roof an play white noise, would the whole roof become some giant speaker emitting white noise in turn blocking outside noise since the roof is the main culprit for noise coming in, the walls are fairly thin.
I know this is a weird physics question but I need my peace to work on my projects, so I'd appreciate any information
| Yes, you could make a big surface like your metal roof radiate sound. If it's effective, or how effective, depends on the details such as density, thickness, total area and geometry.
Regarding the cancellation of external noise, I doubt it. Maybe you could add some layers, like a thick curtain, to add disipation for medium or high frequencies. For low frequencies there's not much you can do, maybe a sound trap with a Helmholtz resonator un s corner.
| {
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Do the quark types differ from each other in ways other than charge and mass? I've read things online here and there that seemed to hint that there's more to quark type than mass and charge. Is this true?
For clarity's sake, I'm not asking about properties individual quarks have other than mass and charge, such as spin and color charge. I'm asking about the properties of quark types.
| Yes, that is true, and well-known.
Quarks come in 6 different families, called flavors: up, down, top, bottom, charm, and strange. Each has a corresponding antiquark family, each of which has the opposite electric charge of the quark family.
Each quark in turn comes in three color charges: red, blue or green, which are complimented by the the anticolors (antired, antiblue or antigreen).
Quarks also have spin 1/2, which can point up or down.
All combinations are possible, and determine exactly how the quark interacts with other quarks: red antibottom spin up, antigreen strange spin down, etc.
| {
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Question about instant axis of rotation The given question is:
A man is rotating a stone of mass 10 kg tied at the end of a light rope in a circle of radius 1m. To
do this, he continuously moves his hand in a circle of radius 0.6 m. Assume, both circular motions
to be occurring in the same horizontal plane. What is the maximum speed with which he can throw
the stone, if he can exert a pull not exceeding 1250 N on the string?
Now I am lacking some conceptual clarity in this area, So could someone explain the logic behind this question so that I can also get the conceptual clarity I need in this area? The answer given is 10m/s.
| If the motions are in the same horizontal plane, then the 1250 N is the maximum horizontal centripetal force. Knowing this, you can find the corresponding tangential speed.
| {
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How do geodesics explain two identical balls thrown up at the different speeds? As stated in the title, two identical balls, both thrown directly upward, but at different speeds. The slower ball will reverse direction at a lower height than the faster ball. But the curvature of spacetime that they are passing through would be nearly identical.
How do geodesics explain these two paths?
Another way of looking at the same issue: Two identical balls dropped, but from different heights. Both balls travel straight down but hit the ground at different velocities. Each ball should be passing through the same spacetime curvature at the point of impact. Since neither ball experienced a force or acceleration and their motion is purely a product of the curvature of spacetime, why are they traveling a different speeds at impact?
| Two worldlines starting at the same event in spacetime but having different velocities are going in different “directions” in spacetime, even if they are going in the same direction in space. So their geodesics are different.
Remember, most worldlines here on Earth are nearly parallel, since they travel at very small fractions of c relative to each other. So the extent in the time domain is much larger than in the spatial domains.
For a sense of scale, consider a baseball pitch. It travels about 18 m in about 0.5 s, for a velocity around 37 m/s which is 1.2 e-7 c. During the flight it travels 18 m in space and 1.5 e8 m in time. This means there is a spacetime "angle" of about 1.2 E-7 radians between the earth and the baseball, which is "nearly parallel". Now, if we consider the height, a baseball goes up about 0.3 m during the flight. If we approximate that as a spacetime circular segment with a chord length of 1.5 e8 m and a height of 0.3 m, then we find a radius of curvature of 9.4 e15 m, which compares closely with c^2/g of 9.2 e15 m (the discrepancy being due to the approximation as a circle). So, because the baseball pitch is so long in time, the small difference in direction in spacetime can lead to the observed difference in the Newtonian path.
| {
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Electric field above a ring of charge To find the electric field at a point $p$ which is at a distance $h$ above the center of a ring of total charge $q$ with radius $r$, one can integrate the charge density over the circumference of the ring and get:
$$E = \frac{qh}{4\pi\epsilon_o(r^2+h^2)^{\frac{3}{2}}}$$
Another approach is to sum up the total charge on the circumference and multiply it by the distance between each point on the circumference and point $p$. The distance from $p$ to any point on the circumference is constant and is equal to:
$$\sqrt{r^2+h^2}$$
Since the horizontal components of the field cancel out, the field can be calculated as:
$$E = \frac{q}{4\pi\epsilon_o d^2} = \frac{q}{4\pi\epsilon_o\,(r^2+h^2)}$$
The two approaches yield different results, so the second must be wrong. But where?
|
Firstly the first method you have shown is correct and the second is wrong .
As you can clearly see in the diagram the force due to each elemental charged particle makes an angle $\theta$ with the axis so the net force is $F\cos\theta$ along the axis due to each particle and the net force due to 2 diametrically opposite particle as shown in the diagram is $2F\cos\theta$ along the axis. The value of $\cos \theta$ is obtained as $\frac{d}{\sqrt{R^2+d^2}}$. In your second derivation you have missed out the $\cos\theta$ term. Then$$F=\frac{Kdq}{R^2+d^2}$$
$$F \cos\theta = \frac{Kdq\cdot d}{(R^2+d^2)(\sqrt{(R^2+d^2)}}$$
$$ E = \frac{KQ\cdot d}{4\pi \epsilon_0(R^2+d^2)^\frac{3}{2}}$$
| {
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Why does a fluid follow the wing? When air is moving above and under a wing that is curved, why does the air at the top of the wing follow the wings shape and go downwards when it could just go in a straight line? It doesn't make sense to me.
| When a wing flows through air, if the air strikes the the top surface of the wing, and then moves off in a straight line, there would be lower pressure on the top surface of the wing. For this not to happen, the wind must flow around the surface of the wing.
You can also think about it as if the air above the wing has higher pressure than the air just on its surface due to the motion of the air, and so it gets pushed down onto the wing.
| {
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Are there any quantum mechanical systems where the position vector lives in a finite dimensional Hilbert Space? I don't have much to add to the title. Are there any quantum mechanical systems where the position vector lives in a finite dimensional Hilbert Space? If so, please provide example(s).
| We know that for any system, the position and momentum satisfy the commutation relation
$$[\hat{x},\hat{p}]=i\hbar $$
Take the trace of the both hand side
$$\text{Tr}(\hat{x}\hat{p}-\hat{p}\hat{x})=i\hbar\text{Tr} (I)$$
$$0=i\hbar \text{Tr}\ (I)\ \ !!! $$
It's clear that such a commutation relation can't be satisfied for finite-dimensional cases. Thus the operator these operators $\hat{x}$ and $\hat{p}$ live in infinite dimensional space.
| {
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Can we feel heat in outer space? Is there air outside of earth atmosphere? If not, could we feel heat coming from sun?
| We can most definitely feel heat in space.
As @aystack said, on earth heat transfer is through contact, convection or radiation. Convection relies on the movement of air or some other fluid, so that will only work inside a spacecraft or spacesuit. Transfer of heat through contact means that, if you touch something that is hot or cold, you will definitely feel it.
Radiative transfer of heat in space works in the same way in which you feel the heat of a roaring fire, or an electric bar heater. Photons travel from the source to us, enabling us to feel the heat. Photons travel through vacuum even easier than through air - if they didn't, we would not even be able to see the sun or stars.
It is because the photons travel from the sun through vacuum that we are able to feel the heat of the sun here on earth. Spacecraft also have to make sure that they do not overheat when the sun shines on them. With some spacecraft that means installing massive sun-shields. For example, the James Webb telescope, due for launch in November, uses a 5-layer, tennis court-size sunshield to keep cool. This shield is made of reflective material, so the sunlight is reflected away from the telescope.
The Parker Solar Probe, which flies to within about 6 million km of the sun (roughly a 10th of Mercury's distance), carries a massive carbon-composite shield that will reach temperatures of 1350C.
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Conservation of Angular Momentum when potential energy is function of position I have read that the angular momentum $(L)$ of a body is conserved if its potential energy is a function of solely its position vector. For example, the motion of planets on their orbits. I have two questions regarding this concept:
*
*How do we prove this?
*Is the converse also true?
| Angular momentum is conserved if your system is invariant under any arbitrary rotation.
To prove that angular momentum is conserved, note that if you calculate the force from the potential, the force itself will be parallel to the position vector.
\begin{equation}
\mathbf{F}(\mathbf{r}) = -\mathbf{\nabla} \Phi(r) =- \frac{d \Phi(r)}{dr} \frac{dr}{d \mathbf{r}},
\end{equation}
where only the chain-rule was used to find the derivative. The first term in the product does not depend on $\mathbf{r}$, only on $r$ (it depends only on the magnitude of $\mathbf{r}$). The second term is the radial unit vector $\mathbf{\hat{r}}$. Now note, that the force has a magnitude and a direction. The magnitude is contained in the first term, the direction is contained in the second term. If you calculate the torque that this field exerts on an object, you find that it is zero:
\begin{equation}
\mathbf{M} = \mathbf{r} \times \mathbf{F} = - \mathbf{r} \times \left(\frac{d \Phi(r)}{dr} \frac{dr}{d \mathbf{r}}\right) = -\frac{d \Phi(r)}{dr} \cdot \mathbf{r} \times\mathbf{\hat{r}} = 0
\end{equation}
because $\mathbf{r} \times \mathbf{\hat{r}}$ = 0 (property of cross product).
So $\Phi (r) \rightarrow \mathbf{M}=0$ thus angular momentum is conserved.
I think that there might be some special potentials for which the angular momentum is conserved but the potential itself does not depend only on the magnitude of the position vector. This is why I would say that the converse might not be true.
| {
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Will a stone released from an accelerated train retain the velocity the train had at the instant of release? My books says that since the 2nd law is a local relation, when a stone is released from an accelerated train, there is no horizontal force or acceleration on the stone, if air resistance is neglected. The stone only has vertical force of gravity. Will the stone retain the velocity (or at least a part of it), the train had at the instant the stone was released, due to inertia of motion?
| Yes, the stone's horizontal velocity (which will match that of the train's when it is thrown) will be conserved, so long as no horizontal force is acting on it.
If the stone is thrown up while the train is accelerating forwards, then an observer in the train will witness the stone appearing to accelerate backwards in the time that it is in the air.
| {
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How can spacetime be expanding faster than the speed of light? How can spacetime be expanding faster than the speed of light when the speed of light is the speed limit of the universe?
| The speed of light is the speed of causality. It is defined by the geometry of spacetime itself, and every observer will agree on its value (except for experimental issues that may arise when measuring it).
However, when trying to measure velocities between distant observers you must compare vectors (the four-velocity in this case) that live in very distant regions of spacetime, and there is no intrinsic way of comparing them. The way we usually assign meaning to this comparison is by defining a coordinate system that is centered around us and to compare the coordinate motion of objects with respect to our notion of velocities.
This is a rather intuitive way of doing these comparisons, both from a physical side and from a geometrical side, but it does allow objects to move "faster than light" with respect to this notion. However, it is important to remark that light emitted by them would be seen (by them) travelling at the exact same speed.
| {
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How to predict the photoelectric effect with modern quantum theory? In introduction class to quantum mechanics, the example of the photoelectric effect is often shown to the students to explain how the classical physics fails to explain it. We are told that one can solve the problem by only allowing the light to have discrete energies, proportional to the frequency of the light.
But somehow I can't see what is the connexion between these discrete energies of the photons and the rest of the stuff we learn (wave function, Schrödinger equation). Is it possible to predict the photoelectric effect using Schrödinger's equation on a wave function? If this is the case, I would be very happy to know how.
EDIT: If I understood it well, the weird thing in the photoelectric effect is that even at very high intensity (classically proportional to energy squared) the light can't overcome the work function if the light has a frequency which is too low. So my question is rather : How can one see that light is quantised and that the energy of a quanta depends on its frequency using quantum mechanics?
| The photoelectric effect indicates that either energy levels in matter or light is quantized. Willis Lamb and Marlan Scully authored a relevant paper in 1968, entitled "The Photoelectric Effect without Photons.".
There are several phenomenon, particularly spontaneous emission (which is used in lasers), that require quantization of the electromagnetic field in order to be described.
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What is the effect of normal reaction at round corners? Should we consider the case of collision at corners? or it is a kind of circular motion? $A B C D$ is a rhombus shaped tube with rounded corners (Fig). It lies in a vertical plane. A ball is made to roll from $A$ to $B$ to $C$. Next the same ball is made to roll from $A$ to $D$ to $C .$ In which of the two cases will it take less time?
| The answer is that the path ADC will be the quicker, because the ball undergoes the higher rate of acceleration first, and therefore completes the DC section at a relatively high speed compared with the speed at which it would have completed the corresponding AB section had it taken the other route.
There is no information in the question about the elasticity of the ball or the tube, so you should conclude that the question does not expect you to consider such complications.
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Deriving the equivalent capacitance in a series circuit formula When we derive the formula for the effective capacitance in series, we say:
$$Q/C_{eqv} = Q/C_1 + Q/C_2 + Q/C_3$$ (if there were 3 capacitors in this case). We would then cancel $Q$ to obtain the formula.
I understand why each capacitor has the same charge, but why does the effective capacitor have the same charge as each individual capacitor? I'd expect the effective capacitor to store a total charge of 3Q (in the given example), not Q?
When the capacitor discharges, would the overall amount of charge released not be 3Q (i.e. the overall charge of the capacitors)?
I saw a similar question on here, and it was answered by explaining that the 'inner capacitors' are isolated from the rest of the circuit, and the +Q and -Q charges cancel? But even so, the isolated charges can trigger electron flow from the 'outer capacitors' during discharge.
If anyone can clear up these doubts, I would be grateful.
| Let's imagine a series of three $0.6F$ capacitors, being charged by a $120V$ battery. From this website
Each capacitor will end up with $40V$ across it and a charge of $24C$, from $Q=CV$.
This can happen by charge (electrons) leaving one capacitor, e.g. the right plate of the left capacitor and ending up on the left plate of the second capacitor - (similarly for the other capacitor) - but only $24C$ has flowed through the battery.
(big numbers, but it's just an example)
The battery has charged the combination with $24C$ using $120V$ and so the effective capacitance must be $0.2F$.
Also if the combination were to discharge through a resistor, only $24C$ would flow through it. Charge cannot flow through a capacitor and the only charge flowing through the resistor would be due to electrons leaving the left plate of the left capacitor and a similar number in the wire moving onto the right plate of the right capacitor.
| {
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Issue about rotational and translational kinetic energy of a pendulum Let’s say we have a pendulum that consist of a light string hanging a disk-like object. It is allowed to undergo simple harmonic motion with small oscillations.
My question: Is the energy of the disk pendulum at anytime written as
*
*(a) $$E_\text{total}= \frac12mv^2 + mgh + \frac12Iω^2,$$ where $v$ is tangential velocity of the center of mass of the pendulum and $I$ is the moment of inertia of disk $+ m(\text{length of string})^2,$ or
*(b) $$E_\text{total}= mgh + \frac12Iω^2$$ and $I$ is the moment of inertia of disk $+ m(\text{length of string})^2$?
| Using the parallel axis theorem, a disk pendulum that is allowed to rotate around a fixed point with a distance $l$ from it's center, has total rotational kinetic energy $$E= \frac{1}{2}\omega^2(I_{cm}+ml^2)=\frac{1}{2}I_{cm}\omega^2 +\frac{1}{2}m\omega^2l^2$$ and since we can write the angular velocity in terms of the tangential velocity, i.e., $$\omega=\frac{v}{l}$$ where $v$ is the linear (tangential) velocity of the center of mass of the disk, then the above equation for the kinetic energy can simply be written $$E= \frac{1}{2}I_{cm}\omega^2 +\frac{1}{2}mv^2$$ and the disk has a moment of inertia $$I_{cm}=\frac{mR^2}{2}$$ which is the moment of inertia of thin disk (similar to a cylinder of small height) of radius $R$.
This means we can write its total energy as $$T=\frac{1}{2}I_{cm}\omega^2 +\frac{1}{2}mv^2+mgh$$
So in your above question, option (a) is the correct answer since option (b) ignores this second kinetic energy term altogether. It's also important to note that the second term in this expression has the velocity of the center of mass, which again can be expressed in terms of the angular velocity that the disc has around the fixed point i.e., $$KE_{p}=\frac{1}{2}m\omega^2l^2$$
and to note that the first term is the kinetic energy the disc has about its own axis. i.e., $$KE_{ax}=\frac{1}{2}I_{cm}\omega^2$$
| {
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Where does pseudo force act at? It is known that, to apply Newton's laws in a non-inertial frame, we use the concept of pseudo force. We also know that force is a bound vector. Hence, is there a general way to determine where the pseudo force vector would be located at?
| Like gravity, pseudoforces apply at all points in a body. In mechanics, when we say that a distributed force "acts" at a single point, we mean that performing such a replacement does not change the torque acting on the body as a whole. Whether this is possible depends on the pseudoforce.
*
*In a uniformly accelerating frame, the pseudoforce can be taken to act at the center of mass of the body, with strength $M \mathbf{a}$.
*In a uniformly rotating frame, the centrifugal pseudoforce can be taken to act at the center of mass of the body, with strength $M \omega^2 \mathbf{r}_{\mathrm{cm}}$.
*The Coriolis force generally cannot be treated in this way. For example, there are situations where the total Coriolis force vanishes, but the total Coriolis torque doesn't.
| {
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A bowling ball on an infinitely long track We knew that a after a bowling ball is threw out with a certain velocity to a non smooth track, it first rolls and skids as the translational velocity (decelerates due to friction) of the center of mass is greater than tangential velocity of the point of contact of ball with the floor which has a opposite direction. But after some time torque due to friction causes the angular velocity to increase and eventually the tangential velocity of point of contact achieve same value with translational velocity and have v=0, it starts rolling without slipping, and eventually come to stop.
My question here is after rolling without slipping is achieve how does the translational kinetic energy and rotational kinetic energy change? How does translational kinetic energy decrease while rotational kinetic energy is increase by torque due to friction?
|
How does translational kinetic energy decrease while rotational
kinetic energy is increase by torque due to friction?
Since, you are asking about the situation AFTER rolling without slipping has been achieved, i think it is incorrect to say that rotational KE increases.
Once rolling without slipping has been achieved, the rotational KE will no longer increase.
Both rotational KE and translational KE will start to decrease.
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Confusion about friction during pure rolling Friction comes into play with the relative motion between the surface and the points of the body in contact with it. In a perfectly rolling sphere, the instantaneous velocity of the bottommost point is zero, which means there shouldn't be any friction.
But if there is not friction at all, why do we talk about the work of friction being zero during rolling?
Or, is there some friction after all?
| Zero velocity doesn't mean no friction. The friction acting on a stationary object is called static friction. Friction arises when there is a relative motion or a tendency for the relative motion. So there is a tendency for the relative motion at point of contact in rolling motion. Therefore there is friction, which helps the rolling motion by providing torque. But the rolling friction overcomes this helping static friction, therefore rolling object gradually slows down.
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Do we hear sound at pressure or displacement antinodes? I have read from Young & Geller (2007), College Physics 8th Edition, Pearson Education Inc. (pg 385) that we hear sound at pressure anti-nodes rather than displacement anti-nodes as microphones sense pressure variations.
When we have 2 speakers facing each other to form a standing wave between them, it forms a standing wave with open ends at both speakers. This means that the ends closest to the speaker are displacement antinodes and thus pressure nodes. However, going by the above fact that we detect loudest sounds at pressures antinodes, this would mean that when we place a microphone closest to the speakers, it should detect very soft sounds. Yet, many experiments online showing standing wave formation between 2 speakers seem to show that the loudest sound will occur closest to the speakers.
So where exactly do we hear/ detect sound?
Thank you!
| If you realise that our ears are pressure sensors you can easily reach the answer that we perceive the sound to be louder at pressure anti-nodes.
One more thing to consider is that no speaker or any other physical system is a good candidate for the idealised (and rather simplified) models used at an introductory textbook. I won't say always (although maybe I could when referring to every day physical systems such as speakers), but most often there are evanescent waves close to most (if not all) vibrating bodies (related to acoustic radiation of course). This, as well as the fact that an "open end" (in the ideal sense) does not exist (this means that the terminating impedance cannot become infinite) you will never get perfect cancellation (nodes).
Hope this helps somehow...
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Is entropy of planets and others bodies proportional to the area? Is entropy of planets and others bodies proportional to the area? If yes, how do I prove that? I know it works for a black hole but, and for others? If no, then the entropy is really proportional to the volume in some cases? I have seen this proportionality in some cases in thermodynamics.
| The total entropy is an integral of the entropy density over the body volume. I do not know if it even makes any sense for planets, though.
Black holes are exotic (thoeretical) objects with non verifiable properties.
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Cauchy sequences through examples in Quantum Mechanics (at the level of the rigor of physicists) I have just read the definition of a Cauchy sequence:
A sequence ($\psi_n$) is a Cauchy sequence in a vector space $V$ when $||\psi_n-\psi_m||\to 0$ when $n,m\to\infty$. The limit of every Cauchy sequence $(\psi_n)$ converges to a definite element $\psi\in V$ i.e. $$\lim\limits_{n\to\infty}\psi_n=\psi.$$
But I cannot feel it completely unless I see an example of such a sequence. What is an example of a Cauchy sequence of vectors $(\psi_1,\psi_2,...)$ that we encounter in quantum mechanics?
| The most common way Cauchy sequence appear is as series. For example, if $|0\rangle, |1\rangle, \ldots ,|n\rangle , \ldots$ is an orthonormal sequence and $\sum |a_n|^2 = 1$, we expect $|\psi \rangle = \sum a_n |n\rangle$ to be a well defined state, ie we would like the limit :
$$\lim_{N\to +\infty} \sum_{n=0}^N a_n|n\rangle$$
to exist.
The sequence $(\sum_{n=0}^N a_n|n\rangle)_{N\in\mathbb N}$ is Cauchy, since :
$$\left\|\sum_{n=0}^{N+p}a_n|n\rangle - \sum_{n=0}^{N}a_n|n\rangle \right\|^2 = \left\| \sum_{n=N+1}^{N+p}a_n|n\rangle\right\|^2 = \sum_{n=N+1}^{N+p}|a_n|^2\overset{N\to +\infty}{\longrightarrow}0$$
Therefore, because we work in a (complete) Hilbert space, the limit we wanted does indeed exist.
| {
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Rigorous proof that a net force of zero guarantees zero linear acceleration in rigid bodies I've never found a rigorous proof of this fact.
The center of mass' acceleration is not necessarily the linear acceleration, specially if the body is attached to a pin or another geometric constrain, then the center of mass spins like the rest of the body. So how can we find the linear acceleration of a body?
EDIT: ok, the pin or constrain seems to add an external force and thus is a bad example to ilustrate the zero transltational acceleration derived from zero net force.
Yet, the result is still seemly true.
| real answer: it uses three facts:
*
*the arbirtrary movement of a rigid can be seen as a translation through any of it's points $P$, whose image is $P'$, and a rotation by some axis passing through $P'$. Valid to any point $P$ on the body. (Chasles theorem)
*the center of mass of a rigid body can be seen as a point of the body (with mass zero, only kinematically), meaning it's distance from the rest of points remain unchanged (easily checkable with previous fact and definition of COM).
*If a rigid body movement (continuous in time) is such that it keeps one of its points in the same position, then it must be equivalent to a rotation around some axis that passes through the fixed point.
So here it is: an infinitesimal movement of a rigid body can be understood, in particular, as a translation and a rotation through (an instantaneous axis that passes by) the center of mass. If the net external force is zero, then the total acceleration of the COM is zero (famous result), but, as the center of mass lies on the instanteneous rotation axis, it only has translational acceleration. Thus, the translation acceleration of the system is zero with respect to the COM. If the net external force is not zero, this argument guarantees that the COM will transladate.
If the net external force is zero, we know that the center of mass is not moving (accelerating). From facts 2 and 3, the system will be then rotating about some axis that passes through the COM, and thus the system as a whole must have ZERO translational acceleration.
Thus it is necessary and sufficient that the net external force is null in order for a rigid body to not transladate.
EDIT: this argument only works for static equilibrium, not dynamic, but maybe an analysis over a inertial frame with same speed as the center of mass in the moment of equilibrium may do the trick.
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Internal structure of atomic nuclei When one considers the decay chain of heavy elements, alpha decay forms a significant part of the decay series.
This make me wonder about the internal structure of atomic nuclei. When atomic nuclei are created by fusion is the structure of the two combining elements “scrambled”, or are they stuck together as clusters.
Question: Are atomic nuclei composed of an agglomeration (clustering) of helium nuclei and for odd numbered elements a clustering of helium nuclei and a hydrogen nucleus? Something like big marbles in a bag, where each big marble is a helium nucleus and a small marble is a hydrogen nucleus. Under such a scenario a helium nuclei (alpha particle) could potentially be more easily separated from the other clusters in the nucleus during alpha decay.
| I think yes, because of the high binding energy of alpha particle the matter is clustered inside heavy nuclei. At least there are such models.
You can read about them here:
https://iopscience.iop.org/article/10.1088/1742-6596/111/1/012001/pdf
https://www.ggi.infn.it/talkfiles/slides/talk3742.pdf
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What is the intuition behind the spacetime interval? In an article that I am currently reading (under the Lorentz Invariants sub-heading), it explains that, just as the distance between two points on a Cartesian plane are obviously invariant of the coordinate system, the “spacetime distance” is also invariant. While in Cartesian coordinates $$(x_1-x_2)^2+(y_1-y_2)^2 = (x_1'-x_2')^2+ (y_1'-y_2')^2,$$ the space time analog is $$c^2(t_1-t_2)^2-(x_1-x_2)^2 = c^2(t_1'-t_2')^2- (x_1'-x_2')^2 = s^2$$ where $s^2$ is the spacetime interval.
I am having difficulty in understanding this notion of a spacetime interval and the intuition/derivation for why it can be written in this way and is invariant under Lorentz transformation.
I am aware that similar questions have been asked on this platform but none of them have fully cleared things up for me so far. Any help in providing an intuition or understanding would be appreciated.
| You have two great answers, but you might find it interesting to know that it was once common for spacetime in SR to be described with an imaginary time axis. That allowed people to consider that it was a straightforward Cartesian arrangement, where the calculation of a length was through the usual Pythagorean method of taking the square root of the squares of the component displacements along the four orthogonal axes. The fact that the time axis was iT meant that when you squared the displacement along the time axis you automatically got minus T squared.
The idea of an imaginary time axis also made the Lorenz transformation look like straightforward rotations in a 4D space, so some people thought that would make SR easier to grasp if described in that way. However, it turns out that using an imaginary time axis only works straightforwardly for SR, and causes all kinds of complications in GR, so it dropped out of fashion.
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Can we define operators like $\dfrac{1}{a^\dagger a}$? I was recently reading this paper on Enhancement of Few Photon Optomechanical Effects and could not quite understand eq.(2). The author has written an operator like this:
$$\hat \xi=\dfrac{g_oa^\dagger a}{w_m-g_{cK}a^\dagger a}$$
I don't understand how I am supposed to interpret that number operator in the denominator. My guess is that the operator $\hat \xi$ is defined such that its product with $(w_m-g_{cK}a^\dagger a)$ gives $g_oa^\dagger a$. But I am not sure.
Any help is appreciated.
| In general, $\frac{A}{B}$ is lazy physicist shorthand notation for $B^{-1}A$. You might rightly complain that there is an ordering ambiguity and the expression could also mean $A B^{-1}$. That's completely correct, and this notation is only meaningful if $[A,B^{-1}]=0$.
If we take $A=g_o a^\dagger a$ and $B = w_m - g_{cK} a^\dagger a$, then indeed $[A, B^{-1}]=0$ so there is no ordering ambiguity. To see this, define $\epsilon = g_{cK}/w_m$ and write $B=w_m C$, with $C=1-\epsilon a^\dagger a$. Then (assuming the Taylor series converges -- see J. Murray's answer for more details on this assumption and ways to remove the need for it) we can use the Taylor series (specifically it is a geometric series) to write
\begin{equation}
B^{-1} = \frac{1}{w_m} C^{-1} = \frac{1}{w_m} (1-\epsilon a^\dagger a)^{-1} = \frac{1}{w_m} \sum_{n=0}^\infty \epsilon^n (a^\dagger a)^n
\end{equation}
Since $[a^\dagger a, a^\dagger a]=0$, we have that $A$ commutes with every term in the Taylor series, so $[A,B^{-1}]=0.$
Here's another argument that $[A,B^{-1}]=0$, which doesn't rely on the Taylor series.
First, we note that if $[A,B]=0$, and if $B^{-1}$ exists, then $[A,B^{-1}]=0$. Here's a proof:
\begin{eqnarray}
0 &=& BA - AB\\
&=& A - B^{-1}A B\\
&=& AB^{-1} - B^{-1} A\\
&=& [A, B^{-1}]
\end{eqnarray}
Since it's easy to see that for $A=g_o a^\dagger a$ and $B=w_m - g_{cK} a^\dagger a$, that $[A,B]=0$, it follows that $[A, B^{-1}]=0$.
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Newtonian vs Lagrangian symmetry Suppose we have a ball of mass $m$ in the Earth's gravitational field ($g=const.$). Equation of motion reads as:
$$
ma = -mg
$$
From here we can conclude that we have translational symmetry of the form $x(t) \to x(t) + const$ (we are working in only 1D). However, we cannot see this symmetry from the Lagrangian:
$$
L = \frac{mv^2}{2} - mgx
$$
because the linear term "breaks" this symmetry. Moreover, we also do not have the corresponding conserved quantity (as far as I can see).
Does this mean that we can have symmetries in the Newtonian sense (transformations that map solutions to other solutions) that are not present in the Lagrangian?
| Well, it seems to me that under a translation $x(t) \to x(t) + c$, the Lagrangian goes to $$\mathcal{L} \to \mathcal{L}' = \frac{1}{2}m v^2 - mg(x+c) = \mathcal{L} -mgc.$$
So yes, the Lagrangian may appear to be different, however since it only shifts by a constant, these two Lagrangians ($\mathcal{L}$ and $\mathcal{L}'$) are equivalent and produce the same Euler-Lagrange Equations. Indeed, more generally, two Lagrangians are equivalent if their difference is a total time derivative. i.e. $\mathcal{L}$ and $$\mathcal{L}' = \mathcal{L} + \frac{\text{d}f}{\text{d}t}$$ are equivalent for any $f(t)$.
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Do fermions of different types have the same quantum states available to occupy? I'm not asking whether two fermions of different types can occupy the same quantum state, cf. the Pauli exclusion principle. I'm asking whether fermions of different types would have the same options available if you had one in at a time. An example of what I mean is the fact that muons will occupy much smaller orbitals around nuclei than electrons will. In this case, electrons and muons don't seem to have the same quantum states available for them around a nucleus. Is this the case with any two fermion types?
| The difference in the bound states of electrons or muons and a nucleus originates from their different masses (muons are a lot heavier than electrons). Because the Hamiltonian of the problem depends on the mass, the energy eigenvalues do as well.
In general, fermions (or any two particles) will be able to occupy the same state if all their properties which enter the Hamiltonian are the same. For example, a spin-up and a spin-down electron have the same states in an atom as long as there is no magnetic field. As soon as a magnetic field is switched on, this degeneracy disappears, because the spins have an energy in the magnetic field which depends on their direction.
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Why is the angle between the radial velocities at two instants the same as the angle between the tangential velocities at those same instants?
While it is clear that the angle between $v_r$ and $v_r + \Delta v_r$ is $\Delta \theta$, I cannot see a clear geometric reason as to why the angle between $v_t$ and $v_t + \Delta v_t$ must also be $\Delta \theta$.
The book by Kleppner and Kolenkow uses this to argue that $\Delta v_t \approx v_r \Delta \theta$ and in the limit in which $\Delta t \rightarrow 0$ , $\frac{d v_t}{dt} = v_t \dot{\theta} = r\dot{\theta}^2 $.
| If I'm correct, this is on p37, also this book is a classic, so many people have it, referencing it properly would speed things up.
First, I'm assuming that Vt is the component of velocity tangent to the f(x) that is your objects position, and Vr, well is just radial velocity.
Before diving in, I want you to show how your understanding is wrong of the problem on hand. Vr is not always perpendicular to tangent of a curve, rather the normal to that curve is that, radial is meant as anything to do with respect to the center of one's coordinate system, be it distance, velocity or others. Now you might say that the tangent is to the curve, rather than the conventional meaning (theta direction in the polar coordinates you mentioned).
Let's assume that there is a strait line, say with equation x = const., so if Vt is tangent to the line of motion, it's always the same.
So let's assume that the tangential component, is in fact the component perpendicular to the radial velocity, so the overall speed is root square of the sum of 2 velocities, written as:
$$V(total)^2 = Vr^2 + Vt^2$$
Now, if you think of your tangent and radial velocity like the figure below, everything will fall into place:
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Does relativity mean that the crew of a relativistic rocket would experience less acceleration than in our frame of reference? I have been told regarding a 1 g rocket that "the amount you accelerate would be less due to relativity".
Does that mean that from the crew's time dilated perspective, they would experience less acceleration than we observe in our frame of reference? Could a ship be accelerated at say, 10 g from our frame of reference on Earth, while the crew of the ship only experiences 1 g of acceleration in theirs?
If this were possible, how far can we take this, and how quickly?
| If a ship starts from rest in a frame S, $a'$, the acceleration in the rest frame, S', of the crew is related to the acceleration, $a$, in S, by $a'=\gamma^3 a$.
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Confusion over Feynman’s description of the Wu experiment for parity violation In his lecture on symmetry in physical law, Feynman said:
Using a very strong magnet at a very low temperature, it turns out that a certain isotope of cobalt, which disintegrates by emitting an electron, is magnetic, and if the temperature is low enough that the thermal oscillations do not jiggle the atomic magnets about too much, they line up in the magnetic field. So the cobalt atoms will all line up in this strong field. They then disintegrate, emitting an electron, and it was discovered that when the atoms were lined up in a field whose B vector points upward, most of the electrons were emitted in a downward direction. Therefore if we were to put it in a corresponding experiment in a “mirror,” in which the cobalt atoms would be lined up in the opposite direction, they would spit their electrons up, not down; the action is unsymmetrical. The magnet has grown hairs! The south pole of a magnet is of such a kind that the electrons in a β-disintegration tend to go away from it; that distinguishes, in a physical way, the north pole from the south pole.
This doesn’t make sense to me.
I understood that the Wu experiment found electrons were emitted in the direction of Cobalt atom spin and, therefore, in the direction of the B vector, not opposite it as Feynman writes. Where have I gone wrong?
| The electron charge is negative, so the magnetic moment is anti-aligned with the spin.
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Why do icebergs flip over? Why do icebergs flip over? Are certain shapes of icebergs more "stable" than others, in that it's harder to flip them over? If so, why?
For example, it somehow makes intuitive sense, that a thick iceberg with a certain height (or depth, because 90% of it is below water) would be harder to flip over than a thin one with the same height/depth. But why is this the case?
| I really don't know if this will answer your question but you have to take into account the forces acting over the iceberg. You have pressure from the sea stream, buoyancy from the Archimedes principle, pressure from wind, and weight from gravity acceleration.
Then, why would something flip? For something to flip you must have torque
$$
\vec{\tau}=\vec{r}\times\vec{F}
$$
When you see this equation you have to consider the object as a solid body, not a point in the center of mass. This means that if the force is applied far from the center of mass, there will be more torque. If you consider a long but thin iceberg in the vertical direction you'll see that it is less stable than in the horizontal direction. Think that horizontal forces come from wind and sea streams. The vertical forces, weight, and buoyancy only do torque when the iceberg is already angled. This effect is quite interesting because, at the moment where the iceberg tilts, the torque increases suddenly because there are more forces doing torque.
At the end of the day, the iceberg flips to reach a lower energy state (A configuration where it's more stable). Icebergs change all the time, they melt, lose pieces of ice, and therefore the stable state may change over time. That could mean flipping a couple of times.
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Am trying to determine the source of a low level, low pitched "rumbling" noise that appears to only be heard inside houses in our road, any thoughts? I live in the Olton area within the UK. The noise can be briefly intermittent or continuous for hours and can only be heard indoors - louder at night. One neighbour has reported that his pillow vibrates at night. The local water authority have said that the noise is nothing to do with them, even though their sewers run through pipes that run under ground along the road and that they have storm tanks within the area. WE do have HS2 boring tunnels about 10 mile away, could this be the cause ?
| Arthur,
High frequency sound get absorbed more quickly in dense material than low frequency sound. The fact that the noise is low frequency and can only be heard indoors suggest that it's travelled some distance underground.
However the distance could be small e.g. 10m or many km if the source is loud enough.
You've asked on a physics website, so it's best to test different theories with experiment.
The theory that it's HS2 tunnelling noise travelling along pipes is best tested like this - Try and get a record of tunnelling times from HS2 (e.g. for a particular few days) and see if it matches what you hear.
If it is the above, it seems unlikely that tunnelling so far away would always effect you, i.e. the source of the sound moves as the tunnelling progresses, presumably it wouldn't be near enough to the pipes that had a route under your house before long, in which case it'll cease to be a problem.
It's quite likely to be more local.
Here is the website for Coventry Environmental Health, they have a 'report a noise problem' link Environmental Health, although you might need the Birmingham one.
Best of luck with it.
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Simplest exactly solved model displaying a phase transition? The classical example of an exactly solved model which displays a phase transition is the 2D Ising model. However, all the proofs I've seen of this have been very long and complicated.
So, I wanted to know whether there were any other exactly solved models with phase transition, which were easier to solve, or that the 2D Ising model is the simplest such model that we know of.
| Exactly solvable models in Statistical mechanics by Baxter is the place to look.
The other answers have already pointed the infinite range Ising model and 1D Ising model as exactly solvable (although the latter has phase transition at zero temperature). Besides the infinite and the 1D case, the Ising (and more generally Potts) model is also solvable on a Bethe lattice, as discussed extensively in the book cited above.
There are also offshoots of these models with additional features: e.g., Sherrington-Kirkpatrick model is the infinite range Ising model with random couplings, which is useful for understanding spin glasses.
However, the specific features of phase transitions in these models differ, and thus the model of choice depends on what one is looking for.
Remark: As more exotic case of exactly solvable models one could mention soluton of Kondo and Anderson impurity models by Bethe ansatz.
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How are these two expressions for the canonical partition function equivalent? In Equilibrium Statistical Physics by Plischke and Bergersen the canonical partition function is defined (on page 37, eq. (2.33)) as
$$Z_C = \int \frac{dE}{\delta E} \Omega(E) \exp\{-\beta E\}\tag{1},$$
where $\Omega(E)$ is the accessible phase space volume at energy $E$ for the system under consideration and $\delta E$ is a "tolerance" term which is later set to be proportional to $\sqrt{\langle E \rangle}$ to obtain agreement with classial thermodynamics.
But then in the end of chapter problems, if one looks at the official solutions manual for example, one instead uses
$$Z_C = \frac{1}{h^{3N}} \int d^{3N}q \int d^{3N}p \exp\{-\beta H\}\tag{2},$$
with $H$ the Hamiltonian of the system. But nowhere does the text show the equivalence of these two integrals.
So how are (1) and (2) equivalent? How does one go from the definition (1) to (2)?
| (2) is the canonical partition function of a classical system of $N$ particles with Hamiltonian H. The Boltzmann weight is integrated over the $6N$-dimensional phase space $(q_i,p_i)$. This integral can be cast in the form of (1) by introducing
$$\Omega(E)={1\over h_0^3}\int d^{3N}q\int d^{3N}p\ \!\delta(E-H(q,p))$$
which counts (up to the factor h_0^3) the volume of phase space which is accessible to the system with an energy $E$. It follows that
$$Z_C=\int \Omega(E)e^{-\beta E}dE$$
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When a car accelerates relative to earth, why can't we say earth accelerates relative to car? When a car moves away from a standstill, why do we say that the car has accelerated? Isn't it equally correct to say that the earth has accelerated in the reference frame of the car? What breaks the symmetry here? Do the forces applied to the car have special significance in determining which frame is inertial and which one is not?
Please explain in simple terms.
| Here is my favorite view of what a person in the car would observe. I don't see it represented very often, for some reason.
By the equivalence principles (that e.g. general relativity builds on), the moment the car engine starts applying torque to the wheels, a person inside the car will feel a change in gravity. It will get a new component in the backward direction of the car.
The person in the car is being held up against this new component of gravity by the back of their seat, and the car is being held up against this gravity by the friction between the wheels and the ground. But nothing holds the ground up, so it begins to fall backwards. Which is to say, yes, the ground is accelerating backwards, just as much as an apple falling to the ground is accelerating downwards.
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Find the values of $A$, $B$, and $C$ such that the action is a minimum
A particle is subjected to the potential $V (x) = −F x$, where $F$ is a constant. The particle travels from $x = 0$ to $x = a$ in a time interval $t_0 $. Assume the motion of the particle can be expressed in the form $x(t) = A + B t + C t^2$. Find the values of $A, B$ and $C$ such that the action is a minimum.
I was thinking it can solved using Lagrangian rather than Hamilton. There's no frictional force.
$$L=\frac{1}{2}m\dot{x}^2+Fx$$
$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)-\frac{\partial L}{\partial x}=0\implies m\ddot{x}=F\implies \ddot{x}=\frac{F}{m}$$
Differentiate $x(t)$ twice. $$2C=\frac{F}{m}\implies C=\frac{F}{2m}$$
For finding B I was thinking to integrate $\ddot{x}$ once.
$$\dot{x}=\int \ddot{x} \mathrm dt =\ddot{x}t$$
initial position is 0 so, not writing constant.
$$\dot{x}=\frac{F}{m}$$
Differentiate $x(t)$ once.
$$B+2Ct=\frac{F}{m}$$
$$\implies B=\frac{F}{m}-\frac{2Ft}{2m}=-\frac{Ft}{2m}$$
Again, going to integrate $\ddot{x}$ twice.
$$x=\iint \ddot{x} dt dt=\frac{\ddot{x}t^2}{2}$$
initial velocity and initial position is 0.
$$x=\frac{Ft^2}{2m}$$
$$A+Bt+Ct^2=\frac{Ft^2}{2m}$$
$$A=\frac{Ft^2+Ft-F}{2m}$$
According to my, I think that C is the minimum (I think B is cause, B is negative; negative is less than positive). And, A is maximum.
A person were saying that "It asked you to minimise the action; it told you the particle moved from $0$ to $a$ in time $t_0$; it gave you the equation of the trajectory."
In my work where should I put the interval?
| The Euler-lagrangian equation gives the equations of motion that once solved give you a family of solutions that minimize the action. A unique solution is given by specifying boundary conditions. It is just a case of inputing those boundary conditions.
Wlog let $ x(0)=0 $ and $x(t_0)=a $. Integrating $\ddot{x} = \frac{F}{m}$ gives the general solution $x(t)=\frac{F}{2m}t^2 +Bt + A$, fixing C. Subbing in $x(0)=0$ gives $A=0$ and subbing $x(t_0)=a$ gives $B$ as $B=\frac{a - Ct_0^2}{t_0}$.
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How to obtain position data from acceleration without forward euler? I am doing a investigation into the Wilberforce Pendulum and in order to find the position and rotation at any time I have attached my phone onto the pendulum in order to use Phyphox, a app that finds the acceleration and angular velocity. However, when I put the data into excel and use forward Euler to find the velocity from the acceleration, and the position from the velocity. However, this doesn't really work, and the velocity seems to drift downwards a bit, and then the position seems to drift down much, much more. I also took a video of the pendulum, and I compared this to the results to show that it isn't experimental error.
Do any of you know how to get the position data from acceleration, without this error?
| Since you already have the data you just need a numeric integrator which inherently smooths the data. But if you use a forward Euler, then you are biasing the smoothing (averaging) to previous values causing a bias in the results.
I have been down the road you are going through and here are my suggestions
*
*Use the gyroscopes to measure rotational velocity.
*Use trapezoidal integration to calculate angles. In general, use an integration technique that considers forwards data equally with backwards data in order to avoid bias.
*Numerically adjust the data by applying a DC offset, or a slope to hit a target value at the end of the test. This corrects for the numeric drift you see.
If you want more accurate integration of data, I suggest to use a cubic spline interpolation to do so.
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Why and when can the Earth be considered an inertial reference frame? The question has been asked (e.g., here and here), but I would like to get a more definitive and mathematically formal answer.
The Earth rotates around its axis, around the Sun, and participates in larger scale motions as a part of the Solar system. Yet, we often can get by treating it as an inertial reference frame (e.g., when constructing furniture, cars and buildings). In some cases we do need to take account for the effect of its rotation - e.g., in weather prediction one takes account for the Coriolis force, but we still consider the Solar system as the inertial reference frame.
We do that because:
*
*accelerations that we deal with (notably $g$) are much greater than the accelerations due to the other motions that it is involved in?
*we can neglect the non-inertial forces because all the objects around experience the same accelerations due to these forces?
*something else?
I am looking for a mathematically motivated answer. I also suggest delineating between what is specific to Earth (accidental) and what would apply to all (or most) planets/stellar bodies.
Update
I took the liberty to summarize the opinions expressed so far in my own answer. Yet, there remains non-inertial effects not covered by free fall and the Earth's rotation - those related to the Earth's finite size and responsible for the tidal forces (more specific question is here). Thus, this question still needs a canonical answer.
| Motion near the Earth is basically inertial if an object is not being "pushed" by anything! If you throw a stone it is in inertial motion (apart from air resistance) until it hits the ground.
Motion of the Earth around the Solar System, Galaxy etc. is irrelevant if you are not being pushed, and is completely dominant if you are on the ground!
Artillery is in inertial motion too; the coriolis effect is an artifact only visible to observers on the ground!
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Mandl & Shaw QFT chapter 1 question Page 3 of Mandl & Shaw claims that, given a vector $\pmb{A}(\pmb{x},t)=\pmb{A}_{0}e^{i(\pmb{k}\pmb{\cdot} \pmb{x} - \omega t)}$, $\pmb{\nabla} \pmb{\cdot} \pmb{A} = 0$ (eq. 1.6) implies $\pmb{k} \pmb{\cdot}
\pmb{A} = 0$ (eq. 1.7).
I'm having trouble figuring out why (1.6) implies (1.7).
| You can just take the derivative of the exponential. In particular:
$$
\vec{\nabla}\cdot \vec{A}(t,\vec{x})=\vec{A}_0 \cdot \vec{\nabla} e^{i(\vec{k}\cdot\vec{x}-\omega t)} = i \vec{k}\cdot\vec{A}_0e^{i(\vec{k}\cdot\vec{x}-\omega t)} = i \vec{k}\cdot\vec{A}(t,\vec{x}) = 0\,.
$$
There you go!
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Bar released from upright position rotates about a lower hinge before being caught by a steel cable, how to calculate the tension on the cable? I wish to check the safety factor of a steel cable that I've designed to catch a heavy bar that rotates freely due to the gravitational force acting on it.
From the upright position, I'm assuming that the bar has a rotational velocity of 0rad/s before rotating about a lower hinge (hence, gaining rotational kinetic energy), and comes to a stop when the cable goes from slack (sagging cable) to taut (straight - full length).
I've currently modelled the cable as a spring where the rotational kinetic energy is converted to elastic potential energy within the cable upon a small extension. This may be where I've made a mistake since cables don't alway follow Hooke's law and the elastic extension I've found doesn't seem to account for the momentum. Based on this, I'm guessing that I've gone about this problem the wrong way. I'll attach my working with some example numbers below.
Any help or guidance on this will be hugely appreciated!
| The cable will exert an impulse which changes the bar's angular momentum. The maximum force will depend on how much the cable can stretch. Treating it like a spring (with a high k) is not unreasonable (but measuring the k could be a problem), and there is likely to some bounce.
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Spin without quantum mechanics? In Emergence of spin from special relativity some answers discuss how spin can arise in non-relativistic quantum mechanics (let's not enter into those details here). However it is also argued that you do not even need quantum mechanics as there are some relativistic constructions.
Roger Penrose & Wolfgang Rindler's "Spinors and Space-Time" construct some spinorial space-time. What does this imply for spin? Can you have spin without quantum mechanics? Spin more precisely in the context of describing the spin of particles (if that makes any sense without quantumness).
Disclaimer: I guess spinors can be an interesting mathematical tool (as described by the comments) but I hope there will be some kind of correspondence between the not-quantum and quantum. Does the quantum mechanical spin lead to some classical spinor quantity at the not-quantum limit? I guess not.
| In general relativity textbooks, it will be mentioned that general covariance can be achieved easily if the equations are tensorial. But tensor equations are not the only equations that have general covariance. Spinor equations also satisfy general covariance. In curved spacetime, spinors are defined using fiber bundles.
Quoting Robert M. Wald from chapter 13 of General Relativity called Spinors
Spinors arise most naturally in the context of quantum theory..... However, we should emphasize that the notion of spinors has proven to be an extremely powerful tool for analyzing purely classical problems. Perhaps the most dramatic example of this is Witten's (1981) spinorial proof of the positive mass conjecture. In section 13.2 we shall give further examples of this by deriving a useful spinorial decomposition of the curvature tensor and obtaining the existence and properties of the principal null directions of the Weyl tensor in a manner far simpler than can be achieved by tensor methods.
You can find more information in that chapter. Also check this and this which seem more intuitive than Wald.
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Riemann curvature tensor is the only tensor that you can write down that has two derivatives of the metric tensor Is the statement in the main question correct? Can someone send me a proof (link, pdf file etc.)
| You can prove this by moving to Riemann normal coordinates at a point $p$. In these coordinates
$$
g_{ij}(p) = \eta_{ij} , \qquad \partial_k g_{ij}(p) = 0 , \qquad \partial_k \partial_l g_{ij}(p) = \frac{1}{3} [ R_{ikjl}(p) + R_{jkil}(p) ] .
$$
It is then clear that anything that contains 2 metric derivatives can be re-written in terms of the Riemann tensor.
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Mechanism of Relativistic Momentum The formula for relativistic momentum is $\vec{p}=\gamma m\vec{v}$.
To derive this formula, one analyzes a collision while assuming the principle of relativity and the conservation of momentum principle are correct:
https://www.feynmanlectures.caltech.edu/I_16.html
I'm fine with all that. What I want to know is the mechanism which causes this. At speed $0.0001c$, force applied is approximately proportional to acceleration. At speed $0.89c$, the same force applied changes the speed of the object very little. Acceleration due to a force is a precise function of the mass and the pre-existing velocity. What is causing this, is there something unseen which 'acts' upon the moving object to render this phenomenon? Does it have something to do with the Higgs field? Is it because of time dilation? Thanks.
| Using your arm to accelerate a block of iron 1: to low speed 2: to high speed:
Low speed: afterwards iron atoms have gained momentum
High speed: afterwards iron atoms and fat molecules have gained momentum (there also occurs a conversion of fat molecules to something else)
Extreme speed case: before the acceleration you are a 300kg guy, then you use your legs to to accelerate yourself and an iron block to extreme speed, afterwards your mass is zero, another observer sees only iron that has been accelerated to high speed.
Let's say the mass of iron block is 1kg. And in the extreme case the final speed of the block is v. The momentum of the block will be:
$301 kg * v.$
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How come black body have more emissivity and more absorbtivity(a) at same time? We have two definitions to look at
Absorbtivity(a): the ratio of absorbed energy and incident energy on a body
$a_{BlackBody} = 1$
so if i have a tourch light that gives red light, in a dark room I point this tourch light at this black body, then according to above definition, I should see black color
Good absorbers are good emitters from kirchoffs law, and also stefans law
$$\frac{d\theta}{dt} = \sigma AeT^4$$
Now according to that statement, if we reconduct same experiment, when I point my red torch light towards black body it should now emmit red color
Aren't both definitions contradictory to each other or did I misinterpret something
| The light that the black body emits is defined by the temperature of that body. When you shine red torchlight and the body is cold, it will still look black, but as it heats up, it may start glowing red. It still will formally be called a "black body" though, since the term is reserved for a body that absorbs all the light that fall onto it.
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Eigenvalues of Product of 2 hermitian operators Let $A$ and $B$ be two Hermitian operators. Let $C$ be another operator such that $C = AB$. What can we say about Eigenvalues of $C$? Will they be real/imaginary/complex? What I did was to search for examples. The following were examples (in matrix representation) I looked for:
$ A =
\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}$ and $ B = \begin{bmatrix}
0 & 1\\
1 & 0
\end{bmatrix}$ to get a hermitian matrix and so real eigenvalues.
Next I tried:
$ A = \begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}$ and $ B = \begin{bmatrix}
0 & 1\\
1 & 0
\end{bmatrix}$ to get Anti-Hermitian matrix and so imaginary eigenvalues.
Is there a more concrete way of solving this? Can we have a general complex number as eigenvalues for the product of the Hermitian Matrices?
| TL;DR: Assuming that $A,B$ are self-adjoint, the product $AB$ does not need to be diagonalizable. And if $AB$ is diagonalizable, the eigenvalues need not be real or imaginary.
Example 1: $AB$ is not diagonalizable:
$$A~=~\begin{pmatrix} 0 & 1 \cr 1 & 0 \end{pmatrix} \quad\wedge\quad B~=~\begin{pmatrix} 1 & 0 \cr 0 & 0 \end{pmatrix}\quad\Rightarrow\quad AB~=~\begin{pmatrix} 0 & 0 \cr 1 & 0 \end{pmatrix}.$$
Example 2: $AB$ has complex eigenvalues:
$$A~=~\begin{pmatrix} 0 & 1 \cr 1 & 0 \end{pmatrix}\quad\wedge\quad B~=~\begin{pmatrix} 0 & b \cr b^{\ast} & 0 \end{pmatrix}\quad\Rightarrow\quad AB~=~\begin{pmatrix} b^{\ast} & 0 \cr 0 & b \end{pmatrix}. $$
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In freestyle swimming does flutter kick recovery causes sinking? As per my understanding in flutter kick, we kick downwards and the water pushes the swimmer up. However, in flutter kick the swimmer is also actually kicking the water upwards when recovering from the kick. We can contrast this with a dolphin kick where swimmer essentially glides back in recovery and there is little upward kicking. Another example is an oar on boat pushing water where in recovery one goes through air.
So swimmer is pushing the water down and up both. And it looks like the velocity of down and upwards kick is the same. So why does the fast flutter kick still help the swimmer to stay afloat ?
| The propulsion required to move forward is created by rhythmic upward and downward movement of the legs.To answer your question ,when one leg moves up the other leg goes down hence being unable to cancel the forces.In detail there are two phases of one flutter and its recovery.
Downbeat:-
In the first half of the downbeat, the downward movement is initiated by a slight flexion of the leg at the hip.
Shortly after that, the knee also bends a little. The foot goes into plantar flexion (meaning the toes are pointed, both by muscle contraction and by the pressure of the water against the foot as it moves downwards.
During this phase, the upper side of the foot is facing downwards and a little backward. For this reason, while the foot is moving downwards, some water is pushed back. This is how propulsion is created in the flutter kick.
In the second half of the downbeat, the hip is locked in place while the knee stretches. The toes are still pointed. This phase isn’t propulsive but prepares the leg for its upward movement.
Upbeat:-
The upward movement of the leg begins while the knee is still stretching. As the thigh moves upwards, the pressure of the water against the lower leg causes the leg to straighten.
The pressure of the water on the ball of the foot and on the toes brings the foot to a neutral intermediate position. This phase of the flutter kick is not propulsive either.
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Confusion about Newton's third law The question:
A toy rocket consists of a container of water and compressed air.Water is pushed vertically downwards through a nozzle by the compressed air. The rocket moves vertically upwards. The nozzle has a circular cross-section of radius 7.5mm. The density of the water is 1000kgm–3. Assume that the water leaving the nozzle has the shape of a cylinder of radius 7.5mm and has a constant speed of 13ms–1 relative to the rocket. Given t the mass of water leaving the nozzle in the first 0.20s after the rocket launch
is 0.46kg, find the force exerted on the this mass of water by the rocket.
My Problem:
My confusion is that will the force calculated using the formula "change in momentum over time" only be the force exerted by the rocket because the weight of the water is being balanced by some other force, as suggested by the constant speed statement or is there another reason? I thought this formula gives the resultant force, so if my logic is incorrect wouldn't the weight have to be subtracted from the answer to get the action reaction force?
| In this simple model for a rocket of mass M you have a constant mass flow $\dot{M}$ (i.e. the mass of the water that leaves the rocket per second) away from the rocket which you can calculate with the data given in the exercise. The matter that leaves the rocket is assumed to have a velocity $v$ relative to the rocket. This leads to a resultant force
$F = \dot{M} v$
I think that the concerns about the mass of the rocket that you are referring to in the end of your question is only relevant when you are interested in the resulting accelaration. For this you will have to integrate:
$M(t) = M(0) + \dot{M} t $
and then calculate
$a(t) = F(t)/M(t) = \frac{\dot{M}}{ (M(0) + \dot{M} t )} v = \frac{v}{ M(0)/\dot{M} + t}$
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If the probability of a point (photon) hitting another point (electron) is zero why do they collide? If the probability of a point (photon) hitting another point (electron) is zero why do they collide? To have a probability greater than zero almost one of them should be not a point. Correct me, please if I am wrong.
| The photon interacts with the electron, but it does not collide with it.
You can think of an electron as having a cloud of virtual particles, which are basically excitations in the quantum EM field. The quanta of this field are called photons
Take for example, this simple interaction between 2 electrons:
As shown with a Feynman diagram.
Explanation of the diagram
The 2 solid external lines on either side of the diagram are the electrons and the internal line in the middle of the diagram (the wavy line) is the virtual photon. This photon is being exchanged between the 2 electrons. This can be thought of as an exchange of momentum between the electrons.
Answer
Since every particle that has an electric charge has a cloud of virtual photons, the photon exchanged between the particles gets absorbed into the cloud of virtual particles of the particle that it was propagating towards.
And since the EM field is not in a fixed position or a point, but rather it is a field that expands out from the electron throughout spacetime. The photon interacts with this field rather than the election itself. Therefore, the exchanged photon gets absorbed into this field and the particles travel on their merry way, away from each other (in this case).
This whole process can be calculated mathematically with the S-Matrix or using the Feynman rules if you want to work smart.
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Provided a unit vector and Force, how can I calculate it's components? Say I have a $F=kQ_{1}Q_{2}/r^{2}$ and a direction vector $(x, y, z).$ How can I find the component forces $F_{x}$, $F_{y}$, and $F_{z}$?
| It's somewhat unclear from your question, but I interpreted $F$ to be just the magnitude of the force (a scalar), and you want to construct a force of that magnitude pointing along the given direction vector.
If the direction vector is a unit vector (a vector of length 1), then all you have to do is scale (resize) it. So it's just:
$$\vec F = F \cdot (x, y, z) = (Fx, Fy, Fz)$$
If the direction vector is not a unit vector, then you have to make it into one first:
$$\text{let }\space \vec d = (x, y, z)$$
Then it's magnitude squared is
$$d^2= \vec d \vec d = x^2 + y^2 + z^2 \space\space\text{, and}\\
d = \sqrt{x^2 + y^2 + z^2}
$$
so
$$\vec F = F \cdot \frac{\vec d}{d} = (\frac{Fx}{d}, \frac{Fy}{d}, \frac{Fz}{d})$$
| {
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Why the photomultiplier experiment proves that photon is a particle, not a wave? In QED Feynman describes the photomultiplier experiment as a proof that a photon behaves as a particle. The logic is as follows: with monochromatic single photon light source a photomultiplier coupled with a speaker is either silent or responds with sounds of the same loudness. Hence light is a particle which either hits the multipliers plate or not.
However,the multiplier's action is based on the electrons, which are particles. Why there cannot be an alternative explanation, that light is a wave, whose energy either suffices to knock out a single electron,that initiates the cascade, or does not and in that case the photomultiplier remains silent.
| Term photomultiplier refers more to the principle of detection than to what we count, which is the central point of the question here. In this respect it is better to differentiate between a single-photon detector and a photon counter, which are doing somewhat different job (although the former is usually a part of the latter).
Single photon detection is not the same as photon counting. The counting error typically increases as $$\sigma_n\propto\frac{1}{\sqrt{n}}$$ where $n$ is the number of photons detected. Thus, indeed, detecting with certainty presence of a single photon is nearly impossible. However, detecting the arrival of the 1001th photon after we counted a thousand of them can be done with high precision.
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How seriously can we take the success of the Standard Model when it has so many input parameters? The Standard Model of particle physics is immensely successful. However, it has many experimentally fitted input parameters (e.g. the fermion masses, mixing angles, etc). How seriously can we take the success of the Standard Model when it has so many input parameters?
On face value, if a model has many input parameters, it can fit a large chunk of data. Are there qualitative and more importantly, quantitative, predictions of the Standard Model that are independent of these experimentally fitted parameters? Again, I do not doubt the success of the SM but this is a concern I would like to be addressed and be demystified.
| If you have $n$ input parameters in a deterministic theory you can perfectly fit at most $n$ data points just by adjusting those parameters. In a probabalistic theory that is more subtle, but there is a similar association. Regardless of how many parameters the standard model needs, it is a lot less than what would be necessary to fit the 1 petabyte of data collected at the LHC per second.
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Where should the reference point be considered during torque measurement? This is an extremely silly and wierd question.
https://en.wikipedia.org/wiki/Varignon%27s_theorem_(mechanics)
While reading about Varignon's Theorem in wikipedia I noticed this sentence,
"If many concurrent forces are acting on a body, then the algebraic sum of torques of all the forces about a point in the plane of the forces is equal to the torque of their resultant about the same point."
Well what does "a point" specify here? Can it be any point even outside the object which we are dealing with?
Another thing, lets say we are dealing with 2 concurrent forces about a point C and the two forces are acting on A. Now according to the above theorem shouldn't the above scenario be looked as
$$F_1
*AC+F_2*AC=F*CC=F*0=0?$$
Perhaps I misunderstood the theorem. As for the 1st question, I have never thought of any reference point in case of rotational motion other than the center of mass of the object and mostly dealt with uniform shapes.
| In general, the sum of vectors gives their resultant. But, for that to be true they must be similar and related: the sum of electric fields at a point, the sum of forces acting on a mass, and the sum of torques about a point. In the case of torques, the sum produces the rate of change of the angular momentum measured relative to that point. Some situations suggest a choice of point. If you have a fixed axle of rotation, the point should be on that axle. For a freely moving mass, the center of mass is a good choice. In a statics problem choose a point that reduces the number of torques.
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Can we bend a light ray into any closed loop? Suppose we have a medium with varying refractive index and a source of light inside that medium emitting rays. Is it possible to bend the ray into any closed loop?
As the medium has varying refractive index, is it possible?
And if possible, how will it look like if anyone stand in the path of the ray?
| I think so, and not just only loops but also knots. Any optic cable tied into a loop or a knot will probably do. It's easy to imagine theoretically, but probably much more difficult to set up practically as there is the problem of getting the light into the loop into the first place.
| {
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Do electromagnetic waves contain electrons? I understand that EM waves are oscillating electric and magnetic fields. But doesn't this mean that the wave itself contains charged particles that generate the fields?
| Every body above zero Kelvin emits photons. Electromagnetic radiation is the sum of all photons emitted by excited subatomic particles. It is basically thermal radiation.
Electromagnetic waves are a special form of EM radiation. The synchronous and periodic acceleration of electrons on the surface of a conductor results in radiation of polarized photons. This radiation has a periodic maximum of photons - it is a propagating wave.
The point is that each photon is an indivisible particle, moves in vacuum with the same velocity c, and has an oscillating electric field component and an oscillating magnetic field component. Photons support themselves with their oscillating fields. They don't need any environment to exist. On the contrary, every matter on the way of the photon leads to the absorption of a photon by subatomic particles.
| {
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Mixed symmetry of rank $3$ tensor I have rank 3 tensor $T_{ijk}$ with following properties:
$T_{ijk}=T_{jik}$
$T_{ijk}=-T_{kji}$
Is it true that there is the only one tensor of rank 3 with those properties and it is $T_{ijk}=0$.
I'm starting from the following
$T_{ijk}=-T_{kij}=-T_{ikj}=T_{jki}=T_{kji}=-T_{ijk}$
$\Rightarrow 2T_{ijk}=0$
$\Rightarrow T_{ijk}=0$
Where am I making a mistake? Because the result is strange enough. Does this mean that from the point of view of Young diagrams, the hook is not interesting?
| The question and accepted answer here miss the main point, which is that for tensors of mixed symmetry, the indices themselves do not obey that symmetry, which is somewhat counterintuitive. This is explained in the comments by Michael Seifert here.
Tensors of mixed symmetry are generated by applying Young symmetrizers. In your case, you want to generate a tensor with symmetry
1 2 3
which means that to an arbitrary tensor $T_{ijk}$, you apply the symmetrizers
$$[e-(13)][e+(12)]T$$
where $e$ is the identity permutation, so that you get the desired tensor
$$T_{ijk}+T_{jik}-T_{kji}-T_{jki}$$
which is non-trivial.
For your method, you correctly arrived at the conclusion that the only tensor with $T_{ijk}=T_{jik}$ and $T_{ijk}=-T_{kji}$ is the zero tensor. It is not the right way to generate tensors of mixed symmetry, for the reasons mentioned at the beginning of this answer.
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Is pulling a string instantaneous at both ends? Why is it or isn't it? This is a question that has bothered me for quite some time but I don't have a clue where to start in researching it.
Let's say we have a string which is arbitrarily long, light enough to not take much effort to pull but rigid enough that it would not break or stretch. If I now stand at one end of the string and pull on it, does the other end move instantaneously? If so, could we use such a method to send a message faster than light? I'm not referring to the speed of the rope itself, but to the time between pulling one end and the other end moving accordingly.
My knowledge of FTL messaging/travel (basically gen-ed physics and the movie Interstellar) tells me the answer is no, and there's a whole lot more going on at the atomic level, but I don't know exactly what forces are at play.
| No. When you pull really fast on one end of the string, that pull gets transmitted along its length to the other end at the speed of sound waves in the string. This may be several thousand feet per second, but is nowhere near the speed of light.
| {
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Why do rain drops fall with a constant velocity? While reading my physics book. I came across a line that says that:
Rain drop falls with a constant velocity because the weight(which is the force of gravity acting on body) of the drop is balanced by the sum of the buoyant force and force due to friction(or viscosity )of air. Thus the net force on the drop is zero so it falls down with a constant velocity.
I was not satisfied by the explanation So I searched the internet which too had similar explanations:
The falling drop increases speed until the resistance of the air equals the pull of gravity, at which point the drop begins to fall at a constant speed, its terminal velocity.
My confusion regarding the matter is that if the net force acting on a body (here the rain drop) is zero then it should remain suspended in air rather than falling towards the earth. So how come the rain drop keeps falling when net force acting on it becomes zero? How the air resistance and other forces stops the rain drop from acquiring accelerated downward motion?
| For intuition, think of curling (except the rain falls down, curling stones move horizontally). Once the stone is released, there are no horizontal forces on the stone (except a slight friction) yet it moves at a roughly constant speed (the slight friction slows it down over time). The player first accelerates the stone, just like gravity first accelerates the raindrops. Then, if the net force is zero, the velocity doesn't change. Whether this is due to a near-lack of forces (a released curling stone), or forces adding up to zero (raindrops), is irrelevant. A net force of zero preserves the prior state: either standing still or a constant speed.
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Would a pressurized container move by itself if opposite edges have different size surface area? If inside a closed container there is gas with higher pressure than outside the container, and one edge of the container has a larger surface area than the opposite side, would the container move by itself? Wouldn't there be a net force in one direction since one surface area is bigger than the opposite side while the pressure is the same?
| Unfortunately, your thinking is not correct. If we treat the whole box as a system, then box will move if and only if $\sum F_{ext}\neq0$.
However in your setup, it's shown that internal pressure (or intenral forces) is somehow moving the box, which cannot be possible. The internal forces must cancel each other out in this case.
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Eccentricity of planets based on distance from Sun Are the orbits of inner-solar system planets more circular than outer planets? Or is it the other way around? What's the reason for this? We were taught in our high school Physics class that outer planets had more circular orbits, but some sources online and even on SE state otherwise.
| The degree to which an orbit deviates from a perfect circle is measured by its orbital eccentricity. An eccentricity of $0$ is a perfect circle; an ellipse has an eccentricity between $0$ and $1$ - the higher the eccentricity, the more "elliptical" the ellipse becomes; an eccentricity of $1$ is an open parabolic orbit and an eccentricity greater than $1$ is an open hyperbolic orbit.
According to Wikipedia the current orbital eccentricities of the planets of the solar system are:
*
*Mercury $0.2056$
*Venus $0.0068$
*Earth $0.0167$
*Mars $0.0934$
*Jupiter $0.0484$
*Saturn $0.0541$
*Uranus $0.0472$
*Neptune $0.0086$
so in order of increasing orbital eccentricity the planets are Venus, Neptune, Earth, Uranus, Jupiter, Saturn, Mars, Mercury. There is no obvious correlation between orbital eccentricity and distance from the Sun.
Note that these values are current values - we know that the orbital eccentricities of the planets do vary slightly over time scales of tens of thousands of years. In $30,000$ years' time the Earth's orbit will be less eccentric than that of Venus.
Some Trans Neptunian Objects such as Pluto ($0.2488$), Eris ($0.4407$) and Sedna ($0.8549$) have higher eccentricities than any of the planets. Many comets have orbital eccentricities between $0.9$ and $1$, and interstellar objects such as 'Oumuamua have orbital eccentricities greater than $1$ because they are not gravitationally bound to the Sun.
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Inverse Laplacian I have seen the following operator somewhere in a paper on cosmology
$$
\frac{\partial_i \partial_j}{\nabla^2} - \frac{1}{3} \delta_{ij}.
$$
What is the definition of the inverse Laplacian? What is meant by this misleading notation? Is this the inverse Laplacian? If yes, what is then $\frac{1}{\nabla^2}$?
| Let's write an arbitrary function $f:\,\Bbb R^3\mapsto\Bbb R$ as a Fourier transform:$$f(\vec{r})=(2\pi)^{-3/2}\int_{\Bbb R^3}\tilde{f}(\vec{k})e^{i\vec{k}\cdot\vec{r}}d^3\vec{k},\,f(\vec{k}):=(2\pi)^{-3/2}\int_{\Bbb R^3}f(\vec{r})e^{-i\vec{k}\cdot\vec{r}}d^3\vec{r}.$$(I've restricted to a $3$-dimensional space because I'm confident that's why $\frac13$ comes up in the expression you encountered.) The most obvious definition of $\frac{1}{\nabla^2}f$ is$$\frac{1}{\nabla^2}f:=(2\pi)^{-3/2}\int_{\Bbb R^n}\frac{\tilde{f}(\vec{k})}{-k^2}e^{i\vec{k}\cdot\vec{r}}d^3\vec{k}=(2\pi)^{-3}\int_{(\Bbb R^3)^2}\frac{f(\vec{R})e^{i\vec{k}\cdot(\vec{r}-\vec{R})}}{-k^2}d^3\vec{k}d^3\vec{R}.$$This coincides with @Vincent's definition of $(\nabla^2)^{-1}$ provided$$(2\pi)^{-3}\int_{\Bbb R^3}\frac{e^{i\vec{k}\cdot\vec{z}}}{k^2+m^2}d^3\vec{k}=\frac{e^{-mz}}{4\pi z}$$is taken as $m\to0^+$ (strictly speaking, this is a distributional limit). By Fourier inversion, this conjecture is equivalent to$$\int_{\Bbb R^3}\frac{e^{-mz-i\vec{k}\cdot\vec{z}}d^3\vec{z}}{4\pi z}=\frac{1}{k^2+m^2}.$$Indeed, spherical polars rewrite the LHS as$$\begin{align}\frac12\int_0^\pi d\theta\sin\theta\int_0^\infty ze^{-(m+ik\cos\theta)z}dz&=\frac12\int_0^\pi\frac{\sin\theta}{(m+ik\cos\theta)^2}d\theta\\&=\frac{1}{2ik}\left[\frac{1}{m+ik\cos\theta}\right]_0^\pi\\&=\frac{1}{k^2+m^2}.\end{align}$$This is quite a complicated manipulation, so it pays to use dimensional analysis as a power-counting sanity check.
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Sustaining a current in conductor placed in external electric field Consider the following arrangement-
We have a conducting sphere and a positively charged infinite sheet on the left. The field creates induced charges and the net electric field inside the conductor is zero after a very short time. During this short time, there is a current in the conductor as electrons as dragged opposite to the external electric field.
My teacher says that if we want to sustain this brief current we should connect it with a conducting wire making a closed loop. The electrons will flow anticlockwise giving a steady current, which I think is wrong.
When we connect the conducting wire, the sphere and the wire become one complete metal and after a short time again, there will be induced charges in this big metal and electrostatic condition will be reached. The potential inside will be the same everywhere so how will the current flow in a closed loop?
| There is really not much to add to the comments. Why are the charges separated on the two sides of the sphere? because there is an electric field that pushes the negative charges to the left and the positive to the right, right? This field is produced by the charge on that infinite sheet. This field exists not only in the space occupied by the sphere but everywhere. If you have another piece of metal in this space you will see the same effect, negative charges pushed to the left hand side, the side nearest to the infinite sheet. So another piece of metal is also the wire. The same effect as in the sphere is in the wire, the external field keeps the charges separated. You also can think, in a more abstract way, in terms of potential. Why the charges on the sphere do not get together? Because there is no potential difference between the two sides. The field of the charges on the sphere and the field of the infinite sheet compensate so that the surface of the sphere is equipotential. Connecting two points having the same potential with a wire does not result in any current flow.
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Which topological orders described by TQFT and tensor category theories are not known to be microscopically realizable? Topological order refers to long-range-entangled phases of matter that cannot be smoothly deformed into ordinary phases characterized by Landau’s symmetry breaking theory. A large number of topological orders are described and classified by topological quantum field theory and unitary modular tensor category theory [or unitary braided fusion category], the latter of which describes the rules governing the fusion and braiding process of topological excitations (anyons).
What is not clear to me is whether or not all of these phases are realizable in microscopic systems with local interactions, i.e. does there always exist a locally interacting Hamiltonian that have ground states and low energy excitations described by these macroscopic theories (TQFT and UMTC/UBFC)? The string-net models answer this question affirmatively for the case of "doubled topological order", but that is just a small subclass of topological order, and the answer for the general case is still missing.
[As a useful comparison with symmetry-breaking phases, UMTC or UBFC takes the role of group theory, while TQFT plays the role of some kind of effective field theory like Landau-Ginzberg. But to establish their existence, seems that we are still missing a microscopic Hamiltonian]
In particular, I want to ask: is there an example of an important topological order consistently described by TQFT or UMTC but not yet known to be microscopically realizable?
| Any (2+1)-D TQFTs described by a unitary modular tensor category can be realized as the boundary of a Walker-Wang model. Moreover it is believed that any such Walker-Wang model is trivial in the bulk; hence there should exist a bulk disentangling unitary. Applying this unitary will give you a microscopic 2D lattice Hamiltonian. Of course turning this into a concrete construction is still challenging.
https://arxiv.org/abs/1104.2632
| {
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Graph for Coulomb Force vs $1/r$ My teacher told me that the graph for the coulomb force $F$ vs $1/r$ where $r$ is the distance between the 2 charges should be parabolic but I can't seem to understand why. I am aware that equations of the form $y^2=4ax$ are parabolic but why should $F$ vs $1/r$ graph be parabolic?
| You are right that functions like $y = ax^2$ have a parabolic graph of $y$ vs $x$.
The force between two charges is
$$F = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$$
So the graph of $F$ vs $r$ is not parabolic. But you are not graphing that.
You are graphing $F$ vs $1/r$. You can see how the equations can be made to be similar. $$F \to y, \space \frac{q_1q_2}{4\pi\epsilon_0}\to a, \space and \space1/r \to x$$
The confusing part might be $1/r \to x$. Why do that? What does it mean?
The usual function $F = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$ tells you how $F$ changes as distance changes. As $r$ gets big, $F$ gets small.
But suppose you wanted a measure of "closeness" instead of "distance". $F$ gets big as the charges get close together. One way to do that is to use $1/r$ as a measure of closeness. Inventing the terminology $x = 1/r$ for closeness, we get
$$F = \frac{q_1q_2}{4\pi\epsilon_0}x^2$$
Now you can see exactly how $F$ gets big as closeness gets big. The mental shift can be confusing, but sometimes you learn things by transforming variables like that.
| {
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Does "lifetime of up quark" have a physical meaning? I saw this question about the lifetime of an up quark.
As far as I know, free quarks are never observed in experiments. Then what is the significance of a statement like "the lifetime of an up quark is X units"?
I am looking for a physical explanation without involving much mathematics. I am not very familiar with the mathematical formulation of QCD, but I know about Feynman diagrams.
| No, "lifetime of an up quark" is utterly meaningless (at least here, but I'd be hard pressed to find legitimate contexts for it...).
The lifetime discussed is that of a neutral pion, decaying by the F diagram (sorry)
In words, the pion "resolves" to virtual states of its valence quarks, u or d, which then couple to two real photons, to which the pion thus decays with a given width (/lifetime) thus computed.
The lifetime discussion never applied to the quarks, but only to the size of the amplitude represented by this diagram/process. This size eventually determines the probability of decay per unit of time, related to the lifetime.
| {
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Why can basis vectors change direction? I thought that basis vectors were of magnitude one and located at the origin and were each linearly independent, so how in things like polar coordinates can the basis vectors be moving?
| "Coordinates" is a more general term that "basis vectors". Basis vectors apply only to vector spaces, while coordinates apply to any manifold (and, if one uses the term loosely enough, pretty much any space). Basis vectors provide a coordinate system by simply taking their coefficients, but not all coordinate systems correspond to a set of basis vectors. Also, while basis vectors do have to be independent, they do not have to have magnitude $1$, and in fact vector spaces do not have to have any norm at all to have basis vectors, and if a space doesn't have a norm, then "magnitude $1$" is meaningless. Additionally, vectors aren't really "located" at the origin. The are often represented as displacements from the origin to another point in affine space, but that's not quite the same thing as being "at" the origin.
Polar coordinates are not based on basis vectors, although we can use them to define local basis vectors for each point.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Could a non-photon massless particle travel at a speed other than $c$? The speed of light is given as $c=\frac{1}{\sqrt{ε_0μ_0}}$ which is in terms of the electric and magnetic constants.
Hypothetically, another massless particle could exist which does not interact with the electronic or magnetic fields. Postulate that a "foo" field and a "bar" field exist which were mathematically analogous to the electric and magnetic fields, the "foobaron" particle might have a speed given by $c_{fb}=\frac{1}{\sqrt{f_0b_0}}$, derived in exactly the same way that $c$ is derived, but using different, independent fields.
If such a particle were to exist, must $c=c_{fb}$? Why?
| In special relativity, the relativistic factor
$$
\gamma = \frac1{\sqrt{1-v^2/c^2}} = \frac{E_\text{total}}{mc^2}
$$
is just as closely related to the particle’s total energy as to the particle’s speed. Any object whose total energy is very much larger than its rest mass will be traveling near $c$. Objects whose rest mass is identically zero are a special case of this limit.
The speed limit $c$ is not a property of electromagnetism; it is a property of spacetime. If a massless particle were traveling at less than $c$, I could hop into my massive rocket ship and catch up to it. In its rest frame, it would not be a massless particle. A contradiction, because mass is invariant under Lorentz boosts.
| {
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Why do metallic objects reflect radar? We learn that EM waves cause the electrons in a conductor to move around. For example, air to ground radar shows the ocean as having few returns compared to land. Water molecules absorb the energy. Doesn't the induced motion of electrons in the metal in chaotic eddy currents dissipate the energy? I read this:
https://www.researchgate.net/post/Why-are-Microwaves-reflected-by-metals
but was not enlightened.
| Because metals are electrically conductive, an incoming radar pulse induces a current to flow in the surface of the metal. That current flow then radiates a replica of the original wave, moving in the opposite direction.
| {
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Why I feel cool in a region with lot of trees inspite of humidity by transporation? Today I went for a walk and entered a region with lot of trees, It feel a sudden chill, a cool feeling (comparatively)
Why, Although trees perform transpiration that cause moisture so we should feel warm but I feel cool?
| Trees draw liquid water from the ground and transport it to the leaves. This water then evaporates through the process of transpiration. Turning liquid water into water vapor requires a significant amount of energy due to water's high latent heat of vaporization. The process of transpiration absorbs heat energy from the surrounding area, resulting in a cooling effect - this is known as transpiration cooling. It's a form of evaporative cooling, much like how a swamp cooler or human sweat has a cooling effect.
| {
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Is the intensity of light dependent on number of photons per unit area? I was learning about the photoelectric effect of light and there it says more the intensity of light, the more number of electrons will be ejected from the metal surface given that the frequency of light is more than its threshold frequency.
Now what does intensity basically means?? Is it dependent on the number of photons per unit area over which the light falls..
Let us consider two light rays, $X$ and $Y$ which have same frequency and same wavelength and it is said that $X$ has more intensity than $Y$. The light rays fall over an area, $dA$. Does that mean that light ray $X$ will have more photons falling on that area $dA$ as compared to light ray $Y$ ??
| Yes
light ray X will have more photons falling on that area dA as compared to light ray Y
(per second).
The energy of each photon depends on its frequency from $E=hf$.
If $X$ and $Y$ have the same frequency, then more photons per second are falling on the area, from light $X$.
This matches with the classical definition of Intensity (power per unit area)
| {
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Pauli exclusion principle and antimatter Have the Pauli exclusion principle been proven to apply to antimatter experimentally?
| While CERN has created anti-hydrogen, there has never been a synthesized multi-antielectron anti-atom. So there has never been a large enough anti-atom (antinucleons with antielectrons) to experimentally test the PEP for antimatter.
But the Pauli exclusion principle applies to all spin $\frac{1}{2}$ particles (all fermions) and one would be shocked that in anti-atoms containing several anti-electrons, the PEP was not obeyed.
| {
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Pipe Open or closed from a Standing Wave Equation Here is the full question:
The part in the brackets are the answers. I am still terribly confused.
Lets say we are given an equation for a standing wave in a pipe:
$$y(x,t) = A\sin\left(\frac{3\pi x}{L}\right)\sin(\omega t).$$
Is this enough information to know whether it is an open-open or open-closed pipe system? Obviously at $x = 0$, we have a node, so at a minimum there is one closed end. $k = \frac{2\pi}{\lambda} = \frac{3\pi}{L} \rightarrow \lambda = \frac{2}{3}L$.
I know that a one-sided open system $\lambda_n = \frac{2L}{n}$ which would work if $n = 3$. Hence it is an open-closed system? I am super confused about what $L$ is here. Is this the length of the tube?
| The question in isolation is not well posed for a single answer; in fact, there are an infinite number of "correct" answers that could be given. Further information could be provided (or assumed by the questioner) that would lead to a particular solution.
Analysis
First, I will assume that $y$ denotes the particle displacement within the tube (and not the pressure). Then, as you comment, we may say that the tube is closed at $x=0$. However, the length may be any value of $x$ such that $\sin(3\pi x/L)=0$ if the tube is closed at the far end, or any value of $x$ such that $\cos(3\pi x/L)=0$ (anti-node) if the tube is open at the far end. Why don't we calculate all of these distances, just for fun!
Closed Far End
We are looking for
$$
\sin\left( \frac{3\pi x}{L} \right) = 0 \hspace{15mm}\Rightarrow\hspace{15mm} \frac{3\pi x}{L} = n\pi,
$$
where $n$ is any integer. Thus, we may conclude that the tube may be any of the following lengths:
$$
\left\{ \frac{nL}{3}: n\in\mathbb{Z} \right\}.
$$
We are only interested in positive values of the length, and so the lowest possible length of a closed-closed tube would be $n=1$, and so the tube length is $L/3$.
Open Far End
In this case we are looking for
$$
\cos\left( \frac{3\pi x}{L} \right) = 0 \hspace{15mm}\Rightarrow\hspace{15mm} \frac{3\pi x}{L} = \frac{2n-1}{2}\pi,
$$
where $n$ is again any integer. Thus, we may conclude that the tube may be any of the following lengths:
$$
\left\{ \frac{2n-1}{6}L: n\in\mathbb{Z} \right\}.
$$
Again, we are only interested in positive lengths, and so the smallest tube would be one where $n=1$, which yields a tube length of $L/6$. The answer they provided is the case where $n=3$.
One set of information they could have provided to force a specific answer is how many nodes and anti-nodes are present in the tube (3 and 3 in the case given by the answer). I am sure there are other options as well.
| {
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Intuition for Stress and the Cauchy Stress Tensor I'm struggling to get an intuitive understanding of what exactly Stress is, particularly the "direction" associated with it.
In the case of a 1 dimensional bar with just uniaxial loading, the way stress was explained to me was just $\pm\frac{F}{A}$ where $F$ is the force applied to either end, A is the cross sectional area, and the sign refers to tension or compression. This explanation is fine as a formula, but I don't see how it relates to "internal forces".
I've found other sources explaining it with an "imaginary cut" through the material, ignoring one side of the cut, and imposing equilibrium on the other piece. Why can either side be "ignored"? If stress is the internal force per unit of surface, why doesn't the neglected part contribute to the stress? (after all, the neglected part shares the surface).
In the more general case using the Stress Tensor,
$$T=\begin{bmatrix}
\sigma_{xx} & \tau_{xy} & \tau_{xz}\\
\tau_{yx} & \sigma_{yy} & \tau_{yz}\\
\tau_{zx} & \tau_{zy} & \sigma_{zz}\\
\end{bmatrix}$$
Do each of the components describe the stress on the surfaces of some infinitesimal volume? If so, which faces do they describe (there are 2 faces normal to each direction)- is it the sum of the stresses on each face?
Any insight on these questions would be greatly appreciated, thanks for reading
|
Do each of the components describe the stress on the surfaces of some
infinitesimal volume?
Essentially, yes.
If so, which faces do they describe (there are 2faces normal to each direction)- is it the sum of the stresses on each
face?
The assumption is the volume is in equilibrium, both translational and rotational. On that basis, the diagonal terms are the applied external normal stresses on the faces of the cube. There are six faces, but the normal stress on each opposite face is equal and opposite for translational equilibrium, so only three are specified in the tensor. If there was only a normal stress on one pair of opposite faces and no applied shear stress on the faced, you would have your uniaxial stress equation. You can see this in the figure in the Wikipedia link supplied by @nicoguaro, except that $\sigma$ is used for shear stress an $e$ is used for normal stress.
The off diagonal terms are the shear stresses on each face. Six are specified but three pairs are identical, e.g., $\tau_{xy}=\tau_{yx}$. This needs to be so in order that there is not rotation of the cube. For example, in terms of @nicoguaro indices, $\sigma_{21}=\sigma_{12}$
Hope this helps.
| {
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Confused about the Pauli exclusion principle I've been struggling to understand this: Let's say I have a gas of one million electrons. Does every single one of those electrons have a different energy (up to the degeneracy from the different momentum components)?
| The Pauli principle was derived from observations of the electrons of atoms. It expresses that two electrons with the same level in the atom are distinguishable. By an external magnetic field the orientations of the spin can be manipulated.
So for an electron gas it is possible to align the electrons a bit in the direction of an external magnetic field. But this alignment of course will be disturbed at any time by the chaotic thermal movement of the gas.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Matrix element and Dirac notation If
$$
T=
\left[
\begin{array}{cccc}
e^{\beta J} & e^{-\beta J} \\
e^{-\beta J} & e^{\beta J} \\
\end{array} \right]
$$
and
$$Z = \sum_{S_i=\pm 1} ... \sum_{S_N=\pm 1} \exp{\beta J(\vec{S_1}\vec{S_2}+\vec{S_2}\vec{S_3}+...+\vec{S_{N-1}}\vec{S_N}+\vec{S_N}\vec{S_1})}
$$
Then why can we say that
$$Z = \sum_{S_i=\pm 1} ... \sum_{S_N=\pm 1} \langle S_1|T|S_2\rangle\langle S_2|T|S_3\rangle...\langle S_N|T|S_1\rangle ?
$$
| Because, with $S_i$ taking values $\pm 1$, we have
$$
\langle S_1|e^{\beta J {S}_i {S}_{i+1}}| S_2\rangle=
\left[\matrix{e^{\beta J}& e^{-\beta J} \cr e^{-\beta J} &e^{\beta J}}\right]_{S_1,S_2}
$$
where the subscript on the matrix means the appropriate matrix entry.
| {
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"timestamp": "2023-03-29T00:00:00",
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Confusion on molecular dynamics (MD) simulation units leading to absurd acceleration values I am trying to code up a very simple MD simulation in order to learn more about it. I am using the Leonard-Jones potential, expressed as $ V=4\epsilon (\frac{\sigma}{r}^{12}-\frac{\sigma}{r}^{6}) $
The pairwise force is therefore $\frac{1}{r}\frac{dV}{dr}=\frac{24\epsilon}{r}((\frac{\sigma}{r})^6-2(\frac{\sigma}{r})^{12})$
I want to use parameter values as listed in Li's publication:
http://li.mit.edu/A/Papers/05/Li05-2.8.pdf
However, when I work through the math, I end up with unreasonable values as shown in the following calculation:
Suppose two argon atoms are separated by 4 angstroms. The ambient temperature is 300K.
In Joules, $\epsilon=119.8k_B T=119.8\cdot 1.3806\cdot 10^{-23} \frac{J}{K} \cdot 300K=4.9619\cdot 10^{-19} J$.
The pairwise force is then $4.9619\cdot 10^{-19}J * \frac{24}{4\cdot 10^{-10} m} ((\frac{3.405Å}{4.0Å})^6-2(\frac{3.405Å}{4.0Å})^{12}) = 2.26\frac{J}{m}=2.71\cdot 10^{-9}N$.
The mass of one a.m.u. in kg is $1.6605\cdot 10^{-27} kg$.
$f=\frac{m}{a}\implies a=\frac{f}{m}=\frac{2.71\cdot 10^{-9} N}{39.948*1.6605\cdot 10^{-27} kg}=4.0854\cdot 10^{16}\frac{m}{s^2}$
My simulation has units of angstroms, so I usually convert the acceleration to angstroms.
$ a=4.0854 \cdot 10^{26} \frac{Å}{s^2}$.
This number is absurd. Even with a timestep of 1ps, the particles will fly out of the bounding box in just a few simulation steps.
Where did I go wrong?
EDIT: Thank you to @Samson for providing the correct pairwise force equation, $\vec{F_{IJ}=\frac{48\epsilon}{\sigma^2}[(\frac{\sigma}{r_{IJ}})^{14}-0.5\cdot(\frac{\sigma}{r_{IJ}})^{8}]}\vec{r_{IJ}}$. Here, $\sigma$ has units of angstroms, $\epsilon$ units of Joules, $r_{IJ}$ units of angstroms. $\vec{r_{IJ}}$ is the vector difference of the pair's positions, i.e. $\vec{r_I}-\vec{r_J}$.
If you use that equation, and handle the units correctly, you get acceleration on the order of $10^{25} \frac{Å}{s^2}$. Regardless of what integrator you use, the total displacement after one timestep will be proportional to the timestep squared. If the timestep is $10^{-13}$ seconds or so, the order of magnitude of displacement in each step will be less than one angstrom, which is an agreeable number to work with. In my implementation, I expressed the time-step as multiples of picoseconds (in some cases, a fraction of a picosecond) and did the unit conversions on paper. This avoids using extreme floating-point numbers and leads to better numerical stability.
| Maybe your numerical solver is instable.
| {
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